The Universe of Discourse

Fri, 01 Jan 2021

My big mistake about dense sets

I made a big mistake in a Math Stack Exchange answer this week. It turned out that I believed something that was completely wrong.

Here's the question, are terminating decimals dense in the reals?. It asks if the terminating decimals (that is, the rational numbers of the form !!\frac m{10^n}!!) are dense in the reals. “Dense in the reals” means that if an adversary names a real number !!r!! and a small distance !!d!!, and challenges us to find a terminating decimal !!t!! that is closer than !!d!! to point !!r!!, we can always do it. For example, is there a terminating decimal !!t!! that is within !!0.0000001!! of !!\sqrt 2!!? There is: !!\frac{14142135}{10^7} = 1.4142135!! is closer than that; the difference is less than !!0.00000007!!.

The answer to the question is ‘yes’ and the example shows why: every real number has a decimal expansion, and if you truncate that expansion far enough out, you get a terminating decimal that is as close as you like to the original number. This is the obvious and straightforward way to prove it, and it's just what the top-scoring answer did.

I thought I'd go another way, though. I said that it's enough to show that for any two terminating decimals, !!a!! and !!b!!, there is another one that lies between them. I remember my grandfather telling me long ago that this was a sufficient condition for a set to be dense in the reals, and I believed him. But it isn't sufficient, as Noah Schweber kindly pointed out.

(It is, of course, necessary, since if !!S!! is a subset of !!\Bbb R!!, and !!a,b\in S!! but no element of !!S!! between these, then there is no element of !!S!! that is less than distance !!\frac{b-a}2!! of !!\frac{a+b}2!!. Both !!a!! and !!b!! are at that distance, and no other point of !!S!! is closer.)

The counterexample that M. Schweber pointed out can be explained quickly if you know what the Cantor middle-thirds set is: construct the Cantor set, and consider the set of midpoints of the deleted intervals; this set of midpoints has the property that between any two there is another, but it is not dense in the reals. I was going to do a whole thing with diagrams for people who don't know the Cantor set, but I think what follows will be simpler.

Consider the set of real numbers between 0 and 1. These can of course be represented as decimals, some terminating and some not. Our counterexample will consist of all the terminating decimals that end with !!5!!, and before that final !!5!! have nothing but zeroes and nines. So, for example, !!0.5!!. To the left and right of !!0.5!!, respectively, are !!0.05!! and !!0.95!!.

In between (and around) these three are: $$\begin{array}{l} \color{darkblue}{ 0.005 }\\ 0.05 \\ \color{darkblue}{ 0.095 }\\ 0.5 \\ \color{darkblue}{ 0.905 }\\ 0.95 \\ \color{darkblue}{ 0.995 }\\ \end{array}$$

(Dark blue are the new ones we added.)

And in between and around these are:

$$\begin{array}{l} \color{darkblue}{ 0.0005 }\\ 0.005 \\ \color{darkblue}{ 0.0095 }\\ 0.05 \\ \color{darkblue}{ 0.0905 }\\ 0.095 \\ \color{darkblue}{ 0.0995 }\\ 0.5 \\ \color{darkblue}{ 0.9005 }\\ 0.905 \\ \color{darkblue}{ 0.9095 }\\ 0.95 \\ \color{darkblue}{ 0.9905 }\\ 0.995 \\ \color{darkblue}{ 0.9995 }\\ \end{array}$$

Clearly, between any two of these there is another one, because around !!0.????5!! we've added !!0.????05!! before and !!0.????95!! after, which will lie between !!0.????5!! and any decimal with fewer !!?!! digits before it terminates. So this set does have the between-each-two-is-another property that I was depending on.

But it should also be clear that this set is not dense in the reals, because, for example, there is obviously no number of this type that is near !!0.7!!.

(This isn't the midpoints of the middle-thirds set, it's the midpoints of the middle-four-fifths set, but the idea is exactly the same.)

Happy New Year, everyone!

[Other articles in category /math] permanent link

Sat, 21 Nov 2020

Testing for divisibility by 19

[ Previously, Testing for divisibility by 7. ]

A couple of nights ago I was keeping Katara company while she revised an essay on The Scarlet Letter (ugh) and to pass the time one of the things I did was tinker with the tests for divisibility rules by 9 and 11. In the course of this I discovered the following method for divisibility by 19:

Double the last digit and add the next-to-last.
Double that and add the next digit over.
Repeat until you've added the leftmost digit.

The result will be a smaller number which is a multiple of 19 if and only if the original number was.

For example, let's consider, oh, I don't know, 2337. We calculate:

  • 7·2+3 = 17
  • 17·2 + 3 = 37
  • 37·2 + 2 = 76

76 is a multiple of 19, so 2337 was also. But if you're not sure about 76 you can compute 2·6+7 = 19 and if you're not sure about that you need more help than I can provide.

I don't claim this is especially practical, but it is fun, not completely unworkable, and I hadn't seen anything like it before. You can save a lot of trouble by reducing the intermediate values mod 19 when needed. In the example above above, after the first step you get to 17, which you can reduce mod 19 to -2, and then the next step is -2·2+3 = -1, and the final step is -1·2+2 = 0.

Last time I wrote about this Eric Roode sent me a whole compendium of divisibility tests, including one for divisibility by 19. It's a little like mine, but in reverse: group the digits in pairs, left to right; multiply each pair by 5 and then add the next pair. Here's 2337 again:

  • 23·5 + 37 = 152
  • 1·5 + 52 = 57

Again you can save a lot trouble by reducing mod 19 before the multiplication. So instead of the first step being 23·5 + 37 you can reduce the 23·5 to 4·5 = 20 and then add the 37 to get 57 right away.

[ Addendum: Of course this was discovered long ago, and in fact Wikipedia mentions it. ]

[ Addendum 20201123: An earlier version of this article claimed that the double-and-add step I described preserves the mod-19 residue. It does not, of course; the doubling step doubles it. It is, however, true that it is zero afterward if and only if it was zero before. ]

[Other articles in category /math] permanent link

Sun, 18 Oct 2020

Newton's Method and its instability

While messing around with Newton's method for last week's article, I built this Desmos thingy:

The red point represents the initial guess; grab it and drag it around, and watch how the later iterations change. Or, better, visit the Desmos site and play with the slider yourself.

(The curve here is !!y = (x-2.2)(x-3.3)(x-5.5)!!; it has a local maximum at around !!2.7!!, and a local minimum at around !!4.64!!.)

Watching the attractor point jump around I realized I was arriving at a much better understanding of the instability of the convergence. Clearly, if your initial guess happens to be near an extremum of !!f!!, the next guess could be arbitrarily far away, rather than a small refinement of the original guess. But even if the original guess is pretty good, the refinement might be near an extremum, and then the following guess will be somewhere random. For example, although !!f!! is quite well-behaved in the interval !![4.3, 4.35]!!, as the initial guess !!g!! increases across this interval, the refined guess !!\hat g!! decreases from !!2.74!! to !!2.52!!, and in between these there is a local maximum that kicks the ball into the weeds. The result is that at !!g=4.3!! the method converges to the largest of the three roots, and at !!g=4.35!!, it converges to the smallest.

This is where the Newton basins come from:

Julia set for the rational function
associated to Newton's method for ƒ:z→z³−1.  The plane is divided
symmetrically into three regions, colored red, green, and blue.  The
boundary of each two regions (red and green, say) is inhabited by a
series of leaf shapes of the third color (blue), and the boundaries
between the main regions (green, say) and the (blue) leaves are
inhabited by smaller leaves again of the other color (red), and so on
ad infinitum.  The boundaries are therefore an infinitely detailed
filigree of smaller and smaller leaves of all three colors.

Here we are considering the function !!f:z\mapsto z^3 -1!! in the complex plane. Zero is at the center, and the obvious root, !!z=1!! is to its right, deep in the large red region. The other two roots are at the corresponding positions in the green and blue regions.

Starting at any red point converges to the !!z=1!! root. Usually, if you start near this root, you will converge to it, which is why all the points near it are red. But some nearish starting points are near an extremum, so that the next guess goes wild, and then the iteration ends up at the green or the blue root instead; these areas are the rows of green and blue leaves along the boundary of the large red region. And some starting points on the boundaries of those leaves kick the ball into one of the other leaves…

Here's the corresponding basin diagram for the polynomial !!y = (x-2.2)(x-3.3)(x-5.5)!! from earlier:

Julia set for
the function (x-2.2)(x-3.3)(x-5.5).  There is a large pink region on
the left, a large yellow region on the right, and in between a large
blue hourglass-shaped region.  The boundary of yellow region is decorated
with pink bobbles that protrude into the blue region, and similarly
the boundary of the pink region is decorated with yellow
bobbles. Further details are in the text below.

The real axis is the horizontal hairline along the middle of the diagram. The three large regions are the main basins of attraction to the three roots (!!x=2.2, 3.3!!, and !!5.5!!) that lie within them.

But along the boundaries of each region are smaller intrusive bubbles where the iteration converges to a surprising value. A point moving from left to right along the real axis passes through the large pink !!2.2!! region, and then through a very small yellow bubble, corresponding to the values right around the local maximum near !!x=2.7!! where the process unexpectedly converges to the !!5.5!! root. Then things settle down for a while in the blue region, converging to the !!3.3!! root as one would expect, until the value gets close to the local minimum at !!4.64!! where there is a pink bubble because the iteration converges to the !!2.2!! root instead. Then as !!x!! increases from !!4.64!! to !!5.5!!, it leaves the pink bubble and enters the main basin of attraction to !!5.5!! and stays there.

If the picture were higher resolution, you would be able to see that the pink bubbles all have tiny yellow bubbles growing out of them (one is 4.39), and the tiny yellow bubbles have even tinier pink bubbles, and so on forever.

(This was generated by the Online Fractal Generator at; the labels were added later by me. The labels’ positions are only approximate.)

[ Addendum: Regarding complex points and !!f : z\mapsto z^3-1!! I said “some nearish starting points are near an extremum”. But this isn't right; !!f!! has no extrema. It has an inflection point at !!z=0!! but this doesn't explain the instability along the lines !!\theta = \frac{2k+1}{3}\pi!!. So there's something going on here with the complex derivative that I don't understand yet. ]

[Other articles in category /math] permanent link

Fixed points and attractors, part 3

Last week I wrote about a super-lightweight variation on Newton's method, in which one takes this function: $$f_n : \frac ab \mapsto \frac{a+nb}{a+b}$$

or equivalently

$$f_n : x \mapsto \frac{x+n}{x+1}$$

Iterating !!f_n!! for a suitable initial value (say, !!1!!) converges to !!\sqrt n!!:

$$ \begin{array}{rr} x & f_3(x) \\ \hline 1.0 & 2.0 \\ 2.0 & 1.667 \\ 1.667 & 1.75 \\ 1.75 & 1.727 \\ 1.727 & 1.733 \\ 1.733 & 1.732 \\ 1.732 & 1.732 \end{array} $$

Later I remembered that a few months back I wrote a couple of articles about a more general method that includes this as a special case:

The general idea was:

Suppose we were to pick a function !!f!! that had !!\sqrt 2!! as a fixed point. Then !!\sqrt 2!! might be an attractor, in which case iterating !!f!! will get us increasingly accurate approximations to !!\sqrt 2!!.

We can see that !!\sqrt n!! is a fixed point of !!f_n!!:

$$ \begin{align} f_n(\sqrt n) & = \frac{\sqrt n + n}{\sqrt n + 1} \\ & = \frac{\sqrt n(1 + \sqrt n)}{1 + \sqrt n} \\ & = \sqrt n \end{align} $$

And in fact, it is an attracting fixed point, because if !!x = \sqrt n + \epsilon!! then

$$\begin{align} f_n(\sqrt n + \epsilon) & = \frac{\sqrt n + \epsilon + n}{\sqrt n + \epsilon + 1} \\ & = \frac{(\sqrt n + \sqrt n\epsilon + n) - (\sqrt n -1)\epsilon}{\sqrt n + \epsilon + 1} \\ & = \sqrt n - \frac{(\sqrt n -1)\epsilon}{\sqrt n + \epsilon + 1} \end{align}$$

Disregarding the !!\epsilon!! in the denominator we obtain $$f_n(\sqrt n + \epsilon) \approx \sqrt n - \frac{\sqrt n - 1}{\sqrt n + 1} \epsilon $$

The error term !!-\frac{\sqrt n - 1}{\sqrt n + 1} \epsilon!! is strictly smaller than the original error !!\epsilon!!, because !!0 < \frac{x-1}{x+1} < 1!! whenever !!x>1!!. This shows that the fixed point !!\sqrt n!! is attractive.

In the previous articles I considered several different simple functions that had fixed points at !!\sqrt n!!, but I didn't think to consider this unusally simple one. I said at the time:

I had meant to write about Möbius transformations, but that will have to wait until next week, I think.

but I never did get around to the Möbius transformations, and I have long since forgotten what I planned to say. !!f_n!! is an example of a Möbius transformation, and I wonder if my idea was to systematically find all the Möbius transformations that have !!\sqrt n!! as a fixed point, and see what they look like. It is probably possible to automate the analysis of whether the fixed point is attractive, and if not to apply one of the transformations from the previous article to make it attractive.

[Other articles in category /math] permanent link

Tue, 13 Oct 2020

Newton's Method but without calculus — or multiplication

Newton's method goes like this: We have a function !!f!! and we want to solve the equation !!f(x) = 0.!! We guess an approximate solution, !!g!!, and it doesn't have to be a very good guess.

The graph of a wiggly
polynomial girve with roots between 2 and 2.5, between 3 and 3.5, and
between 5 and 6.  The middle root is labeled with a question mark.
One point of the curve, not too different from the root, is marked in
blue and labeled “⟨g,f(g)⟩”.

Then we calculate the line !!T!! tangent to !!f!! through the point !!\langle g, f(g)\rangle!!. This line intersects the !!x!!-axis at some new point !!\langle \hat g, 0\rangle!!, and this new value, !!\hat g!!, is a better approximation to the value we're seeking.

The same graph as
before, but with a tangent to the curve drawn through the blue point.
It intersects the x-axis quite close to the root.  The intersection
point is labeled “ĝ”.

In the left
margin, a close-up detail of the same curve as before, this time
showing the tangent line at ⟨ĝ,f(ĝ)⟩.  The diagram is a close-up
because the line intersects the x-axis extremely close to the actual

Analytically, we have:

$$\hat g = g - \frac{f(g)}{f'(g)}$$

where !!f'(g)!! is the derivative of !!f!! at !!g!!.

We can repeat the process if we like, getting better and better approximations to the solution. (See detail at left; click to enlarge. Again, the blue line is the tangent, this time at !!\langle \hat g, f(\hat g)\rangle!!. As you can see, it intersects the axis very close to the actual solution.)

In general, this requires calculus or something like it, but in any particular case you can avoid the calculus. Suppose we would like to find the square root of 2. This amounts to solving the equation $$x^2-2 = 0.$$ The function !!f!! here is !!x^2-2!!, and !!f'!! is !!2x!!. Once we know (or guess) !!f'!!, no further calculus is needed. The method then becomes: Guess !!g!!, then calculate $$\hat g = g - \frac{g^2-2}{2g}.$$ For example, if our initial guess is !!g = 1.5!!, then the formula above tells us that a better guess is !!\hat g = 1.5 - \frac{2.25 - 2}{3} = 1.4166\ldots!!, and repeating the process with !!\hat g!! produces !!1.41421\mathbf{5686}!!, which is very close to the correct result !!1.41421\mathbf{3562}!!. If we want the square root of a different number !!n!! we just substitute it for the !!2!! in the numerator.

This method for extracting square roots works well and requires no calculus. It's called the Babylonian method and while there's no evidence that it was actually known to the Babylonians, it is quite ancient; it was first recorded by Hero of Alexandria about 2000 years ago.

How might this have been discovered if you didn't have calculus? It's actually quite easy. Here's a picture of the number line. Zero is at one end, !!n!! is at the other, and somewhere in between is !!\sqrt n!!, which we want to find.

A line with
the left enpoint marked “0”, the right endpoint marked “n”, and a
point in between marked “square root of n”.

Also somewhere in between is our guess !!g!!. Say we guessed too low, so !!0 \lt g < \sqrt n!!. Now consider !!\frac ng!!. Since !!g!! is too small to be !!\sqrt n!! exactly, !!\frac ng!! must be too large. (If !!g!! and !!\frac ng!! were both smaller than !!\sqrt n!!, then their product would be smaller than !!n!!, and it isn't.)

The previous
illustration, with green points marked “g” and “\frac{n}{g}”. The
first of these it to the left of the square root of n, the second to
its right.

Similarly, if the guess !!g!! is too large, so that !!\sqrt n < g!!, then !!\frac ng!! must be less than !!\sqrt n!!. The important point is that !!\sqrt n!! is between !!g!! and !!\frac ng!!. We have narrowed down the interval in which !!\sqrt n!! lies, just by guessing.

Since !!\sqrt n!! lies in the interval between !!g!! and !!\frac ng!! our next guess should be somewhere in this smaller interval. The most obvious thing we can do is to pick the point halfway in the middle of !!g!! and !!\frac ng!!, So if we guess the average, $$\frac12\left(g + \frac ng\right),$$ this will probably be much closer to !!\sqrt n!! than !!g!! was:

The previous
illustration, but the point exactly midway between g and \frac{n}{g}
is marked in blue.  It is quote close to the point marked “square root
of n”.

This average is exactly what Newton's method would have calculated, because $$\frac12\left(g + \frac ng\right) = g - \frac{g^2-n}{2g}.$$

But we were able to arrive at the same computation with no calculus at all — which is why this method could have been, and was, discovered 1700 years before Newton's method itself.

If we're dealing with rational numbers then we might write !!g=\frac ab!!, and then instead of replacing our guess !!g!! with a better guess !!\frac12\left(g + \frac ng\right)!!, we could think of it as replacing our guess !!\frac ab!! with a better guess !!\frac12\left(\frac ab + \frac n{\frac ab}\right)!!. This simplifies to

$$\frac ab \Rightarrow \frac{a^2 + nb^2}{2ab}$$

so that for example, if we are calculating !!\sqrt 2!!, and we start with the guess !!g=\frac32!!, the next guess is $$\frac{3^2 + 2\cdot2^2}{2\cdot3\cdot 2} = \frac{17}{12} = 1.4166\ldots$$ as we saw before. The approximation after that is !!\frac{289+288}{2\cdot17\cdot12} = \frac{577}{408} = 1.41421568\ldots!!. Used this way, the method requires only integer calculations, and converges very quickly.

But the numerators and denominators increase rapidly, which is good in one sense (it means you get to the accurate approximations quickly) but can also be troublesome because the numbers get big and also because you have to multiply, and multiplication is hard.

But remember how we figured out to do this calculation in the first place: all we're really trying to do is find a number in between !!g!! and !!\frac ng!!. We did that the first way that came to mind, by averaging. But perhaps there's a simpler operation that we could use instead, something even easier to compute?

Indeed there is! We can calculate the mediant. The mediant of !!\frac ab!! and !!\frac cd!! is simply $$\frac{a+c}{b+d}$$ and it is very easy to show that it lies between !!\frac ab!! and !!\frac cd!!, as we want.

So instead of the relatively complicated $$\frac ab \Rightarrow \frac{a^2 + nb^2}{2ab}$$ operation, we can try the very simple and quick $$\frac ab \Rightarrow \operatorname{mediant}\left(\frac ab, \frac{nb}{a}\right) = \frac{a+nb}{b+a}$$ operation.

Taking !!n=2!! as before, and starting with !!\frac 32!!, this produces:

$$ \frac 32 \Rightarrow\frac{ 7 }{ 5 } \Rightarrow\frac{ 17 }{ 12 } \Rightarrow\frac{ 41 }{ 29 } \Rightarrow\frac{ 99 }{ 70 } \Rightarrow\frac{ 239 }{ 169 } \Rightarrow\frac{ 577 }{ 408 } \Rightarrow\cdots$$

which you may recognize as the convergents of !!\sqrt2!!. These are actually the rational approximations of !!\sqrt 2!! that are optimally accurate relative to the sizes of their denominators. Notice that !!\frac{17}{12}!! and !!\frac{577}{408}!! are in there as they were before, although it takes longer to get to them.

I think it's cool that you can view it as a highly-simplified version of Newton's method.

[ Addendum: An earlier version of the last paragraph claimed:

None of this is a big surprise, because it's well-known that you can get the convergents of !!\sqrt n!! by applying the transformation !!\frac ab\Rightarrow \frac{a+nb}{a+b}!!, starting with !!\frac11!!.

Simon Tatham pointed out that this was mistaken. It's true when !!n=2!!, but not in general. The sequence of fractions that you get does indeed converge to !!\sqrt n!!, but it's not usually the convergents, or even in lowest terms. When !!n=3!!, for example, the numerators and denominators are all even. ]

[ Addendum: Newton's method as I described it, with the crucial !!g → g - \frac{f(g)}{f'(g)}!! transformation, was actually invented in 1740 by Thomas Simpson. Both Isaac Newton and Thomas Raphson had earlier described only special cases, as had several Asian mathematicians, including Seki Kōwa. ]

[ Previous discussion of convergents: Archimedes and the square root of 3; 60-degree angles on a lattice. A different variation on the Babylonian method. ]

[ Note to self: Take a look at what the AM-GM inequality has to say about the behavior of !!\hat g!!. ]

[ Addendum 20201018: A while back I discussed the general method of picking a function !!f!! that has !!\sqrt 2!! as a fixed point, and iterating !!f!!. This is yet another example of such a function. ]

[Other articles in category /math] permanent link

Fri, 21 Aug 2020

Mixed-radix fractions in Bengali

[ Previously, Base-4 fractions in Telugu. ]

I was really not expecting to revisit this topic, but a couple of weeks ago, looking for something else, I happened upon the following curiously-named Unicode characters:


Oh boy, more base-four fractions! What on earth does “NUMERATOR ONE LESS THAN THE DENOMINATOR” mean and how is it used?

An explanation appears in the Unicode proposal to add the related “ganda” sign:

     U+09FB (e0 a7 bb): BENGALI GANDA MARK [৻]

(Anshuman Pandey, “Proposal to Encode the Ganda Currency Mark for Bengali in the BMP of the UCS”, 2007.)

Pandey explains: prior to decimalization, the Bengali rupee (rupayā) was divided into sixteen ānā. Standard Bengali numerals were used to write rupee amounts, but there was a special notation for ānā. The sign ৹ always appears, and means sixteenths. Then. Prefixed to this is a numerator symbol, which goes ৴, ৵, ৶, ৷ for 1, 2, 3, 4. So for example, 3 ānā is written ৶৹, which means !!\frac3{16}!!.

The larger fractions are made by adding the numerators, grouping by 4's:

 1, 2, 3
৷৴ ৷৵ ৷৶ 4, 5, 6, 7
৷৷৷৷৴ ৷৷৵ ৷৷৶ 8, 9, 10, 11
৸৴ ৸৵ ৸৶ 12, 13, 14, 15

except that three fours (৷৷৷) is too many, and is abbreviated by the intriguing NUMERATOR ONE LESS THAN THE DENOMINATOR sign ৸ when more than 11 ānā are being written.

Historically, the ānā was divided into 20 gaṇḍā; the gaṇḍā amounts are written with standard (Benagli decimal) numerals instead of the special-purpose base-4 numerals just described. The gaṇḍā sign ৻ precedes the numeral, so 4 gaṇḍā (!!\frac15!! ānā) is wrtten as ৻৪. (The ৪ is not an 8, it is a four.)

What if you want to write 17 rupees plus !!9\frac15!! ānā? That is 17 rupees plus 9 ānā plus 4 gaṇḍā. If I am reading this report correctly, you write it this way:


This breaks down into three parts as ১৭ ৷৷৴ ৻৪. The ১৭ is a 17, for 17 rupees; the ৷৷৴ means 9 ānā (the denominator ৹ is left implicit) and the ৻৪ means 4 gaṇḍā, as before. There is no separator between the rupees and the ānā. But there doesn't need to be, because different numerals are used! An analogous imaginary system in Latin script would be to write the amount as


where the ‘17’ means 17 rupees, the ‘dda’ means 4+4+1=9 ānā, and the ¢4 means 4 gaṇḍā. There is no trouble seeing where the ‘17’ ends and the ‘dda’ begins.

Pandey says there was an even smaller unit, the kaṛi. It was worth ¼ of a gaṇḍā and was again written with the special base-4 numerals, but as if the gaṇḍā had been divided into 16. A complete amount might be written with decimal numerals for the rupees, base-4 numerals for the ānā, decimal numerals again for the gaṇḍā, and base-4 numerals again for the kaṛi. No separators are needed, because each section is written symbols that are different from the ones in the adjoining sections.

[Other articles in category /math] permanent link

Wed, 05 Aug 2020

A maybe-interesting number trick?

I'm not sure if this is interesting, trivial, or both. You decide.

Let's divide the numbers from 1 to 30 into the following six groups:

A 1 2 4 8 16
B 3 6 12 17 24
C 5 9 10 18 20
D 7 14 19 25 28
E 11 13 21 22 26
F 15 23 27 29 30

Choose any two rows. Chose a number from each row, and multiply them mod 31. (That is, multiply them, and if the product is 31 or larger, divide it by 31 and keep the remainder.)

Regardless of which two numbers you chose, the result will always be in the same row. For example, any two numbers chosen from rows B and D will multiply to yield a number in row E. If both numbers are chosen from row F, their product will always appear in row A.

[Other articles in category /math] permanent link

Sat, 01 Aug 2020

How are finite fields constructed?

Here's another recent Math Stack Exchange answer I'm pleased with.

OP asked:

I know this question has been asked many times and there is good information out there which has clarified a lot for me but I still do not understand how the addition and multiplication tables for !!GF(4)!! is constructed?

I've seen [links] but none explicity explain the construction and I'm too new to be told "its an extension of !!GF(2)!!"

The only “reasonable” answer here is “get an undergraduate abstract algebra text and read the chapter on finite fields”. Because come on, you can't expect some random stranger to appear and write up a detailed but short explanation at your exact level of knowledge.

But sometimes Internet Magic Lightning strikes  and that's what you do get! And OP set themselves up to be struck by magic lightning, because you can't get a detailed but short explanation at your exact level of knowledge if you don't provide a detailed but short explanation of your exact level of knowledge — and this person did just that. They understand finite fields of prime order, but not how to construct the extension fields. No problem, I can explain that!

I had special fun writing this answer because I just love constructing extensions of finite fields. (Previously: [1] [2])

For any given !!n!!, there is at most one field with !!n!! elements: only one, if !!n!! is a power of a prime number (!!2, 3, 2^2, 5, 7, 2^3, 3^2, 11, 13, \ldots!!) and none otherwise (!!6, 10, 12, 14\ldots!!). This field with !!n!! elements is written as !!\Bbb F_n!! or as !!GF(n)!!.

Suppose we want to construct !!\Bbb F_n!! where !!n=p^k!!. When !!k=1!!, this is easy-peasy: take the !!n!! elements to be the integers !!0, 1, 2\ldots p-1!!, and the addition and multiplication are done modulo !!n!!.

When !!k>1!! it is more interesting. One possible construction goes like this:

  1. The elements of !!\Bbb F_{p^k}!! are the polynomials $$a_{k-1}x^{k-1} + a_{k-2}x^{k-2} + \ldots + a_1x+a_0$$ where the coefficients !!a_i!! are elements of !!\Bbb F_p!!. That is, the coefficients are just integers in !!{0, 1, \ldots p-1}!!, but with the understanding that the addition and multiplication will be done modulo !!p!!. Note that there are !!p^k!! of these polynomials in total.

  2. Addition of polynomials is done exactly as usual: combine like terms, but remember that the the coefficients are added modulo !!p!! because they are elements of !!\Bbb F_p!!.

  3. Multiplication is more interesting:

    a. Pick an irreducible polynomial !!P!! of degree !!k!!. “Irreducible” means that it does not factor into a product of smaller polynomials. How to actually locate an irreducible polynomial is an interesting question; here we will mostly ignore it.

    b. To multiply two elements, multiply them normally, remembering that the coefficients are in !!\Bbb F_p!!. Divide the product by !!P!! and keep the remainder. Since !!P!! has degree !!k!!, the remainder must have degree at most !!k-1!!, and this is your answer.

Now we will see an example: we will construct !!\Bbb F_{2^2}!!. Here !!k=2!! and !!p=2!!. The elements will be polynomials of degree at most 1, with coefficients in !!\Bbb F_2!!. There are four elements: !!0x+0, 0x+1, 1x+0, !! and !!1x+1!!. As usual we will write these as !!0, 1, x, x+1!!. This will not be misleading.

Addition is straightforward: combine like terms, remembering that !!1+1=0!! because the coefficients are in !!\Bbb F_2!!:

$$\begin{array}{c|cccc} + & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 1 & x & x+1 \\ 1 & 1 & 0 & x+1 & x \\ x & x & x+1 & 0 & 1 \\ x+1 & x+1 & x & 1 & 0 \end{array} $$

The multiplication as always is more interesting. We need to find an irreducible polynomial !!P!!. It so happens that !!P=x^2+x+1!! is the only one that works. (If you didn't know this, you could find out easily: a reducible polynomial of degree 2 factors into two linear factors. So the reducible polynomials are !!x^2, x·(x+1) = x^2+x!!, and !!(x+1)^2 = x^2+2x+1 = x^2+1!!. That leaves only !!x^2+x+1!!.)

To multiply two polynomials, we multiply them normally, then divide by !!x^2+x+1!! and keep the remainder. For example, what is !!(x+1)(x+1)!!? It's !!x^2+2x+1 = x^2 + 1!!. There is a theorem from elementary algebra (the “division theorem”) that we can find a unique quotient !!Q!! and remainder !!R!!, with the degree of !!R!! less than 2, such that !!PQ+R = x^2+1!!. In this case, !!Q=1, R=x!! works. (You should check this.) Since !!R=x!! this is our answer: !!(x+1)(x+1) = x!!.

Let's try !!x·x = x^2!!. We want !!PQ+R = x^2!!, and it happens that !!Q=1, R=x+1!! works. So !!x·x = x+1!!.

I strongly recommend that you calculate the multiplication table yourself. But here it is if you want to check:

$$\begin{array}{c|cccc} · & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & x & x+1 \\ x & 0 & x & x+1 & 1 \\ x+1 & 0 & x+1 & 1 & x \end{array} $$

To calculate the unique field !!\Bbb F_{2^3}!! of order 8, you let the elements be the 8 second-degree polynomials !!0, 1, x, \ldots, x^2+x, x^2+x+1!! and instead of reducing by !!x^2+x+1!!, you reduce by !!x^3+x+1!!. (Not by !!x^3+x^2+x+1!!, because that factors as !!(x^2+1)(x+1)!!.) To calculate the unique field !!\Bbb F_{3^2}!! of order 27, you start with the 27 third-degree polynomials with coefficients in !!{0,1,2}!!, and you reduce by !!x^3+2x+1!! (I think).

The special notation !!\Bbb F_p[x]!! means the ring of all polynomials with coefficients from !!\Bbb F_p!!. !!\langle P \rangle!! means the ring of all multiples of polynomial !!P!!. (A ring is a set with an addition, subtraction, and multiplication defined.)

When we write !!\Bbb F_p[x] / \langle P\rangle!! we are constructing a thing called a “quotient” structure. This is a generalization of the process that turns the ordinary integers !!\Bbb Z!! into the modular-arithmetic integers we have been calling !!\Bbb F_p!!. To construct !!\Bbb F_p!!, we start with !!\Bbb Z!! and then agree that two elements of !!\Bbb Z!! will be considered equivalent if they differ by a multiple of !!p!!.

To get !!\Bbb F_p[x] / \langle P \rangle!! we start with !!\Bbb F_p[x]!!, and then agree that elements of !!\Bbb F_p[x]!! will be considered equivalent if they differ by a multiple of !!P!!. The division theorem guarantees that of all the equivalent polynomials in a class, exactly one of them will have degree less than that of !!P!!, and that is the one we choose as a representative of its class and write into the multiplication table. This is what we are doing when we “divide by !!P!! and keep the remainder”.

A particularly important example of this construction is !!\Bbb R[x] / \langle x^2 + 1\rangle!!. That is, we take the set of polynomials with real coefficients, but we consider two polynomials equivalent if they differ by a multiple of !!x^2 + 1!!. By the division theorem, each polynomial is then equivalent to some first-degree polynomial !!ax+b!!.

Let's multiply $$(ax+b)(cx+d).$$ As usual we obtain $$acx^2 + (ad+bc)x + bd.$$ From this we can subtract !!ac(x^2 + 1)!! to obtain the equivalent first-degree polynomial $$(ad+bc) x + (bd-ac).$$

Now recall that in the complex numbers, !!(b+ai)(d + ci) = (bd-ac) + (ad+bc)i!!. We have just constructed the complex numbers,with the polynomial !!x!! playing the role of !!i!!.

[ Note to self: maybe write a separate article about what makes this a good answer, and how it is structured. ]

[Other articles in category /math/se] permanent link

Fri, 31 Jul 2020

What does it mean to expand a function “in powers of x-1”?

A recent Math Stack Excahnge post was asked to expand the function !!e^{2x}!! in powers of !!(x-1)!! and was confused about what that meant, and what the point of it was. I wrote an answer I liked, which I am reproducing here.

You asked:

I don't understand what are we doing in this whole process

which is a fair question. I didn't understand this either when I first learned it. But it's important for practical engineering reasons as well as for theoretical mathematical ones.

Before we go on, let's see that your proposal is the wrong answer to this question, because it is the correct answer, but to a different question. You suggested: $$e^{2x}\approx1+2\left(x-1\right)+2\left(x-1\right)^2+\frac{4}{3}\left(x-1\right)^3$$

Taking !!x=1!! we get !!e^2 \approx 1!!, which is just wrong, since actually !!e^2\approx 7.39!!. As a comment pointed out, the series you have above is for !!e^{2(x-1)}!!. But we wanted a series that adds up to !!e^{2x}!!.

As you know, the Maclaurin series works here:

$$e^{2x} \approx 1+2x+2x^2+\frac{4}{3}x^3$$

so why don't we just use it? Let's try !!x=1!!. We get $$e^2\approx 1 + 2 + 2 + \frac43$$

This adds to !!6+\frac13!!, but the correct answer is actually around !!7.39!! as we saw before. That is not a very accurate approximation. Maybe we need more terms? Let's try ten:

$$e^{2x} \approx 1+2x+2x^2+\frac{4}{3}x^3 + \ldots + \frac{8}{2835}x^9$$

If we do this we get !!7.3887!!, which isn't too far off. But it was a lot of work! And we find that as !!x!! gets farther away from zero, the series above gets less and less accurate. For example, take !!x=3.1!!, the formula with four terms gives us !!66.14!!, which is dead wrong. Even if we use ten terms, we get !!444.3!!, which is still way off. The right answer is actually !!492.7!!.

What do we do about this? Just add more terms? That could be a lot of work and it might not get us where we need to go. (Some Maclaurin series just stop working at all too far from zero, and no amount of terms will make them work.) Instead we use a different technique.

Expanding the Taylor series “around !!x=a!!” gets us a different series, one that works best when !!x!! is close to !!a!! instead of when !!x!! is close to zero. Your homework is to expand it around !!x=1!!, and I don't want to give away the answer, so I'll do a different example. We'll expand !!e^{2x}!! around !!x=3!!. The general formula is $$e^{2x} \approx \sum \frac{f^{(i)}(3)}{i!} (x-3)^i\tag{$\star$}\ \qquad \text{(when $x$ is close to $3$)}$$

The !!f^{(i)}(x)!! is the !!i!!'th derivative of !! e^{2x}!! , which is !!2^ie^{2x}!!, so the first few terms of the series above are:

$$\begin{eqnarray} e^{2x} & \approx& e^6 + \frac{2e^6}1 (x-3) + \frac{4e^6}{2}(x-3)^2 + \frac{8e^6}{6}(x-3)^3\\ & = & e^6\left(1+ 2(x-3) + 2(x-3)^2 + \frac34(x-3)^3\right)\\ & & \qquad \text{(when $x$ is close to $3$)} \end{eqnarray} $$

The first thing to notice here is that when !!x!! is exactly !!3!!, this series is perfectly correct; we get !!e^6 = e^6!! exactly, even when we add up only the first term, and ignore the rest. That's a kind of useless answer because we already knew that !!e^6 = e^6!!. But that's not what this series is for. The whole point of this series is to tell us how different !!e^{2x}!! is from !!e^6!! when !!x!! is close to, but not equal to !!3!!.

Let's see what it does at !!x=3.1!!. With only four terms we get $$\begin{eqnarray} e^{6.2} & \approx& e^6(1 + 2(0.1) + 2(0.1)^2 + \frac34(0.1)^3)\\ & = & e^6 \cdot 1.22075 \\ & \approx & 492.486 \end{eqnarray}$$

which is very close to the correct answer, which is !!492.7!!. And that's with only four terms. Even if we didn't know an exact value for !!e^6!!, we could find out that !!e^{6.2}!! is about !!22.075\%!! larger, with hardly any calculation.

Why did this work so well? If you look at the expression !!(\star)!! you can see: The terms of the series all have factors of the form !!(x-3)^i!!. When !!x=3.1!!, these are !!(0.1)^i!!, which becomes very small very quickly as !!i!! increases. Because the later terms of the series are very small, they don't affect the final sum, and if we leave them out, we won't mess up the answer too much. So the series works well, producing accurate results from only a few terms, when !!x!! is close to !!3!!.

But in the Maclaurin series, which is around !!x=0!!, those !!(x-3)^i!! terms are !!x^i!! terms intead, and when !!x=3.1!!, they are not small, they're very large! They get bigger as !!i!! increases, and very quickly. (The !! i! !! in the denominator wins, eventually, but that doesn't happen for many terms.) If we leave out these many large terms, we get the wrong results.

The short answer to your question is:

Maclaurin series are only good for calculating functions when !!x!! is close to !!0!!, and become inaccurate as !!x!! moves away from zero. But a Taylor series around !!a!! has its “center” near !!a!! and is most accurate when !!x!! is close to !!a!!.

[Other articles in category /math/se] permanent link

Wed, 08 Jul 2020

Ron Graham has died

Ron Graham has died. He had a good run. When I check out I will probably not be as accomplished or as missed as Graham, even if I make it to 84.

I met Graham once and he was very nice to me, as he apparently was to everyone. I was planning to write up a reminiscence of the time, but I find I've already done it so you can read that if you care.

Graham's little book Rudiments of Ramsey Theory made a big impression on me when I was an undergraduate. Chapter 1, if I remember correctly, is a large collection of examples, which suited me fine. Chapter 2 begins by introducing a certain notation of Erdős and Rado: !!\left[{\Bbb N\atop k}\right]!! is the family of subsets of !!\Bbb N!! of size !!k!!, and

$$\left[{\Bbb N\atop k}\right] \to \left[{\Bbb N\atop k}\right]_r$$

is an abbreviation of the statement that for any !!r!!-coloring of members of !!\left[{\Bbb N\atop k}\right]!! there is always an infinite subset !!S\subset \Bbb N!! for which every member of !!\left[{S\atop k}\right]!! is the same color. I still do not find this notation perspicuous, and at the time, with much less experience, I was boggled. In the midst of my bogglement I was hit with the next sentence, which completely derailed me:

Scan of two lines from _Rudiments of Ramsey Theory_
including the sentence “We will occasionally use this arrow notation unless there is danger
of no confusion.”

After this I could no longer think about the mathematics, but only about the sentence.

Outside the mathematical community Graham is probably best-known for juggling, or for Graham's number, which Wikipedia describes:

At the time of its introduction, it was the largest specific positive integer ever to have been used in a published mathematical proof.

One of my better Math Stack Exchange posts was in answer to the question Graham's Number : Why so big?. I love the phrasing of this question! And that, even with the strange phrasing, there is an answer! This type of huge number is quite typical in proofs of Ramsey theory, and I answered in detail.

The sense of humor that led Graham to write “danger of no confusion” is very much on display in the paper that gave us Graham's number. If you are wondering about Graham's number, check out my post.

[Other articles in category /math] permanent link

Mon, 06 Jul 2020

Weird constants in math problems

Michael Lugo recently considered a problem involving the allocation of swimmers to swim lanes at random, ending with:

If we compute this for large !!n!! we get !!f(n) \sim 0.4323n!!, which agrees with the Monte Carlo simulations… The constant !!0.4323!! is $$\frac{(1-e^{-2})}2.$$

I love when stuff like this happens. The computer is great at doing a quick random simulation and getting you some weird number, and you have no idea what it really means. But mathematical technique can unmask the weird number and learn its true identity. (“It was Old Man Haskins all along!”)

A couple of years back Math Stack Exchange had Expected Number and Size of Contiguously Filled Bins, and although it wasn't exactly what was asked, I ended up looking into this question: We take !!n!! balls and throw them at random into !!n!! bins that are lined up in a row. A maximal contiguous sequence of all-empty or all-nonempty bins is called a “cluster”. For example, here we have 13 balls that I placed randomly into 13 bins:

13 boxes, some with blue balls.  The boxes
contain, respectively, 1, 0, 3, 0, 1, 2, 1, 1, 0, 1, 2, 1, 0 balls.

In this example, there are 8 clusters, of sizes 1, 1, 1, 1, 4, 1, 3, 1. Is this typical? What's the expected cluster size?

It's easy to use Monte Carlo methods and find that when !!n!! is large, the average cluster size is approximately !!2.15013!!. Do you recognize this number? I didn't.

But it's not hard to do the calculation analytically and discover that that the reason it's approximately !!2.15013!! is that the actual answer is $$\frac1{2(e^{-1} - e^{-2})}$$ which is approximately !!2.15013!!.

Math is awesome and wonderful.

(Incidentally, I tried the Inverse Symbolic Calculator just now, but it was no help. It's also not in Plouffe's Miscellaneous Mathematical Constants)

[ Addendum 20200707: WolframAlpha does correctly identify the !!2.15013!! constant. ]

[Other articles in category /math] permanent link

Fri, 29 May 2020

Infinite zeroes with one on the end

I recently complained about people who claim:

you can't have a 1 after an infinite sequence of zeroes, because an infinite sequence of zeroes goes on forever.

When I read something like this, the first thing that usually comes to mind is the ordinal number !!\omega+1!!, which is a canonical representative of this type of ordering. But I think in the future I'll bring up this much more elementary example:

$$ S = \biggl\{-1, -\frac12, -\frac13, -\frac14, \ldots, 0\biggr\} $$

Even a middle school student can understand this, and I doubt they'd be able to argue seriously that it doesn't have an infinite sequence of elements that is followed by yet another final element.

Then we could define the supposedly problematic !!0, 0, 0, \ldots, 1!! thingy as a map from !!S!!, just as an ordinary sequence is defined as a map from !!\Bbb N!!.

[ Related: A familiar set with an unexpected order type. ]

[Other articles in category /math] permanent link

Thu, 13 Feb 2020

Gentzen's rules for natural deduction

Here is Gerhard Gentzen's original statement of the rules of Natural Deduction (“ein Kalkül für ‘natürliche’, intuitionistische Herleitungen”):

from Gentzen's 1934 paper, titled “Die Schlußfiguren-Schemata”.  The
table is laid out in three lines, with the rules for ‘and’ and ‘or’,
then the rules for ‘exists’ and “for all’, and then the rules for
‘implies’, ‘not’, and ‘false’.  The variable names are written in
old-style German black-letter font, but otherwise the presentation is
almost identical to the modern form.

Natural deduction looks pretty much exactly the same as it does today, although the symbols are a little different. But only a little! Gentzen has not yet invented !!\land!! for logical and, and is still using !!\&!!. But he has invented !!\forall!!. The style of the !!\lnot!! symbol is a little different from what we use now, and he has that tent thingy !!⋏!! where we would now use !!\bot!!. I suppose !!⋏!! didn't catch on because it looks too much like !!\land!!. (He similarly used !!⋎!! to mean !!\top!!, but as usual, that doesn't appear in the deduction rules.)

We still use Gentzen's system for naming the rules. The notations “UE” and “OB” for example, stand for “und-Einführung” and “oder-Beseitigung”, which mean “and-introduction” and “or-elimination”.

Gentzen says (footnote 4, page 178) that he got the !!\lor, \supset, \exists!! signs from Russell, but he didn't want to use Russell's signs !!\cdot, \equiv, \sim, ()!! because they already had other meanings in mathematics. He took the !!\&!! from Hilbert, but Gentzen disliked his other symbols. Gentzen objected especially to the “uncomfortable” overbar that Hilbert used to indicate negation (“[Es] stellt eine Abweichung von der linearen Anordnung der Zeichen dar”). He attributes his symbols for logical equivalence (!!\supset\subset!!) and negation to Heyting, and explains that his new !!\forall!! symbol is analogous to !!\exists!!. I find it remarkable how quickly this caught on. Gentzen also later replaced !!\&!! with !!\land!!. Of the rest, the only one that didn't stick was !!\supset\subset!! in place of !!\equiv!!. But !!\equiv!! is much less important than the others, being merely an abbreviation.

Gentzen died at age 35, a casualty of the World War.

Source: Gerhard Gentzen, “Untersuchungen über das logische Schließen I”, pp. 176–210 Mathematische Zeitschrift v. 39, Springer, 1935. The display above appears on page 186.

[ Addendum 20200214: Thanks to Andreas Fuchs for correcting my German grammar. ]

[Other articles in category /math/logic] permanent link

Thu, 06 Feb 2020

Major screwups in mathematics: example 3

[ Previously: “Cases in which some statement S was considered to be proved, and later turned out to be false”. ]

In 1905, Henri Lebesgue claimed to have proved that if !!B!! is a subset of !!\Bbb R^2!! with the Borel property, then its projection onto a line (the !!x!!-axis, say) is a Borel subset of the line. This is false. The mistake was apparently noticed some years later by Andrei Souslin. In 1912 Souslin and Luzin defined an analytic set as the projection of a Borel set. All Borel sets are analytic, but, contrary to Lebesgue's claim, the converse is false. These sets are counterexamples to the plausible-seeming conjecture that all measurable sets are Borel.

I would like to track down more details about this. This Math Overflow post summarizes Lebesgue's error:

It came down to his claim that if !!{A_n}!! is a decreasing sequence of subsets in the plane with intersection !!A!!, the the projected sets in the line intersect to the projection of !!A!!. Of course this is nonsense. Lebesgue knew projection didn't commute with countable intersections, but apparently thought that by requiring the sets to be decreasing this would work.

[Other articles in category /math] permanent link

Sun, 24 Nov 2019

The least common divisor and the greatest common multiple

One day Katara reported that her middle school math teacher had gone into a rage (whether real or facetious I don't know) about some kids’ use of the phrase “greatest common multiple”. “There is no greatest common multiple!” he shouted.

But that got me thinking. There isn't a greatest common multiple, but there certainly is a least common divisor; the least common divisor of !!a!! and !!b!! is !!1!!, for any !!a!! and !!b!!.

The least common multiple and greatest common divisor are beautifully dual, via the identity $$\operatorname{lcm}(a,b)\cdot\gcd(a,b) = ab,$$ so if there's a simple least common divisor I'd expect there would be a greatest common multiple also. I puzzled over this for a minute and realized what is going on. Contrary to Katara's math teacher, there is a greatest common multiple if you understand things properly.

There are two important orderings on the positive integers. One is the familiar total ordering by magnitude: $$0 ≤ 1≤ 2≤ 3≤ 4≤ \ldots.$$ (The closely related !!\lt!! is nearly the same: !!a\lt b!! just means that !!a\le b!! and !!a\ne b!!.)

But there is another important ordering, more relevant here, in which the numbers are ordered not by magnitude but by divisibility. We say that !!a\preceq b!! only when !!b!! is a multiple of !!a!!. This has many of the same familiar and comforting properties that !!\le!! enjoys. Like !!\le!!, it is reflexive. That is, !!a\preceq a!! for all !!a!!, because !!a!! is a multiple of itself; it's !!a\cdot 1!!.

!!\preceq!! is also transitive: Whenever !!a\preceq b!! and !!b\preceq c!!, then also !!a\preceq c!!, because if !!b!! is a multiple of !!a!!, and !!c!! is a multiple of !!b!!, then !!c!! is a multiple of !!a!!.

Like !!\le!!, the !!\preceq!! relation is also antisymmetric. We can have both !!a≤b!! and !!b≤a!!, but only if !!a=b!!. Similarly, we can have both !!a\preceq b!! and !!b\preceq a!! but only if !!a=b!!.

But unlike the familiar ordering, !!\preceq!! is only a partial ordering. For any given !!a!! and !!b!! it must be the case that either !!a\le b!! or !!b \le a!!, with both occurring only in the case where !!a=b!!. !!\preceq!! does not enjoy this property. For any given !!a!! or !!b!!, it might be the case that !!a\preceq b!! or !!b\preceq a!!, or both (if !!a=b!!) or neither. For example, neither of !!15\preceq 10!! and !!10 \preceq 15!! holds, because neither of !!10!! and !!15!! is a multiple of the other.

The total ordering !!≤!! lines up the numbers in an infinite queue, first !!0!!, then !!1!!, and so on to infinity, each one followed directly by its successor.

described in the previous paragraph

The partial ordering !!\preceq!! is less rigid and more complex. It is not anything like an infinite queue, and it is much harder to draw. A fragment is shown below:

one node
for thirty of the first 42 positive integers, arranged in a complex
graph; further description in the main text

The nodes here have been arranged so that each number is to the left of all the numbers it divides. The number !!1!!, which divides every other number, is leftmost. The next layer has all the prime numbers, each one with a single arrow pointing to its single divisor !!1!!, although I've omitted most of them from the picture. The second layer has all the numbers of the form !!p^2!! and !!pq!! where !!p!! and !!q!! are primes: !!4, 6, 9, 10, 14, \ldots!!, each one pointing to its prime divisors in the first layer. The third layer should have all the numbers of the form !!p^3, p^2q, !! and !!pqr!!, but I have shown only !!8, 12, 18, !! and !!20!!. The complete diagram would extend forever to the right.

In this graph, “least” and “greatest” have a sort-of-geometric interpretation, “least” roughly corresponding to “leftmost”. But you have to understand “leftmost” in the correct way: it really means “leftmost along the arrows”. In the picture !!11!! is left of !!4!!, but that doesn't count because it's not left of 4 along an arrow path. In the !!\preceq!! ordering, !!11\not \preceq 4!! and !!4\not \preceq 11!!. Similarly “greatest” means “rightmost, but only along the arrows”.

A divisor of a number !!n!! is exactly a node that can be reached from !!n!! by following the arrows leftward. A common divisor of !!n!! and !!m!! is a node that can be reached from both !!n!! and !!m!!. And the greatest common divisor is the rightmost such node.

Just the six nodes 1, 2, 4, 8,
12, 16 from the previous diagram.  1 is left of 2, which is left of 4,
which is left of 8 and 12.  8 is left of 16.

For example, the greatest common divisor of !!12!! and !!16!! is the rightmost node that can be reached by following arrows from both !!12!! and from !!16!!; that's !!4!!. The number !!8!! is right of !!4!! but can't be reached from !!12!!. The number !!12!! is right of !!4!! but can't be reached from !!16!!. The number !!2!! can be reached from both, but !!4!! is right of !!2!!. So the greatest common divisor is !!4!!.

Is there a least common divisor of !!n!! and !!m!!? Sure there is, it's the leftmost node that can be reached from both !!n!! and !!m!!. Clearly, it's always !!1!!, regardless of what !!n!! and !!m!! are.

Going in the other direction, a multiple of !!n!! is a node from which you can reach !!n!! by following the arrows. A common multiple of !!n!! and !!m!! is one from which you can reach both !!n!! and !!m!!, and the least common multiple of !!n!! and !!m!! is the leftmost such node.

But what about the greatest common multiple, angrily denied by Katara's math teacher? It should be the rightmost node from which you can reach both !!n!! and !!m!!. The diagram extends infinitely far to the right, so surely there isn't a rightmost such node?

Not so! There is a rightmost node! !!0!! is a multiple of !!n!! for every !!n!!, we have !!n \preceq 0!! for every !!n!!, and so zero is to the right of every other node. The diagram actually looks like this:

The same
as the previous, but with zero added on the right margin.  There are
no explicit connections with any of the other nodes, because in every
case there are (infinite many) nodes on the path between, which are
omitted from the diagram.

where I left out an infinite number of nodes in between !!0!! and the rest of the diagram.

It's a bit odd to think of !!0!! by itself there, all the way on the right, because in every other case, we have !!m \preceq n!! only when !!m≤ n!!. !!n=0!! is the only exception to that. But when you're thinking about divisibility, that's how you have to think about it: !!1!! is least, as you expect, not because it's smallest but because it's a divisor of every number. And dually, !!0!! is greatest, because it's a multiple of every number.

So that's the answer. Consideration of the seemingly silly question of the greatest common multiple has led us to a better understanding of the multiplicative structure of the integers, and we see that there is a greatest common multiple, which is exactly dual to the least common divisor, just as the least common multiple is dual to the greatest common divisor.

[Other articles in category /math] permanent link

Fri, 11 Oct 2019

More fair cake-cutting

In a recent article about fair cake division, I said:

Grandma can use the same method … to divide a regular 17-gonal cake into 23 equally-iced equal pieces.

I got to wondering what that would look like, and the answer is, not very interesting. A regular 17-gon is pretty close to a circle, and the 23 pieces, which are quite narrow, look very much like equal wedges of a circle:

A 17-gon divided into 23 equal pieces, as described in the
previous paragraph

This is generally true, and it becomes more nearly so both as the number of sides of the polygon increases (it becomes more nearly circular) and as the number of pieces increases (the very small amount of perimeter included in each piece is not very different from a short circular arc).

Recall my observation from last time that even in the nearly extreme case of a square divided into three slices, the central angles deviate from equality by only a few percent.

Of particular interest to me is this series of demonstrations of how to cut four pieces from a cake with an odd number of sides:

I think this shows that the whole question is a little bit silly: if you just cut the cake into equiangular wedges, the resulting slices are very close in volume and in frosting. If the nearly-horizontal cuts in the pentagon above had been perfectly straight and along the !!y!!-axis, they would have intersected the pentagon only 3% of a radius-length lower than they should have.

Some of the simpler divisions of simpler cakes are interesting. A solution to the original problem (of dividing a square cake into nine pieces) is highlighted.

The method as given works regardless of where you make the first cut. But the results do not look very different in any case:

The original SVG files are also available, as is the program that wrote them.

[Other articles in category /math] permanent link

Thu, 10 Oct 2019

Incenters of chocolate-iced cakes

A MathOverflow post asks:

Puzzle 1: Grandma made a cake whose base was a square of size 30 by 30 cm and the height was 10 cm. She wanted to divide the cake fairly among her 9 grandchildren. How should she cut the cake?

Okay, this is obvious.

Puzzle 2: Grandma made a cake whose base was a square of size 30 by 30 cm and the height was 10 cm. She put chocolate icing on top of the cake and on the sides, but not on the bottom. She wanted to divide the cake fairly among her 9 grandchildren so that each child would get an equal amount of the cake and the icing. How should she cut the cake?

This one stumped me; the best I could do was to cut the cake into 27 slabs, each !!\frac{10}3×10×10!! cm, and each with between 1 and 5 units of icing. Then we can give three slabs to each grandkid, taking care that each kid's slabs have a total of 7 units of icing. This seems like it might work for an actual cake, but I suspected that it wasn't the solution that was wanted, because the problem seems like a geometry problem and my solution is essentially combinatorial.

Indeed, there is a geometric solution, which is more interesting, and which cuts the cake into only 9 pieces.

I eventually gave up and looked at the answer, which I will discuss below. Sometimes when I give up I feel that if I had had thought a little harder or given up a little later, I would have gotten the answer, but not this time. It depends on an elementary property of squares that I had been completely unaware of.

This is your last chance to avoid spoilers.

The solution given is this: Divide the perimeter of the square cake into 9 equal-length segments, each of length !!\frac{120}{9}!! cm. Some of these will be straight and others may have right angles; it does not matter. Cut from the center of the cake to the endpoints of these segments; the resulting pieces will satisfy the requirements.

“Wat.” I said. “If the perimeter lengths are equal, then the areas are equal? How can that be?”

This is obviously true for two pieces; if you cut the square from the center into two pieces into two parts that divide the perimeter equally, then of course they are the same size and shape. But surely that is not the case for three pieces?

I could not believe it until I got home and drew some pictures on graph paper. Here Grandma has cut her cake into three pieces in the prescribed way:

A square with vertices
at ‹±3, ±3› and center at ‹0,0›. Three regions are marked on it: a
blue kite
with vertices
‹0,0›, ‹1,3›, ‹-3,3›, ‹-3,-1›; a pink irregular quadrilateral
with vertices ‹0,0›, ‹3,-3›, ‹3,3›, ‹1,3›; a green irregular
quadrilateral congruent to the pink one with vertices ‹0,0›, ‹-3,-1›,
‹-3,-3›, ‹3,-3›.  The three regions completely partition the square
with no overlap and nothing left over.

The three pieces are not the same shape! But each one contains one-third of the square's outer perimeter, and each has an area of 12 square units. (Note, by the way, that although the central angles may appear equal, they are not; the blue one is around 126.9° and the pink and green ones are only 116.6°.)

And indeed, any piece cut by Grandma from the center that includes one-third of the square's perimeter will have an area of one-third of the whole square:

A square with vertices
    at ‹±3, ±3› and center at ‹0,0›. Marked on it is an orange trapezoid
    with vertices ‹0,0›, ‹-3,-2›, ‹-3,3›, ‹0,3›. Also a pink pentagon with
    vertices ‹0,0›, ‹2,-3›, ‹3,-3›, ‹3,3›, ‹2,3›.  Both polygons
    include 8 units of the square’s 24 perimeter units, and both
    have an area of 12 square units.

The proof that this works is actually quite easy. Consider a triangle !!OAB!! where !!O!! is the center of the square and !!A!! and !!B!! are points on one of the square's edges.

The triangle's area is half its height times its base. The base is of course the length of the segment !!AB!!, and the height is the length of the perpendicular from !!O!! to the edge of the square. So for any such triangle, its area is proportional to the length of !!AB!!.

No two of the five triangles below are congruent, but each has the same base and height, and so each has the same area.

Five wedges
radiate downward in different directions from the center of the
square, each arriving at a different part of the edge but each with a
base of 1 unit.

Since the center of the square is the same distance from each of the four edges, the same is true for any two triangles, regardless of which edge they arise from: the area of each triangle is proportional to the length of the square's perimeter at its base. Any piece Grandma cuts in this way, from the center of the cake to the edge, is a disjoint union of triangular pieces of this type, so the total area of any such piece is also proportional to the length of the square's perimeter that it includes.

That's the crucial property of the square that I had not known before: if you make cuts from the center of a square, the area of the piece you get is proportional to the length of the perimeter that it contains. Awesome!

Here Grandma has used the same method to cut a pair of square cakes into ten equal-sized pieces that all the have same amount of icing.

A 10×10 square divided
    into five pieces from the center.  The pieces are three
    different shapes, but each piece contains 8 units
    of the square's perimeter and has an area of 20 square units.

The crucial property here was that the square’s center is the same distance from each of its four edges. This is really obvious, but not every polygon has an analogous point. The center of a regular polygon always has this property, and every triangle has a unique point, called its incenter, which enjoys this property. So Grandma can use the same method to divide a triangular cake into 7 equally-iced equal pieces, if she can find its incenter, or to divide a regular 17-gonal cake into 23 equally-iced equal pieces.

Not every polygon does have an incenter, though. Rhombuses and kites always do, but rectangles do not, except when they are square. If Grandma tries this method with a rectangular sheet cake, someone will get shortchanged. I learned today that polygons that have incenters are known as tangential polygons. They are the only polygons in which can one inscribe a single circle that is tangent to every side of the polygon. This property is easy to detect: these are exactly the polygons in which all the angle bisectors meet at a single point. Grandma should be able to fairly divide the cake and icing for any tangential polygon.

I have probably thought about this before, perhaps in high-school geometry but perhaps not since. Suppose you have two lines, !!m!! and !!n!!, that cross at an acute angle at !!P!!, and you consider the set of points that are equidistant from both !!m!! and !!n!!. Let !!\ell!! be a line through !!P!! which bisects the angle between !!m!! and !!n!!; clearly any point on !!\ell!! will be equidistant from !!m!! and !!n!! by a straightforward argument involving congruent triangles.

Now consider a triangle !!△ABC!!. Let !!P!! be the intersection of the angle bisectors of angles !!∠ A!! and !!∠ B!!.

triangle, as described, with the two angle bisectors drawn and their
intersection at P.  Centered at P is a circle that is tangent to all
of AB, BC, and CA at the same time.

!!P!! is the same distance from both !!AB!! and !!AC!! because it is on the angle bisector of !!∠ A!!, and similarly it is the same distance from both !!AB!! and !!BC!! because it is on the angle bisector of !!∠ B!!.

So therefore !!P!! is the same distance from both !!AC!! and !!BC!! and it must be on the angle bisector of angle !!∠ C!! also. We have just shown that a triangle's three angle bisectors are concurrent! I knew this before, but I don't think I knew a proof.

[ Addendum 20191011: Many illustrated examples. ]

[Other articles in category /math] permanent link

Tue, 17 Sep 2019

Breaking pills

Suppose you have a bottle that contains !!N!! whole pills. Each day you select a pill at random from the bottle. If it is a whole pill you eat half and put the other half back in the bottle. If it is a half pill, you eat it. How many half-pills can you expect to have in the bottle the day after you break the last whole pill?

Let's write !!H(N)!! for the expected number of half-pills left, if you start with !!N!!. It's easily seen that !!H(N) = 0, 1, \frac32!! for !!N=0,1,2!!, and it's not hard to calculate that !!H(3) = \frac{11}{6}!!.

For larger !!N!! it's easy to use Monte Carlo simulation, and find that !!H(30)!! is almost exactly !!4!!. But it's also easy to use dynamic programming and compute that $$H(30) = \frac{9304682830147}{2329089562800}$$ exactly, which is a bit less than 4, only !!3.994987!!. Similarly, the dynamic programming approach tells us that $$H(100) = \frac{14466636279520351160221518043104131447711}{2788815009188499086581352357412492142272}$$ which is about !!5.187!!.

(I hate the term “dynamic programming”. It sounds so cool, but then you find out that all it means is “I memoized the results in a table”. Ho hum.)

As you'd expect for a distribution with a small mean, you're much more likely to end with a small number of half-pills than a large number. In this graph, the red line shows the probability of ending with various numbers of half-pills for an initial bottle of 100 whole pills; the blue line for an initial bottle of 30 whole pills, and the orange line for an initial bottle of 5 whole pills. The data were generated by this program.

screenshot of the graph
described above.  In each case, the probability starts relatively
high, then drops rapidly to nearly zero.

The !!E!! function appears to increase approximately logarithmically. It first exceeds !!2!! at !!N=4!!, !!3!! at !!N=11!!, !!4!! at !!N=31!!, and !!5!! at !!N=83!!. The successive ratios of these !!N!!-values are !!2.75, 2.81,!! and !!2.68!!. So we might guess that !!H(N)!! first exceeds 6 around !!N=228!! or so, and indeed !!H(226) < 6 < H(227)!!. So based on purely empirical considerations, we might guess that $$H(N) \approx \frac{\log{\frac{15}{22}N}}{\log 2.75}.$$

(The !!\frac{15}{22}!! is a fudge factor to get the curves to line up properly.)

I don't have any theoretical justification for this, but I think it might be possible to get a bound.

I don't think modeling this as a Markov process would work well. There are too many states, and it is not ergodic.

[ Addendum 20190919: Ben Handley informs me that !!H(n)!! is simply the harmonic numbers, !!H(n) = \sum_1^n \frac1n!!. I feel a little foolish that I did not notice that the first four terms matched. The appearance of !!H(3)=\frac{11}6!! should have tipped me off. Thank you, M. Handley. ]

[ Addendum 20190920: I was so fried when I wrote this that I also didn't notice that the denominator I guessed, !!2.75!!, is almost exactly !!e!!. (The correct value is !!e!!.) I also suspect that the !!\frac{15}{22}!! is just plain wrong, and ought to be !!e^\gamma!! or something like that, but I need to take a closer look. ]

[ Addendum 20191004: More about this ]

[Other articles in category /math] permanent link

Sat, 14 Sep 2019

Convergence of Taylor series

[ (Previously) ]

Wikipedia has an animation that shows the convergence of the Maclaurin series for the exponential function:

graphic, which focuses on x between -3 and +3, and y between -5 and
+20.  It displays the curve
y=exp(x) overlaid with its sequence of Maclaurin series
approximations, one after the other.

This makes it look a lot better-behaved than it is. First, because it only shows a narrow region around the center of convergence. But also, it emphasizes the right-hand quadrant, where the convergence is monotone.

If you focus on the left-hand side instead, you see the appoximations thrashing back and forth like a screen door in a hurricane:

The same,
but it focuses on x between -6 and 0, and y between -5 and 1.

This is the same as Wikipedia's animation, just focused on a slightly different region of the plane. But how different it looks!

(Individual frames are available)

[Other articles in category /math] permanent link

Sun, 08 Sep 2019

The exponential function is a miracle

The Maclaurin series for the exponential function converges for every complex number !!x!!: $$1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \ldots = e^{-x} $$

Say that !!x!! is any reasonably large number, such as 5. Then !!e^{-x}!! is close to zero, But the terms of the series are not close to zero. For !!x=5!! we have: $$ 1 - 5 + 12.5 - 20.83 + 26.04 - 26.04 + 21.7 - 15.5 + 9.69 - 5.38 + \ldots \approx {\Large 0}$$

Somehow all these largish random numbers manage to cancel out almost completely. And the larger we make !!x!!, the more of these largish random numbers there are, the larger they are, and yet the more exactly they cancel out. For even as small an argument as !!x=20!!, the series begins with 52 terms that vary between 1 and forty-three million, and these somehow cancel out almost entirely. The sum of these 52 numbers is !!-0.4!!.

In this graph, the red lines are the various partial sums (!!1-x, 1-x+\frac{x^2}2, !! etc.) and the blue line is the total sum !!e^{-x}!!.

As you can see, each red line is a very bad approximation to the blue one, except within a rather narrow region. And yet somehow, it all works out in the end.

[ (Addendum) ]

[Other articles in category /math] permanent link

Wed, 07 Aug 2019

Technical devices for reducing the number of axioms

In a recent article, I wrote:

I guessed it was a mere technical device, similar to the one that we can use to reduce five axioms of group theory to three. …

The fact that you can discard two of the axioms is mildly interesting, but of very little practical value in group theory.

There was a sub-digression, which I removed, about a similar sort of device that does have practical value. Suppose you have a group !!\langle G, \ast \rangle!! with a nonempty subset !!H\subset G!!, and you want to show that !!\langle H, \ast \rangle!! is a subgroup of !!G!!. To do this is it is sufficient to show three things:

  1. !!H!! is closed under !!\ast!!
  2. !!G!!'s identity element is in !!H!!
  3. For each element !!h!! of !!H!!, its inverse !!h^{-1}!! is also in !!H!!

Often, however, it is more convenient to show instead:

For each !!a, b\in H!!, the product !!ab^{-1}!! is also in !!H!!

which takes care of all three at once.

[Other articles in category /math] permanent link

Tue, 23 Jul 2019

Obtuse axiomatization of category theory

About ten years ago I started an article, addressed to my younger self, reviewing various books in category theory. I doubt I will ever publish this. But it contained a long, plaintive digression about Categories, Allegories by Peter Freyd and Andre Scedrov:

I keep this one around on the shelf just so that I can pick it up ever few months and marvel at its opacity.

It is a superb example of the definition-theorem-remark style of mathematics textbooks. I have heard that this was a style pioneered by a book you are already familiar with, John Kelley's General Topology of 1955. If so, all I can say is, sometimes it works, and sometimes it doesn't. It worked for Kelley in 1955 writing about topology.

Here is an example of what is wrong with this book.

Everyone who knows anything about category theory knows that a category is a sort of abstraction of a domain of mathematical objects, like sets, groups, or topological spaces. A category has a bunch of "objects", which are the sets, the groups, or the topological spaces, and it has a bunch of "morphisms", which are maps between the objects that preserve the objects' special structure, be it algebraic, topological, or whatever. If the objects are sets, the morphisms are simply functions. If the objects are groups, the morphisms are group homomorphisms; if the objects are topological spaces, the morphisms are continuous maps. The basic point of category theory is to study the relationships between these structure-preserving maps, independent of the underlying structure of the objects themselves. We ignore the elements of the sets or groups, and the points in the topological spaces, and instead concentrate on the relationships between whole sets, groups, and spaces, by way of these "morphisms".

Here is the opening section of Categories, Allegories:


The theory of CATEGORIES is given by two unary operations and a binary partial operation. In most contexts lower-case variables are used for the 'individuals' which are called morphisms or maps. The values of the operations are denoted and pronounced as:
!!□x!! the source of !!x!!,
!!x□!! the target of !!x!!,
!!xy!! the composition of !!x!! and !!y!!,
The axioms:
!!xy!! is defined iff !!x□ = □y!!,
!!(□x)□ = □x!! and !!□(x□) = x□!!,
!!(□x)x = x!! and !!x(x□) = x!!,
!!□(xy) = □(x(□y))!! and !!(xy)□ = ((x□)y)□!!,
!!x(yz) = (xy)z!!.

In light of my capsule summary of category theory, can you figure out what is going on here? Even if you already know what is supposed to be going on you may not be able to make much sense of this. What to make of the axiom that !!□(xy) = □(x(□y))!!, for example?

The explanation is that Freyd has presented a version of category theory in which the objects are missing. Since every object !!X!! in a category is associated with a unique identity morphism !!{\text{id}}_X!! from !!X!! to itself, Freyd has identified each object with its identity morphism. If !!x:C\to D!!, then !!□x!! is !!{\text{id}}_C!! and !!x□!! is !!{\text{id}}_D!!. The axiom !!(□x)□ = □x!! is true because both sides are equal to !!{\text{id}}_C!!.

Still, why phrase it this way? And what about that !!□(x(□y))!! thing? I guessed it was a mere technical device, similar to the one that we can use to reduce five axioms of group theory to three. Normally, one defines a group to have an identity element !!e!! such that !!ex=xe=x!! for all !!x!!, and each element !!x!! has an inverse !!x^{-1}!! such that !!xx^{-1} = x^{-1}x = e!!. But if you are trying to be clever, you can observe that it is sufficient for there to be a left identity and a left inverse:

There must be an identity !!e!! such that !!ex=x!! for all !!x!!, and for each !!x!! there must be an !!x^{-1}!! such that !!x^{-1}x=e!!.

We no longer require !!xe=x!! or !!xx^{-1}=e!!, but it turns out that you can prove these anyway, from what is left. The fact that you can discard two of the axioms is mildly interesting, but of very little practical value in group theory.

I thought that probably the !!□(x(□y))!! thing was some similar bit of “cleverness”, and that perhaps by adopting this one axiom Freyd was able to reduce his list of axioms. For example, from that mysterious fourth axiom !!□(xy) = □(x(□y))!! you can conclude that !!xy!! is defined if and only if !!x(□y)!! is, and therefore, by the first axiom, that !!x□ = □y!! if and only if !!x□ = □(□y)!!, so that !!□y = □(□y)!!. So perhaps the phrasing of the axiom was chosen to allow Freyd to dispense with an additional axiom stating that !!□y = □(□y)!!.

Today I tinkered with it a little bit and decided I think not.

Freyd has:

$$\begin{align} xy \text{ is defined if and only if } x□ & = □y \tag{1} \\ (□x)□ & = □x \tag{2} \\ (□x)x & = x \tag{3} \\ □(xy) & = □(x(□y)) \tag{4} \end{align} $$

and their duals. Also composition is associative, which I will elide.

In place of 4, let's try this much more straightforward axiom:

$$ □(xy) = □x\tag{$4\star$} $$

I can now show that !!1, 2, 3, 4\star!! together imply !!4!!.

First, a lemma: !!□(□x) = □x!!. Axiom !!3!! says !!(□x)x = x!!, so therefore !!□((□x)x) = □x!!. By !!4\star!!, the left-hand side reduces to !!□(□x)!!, and we are done.

Now I want to show !!4!!, that !!□(xy) = □(x(□y))!!. Before I can even discuss the question I first need to show that !!x(□y)!! is defined whenever !!xy!! is; that is, whenever !!x□ = □y!!. But by the lemma, !!□y=□(□y)!!, so !!x□ = □(□y)!!, which is just what we needed.

At this point, !!4\star!! implies !!4!! directly: both sides of !!4!! have the form !!□(xz)!!, and !!4\star!! tells us that both are equal to !!□x!!.

Conversely, !!4!! implies !!4\star!!. So why didn't Freyd use !!4\star!! instead of !!4!!? I emailed him to ask, but he's 83 so I may not get an answer. Also, knowing Freyd, there's a decent chance I won't understand the answer if I do get one.

My plaintive review of this book continued:

Another, briefer complaint about this book: Early on, no later than page 13, Freyd begins to allude to "Lazard sheaves". These are apparently an important example. Freyd does not define or explain what "Lazard sheaves" are. Okay, you are expected to do some background reading, perhaps. Fair enough. But you are doomed, because "Lazard sheaves" is Freyd's own private coinage, and you will not be able to look it up under that name.

Apparently some people like this book. I don't know why, and perhaps I never will.

[Other articles in category /math] permanent link

Tue, 25 Jun 2019

Projection morphisms are (not) epic

I don't have a good catalog in my head of basic theorems of category theory. Every time I try to think about category theory, I get stuck on really basic lemmas like “can I assume that a product !!1×A!! is canonically isomorphic to !!A!!?” Or “Suppose !!f:A→B!! is both monic and epic. Must it be an isomorphism?” Then I get sidetracked trying to prove those lemmas, or else I assume they are true, go ahead, and even if I'm right I start to lose my nerve.

So for years I've wanted to make up a book of every possible basic theorem of category theory, in order from utterly simple to most difficult, and then prove every theorem. Then I'd know all the answers, or if I didn't, I could just look in the book. There would be a chapter on products with a long list of plausible-seeming statements:

  1. If !!P_1!! and !!P_2!! are both products of some !!A!! and !!B!!, then !!P_1\cong P_2!!
  2. !!A×B \cong B×A!!
  3. !!(A×B)×C \cong A×(B×C)!!
  4. !!1×A\cong A!!
  5. !!0×A\cong 0!!
  6. !!A×A\cong A!! if and only if !!A!! is initial
  7. If !!f!! and !!g!! are both monic, then so is !!f×g!!
  8. If !!f×g!! is monic, so are !!f!! and !!g!!
  9. (etc…)

and each one would either be annotated, Snopes-style, with “True”, or with a brief description of a counterexample.

On Sunday I thought I'd give it a shot, and I started with:

Suppose !!A×B!! is a product with projection morphism !!π_1:A×B→A!!. Then !!π_1!! is epic.

This seems very plausible, because how could the product possibly work if its left-hand component couldn't contain any possible element of !!A!!?

I struggled with this for rather a long time but I just got more and more stuck. To prove that !!π_1!! is an epimorphism I need to have !!g,h:A→C!! and then show that !!g ∘ π_1 = h ∘ π_1!! only when !!g=h!!. But !!π_1!! being a projection arrow doesn't help with this, because products are all about arrows into !!A!! and !!B!! and here I need to show something about arrows out of !!A!!.

And there's no hope that I could get any leverage by introducing some arrows into !!A!! and !!B!!, because there might not be any arrows into !!B!!. (Other than !!\text{id}_B!!, of course, but then you need an arrow !!B→A!! and that might not exist either.) Or what if I consider how the arrows from !!A×B!! factor through !!C×B!! — ah ah ah, not so fast! !!C×B!! might not exist!

I eventually gave up and looked it up online, and discovered that, in fact, the claim is not true in general. It's not even true in Set. The left projection !!π_1: X×\emptyset → X!! is not epic. (Which answers my rhetorical question above that asks “how could the product possibly work if…”)

So, uh, victory, I guess? I set out to prove something that is false, so failing to produce a proof is the best possible outcome.

And I can make lemonade out of the lemons. I couldn't prove the theorem, and my ideas about why not were basically right. Now I ought to be able to look carefully at what additional tools I might be able to use to make the proof go through anyway, and those then become part of the statement of the theorem, which then would become something like “If all binary products exist in a category with an initial object, then projection morphisms are epic.”

[Other articles in category /math] permanent link

Sat, 08 Jun 2019

The inside-outness of category theory

I have pondered category theory periodically for the past 35 years, but not often enough to really stay comfortable with it. Today I was pondering again. I wanted to prove that !!1×A \cong A!! and I was having trouble. I eventually realized my difficulty: my brain had slipped out of category theory mode so that the theorem I was trying to prove did not even make sense.

In most of mathematics, !!1\times A!! would denote some specific entity and we would then show that that entity had a certain property. For example, in set theory we might define !!1\times A!! to be some set, say the set of all Kuratowski pairs !!\langle \varnothing, a\rangle!! where !!a\in A!!:

$$ 1×A =_{\text{def}} \{ z : \exists a\in A : z = \{\{\varnothing\}, \{\varnothing, a\}\} \} $$

and then we would explicitly construct a bijection !!f:A\leftrightarrow 1×A!!:

$$ f(a) = \{\{\varnothing\}, \{\varnothing, a\}\}$$

In category theory, this is not what we do. Everything is less concrete. !!\times!! looks like an operator, one that takes two objects and yields a third. It is not. !!1\times A!! does not denote any particular entity with any particular construction. (Nor does !!1!!, for that matter.) Instead, it denotes an unspecified entity, which happens to have a certain universal property, perhaps just one of many such entities with that property, and there is no canonical representative of the class. It's a mistake to think of it as a single thing, and it's also a mistake to ask the question the way I did ask it. You can't show that !!1×A!! has any specific property, because it's not a specific thing. All you can do is show that anything with the one required universal property must also have the other property. We should rephrase the question like this:

Let !!1×A!! be a product of !!1!! and !!A!!. Then !!1×A\cong A!!.

Maybe a better phrasing is:

Let !!1×A!! be some object that is a product of !!1!! and !!A!!. Then !!1×A\cong A!!.

The notation is still misleading, because it looks like !!1×A!! denotes the result of some operation, and it isn't. We can do a little better:

Let !!B!! be a product of !!1!! and !!A!!. Then !!B\cong A!!.

That it, that's the real theorem. It seems at first to be more difficult — where do we start? But it's actually easier! Because now it's enough to simply prove that !!A!! itself is a product of !!1!! and !!A!!, which is easily done: its projection morphisms are evidently !!! !! and !!{\text{id}}_A!!. And by a previous theorem that all products are isomorphic, any other product, such as !!B!!, must be isomorphic to this one, which is !!A!! itself.

(We can similarly imagine that any theorem that mentions !!1!! is preceded by the phrase “Let !!1!! be some terminal object.”)

[Other articles in category /math] permanent link

Fri, 17 May 2019

What's the difference between 0/0 and 1/0?

Last year a new Math Stack Exchange user asked What's the difference between !!\frac00!! and !!\frac10!!?.

I wrote an answer I thought was pretty good, but the question was downvoted and deleted as “not about mathematics”. This is bullshit, but what can I do?

I can repatriate my answer here, anyway.

This long answer has two parts. The first one is about the arithmetic, and is fairly simple, and is not very different from the other answers here: neither !!\frac10!! nor !!\frac00!! has any clear meaning. But your intuition is a good one: if one looks at the situation more carefully, !!\frac10!! and !!\frac00!! behave rather differently, and there is more to the story than can be understood just from the arithmetic part. The second half of my answer tries to go into these developments.

The notation !!\frac ab!! has a specific meaning:

The number !!x!! for which $$x\cdot b=a.$$

Usually this is simple enough. There is exactly one number !!x!! for which !!x\cdot 7=21!!, namely !!3!!, so !!\frac{21}7=3!!. There is exactly one number !!x!! for which !!x\cdot 4=7!!, namely !!\frac74!!, so !!\frac74\cdot4=7!!.

But when !!b=0!! we can't keep the promise that is implied by the word "the" in "The number !!x!! for which...". Let's see what goes wrong. When !!b=0!! the definition says:

The number !!x!! for which $$x\cdot 0=a.$$

When !!a\ne 0!! this goes severely wrong. The left-hand side is zero and the right-hand size is not, so there is no number !!x!! that satisfies the condition. Suppose !!x!! is the ugliest gorilla on the dairy farm. But the farm has no gorillas, only cows. Any further questions you have about !!x!! are pointless: is !!x!! a male or female gorilla? Is its fur black or dark gray? Does !!x!! prefer bananas or melons? There is no such !!x!!, so the questions are unanswerable.

When !!a!! and !!b!! are both zero, something different goes wrong:

The number !!x!! for which $$x\cdot 0=0.$$

It still doesn't work to speak of "The number !!x!! for which..." because any !!x!! will work. Now it's like saying that !!x!! is ‘the’ cow from the dairy farm, But there are many cows, so questions about !!x!! are still pointless, although in a different way: Does !!x!! have spots? I dunno man, what is !!x!!?

Asking about this !!x!!, as an individual number, never makes sense, for one reason or the other, either because there is no such !!x!! at all (!!\frac a0!! when !!a≠0!!) or because the description is not specific enough to tell you anything (!!\frac 00!!).

If you are trying to understand this as a matter of simple arithmetic, using analogies about putting cookies into boxes, this is the best you can do. That is a blunt instrument, and for a finer understanding you need to use more delicate tools. In some contexts, the two situations (!!\frac00!! and !!\frac10!!) are distinguishable, but you need to be more careful.

Suppose !!f!! and !!g!! are some functions of !!x!!, each with definite values for all numbers !!x!!, and in particular !!g(0) = 0!!. We can consider the quantity $$q(x) = \frac{f(x)}{g(x)}$$ and ask what happens to !!q(x)!! when !!x!! gets very close to !!0!!. The quantity !!q(0)!! itself is undefined, because at !!x=0!! the denominator is !!g(0)=0!!. But we can still ask what happens to !!q!! when !!x!! gets close to zero, but before it gets all the way there. It's possible that as !!x!! gets closer and closer to zero, !!q(x)!! might get closer and closer to some particular number, say !!Q!!; we can ask if there is such a number !!Q!!, and if so what it is.

It turns out we can distinguish quite different situations depending on whether the numerator !!f(0)!! is zero or nonzero. When !!f(0)\ne 0!!, we can state decisively that there is no such !!Q!!. For if there were, it would have to satisfy !!Q\cdot 0=f(0)!! which is impossible; !!Q!! would have to be a gorilla on the dairy farm. There are a number of different ways that !!q(x)!! can behave in such cases, when its denominator approaches zero and its numerator does not, but all of the possible behaviors are bad: !!q(x)!! can increase or decrease without bound as !!x!! gets close to zero; or it can do both depending on whether we approach zero from the left or the right; or it can oscillate more and more wildly, but in no case does it do anything like gently and politely approaching a single number !!Q!!.

But if !!f(0) = 0!!, the answer is more complicated, because !!Q!! (if it exists at all) would only need to satisfy !!Q\cdot 0=0!!, which is easy. So there might actually be a !!Q!! that works; it depends on further details of !!f!! and !!g!!, and sometimes there is and sometimes there isn't. For example, when !!f(x) = x^2+2x!! and !!g(x) = x!! then !!q(x) = \frac{x^2+2x}{x}!!. This is still undefined at !!x=0!! but at any other value of !!x!! it is equal to !!x+2!!, and as !!x!! approaches zero, !!q(x)!! slides smoothly in toward !!2!! along the straight line !!x+2!!. When !!x!! is close to (but not equal to) zero, !!q(x)!! is close to (but not equal to) !!2!!; for example when !!x=0.001!! then !!q(x) = \frac{0.002001}{0.001} = 2.001!!, and as !!x!! gets closer to zero !!q(x)!! gets even closer to !!2!!. So the number !!Q!! we were asking about does exist, and is in fact equal to !!2!!. On the other hand if !!f(x) = x!! and !!g(x) = x^2!! then there is still no such !!Q!!.

The details of how this all works, when there is a !!Q!! and when there isn't, and how to find it, are very interesting, and are the basic idea that underpins all of calculus. The calculus part was invented first, but it bothered everyone because although it seemed to work, it depended on an incoherent idea about how division by zero worked. Trying to frame it as a simple matter of putting cookies into boxes was no longer good enough. Getting it properly straightened out was a long process that took around 150 years, but we did eventually get there and now I think we understand the difference between !!\frac10!! and !!\frac00!! pretty well. But to really understand the difference you probably need to use the calculus approach, which may be more delicate than what you are used to. But if you are interested in this question, and you want the full answer, that is definitely the way to go.

[Other articles in category /math] permanent link

Thu, 25 Apr 2019

Sometimes it matters how you get there

Katara was given the homework exercise of rationalizing the denominator of $$\frac1{\sqrt2+\sqrt3+\sqrt5}$$ which she found troublesome. You evidently need to start by multiplying the numerator and denominator by !!-\sqrt2 + \sqrt 3 + \sqrt 5!!, obtaining $$ \frac1{(\sqrt2+\sqrt3+\sqrt5)}\cdot \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{-\sqrt2 + \sqrt 3 + \sqrt 5} = \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{(-2 +3 + 5 + 2\sqrt{15})} = \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{6 + 2\sqrt{15}} $$ and then you go from there, multiplying the top and bottom by !!6 - 2\sqrt{15}!!. It is a mess.

But when I did it, it was much quicker. Instead of using !!-\sqrt2 + \sqrt 3 + \sqrt 5!!, I went with !!\sqrt2 + \sqrt 3 - \sqrt 5!!, not for any reason, but just at random. This got me: $$ \frac1{\sqrt2+\sqrt3+\sqrt5}\cdot \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{\sqrt2 + \sqrt 3 - \sqrt 5} = \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{(2 +3 - 5 + 2\sqrt{6})} = \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{2\sqrt{6}} $$

with the !!2+3-5!! vanishing in the denominator. Then the next step is quite easy; just get rid of the !!\sqrt6!!: $$ \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{2\sqrt{6}}\cdot \frac{\sqrt6}{\sqrt6} = \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12} $$

which is correct.

I wish I could take credit for this, but it was pure dumb luck.

[Other articles in category /math] permanent link

Wed, 02 Jan 2019

More about the happy numeric coincidence

[ Previously ]

Observing three small examples where the digital sum of !!n^k!! was equal to !!n!!, I said:

I supposed that there were few other examples, probably none … Still I hoped there might be one or two…

but in fact there are a great many; for example the digital sum of !!217^{22}!! is !!217!!.

Dave McKee asked what my intuition was. It was something like this. Fix !!n!! and consider the sequence !!n^2, n^3, …!!. The elements become increasingly sparse, and for their digital sum to equal !!n!! requires increasingly improbable coincidences. For example, taking !!n=3!!, we would need to have !!k!! such that $$3^k = 2·10^p + 1.$$ While I can't immediately rule out this possibility, it seems extremely unlikely. This also resembles the Catalan conjecture, which is that the only nontrivial solution of $$ \left\lvert2^i - 3^j\right\rvert = 1$$ is !!i=3, j=2!!. The Catalan conjecture is indeed true, but it was an open problem for 158 years.

[ Addendum 20190205: I have since learned that the Catalan conjecture actually concerns the more general claim about the equation !! \left\lvert a^i - b^j\right\rvert = 1!!. The special case of !!a=2, b=3!! turns out to be elementary. ]

I was quite mistaken, so what went wrong? First, there is one nontrivial solution to the Catalan conjecture, and here we have not one sequence of !!n^k!! but an infinite family, each of which might have an exceptional solution or two. And second, the case of !!n=3!! is atypically bleak. For !!3^k!! to have a digital sum of !!3!! is a tough order because there are very few sequences of digits whose sum is 3. But the number of sequences of !!d!! digits whose sum is !!n!! grows quite rapidly with !!n!!, and for small !!n!! is very misleading.

[Other articles in category /math] permanent link

Tue, 01 Jan 2019

A happy numeric coincidence

A couple of days ago I was pleased to notice the following coincidence:

!!9^2 = 81!!and !!8 + 1 = 9!!
!!8^3 = 512!!and !!5 + 1 + 2 = 8!!
!!7^4 = 2401!!and !!2 + 4 + 0 + 1 = 7!!

I supposed that there were few other examples, probably none, and that at least I could prove that there were only a finite number of examples. My expected proof of this didn't work, but I still supposed that there would be no further examples. Still I hoped there might be one or two, so I set the computer to look for them if there were.

My first run produced:

\begin{array}{rcr} 17^3 &=& 4\;913 \\ 18^3 &=& 5\;832\\ 22^4 &=& 234\;256\\ 25^4 &=& 390\;625\\ 26^3 &=& 17\;576\\ 27^3 &=& 19\;683\\ 28^4 &=& 614\;656\\ 36^4 &=& 1\;679\;616\\ \end{array}

Well, that was a happy surprise.

Wait a minute, though:

\begin{array}{rcr} 18 ^{ 6 } &=& 34\;012\;224 \\ 18 ^{ 7 } &=& 612\;220\;032 \\ 27 ^{ 7 } &=& 10\;460\;353\;203 \\ 28 ^{ 5 } &=& 17\;210\;368 \\ 31 ^{ 7 } &=& 27\;512\;614\;111 \\ 34 ^{ 7 } &=& 52\;523\;350\;144 \\ 35 ^{ 5 } &=& 52\;521\;875 \\ 36 ^{ 5 } &=& 60\;466\;176 \\ 43 ^{ 7 } &=& 271\;818\;611\;107 \\ 45 ^{ 6 } &=& 8\;303\;765\;625 \\ 46 ^{ 5 } &=& 205\;962\;976 \\ 46 ^{ 8 } &=& 20\;047\;612\;231\;936 \\ 53 ^{ 7 } &=& 1\;174\;711\;139\;837 \\ 54 ^{ 6 } &=& 24\;794\;911\;296 \\ 54 ^{ 8 } &=& 72\;301\;961\;339\;136 \\ 54 ^{ 9 } &=& 3\;904\;305\;912\;313\;344 \\ 58 ^{ 7 } &=& 2\;207\;984\;167\;552 \\ 63 ^{ 8 } &=& 248\;155\;780\;267\;521 \\ 64 ^{ 6 } &=& 68\;719\;476\;736 \\ 68 ^{ 7 } &=& 6\;722\;988\;818\;432 \\ 71 ^{ 9 } &=& 45\;848\;500\;718\;449\;031 \\ 81 ^{ 9 } &=& 150\;094\;635\;296\;999\;121 \\ 82 ^{ 10 } &=& 13\;744\;803\;133\;596\;058\;624 \\ 85 ^{ 10 } &=& 19\;687\;440\;434\;072\;265\;625 \\ 94 ^{ 10 } &=& 53\;861\;511\;409\;489\;970\;176 \\ 97 ^{ 10 } &=& 73\;742\;412\;689\;492\;826\;049 \\ 106 ^{ 10 } &=& 179\;084\;769\;654\;285\;362\;176 \\ 117 ^{ 10 } &=& 480\;682\;838\;924\;478\;847\;449 \\ \end{array}

Oh my.

\begin{array}{rcr} 20 ^{ 13 } &=& 81\;920 & · 10^{12} \\ 40 ^{ 13 } &=& 671\;088\;640 & · 10^{12} \\ 80 ^{ 17 } &=& 225\;179\;981\;368\;524\;800 & · 10^{15} \\ \hline 80 ^{ 19 } &=& 1\;441\;151\;880\;758\;558\;720 & · 10^{18} \\ 86 ^{ 13 } &=& 14\;076\;019\;706\;120\;526\;112\;710\;656 \\ 90 ^{ 19 } &=& 13\;508\;517\;176\;729\;920\;890 & · 10^{18} \\ \hline 90 ^{ 20 } &=& 1\;215\;766\;545\;905\;692\;880\;100 & · 10^{18} \\ 90 ^{ 21 } &=& 109\;418\;989\;131\;512\;359\;209 & · 10^{21} \\ 90 ^{ 22 } &=& 9\;847\;709\;021\;836\;112\;328\;810 & · 10^{21} \\ \hline 90 ^{ 28 } &=& 5\;233\;476\;330\;273\;605\;372\;135\;115\;210 & · 10^{27} \\ 91 ^{ 14 } &=& 2\;670\;419\;511\;272\;061\;205\;254\;504\;361 \\ 98 ^{ 11 } &=& 8\;007\;313\;507\;497\;959\;524\;352 \\ \hline 103 ^{ 13 } &=& 146\;853\;371\;345\;156\;431\;381\;127\;623 \\ 104 ^{ 13 } &=& 166\;507\;350\;731\;038\;802\;170\;609\;664 \\ 106 ^{ 13 } &=& 213\;292\;826\;014\;568\;334\;917\;410\;816 \\ \hline 107 ^{ 11 } &=& 21\;048\;519\;522\;998\;348\;950\;643 \\ 107 ^{ 13 } &=& 240\;984\;500\;018\;808\;097\;135\;911\;707 \\ 107 ^{ 15 } & = & 2\;759\;031\;540\;715\;333\;904\;109\;053\;133 & & \\ & & 443 & & \\ \hline 108 ^{ 11 } &=& 23\;316\;389\;970\;546\;096\;340\;992 \\ 108 ^{ 12 } &=& 2\;518\;170\;116\;818\;978\;404\;827\;136 \\ 118 ^{ 14 } &=& 101\;472\;439\;712\;019\;470\;540\;189\;876\;224 \\ \hline 126 ^{ 13 } &=& 2\;017\;516\;459\;574\;609\;153\;391\;845\;376 \\ 127 ^{ 14 } &=& 283\;956\;682\;347\;124\;706\;942\;551\;243\;009 \\ 133 ^{ 16 } & = & 9\;585\;753\;470\;490\;322\;141\;591\;520\;062 & & \\ & & 265\;281 & & \\ \hline 134 ^{ 13 } &=& 4\;491\;199\;828\;872\;408\;503\;792\;328\;704 \\ 134 ^{ 15 } & = & 80\;643\;984\;127\;232\;967\;094\;095\;054\;209 & & \\ & & 024 & & \\ 135 ^{ 13 } &=& 4\;946\;966\;739\;525\;117\;513\;427\;734\;375 \\ \hline 135 ^{ 14 } &=& 667\;840\;509\;835\;890\;864\;312\;744\;140\;625 \\ 136 ^{ 15 } & = & 100\;712\;557\;719\;971\;285\;024\;106\;952\;523 & & \\ & & 776 & & \\ 140 ^{ 25 } &=& 449\;987\;958\;058\;483\;731\;145\;152\;266\;240 & · 10^{24} \\ \hline 142 ^{ 16 } & = & 27\;328\;356\;228\;554\;426\;163\;172\;505\;624 & & \\ & & 313\;856 & & \\ 143 ^{ 17 } & = & 4\;372\;327\;021\;734\;283\;642\;004\;853\;327 & & \\ & & 592\;915\;343 & & \\ 152 ^{ 15 } & = & 534\;138\;422\;146\;939\;893\;094\;821\;310\;496 & & \\ & & 768 & & \\ \hline 154 ^{ 14 } & = & 4\;219\;782\;742\;781\;494\;680\;756\;610\;809 & & \\ & & 856 & & \\ 154 ^{ 15 } & = & 649\;846\;542\;388\;350\;180\;836\;518\;064\;717 & & \\ & & 824 & & \\ 155 ^{ 19 } & = & 413\;335\;079\;574\;020\;313\;162\;122\;296\;733 & & \\ & & 856\;201\;171\;875 & & \\ \hline 157 ^{ 19 } & = & 527\;343\;255\;303\;841\;790\;870\;720\;812\;082 & & \\ & & 050\;804\;460\;293 & & \\ 160 ^{ 28 } & = & 51\;922\;968\;585\;348\;276\;285\;304\;963\;292 & & \\ & & 200\;960 & · 10^{27} & \\ 163 ^{ 16 } & = & 248\;314\;265\;639\;726\;167\;358\;751\;235\;626 & & \\ & & 296\;641 & & \\ \hline 169 ^{ 16 } & = & 442\;779\;263\;776\;840\;698\;304\;313\;192\;148 & & \\ & & 785\;281 & & \\ 169 ^{ 22 } & = & 10\;315\;908\;977\;942\;302\;627\;204\;470\;186 & & \\ & & 314\;316\;211\;062\;255\;002\;161 & & \\ 170 ^{ 31 } & = & 1\;392\;889\;173\;388\;510\;144\;614\;180\;174 & & \\ & & 894\;677\;204\;330 & · 10^{30} & \\ \hline 170 ^{ 33 } & = & 40\;254\;497\;110\;927\;943\;179\;349\;807\;054 & & \\ & & 456\;171\;205\;137 & · 10^{33} & \\ 171 ^{ 17 } & = & 91\;397\;407\;411\;741\;874\;683\;083\;843\;738 & & \\ & & 640\;173\;291 & & \\ 171 ^{ 19 } & = & 2\;672\;551\;590\;126\;744\;157\;608\;054\;674 & & \\ & & 761\;577\;307\;202\;131 & & \\ \hline 172 ^{ 18 } & = & 17\;358\;494\;027\;033\;103\;736\;099\;033\;196 & & \\ & & 316\;709\;617\;664 & & \\ 173 ^{ 19 } & = & 3\;333\;311\;951\;341\;729\;629\;204\;978\;703 & & \\ & & 084\;632\;004\;627\;637 & & \\ 181 ^{ 18 } & = & 43\;472\;473\;122\;830\;653\;562\;489\;222\;659 & & \\ & & 449\;707\;872\;441 & & \\ \hline 181 ^{ 19 } & = & 7\;868\;517\;635\;232\;348\;294\;810\;549\;301 & & \\ & & 360\;397\;124\;911\;821 & & \\ 181 ^{ 20 } & = & 1\;424\;201\;691\;977\;055\;041\;360\;709\;423 & & \\ & & 546\;231\;879\;609\;039\;601 & & \\ 189 ^{ 19 } & = & 17\;896\;754\;443\;176\;031\;520\;198\;514\;559 & & \\ & & 819\;163\;143\;441\;509 & & \\ \hline 193 ^{ 22 } & = & 191\;540\;580\;003\;116\;921\;429\;323\;712\;183 & & \\ & & 642\;218\;614\;831\;262\;597\;249 & & \\ 199 ^{ 21 } & = & 1\;887\;620\;149\;539\;230\;539\;058\;375\;534 & & \\ & & 310\;517\;606\;114\;631\;604\;199 & & \\ 207 ^{ 20 } & = & 20\;864\;448\;472\;975\;628\;947\;226\;005\;981 & & \\ & & 267\;194\;447\;042\;584\;001 & & \\ \hline 211 ^{ 25 } & = & 12\;795\;621\;425\;112\;113\;141\;935\;148\;247 & & \\ & & 655\;082\;376\;252\;275\;523\;500\;373\;035\;251 & & \\ 217 ^{ 22 } & = & 2\;524\;144\;100\;572\;738\;110\;818\;511\;483 & & \\ & & 976\;079\;134\;636\;146\;367\;057\;489 & & \\ 221 ^{ 25 } & = & 40\;719\;913\;064\;560\;249\;818\;128\;041\;081 & & \\ & & 360\;346\;218\;088\;750\;603\;039\;104\;925\;501 & & \\ \hline 225 ^{ 22 } & = & 5\;597\;774\;487\;475\;881\;147\;025\;802\;420 & & \\ & & 102\;991\;163\;730\;621\;337\;890\;625 & & \\ 234 ^{ 23 } & = & 3\;104\;307\;401\;943\;398\;225\;947\;002\;752 & & \\ & & 118\;451\;297\;846\;365\;869\;366\;575\;104 & & \\ 236 ^{ 25 } & = & 210\;281\;019\;656\;164\;214\;863\;109\;519\;134 & & \\ & & 145\;127\;118\;463\;502\;897\;144\;582\;373\;376 & & \\ \hline 250 ^{ 40 } & = & 827\;180\;612\;553\;027\;674\;871\;408\;692\;069 & & \\ & & 962\;853\;565\;812\;110\;900\;878\;906\;250 & · 10^{39} & \\ 265 ^{ 28 } & = & 70\;938\;903\;323\;020\;164\;041\;464\;952\;207 & & \\ & & 191\;804\;150\;246\;813\;586\;391\;508\;579\;254 & & \\ & & 150\;390\;625 & & \\ \end{array}

Holy cow.

I should have been able to foresee some of those, like !!20^{13} = 8192·10^{13}!!, which in hindsight is obvious. But seriously, !!163^{16}!!?

(It rather looks like !!265^{28}!! might be the last one where !!n!! is not a multiple of !!10!!, however. Or maybe that is a misleading artifact of the calculating system I am using? I'm not sure yet.)

Happy new year, all.

[ Addendum 20190102: Several readers have confirmed that there are many examples past !!265^{28}!!. I am probably running into some mismatch between the way the computer represents numbers and the way they actually work. ]

[ Addendum 20190102: Dave McKee asked why I thought there would be few examples. ]

[Other articles in category /math] permanent link

Sun, 02 Dec 2018

Necklaces and bracelets

There are combinatorial objects called necklaces and bracelets and I can never remember which is which.

Both are finite sequences of things (typically symbols) where the start point is not important. So the bracelet ABCDE is considered to be the same as the bracelets BCDEA, CDEAB, DEABC, and EABCD.

One of the two also disregards the direction you go, so that ABCDE is also considered the same as EDCBA (and so also DCBAE, etc.). But which? I have to look it up every time.

I have finally thought of a mnemonic. In a necklace, the direction is important, because to reverse an actual necklace you have to pull it off over the wearer's head, turn it over, and put it back on. But in a bracelet the direction is not important, because it could be on either wrist, and a bracelet on the left wrist is the same as the reversed bracelet on the right wrist.

Okay, silly, maybe, but I think it's going to work.

[Other articles in category /math] permanent link

Thu, 29 Nov 2018

How many kinds of polygonal loops? (part 2)

I recently asked about these:

Four displays, each with five dark gray dots arranged at the vertices
of a regular pentagon.  In each display the dots are connected with
purple lines, but each in a different order.  If the dots were
numbered 0-1-2-3-4 in clockwise order, the four figures have purple
lines connecting them, respectively, in the orders
0-1-2-3-4-0, 0-1-3-2-4-0, 0-2-1-4-3-0, and 0-2-4-1-3-0.  The first of
these is a plain pentagon, and the last is a five-pointed star.  The
middle two are less symmetric.

And I said I thought there were nine analogous figures with six points.

Rahul Narain referred me to a recent discussion of almost this exact question on Math Stackexchange. (Note that the discussion there considers two figures different if they are reflections of one another; I consider them the same.) The answer turns out to be OEIS A000940. I had said:

… for !!N=6!!, I found it hard to enumerate. I think there are nine shapes but I might have missed one, because I know I kept making mistakes in the enumeration …

I missed three. The nine I got were:

This time
there are nine displays, each with six dots.  The connections are,
012345 (a hexagon), 015432, 032145 (two diamonds), 
015234 (highly irregular), 014523 (three triangles that share a vertex
in the center), 013254, 023154, 031254, and 035142.

And the three I missed are:

more hexagons, this time connected as follows:
014253, 013524, and 015342

I had tried to break them down by the arrangement of the outside ring of edges, which can be described by a composition. The first two of these have type !!1+1+1+1+2!! (which I missed completely; I thought there were none of this type) and the other has type !!1+2+1+2!!, the same as the !!015342!! one in the lower right of the previous diagram.

I had ended by saying:

I would certainly not trust myself to hand-enumerate the !!N=7!! shapes.

Good call, Mr. Dominus! I considered filing this under “oops” but I decided that although I had gotten the wrong answer, my confidence in it had been adequately low. On one level it was a mistake, but on a higher and more important level, I did better.

I am going to try the (Cauchy-Frobenius-)Burnside-(Redfield-)Pólya lemma on it next and see if I can get the right answer.

Thanks again to Rahul Narain for bringing this to my attention.

[Other articles in category /math] permanent link

How many kinds of polygonal loops?

Take !!N!! equally-spaced points on a circle. Now connect them in a loop: each point should be connected to exactly two others, and each point should be reachable from the others. How many geometrically distinct shapes can be formed?

For example, when !!N=5!!, these four shapes can be formed:

Four displays, each with five dark gray dots arranged at the vertices
of a regular pentagon.  In each display the dots are connected with
purple lines, but each in a different order.  If the dots were
numbered 0-1-2-3-4 in clockwise order, the four figures have purple
lines connecting them, respectively, in the orders
0-1-2-3-4-0, 0-1-3-2-4-0, 0-2-1-4-3-0, and 0-2-4-1-3-0.  The first of
these is a plain pentagon, and the last is a five-pointed star.  The
middle two are less symmetric.

(I phrased this like a geometry problem, but it should be clear it's actually a combinatorics problem. But it's much easier to express as a geometry problem; to talk about the combinatorics I have to ask you to consider a permutation !!P!! where !!P(i±1)≠P(i)±1!! blah blah blah…)

For !!N<5!! it's easy. When !!N=3!! it is always a triangle. When !!N=4!! there are only two shapes: a square and a bow tie.

But for !!N=6!!, I found it hard to enumerate. I think there are nine shapes but I might have missed one, because I know I kept making mistakes in the enumeration, and I am not sure they are all corrected:

This time
there are nine displays, each with six dots.  The connections are,
012345 (a hexagon), 015432, 032145 (two diamonds), 
015234 (highly irregular), 014523 (three triangles that share a vertex
in the center), 013254, 023154, 031254, and 035142.

It seems like it ought not to be hard to generate and count these, but so far I haven't gotten a good handle on it. I produced the !!N=6!! display above by considering the compositions of the number !!6!!:

Composition How many
6 1
2+4 1
3+3 1
1+2+3 1
2+2+2 2
1+1+2+2 1
1+2+1+1 1
1+1+1+1+1+1 1
Total9 (?)

(Actually it's the compositions, modulo bracelet symmetries — that is, modulo the action of the dihedral group.)

But this is fraught with opportunities for mistakes in both directions. I would certainly not trust myself to hand-enumerate the !!N=7!! shapes.

[ Addendum: For !!N=6!! there are 12 figures, not 9. For !!N=7!!, there are 39. Further details. ]

[Other articles in category /math] permanent link

Sun, 25 Nov 2018

60-degree angles on a lattice

A couple of years back I was thinking about how to draw a good approximation to an equilateral triangle on a piece of graph paper. There are no lattice points that are exactly the vertices of an equilateral triangle, but you can come close, and one way to do it is to find integers !!a!! and !!b!! with !!\frac ba\approx \sqrt 3!!, and then !!\langle 0, 0\rangle, \langle 2a, 0\rangle,!! and !!\langle a, b\rangle!! are almost an equilateral triangle.

But today I came back to it for some reason and I wondered if it would be possible to get an angle closer to 60°, or numbers that were simpler, or both, by not making one of the sides of the triangle perfectly horizontal as in that example.

So okay, we want to find !!P = \langle a, b\rangle!! and !!Q = \langle c,d\rangle!! so that the angle !!\alpha!! between the rays !!\overrightarrow{OP}!! and !!\overrightarrow{OQ}!! is as close as possible to !!\frac\pi 3!!.

The first thing I thought of was that the dot product !!P\cdot Q = |P||Q|\cos\alpha!!, and !!P\cdot Q!! is super-easy to calculate, it's just !!ac+bd!!. So we want $$\frac{ad+bc}{|P||Q|} = \cos\alpha \approx \frac12,$$ and everything is simple, except that !!|P||Q| = \sqrt{a^2+b^2}\sqrt{c^2+d^2}!!, which is not so great.

Then I tried something else, using high-school trigonometry. Let !!\alpha_P!! and !!\alpha_Q!! be the angles that the rays make with the !!x!!-axis. Then !!\alpha = \alpha_Q - \alpha_P = \tan^{-1} \frac dc - \tan^{-1} \frac ba!!, which we want close to !!\frac\pi3!!.

Taking the tangent of both sides and applying the formula $$\tan(q-p) = \frac{\tan q - \tan p}{1 + \tan q \tan p}$$ we get $$ \frac{\frac dc - \frac ba}{1 + \frac dc\frac ba} \approx \sqrt3.$$ Or simplifying a bit, the super-simple $$\frac{ad-bc}{ac+bd} \approx \sqrt3.$$

After I got there I realized that my dot product idea had almost worked. To get rid of the troublesome !!|P||Q|!! you should consider the cross product also. Observe that the magnitude of !!P\times Q!! is !!|P||Q|\sin\alpha!!, and is also $$\begin{vmatrix} a & b & 0 \\ c & d & 0 \\ 1 & 1 & 1 \end{vmatrix} = ad - bc$$ so that !!\sin\alpha = \frac{ad-bc}{|P||Q|}!!. Then if we divide, the !!|P||Q|!! things cancel out nicely: $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{ad-bc}{ac+bd}$$ which we want to be as close as possible to !!\sqrt 3!!.

Okay, that's fine. Now we need to find some integers !!a,b,c,d!! that do what we want. The usual trick, “see what happens if !!a=0!!”, is already used up, since that's what the previous article was about. So let's look under the next-closest lamppost, and let !!a=1!!. Actually we'll let !!b=1!! instead to keep things more horizonal. Then, taking !!\frac74!! as our approximation for !!\sqrt3!!, we want

$$\frac{ad-c}{ac+d} = \frac74$$

or equivalently $$\frac dc = \frac{7a+4}{4a-7}.$$

Now we just tabulate !!7a+4!! and !!4a-7!! looking for nice fractions:

!!a!!!!d =!!

Each of these gives us a !!\langle c,d\rangle!! point, but some are much better than others. For example, in line 3, we have take !!\langle 5,25\rangle!! but we can use !!\langle 1,5\rangle!! which gives the same !!\frac dc!! but is simpler. We still get !!\frac{ad-bc}{ac+bd} = \frac 74!! as we want.

Doing this gives us the two points !!P=\langle 3,1\rangle!! and !!Q=\langle 1, 5\rangle!!. The angle between !!\overrightarrow{OP}!! and !!\overrightarrow{OQ}!! is then !!60.255°!!. This is exactly the same as in the approximately equilateral !!\langle 0, 0\rangle, \langle 8, 0\rangle,!! and !!\langle 4, 7\rangle!! triangle I mentioned before, but the numbers could not possibly be easier to remember. So the method is a success: I wanted simpler numbers or a better approximation, and I got the same approximation with simpler numbers.

To draw a 60° angle on graph paper, mark !!P=\langle 3,1\rangle!! and !!Q=\langle 1, 5\rangle!!, draw lines to them from the origin with a straightedge, and there is your 60° angle, to better than a half a percent.

A graph of the lines 3y=x and y=5x, with the points
(3,1) and (1,5) marked, demonstrating that the angle between the
lines is very close to 60 degrees.

There are some other items in the table (for example row 18 gives !!P=\langle 18,1\rangle!! and !!Q=\langle 1, 2\rangle!!) but because of the way we constructed the table, every row is going to give us the same angle of !!60.225°!!, because we approximated !!\sqrt3\approx\frac74!! and !!60.225° = \tan^{-1}\frac74!!. And the chance of finding numbers better than !!\langle 3,1\rangle!! and !!\langle 1, 5\rangle!! seems slim. So now let's see if we can get the angle closer to exactly !!60°!! by using a better approximation to !!\sqrt3!! than !!\frac 74!!.

The next convergents to !!\sqrt 3!! are !!\frac{19}{11}!! and !!\frac{26}{15}!!. I tried the same procedure for !!\frac{19}{11}!! and it was a bust. But !!\frac{26}{15}!! hit the jackpot: !!a=4!! gives us !!15a-26 = 34!! and !!26a-15=119!!, both of which are multiples of 17. So the points are !!P=\langle 4,1\rangle!! and !!Q=\langle 2, 7\rangle!!, and this time the angle between the rays is !!\tan^{-1}\frac{26}{15} = 60.018°!!. This is as accurate as anyone drawing on graph paper could possibly need; on a circle with a one-mile radius it is an error of 20 inches.

A very similar plot, this time with the lines 4y=x and 7y=2x

Of course, the triangles you get are no longer equilateral, not even close. That first one has sides of !!\sqrt{10}, \sqrt{20}, !! and !!\sqrt{26}!!, and the second one has sides of !!\sqrt{17}, \sqrt{40}, !! and !!\sqrt{53}!!. But! The slopes of the lines are so simple, it's easy to construct equilateral triangles with a straightedge and a bit of easy measuring. Let's do it on the !!60.018°!! angle and see how it looks.

!!\overrightarrow{OP}!! has slope !!\frac14!!, so the perpendicular to it has slope !!-4!!, which means that you can draw the perpendicular by placing the straightedge between !!P!! and some point !!P+x\langle -1, 4\rangle!!, say !!\langle 2, 9\rangle!! as in the picture. The straightedge should have slope !!-4!!, which is very easy to arrange: just imagine the little squares grouped into stacks of four, and have the straightedge go through opposite corners of each stack. The line won't necessarily intersect !!\overrightarrow{OQ}!! anywhere great, but it doesn't need to, because we can just mark the intersection, wherever it is:

Same as the previous plot, but now there is another line y-1=-4(x-4)
drawn through (4, 1) and its intersection with the other line, near
(2¼, 8).

Let's call that intersection !!A!! for “apex”.

The point opposite to !!O!! on the other side of !!P!! is even easier; it's just !!P'=2P =\langle 8, 2\rangle!!. And the segment !!P'A!! is the third side of our equilateral triangle:

The previous diagram, with the third side of the equilateral triangle drawn in.

This triangle is geometrically similar to a triangle with vertices at !!\langle 0, 0\rangle, \langle 30, 0\rangle,!! and !!\langle 15, 26\rangle!!, and the angles are just close to 60°, but it is much smaller.


[Other articles in category /math] permanent link

Mon, 19 Nov 2018

I love the dodecahedron

I think I forgot to mention that I was looking recently at hamiltonian cycles on a dodecahedron. The dodecahedron has 30 edges and 20 vertices, so a hamiltonian path contains 20 edges and omits 10. It turns out that it is possible to color the edges of the dodecahedron in three colors so that:

  • Every vertex is incident to one edge of each color
  • The edges of any two of the three colors form a hamiltonian cycle

Dear visually impaired people: I try to provide clear
descriptions of illustrations, but here I was stumped about what I
could say that would be helpful that I had not already said.  I would
be grateful for your advice and suggestions.


(In this presentation, I have taken one of the vertices and sent it away to infinity. The three edges with arrowheads are all attached to that vertex, off at infinity, and the three faces incident to it have been stretched out to cover the rest of the plane.)

Every face has five edges and there are only three colors, so the colors can't be distributed evenly around a face. Each face is surrounded by two edges of one color, two of a second color, and only one of the last color. This naturally divides the 12 faces into three classes, depending on which color is assigned to only one edge of that face.

This is
the same picture as before, but each of the 12 regions of the plane
has been annotated with a large colored circle, the same as the color
of which the region has only one boundary edge.

Each class contains two pairs of two adjacent pentagons, and each adjacent pair is adjacent to the four pairs in the other classes but not to the other pair in its own class.

Each pair shares a single edge, which we might call its “hinge”, shown here as dotted lines. Each pair has 8 vertices, of which two are on its hinge, four are adjacent to the hinge, and two are not near of the hinge. These last two vertices are always part of the hinges of the pairs of a different class.

I could think about this for a long time, and probably will. There is plenty more to be seen, but I think there is something else I was supposed to be doing today, let me think…. Oh yes! My “job”! So I will leave you to go on from here on your own.

[ Addendum 20181203: David Eppstein has written a much longer and more detailed article about triply-Hamiltonian edge colorings, using this example as a jumping-off point. ]

[Other articles in category /math] permanent link

Sat, 17 Nov 2018

How do you make a stella octangula?

Yesterday Katara asked me out of nowhere “When you make a stella octangula, do you build it up from an octahedron or a tetrahedron?” Stella octangula was her favorite polyhedron eight years ago.

“Uh,” I said. “Both?”

Then she had to make one to see what I meant. You can start with a regular octahedron:

a regular octahedron made of six steel
ball bearings and twelve blue and green magnetic struts

Then you erect spikes onto four of the octahedron's faces; this produces a regular tetrahedron:

A very similar octahedron, this time
with an orange tripod attached to four of its eight faces, forming an
orange tetrahedron with a blue and green octahedron embedded in it

Then you erect spikes onto the other four of the octahedron's faces. Now you have a stella octangula.

The octahedron from before, but with
red tripods attached to its other four faces, making eight tripods in
all.  The final result looks like interpenetrating red and orange
tetrahedra, with the original octahedron in their intersection

So yeah, both. Or instead of starting with a unit octahedron and erecting eight spikes of size 1, you can start with a unit tetrahedron and erect four spikes of size ½. It's both at once.

[Other articles in category /math] permanent link

Tue, 13 Nov 2018

Counting paths through polyhedra

A while back someone asked on math stack exchange how many paths there were of length !!N!! from one vertex of a dodecahedron to the opposite vertex. The vertices are distance 5 apart, so for !!N<5!! the answer is zero, but the paths need not be simple, so the number grows rapidly with !!N!!; there are 58 million paths of length 19.

This is the kind of thing that the computer answers easily, so that's where I started, and unfortunately that's also where I finished, saying:

I'm still working out a combinatorial method of calculating the answer, and I may not be successful.

Another user reminded me of this and I took another whack at it. I couldn't remember what my idea had been last year, but my idea this time around was to think of the dodecahedron as the Cayley graph for a group, and then the paths are expressions that multiply out to a particular group element.

I started by looking at a tetrahedron instead of at a dodecahedron, to see how it would work out. Here's a tetrahedron.

Let's say we're counting paths from the center vertex to one of the others, say the one at the top. (Tetrahedra don't have opposite vertices, but that's not an important part of the problem.) A path is just a list of edges, and each edge is labeled with a letter !!a!!, !!b!!, or !!c!!. Since each vertex has exactly one edge with each label, every sequence of !!a!!'s, !!b!!'s, and !!c!!'s represents a distinct path from the center to somewhere else, although not necessarily to the place we want to go. Which of these paths end at the bottom vertex?

The edge labeling I chose here lets us play a wonderful trick. First, since any edge has the same label at both ends, the path !!x!! always ends at the same place as !!xaa!!, because the first !!a!! goes somewhere else and then the second !!a!! comes back again, and similarly !!xbb!! and !!xcc!! also go to the same place. So if we have a path that ends where we want, we can insert any number of pairs !!aa, bb, !! or !!cc!! and the new path will end at the same place.

But there's an even better trick available. For any starting point, and any letters !!x!! and !!y!!, the path !!xy!! always ends at the same place as !!yx!!. For example, if we start at the middle and follow edge !!b!!, then !!c!!, we end at the lower left; similarly if we follow edge !!c!! and then !!b!! we end at the same place, although by a different path.

Now suppose we want to find all the paths of length 7 from the middle to the top. Such a path is a sequence of a's, b's, and c's of length 7. Every such sequence specifies a different path out of the middle vertex, but how can we recognize which sequences end at the top vertex?

Since !!xy!! always goes to the same place as !!yx!!, the order of the seven letters doesn't matter. A complicated-seeming path like abacbcb must go to the same place as !!aabbbcc!!, the same path with the letters in alphabetical order. And since !!xx!! always goes back to where it came from, the path !!aabbbcc!! does to the same place as !!b!!

Since the paths we want are those that go to the same place as the trivial path !!c!!, we want paths that have an even number of !!a!!s and !!b!!s and an odd number of !!c!!s. Any path fitting that description will go to same place as !!c!!, which is the top vertex. It's easy to enumerate such paths:

How many?

Here something like “cccbbbb” stands for all the paths that have three c's and four b's, in some order; there are !!\frac{7!}{4!3!} = 35!! possible orders, so 35 paths of this type. If we wanted to consider paths of arbitrary length, we could use Burnside's lemma, but I consider the tetrahedron to have been sufficiently well solved by the observations above (we counted 547 paths by hand in under 60 seconds) and I don't want to belabor the point.

Okay! Easy-peasy!

Now let's try cubes:

Here we'll consider paths between two antipodal vertices in the upper right and the lower left, which I've colored much darker gray than the other six vertices.

The same magic happens as in the tetrahedron. No matter where we start, and no matter what !!x!! and !!y!! are, the path !!xy!! always gets us to the same place as !!yx!!. So again, if some complicated path gets us where we want to go, we can permute its components into any order and get a different path of the same langth to the same place. For example, starting from the upper left, bcba, abcb, and abbc all go to the same place.

And again, because !!xx!! always make a trip along one edge and then back along the same edge, it never goes anywhere. So the three paths in the previous paragraph also go to the same place as ac and ca and also aa bcba bb aa aa aa aa bb cc cc cc bb.

We want to count paths from one dark vertex to the other. Obviously abc is one such, and so too must be bac, cba, acb, and so forth. There are six paths of length 3.

To get paths of length 5, we must insert a pair of matching letters into one of the paths of length 3. Without loss of generality we can assume that we are inserting aa. There are 20 possible orders for aaabc, and three choices about which pair to insert, for a total of 60 paths.

To get paths of length 7, we must insert two pairs. If the two pairs are the same, there are !!\frac{7!}{5!} = 42!! possible orders and 3 choices about which letters to insert, for a total of 126. If the two pairs are different, there are !!\frac{7!}{3!3!} = 140!! possible orders and again 3 choices about which pairs to insert, for a total of 420, and a grand total of !!420+126 = 546!! paths of length 7. Counting the paths of length 9 is almost as easy. For the general case, again we could use Burnside's lemma, or at this point we could look up the unusual sequence !!6, 60, 546!! in OEIS and find that the number of paths of length !!2n+1!! is already known to be !!\frac34(9^n-1)!!.

So far this technique has worked undeservedly well. The original problem wanted to use it to study paths on a dodecahedron. Where, unfortunately, the magic property !!xy=yx!! doesn't hold. It is possible to label the edges of the dodecahedron so that every sequence of labels determines a unique path:

but there's nothing like !!xy=yx!!. Well, nothing exactly like it. !!xy=yx!! is equivalent to !!(xy)^2=1!!, and here instead we have !!(xy)^{10}=1!!. I'm not sure that helps. I will probably need another idea.

The method fails similarly for the octahedron — which is good, because I can use the octahedron as a test platform to try to figure out a new idea. On an octahedron we need to use four kinds of labels because each vertex has four edges emerging from it:

Here again we don't have !!(xy)^2=1!! but we do have !!(xy)^3 = 1!!. So it's possible that if I figure out a good way to enumerate paths on the octahedron I may be able to adapt the technique to the dodecahedron. But the octahedron will be !!\frac{10}3!! times easier.

Viewed as groups, by the way, these path groups are all examples of Coxeter groups. I'm not sure this is actually a useful observation, but I've been wanting to learn about Coxeter groups for a long time and this might be a good enough excuse.

[Other articles in category /math] permanent link

A puzzle about representing numbers as a sum of 3-smooth numbers

I think this would be fun for a suitably-minded bright kid of maybe 12–15 years old.

Consider the following table of numbers of the form !!2^i3^j!!:

1 3 9 27 81 243
2 6 18 54 162
4 12 36 108
8 24 72 216
16 48 144
32 96
64 192

Given a number !!n!!, is is possible to represent !!n!! as a sum of entries from the table, with the following constraints:

  • No more than one entry from any column
  • An entry may only be used if it is in a strictly higher row than any entry to its left.

For example, one may not represent !!23 = 2 + 12 + 9!!, because the !!12!! is in a lower row than the !!2!! to its left.

1 3 9 27
2 6 18 54
4 12 36 108

But !!23 = 8 + 6 + 9!! is acceptable, because 6 is higher than 8, and 9 is higher than 6.

1 3 9 27
2 6 18 54
4 12 36 108
8 24 72 216

Or, put another way: can we represent any number !!n!! in the form $$n = \sum_i 2^{a_i}3^{b_i}$$ where the !!a_i!! are strictly decreasing and the !!b_i!! are strictly increasing?


maxpow3 1 = 1 maxpow3 2 = 1 maxpow3 n = 3 * maxpow3 (n `div` 3) rep :: Integer -> [Integer] rep 0 = [] rep n = if even n then map (* 2) (rep (n `div` 2)) else (rep (n - mp3)) ++ [mp3] where mp3 = maxpow3 n

Sadly, the representation is not unique. For example, !!8+3 = 2+9!!, and !!32+24+9 = 32+6+27 = 8+12=18+27!!.

[Other articles in category /math] permanent link

Mon, 24 Sep 2018

Zipfian regression

A long time ago, I wrote up a blog article about how to derive the linear regression formulas from first principles. Then I decided it was not of general interest, so I didn't publish it. (Sometime later I posted it to math stack exchange, so the effort wasn't wasted.)

The basic idea is, you have some points !!(x_i, y_i)!!, and you assume that they can be approximated by a line !!y=mx+b!!. You let the error be a function of !!m!! and !!b!!: $$\varepsilon(m, b) = \sum (mx_i + b - y_i)^2$$ and you use basic calculus to find !!m!! and !!b!! for which !!\varepsilon!! is minimal. Bing bang boom.

I knew this for a long time but it didn't occur to me until a few months ago that you could use basically the same technique to fit any other sort of curve. For example, suppose you think your data is not a line but a parabola of the type !!y=ax^2+bx+c!!. Then let the error be a function of !!a, b, !! and !!c!!:

$$\varepsilon(a,b,c) = \sum (ax_i^2 + bx_i + c - y_i)^2$$

and again minimize !!\varepsilon!!. You can even get a closed form as you can with ordinary linear regression.

I especially wanted to try fitting hyperbolas to data that I expected to have a Zipfian distribution. For example, take the hundred most popular names for girl babies in Illinois in 2017. Is there a simple formula which, given an ordinal number like 27, tells us approximately how many girls were given the 27th most popular name that year? (“Scarlett”? Seriously?)

I first tried fitting a hyperbola of the form !!y = c + \frac ax!!. We could, of course, take !!y_i' = \frac 1{y_i}!! and then try to fit a line to the points !!\langle x_i, y_i'\rangle!! instead. But this will distort the measurement of the error. It will tolerate gross errors in the points with large !!y!!-coordinates, and it will be extremely intolerant of errors in points close to the !!x!!-axis. This may not be what we want, and it wasn't what I wanted. So I went ahead and figured out the Zipfian regression formulas:

$$ \begin{align} a & = \frac{HY-NQ}D \\ c & = \frac{HQ-JY}D \end{align} $$


$$\begin{align} H & = \sum x_i^{-1} \\ J & = \sum x_i^{-2} \\ N & = \sum 1\\ Q & = \sum y_ix_i^{-1} \\ Y & = \sum y_i \\ D & = H^2 - NJ \end{align} $$

When I tried to fit this to some known hyperbolic data, it worked just fine. For example, given the four points !!\langle1, 1\rangle, \langle2, 0.5\rangle, \langle3, 0.333\rangle, \langle4, 0.25\rangle!!, it produces the hyperbola $$y = \frac{1.00018461538462}{x} - 0.000179487179486797.$$ This is close enough to !!y=\frac1x!! to confirm that the formulas work; the slight error in the coefficients is because we used !!\bigl\langle3, \frac{333}{1000}\bigr\rangle!! rather than !!\bigl\langle3, \frac13\bigr\rangle!!.

Unfortunately these formulas don't work for the Illinois baby data. Or rather, the hyperbola fits very badly. The regression produces !!y = \frac{892.765272442475}{x} + 182.128894972025:!!

A graph showing
the actual data (as blue dots) and the best-fit hyperbola (as a red
line).  They do not match well.  The hyperbola drops down much too
quickly, then makes a sharp turn to catch up with the right tail it
has overshot. It levels off at 182.12, instead of at around 100.

I think maybe I need to be using some hyperbola with more parameters, maybe something like !!y = \frac a{x-b} + c!!.

In the meantime, here's a trivial script for fitting !!y = \frac ax + c!! hyperbolas to your data:

while (<>) {
  my ($x, $y) = split;
  ($x, $y) = ($., $x) if not defined $y;
  $H += 1/$x;
  $J += 1/($x*$x);
  $N += 1;
  $Q += $y/$x;
  $Y += $y;

my $D = $H*$H - $J*$N;
my $c = ($Q*$H - $J*$Y)/$D;
my $a = ($Y*$H - $Q*$N)/$D;

print "y = $a / x + $c\n";

[ Addendum 20180925: Shreevatsa R. asked a related question on StackOverflow and summarized the answers. The problem is more complex than it might first appear. Check it out. ]

[Other articles in category /math] permanent link

Fri, 14 Sep 2018

How not to remember the prime numbers under 1,000

A while back I said I wanted to memorize all the prime numbers under 1,000, because I am tired of getting some number like 851 or 857, or even 307, and then not knowing whether it is prime.

The straightforward way to deal with this is: just memorize the list. There are only 168 of them, and I have the first 25 or so memorized anyway.

But I had a different idea also. Say that a set of numbers from !!10n!! to !!10n+9!! is a “decade”. Each decade contains at most 4 primes, so 4 bits are enough to describe the primes in a single decade. Assign a consonant to each of the 16 possible patterns, say “b” when none of the four numbers is a prime, “d” when only !!10n+1!! is prime, “f” when only !!10n+3!! is, and so on.

Now memorizing the primes in the 90 decades is reduced to memorizing 90 consonants. Inserting vowels wherever convenient, we have now turned the problem into one of memorizing around 45 words. A word like “potato” would encode the constellation of primes in three consecutive decades. 45 words is only a few sentences, so perhaps we could reduce the list of primes to a short and easily-remembered paragraph. If so, memorizing a few sentences should be much easier than memorizing the original list of primes.

The method has several clear drawbacks. We would have to memorize the mapping from consonants to bit patterns, but this is short and maybe not too difficult.

More significant is that if we're trying to decide if, say, 637 is prime, we have to remember which consonant in which word represents the 63rd decade. This can be fixed, maybe, by selecting words and sentences of standard length. Say there are three sentences and each contains 30 consonants. Maybe we can arrange that words always appear in patterns, say four words with 1 or 2 consonants each that together total 7 consonants, followed by a single long word with three consonants. Then each sentence can contain three of these five-word groups and it will be relatively easy to locate the 23rd consonant in a sentence: it is early in the third group.

Katara and I tried this, with not much success. But I'm not ready to give up on the idea quite yet. A problem we encountered early on is that we remember consonants not be how words are spelled but by how they sound. So we don't want a word like “hammer” to represent the consonant pattern h-m-m but rather just h-m.

Another problem is that some constellations of primes are much more common than others. We initially assigned consonants to constellations in order. This assigned letter “b” to the decades that contain no primes. But this is the most common situation, so the letter “b” tended to predominate in the words we needed for our mnemonic. We need to be careful to assign the most common constellations to the best letters.

Some consonants in English like to appear in clusters, and it's not trivial to match these up with the common constellations. The mapping from prime constellations to consonants must be carefully chosen to work with English. We initially assigned “s” to the constellation “☆•☆☆” (where !!10n+1, 10n+7,!! and !!10n+9!! are prime but !!10n+3!! is not) and “t” to the constellation “☆☆••” (where !!10n+1!! and !!10n+3!! are prime but !!10n+7!! and !!10n+9!! are not) but these constellations cannot appear consecutively, since at least one of !!10n+7, 10n+9, 10n+11!! is composite. So any word with “s” and “t” with no intervening consonants was out of play. This eliminated a significant fraction of the entire dictionary!

I still think it could be made to work, maybe. If you're interested in playing around with this, the programs I wrote are available on Github. The mapping from decade constellations to consonant clusters is in

[Other articles in category /math] permanent link

Mon, 27 Aug 2018

Linear mappings of projective space

[ Epistemic status: uninformed musings; anything and everything in here might be wrong or ill-conceived. ]

Suppose !!V!! is some vector space, and let !!V_n!! be the family of all !!n!!-dimensional subspaces of !!V!!. In particular !!V_1!! is the family of all one-dimensional subspaces of !!V!!. When !!V!! is euclidean space, !!V_1!! is the corresponding projective space.

if !!L!! is a nonsingular linear mapping !!V\to V!!, then !!L!! induces a mapping !!L_n!! from !!V_n\to V_n!!. (It does not make sense to ask at this point if the induced mapping is linear because we do not have any obvious linear structure on the projective space. Or maybe there is but I have not heard of it.)

The eigenspaces of !!V!! are precisely the fixed points of !!L_n!!.

(Wrong! Any subspace generated by an !!n!!-set of eigenvectors is a fixed point of !!L_n!!. But such a subspace is not in general an eigenspace. (Note this includes the entire space as a special case.) The converse, however, does hold, since any eigenspace is generated by a subset of eigenvectors.)

Now it is an interesting and useful fact that for typical mappings, say from !!\Bbb R\to\Bbb R!!, fixed points tend to be attractors or repellers. (For example, see this earlier article.) This is true of !!L_1!! also. One-dimensional eigenspaces whose eigenvalues are larger than !!1!! are attractors of !!L_1!!, and those whose eigenvalues are smaller than !!1!! are repellers, and this is one reason why the eigenspaces are important: if !!L!! represents the evolution of state space, then vectors in !!V!! will tend to evolve toward being eigenvectors.

So consider, say, the projective plane !!\Bbb P^2!!, under the induced mapping of some linear operator on !!\Bbb R^3!!. There will be (typically) three special points in !!\Bbb P^2!! and other points will typically tend to gravitate towards one or more of these. Isn't this interesting? Is the three-dimensional situation more interesting than the two-dimensional one? What if a point attracts in one dimension and repels in the other? What can the orbits look like?

Or consider the Grassmanian space !!Gr(2, \Bbb R^3)!! of all planes in !!\Bbb R^3!!. Does a linear operator on !!\Bbb R^3!! tend to drive points in !!Gr(2, \Bbb R^3)!! toward three fixed points? (In general, I suppose !!Gr(k, \Bbb R^n)!! will have !!n\choose k!! fixed points, some of which might attract and some repel.) Is there any geometric intuition for this?

I have been wanting to think about Grassmanian spaces more for a long time now.

[Other articles in category /math] permanent link

Thu, 02 Aug 2018

How to explain infinity to kids

A professor of mine once said to me that all teaching was a process of lying, and then of replacing the lies with successively better approximations of the truth. “I say it's like this,” he said, “and then later I say, well, it's not actually like that, it's more like this, because the real story is too complicated to explain all at once.” I wouldn't have phrased it like this, but I agree with him in principle. One of the most important issues in pedagogical practice is deciding what to leave out, and for how long.

Kids inevitably want to ask about numerical infinity, and many adults will fumble the question, mumbling out some vague or mystical blather. Mathematics is prepared to offer a coherent and carefully-considered answer. Unfortunately, many mathematically-trained people also fumble the question, because mathematics is prepared to offer too many answers. So the mathematical adult will often say something like “well, it's a lot of things…” which for this purpose is exactly not what is wanted. When explaining the concept for the very first time, it is better to give a clear and accurate partial explanation than a vague and imprecise overview. This article suggests an answer that is short, to the point, and also technically correct.

In mathematics “infinity” names a whole collection of not always closely related concepts from analysis, geometry, and set theory. Some of the concepts that come under this heading are:

  • The !!+\infty!! and !!-\infty!! one meets in real analysis, which can be seen either as a convenient fiction (with “as !!x!! goes to infinity” being only a conventional rephrasing of “as !!x!! increases without bound”) or as the endpoints in the two-point compactification of !!\Bbb R!!.

  • The !!\infty!! one meets in complex analysis, which is a single point one adds to compactify the complex plane into the Riemann sphere.

  • The real version of the preceding, the “point at infinity” on the real projective line.

  • The entire “line at infinity” that bounds the real projective plane.

  • The vast family of set-theoretic infinite cardinals !!\aleph_0, \aleph_1, \ldots!!.

  • The vast family of set-theoretic infinite ordinals, !!\omega, \omega+1, \ldots, \epsilon_0, \ldots, \Omega, \ldots!!.

I made a decision ahead of time that when my kids first asked what infinity was, I would at first adopt the stance that “infinity” referred specifically to the set-theoretic ordinal !!\omega!!, and that the two terms were interchangeable. I could provide more details later on. But my first answer to “what is infinity” was:

It's the smallest number you can't count to.

As an explanation of !!\omega!! for kids, I think this is flawless. It's brief and it's understandable. It phrases the idea in familiar terms: counting. And it is technically unimpeachable. !!\omega!! is, in fact, precisely the unique smallest number you can't count to.

How can there be a number that you can't count to? Kids who ask about infinity are already quite familiar with the idea that the sequence of natural numbers is unending, and that they can count on and on without bound. “Imagine taking all the numbers that you could reach by counting,” I said. “Then add one more, after all of them. That is infinity.” This is a bit mind-boggling, but again it is technically unimpeachable, and the mind-bogglyness of it is nothing more than the intrinsic mind-bogglyness of the concept of infinity itself. None has been added by vagueness or imprecise metaphor. When you grapple with this notion, you are grappling with the essence of the problem of the completed infinity.

In my experience all kids make the same move at this point, which is to ask “what comes after infinity?” By taking “infinity” to mean !!\omega!!, we set ourselves up for an answer that is much better than the perplexing usual one “nothing comes after infinity”, which, if infinity is to be considered a number, is simply false. Instead we can decisively say that there is another number after infinity, which is called “infinity plus one”. This suggests further questions. “What comes after infinity plus one?” is obvious, but a bright kid will infer the existence of !!2\cdot\omega!!. A different bright kid might ask about !!\omega-1!!, which opens a different but fruitful line of discussion: !!\omega!! is not a successor ordinal, it is a limit ordinal.

Or the kid might ask if infinity plus one isn't equal to infinity, in which case you can discuss the non-commutativity of ordinal addition: if you add the “plus one” at the beginning, it is the same (except for !!\omega!! itself, the picture has just been shifted over on the page):

But if you add the new item at the other end, it is a different picture:

Before there was only one extra item on the right, and now there are two. The first picture exemplifies the Dedekind property, which is an essential feature of infinity. But the existence of an injection from !!\omega+1!! to !!\omega!! does not mean that every such map is injective.

Just use !!\omega!!. Later on the kid may ask questions that will need to be answered with “Earlier, I did not tell you the whole story.” That is all right. At that point you can reveal the next thing.

[Other articles in category /math] permanent link

Thu, 26 Jul 2018

Testing for divisibility by 7

There are well-known tests if a number (represented as a base-10 numeral) is divisible by 2, 3, 5, 9, or 11. What about 7?

Let's look at where the divisibility-by-9 test comes from. We add up the digits of our number !!n!!. The sum !!s(n)!! is divisible by !!9!! if and only if !!n!! is. Why is that?

Say that !!d_nd_{n-1}\ldots d_0!! are the digits of our number !!n!!. Then

$$n = \sum 10^id_i.$$

The sum of the digits is

$$s(n) = \sum d_i$$

which differs from !!n!! by $$\sum (10^i-1)d_i.$$ Since !!10^i-1!! is a multiple of !!9!! for every !!i!!, every term in the last sum is a multiple of !!9!!. So by passing from !!n!! to its digit sum, we have subtracted some multiple of !!9!!, and the residue mod 9 is unchanged. Put another way:

$$\begin{align} n &= \sum 10^id_i \\ &\equiv \sum 1^id_i \pmod 9 \qquad\text{(because $10 \equiv 1\pmod 9$)} \\ &= \sum d_i \end{align} $$

The same argument works for the divisibility-by-3 test.

For !!11!! the analysis is similar. We add up the digits !!d_0+d_2+\ldots!! and !!d_1+d_3+\ldots!! and check if the sums are equal mod 11. Why alternating digits? It's because !!10\equiv -1\pmod{11}!!, so $$n\equiv \sum (-1)^id_i \pmod{11}$$

and the sum is zero only if the sum of the positive terms is equal to the sum of the negative terms.

The same type of analysis works similarly for !!2, 4, 5, !! and !!8!!. For !!4!! we observe that !!10^i\equiv 0\pmod 4!! for all !!i>1!!, so all but two terms of the sum vanish, leaving us with the rule that !!n!! is a multiple of !!4!! if and only if !!10d_1+d_0!! is. We could simplify this a bit: !!10\equiv 2\pmod 4!! so !!10d_1+d_0 \equiv 2d_1+d_0\pmod 4!!, but we don't usually bother. Say we are investigating !!571496!!; the rule tells us to just consider !!96!!. The "simplified" rule says to consider !!2\cdot9+6 = 24!! instead. It's not clear that that is actually easier.

This approach works badly for divisibility by 7, because !!10^i\bmod 7!! is not simple. It repeats with period 6.

$$\begin{array}{c|cccccc|ccc} i & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ %10^i & 1 & 10 & 100 & 1000 & 10000 & \ldots \\ 10^i\bmod 7 & 1 & 3 & 2 & 6 & 4 & 5 & 1 & 3 & 2 & 6 & 4 &\ldots \\ \end{array} $$

The rule we get from this is:

Take the units digit. Add three times the ones digit, twice the hundreds digit, six times the thousands digit… (blah blah blah) and the original number is a multiple of !!7!! if and only if the sum is also.

For example, considering !!12345678!! we must calculate $$\begin{align} 12345678 & \Rightarrow & 3\cdot1 + 1\cdot 2 + 5\cdot 3 + 4\cdot 4 + 6\cdot 5 + 2\cdot6 + 3\cdot 7 + 1\cdot 8 & = & 107 \\\\ 107 & \Rightarrow & 2\cdot1 + 3\cdot 0 + 1\cdot7 & = & 9 \end{align} $$ and indeed !!12345678\equiv 107\equiv 9\pmod 7!!.

My kids were taught the practical divisibility tests in school, or perhaps learned them from YouTube or something like that. Katara was impressed by my ability to test large numbers for divisibility by 7 and asked how I did it. At first I didn't think about my answer enough, and just said “Oh, it's not hard, just divide by 7 and look at the remainder.” (“Just count the legs and divide by 4.”) But I realized later that there are several tricks I was using that are not obvious. First, she had never learned short division. When I was in school I had been tormented extensively with long division, which looks like this:

This was all Katara had been shown, so when I said “just divide by 7” this is what she was thinking of. But you only need long division for large divisors. For simple divisors like !!7!!, I was taught short division, an easier technique:

Yeah, I wrote 4 when I meant 3. It doesn't matter, we don't care about the quotient anyway.

But that's one of the tricks I was using that wasn't obvious to Katara: we don't care about the quotient anyway, only the remainder. So when I did this in my head, I discarded the parts of the calculation that were about the quotient, and only kept the steps that pertained to the remainder. The way I was actually doing this sounded like this in my mind:

7 into 12 leaves 5. 7 into 53 leaves 4. 7 into 44 leaves 2. 7 into 25 leaves 4. 7 into 46 leaves 4. 7 into 57 leaves 5. 7 into 58 leaves 9. The answer is 9.

At each step, we consider only the leftmost part of the number, starting with !!12!!. !!12\div 7 !! has a remainder of 5, and to this 5 we append the next digit of the dividend, 3, giving 53. Then we continue in the same way: !!53\div 7!! has a remainder of 4, and to this 4 we append the next digit, giving 44. We never calculate the quotient at all.

I explained the idea with a smaller example, like this:

Suppose you want to see if 1234 is divisible by 7. It's 1200-something, so take away 700, which leaves 500-something. 500-what? 530-something. So take away 490, leaving 40-something. 40-what? 44. Now take away 42, leaving 2. That's not 0, so 1234 is not divisible by 7.

This is how I actually do it. For me this works reasonably well up to 13, and after that it gets progressively more difficult until by 37 I can't effectively do it at all. A crucial element is having the multiples of the divisor memorized. If you're thinking about the mod-13 residue of 680-something, it is a big help to know immediately that you can subtract 650.

A year or two ago I discovered a different method, which I'm sure must be ancient, but is interesting because it's quite different from the other methods I described.

Suppose that the final digit of !!n!! is !!b!!, so that !!n=10a+b!!. Then !!-2n = -20a-2b!!, and this is a multiple of !!7!! if and only if !!n!! is. But !!-20a\equiv a\pmod7 !!, so !!a-2b!! is a multiple of !!7!! if and only if !!n!! is. This gives us the rule:

To check if !!n!! is a multiple of 7, chop off the last digit, double it, and subtract it from the rest of the number. Repeat until the answer becomes obvious.

For !!1234!! we first chop off the !!4!! and subtract !!2\cdot4!! from !!123!! leaving !!115!!. Then we chop off the !!5!! and subtract !!2\cdot5!! from !!11!!, leaving !!1!!. This is not a multiple of !!7!!, so neither is !!1234!!. But with !!1239!!, which is a multiple of !!7!!, we get !!123-2\cdot 9 = 105!! and then !!10-2\cdot5 = 0!!, and we win.

In contrast to the other methods in this article, this method does not tell you the remainder on dividing because it does not preserve the residue mod 7. When we started with !!1234!! we ended with !!1!!. But !!1234\not\equiv 1\pmod 7!!; rather !!1234\equiv 2!!. Each step in this method multiplies the residue by -2, or, if you prefer, by 5. The original residue was 2, so the final residue is !!2\cdot-2\cdot-2 = 8 \equiv 1\pmod 7!!. (Or, if you prefer, !!2\cdot 5\cdot 5= 50 \equiv 1\pmod 7!!.) But if we only care about whether the residue is zero, multiplying it by !!-2!! doesn't matter.

There are some shortcuts in this method too. If the final digit is !!7!!, then rather than doubling it and subtracting 14 you can just chop it off and throw it away, going directly from !!10a+7!! to !!a!!. If your number is !!10a+8!! you can subtract !!7!! from it to make it easier to work with, getting !!10a+1!! and then going to !!a-2!! instead of to !!a-16!!. Similarly when your number ends in !!9!! you can go to !!a-4!! instead of to !!a-18!!. And on the other side, if it ends in !!4!! it is easier to go to !!a-1!! instead of to !!a-8!!.

But even with these tricks it's not clear that this is faster or easier than just doing the short division. It's the same number of steps, and it seems like each step is about the same amount of work.

Finally, I once wowed Katara on an airplane ride by showing her this:

To check !!1429!! using this device, you start at ⓪. The first digit is !!1!!, so you follow one black arrow, to ①, and then a blue arrow, to ③. The next digit is !!4!!, so you follow four black arrows, back to ⓪, and then a blue arrow which loops around to ⓪ again. The next digit is !!2!!, so you follow two black arrows to ② and then a blue arrow to ⑥. And the last digit is 9 so you then follow 9 black arrows to ① and then stop. If you end where you started, at ⓪, the number is divisible by 7. This time we ended at ①, so !!1429!! is not divisible by 7. But if the last digit had been !!1!! instead, then in the last step we would have followed only one black arrow from ⑥ to ⓪, before we stopped, so !!1421!! is a multiple of 7.

This probably isn't useful for mental calculations, but I can imagine that if you were stuck on a long plane ride with no calculator and you needed to compute a lot of mod-7 residues for some reason, it could be quicker than the short division method. The chart is easy to construct and need not be memorized. The black arrows obviously point from !!n!! to !!n+1!!, and the blue arrows all point from !!n!! to !!10n!!.

I made up a whole set of these diagrams and I think it's fun to see how the conventional divisibility rules turn up in them. For example, the rule for divisibility by 3 that says just add up the digits:

Or the rule for divisibility by 5 that says to ignore everything but the last digit:

[ Addendum 20201122: Testing for divisibility by 19. ]

[Other articles in category /math] permanent link

Mon, 23 Jul 2018

Operations that are not quite associative

The paper “Verification of Identities” (1997) of Rajagopalan and Schulman discusses fast ways to test whether a binary operation on a finite set is associative. In general, there is no method that is faster than the naïve algorithm of simply checking whether $$a\ast(b\ast c) = (a\ast b)\ast c$$ for all triples !!\langle a, b, c\rangle!!. This is because:

For every !!n≥3!!, there exists an operation with just one nonassociative triple.

(Page 3.)

But R&S do not give an example. I now have a very nice example, and I think the process that led me to it is a good example of Lower Mathematics at work.

Let's say that an operation !!\ast!! is “good” if it is associative except in exactly one case. We want to find a “good” operation. My first idea was that if we could find a primitive good operation on a set of 3 elements, we could probably extend it to give a good operation on a larger set. Say the set is !!\{a_0, a_1, a_2, b_0, b_1, \ldots, b_{n-4}\}!!. We just need to define the extended operation so that if either of the operands is !!b_i!!, it is associative, and if both operands are !!a_i!! then it is the same as the primitive good operation we already found. The former part is easy: just make it constant, say !!x\ast y = b_0!! except when !!x,y\in\{a_0, a_1, a_2\}!!.

So now all we need to do is find a single good primitive operation, and I did not expend any thought on this at all. There are only !!3^9=19683!! binary operations, and we could quite easily write a program to check them all. In fact, we can do better: generate a binary operation at random and check it. If it's not the primitive good operation we want, just throw it away and try again. This could take longer to run than the exhaustive search, if there happen to be very few good operations and if the program is unlucky, but the program is easier to write, and the run time will be utterly insignificant either way.

I wrote the program, which instantly produced:

$$ \begin{array}{c|ccc} \ast & 0 & 1 & 2 \\ \hline 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 2 & 0 & 1 & 2 \end{array} $$

This is associative except for the case !!1\ast(2\ast 1) \ne (1\ast 2)\ast 1!!. This solves the problem. But confirming that this is a good operation requires manually checking 27 cases, or a perhaps not-immediately-obvious case analysis.

But on a later run of the program, I got lucky and it found a simpler operation, which I can explain without a table:

!!a\ast b = 0!!, except !!2\ast 1=2!!.

Now we don't have to check 27 cases. The operation is simpler, so the proof is too: We know !!b\ast c\ne 1!!, so !!a\ast(b\ast c) !! must be !!0!!. And the only way !!(a \ast b)\ast c \ne 0!! can occur is when !!a\ast b=2!! and !!c=1!!, so !!a=2, b=1, c=1!!.

Now we can even dispense with the construction that extended the operation from !!3!! to !!n!! elements, because the description of the extended operation is the same. We wanted to extend it to be constant whenever !!a!! or !!b!! was larger than !!2!!, and that's what the description already says!

So that reduces the whole thing to about two sentences, which explain the example and why it works. But when reduced in that way, you see how the example works but not how you might go about finding it. I think the interesting part is to see how to get there, and quite a lot of mathematical education tends to over-emphasize analysis (“how can we show this is a good operation?”) at the expense of synthesis (“how can we find a good operation?”).

The exhaustive search would probably have produced the simple operation early on in its run, so there is something to be said for that approach too.

[Other articles in category /math] permanent link

Mon, 18 Jun 2018

Yesterday's metric has a name

Yesterday I presented as a counterexample the topology induced by the following metric:

A circle, with the
center marked.  A shortest-distance path is drawn in blue between two
blue points on the same radius, and in red between two red points on different
radii.  The blue path goes straight from one blue point to the other.
The red path goes from one point straight to the origin,
then straight to the other point.

I asked:

It seems like this example could be useful in other circumstances too. Does it have a name?

Several Gentle Readers have written in to tell me that that this metric is variously known as the British Rail metric, French Metro metric, or SNCF metric. (SNCF = Société nationale des chemins de fer français, the French national railway company). In all cases the conceit is the same (up to isomorphism): to travel to a destination on another railway line one must change trains in London / Paris, where all the lines converge.

Wikipedia claims this is called the post office metric, again I suppose because all the mail comes to the central post office for sorting. I have not seen it called the FedEx metric, but it could have been, with the center of the disc in Memphis.

[ Addendum 20180621: Thanks for Brent Yorgey for correcting my claim that the FedEx super hub is in Nashville. It is in Memphis ]

[Other articles in category /math] permanent link

Sun, 17 Jun 2018

Bounded does not imply totally bounded

I somehow managed to miss the notion of totally bounded when I was learning topology, and it popped up on stack exchange recently. It is a stronger version of boundedness for metric spaces: a space !!M!! is totally bounded if, for any chosen !!\epsilon!!, !!M!! can be covered by a finite family of balls of radius !!\epsilon!!.

This is a strictly stronger property than ordinary boundedness, so the question immediately comes up: what is an example of a space that is bounded but not totally bounded. Many examples are well-known. For example, the infinite-dimensional unit ball is bounded but not totally bounded. But I didn't think of this right away.

Instead I thought of the following rather odd example: Let !!S!! be the closed unit disc and assign each point a polar coordinate !!\langle r,\theta\rangle!! as usual. Now consider the following metric:

$$ d(\langle r_1, \theta_1\rangle, \langle r_2, \theta_2\rangle) = \begin{cases} r_1, & \qquad \text{ if $r_2 = 0$} \\ \lvert r_1 - r_2 \rvert, & \qquad\text{ if $\theta_1 = \theta_2$} \\ r_1 + r_2 & \qquad\text{ otherwise} \\ \end{cases} $$

The idea is this: you can travel between points only along the radii of the disc. To get from !!p_1!! to !!p_2!! that are on different radii, you must go through the origin:

A circle, with the
center marked.  A shortest-distance path is drawn in blue between two
blue points on the same radius, and in red between two red points on different
radii.  The blue path goes straight from one blue point to the other.
The red path goes from one point straight to the origin,
then straight to the other point.

Now clearly when !!\epsilon < \frac12!!, the !!\epsilon!!-ball that covers each point point !!\left\langle 1, \theta\right\rangle!! lies entirely within one of the radii, and so an uncountable number of such balls are required to cover the disc.

It seems like this example could be useful in other circumstances too. Does it have a name?

[ Addendum 2018-07-18: Several Gentle Readers have informed me that this metric has not just one name, but several. ]

[Other articles in category /math] permanent link

Mon, 07 May 2018

Katara constructs finite projective planes

This weekend I got a very exciting text message from Katara:

I have a math question for you

Oh boy! I hope it's one I can answer.


there's this game called spot it where you have cards with 8 symbols on them like so

The back of the 'Spot It' box shows four white circular cards,
each with 8 little pictures on it. A red double-headed arrow between
the first two cards shows that both cards have a picture of a
ladybug.  Another arrow between the second and third cards shows that
both cards have a picture of a red heart.  The next arrow connects the
trees on cards 2 and 3, and a fourth arrow shows that cards 1 and 4
both have apurple cat.  Even though part of card 3 is cut off, we can
see that they share a pair of lips. Cards 2 and 4 both have a gray
padlock picture, even though there's no arrow pointing it out.

and the goal is to find the one matching symbol on your card and the one in the middle

how is it possible that given any pair of cards, there is exactly one matching symbol

Well, whatever my failings as a dad, this is one problem I can solve. I went a little of overboard in my reply:

You need a particular kind of structure called a projective plane.

They only exist for certain numbers of symbols

A simpler example has 7 cards with 3 symbols each.

One thing that's cool about it is that the symbols and the cards are "dual": say you give each round card a name. Then make up a new deck of square cards. There's one square card for each symbol. So there's a square"Ladybug" card. The Ladybug card has on it the names of the round cards that have the Ladybug. Now you can play Spot with the square cards instead of the round ones: each two square cards have exactly one name in common.

In a geometric plane any two points lie on exactly one common line and any two lines intersect in exactly one common point. This is a sort of finite model of that, with cards playing the role of lines and symbols playing the role of points. Or vice versa, it doesn't matter.

More than you wanted to know 😂

ah thank you, I'm pretty sure I understand, sorry for not responding, my phone was charging

I still couldn't shut up about the finite projective planes:

No problem! No response necessary.

It is known that all finite projective planes have n²+n+1 points for some n. So I guess the Spot deck has either 31, 57, or 73 cards and the same number of symbols. Is 57 correct?

Must be 57 because I see from your picture that each card has 8 symbols.

Katara was very patient:

I guess, I would like to talk about this some more when i get home if that's okay

Any time.

(The game costs $13.)

Anyway this evening I cut up some index cards, and found a bunch of stickers in a drawer, and made Katara a projective plane of order 3. This has 13 cards, each with 4 different stickers, and again, every two cards share exactly one sticker. She was very pleased and wanted to know how to make them herself.

Each set of cards has an order, which is a non-negative integer. Then there must be !!n^2 + n + 1!! cards, each with !!n+1!! stickers or symbols. When !!n!! is a prime power, you can use field theory to construct a set of cards from the structure of the (unique) field of order !!n!!.

Fields to projective planes

Order 2

I'll describe the procedure using the plane of order !!n=2!!, which is unusually simple. There will be !!2^2+2+1 = 7!! cards, each with !!3!! of the !!7!! symbols.

Here is the finite field of order 2, called !!GF(2)!!:

+ 0 1
0 0 1
1 1 0
× 0 1
0 0 0
1 0 1
  • The stickers correspond to ordered triples of elements of !!GF(2)!!, except that !!\langle 0,0,0\rangle!! is always omitted. So they are:

    $$\require{cancel}\begin{array}{cc} \cancel{\langle 0,0,0\rangle} & \langle 1,0,0\rangle \\ \langle 0,0,1\rangle & \langle 1,0,1\rangle \\ \langle 0,1,0\rangle & \langle 1,1,0\rangle \\ \langle 0,1,1\rangle & \langle 1,1,1\rangle \\ \end{array} $$

    Of course, you probably do not want to use these symbols exactly. You might decide that !!\langle 1,0,0\rangle!! is a sticker with a picture of a fish, and !!\langle 0,1,0\rangle!! is a sticker with a ladybug.

  • Each card will have !!n+1 = 3!! stickers. To generate a card, pick any two stickers that haven't appeared together before and put them on the card. Say these stickers correspond to the triples !!\langle a,b,c\rangle!! and !!\langle x,y,z\rangle!!. To find the triple for the third sticker on the card, just add the first two triples componentwise, obtaining !!\langle a+x,b+y,c+z\rangle!!. Remember that the addition must be done according to the !!GF(2)!! addition table above! So for example if a card has !!\langle 1,0,1\rangle!! and !!\langle 0,1,1\rangle!!, its third triple will be

    $$\begin{align} \langle 1,0,1 \rangle + \langle 0,1,1 \rangle & = \\ \langle 1+0,0+1,1+1 \rangle & = \\ \langle 1,1,0 \rangle \end{align} $$

    Observe that it doesn't matter which two triples you add; you always get the third one!

Okay, well, that was simple.

Larger order

After Katara did the order 2 case, which has 7 cards, each with 3 of the 7 kinds of stickers, she was ready to move on to something bigger. I had already done the order 3 deck so she decided to do order 4. This has !!4^2+4+1 = 21!! cards each with 5 of the 21 kinds of stickers. The arithmetic is more complicated too; it's !!GF(2^2)!! instead of !!GF(2)!!:

+ 0 1 2 3
0 0 1 2 3
1 1 0 3 2
2 2 3 0 1
3 3 2 1 0
× 0 1 2 3
0 0 0 0 0
1 0 1 2 3
2 0 2 3 1
3 0 3 1 2

When the order !!n!! is larger than 2, there is another wrinkle. There are !!4^3 = 64!! possible triples, and we are throwing away !!\langle 0,0,0\rangle!! as usual, so we have 63. But we need !!4^2+4+1 = 21!!, not !!63!!.

Each sticker is represented not by one triple, but by three. The triples !!\langle a,b,c\rangle, \langle 2a,2b,2c\rangle,!! and !!\langle 3a,3b,3c\rangle!! must be understood to represent the same sticker, all the multiplications being done according to the table above. Then each group of three triples corresponds to a sticker, and we have 21 as we wanted.

Each triple must have a leftmost non-zero entry, and in each group of three similar triples, there will be one where this leftmost non-zero entry is a !!1!!; we will take this as the canonical representative of its class, and it can wear a costume or a disguise that makes it appear to begin with a !!2!! or a !!3!!.

We might assign stickers to triples like this:

$$ \begin{array}{rl} \cancel{\langle 0,0,0\rangle} & \\ \langle 0,0,1 \rangle & \text{apple} \\ \hline \langle 0,1,0 \rangle & \text{bicycle} \\ \langle 0,1,1 \rangle & \text{carrot} \\ \langle 0,1,2 \rangle & \text{dice} \\ \langle 0,1,3 \rangle & \text{elephant} \\ \hline \langle 1,0,0 \rangle & \text{frog} \\ \langle 1,0,1 \rangle & \text{goat} \\ \langle 1,0,2 \rangle & \text{hat} \\ \langle 1,0,3 \rangle & \text{igloo} \\ \langle 1,1,0 \rangle & \text{jellyfish} \\ \langle 1,1,1 \rangle & \text{kite} \\ \langle 1,1,2 \rangle & \text{ladybug} \\ \langle 1,1,3 \rangle & \text{mermaid} \\ \langle 1,2,0 \rangle & \text{nose} \\ \langle 1,2,1 \rangle & \text{octopus} \\ \langle 1,2,2 \rangle & \text{piano} \\ \langle 1,2,3 \rangle & \text{queen} \\ \langle 1,3,0 \rangle & \text{rainbow} \\ \langle 1,3,1 \rangle & \text{shoe} \\ \langle 1,3,2 \rangle & \text{trombone} \\ \langle 1,3,3 \rangle & \text{umbrella} \\ \end{array} $$

We can stop there, because everything after !!\langle 1,3,3 \rangle!! begins with a !!2!! or a !!3!!, and so is some other triple in disguise. For example what sticker goes with !!\langle 0,2,3 \rangle!!? That's actually !!\langle 0,1,2 \rangle!! in disguise, it's !!2·\langle 0,1,2 \rangle!!, which is “dice”. Okay, how about !!\langle 3,3,1 \rangle!!? That's the same as !!3\cdot\langle 1,1,2 \rangle!!, which is “ladybug”. There are !!21!!, as we wanted. Note that the !!21!! naturally breaks down as !!1+4+4^2!!, depending on how many zeroes are at the beginning; that's where that comes from.

Now, just like before, to make a card, we pick two triples that have not yet gone together, say !!\langle 0,0,1 \rangle!! and !!\langle 0,1,0 \rangle!!. We start adding these together as before, obtaining !!\langle 0,1,1 \rangle!!. But we must also add together the disguised versions of these triples, !!\langle 0,0,2 \rangle!! and !!\langle 0,0,3 \rangle!! for the first, and !!\langle 0,2,0 \rangle!! and !! \langle 0,3,0 \rangle!! for the second. This gets us two additional sums, !!\langle 0,2,3 \rangle!!, which is !!\langle 0,1,2 \rangle!! in disguise, and !!\langle 0,3,2 \rangle!!, which is !!\langle 0,1,3 \rangle!! in disguise.

It might seem like it also gets us !!\langle 0,2,2 \rangle!! and !!\langle 0,3,3 \rangle!!, but these are just !!\langle 0,1,1 \rangle!! again, in disguise. Since there are three disguises for !!\langle 0,0,1 \rangle!! and three for !!\langle 0,1,0 \rangle!!, we have nine possible sums, but it turns out the the nine sums are only three different triples, each in three different disguises. So our nine sums get us three additional triples, and, including the two we started with, that makes five, which is exactly how many we need for the first card. The first card gets the stickers for triples !!\langle 0,0,1 \rangle, \langle 0,1,0 \rangle \langle 0,1,1 \rangle \langle 0,1,2 \rangle,!! and !!\langle 0,1,3 \rangle,!! which are apple, bicycle, carrot, dice, and elephant.

That was anticlimactic. Let's do one more. We don't have a card yet with ladybug and trombone. These are !!\langle 1,1,2 \rangle!! and !!\langle 1,3,2 \rangle!!, and we must add them together, and also the disguised versions:

$$\begin{array}{c|ccc} & \langle 1,1,2 \rangle & \langle 2,2,3 \rangle & \langle 3,3,1 \rangle \\ \hline \langle 1,3,2 \rangle & \langle 0,2,0 \rangle & \langle 3,1,1 \rangle & \langle 2,0,3 \rangle \\ \langle 2,1,3 \rangle & \langle 3,0,1 \rangle & \langle 0,3,0 \rangle & \langle 1,2,2 \rangle \\ \langle 3,2,1 \rangle & \langle 2,3,3 \rangle & \langle 1,0,2 \rangle & \langle 0,1,0 \rangle \\ \end{array}$$

These nine results do indeed pick out three triples in three disguises each, and it's easy to select the three of these that are canonical: they have a 1 in the leftmost nonzero position, so the three sums are !!\langle 0,1,0 \rangle,!! !!\langle 1,0,2 \rangle,!! and !!\langle 1,2,2 \rangle!!, which are bicycle, hat, and piano. So the one card that has a ladybug and a trombone also has a bicycle, a hat, and a piano, which should not seem obvious. Note that this card does have the required single overlap with the other card we constructed: both have bicycles.

Well, that was fun. Katara did hers with colored dots instead of stickers:

Katara's set of 21 cards.  Each
has five colored dots of 21 total different colors, and each colored
dot is labeled with a letter of the alphabet in the same color.  There
are some mistakes that are crossed out and corrected, and a couple of
mistakes that are not yet crossed out or corrected.

The ABCDE card is in the upper left; the bicycle-hat-ladybug-piano-trombone one is the second row from the bottom, second column from the left. The colors look bad in this picture; the light is too yellow and so all the blues and purples look black.

After I took this picture, we checked these cards and found a couple of calculation errors, which we corrected. A correct set of cards is:

$$ \begin{array}{ccc} \text{abcde} & \text{bhlpt} & \text{dgmpr} \\ \text{afghi} & \text{bimqu} & \text{dhjou} \\ \text{ajklm} & \text{cfkpu} & \text{diknt} \\ \text{anopq} & \text{cgjqt} & \text{efmot} \\ \text{arstu} & \text{chmns} & \text{eglnu} \\ \text{bfjnr} & \text{cilor} & \text{ehkqr} \\ \text{bgkos} & \text{dflqs} & \text{eijps} \\ \end{array} $$

Fun facts about finite projective planes:

  • This construction always turns a finite field of order !!n!! into a finite projective plane of order !!n!!.

  • A finite field of order !!n!! exists exactly when !!n!! is a prime power and then there is exactly one finite field. So this construction gives us finite projective planes of orders !!1,2,3,4,5,7,8,9,11,13,16!!, but not of orders !!6,10,12,14,15!!. Do finite projective planes of those latter orders exist?

  • Is this method the only way to construct a finite projective plane? Yes, when !!n<9!!. But there are four non-isomorphic projective planes of order !!9!!, and this only constructs one of them.

    What about for !!n≥11!!? Nobody knows.

[Other articles in category /math] permanent link

Wed, 18 Apr 2018

Lower mathematics solves an easy problem

[ Warning: this article is mathematically uninteresting. ]

I woke up in the middle of the night last night and while I was waiting to go back to sleep, I browsed math Stack Exchange. At four in the morning I am not at my best, but sometimes I can learn something and sometimes I can even contribute.

The question that grabbed my attention this time was Arithmetic sequence where every term is prime?. OP wants to know if the arithmetic sequence $$d\mapsto a+d b$$ contains composite elements for every fixed positive integers !!a,b!!.

Now of course the answer is yes, or the counterexample would give us a quick and simple method for constructing prime numbers, and finding such has been an open problem for thousands of years. OP was certainly aware of this, but had not been able to find a simple proof. Their searching was confounded by more advanced matters relating to the Green-Tao theorem and such like, which, being more interesting, are much more widely discussed.

There are a couple of remarkable things about the answers that were given. First, even though the problem is easy, the first two answers posted were actually wrong, and another (quickly deleted) was so complicated that I couldn't tell if it was right or not.

One user immediately commented:

What happens if !!d=a!!?

which is very much to the point; when !!d=a!! then the element of the sequence is !!a+ab!! which is necessarily composite…

…unless !!a=1!!. So the comment does not quite take care of the whole question. A second user posted an answer with this same omission, and had to correct it later.

I might not have picked up on this case either, during the daytime. But at 4 AM I was not immediately certain that !!a+ab!! was composite and I had to think about it. I factored it to get !!a(1+b)!! and then I saw that if !!a=1!! or !!1+b=1!! then we lose. (!!1+b=1!! is impossible. !!a+ab!! might of course be composite even if !!a=1!!, but further argumentation is needed.)

So I did pick up on this, and gave a complete answer, of which the important part is:

Just take !!d=kb+k+a!! for any !!k!! whatever. Then the element is $$kb^2+(k+a)b+a = (kb+a)(b+a)$$ which is composite.

Okay, fine. But OP asked how I came up with that and if it was pure “insight”, so I thought I'd try to reconstruct how I got there at 4 AM. The problem is simple enough that I think I can remember most of how I got to the answer.

As I've mentioned before, I am not a pure insight kind of person. While better mathematicians are flitting swiftly from peak to peak, I plod along in the dark and gloomy valleys. I did not get !!d=kb+k+a!! in a brilliant flash of inspiration.

Instead, my thought process, as well as I can remember it, went like this:

  • “That doesn't work when !!a=1!!. But there must be composite numbers in sequences of the form !!d b+1!!. I wonder what they look like?”

  • “I guess I'll try an example. How about !!4d+1!!?”

  • “Hmm, it starts at 5, 5 is composite… No it isn't.”

  • “But it's increasing by 4 each time so it must hit each mod-5 residue class in some order.”

I should cut in at this point to add that my thinking was nowhere near this articulate or even verbal. The thing about the sequence hitting all the residue classes was more like a feeling in my body, like when I am recognizing a familiar place. When !!a!! and !!b!! are relatively prime, that means that when you are taking steps of size !!b!!, you hit all the !!a!!'s and don't skip any; that's what relatively prime is all about.

So maybe that counts as “insight”? Or “intuition about relative primality”? I think that description makes it sound much more impressive than it really is. I do not want a lot of credit for this. Maybe a better way to describe it is that I had been in this familiar place many times before, and I recognized it again.

Anyway I continued something like this:

  • “If it hits every residue class over and over in sequence it must hit residue class !!0!! infinitely often. How does that go? It hits !!5!!, so it hits !!5+4·5 = 25!! also.”

That was good enough for me; I did not even consider the next hit, !!45!!, perhaps because that number was too big for me to calculate at that moment.

I didn't use the phrase “residue class” either. That's just my verbal translation of my 4AM nonverbal thinking. At the time it was more like: there are some good things to hit and some bad ones, and the good ones are evenly spaced out, so if we hit each position in the even spacing-ness we must periodically hit some of the good things.

  • “Okay, then the first element of !!1+d b!! is !!1+b!!, and then every !!1+b!! steps after that it hits another multiple of !!1+b!!, so we want every !!1+b!! elements, so take !!d = 1 + k(1+b)!!.”

Then I posted the answer, saying that when !!a=1!! you take !!d=1+k(1+b)!! and the sequence element is !!1+b+kb+kb^2 = (1+b)(1+kb)!!.

Then I realized that I had the same feeling in my body even when !!a≠1!!, because it only depended on the way the residue classes repeated, and changing !!a!! doesn't affect that, it just slides everything left or right by a constant amount. So I went back to edit the !!1+k(1+b)!! to be !!a+k(1+b)!! instead.

I have no particular conclusion to draw about this.

[Other articles in category /math] permanent link

Mon, 16 Apr 2018

A familiar set with an unexpected order type

I dreamed this one up in high school and I recommend it as an exercise for kids at an appropriate level.

Consider the set of all Roman numerals

$${ \text{I}, \text{II}, \text{III}, \text{IV}, \text{V}, \ldots, \text{XIII}, \text{XIV}, \text{XV}, \text{XVI}, \ldots, \\ \text{XXXVIII}, \text{XXXIX}, \text{XL}, \text{XLI}, \ldots, \text{XLIX}, \text{L},\ldots,\\ \text{C}, \ldots , \text{D}, \ldots, \text{M}, \ldots, \text{MM}, \ldots, \text{MMM}, \ldots, \text{MMMM}, \ldots, \text{MMMMM}, \ldots }$$

where we allow an arbitrarily large number of M's on the front, so that every number has a unique representation. For example the number 10,067 is represented by !!\text{MMMMMMMMMMLXVII}!!.

Now sort the list into alphabetical order.

It is easy to show that it begins with !!\text{C}, \text{CC}, \text{CCC}, \text{CCCI}, \ldots!! and ends !!\text{XXXVII}, \text{XXXVIII}!!. But it's still an infinite list! Instead of being infinite at one end or the other, or even both, like most infinite lists, it's infinite in the middle.

Of course once you have the idea it's easy to think of more examples (!!\left\{ \frac1n\mid n\in\Bbb Z, n\ne 0\right\}!! for instance) but I hadn't seen anything like this before and I was quite pleased.

[Other articles in category /math] permanent link

Sun, 15 Apr 2018

On the smallest natural number

The earliest known mathematics book printed in Europe is an untitled arithmetic text published in Treviso in 1478, Originally written in Venetian dialect.

The Treviso Arithmetic states unequivocally:

Number is a multitude brought together or assembled from several units, and always from two at least, as in the case of 2, which is the first and the smallest number.

(original Venetian)

And a little later:

Of [the digits] the first figure, 1, is not called a number but the source of number.

(original Venetian)

(English translations are from David Eugene Smith, A Source Book in Mathematics (1959). A complete translation appears in Frank J. Swetz, Capitalism and Arithmetic The New Math of the Fifteenth Century (1987).)

Complete text (in Venetian)

By the way, today is the 311th birthday of Leonhard Euler.

[Other articles in category /math] permanent link

Fri, 13 Apr 2018

Exciting homology group news!

As an undergraduate I wondered and wondered about how manifolds and things are classified in algebraic topology, but I couldn't find any way into the subject. All the presentations I found were too abstract and I never came out of it with any concrete idea of how you would actually calculate any specific fundamental groups. I knew that the fundamental group of the circle was !!\Bbb Z!! and the group of the torus was !!\Bbb Z^2!! and I understood basically why, but I didn't know how you would figure this out without geometric intuition.

This was fixed for me in the very last undergrad math class I took, at Columbia University with Johan Tysk. That was the lowest point of my adult life, but the algebraic topology was the one bright spot in it. I don't know what might have happened to me if I hadn't had that class to sustain my spirit. And I learned how to calculate homotopy groups!

(We used Professor Tysk's course notes, supplemented by William Massey's introduction to algebraic topology. I didn't buy a copy of Massey and I haven't read it all, but I think I can recommend it for this purpose. The parts I have read seemed clear and direct.)

Anyway there things stood for a long time. Over the next few decades I made a couple of superficial attempts to find out about homology groups, but again the presentations were too abstract. I had been told that the homology approach was preferred to the homotopy approach because the groups were easier to actually calculate. But none of the sources I found seemed to tell me how to actually calculate anything concrete.

Then a few days ago I was in the coffee shop working on a geometry problem involving an icosi-dodecahedron, and the woman next to me asked me what I was doing. Usually when someone asks me this in a coffee shop, they do not want to hear the answer, and I do not want to give it, because if I do their eyes will glaze over and then they will make some comment that I have heard before and do not want to hear again. But it transpired that this woman was a math postdoc at Penn, and an algebraic topologist, so I could launch into an explanation of what I was doing, comfortable in the knowledge that if I said something she didn't understand she would just stop me and ask a question. Yay, fun!

Her research is in “persistent homology”, which I had never heard of. So I looked that up and didn't get very far, also because I still didn't know anything about homology. (Also, as she says, the Wikipedia article is kinda crappy.) But I ran into her again a couple of days later and she explained the persistent part, and I know enough about what homology is that the explanation made sense.

Her research involves actually calculating actual homology groups of actual manifolds on an actual computer, so I was inspired to take another crack at understanding homology groups. I did a couple of web searches and when I searched for “betti number tutorial” I hit paydirt: these notes titled “persistent homology tutorial” by Xiaojin Zhu of the University of Wisconsin at Madison. They're only 37 slides long, and I could skip the first 15. Then slide 23 gives the magic key. Okay! I have not yet calculated any actual homology groups, so this post might be premature, but I expect I'll finish the slides in a couple of days and try my hand at the calculations and be more or less successful. And the instructions seem clear enough that I can imagine implementing a computer algorithm to calculate the homology groups for a big ugly complex, as this math postdoc does.

I had heard before that the advantage of the homology approach over the homotopy approach is that the homologies are easier to actually calculate with, and now I see why. I could have programmed a computer to do homotopy group calculations, but the output would in general have been some quotient of a free group given by a group presentation, and this is basically useless as far as further computation goes. For example the question of whether two differently-presented groups are isomorphic is undecidable, and I think similar sorts of questions, such as whether the group is abelian, or whether it is infinite, are similarly undecidable.

Sometimes you get a nice group, but usually you don't. For example the homotopy group of the Klein bottle is the quotient of the free group on two generators under the smallest equivalence relation in which !!aba = b!!; that is:

$$\langle a, b\mid abab^{-1}\rangle$$

which is not anything I have seen in any other context. Even the question of whether two given group elements are equal is in general undecidable. So you get an answer, but then you can't actually do anything with it once you have it. (“You're in a balloon!”) The homology approach throws away a lot of information, enough to render the results comprehensible, but it also leaves enough to do something with.

[Other articles in category /math] permanent link

Mon, 19 Mar 2018

Funky coordinate systems

I had a fun idea this morning. As a kid I was really interested in polar coordinates and kind of disappointed that there didn't seem to be any other coordinate systems to tinker with. But this morning I realized there were a lot.

Let !!F(c)!! be some parametrized family of curves that partition the plane, or almost all of the plane, say except for a finite number of exceptions. If you have two such families !!F_1(c)!! and !!F_2(c)!!, and if each curve in !!F_1!! intersects each curve in !!F_2!! in exactly one point (again with maybe a few exceptions) then you have a coordinate system: almost every point !!P!! lies on !!F_1(a)!! and !!F_2(b)!! for some unique choice of !!\langle a, b\rangle!!, and these are its coordinates in the !!F_1–F_2!! system.

For example, when !!F_1(c)!! is the family of lines !!x=c!! and !!F_2(c)!! is the family of lines !!y=c!! then you get ordinary Cartesian coordinates, and when !!F_1(c)!! is the family of circles !!x^2+y^2=c!! and !!F_2(c)!! is the family !!y=cx!! (plus also !!x=0!!) you get standard polar coordinates, which don't quite work because the origin is in every member of !!F_2!!, but it's the only weird exception.

But there are many other families that work. To take a particularly simple example you can pick some constant !!k!! and then take

$$\begin{align} F_1(c): && x & =c \\ F_2(c): && y & =kx+c. \end{align} $$

This is like Cartesian coordinates except the axes are skewed. I did know about this when I was a kid but I considered it not sufficiently interesting.

For a more interesting example, try

$$\begin{align} F_1(c): && x^2-y^2 & =c \\ F_2(c): && xy & =c \end{align} $$

which looks like this:

Two families of hyperbolas, as in the displayed equation
immediately preceding. The hyperbolas !!x^2-y^2 = c!! (in blue) and !!xy=c!! (in red)

I've seen that illustration before but I don't think I thought of using it as a coordinate system. Well, okay, every pair of hyperbolas intersects in two points, not one. So it's a parametrization of the boundary of real projective space or something, fine. Still fun!

In the very nice cases (such as the hyperbolas) each pair of curves is orthogonal at their point of intersection, but that's not a requirement, as with the skew Cartesian system. I'm pretty sure that if you have one family !!F!! you can construct a dual family !!F'!! that is orthogonal to it everywhere by letting !!F'!! be the paths of gradient descent or something. I'm not sure what the orthogonality is going to be important for but I bet it's sometimes useful.

You can also mix and match families, so for example take:

$$\begin{align} F_1(c): && x & =c \\ F_2(c): && xy & =c \end{align} $$

Some examples work better than others. The !!xy=c!! hyperbolas are kind of a mess when !!c=0!!, and they don't go together with the !!x^2+y^2=c!! circles in the right way at all: each circle intersects each hyperbola in four points. But it occurs to me that as with the projective plane thingy, we don't have to let that be a problem. Take !!S!! to be the quotient space of the plane where two points are identified if their !!F_1–F_2!!-coordinates are the same and then investigate !!S!!. Or maybe go more directly and take !!S = F_1 \times F_2!! (literally the Cartesian product), and then topologize !!S!! in some reasonably natural way. Maybe just give it the product topology. I dunno, I have to think about it.

(I was a bit worried about how to draw the hyperbola picture, but I tried Google Image search for “families of orthogonal hyperbolas”, and got just what I needed. Truly, we live in an age of marvels!)

[Other articles in category /math] permanent link

Sun, 11 Mar 2018

Quick and dirty prime counting

I've been thinking for a while that I probably ought to get around to memorizing all the prime numbers under 1,000, so that I don't have to wonder about things like 893 all the time, and last night in the car I started thinking about it again, and wondered how hard it would be. There are 25 primes under 100, so presumably fewer than 250 under 1,000, which is not excessive. But I wondered if I could get a better estimate.

The prime number theorem tells us that the number of primes less than !!n!! is !!O(\frac n{\log n})!! and I think the logarithm is a natural one, but maybe there is some constant factor in there or something, I forget and I did not want to think about it too hard because I was driving. Anyway I cannot do natural logarithms in my head.

Be we don't need to do any actual logarithms. Let's estimate the fraction of primes up to !!n!! as !!\frac 1{c\log n}!! where !!c!! is unknown and the base of the logarithm is then unimportant. The denominator scales linearly with the power of !!n!!, so the difference between the denominators for !!n=10!! and !!n=100!! is the same as the difference between the denominators for !!n=100!! and !!n=1000!!.

There are 4 primes less than 10, or !!\frac25!!, so the denominator is 2.5. And there are 25 primes less than 100, so the denominator here is 4. The difference is 1.5, so the denominator for !!n=1000!! ought to be around 5.5, and that means that about !!\frac2{11}!! of the numbers up to 1000 are prime. This yields an estimate of 182.

I found out later that the correct number is 186, so I felt pretty good about that.

[ Addendum: The correct number is 168, not 186, so I wasn't as close as I thought. ]

[Other articles in category /math] permanent link

Wed, 07 Feb 2018

The many faces of the Petersen graph

(Actually the Petersen graph cannot really be said to have faces, as it is nonplanar. HA! HA! I MAKE JOKE!​!1!)

This article was going to be about how GraphViz renders the Petersen graph, but instead it turned out to be about how GraphViz doesn't render the Petersen graph. The GraphViz stuff will be along later.

Here we have the Petersen graph, which, according to Donald Knuth, “serves as a counterexample to many optimistic predictions about what might be true for graphs in general.” It is not that the Petersen graph is stubborn! But it marches to the beat of a different drummer. If you have not met it before, prepare to be delighted.

The Petersen
graph has two sets of five vertices each.  Each set is connected into
a pentagonal ring.  There are five more edges between vertices in
opposite rings, but instead of being connected 0–0 1–1 2–2 3–3 4–4,
they are connected 0–0 1–2 2–4 3–1 4–3.

This is the basic structure: a blue 5-cycle, and a red 5-cycle. Corresponding vertices in the two cycles are connected by five purple edges. But there is a twist! Notice that the vertices in the red cycle are connected in the order 1–3–5–2–4.

There are different ways to lay out the Petersen graph that showcase its many interesting properties. For example, the standard presentation, above, demonstrates that the Petersen graph is nonplanar, since it obviously contracts to !!K_5!!. The presentation below obscures this, but it is good for seeing that the graph has diameter only 2:

Wait, what? Where did the pentagons go?

Try this instead:

The Petersen
graph laid out as a tree, with a root attached to three level-1 nodes,
each attached to 2 level-2 nodes.  The six level-2 nodes are then
connected into a ring so that each level-2 node is at distance 1 or
distance 2 from each other level-2 node.

Again the red vertices are connected in the order 1–3–5–2–4.

Okay, that is indeed the Petersen graph, but how does it help us see that the graph has diameter 2? Color the nodes by how far down they are from the root:

  • Obviously, the root node (black) has distance at most 2 to every other node, because the tree has only depth 2.

  • Each of the three second-level nodes (red) is distance 2 from the other two, via a path through the root.

  • The six third-level nodes (blue) are linked in a 6-cycle (dotted lines), so that each third-level node is at most two steps away along the cycle from the others, except for the one furthest away, but that is its sibling in the tree, and it has a path of length 2 through their common parent.

  • And since each third-level node (say, the one with the red ring) is connected by a dotted edge (orange) to cousins in both of the other branches of the tree, it's only distance 2 from both of its red uncle nodes.

Looking at the pentagonal version, you would not suspect the Petersen graph of also having a sixfold symmetry, but it does. We'll get there in two steps. Again, here's a version where it's not so easy to see that it's actually the Petersen graph, but whatever it is, it is at least clear that it has an automorphism of order six (give it a one-sixth turn):


The represents three vertices, one in each color. In the picture they are superimposed, but in the actual graph, no pair of the three is connected by an edge. Instead, each of the three is connected not to the others but to a tenth vertex that I omitted from the diagram entirely.

Let's pull apart the three vertices and reveal the hidden tenth vertex and its three edges:


Here is the same drawing, recolored to match the tree diagram from before; the outer hexagon is just the 6-cycle formed by the six blue leaf nodes:


But maybe it's easier to see if we look for red and blue pentagons. There are a couple of ways to do that:

narf   narf

As always, the red vertices are connected in the order 1–3–5–2–4.

Finally, here's a presentation you don't often see. It demonstrates that the Petersen graph also has fourfold symmetry:


Again, and represent single vertices stretched out into dumbbell shapes. The diagram only shows 14 of the 15 edges; the fifteenth connects the two dumbbells.

The pentagons are deeply hidden here. Can you find them? (Spoiler)

Even though this article was supposed to be about GraphViz, I found it impossible to get it to render the diagrams I wanted it to, and I had to fall back on Inkscape. Fortunately Inkscape is a ton of fun.

[Other articles in category /math] permanent link

Mon, 29 Jan 2018

The rubber duck explains coherence spaces

I've spent a chunk of the past week, at least, trying to understand the idea of a coherence space (or coherent space). This appears in Jean-Yves Girard's Proofs and Types, and it's a model of a data type. For example, the type of integers and the type of booleans can be modeled as coherence spaces.

The definition is one of those simple but bafflingly abstract ones that you often meet in mathematics: There is a set !!\lvert \mathcal{A}\rvert!! of tokens, and the points of the coherence space !!\mathcal{A}!! (“cliques”) are sets of tokens. The cliques are required to satisfy two properties:

  1. If !!a!! is a clique, and !!a'\subset a!!, then !!a'!! is also a clique.
  2. Suppose !!\mathcal M!! is some family of cliques such that !!a\cup a'!! is a clique for each !!a, a'\in \mathcal M!!. Then !!\bigcup {\mathcal M}!! is also a clique.

To beginning math students it often seems like these sorts of definitions are generated at random. Okay, today we're going to study Eulerian preorders with no maximum element that are closed under finite unions; tomorrow we're going to study semispatulated coalgebras with countably infinite signatures and the weak Cosell property. Whatever, man.

I have a long article about this in progress, but I'll summarize: they are never generated at random. The properties are carefully chosen because we have something in mind that we are trying to model, and until you understand what that thing is, and why someone thinks those properties are the important ones, you are not going to get anywhere.

So when I see something like this I must stop immediately and ask ‘wat’. I can try to come up with the explanation myself, or I can read on hoping for an explanation, or I can do some of each, but I am not going to progress very far until I understand what it is about. And I'm not sure anyone short of Alexander Grothendieck would have any more success trying to move on with the definition itself and nothing else.

Girard explains shortly after:

The aim is to interpret a type by a coherence space !!\mathcal{A}!!, and a term of this type as a point [clique] of !!\mathcal{A}!!, infinite in general…

Okay, fine. I understand the point of the project, although not why the definition is what it is. I know a fair amount about types. And Girard has given two examples: booleans and integers. But these examples are unusually simple, because none of the cliques has more than one element, and so the examples are not as illuminating as they might be.

Some of the ways I tried to press onward were:

  1. Read ahead and see if there is more explanation. I tried this but I still wasn't getting it. The next section seemed clear: the cliques define a “coherence” relation on the tokens, from which the cliques can be recovered. Consider a graph, where the vertices are tokens and there is an edge !!a—b!! exactly when !!\{a, a'\}!! is a clique; we say that !!a!! and !!a'!! are coherent. Then the cliques of the coherence space are exactly the cliques of the graph; hence the name. The graph is called the web of the space, and from the web one can recover the original space.

    But after that part came stable functions, which I couldn't figure out, and I got stuck again.

  2. Read ahead and see if there is a more complicated specific example. There wasn't.

  3. Read ahead and see if any of the derived concepts are familiar, and if so then work backward. For instance, if I had been able to recognize that I already knew what stable functions were, I might have been able to leverage tha tinto an understanding of what was going on with the coherence spaces. But for me they were just another problem of the same sort: what is a stable function supposed to be modeling?

  4. Read someone else's explanation instead. I tried several without much success. They all seemed to be written for someone who already had a clue what was going on. (That is a large part of the reason I have written up this long and clueless explanation.)

  5. Try to construct some examples and see if they make sense in the context of what comes later. For example, I know what the coherence space of booleans looks like because Girard showed me. Can I figure out the structure of the coherence space for the type of “wrapped booleans”?

    -- (Haskell)
    data WrappedBoolean = W Bool

    Can I figure it out for the type of pairs of booleans?

    -- (Haskell)
    type BooleanPair = (Bool, Bool)

None of this was working. I had several different ideas about what the coherence spaces might look like for other types, but none of them seemed to fit with what Girard was doing. I couldn't come up with any consistent story.

So I prepared to ask on StackExchange, and I spent about an hour writing up my question, explaining all the things I had tried and what the problems were with each one. And as I drew near to the end of this, the clouds parted! I never had to post the question. I was in the middle of composing this paragraph:

In section 8.4 Girard defines a direct product of coherence spaces, but it doesn't look like the direct product I need to get a product type; it looks more like a disjoint union type. If the coherence space for Pairbool is the square of the coherence space for !!{{\mathcal B}ool}!!, how? It has 4 2-cliques, but if those are the total elements of !!{{\mathcal B}ool}^2!!, then what are do the 1-cliques mean?

I decided I hadn't made enough of an effort to understand the direct product. So even though I couldn't see how it could possibly give me anything like what I wanted, I followed its definition for !!{{\mathcal B}ool}^2!! — and the light came on.

Here's the puzzling coproduct-like definition of the product of two coherence spaces, from page 61:

If !!{\mathcal A}!! and !!{\mathcal B}!! are two coherence spaces, we define !!{\mathcal A}\&{\mathcal B}!! by:

!!|{\mathcal A}\&{\mathcal B}| = |{\mathcal A}| + |{\mathcal B}| = \{1\}×|{\mathcal A}| \cup \{2\}×|{\mathcal B}|!!

That is, the tokens in the product space are literally the disjoint union of the tokens in the component spaces.

And the edges in the product's web are whatever they were in !!{\mathcal A}!!'s web (except lifted from !!|{\mathcal A}|!! to !!\{1\}×|{\mathcal A}|!!), whatever they were in !!{\mathcal B}!!'s web (similarly), and also there is an edge between every !!\langle1, {\mathcal A}\rangle!! and each !!\langle2, {\mathcal B}\rangle!!. For !!{{\mathcal B}ool}^2!! the web looks like this:

There is no edge between !!\langle 1, \text{True}\rangle!! and !!\langle 1, \text{False}\rangle!! because in !!{{\mathcal B}ool}!! there is no edge between !!\text{True}!! and !!\text{False}!!.

This graph has nine cliques. Here they are ordered by set inclusion:

(In this second diagram I have abbreviated the pair !!\langle1, \text{True}\rangle!! to just !!1T!!. The top nodes in the diagram are each labeled with a set of two ordered pairs.)

What does this mean? The ordered pairs of booleans are being represented by functions. The boolean pair !!\langle x, y\rangle!! is represented by the function that takes as its argument a number, either 1 or 2, and then returns the corresponding component of the pair: the first component !!x!! if the argument was 1, and the second component !!y!! if the argument was 2.

The nodes in the bottom diagram represent functions. The top row are fully-defined functions. For example, !!\{1F, 2T\}!! is the function with !!f(1) = \text{False}!! and !!f(2) = \text{True}!!, representing the boolean pair !!\langle\text{False}, \text{True}\rangle!!. Similarly if we were looking at a space of infinite lists, we could consider it a function from !!\Bbb N!! to whatever the type of the lists elements was. Then the top row of nodes in the coherence space would be infinite sets of pairs of the form !!\langle n, \text{(list element)}\rangle!!.

The lower nodes are still functions, but they are functions about which we have only incomplete information. The node !!\{2T\}!! is a function for which !!f(2) = \text{True}!!. But we don't yet know what !!f(1)!! is because we haven't yet tried to compute it. And the bottommost node !!\varnothing!! is a function where we don't know anything at all — yet. As we test the function on various arguments, we move up the graph, always following the edges. The lower nodes are approximations to the upper ones, made on the basis of incomplete information about what is higher up.

Now the importance of finite approximants on page 56 becomes clearer. !!{{\mathcal B}ool}^2!! is already finite. But in general the space is infinite because the type is functions on an infinite domain, or infinite lists, or something of that sort. In such a space we can't get all the way to the top row of nodes because to do that we would have to call the function on all its possible arguments, or examine every element of the list, which is impossible. Girard says “Above all, there are enough finite approximants to a.” I didn't understand what he meant by “enough”. But what he means is that each clique !!a!! is the union of its finite approximants: each bit of information in the function !!a!! is obtainable from some finite approximation of !!a!!. The “stable functions” of section 8.3 start to become less nebulous also.

I had been thinking that the !!\varnothing!! node was somehow like the !!\bot!! element in a Scott domain, and then I struggled to identify anything like !!\langle \text{False}, \bot\rangle!!. It looks at first like you can do it somehow, because there are the right number of nodes at the middle level. Trouble arises in other coherence spaces.
For the WrappedBoolean type, for example, the type has four elements: !! W\ \text{True}, W\ \text{False}, W\ \bot,!! and !!\bot!!. I think the coherence space for WrappedBoolean is just like the one for !!{{\mathcal B}ool}!!:

Presented with a value from WrappedBoolean, you don't initially know what it is. Then you examine it, and you know whether it is !!W\ \text{True}!! or !!W\ \text{False}!!. You are now done.

I think there isn't anything like !!\bot!! or !!W\ \bot!! in the coherence space. Or maybe they they are there but sharing the !!\varnothing!! node. But I think more likely partial objects will appear in some other way.

Whew! Now I can move along.

(If you don't understand why “rubber duck”, Wikipedia explains:

Many programmers have had the experience of explaining a problem to someone else, possibly even to someone who knows nothing about programming, and then hitting upon the solution in the process of explaining the problem.

[“Rubber duck”] is a reference to a story in the book The Pragmatic Programmer in which a programmer would carry around a rubber duck and debug their code by forcing themselves to explain it, line-by-line, to the duck.

I spent a week on this but didn't figure it out until I tried formulating my question for StackExchange. The draft question, never completed, is here if for some reason you want to see what it looked like.)

[Other articles in category /math/logic] permanent link

Mon, 22 Jan 2018

The gods of Stackexchange karma are fickle

A few years ago an se.math user asked why their undistinguished answer to some humdrum question had gotten so many upvotes. I replied:

It's the gods of Stackexchange Karma evening your score for that really clever answer you posted two weeks ago that still has 0 upvotes.

If you're going to do Stack Exchange, I think it's important not to stress out about the whys of the votes, and particularly important not to take them personally. The karma gods do not always show the most refined taste. As Brandon Tartikoff once said, “All hits are flukes”.

Today I got an unusually flukey hit. But first, here are some nice examples in the opposite direction, posts that I put a lot of trouble and effort into, which were clear and useful, helpful to the querent, and which received no upvotes at all.

Here the querent asked, suppose I have several nonstandard dice with various labeling on the faces. Player A rolls some of the dice, and Player B rolls a different set of dice. How do I calculate things like “Player A has an X% chance of rolling a higher total”?

Mathematicians are not really the right people to ask this to, because many of them will reply obtusely, informing you that it depends on how many dice are rolled and on how their faces are actually labeled, and that the question did not specify these, but that if it had, the problem would be trivial. (I thought there were comments to that effect on this question, but if there were they have been deleted. In any case nobody else answered.) But this person was writing a computer game and wanted to understand how to implement a computer algorithm for doing the calculation. There is a lot one can do to help this person. I posted an answer that I thought was very nice. The querent liked it, but it got zero upvotes. This happens quite often. Which is fine, because the fun of doing it and the satisfaction of helping someone are reward enough.

Like that one, many of these questions are ignored because they aren't mathematically interesting. Or sometimes the mathematics is simple but the computer implementation is not.

Sometimes I do them just because I want to know the particular answer. (How much of an advantage does Player A get from being allowed to add 1 to their total?)

Sometimes there's an interesting pedagogical problem, such as: how do I give a hint that will point in the right direction without giving away the whole secret? Or: how do I wade through this person's confused explanation and understand what they are really asking, or what they are really confused about? That last person was given a proof that glossed over six or seven steps in the reasoning, focusing instead on the technically interesting induction proof in the middle. The six or seven steps are straightforward, if you already have enough practice with logical reasoning about quantified statements. But this querent didn't follow the reasoning that led up to the induction, so they didn't understand why the induction was useful or what it was for.

Some people are just confused about what the question is. That person has a complicated-seeming homework exercise about the behavior of the logistic map, and need helps interpreting it from someone who has a better idea what is going on. The answer didn't require any research effort, and it's mathematically uninteresting, but it did require attention from someone who has a better idea of what is going on. The querent was happy, I was happy, and nobody else noticed.

I never take the voting personally, because the gods of Stackexchange karma are so fickle, and today I got a reminder of that. Today's runaway hit was for a complete triviality that I knocked off in two minutes. It currently has 94 upvotes and seems likely to get a gold medal. (That gold medal, plus $4.25, will get me a free latte at Starbuck's!)

The question asks how to find a prime factor of 7,999,973 without using a calculator.

It's one of those easy-if-you-happen-to-know-the-trick things, and I just happened to get there before one of the (many) other people who happens to know the trick.

As we used to say in the system administration biz, some days you're an idiot if you can't explain how to do real-time robot arm control in Unix, other days you're a genius if you fix their terminal by plugging it in.

[ Addendum 20180123: I got the gold medal. Woo-hoo, free latte! ]

[Other articles in category /math] permanent link

Fri, 22 Dec 2017

Orthogonal polygons

A couple of years ago I wrote here about some interesting projects I had not finished. One of these was to enumerate and draw orthogonal polygons.

An orthogonal polygon is simply one whose angles are all right angles. All rectangles are orthogonal polygons, but there are many other types. For example, here are examples of orthogonal decagons:

orthogonal decagons

If you ignore the lengths of the edges, and pay attention only to the direction that the corners turn, the orthogonal polygons fall into types. The rectangle is the only type with four sides. There is also only one type with six sides; it is an L-shaped hexagon. There are four types with eight sides, and the illustration shows the eight types with ten sides.

Contributing to OEIS was a life goal of mine and I was thrilled when I was able to contribute the sequence of the number of types of orthogonal !!2n!!-gons.

Enumerating the types is not hard. For !!2n!!-gons, there is one type for each “bracelet” of !!n-2!! numbers whose sum is !!n+2!!.[1]

In the illustration above, !!n=5!! and each type is annotated with its !!5-2=3!! numbers whose sum is !!n+2=7!!. But the number of types increases rapidly with the number of sides, and it soons becomes infeasible to draw them by hand as I did above. I had wanted to write a computer program that would take a description of a type (the sequence) and render a drawing of one of the polygons of that type.

The tricky part is how to keep the edges from crossing, which is not allowed. I had ideas for how to do this, but it seemed troublesome, and also it seemed likely to produce ugly, lopsided examples, so I did not implement it. And eventually I forgot about the problem.

But Brent Yorgey did not forget, and he had a completely different idea. He wrote a program to convert a type description to a set of constraints on the !!x!! and !!y!! coordinates of the vertices, and fed the constraints to an SMT solver, which is a system for finding solutions to general sets of constraints. The outcome is as handsome as I could have hoped. Here is M. Yorgey's program's version of the hand-drawn diagram above:

M. Yorgey rendered beautiful pictures of all types of orthogonal polygons up to 12 sides. Check it out on his blog.

[1] “Bracelet” is combinatorist jargon for a sequence of things where the ends are joined together: you can start in the middle, run off one end and come back to the other end, and that is considered the same bracelet. So for example ABCD and BCDA and CDAB are all the same bracelet. And also, it doesn't matter which direction you go, so that DCBA, ADCB, and BADC are also the same bracelet again. Every bracelet made from from A, B, C, and D is equivalent to exactly one of ABCD, ACDB, or ADBC.

[ Addendum 20180202: M. Yorgey has more to say about it on his blog. ]

[Other articles in category /math] permanent link

Fri, 15 Dec 2017

Wasteful and frugal proofs in Ramsey theory

This question asks how to show that, among any 11 integers, one can find a subset of exactly six that add up to a multiple of 6. Let's call this “Ebrahimi’s theorem”.

This was the last thing I read before I put away my phone and closed my eyes for the night, and it was a race to see if I would find an answer before I fell asleep. Sleep won the race this time. But the answer is not too hard.

  1. First, observe that among any five numbers there are three that sum to a multiple of 3: Consider the remainders of the five numbers upon division by 3. There are three possible remainders. If all three remainders are represented, then the remainders are !!\{0,1,2\}!! and the sum of their representatives is a multiple of 3. Otherwise there is some remainder with three representatives, and the sum of these three is a multiple of 3.

  2. Now take the 11 given numbers. Find a group of three whose sum is a multiple of 3 and set them aside. From the remaining 8 numbers, do this a second time. From the remaining 5 numbers, do it a third time.

  3. We now have three groups of three numbers that each sum to a multiple of 3. Two of these sums must have the same parity. The six numbers in those two groups have an even sum that is a multiple of 3, and we win.

Here is a randomly-generated example:

$$3\quad 17\quad 35\quad 42\quad 44\quad 58\quad 60\quad 69\quad 92\quad 97\quad 97$$

Looking at the first 5 numbers !!3\ 17\ 35\ 42\ 44!! we see that on division by 3 these have remainders !!0\ 2\ 2\ 0\ 2!!. The remainder !!2!! is there three times, so we choose those three numbers !!\langle17\ 35\ 44\rangle!!, whose sum is a multiple of 3, and set them aside.

Now we take the leftover !!3!! and !!42!! and supplement them with three more unused numbers !!58\ 60\ 69!!. The remainders are !!0\ 0\ 1\ 0\ 0!! so we take !!\langle3\ 42\ 60\rangle!! and set them aside as a second group.

Then we take the five remaining unused numbers !!58\ 69\ 92\ 97\ 97!!. The remainders are !!1\ 0\ 2\ 1\ 1!!. The first three !!\langle 58\ 69\ 92\rangle!!have all different remainders, so let's use those as our third group.

The three groups are now !! \langle17\ 35\ 44\rangle, \langle3\ 42\ 60\rangle, \langle58\ 69\ 92\rangle!!. The first one has an even sum and the second has an odd sum. The third group has an odd sum, which matches the second group, so we choose the second and third groups, and that is our answer:

$$3\qquad 42\qquad 60\qquad 58 \qquad 69 \qquad 92$$

The sum of these is !!324 = 6\cdot 54!!.

This proves that 11 input numbers are sufficient to produce one output set of 6 whose sum is a multiple of 6. Let's write !!E(n, k)!! to mean that !!n!! inputs are enough to produce !!k!! outputs. That is, !!E(n, k)!! means “any set of !!n!! numbers contains !!k!! distinct 6-element subsets whose sum is a multiple of 6.” Ebrahimi’s theorem, which we have just proved, states that !!E(11, 1)!! is true, and obviously it also proves !!E(n, 1)!! for all larger !!n!!.

I would like to consider the following questions:

  1. Does this proof suffice to show that !!E(10, 1)!! is false?
  2. Does this proof suffice to show that !!E(11, 2)!! is false?

I am specifically not asking whether !!E(10, 1)!! or !!E(11, 2)!! are actually false. There are easy counterexamples that can be found without reference to the proof above. What I want to know is if the proof, as given, contains nontrivial information about these questions.

The reason I think this is interesting is that I think, upon more careful examination, that I will find that the proof above does prove at least one of these, perhaps with a very small bit of additional reasoning. But there are many similar proofs that do not work this way. Here is a famous example. Let !!W(n, k)!! be shorthand for the following claim:

Let the integers from 1 to !!n!! be partitioned into two sets. Then one of the two sets contains !!k!! distinct subsets of three elements of the form !!\{a, a+d, a+2d\}!! for integers !!a, d!!.


Van der Waerden's theorem: !!W(325, 1)!! is true.

!!W()!!, like !!E()!!, is monotonic: van der Waerden's theorem trivially implies !!W(n, 1)!! for all !!n!! larger than 325. Does it also imply that !!W(n, 1)!! is false for smaller !!n!!? No, not at all; this is actually untrue. Does it also imply that !!W(325, k)!! is false for !!k>1!!? No, this is false also.

Van der Waerden's theorem takes 325 inputs (the integers) and among them finds one output (the desired set of three). But this is extravagantly wasteful. A better argument shows that only 9 inputs were required for the same output, and once we know this it is trivial that 325 inputs will always produce at least 36 outputs, and probably a great many more.

Proofs of theorems in Ramsey theory are noted for being extravagant in exactly this way. But the proof of Ebrahimi's theorem is different. It is not only frugal, it is optimally so. It uses no more inputs than are absolutely necessary.

What is different about these cases? What is the source the frugality of the proof of Ebrahimi’s theorem? Is there a way that we can see from examination of the proof that it will be optimally frugal?

Ebrahimi’s theorem shows !!E(11, 1)!!. Suppose instead we want to show !!E(n, 2)!! for some !!n!!. From Ebrahimi’s theorem itself we immediately get !!E(22, 2)!! and indeed !!E(17, 2)!!. Is this the best we can do? (That is, is !!E(16, 2)!! false?) I bet it isn't. If it isn't, what went wrong? Or rather, what went right in the !!k=1!! case that stopped working when !!k>1!!?

I don't know.

[Other articles in category /math] permanent link

Wed, 22 Nov 2017

Mathematical pettifoggery and pathological examples

Last week I said:

Mathematicians do tend to be the kind of people who quibble and pettifog over the tiniest details. This is because in mathematics, quibbling and pettifogging does work.

This example is technical, but I think I can explain it in a way that will make sense even for people who have no idea what the question is about. Don't worry if you don't understand the next paragraph.

In this math SE question: a user asks for an example of a connected topological space !!\langle X, \tau\rangle!! where there is a strictly finer topology !!\tau'!! for which !!\langle X, \tau'\rangle!! is disconnected. This is a very easy problem if you go about it the right way, and the right way follows a very typical pattern which is useful in many situations.

The pattern is “TURN IT UP TO 11!​!” In this case:

  1. Being disconnected means you can find some things with a certain property.
  2. If !!\tau'!! is finer than !!\tau!!, that means it has all the same things, plus even more things of the same type.
  3. If you could find those things in !!\tau!!, you can still find them in !!\tau'!! because !!\tau'!! has everything that !!\tau!! does.
  4. So although perhaps making !!\tau!! finer could turn a connected space into a disconnected one, by adding the things you needed, it definitely can't turn a disconnected space into a connected one, because the things will still be there.
  5. So a way to look for a connected space that becomes disconnected when !!\tau!! becomes finer is:

    Start with some connected space. Then make !!\tau!! fine as you possibly can and see if that is enough.

  6. If that works, you win. If not, you can look at the reason it didn't work, and maybe either fix up the space you started with, or else use that as the starting point in a proof that the thing you're looking for doesn't exist.

I emphasized the important point here. It is: Moving toward finer !!\tau!! can't hurt the situation and might help, so the first thing to try is to turn the fineness knob all the way up and see if that is enough to get what you want. Many situations in mathematics call for subtlety and delicate reasoning, but this is not one of those.

The technique here works perfectly. There is a topology !!\tau_d!! of maximum possible fineness, called the “discrete” topology, so that is the thing to try first. And indeed it answers the question as well as it can be answered: If !!\langle X, \tau\rangle!! is a connected space, and if there is any refinement !!\tau'!! for which !!\langle X, \tau'\rangle!! is disconnected, then !!\langle X, \tau_d\rangle!! will be disconnected. It doesn't even matter what connected space you start with, because !!\tau_d!! is always a refinement of !!\tau!!, and because !!\langle X, \tau_d\rangle!! is always disconnected, except in trivial cases. (When !!X!! has fewer than two points.)

Right after you learn the definition of what a topology is, you are presented with a bunch of examples. Some are typical examples, which showcase what the idea is really about: the “open sets” of the real line topologize the line, so that topology can be used as a tool for studying real analysis. But some are atypical examples, which showcase the extreme limits of the concept that are as different as possible from the typical examples. The discrete space is one of these. What's it for? It doesn't help with understanding the real numbers, that's for sure. It's a tool, it's the knob on the topology machine that turns the fineness all the way up.[1] If you want to prove that the machine does something or other for the real numbers, one way is to show that it always does that thing. And sometimes part of showing that it always does that thing is to show that it does that even if you turn the knob all the way to the right.

So often the first thing a mathematician will try is:

What happens if I turn the knob all the way to the right? If that doesn't blow up the machine, nothing will!

And that's why, when you ask a mathematician a question, often the first thing they will say is “ťhat fails when !!x=0!!” or “that fails when all the numbers are equal” or “ťhat fails when one number is very much bigger than the other” or “that fails when the space is discrete” or “that fails when the space has fewer than two points.” [2]

After the last article, Kyle Littler reminded me that I should not forget the important word “pathological”. One of the important parts of mathematical science is figuring out what the knobs are, how far they can go, what happens if you turn them all the way up, and what are the limits on how they can be set if we want the machine to behave more or less like the thing we are trying to study.

We have this certain knob for how many dents and bumps and spikes we can put on a sphere and have it still be a sphere, as long as we do not actually puncture or tear the surface. And we expected that no matter how far we turned this knob, the sphere would still divide space into two parts, a bounded inside and an unbounded outside, and that these regions should behave basically the same as they do when the sphere is smooth.[3]

But no, we are wrong, the knob goes farther than we thought. If we turn it to the “Alexander horned sphere” setting, smoke starts to come out of the machine and the red lights begin to blink.[4] Useful! Now if someone has some theory about how the machine will behave nicely if this and that knob are set properly, we might be able to add the useful observation “actually you also have to be careful not to turn that “dents bumps and spikes” knob too far.”

The word for these bizarre settings where some of the knobs are in the extreme positions is “pathological”. The Alexander sphere is a pathological embedding of !!S^2!! into !!\Bbb R^3!!.

[1] The leftmost setting on that knob, with the fineness turned all the way down, is called the “indiscrete topology” or the “trivial topology”.

[2] If you claim that any connected space can be disconnected by turning the “fineness” knob all the way to the right, a mathematican will immediately turn the “number of points” knob all the way to the left, and say “see, that only works for spaces with at least two points”. In a space with fewer than two points, even the discrete topology is connected.

[3]For example, if you tie your dog to a post outside the sphere, and let it wander around, its leash cannot get so tangled up with the sphere that you need to walk the dog backwards to untangle it. You can just slip the leash off the sphere.

[4] The dog can get its leash so tangled around the Alexander sphere that the only way to fix it is to untie the dog and start over. But if the “number of dimensions” knob is set to 2 instead of to 3, you can turn the “dents bumps and spikes” knob as far as you want and the leash can always be untangled without untying or moving the dog. Isn't that interesting? That is called the Jordan curve theorem.

[Other articles in category /math] permanent link

An instructive example of expected value

I think this example is very illuminating of something, although I'm not sure yet what.

Suppose you are making a short journey somewhere. You leave two minutes later than planned. How does this affect your expected arrival time? All other things being equal, you should expect to arrive two minutes later than planned. If you're walking or driving, it will probably be pretty close to two minutes no matter what happens.

Now suppose the major part of your journey involves a train that runs every hour, and you don't know just what the schedule is. Now how does your two minutes late departure affect your expected arrival time?

The expected arrival time is still two minutes later than planned. But it is not uniformly distributed. With probability !!\frac{58}{60}!!, you catch the train you planned to take. You are unaffected by your late departure, and arrive at the same time. But with probability !!\frac{2}{60}!! you miss that train and have to take the next one, arriving an hour later than you planned. The expected amount of lateness is

$$0 \text{ minutes}·\frac{58}{60} + 60 \text{ minutes}·\frac{2}{60} = 2 \text{ minutes}$$

the same as before.

[ Addendum: Richard Soderberg points out that one thing illuminated by this example is that the mathematics fails to capture the emotional pain of missing the train. Going in a slightly different direction, I would add that the expected value reduces a complex situation to a single number, and so must necessarily throw out a lot of important information. I discussed this here a while back in connection with lottery tickets.

But also I think this failure of the expected value is also a benefit: it does capture something interesting about the situation that might not have been apparent before: Considering the two minutes as a time investment, there is a sense in which the cost is knowable; it costs exactly two minutes. Yes, there is a chance that you will be hit by a truck that you would not have encountered had you left on time. But this is exactly offset by the hypothetical truck that passed by harmlessly two minutes before you arrived on the scene but which would have hit you had you left on time. ]

[Other articles in category /math] permanent link

Mon, 20 Nov 2017

Mathematical jargon for quibbling

Mathematicians tend not to be the kind of people who shout and pound their fists on the table. This is because in mathematics, shouting and pounding your fist does not work. If you do this, other mathematicians will just laugh at you. Contrast this with law or politics, which do attract the kind of people who shout and pound their fists on the table.

However, mathematicians do tend to be the kind of people who quibble and pettifog over the tiniest details. This is because in mathematics, quibbling and pettifogging does work.

Mathematics has a whole subjargon for quibbling and pettifogging, and also for excluding certain kinds of quibbles. The word “nontrivial” is preeminent here. To a first approximation, it means “shut up and stop quibbling”. For example, you will often hear mathematicians having conversations like this one:

A: Mihăilescu proved that the only solution of Catalan's equation !!a^x - b^y = 1!! is !!3^2 - 2^3!!.

B: What about when !!a!! and !!b!! are consecutive and !!x=y=1!!?

A: The only nontrivial solution.

B: Okay.

Notice that A does not explain what “nontrivial” is supposed to mean here, and B does not ask. And if you were to ask either of them, they might not be able to tell you right away what they meant. For example, if you were to inquire specifically about !!2^1 - 1^y!!, they would both agree that that is also excluded, whether or not that solution had occurred to either of them before. In this example, “nontrivial” really does mean “stop quibbling”. Or perhaps more precisely “there is actually something here of interest, and if you stop quibbling you will learn what it is”.

In some contexts, “nontrivial” does have a precise and technical meaning, and needs to be supplemented with other terms to cover other types of quibbles. For example, when talking about subgroups, “nontrivial” is supplemented with “proper”:

If a nontrivial group has no proper nontrivial subgroup, then it is a cyclic group of prime order.

Here the “proper nontrivial” part is not merely to head off quibbling; it's the crux of the theorem. But the first “nontrivial” is there to shut off a certain type of quibble arising from the fact that 1 is not considered a prime number. By this I mean if you omit “proper”, or the second “nontrivial”, the statement is still true, but inane:

If a nontrivial group has no subgroup, then it is a cyclic group of prime order.

(It is true, but vacuously so.) In contrast, if you omit the first “nontrivial”, the theorem is substantively unchanged:

If a group has no proper nontrivial subgroup, then it is a cyclic group of prime order.

This is still true, except in the case of the trivial group that is no longer excluded from the premise. But if 1 were considered prime, it would be true either way.

Looking at this issue more thoroughly would be interesting and might lead to some interesting conclusions about mathematical methodology.

  • Can these terms be taxonomized?
  • How do mathematical pejoratives relate? (“Abnormal, irregular, improper, degenerate, inadmissible, and otherwise undesirable”) Kelley says we use these terms to refer to “a problem we cannot handle”; that seems to be a different aspect of the whole story.
  • Where do they fit in Lakatos’ Proofs and Refutations theory? Sometimes inserting “improper” just heads off a quibble. In other cases, it points the way toward an expansion of understanding, as with the “improper” polyhedra that violate Euler's theorem and motivate the introduction of the Euler characteristic.
  • Compare with the large and finely-wrought jargon that distinguishes between proofs that are “elementary”, “easy”, “trivial”, “straightforward”, or “obvious”.
  • Is there a category-theoretic formulation of what it means when we say “without loss of generality, take !!x\lt y!!”?

[ Addendum: Kyle Littler reminds me that I should not forget “pathological”. ]

[Other articles in category /math] permanent link

Wed, 15 Nov 2017

Down with the negation sign!

[ Credit where it is due: This was entirely Darius Bacon's idea. ]

In connection with “Recognizing when two arithmetic expressions are essentially the same”, I had several conversations with people about ways to normalize numeric expressions. In that article I observed that while everyone knows the usual associative law for addition $$ (a + b) + c = a + (b + c)$$ nobody ever seems to mention the corresponding law for subtraction: $$ (a+b)-c = a + (b-c).$$

And while everyone “knows” that subtraction is not associative because $$(a - b) - c ≠ a - (b-c)$$ nobody ever seems to observe that there is an associative law for subtraction: $$\begin{align} (a - b) + c & = a - (b - c) \\ (a -b) -c & = a-(b+c).\end{align}$$

This asymmetry is kind of a nuisance, and suggests that a more symmetric notation might be better. Darius Bacon suggested a simple change that I think is an improvement:

Write the negation of !!a!! as $$a\star$$ so that one has, for all !!a!!, $$a+a\star = a\star + a = 0.$$

The !!\star!! operation obeys the following elegant and simple laws: $$\begin{align} a\star\star & = a \\ (a+b)\star & = a\star + b\star \end{align} $$

Once we adopt !!\star!!, we get a huge payoff: We can eliminate subtraction:

Instead of !!a-b!! we now write !!a+b\star!!.

The negation of !!a+b\star!! is $$(a+b\star)\star = a\star + b{\star\star} = a\star +b.$$

We no longer have the annoying notational asymmetry between !!a-b!! and !!-b + a!! where the plus sign appears from nowhere. Instead, one is !!a+b\star!! and the other is !!b\star+a!!, which is obviously just the usual commutativity of addition.

The !!\star!! is of course nothing but a synonym for multiplication by !!-1!!. But it is a much less clumsy synonym. !!a\star!! means !!a\cdot(-1)!!, but with less inkjunk.

In conventional notation the parentheses in !!a(-b)!! are essential and if you lose them the whole thing is ruined. But because !!\star!! is just a special case of multiplication, it associates with multiplication and division, so we don't have to worry about parentheses in !!(a\star)b = a(b\star) = (ab)\star!!. They are all equal to just !!ab\star!!. and you can drop the parentheses or include them or write the terms in any order, just as you like, just as you would with !!abc!!.

The surprising associativity of subtraction is no longer surprising, because $$(a + b) - c = a + (b - c)$$ is now written as $$(a + b) + c\star = a + (b + c\star)$$ so it's just the usual associative law for addition; it is not even disguised. The same happens for the reverse associative laws for subtraction that nobody mentions; they become variations on $$ \begin{align} (a + b\star) + c\star & = a + (b\star + c\star) \\ & = a + (b+c)\star \end{align} $$ and such like.

The !!\star!! is faster to read and faster to say. Instead of “minus one” or “negative one” or “times negative one”, you just say “star”.

The !!\star!! is just a number, and it behaves like a number. Its role in an expression is the same as any other number's. It is just a special, one-off notation for a single, particularly important number.

Open questions:

  1. Do we now need to replace the !!\pm!! sign? If so, what should we replace it with?

  2. Maybe the idea is sound, but the !!\star!! itself is a bad choice. It is slow to write. It will inevitably be confused with the * that almost every programming language uses to denote multiplication.

  3. The real problem here is that the !!-!! symbol is overloaded. Instead of changing the negation symbol to !!\star!! and eliminating the subtraction symbol, what if we just eliminated subtraction? None of the new notation would be incompatible with the old notation: !!-(a+-b) = b+-a!! still means exactly what it used to. But you are no longer allowed to abbreviate it to !!-(a-b) = b-a!!.

    This would fix the problem of the !!\star!! taking too long to write: we would just use !!-!! in its place. It would also fix the concern of point 2: !!a\pm b!! now means !!a+b!! or !!a+-b!! which is not hard to remember or to understand. Another happy result: notations like !!-1!! and !!-2!! do not change at all.

Curious footnote: While I was writing up the draft of this article, it had a reminder in it: “How did you and Darius come up with this?” I went back to our email to look, and I discovered the answer was:

  1. Darius suggested the idea to me.
  2. I said, “Hey, that's a great idea!”

I wish I could take more credit, but there it is. Hmm, maybe I will take credit for inspiring Darius! That should be worth at least fifty percent, perhaps more.

[ This article had some perinatal problems. It escaped early from the laboratory, in a not-quite-finished state, so I apologize if you are seeing it twice. ]

[Other articles in category /math] permanent link

Tue, 07 Nov 2017

A modern translation of the 1+1=2 lemma

A while back I blogged an explanation of the “!!1+1=2!!” lemma from Whitehead and Russell's Principia Mathematica:

W. Ethan Duckworth of the Department of Mathematics and Statistics at Loyola University translated this into modern notation and has kindly given me permission to publish it here:

I think it is interesting and instructive to compare the two versions. One thing to notice is that there is no perfect translation. As when translating between two natural languages (German and English, say), the meaning cannot be preserved exactly. Whitehead and Russell's language is different from the modern language not only because the notation is different but because the underlying concepts are different. To really get what Principia Mathematica is saying you have to immerse yourself in the Principia Mathematica model of the world.

The best example of this here is the symbol “1”. In the modern translation, this means the number 1. But at this point in Principia Mathematica, the number 1 has not yet been defined, and to use it here would be circular, because proposition ∗54.43 is an important step on the way to defining it. In Principia Mathematica, the symbol “1” represents the class of all sets that contain exactly one element.[1] Following the definition of ∗52.01, in modern notation we would write something like:

$$1 \equiv_{\text{def}} \{x \mid \exists y . x = \{ y \} \}$$

But in many modern universes, that of ZF set theory in particular, there is no such object.[2] The situation in ZF is even worse: the purported definition is meaningless, because the comprehension is unrestricted.

The Principia Mathematica notation for !!|A|!!, the cardinality of set !!A!!, is !!Nc\,‘A!!, but again this is only an approximate translation. The meaning of !!Nc\,‘A!! is something close to

the unique class !!C!! such that !!x\in C!! if and only if there exists a one-to-one relation between !!A!! and !!x!!.

(So for example one might assert that !!Nc\,‘\Lambda = 0!!, and in fact this is precisely what proposition ∗101.1 does assert.) Even this doesn't quite capture the Principia Mathematica meaning, since the modern conception of a relation is that it is a special kind of set, but in Principia Mathematica relations and sets are different sorts of things. (We would also use a one-to-one function, but here there is no additional mismatch between the modern concept and the Principia Mathematica one.)

It is important, when reading old mathematics, to try to understand in modern terms what is being talked about. But it is also dangerous to forget that the ideas themselves are different, not just the language.[3] I extract a lot of value from switching back and forth between different historical views, and comparing them. Some of this value is purely historiological. But some is directly mathematical: looking at the same concepts from a different viewpoint sometimes illuminates aspects I didn't fully appreciate. And the different viewpoint I acquire is one that most other people won't have.

One of my current low-priority projects is reading W. Burnside's important 1897 book Theory of Groups of Finite Order. The value of this, for me, is not so much the group-theoretic content, but in seeing how ideas about groups have evolved. I hope to write more about this topic at some point.

[1] Actually the situation in Principia Mathematica is more complicated. There is a different class 1 defined at each type. But the point still stands.

[2] In ZF, if !!1!! were to exist as defined above, the set !!\{1\}!! would exist also, and we would have !!\{1\} \in 1!! which would contradict the axiom of foundation.

[3] This was a recurring topic of study for Imre Lakatos, most famously in his little book Proofs and Refutations. Also important is his article “Cauchy and the continuum: the significance of nonstandard analysis for the history and philosophy of mathematics.” Math. Intelligencer 1 (1978), #3, p.151–161, which I discussed here earlier, and which you can read in its entireity by paying the excellent people at Elsevier the nominal and reasonable—nay, trivial—sum of only US$39.95.

[Other articles in category /math] permanent link

Tue, 31 Oct 2017

The Blind Spot and the cut rule

[ The Atom and RSS feeds have done an unusually poor job of preserving the mathematical symbols in this article. It will be much more legible if you read it on my blog. ]

Lately I've been enjoying The Blind Spot by Jean-Yves Girard, a very famous logician. (It is translated from French; the original title is Le Point Aveugle.) This is an unusual book. It is solidly full of deep thought and technical detail about logic, but it is also opinionated, idiosyncratic and polemical. Chapter 2 (“Incompleteness”) begins:

It is out of question to enter into the technical arcana of Gödel’s theorem, this for several reasons:

(i) This result, indeed very easy, can be perceived, like the late paintings of Claude Monet, but from a certain distance. A close look only reveals fastidious details that one perhaps does not want to know.

(ii) There is no need either, since this theorem is a scientific cul-de-sac : in fact it exposes a way without exit. Since it is without exit, nothing to seek there, and it is of no use to be expert in Gödel’s theorem.

(The Blind Spot, p. 29)

He continues a little later:

Rather than insisting on those tedious details which «hide the forest», we shall spend time on objections, from the most ridiculous to the less stupid (none of them being eventually respectable).

As you can see, it is not written in the usual dry mathematical-text style, presenting the material as a perfect and aseptic distillation of absolute truth. Instead, one sees the history of logic, the rise and fall of different theories over time, the interaction and relation of many mathematical and philosophical ideas, and Girard's reflections about it all. It is a transcription of a lecture series, and reads like one, including all of the speaker's incidental remarks and offhand musings, but written down so that each can be weighed and pondered at length. Instead of wondering in the moment what he meant by some intriguing remark, then having to abandon the thought to keep up with the lecture, I can pause and ponder the significance. Girard is really, really smart, and knows way more about logic than I ever will, and his offhand remarks reward this pondering. The book is profound in a way that mathematics books often aren't. I wanted to provide an illustrative quotation, but to briefly excerpt a profound thought is to destroy its profundity, so I will have to refrain.[1])

The book really gets going with its discussion of Gentzen's sequent calculus in chapter 3.

Between around 1890 (when Peano and Frege began to liberate logic from its medieval encrustations) and 1935 when the sequent calculus was invented, logical proofs were mainly in the “Hilbert style”. Typically there were some axioms, and some rules of deduction by which the axioms could be transformed into other formulas. A typical example consists of the axioms $$A\to(B\to A)\\ (A \to (B \to C)) \to ((A \to B) \to (A \to C)) $$ (where !!A, B, C!! are understood to be placeholders that can be replaced by any well-formed formulas) and the deduction rule modus ponens: having proved !!A\to B!! and !!A!!, we can deduce !!B!!.

In contrast, sequent calculus has few axioms and many deduction rules. It deals with sequents which are claims of implication. For example: $$p, q \vdash r, s$$ means that if we can prove all of the formulas on the left of the ⊢ sign, then we can conclude some of the formulas on the right. (Perhaps only one, but at least one.)

A typical deductive rule in sequent calculus is:

$$ \begin{array}{c} Γ ⊢ A, Δ \qquad Γ ⊢ B, Δ \\ \hline Γ ⊢ A ∧ B, Δ \end{array} $$

Here !!Γ!! and !!Δ!! represent any lists of formulas, possibly empty. The premises of the rule are:

  1. !!Γ ⊢ A, Δ!!: If we can prove all of the formulas in !!Γ!!, then we can conclude either the formula !!A!! or some of the formulas in !!Δ!!.

  2. !!Γ ⊢ B, Δ!!: If we can prove all of the formulas in !!Γ!!, then we can conclude either the formula !!B!! or some of the formulas in !!Δ!!.

From these premises, the rule allows us to deduce:

!!Γ ⊢ A ∧ B, Δ!!: If we can prove all of the formulas in !!Γ!!, then we can conclude either the formula !!A \land B!! or some of the formulas in !!Δ!!.

The only axioms of sequent calculus are utterly trivial:

$$ \begin{array}{c} \phantom{A} \\ \hline A ⊢ A \end{array} $$

There are no premises; we get this deduction for free: If can prove !!A!!, we can prove !!A!!. (!!A!! here is a metavariable that can be replaced with any well-formed formula.)

One important point that Girard brings up, which I had never realized despite long familiarity with sequent calculus, is the symmetry between the left and right sides of the turnstile ⊢. As I mentioned, the interpretation of !!Γ ⊢ Δ!! I had been taught was that it means that if every formula in !!Γ!! is provable, then some formula in !!Δ!! is provable. But instead let's focus on just one of the formulas !!A!! on the right-hand side, hiding in the list !!Δ!!. The sequent !!Γ ⊢ Δ, A!! can be understood to mean that to prove !!A!!, it suffices to prove all of the formulas in !!Γ!!, and to disprove all the formulas in !!Δ!!. And now let's focus on just one of the formulas on the left side: !!Γ, A ⊢ Δ!! says that to disprove !!A!!, it suffices to prove all the formulas in !!Γ!! and disprove all the formulas in !!Δ!!.

The all-some correspondence, which had previously caused me to wonder why it was that way and not something else, perhaps the other way around, has turned into a simple relationship about logical negation: the formulas on the left are positive, and the ones on the right are negative.[2]) With this insight, the sequent calculus negation laws become not merely simple but trivial:

$$ \begin{array}{cc} \begin{array}{c} Γ, A ⊢ Δ \\ \hline Γ ⊢ \lnot A, Δ \end{array} & \qquad \begin{array}{c} Γ ⊢ A, Δ \\ \hline Γ, \lnot A ⊢ Δ \end{array} \end{array} $$

For example, in the right-hand deduction: what is sufficient to prove !!A!! is also sufficient to disprove !!¬A!!.

(Compare also the rule I showed above for ∧: It now says that if proving everything in !!Γ!! and disproving everything in !!Δ!! is sufficient for proving !!A!!, and likewise sufficient for proving !!B!!, then it is also sufficient for proving !!A\land B!!.)

But none of that was what I planned to discuss; this article is (intended to be) about sequent calculus's “cut rule”.

I never really appreciated the cut rule before. Most of the deductive rules in the sequent calculus are intuitively plausible and so simple and obvious that it is easy to imagine coming up with them oneself.

But the cut rule is more complicated than the rules I have already shown. I don't think I would have thought of it easily:

$$ \begin{array}{c} Γ ⊢ A, Δ \qquad Λ, A ⊢ Π \\ \hline Γ, Λ ⊢ Δ, Π \end{array} $$

(Here !!A!! is a formula and !!Γ, Δ, Λ, Π!! are lists of formulas, possibly empty lists.)

Girard points out that the cut rule is a generalization of modus ponens: taking !!Γ, Δ, Λ!! to be empty and !!Π = \{B\}!! we obtain:

$$ \begin{array}{c} ⊢ A \qquad A ⊢ B \\ \hline ⊢ B \end{array} $$

The cut rule is also a generalization of the transitivity of implication:

$$ \begin{array}{c} X ⊢ A \qquad A ⊢ Y \\ \hline X ⊢ Y \end{array} $$

Here we took !!Γ = \{X\}, Π = \{Y\}!!, and !!Δ!! and !!Λ!! empty.

This all has given me a much better idea of where the cut rule came from and why we have it.

In sequent calculus, the deduction rules all come in pairs. There is a rule about introducing ∧, which I showed before. It allows us to construct a sequent involving a formula with an ∧, where perhaps we had no ∧ before. (In fact, it is the only way to do this.) There is a corresponding rule (actually two rules) for getting rid of ∧ when we have it and we don't want it:

$$ \begin{array}{cc} \begin{array}{c} Γ ⊢ A\land B, Δ \\ \hline Γ ⊢ A, Δ \end{array} & \qquad \begin{array}{c} Γ ⊢ A\land B, Δ \\ \hline Γ ⊢ B, Δ \end{array} \end{array} $$

Similarly there is a rule (actually two rules) about introducing !!\lor!! and a corresponding rule about eliminating it.

The cut rule seems to lie outside this classification. It is not paired.

But Girard showed me that it is part of a pair. The axiom

$$ \begin{array}{c} \phantom{A} \\ \hline A ⊢ A \end{array} $$

can be seen as an introduction rule for a pair of !!A!!s, one on each side of the turnstile. The cut rule is the corresponding rule for eliminating !!A!! from both sides.

Sequent calculus proofs are much easier to construct than Hilbert-style proofs. Suppose one wants to prove !!B!!. In a Hilbert system the only deduction rule is modus ponens, which requires that we first prove !!A\to B!! and !!A!! for some !!A!!. But what !!A!! should we choose? It could be anything, and we have no idea where to start or how big it could be. (If you enjoy suffering, try to prove the simple theorem !!A\to A!! in the Hilbert system I described at the beginning of the article. (Solution)

In sequent calculus, there is only one way to prove each kind of thing, and the premises in each rule are simply related to the consequent we want. Constructing the proof is mostly a matter of pushing the symbols around by following the rules to their conclusions. (Or, if this is impossible, one can conclude that there is no proof, and why.[3]) Construction of proofs can now be done entirely mechanically!

Except! The cut rule does require one to guess a formula: If one wants to prove !!Γ, Λ ⊢ Δ, Π!!, one must guess what !!A!! should appear in the premises !!Γ, A ⊢ Δ!! and !!Λ ⊢ A, Π!!. And there is no constraint at all on !!A!!; it could be anything, and we have no idea where to start or how big it could be.

The good news is that Gentzen, the inventor of sequent calculus, showed that one can dispense with the cut rule: it is unnecessary:

In Hilbert-style systems, based on common sense, the only rule is (more or less) the Modus Ponens : one reasons by linking together lemmas and consequences. We just say that one can get rid of that : it is like crossing the English Channel with fists and feet bound !

(The Blind Spot, p. 61)

Gentzen's demonstration of this shows how one can take any proof that involves the cut rule, and algorithmically eliminate the cut rule from it to obtain a proof of the same result that does not use cut. Gentzen called this the “Hauptsatz” (“principal theorem”) and rightly so, because it reduces construction of logical proofs to an algorithm and is therefore the ultimate basis for algorithmic proof theory.

The bad news is that the cut-elimination process can super-exponentially increase the size of the proof, so it does not lead to a practical algorithm for deciding provability. Girard analyzed why, and what he discovered amazed me. The only problem is in the contraction rules, which had seemed so trivial and innocuous—uninteresting, even—that I had never given them any thought:

$$ \begin{array}{cc} \begin{array}{c} Γ, A, A ⊢ Δ \\ \hline Γ, A ⊢ Δ \end{array} & \qquad \begin{array}{c} Γ ⊢ A, A, Δ \\ \hline Γ ⊢ A, Δ \end{array} \end{array} $$

And suddenly Girard's invention of linear logic made sense to me. In linear logic, contraction is forbidden; one must use each formula in one and only one deduction.

Previously it had seemed to me that this was a pointless restriction. Now I realized that it was no more of a useless hair shirt than the intuitionistic rejection of the law of the proof by contradiction: not a stubborn refusal to use an obvious tool of reasoning, but a restriction of proofs to produce better reasoning. With the rejection of contraction, cut-elimination no longer explodes proof size, and automated theorem proving becomes practical:

This is why its study is, implicitly, at the very heart of these lectures.

(The Blind Spot, p. 63)

The book is going to get into linear logic later in the next chapter. I have read descriptions of linear logic before, but never understood what it was up to. (It has two logical and operators, and two logical or operators; why?) But I am sure Girard will explain it marvelously.

  1. In place of a profound excerpt, I will present the following, which isn't especially profound but struck a chord for me: “By the way, nothing is more arbitrary than a modal logic « I am done with this logic, may I have another one ? » seems to be the motto of these guys.” (p. 24)
  2. Compare my alternative representation of arithmetic expressions that collapses addition and subtraction, and multiplication and division.
  3. A typically Girardian remark is that analytic tableaux are “the poor man's sequents”.

A brief but interesting discussion of The Blind Spot on Hacker News.

Review by Felipe Zaldivar.

[Other articles in category /math/logic] permanent link

Sun, 15 Oct 2017

Counting increasing sequences with Burnside's lemma

[ I started this article in March and then forgot about it. Ooops! ]

Back in February I posted an article about how there are exactly 715 nondecreasing sequences of 4 digits. I said that !!S(10, 4)!! was the set of such sequences and !!C(10, 4)!! was the number of such sequences, and in general $$C(d,n) = \binom{n+d-1}{d-1} = \binom{n+d-1}{n}$$ so in particular $$C(10,4) = \binom{13}{4} = 715.$$

I described more than one method of seeing this, but I didn't mention the method I had found first, which was to use the Cauchy-Frobenius-Redfeld-Pólya-Burnside counting lemma. I explained the lemma in detail some time ago, with beautiful illustrated examples, so I won't repeat the explanation here. The Burnside lemma is a kind of big hammer to use here, but I like big hammers. And the results of this application of the big hammer are pretty good, and justify it in the end.

To count the number of distinct sequences of 4 digits, where some sequences are considered “the same” we first identify a symmetry group whose orbits are the equivalence classes of sequences. Here the symmetry group is !!S_4!!, the group that permutes the elements of the sequence, because two sequences are considered “the same” if they have exactly the same digits but possibly in a different order, and the elements of !!S_4!! acting on the sequences are exactly what you want to permute the elements into some different order.

Then you tabulate how many of the 10,000 original sequences are left fixed by each element !!p!! of !!S_4!!, which is exactly the number of cycles of !!p!!. (I have also discussed cycle classes of permutations before.) If !!p!! contains !!n!! cycles, then !!p!! leaves exactly !!10^n!! of the !!10^4!! sequences fixed.

Cycle class Number
of cycles
How many
left fixed
41 10,000
3 6 1,000

2 3 + 8 = 11 100
1 6 10
  24 17,160

(Skip this paragraph if you already understand the table. The four rows above are an abbreviation of the full table, which has 24 rows, one for each of the 24 permutations of order 4. The “How many permutations?” column says how many times each row should be repeated. So for example the second row abbreviates 6 rows, one for each of the 6 permutations with three cycles, which each leave 1,000 sequences fixed, for a total of 6,000 in the second row, and the total for all 24 rows is 17,160. There are two different types of permutations that have two cycles, with 3 and 8 permutations respectively, and I have collapsed these into a single row.)

Then the magic happens: We average the number left fixed by each permutation and get !!\frac{17160}{24} = 715!! which we already know is the right answer.

Now suppose we knew how many permutations there were with each number of cycles. Let's write !!\def\st#1#2{\left[{#1\atop #2}\right]}\st nk!! for the number of permutations of !!n!! things that have exactly !!k!! cycles. For example, from the table above we see that $$\st 4 4 = 1,\quad \st 4 3 = 6,\quad \st 4 2 = 11,\quad \st 4 1 = 6.$$

Then applying Burnside's lemma we can conclude that $$C(d, n) = \frac1{n!}\sum_i \st ni d^i .\tag{$\spadesuit$}$$ So for example the table above computes !!C(10,4) = \frac1{24}\sum_i \st 4i 10^i = 715!!.

At some point in looking into this I noticed that $$\def\rp#1#2{#1^{\overline{#2}}}% \def\fp#1#2{#1^{\underline{#2}}}% C(d,n) = \frac1{n!}\rp dn$$ where !!\rp dn!! is the so-called “rising power” of !!d!!: $$\rp dn = d\cdot(d+1)(d+2)\cdots(d+n-1).$$ I don't think I had a proof of this; I just noticed that !!C(d, 1) = d!! and !!C(d, 2) = \frac12(d^2+d)!! (both obvious), and the Burnside's lemma analysis of the !!n=4!! case had just given me !!C(d, 4) = \frac1{24}(d^4 +6d^3 + 11d^2 + 6d)!!. Even if one doesn't immediately recognize this latter polynomial it looks like it ought to factor and then on factoring it one gets !!d(d+1)(d+2)(d+3)!!. So it's easy to conjecture !!C(d, n) = \frac1{n!}\rp dn!! and indeed, this is easy to prove from !!(\spadesuit)!!: The !!\st n k!! obey the recurrence $$\st{n+1}k = n \st nk + \st n{k-1}\tag{$\color{green}{\star}$}$$ (by an easy combinatorial argument1) and it's also easy to show that the coefficients of !!\rp nk!! obey the same recurrence.2

In general !!\rp nk = \fp{(n+k-1)}k!! so we have !!C(d, n) = \rp dn = \fp{(n+d-1)}n = \binom{n+d-1}d = \binom{n+d-1}{n-1}!! which ties the knot with the formula from the previous article. In particular, !!C(10,4) = \binom{13}9!!.

I have a bunch more to say about this but this article has already been in the oven long enough, so I'll cut the scroll here.

[1] The combinatorial argument that justifies !!(\color{green}{\star})!! is as follows: The Stirling number !!\st nk!! counts the number of permutations of order !!n!! with exactly !!k!! cycles. To get a permutation of order !!n+1!! with exactly !!k!! cycles, we can take one of the !!\st nk!! permutations of order !!n!! with !!k!! cycles and insert the new element into one of the existing cycles after any of the !!n!! elements. Or we can take one of the !!\st n{k-1}!! permutations with only !!k-1!! cycles and add the new element in its own cycle.)

[2] We want to show that the coefficients of !!\rp nk!! obey the same recurrence as !!(\color{green}{\star})!!. Let's say that the coefficient of the !!n^i!! term in !!\rp nk!! is !!c_i!!. We have $$\rp n{k+1} = \rp nk\cdot (n+k) = \rp nk \cdot n + \rp nk \cdot k $$ so the coefficient of the the !!n^i!! term on the left is !!c_{i-1} + kc_i!!.

[Other articles in category /math] permanent link

Sun, 27 Aug 2017

Miscellanea about 24 puzzles

This is a collection of leftover miscellanea about twenty-four puzzles. In case you forgot what that is:

The puzzle «4 6 7 9 ⇒ 24» means that one should take the numbers 4, 6, 7, and 9, and combine them with the usual arithmetic operations of addition, subtraction, multiplication, and division, to make the number 24. In this case the unique solution is $$6\times\frac{7 + 9}{4}.$$ When the target number is 24, as it often is, we omit it and just write «4 6 7 9».

Prior articles on this topic:

How many puzzles have solutions?

For each value of !!T!!, there are 715 puzzles «a b c d ⇒ T». (I discussed this digression in two more earlier articles: [1] [2].) When the target !!T = 24!!, 466 of the 715 puzzles have solutions. Is this typical? Many solutions of «a b c d» puzzles end with a multiplication of 6 and 4, or of 8 and 3, or sometimes of 12 and 2—so many that one quickly learns to look for these types of solutions right away. When !!T=23!!, there won't be any solutions of this type, and we might expect that relatively few puzzles with prime targets have solutions.

This turns out to be the case:

ALT attribute suggestions welcome!

The x-axis is the target number !!T!!, with 0 at the left, 300 at right, and vertical guide lines every 25. The y axis is the number of solvable puzzles out of the maximum possible of 715, with 0 at the bottom, 715 at the top, and horizontal guide lines every 100.

Dots representing prime number targets are colored black. Dots for numbers with two prime factors (4, 6, 9, 10, 14, 15, 21, 22, etc.) are red; dots with three, four, five, six, and seven prime factors are orange, yellow, green, blue, and purple respectively.

Two countervailing trends are obvious: Puzzles with smaller targets have more solutions, and puzzles with highly-composite targets have more solutions. No target number larger than 24 has as many as 466 solvable puzzles.

These are only trends, not hard rules. For example, there are 156 solvable puzzles with the target 126 (4 prime factors) but only 93 with target 128 (7 prime factors). Why? (I don't know. Maybe because there is some correlation with the number of different prime factors? But 72, 144, and 216 have many solutions, and only two different prime factors.)

The smallest target you can't hit is 417. The following numbers 418 and 419 are also impossible. But there are 8 sets of four digits that can be used to make 416 and 23 sets that can be used to make 420. The largest target that can be hit is obviously !!6561 = 9⁴!!; the largest target with two solutions is !!2916 = 4·9·9·9 = 6·6·9·9!!.

(The raw data are available here).

There is a lot more to discover here. For example, from looking at the chart, it seems that the locally-best target numbers often have the form !!2^n3^m!!. What would we see if we colored the dots according to their largest prime factor instead of according to their number of prime factors? (I tried doing this, and it didn't look like much, but maybe it could have been done better.)

Making zero

As the chart shows, 705 of the 715 puzzles of the type «a b c d ⇒ 0», are solvable. This suggests an interesting inverse puzzle that Toph and I enjoyed: find four digits !!a,b,c, d!! that cannot be used to make zero. (The answers).

Identifying interesting or difficult problems

(Caution: this section contains spoilers for many of the most interesting puzzles.)

I spent quite a while trying to get the computer to rank puzzles by difficulty, with indifferent success.


Seven puzzles require the use of fractions. One of these is the notorious «3 3 8 8» that I mentioned before. This is probably the single hardest of this type. The other six are:

    «1 3 4 6»
    «1 4 5 6»
    «1 5 5 5»
    «1 6 6 8»
    «3 3 7 7»
    «4 4 7 7»

(Solutions to these (formatted image); solutions to these (plain text))

«1 5 5 5» is somewhat easier than the others, but they all follow pretty much the same pattern. The last two are pleasantly symmetrical.

Negative numbers

No puzzles require the use of negative intermediate values. This surprised me at first, but it is not hard to see why. Subexpressions with negative intermediate values can always be rewritten to have positive intermediate values instead.

For instance, !!3 × (9 + (3 - 4))!! can be rewritten as !!3 × (9 - (4 - 3))!! and !!(5 - 8)×(1 -9)!! can be rewritten as !!(8 - 5)×(9 -1)!!.

A digression about tree shapes

In one of the earlier articles I asserted that there are only two possible shapes for the expression trees of a puzzle solution:

(((a # b) # c) # d) ((a # b) # (c # d))
Form A Form B

(Pink square nodes contain operators and green round nodes contain numbers.)

Lindsey Kuper pointed out that there are five possible shapes, not two. Of course, I was aware of this (it is a Catalan number), so what did I mean when I said there were only two? It's because I had the idea that any tree that wasn't already in one of those two forms could be put into form A by using transformations like the ones in the previous section.

For example, the expression !!(4×((1+2)÷3))!! isn't in either form, but we can commute the × to get the equivalent !!((1+2)÷3)×4!!, which has form A. Sometimes one uses the associative laws, for example to turn !!a ÷ (b × c)!! into !!(a ÷ b) ÷ c!!.

But I was mistaken; not every expression can be put into either of these forms. The expression !!(8×(9-(2·3))!! is an example.

Unusual intermediate values

The most interesting thing I tried was to look for puzzles whose solutions require unusual intermediate numbers.

For example, the puzzle «3 4 4 4» looks easy (the other puzzles with just 3s and 4s are all pretty easy) but it is rather tricky because its only solution goes through the unusual intermediate number 28: !!4 × (3 + 4) - 4!!.

I ranked puzzles as follows: each possible intermediate number appears in a certain number of puzzle solutions; this is the score for that intermediate number. (Lower scores are better, because they represent rarer intermediate numbers.) The score for a single expression is the score of its rarest intermediate value. So for example !!4 × (3 + 4) - 4!! has the intermediate values 7 and 28. 7 is extremely common, and 28 is quite unusual, appearing in only 151 solution expressions, so !!4 × (3 + 4) - 4!! receives a fairly low score of 151 because of the intermediate 28.

Then each puzzle received a difficulty score which was the score of its easiest solution expression. For example, «2 2 3 8» has two solutions, one (!!(8+3)×2+2!!) involving the quite unusual intermediate value 22, which has a very good score of only 79. But this puzzle doesn't count as difficult because it also admits the obvious solution !!8·3·\frac22!! and this is the solution that gives it its extremely bad score of 1768.

Under this ranking, the best-scoring twenty-four puzzles, and their scores, were:

      «1 2 7 7» 3
    * «4 4 7 7» 12
    * «1 4 5 6» 13
    * «3 3 7 7» 14
    * «1 5 5 5» 15
      «5 6 6 9» 23
      «2 5 7 9» 24
      «2 2 5 8» 25
      «2 5 8 8» 45
      «5 8 8 8» 45
      «2 2 2 9» 47
    * «1 3 4 6» 59
    * «1 6 6 8» 59
      «2 4 4 9» 151
      «3 4 4 4» 151
    * «3 3 8 8» 152
      «6 8 8 9» 152
      «2 2 2 7» 155
      «2 2 5 7» 155
      «2 3 7 7» 155
      «2 4 7 7» 155
      «2 5 5 7» 155
      «2 5 7 7» 156
      «4 4 8 9» 162

(Something is not quite right here. I think «2 5 7 7» and «2 5 5 7» should have the same score, and I don't know why they don't. But I don't care enough to do it over.)

Most of these are at least a little bit interesting. The seven puzzles that require the use of fractions appear; I have marked them with stars. The top item is «1 2 7 7», whose only solution goes through the extremely rare intermediate number 49. The next items require fractions, and the one after that is «5 6 6 9», which I found difficult. So I think there's some value in this procedure.

But is there enough value? I'm not sure. The last item on the list, «4 4 8 9», goes through the unusual number 36. Nevertheless I don't think it is a hard puzzle.

(I can also imagine that someone might see the answer to «5 6 6 9» right off, but find «4 4 8 9» difficult. The whole exercise is subjective.)

Solutions with unusual tree shapes

I thought about looking for solutions that involved unusual sequences of operations. Division is much less common than the other three operations.

To get it right, one needs to normalize the form of expressions, so that the shapes !!(a + b) + (c + d)!! and !!a + (b + (c + d))!! aren't counted separately. The Ezpr library can help here. But I didn't go that far because the preliminary results weren't encouraging.

There are very few expressions totaling 24 that have the form !!(a÷b)÷(c÷d)!!. But if someone gives you a puzzle with a solution in that form, then !!(a×d)÷(b×c)!! and !!(a×d) ÷ (b÷c)!! are also solutions, and one or another is usually very easy to see. For example, the puzzle «1 3 8 9» has the solution !!(8÷1)÷(3÷9)!!, which has an unusual form. But this is an easy puzzle; someone with even a little experience will find the solution !!8 × \frac93 × 1!! immediately.

Similarly there are relatively few solutions of the form !!a÷((b-c)÷d)!!, but they can all be transformed into !!a×d÷(b-c)!! which is not usually hard to find. Consider $$\frac 8{\left(\frac{6 - 4}6\right)}.$$ This is pretty weird-looking, but when you're trying to solve it one of the first things you might notice is the 8, and then you would try to turn the rest of the digits into a 3 by solving «4 6 6 ⇒ 3», at which point it wouldn't take long to think of !!\frac6{6-4}!!. Or, coming at it from the other direction, you might see the sixes and start looking for a way to make «4 6 8 ⇒ 4», and it wouldn't take long to think of !!\frac8{6-4}!!.

Ezpr shape

Ezprs (see previous article) correspond more closely than abstract syntax trees do with our intuitive notion of how expressions ought to work, so looking at the shape of the Ezpr version of a solution might give better results than looking at the shape of the expression tree. For example, one might look at the number of nodes in the Ezpr or the depth of the Ezpr.


When trying to solve one of these puzzles, there are a few things I always try first. After adding up the four numbers, I then look for ways to make !!8·3, 6·4,!! or !!12·2!!; if that doesn't work I start branching out looking for something of the type !!ab\pm c!!.

Suppose we take a list of all solvable puzzles, and remove all the very easy ones: the puzzles where one of the inputs is zero, or where one of the inputs is 1 and there is a solution of the form !!E×1!!.

Then take the remainder and mark them as “easy” if they have solutions of the form !!a+b+c+d, 8·3, 6·4,!! or !!12·2!!. Also eliminate puzzles with solutions of the type !!E + (c - c)!! or !!E×\left(\frac cc\right)!!.

How many are eliminated in this way? Perhaps most? The remaining puzzles ought to have at least intermediate difficulty, and perhaps examining just those will suggest a way to separate them further into two or three ranks of difficulty.

I give up

But by this time I have solved so many twenty-four puzzles that I am no longer sure which ones are hard and which ones are easy. I suspect that I have seen and tried to solve most of the 466 solvable puzzles; certainly more than half. So my brain is no longer a reliable gauge of which puzzles are hard and which are easy.

Perhaps looking at puzzles with five inputs would work better for me now. These tend to be easy, because you have more to work with. But there are 2002 puzzles and probably some of them are hard.

Close, but no cigar

What's the closest you can get to 24 without hitting it exactly? The best I could do was !!5·5 - \frac89!!. Then I asked the computer, which confirmed that this is optimal, although I felt foolish when I saw the simpler solutions that are equally good: !!6·4 \pm\frac 19!!.

The paired solutions $$5 × \left(4 + \frac79\right) < 24 < 7 × \left(4 - \frac59\right)$$ are very handsome.

Phone app

The search program that tells us when a puzzle has solutions is only useful if we can take it with us in the car and ask it about license plates. A phone app is wanted. I built one with Code Studio.

Code Studio is great. It has a nice web interface, and beginners can write programs by dragging blocks around. It looks very much like MIT's scratch project, which is much better-known. But Code Studio is a much better tool than Scratch. In Scratch, once you reach the limits of what it can do, you are stuck, and there is no escape. In Code Studio when you drag around those blocks you are actually writing JavaScript underneath, and you can click a button and see and edit the underlying JavaScript code you have written.

Suppose you need to convert A to 1 and B to 2 and so on. Scratch does not provide an ord function, so with Scratch you are pretty much out of luck; your only choice is to write a 26-way if-else tree, which means dragging around something like 104 stupid blocks. In Code Studio, you can drop down the the JavaScript level and type in ord to use the standard ord function. Then if you go back to blocks, the ord will look like any other built-in function block.

In Scratch, if you want to use a data structure other than an array, you are out of luck, because that is all there is. In Code Studio, you can drop down to the JavaScript level and use or build any data structure available in JavaScript.

In Scratch, if you want to initialize the program with bulk data, say a precomputed table of the solutions of the 466 twenty-four puzzles, you are out of luck. In Code Studio, you can upload a CSV file with up to 1,000 records, which then becomes available to your program as a data structure.

In summary, you spend a lot of your time in Scratch working around the limitations of Scratch, and what you learn doing that is of very limited applicability. Code Studio is real programming and if it doesn't do exactly what you want out of the box, you can get what you want by learning a little more JavaScript, which is likely to be useful in other contexts for a long time to come.

Once you finish your Code Studio app, you can click a button to send the URL to someone via SMS. They can follow the link in their phone's web browser and then use the app.

Code Studio is what Scratch should have been. Check it out.


Thanks to everyone who contributed to this article, including:

  • my daughters Toph and Katara
  • Shreevatsa R.
  • Dr. Lindsey Kuper
  • Darius Bacon
  • everyone else who emailed me

[Other articles in category /math] permanent link

Sun, 20 Aug 2017

Recognizing when two arithmetic expressions are essentially the same

[ Warning: The math formatting in the RSS / Atom feed for this article is badly mutilated. I suggest you read the article on my blog. ]

In this article, I discuss “twenty-four puzzles”. The puzzle «4 6 7 9 ⇒ 24» means that one should take the numbers 4, 6, 7, and 9, and combine them with the usual arithmetic operations of addition, subtraction, multiplication, and division, to make the number 24. In this case the unique solution is !!6·\frac{7 + 9}{4}!!.

When the target number after the is 24, as it often is, we omit it and just write «4 6 7 9». Every example in this article has target number 24.

This is a continuation of my previous articles on this topic:

My first cut at writing a solver for twenty-four puzzles was a straightforward search program. It had a couple of hacks in it to cut down the search space by recognizing that !!a+E!! and !!E+a!! are the same, but other than that there was nothing special about it and I've discussed it before.

It would quickly and accurately report whether any particular twenty-four puzzle was solvable, but as it turned out that wasn't quite good enough. The original motivation for the program was this: Toph and I play this game in the car. Pennsylvania license plates have three letters and four digits, and if we see a license plate FBV 2259 we try to solve «2 2 5 9». Sometimes we can't find a solution and then we wonder: it is because there isn't one, or is it because we just didn't get it yet? So the searcher turned into a phone app, which would tell us whether there was solution, so we'd know whether to give up or keep searching.

But this wasn't quite good enough either, because after we would find that first solution, say !!2·(5 + 9 - 2)!!, we would wonder: are there any more? And here the program was useless: it would cheerfully report that there were three, so we would rack our brains to find another, fail, ask the program to tell us the answer, and discover to our disgust that the three solutions it had in mind were:

$$ 2 \cdot (5 + (9 - 2)) \\ 2 \cdot (9 + (5 - 2)) \\ 2 \cdot ((5 + 9) - 2) $$

The computer thinks these are different, because it uses different data structures to represent them. It represents them with an abstract syntax tree, which means that each expression is either a single constant, or is a structure comprising an operator and its two operand expressions—always exactly two. The computer understands the three expressions above as having these structures:

It's not hard to imagine that the computer could be taught to understand that the first two trees are equivalent. Getting it to recognize that the third one is also equivalent seems somewhat more difficult.

Commutativity and associativity

I would like the computer to understand that these three expressions should be considered “the same”. But what does “the same” mean? This problem is of a kind I particularly like: we want the computer to do something, but we're not exactly sure what that something is. Some questions are easy to ask but hard to answer, but this is the opposite: the real problem is to decide what question we want to ask. Fun!

Certainly some of the question should involve commutativity and associativity of addition and multiplication. If the only difference between two expressions is that one has !!a + b!! where the other has !!b + a!!, they should be considered the same; similarly !!a + (b + c)!! is the same expression as !!(a + b) + c!! and as !!(b + a) + c!! and !!b + (a + c)!! and so forth.

The «2 2 5 9» example above shows that commutativity and associativity are not limited to addition and multiplication. There are commutative and associative properties of subtraction also! For example, $$a+(b-c) = (a+b)-c$$ and $$(a+b)-c = (a-c)+b.$$ There ought to be names for these laws but as far as I know there aren't. (Sure, it's just commutativity and associativity of addition in disguise, but nobody explaining these laws to school kids ever seems to point out that subtraction can enter into it. They just observe that !!(a-b)-c ≠ a-(b-c)!!, say “subtraction isn't associative”, and leave it at that.)

Closely related to these identities are operator inversion identities like !!a-(b+c) = (a-b)-c!!, !!a-(b-c) = (a-b)+c!!, and their multiplicative analogues. I don't know names for these algebraic laws either.

One way to deal with all of this would to build a complicated comparison function for abstract syntax trees that tried to transform one tree into another by applying these identities. A better approach is to recognize that the data structure is over-specified. If we want the computer to understand that !!(a + b) + c!! and !!a + (b + c)!! are the same expression, we are swimming upstream by using a data structure that was specifically designed to capture the difference between these expressions.

Instead, I invented a data structure, called an Ezpr (“Ez-pur”), that can represent expressions, but in a somewhat more natural way than abstract syntax trees do, and in a way that makes commutativity and associativity transparent.

An Ezpr has a simplest form, called its “canonical” or “normal” form. Two Ezprs represent essentially the same mathematical expression if they have the same canonical form. To decide if two abstract syntax trees are the same, the computer converts them to Ezprs, simplifies them, and checks to see if resulting canonical forms are identical.

The Ezpr

Since associativity doesn't matter, we don't want to represent it. When we (humans) think about adding up a long column of numbers, we don't think about associativity because we don't add them pairwise. Instead we use an addition algorithm that adds them all at once in a big pile. We don't treat addition as a binary operation; we normally treat it as an operator that adds up the numbers in a list. The Ezpr makes this explicit: its addition operator is applied to a list of subexpressions, not to a pair. Both !!a + (b + c)!! and !!(a + b) + c!! are represented as the Ezpr

    SUM [ a b c - ]

which just says that we are adding up !!a!!, !!b!!, and !!c!!. (The - sign is just punctuation; ignore it for now.)

Similarly the Ezpr MUL [ a b c ÷ ] represents the product of !!a!!, !!b!!, and !!c!!. (Please ignore the ÷ sign for the time being.)

To handle commutativity, we want those [ a b c ] lists to be bags. Perl doesn't have a built-in bag object, so instead I used arrays and required that the array elements be in sorted order. (Exactly which sorted order doesn't really matter.)

Subtraction and division

This doesn't yet handle subtraction and division, and the way I chose to handle them is the only part of this that I think is at all clever. A SUM object has not one but two bags, one for the positive and one for the negative part of the expression. An expression like !!a - b + c - d!! is represented by the Ezpr:

SUM [ a c - b d ]

and this is also the representation of !!a + c - b - d!!, of !!c + a - d - b!!, of !!c - d+ a-b!!, and of any other expression of the idea that we are adding up !!a!! and !!c!! and then deducting !!b!! and !!d!!. The - sign separates the terms that are added from those that are subtracted.

Either of the two bags may be empty, so for example !!a + b!! is just SUM [ a b - ].

Division is handled similarly. Here conventional mathematical notation does a little bit better than in the sum case: MUL [ a c ÷ b d ] is usually written as !!\frac{ac}{bd}!!.

Ezprs handle the associativity and commutativity of subtraction and division quite well. I pointed out earlier that subtraction has an associative law !!(a + b) - c = a + (b - c)!! even though it's not usually called that. No code is required to understand that those two expressions are equal if they are represented as Ezprs, because they are represented by completely identical structures:

        SUM [ a b - c ]

Similarly there is a commutative law for subtraction: !!a + b - c = a - c + b!! and once again that same Ezpr does for both.

Ezpr laws

Ezprs are more flexible than binary trees. A binary tree can represent the expressions !!(a+b)+c!! and !!a+(b+c)!! but not the expression !!a+b+c!!. Ezprs can represent all three and it's easy to transform between them. Just as there are rules for building expressions out of simpler expressions, there are a few rules for combining and manipulating Ezprs.

Lifting and flattening

The most important transformation is lifting, which is the Ezpr version of the associative law. In the canonical form of an Ezpr, a SUM node may not have subexpressions that are also SUM nodes. If you have

  SUM [ a SUM [ b c - ] - … ]

you should lift the terms from the inner sum into the outer one:

  SUM [ a b c - … ]

effectively transforming !!a+(b+c)!! into !!a+b+c!!. More generally, in

   SUM [ a SUM [ b - c ]
       - d SUM [ e - f ] ]

we lift the terms from the inner Ezprs into the outer one:

   SUM [ a b f - c d e ]

This effectively transforms !!a + (b - c) - d - (e - f))!! to !!a + b + f - c - d - e!!.

Similarly, when a MUL node contains another MUL, we can flatten the structure.

Say we are converting the expression !!7 ÷ (3 ÷ (6 × 4))!! to an Ezpr. The conversion function is recursive and the naïve version computes this Ezpr:

      MUL [ 7 ÷ MUL [ 3 ÷ MUL [ 6 4 ÷ ] ] ]

But then at the bottom level we have a MUL inside a MUL, so the 4 and 6 in the innermost MUL are lifted upward:

      MUL [ 7 ÷ MUL [ 3 ÷ 6 4 ] ]

which represents !!\frac7{\frac{3}{6\cdot 4}}!!. Then again we have a MUL inside a MUL, and again the subexpressions of the innermost MUL can be lifted:

      MUL [ 7 6 4 ÷ 3 ]

which we can imagine as !!\frac{7·6·4}3!!.

The lifting only occurs when the sub-node has the same type as its parent; we may not lift terms out of a MUL into a SUM or vice versa.

Trivial nodes

The Ezpr SUM [ a - ] says we are adding up just one thing, !!a!!, and so it can be eliminated and replaced with just !!a!!. Similarly SUM [ - a ] can be replaced with the constant !!-a!!, if !!a!! is a constant. MUL can be handled similarly.

An even simpler case is SUM [ - ] which can be replaced by the constant 0; MUL [ ÷ ] can be replaced with 1. These sometimes arise as a result of cancellation.


Consider the puzzle «3 3 4 6». My first solver found 49 solutions to this puzzle. One is !!(3 - 3) + (4 × 6)!!. Another is !!(4 + (3 - 3)) × 6!!. A third is !!4 × (6 + (3 - 3))!!.

I think these are all the same: the solution is to multiply the 4 by the 6, and to get rid of the threes by subtracting them to make a zero term. The zero term can be added onto the rest of expression or to any of its subexpressions—there are ten ways to do this—and it doesn't really matter where.

This is easily explained in terms of Ezprs: If the same subexpression appears in both of a node's bags, we can drop it. For example, the expression !!(4 + (3 -3)) × 6!! starts out as

    MUL [ 6 SUM [ 3 4 - 3 ] ÷ ]

but the duplicate threes in SUM [ 3 4 - 3 ] can be canceled, to leave

    MUL [ 6 SUM [ 4 - ] ÷ ]

The sum is now trivial, as described in the previous section, so can be eliminated and replaced with just 4:

    MUL [ 6 4 ÷ ]

This Ezpr records the essential feature of each of the three solutions to «3 3 4 6» that I mentioned: they all are multiplying the 6 by the 4, and then doing something else unimportant to get rid of the threes.

Another solution to the same puzzle is !!(6 ÷ 3) × (4 × 3)!!. Mathematically we would write this as !!\frac63·4·3!! and we can see this is just !!6×4!! again, with the threes gotten rid of by multiplication and division, instead of by addition and subtraction. When converted to an Ezpr, this expression becomes:

    MUL [ 6 4 3 ÷ 3 ]

and the matching threes in the two bags are cancelled, again leaving

    MUL [ 6 4 ÷ ]

In fact there aren't 49 solutions to this puzzle. There is only one, with 49 trivial variations.

Identity elements

In the preceding example, many of the trivial variations on the !!4×6!! solution involved multiplying some subexpression by !!\frac 33!!. When one of the input numbers in the puzzle is a 1, one can similarly obtain a lot of useless variations by choosing where to multiply the 1.

Consider «1 3 3 5»: We can make 24 from !!3 × (3 + 5)!!. We then have to get rid of the 1, but we can do that by multiplying it onto any of the five subexpressions of !!3 × (3 + 5)!!:

$$ 1 × (3 × (3 + 5)) \\ (1 × 3) × (3 + 5) \\ 3 × (1 × (3 + 5)) \\ 3 × ((1 × 3) + 5) \\ 3 × (3 + (1×5)) $$

These should not be considered different solutions. Whenever we see any 1's in either of the bags of a MUL node, we should eliminate them. The first expression above, !!1 × (3 × (3 + 5))!!, is converted to the Ezpr

 MUL [ 1 3 SUM [ 3 5 - ] ÷ ]

but then the 1 is eliminated from the MUL node leaving

 MUL [ 3 SUM [ 3 5 - ] ÷ ]

The fourth expression, !!3 × ((1 × 3) + 5)!!, is initially converted to the Ezpr

 MUL [ 3 SUM [ 5 MUL [ 1 3 ÷ ] - ] ÷ ]

When the 1 is eliminated from the inner MUL, this leaves a trivial MUL [ 3 ÷ ] which is then replaced with just 3, leaving:

 MUL [ 3 SUM [ 5 3 - ] ÷ ]

which is the same Ezpr as before.

Zero terms in the bags of a SUM node can similarly be dropped.

Multiplication by zero

One final case is that MUL [ 0 … ÷ … ] can just be simplified to 0.

The question about what to do when there is a zero in the denominator is a bit of a puzzle. In the presence of division by zero, some of our simplification rules are questionable. For example, when we have MUL [ a ÷ MUL [ b ÷ c ] ], the lifting rule says we can simplify this to MUL [ a c ÷ b ]—that is, that !!\frac a{\frac bc} = \frac{ac}b!!. This is correct, except that when !!b=0!! or !!c=0!! it may be nonsense, depending on what else is going on. But since zero denominators never arise in the solution of these puzzles, there is no issue in this application.


The Ezpr module is around 200 lines of Perl code, including everything: the function that converts abstract syntax trees to Ezprs, functions to convert Ezprs to various notations (both MUL [ 4 ÷ SUM [ 3 - 2 ] ] and 4 ÷ (3 - 2)), and the two versions of the normalization process described in the previous section. The normalizer itself is about 35 lines.

Associativity is taken care of by the Ezpr structure itself, and commutativity is not too difficult; as I mentioned, it would have been trivial if Perl had a built-in bag structure. I find it much easier to reason about transformations of Ezprs than abstract syntax trees. Many operations are much simpler; for example the negation of SUM [ A - B ] is simply SUM [ B - A ]. Pretty-printing is also easier because the Ezpr better captures the way we write and think about expressions.

It took me a while to get the normalization tuned properly, but the results have been quite successful, at least for this problem domain. The current puzzle-solving program reports the number of distinct solutions to each puzzle. When it reports two different solutions, they are really different; when it fails to support the exact solution that Toph or I found, it reports one essentially the same. (There are some small exceptions, which I will discuss below.)

Since there is no specification for “essentially the same” there is no hope of automated testing. But we have been using the app for several months looking for mistakes, and we have not found any. If the normalizer failed to recognize that two expressions were essentially similar, we would be very likely to notice: we would be solving some puzzle, be unable to find the last of the solutions that the program claimed to exist, and then when we gave up and saw what it was we would realize that it was essentially the same as one of the solutions we had found. I am pretty confident that there are no errors of this type, but see “Arguable points” below.

A harder error to detect is whether the computer has erroneously conflated two essentially dissimilar expressions. To detect this we would have to notice that an expression was missing from the computer's solution list. I am less confident that nothing like this has occurred, but as the months have gone by I feel better and better about it.

I consider the problem of “how many solutions does this puzzle really have to have?” been satisfactorily solved. There are some edge cases, but I think we have identified them.

Code for my solver is on Github. The Ezpr code is in the Ezpr package in the file. This code is all in the public domain.

Some examples

The original program claims to find 35 different solutions to «4 6 6 6». The revised program recognizes that these are of only two types:

!!4 × 6 × 6 ÷ 6!!MUL [ 4 6 - ]
!!(6 - 4) × (6 + 6)!!MUL [ SUM [ 6 - 4 ] SUM [ 6 6 - ] ÷ ]

Some of the variant forms of the first of those include:

$$ 6 × (4 + (6 - 6)) \\ 6 + ((4 × 6) - 6) \\ (6 - 6) + (4 × 6) \\ (6 ÷ 6) × (4 × 6) \\ 6 ÷ ((6 ÷ 4) ÷ 6) \\ 6 ÷ (6 ÷ (4 × 6)) \\ 6 × (6 × (4 ÷ 6)) \\ (6 × 6) ÷ (6 ÷ 4) \\ 6 ÷ ((6 ÷ 6) ÷ 4) \\ 6 × (6 - (6 - 4)) \\ 6 × (6 ÷ (6 ÷ 4)) \\ \ldots

In an even more extreme case, the original program finds 80 distinct expressions that solve «1 1 4 6», all of which are trivial variations on !!4·6!!.

Of the 715 puzzles, 466 (65%) have solutions; for 175 of these the solution is unique. There are 3 puzzles with 8 solutions each («2 2 4 8», «2 3 6 9», and «2 4 6 8»), one with 9 solutions («2 3 4 6»), and one with 10 solutions («2 4 4 8»).

The 10 solutions for «2 4 4 8» are as follows:

!!4 × 8 - 2 × 4 !!SUM [ MUL [ 4 8 ÷ ] - MUL [ 2 4 ÷ ] ]
!!4 × (2 + 8 - 4) !!MUL [ 4 SUM [ 2 8 - 4 ] ÷ ]
!!(8 - 4) × (2 + 4) !!MUL [ SUM [ 8 - 4 ] SUM [ 2 4 - ] ÷ ]
!!4 × (4 + 8) ÷ 2 !!MUL [ 4 SUM [ 4 8 - ] ÷ 2 ]
!!(4 - 2) × (4 + 8) !!MUL [ SUM [ 4 - 2 ] SUM [ 4 8 - ] ÷ ]
!!8 × (2 + 4/4) !!MUL [ 8 SUM [ 1 2 - ] ÷ ]
!!2 × 4 × 4 - 8 !!SUM [ MUL [ 2 4 4 ÷ ] - 8 ]
!!8 + 2 × (4 + 4) !!SUM [ 8 MUL [ 2 SUM [ 4 4 - ] ÷ ] - ]
!!4 + 4 + 2 × 8 !!SUM [ 4 4 MUL [ 2 8 ÷ ] - ]
!!4 × (8 - 4/2) !!MUL [ 4 SUM [ 8 - MUL [ 4 ÷ 2 ] ] ÷ ]

A complete listing of every essentially different solution to every «a b c d» puzzle is available here. There are 1,063 solutions in all.

Arguable points

There are a few places where we have not completely pinned down what it means for two solutions to be essentially the same; I think there is room for genuine disagreement.

  1. Any solution involving !!2×2!! can be changed into a slightly different solution involving !!2+2!! instead. These expressions are arithmetically different but numerically equal. For example, I mentioned earlier that «2 2 4 8» has 8 solutions. But two of these are !! 8 + 4 × (2 + 2)!! and !! 8 + 4 × 2 × 2!!. I am willing to accept these as essentially different. Toph, however, disagrees.

  2. A similar but more complex situation arises in connection with «1 2 3 7». Consider !!3×7+3!!, which equals 24. To get a solution to «1 2 3 7», we can replace either of the threes in !!3×7+3!! with !!(1+2)!!, obtaining !!((1 + 2) × 7) + 3!! or !! (3×7)+(1 +2)!!. My program considers these to be different solutions. Toph is unsure.

It would be pretty easy to adjust the normalization process to handle these the other way if the user wanted that.

Some interesting puzzles

«1 2 7 7» has only one solution, quite unusual. (Spoiler) «2 2 6 7» has two solutions, both somewhat unusual. (Spoiler)

Somewhat similar to «1 2 7 7» is «3 9 9 9» which also has an unusual solution. But it has two other solutions that are less surprising. (Spoiler)

«1 3 8 9» has an easy solution but also a quite tricky solution. (Spoiler)

One of my neighbors has the license plate JJZ 4631. «4 6 3 1» is one of the more difficult puzzles.

What took so long?

Back in March, I wrote:

I have enough material for at least three or four more articles about this that I hope to publish here in the coming weeks.

But the previous article on this subject ended similarly, saying

I hope to write a longer article about solvers in the next week or so.

and that was in July 2016, so don't hold your breath.

And here we are, five months later!

This article was a huge pain to write. Sometimes I sit down to write something and all that comes out is dreck. I sat down to write this one at least three or four times and it never worked. The tortured Git history bears witness. In the end I had to abandon all my earlier drafts and start over from scratch, writing a fresh outline in an empty file.

But perseverance paid off! WOOOOO.

[ Addendum 20170825: I completely forgot that Shreevatsa R. wrote a very interesting article on the same topic as this one, in July of last year soon after I published my first article in this series. ]

[ Addendum 20170829: A previous version of this article used the notations SUM [ … # … ] and MUL [ … # … ], which I said I didn't like. Zellyn Hunter has persuaded me to replace these with SUM [ … - … ] and MUL [ … ÷ … ]. Thank you M. Hunter! ]

[ Yet more on this topic! ]

[Other articles in category /math] permanent link

Tue, 08 Aug 2017

That time I met Erdős

I should have written about this sooner, by now it has been so long that I have forgotten most of the details.

I first encountered Paul Erdős in the middle 1980s at a talk by János Pach about almost-universal graphs. Consider graphs with a countably infinite set of vertices. Is there a "universal" graph !!G!! such that, for any finite or countable graph !!H!!, there is a copy of !!H!! inside of !!G!!? (Formally, this means that there is an injection from the vertices of !!H!! to the vertices of !!G!! that preserves adjacency.) The answer is yes; it is quite easy to construct such a !!G!! and in fact nearly all random graphs have this property.

But then the questions become more interesting. Let !!K_\omega!! be the complete graph on a countably infinite set of vertices. Say that !!G!! is “almost universal” if it includes a copy of !!H!! for every finite or countable graph !!H!! except those that contain a copy of !!K_\omega!!. Is there an almost universal graph? Perhaps surprisingly, no! (Sketch of proof.)

I enjoyed the talk, and afterward in the lobby I got to meet Ron Graham and Joel Spencer and talk to them about their Ramsey theory book, which I had been reading, and about a problem I was working on. Graham encouraged me to write up my results on the problem and submit them to Mathematics Magazine, but I unfortunately never got around to this. Graham was there babysitting Erdős, who was one of Pách's collaborators, but I did not actually talk to Erdős at that time. I think I didn't recognize him. I don't know why I was able to recognize Graham.

I find the almost-universal graph thing very interesting. It is still an open research area. But none of this was what I was planning to talk about. I will return to the point. A couple of years later Erdős was to speak at the University of Pennsylvania. He had a stock speech for general audiences that I saw him give more than once. Most of the talk would be a description of a lot of interesting problems, the bounties he offered for their solutions, and the progress that had been made on them so far. He would intersperse the discussions with the sort of Erdősism that he was noted for: referring to the U.S. and the U.S.S.R. as “Sam” and “Joe” respectively; his ever-growing series of styles (Paul Erdős, P.G.O.M., A.D., etc.) and so on.

One remark I remember in particular concerned the $3000 bounty he offered for proving what is sometimes known as the Erdős-Túran conjecture: if !!S!! is a subset of the natural numbers, and if !!\sum_{n\in S}\frac 1n!! diverges, then !!S!! contains arbitrarily long arithmetic progressions. (A special case of this is that the primes contain arbitrarily long arithmetic progressions, which was proved in 2004 by Green and Tao, but which at the time was a long-standing conjecture.) Although the $3000 was at the time the largest bounty ever offered by Erdős, he said it was really a bad joke, because to solve the problem would require so much effort that the per-hour payment would be minuscule.

I made a special trip down to Philadelphia to attend the talk, with the intention of visiting my girlfriend at Bryn Mawr afterward. I arrived at the Penn math building early and wandered around the halls to kill time before the talk. And as I passed by an office with an open door, I saw Erdős sitting in the antechamber on a small sofa. So I sat down beside him and started telling him about my favorite graph theory problem.

Many people, preparing to give a talk to a large roomful of strangers, would have found this annoying and intrusive. Some people might not want to talk about graph theory with a passing stranger. But most people are not Paul Erdős, and I think what I did was probably just the right thing; what you don't do is sit next to Erdős and then ask how his flight was and what he thinks of recent politics. We talked about my problem, and to my great regret I don't remember any of the mathematical details of what he said. But he did not know the answer offhand, he was not able solve it instantly, and he did say it was interesting. So! I had a conversation with Erdős about graph theory that was not a waste of his time, and I think I can count that as one of my lifetime accomplishments.

After a little while it was time to go down to the auditorium for the the talk, and afterward one of the organizers saw me, perhaps recognized me from the sofa, and invited me to the guest dinner, which I eagerly accepted. At the dinner, I was thrilled because I secured a seat next to Erdős! But this was a beginner mistake: he fell asleep almost immediately and slept through dinner, which, I learned later, was completely typical.

[Other articles in category /math] permanent link

Thu, 15 Jun 2017

Base-4 fractions in Telugu

Rik Signes brought to my attention that since version 5.1 Unicode has contained the following excitingly-named characters:



I looked into this a little and found out what they are for. It makes a lot of sense! The details were provided by “Telugu Measures and Arithmetic Marks” by Nāgārjuna Venna.

Telugu is the third-most widely spoken language in India, spoken mostly in the southeast part of the country. Traditional Telugu units of measurement are often divided into four or eight subunits. For example, the tūmu is divided into four kuṁcamulu, the kuṁcamulu, into four mānikalu, and the mānikalu into four sōlalu.

These days they mainly use liters like everyone else. But the traditional measurements are mostly divided into fours, so amounts are written with a base-10 integer part and a base-4 fractional part. The characters above are the base-4 fractional digits.

To make the point clearer, I hope, let's imagine that we are using the Telugu system, but with the familar western-style symbols 0123456789 instead of the Telugu digits ౦౧౨౩౪౫౬౭౮౯. (The Telugu had theirs first of course.) And let's use 0-=Z as our base-four fractional digits, analogous to Telugu ౦౼౽౾. (As in Telugu, we'll use the same zero symbol for both the integer and the fractional parts.) Then to write the number of gallons (7.4805195) in a cubic foot, we say


which is 7 gallons plus one (-) quart plus three (Z) cups plus two (=) quarter-cups plus three (Z) tablespoons plus zero (0) drams, a total of 7660 drams almost exactly. Or we could just round off to 7.=, seven and a half gallons.

(For the benefit of readers who might be a bit rusty on the details of these traditional European measurements, I should mention that there are four drams in a tablespoon, four tablespoons in a quarter cup, four quarter cups in a cup, four cups in a quart, and four quarts in a gallon, so 4⁵ = 1024 drams in a gallon and 7.4805195·4⁵ = 7660.052 drams in a cubic foot. Note also that these are volume (fluid) drams, not mass drams, which are different.)

We can omit the decimal point (as the Telegu did) and write


and it is still clear where the integer part leaves off and the fraction begins, because we are using special symbols for the fractional part. But no, this isn't quite enough, because if we wrote 20ZZ= it might not be clear whether we meant 20.ZZ= or 2.0ZZ=.

So the system has an elaboration. In the odd positions, we don't use the 0-=Z symbols; we use Q|HN instead. And we don't write 7-Z=Z0, we write


This is always unambiguous: 20.ZZ= is actually written 20NZH and 2.0ZZ= is written 2QZN=, quite different.

This is all fanciful in English, but Telugu actually did this. Instead of 0-=Z they had ౦౼౽౾ as I mentioned before. And instead of Q|HN they had ౸౹౺౻. So if the Telugu were trying to write 7.4805195, where we had 7|ZHZQ they might have written ౭౹౾౺౾౸. Like us, they then appended an abbreviation for the unit of measurement. Instead of “gal.” for gallon they might have put ఘ (letter “gha”), so ౭౹౾౺౾౸ఘ. It's all reasonably straightforward, and also quite sensible. If you have ౭౹౾౺ tūmu, you can read off instantly that there are ౺ (two) sōlalu left over, just as you can see that $7.43 has three pennies left over.

Notice that both sets of Telugu fraction digits are easy to remember: the digits for 3 have either three horizonal strokes ౾ or three vertical strokes ౻, and the others similarly.

I have an idea that the alternating vertical-horizontal system might have served as an error-detection mechanism: if a digit is omitted, you notice right away because the next symbol is wrong.

I find this delightful. A few years back I read all of The Number Concept: Its Origin and Development (1931) by Levi Leonard Conant, hoping to learn something really weird, and I was somewhat disappointed. Conant spends most of his book describing the number words and number systems used by dozens of cultures and almost all of them are based on ten, and a few on five or twenty. (“Any number system which passes the limit 10 is reasonably sure to have either a quinary, a decimal, or a vigesimal structure.”) But he does not mention Telugu!

[ Addendum 20200821: Bengali currency amounts do something similar, but the sections alternate between base-10 and base-4 numerals! ]

[Other articles in category /math] permanent link

Sun, 05 Mar 2017

Solving twenty-four puzzles

Back in July, I wrote:

Lately my kids have been interested in puzzles of this type: You are given a sequence of four digits, say 1,2,3,4, and your job is to combine them with ordinary arithmetic operations (+, -, ×, and ÷) in any order to make a target number, typically 24. For example, with 1,2,3,4, you can go with $$((1+2)+3)×4 = 24$$ or with $$4×((2×3)×1) = 24.$$

I said I had found an unusually difficult puzzle of this type, which is to make 2,5,6,6 total to 17. This is rather difficult. (I will reveal the solution later in this article.) Several people independently wrote to advise me that it is even more difficult to make 3,3,8,8 total to 24. They were right; it is amazingly difficult. After a couple of weeks I finally gave up and asked the computer, and when I saw the answer I didn't feel bad that I hadn't gotten it myself. (The solution is here if you want to give up without writing a program.)

From now on I will abbreviate the two puzzles of the previous paragraph as «2 5 6 6 ⇒ 17» and «3 3 8 8 ⇒ 24», and others similarly.

The article also inspired a number of people to write their own solvers and send them to me, and comparing them was interesting. My solver followed the tree search technique that I described in chapter 5 of Higher-Order Perl, and which has become so familiar to me that by now I can implement it without thinking about it very hard:

  1. Invent a data structure that represents the state of a possibly-incomplete search. This is just a list of the stuff one needs to keep track of while searching. (Let's call this a node.)

  2. Build a function which recognizes when a node represents a successful search.

  3. Build a function which takes a node, computes all the ways the search could proceed from that point, and returns a list of nodes for those slightly-more-advanced searches.

  4. Initialize a queue with a node representing a search that has just begun.

  5. Do this:

      until ( queue.is_empty() ) {
        current_node = queue.get_next()
        if ( is_successful( current_node ) ) { print the solution }
        queue.push( slightly_more_complete_searches( current_node ) )

This is precisely a breadth-first search. To make it into depth-first search, replace the queue with a stack. To make a heuristically directed search, replace get_next with a function that looks at the queue and chooses the best-looking node from which to proceed. Many other variations are possible, which is the advantage of this synthetic approach over letting the search arise organically from a recursive searcher. (Higher-Order Perl says “Recursive functions naturally perform depth-first searches.” (page 203)) In Python or Ruby one would be able to use yield and would not have to manage the queue explicitly, but in this case the queue management is trivial.

In my solver, each node contains a list of available expressions, annotated with its numerical value. Initially, the expressions are single numbers and the values are the same, say

    [ [ "2" => 2 ], [ "3" => 3 ], [ "4" => 4 ], [ "6" => 6 ] ]

Whether you represent expressions as strings or as something more structured depends on what you need to do with them at the end. If you just need to print them out, strings are good enough and are easy to handle.

A node represents a successful search if it contains only a single expression and if the expression's value is the target sum, say 24:

    [ [ "(((6÷2)+3)×4)" => 24 ] ]

From a node, the search should proceed by selecting two of the expressions, removing them from the node, selecting a legal operation, combining the two expressions into a single expression, and inserting the result back into the node. For example, from the initial node shown above, the search might continue by subtracting the fourth expression from the second:

    [ [ "2" => 2 ], [ "4" => 4 ], [ "(3-6)" => -3 ] ]

or by multiplying the second and the third:

    [ [ "2" => 2 ], [ "(3×4)" => 12 ], [ "6" => 6 ] ]

When the program encounters that first node it will construct both of these, and many others, and put them all into the queue to be investigated later.


    [ [ "2" => 2 ], [ "(3×4)" => 12 ], [ "6" => 6 ] ]

the search might proceed by dividing the first expression by the third:

    [ [ "(3×4)" => 12 ], [ "(2÷6)" => 1/3 ] ]

Then perhaps by subtracting the first from the second:

    [ [ "((2÷6)-(3×4))" => -35/3 ] ]

From here there is no way to proceed, so when this node is removed from the queue, nothing is added to replace it. Had it been a winner, it would have been printed out, but since !!-\frac{35}3!! is not the target value of 24, it is silently discarded.

To solve a puzzle of the «a b c d ⇒ t» sort requires examining a few thousand nodes. On modern hardware this takes approximately zero seconds.

The actual code for my solver is a lot of Perl gobbledygook that may not be of general interest so I will provide a link for people who are interested in deciphering it. It also represents my second attempt: I lost the code that I described in the earlier article and had to rewrite it. It is rather bigger than I would have liked.

My puzzle solver in Perl.

Stuff goes wrong

People showed me a lot of programs to solve this, and many didn't work. There are a few hard cases that several of them get wrong.


Some puzzles require that some subexpressions have fractional values. Many of the programs people showed me used integer arithmetic (sometimes implicitly and unintentionally) and failed to solve those puzzles. We can detect this by asking for a solution to «2 5 6 6 ⇒ 17», which requires a fraction. The solution is !!6×(2+(5÷6))!!. A program using integer arithmetic will calculate !!5÷6 = 0!! and fail to recognize the solution.

Several people on Twitter made this mistake and then mistakenly claimed that there was no solution at all. Usually it was possible to correct their programs by changing

        inputs = [ 2, 2, 5, 6 ]


        inputs = [ 2.0, 2.0, 5.0, 6.0 ]

or something like that.

Some people also surprised me by claiming that I had lied when I stated that the puzzle could be solved without any “underhanded tricks”, and that the use of intermediate fractions was itself an underhanded trick. Your Honor, I plead not guilty. I originally described the puzzle this way:

You are given a sequence of four digits, say 1,2,3,4, and your job is to combine them with ordinary arithmetic operations (+, -, ×, and ÷) in any order to make a target number, typically 24.

The objectors are implicitly claiming that when you combine 5 and 6 with the “ordinary arithmetic operation” of division, you get something other than !!\frac56!!. This is an indefensible claim.

I wasn't even trying to be tricky! It never occurred to me that fractions were something that some people would consider underhanded, and now that it has been suggested, I reject the suggestion. Folks, the result of division can be a fraction. Fractions are not some sort of obscure mathematical pettifoggery. They have been with us for at least 3,500 years now, so it is time everyone got used to them.

Floating-point error

Some programs used floating-point arithmetic to deal with the fractions and then fell foul of floating-point error. I will defer discussion of this to a future article.

I've complained about floating-point numbers on this blog before. ( 1 2 3 4 5 ) God, how I loathe them.

[ Addendum 20170825: Looking back on our old discussion from July 2016, I see that Lindsey Kuper said to me:

One nice thing about using Racket or Scheme is that it handles the numeric stuff so nicely. If you weren't careful, I could imagine in Python a solution failing because it evaluated to 16.99999999999999997 or something.

Good call, Dr. Kuper! ]

Expression construction

A more subtle error that several programs made was to assume that all expressions can be constructed by combining a previous expression with a single input number. For example, to solve «2 3 5 7 ⇒ 24», you multiply 3 by 7 to get 21, then add 5 to get 26, then subtract 2 to get 24.

But not every puzzle can be solved this way. Consider «2 3 5 7 ⇒ 41». You start by multiplying 2 by 3 to get 6, but if you try to combine the 6 with either 5 or 7 at this point you will lose. The only solution is to put the 6 aside and multiply 5 by 7 to get 35. Then add the 6 and the 35 to get 41.

Another way to put this is that an unordered binary tree with 4 leaves can take two different shapes. (Imagine filling the green circles with numbers and the pink squares with operators.)

The right-hand type of structure is sometimes necessary, as with «2 3 5 7 ⇒ 41». But several of the proposed solutions produced only expressions with structures like that on the left.

Here's Sebastian Fischer's otherwise very elegant Haskell solution, in its entirety:

    import Data.List ( permutations )

    solution = head
      [ (a,x,(b,y,(c,z,d)))
        | [a,b,c,d] <- permutations [2,5,6,6],
           ops <- permutations [((+),'+'),((-),'-'),((*),'*'),((/),'/')],
           let [u,v,w] = map fst $ take 3 ops,
           let [x,y,z] = map snd $ take 3 ops,
           (a `u` (b `v` (c `w` d))) == 17

You can see the problem in the last line. a, b, c, and d are numbers, and u, v, and w are operators. The program evaluates an expression to see if it has the value 17, but the expression always has the left-hand shape. (The program has another limitation: it never uses the same operator twice in the expression. That second permutations should be (sequence . take 3 . repeat) or something. It can still solve «2 5 6 6 ⇒ 17», however.)

Often the way these programs worked was to generate every possible permutation of the inputs and then apply the operators to the input lists stackwise: pop the first two values, combine them, push the result, and repeat. Here's a relevant excerpt from a program by Tim Dierks, this time in Python:

  for ordered_values in permutations(values):
    for operations in product(ops, repeat=len(values)-1):
      result, formula = calc_result(ordered_values, operations)

Here the expression structure is implicit, but the current result is always made by combining one of the input numbers with the old result.

I have seen many people get caught by this and similar traps in the past. I once posed the problem of enumerating all the strings of balanced parentheses of a given length, and several people assumed that all such strings have the form ()S, S(), or (S), where S is a shorter string of the same type. This seems plausible, and it works up to length 6, but (())(()) does not have that form.

Division by zero

A less common error exhibited by some programs was a failure to properly deal with division by zero. «2 5 6 6 ⇒ 17» has a solution, and if a program dies while checking !!2+(5÷(6-6))!! and doesn't find the solution, that's a bug.

Programs that worked

Ingo Blechschmidt (Haskell)

Ingo Blechschmidt showed me a solution in Haskell. The code is quite short. M. Blechschmidt's program defines a synthetic expression type and an evaluator for it. It defines a function arb which transforms an ordered list of numbers into a list of all possible expressions over those numbers. Reordering the list is taken care of earlier, by Data.List.permutations.

By “synthetic expression type” I mean this:

    data Exp a
        = Lit  a
        | Sum  (Exp a) (Exp a)
        | Diff (Exp a) (Exp a)
        | Prod (Exp a) (Exp a)
        | Quot (Exp a) (Exp a)
        deriving (Eq, Show)

Probably 80% of the Haskell programs ever written have something like this in them somewhere. This approach has a lot of boilerplate. For example, M. Blechschmidt's program then continues:

    eval :: (Fractional a) => Exp a -> a
    eval (Lit x) = x
    eval (Sum  a b) = eval a + eval b
    eval (Diff a b) = eval a - eval b
    eval (Prod a b) = eval a * eval b
    eval (Quot a b) = eval a / eval b

Having made up our own synonyms for the arithmetic operators (Sum for !!+!!, etc.) we now have to explain to Haskell what they mean. (“Not expressions, but an incredible simulation!”)

I spent a while trying to shorten the code by using a less artificial expression type:

    data Exp a
        = Lit  a
        | Op ((a -> a -> a), String) (Exp a) (Exp a)

but I was disappointed; I was only able to cut it down by 18%, from 34 lines to 28. I hope to discuss this in a future article. By the way, “Blechschmidt” is German for “tinsmith”.

Shreevatsa R. (Python)

Shreevatsa R. showed me a solution in Python. It generates every possible expression and prints it out with its value. If you want to filter the voluminous output for a particular target value, you do that later. Shreevatsa wrote up an extensive blog article about this which also includes a discussion about eliminating duplicate expressions from the output. This is a very interesting topic, and I have a lot to say about it, so I will discuss it in a future article.

Jeff Fowler (Ruby)

Jeff Fowler of the Recurse Center wrote a compact solution in Ruby that he described as “hot garbage”. Did I say something earlier about Perl gobbledygook? It's nice that Ruby is able to match Perl's level of gobbledygookitude. This one seems to get everything right, but it fails mysteriously if I replace the floating-point constants with integer constants. He did provide a version that was not “egregiously minified” but I don't have it handy.

Lindsey Kuper (Scheme)

Lindsey Kuper wrote a series of solutions in the Racket dialect of Scheme, and discussed them on her blog along with some other people’s work.

M. Kuper's first draft was 92 lines long (counting whitespace) and when I saw it I said “Gosh, that is way too much code” and tried writing my own in Scheme. It was about the same size. (My Perl solution is also not significantly smaller.)

Martin Janecke (PHP)

I saved the best for last. Martin Janecke showed me an almost flawless solution in PHP that uses a completely different approach than anyone else's program. Instead of writing a lot of code for generating permutations of the input, M. Janecke just hardcoded them:

    $zahlen = [
      [2, 5, 6, 6],
      [2, 6, 5, 6],
      [2, 6, 6, 5],
      [5, 2, 6, 6],
      [5, 6, 2, 6],
      [5, 6, 6, 2],
      [6, 2, 5, 6],
      [6, 2, 6, 5],
      [6, 5, 2, 6],
      [6, 5, 6, 2],
      [6, 6, 2, 5],
      [6, 6, 5, 2]

Then three nested loops generate the selections of operators:

 $operatoren = [];
 foreach (['+', '-', '*', '/'] as $x) {
   foreach (['+', '-', '*', '/'] as $y) {
     foreach (['+', '-', '*', '/'] as $z) {
       $operatoren[] = [$x, $y, $z];

Expressions are constructed from templates:

        $klammern = [
          '%d %s %d %s %d %s %d',
          '(%d %s %d) %s %d %s %d',
          '%d %s (%d %s %d) %s %d',
          '%d %s %d %s (%d %s %d)',
          '(%d %s %d) %s (%d %s %d)',
          '(%d %s %d %s %d) %s %d',
          '%d %s (%d %s %d %s %d)',
          '((%d %s %d) %s %d) %s %d',
          '(%d %s (%d %s %d)) %s %d',
          '%d %s ((%d %s %d) %s %d)',
          '%d %s (%d %s (%d %s %d))'

(I don't think those templates are all necessary, but hey, whatever.) Finally, another set of nested loops matches each ordering of the input numbers with each selection of operators, uses sprintf to plug the numbers and operators into each possible expression template, and uses @eval to evaluate the resulting expression to see if it has the right value:

   foreach ($zahlen as list ($a, $b, $c, $d)) {
     foreach ($operatoren as list ($x, $y, $z)) {
       foreach ($klammern as $vorlage) {
         $term = sprintf ($vorlage, $a, $x, $b, $y, $c, $z, $d);
         if (17 == @eval ("return $term;")) {
           print ("$term = 17\n");

If loving this is wrong, I don't want to be right. It certainly satisfies Larry Wall's criterion of solving the problem before your boss fires you. The same approach is possible in most reasonable languages, and some unreasonable ones, but not in Haskell, which was specifically constructed to make this approach as difficult as possible.

M. Janecke wrote up a blog article about this, in German. He says “It's not an elegant program and PHP is probably not an obvious choice for arithmetic puzzles, but I think it works.” Indeed it does. Note that the use of @eval traps the division-by-zero exceptions, but unfortunately falls foul of floating-point roundoff errors.


Thanks to everyone who discussed this with me. In addition to the people above, thanks to Stephen Tu, Smylers, Michael Malis, Kyle Littler, Jesse Chen, Darius Bacon, Michael Robert Arntzenius, and anyone else I forgot. (If I forgot you and you want me to add you to this list, please drop me a note.)

Coming up

I have enough material for at least three or four more articles about this that I hope to publish here in the coming weeks.

But the previous article on this subject ended similarly, saying

I hope to write a longer article about solvers in the next week or so.

and that was in July 2016, so don't hold your breath.

[ Addendum 20170820: the next article is ready. I hope you weren't holding your breath!  ]

[ Addendum 20170828: yet more about this ]

[Other articles in category /math] permanent link

Tue, 07 Feb 2017

How many 24 puzzles are there?

[ Note: The tables in this article are important, and look unusually crappy if you read this blog through an aggregator. The properly-formatted version on my blog may be easier to follow. ]

A few months ago I wrote about puzzles of the following type: take four digits, say 1, 2, 7, 7, and, using only +, -, ×, and ÷, combine them to make the number 24. Since then I have been accumulating more and more material about these puzzles, which will eventually appear here. But meantime here is a delightful tangent.

In the course of investigating this I wrote programs to enumerate the solutions of all possible puzzles, and these programs were always much faster than I expected at first. It appears as if there are 10,000 possible puzzles, from «0,0,0,0» through «9,9,9,9». But a moment's thought shows that there are considerably fewer, because, for example, the puzzles «7,2,7,1», «1,2,7,7», «7,7,2,1», and «2,7,7,1» are all the same puzzle. How many puzzles are there really?

A back-of-the-envelope estimate is that only about 1 in 24 puzzles is really distinct (because there are typically 24 ways to rearrange the elements of a puzzle) and so there ought to be around !!\frac{10000}{24} \approx 417!! puzzles. This is an undercount, because there are fewer duplicates of many puzzles; for example there are not 24 variations of «1,2,7,7», but only 12. The actual number of puzzles turns out to be 715, which I think is not an obvious thing to guess.

Let's write !!S(d,n)!! for the set of sequences of length !!n!! containing up to !!d!! different symbols, with the duplicates removed: when two sequences are the same except for the order of their symbols, we will consider them the same sequence.

Or more concretely, we may imagine that the symbols are sorted into nondecreasing order, so that !!S(d,n)!! is the set of nondecreasing sequences of length !!n!! of !!d!! different symbols.

Let's also write !!C(d,n)!! for the number of elements of !!S(d,n)!!.

Then !!S(10, 4)!! is the set of puzzles where input is four digits. The claim that there are !!715!! such puzzles is just that !!C(10,4) = 715!!. A tabulation of !!C(\cdot,\cdot)!! reveals that it is closely related to binomial coefficients, and indeed that $$C(d,n)=\binom{n+d-1}{d-1}.\tag{$\heartsuit$}$$

so that the surprising !!715!! is actually !!\binom{13}{9}!!. This is not hard to prove by induction, because !!C(\cdot,\cdot)!! is easily shown to obey the same recurrence as !!\binom\cdot\cdot!!: $$C(d,n) = C(d-1,n) + C(d,n-1).\tag{$\spadesuit$}$$

To see this, observe that an element of !!C(d,n)!! either begins with a zero or with some other symbol. If it begins with a zero, there are !!C(d,n-1)!! ways to choose the remaining !!n-1!! symbols in the sequence. But if it begins with one of the other !!d-1!! symbols it cannot contain any zeroes, and what we really have is a length-!!n!! sequence of the symbols !!1\ldots (d-1)!!, of which there are !!C(d-1, n)!!.

0 0 0 0 1 1 1
0 0 0 1 1 1 2
0 0 0 2 1 1 3
0 0 0 3 1 1 4
0 0 1 1 1 2 2
0 0 1 2 1 2 3
0 0 1 3 1 2 4
0 0 2 2 1 3 3
0 0 2 3 1 3 4
0 0 3 3 1 4 4
0 1 1 1 2 2 2
0 1 1 2 2 2 3
0 1 1 3 2 2 4
0 1 2 2 2 3 3
0 1 2 3 2 3 4
0 1 3 3 2 4 4
0 2 2 2 3 3 3
0 2 2 3 3 3 4
0 2 3 3 3 4 4
0 3 3 3 4 4 4

Now we can observe that !!\binom74=\binom73!! (they are both 35) so that !!C(5,3) = C(4,4)!!. We might ask if there is a combinatorial proof of this fact, consisting of a natural bijection between !!S(5,3)!! and !!S(4,4)!!. Using the relation !!(\spadesuit)!! we have:

$$ \begin{eqnarray} C(4,4) & = & C(3, 4) + & C(4,3) \\ C(5,3) & = & & C(4,3) + C(5,2) \\ \end{eqnarray}$$

so part of the bijection, at least, is clear: There are !!C(4,3)!! elements of !!S(4,4)!! that begin with a zero, and also !!C(4,3)!! elements of !!S(5, 3)!! that do not begin with a zero, so whatever the bijection is, it ought to match up these two subsets of size 20. This is perfectly straightforward; simply match up !!«0, a, b, c»!! (blue) with !!«a+1, b+1, c+1»!! (pink), as shown at right.

But finding the other half of the bijection, between !!S(3,4)!! and !!S(5,2)!!, is not so straightforward. (Both have 15 elements, but we are looking for not just any bijection but for one that respects the structure of the elements.) We could apply the recurrence again, to obtain:

$$ \begin{eqnarray} C(3,4) & = \color{darkred}{C(2, 4)} + \color{darkblue}{C(3,3)} \\ C(5,2) & = \color{darkblue}{C(4,2)} + \color{darkred}{C(5,1)} \end{eqnarray}$$

and since $$ \begin{eqnarray} \color{darkred}{C(2, 4)} & = \color{darkred}{C(5,1)} \\ \color{darkblue}{C(3,3)} & = \color{darkblue}{C(4,2)} \end{eqnarray}$$

we might expect the bijection to continue in that way, mapping !!\color{darkred}{S(2,4) \leftrightarrow S(5,1)}!! and !!\color{darkblue}{S(3,3) \leftrightarrow S(4,2)}!!. Indeed there is such a bijection, and it is very nice.

To find the bijection we will take a detour through bitstrings. There is a natural bijection between !!S(d, n)!! and the bit strings that contain !!d-1!! zeroes and !!n!! ones. Rather than explain it with pseudocode, I will give some examples, which I think will make the point clear. Consider the sequence !!«1, 1, 3, 4»!!. Suppose you are trying to communicate this sequence to a computer. It will ask you the following questions, and you should give the corresponding answers:

  • “Is the first symbol 0?” (“No”)
  • “Is the first symbol 1?” (“Yes”)
  • “Is the second symbol 1?” (“Yes”)
  • “Is the third symbol 1?” (“No”)
  • “Is the third symbol 2?” (“No”)
  • “Is the third symbol 3?” (“Yes”)
  • “Is the fourth symbol 3?” (“No”)
  • “Is the fourth symbol 4?” (“Yes”)

At each stage the computer asks about the identity of the next symbol. If the answer is “yes” the computer has learned another symbol and moves on to the next element of the sequence. If it is “no” the computer tries guessing a different symbol. The “yes” answers become ones and “no” answers become zeroes, so that the resulting bit string is 0 1 1 0 0 1 0 1.

It sometimes happens that the computer figures out all the elements of the sequence before using up its !!n+d-1!! questions; in this case we pad out the bit string with zeroes, or we can imagine that the computer asks some pointless questions to which the answer is “no”. For example, suppose the sequence is !!«0, 1, 1, 1»!!:

  • “Is the first symbol 0?” (“Yes”)
  • “Is the second symbol 0?” (“No”)
  • “Is the second symbol 1?” (“Yes”)
  • “Is the third symbol 1?” (“Yes”)
  • “Is the fourth symbol 1?” (“Yes”)

The bit string is 1 0 1 1 1 0 0 0, where the final three 0 bits are the padding.

We can reverse the process, simply taking over the role of the computer. To find the sequence that corresponds to the bit string 0 1 1 0 1 0 0 1, we ask the questions ourselves and use the bits as the answers:

  • “Is the first symbol 0?” (“No”)
  • “Is the first symbol 1?” (“Yes”)
  • “Is the second symbol 1?” (“Yes”)
  • “Is the third symbol 1?” (“No”)
  • “Is the third symbol 2?” (“Yes”)
  • “Is the fourth symbol 2?” (“No”)
  • “Is the fourth symbol 3?” (“No”)
  • “Is the fourth symbol 4?” (“Yes”)

We have recovered the sequence !!«1, 1, 2, 4»!! from the bit string 0 1 1 0 1 0 0 1.

This correspondence establishes relation !!(\heartsuit)!! in a different way from before: since there is a natural bijection between !!S(d, n)!! and the bit strings with !!d-1!! zeroes and !!n!! ones, there are certainly !!\binom{n+d-1}{d-1}!! of them as !!(\heartsuit)!! says because there are !!n+d-1!! bits and we may choose any !!d-1!! to be the zeroes.

We wanted to see why !!C(5,3) = C(4,4)!!. The detour above shows that there is a simple bijection between

!!S(5,3)!! and the bit strings with 4 zeroes and 3 ones

on one hand, and between

!!S(4,4)!! and the bit strings with 3 zeroes and 4 ones

on the other hand. And of course the bijection between the two sets of bit strings is completely obvious: just exchange the zeroes and the ones.

The table below shows the complete bijection between !!S(4,4)!! and its descriptive bit strings (on the left in blue) and between !!S(5, 3)!! and its descriptive bit strings (on the right in pink) and that the two sets of bit strings are complementary. Furthermore the top portion of the table shows that the !!S(4,3)!! subsets of the two families correspond, as they should—although the correct correspondence is the reverse of the one that was displayed earlier in the article, not the suggested !!«0, a, b, c» \leftrightarrow «a+1, b+1, c+1»!! at all. Instead, in the correct table, the initial digit of the !!S(4,4)!! entry says how many zeroes appear in the !!S(5,3)!! entry, and vice versa; then the increment to the next digit says how many ones, and so forth.

!!S(4,4)!!(bits)(complement bits)!!S(5,3)!!
0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 1 4 4 4
0 0 0 1 1 1 1 0 1 0 0 0 0 0 1 0 1 1 3 4 4
0 0 0 2 1 1 1 0 0 1 0 0 0 0 1 1 0 1 3 3 4
0 0 0 3 1 1 1 0 0 0 1 0 0 0 1 1 1 0 3 3 3
0 0 1 1 1 1 0 1 1 0 0 0 0 1 0 0 1 1 2 4 4
0 0 1 2 1 1 0 1 0 1 0 0 0 1 0 1 0 1 2 3 4
0 0 1 3 1 1 0 1 0 0 1 0 0 1 0 1 1 0 2 3 3
0 0 2 2 1 1 0 0 1 1 0 0 0 1 1 0 0 1 2 2 4
0 0 2 3 1 1 0 0 1 0 1 0 0 1 1 0 1 0 2 2 3
0 0 3 3 1 1 0 0 0 1 1 0 0 1 1 1 0 0 2 2 2
0 1 1 1 1 0 1 1 1 0 0 0 1 0 0 0 1 1 1 4 4
0 1 1 2 1 0 1 1 0 1 0 0 1 0 0 1 0 1 1 3 4
0 1 1 3 1 0 1 1 0 0 1 0 1 0 0 1 1 0 1 3 3
0 1 2 2 1 0 1 0 1 1 0 0 1 0 1 0 0 1 1 2 4
0 1 2 3 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 2 3
0 1 3 3 1 0 1 0 0 1 1 0 1 0 1 1 0 0 1 2 2
0 2 2 2 1 0 0 1 1 1 0 0 1 1 0 0 0 1 1 1 4
0 2 2 3 1 0 0 1 1 0 1 0 1 1 0 0 1 0 1 1 3
0 2 3 3 1 0 0 1 0 1 1 0 1 1 0 1 0 0 1 1 2
0 3 3 3 1 0 0 0 1 1 1 0 1 1 1 0 0 0 1 1 1
1 1 1 1 0 1 1 1 1 0 0 1 0 0 0 0 1 1 0 4 4
1 1 1 2 0 1 1 1 0 1 0 1 0 0 0 1 0 1 0 3 4
1 1 1 3 0 1 1 1 0 0 1 1 0 0 0 1 1 0 0 3 3
1 1 2 2 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 2 4
1 1 2 3 0 1 1 0 1 0 1 1 0 0 1 0 1 0 0 2 3
1 1 3 3 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 2 2
1 2 2 2 0 1 0 1 1 1 0 1 0 1 0 0 0 1 0 1 4
1 2 2 3 0 1 0 1 1 0 1 1 0 1 0 0 1 0 0 1 3
1 2 3 3 0 1 0 1 0 1 1 1 0 1 0 1 0 0 0 1 2
1 3 3 3 0 1 0 0 1 1 1 1 0 1 1 0 0 0 0 1 1
2 2 2 2 0 0 1 1 1 1 0 1 1 0 0 0 0 1 0 0 4
2 2 2 3 0 0 1 1 1 0 1 1 1 0 0 0 1 0 0 0 3
2 2 3 3 0 0 1 1 0 1 1 1 1 0 0 1 0 0 0 0 2
2 3 3 3 0 0 1 0 1 1 1 1 1 0 1 0 0 0 0 0 1
3 3 3 3 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0

Observe that since !!C(d,n) = \binom{n+d-1}{d-1} = \binom{n+d-1}{n} = C(n+1, d-1)!! we have in general that !!C(d,n) = C(n+1, d-1)!!, which may be surprising. One might have guessed that since !!C(5,3) = C(4,4)!!, the relation was !!C(d,n) = C(d+1, n-1)!! and that !!S(d,n)!! would have the same structure as !!S(d+1, n-1)!!, but it isn't so. The two arguments exchange roles. Following the same path, we can identify many similar ‘coincidences’. For example, there is a simple bijection between the original set of 715 puzzles, which was !!S(10,4)!!, and !!S(5,9)!!, the set of nondecreasing sequences of !!0\ldots 4!! of length !!9!!.

[ Thanks to Bence Kodaj for a correction. ]

[ Addendum 20170829: Conway and Guy, in The Book of Numbers, describe the same bijection, but a little differently; see their discussion of the Sweet Seventeen deck on pages 70–71. ]

[ Addendum 20171015: More about this, using Burnside's lemma. ]

[Other articles in category /math] permanent link

Thu, 15 Dec 2016

Let's decipher a thousand-year-old magic square

The Parshvanatha temple in Madhya Pradesh, India was built around 1,050 years ago. Carved at its entrance is this magic square:

The digit signs have changed in the past thousand years, but it's a quick and fun puzzle to figure out what they mean using only the information that this is, in fact, a magic square.

A solution follows. No peeking until you've tried it yourself!

There are 9 one-digit entries
and 7 two-digit entries
so we can guess that the entries are the numbers 1 through 16, as is usual, and the magic sum is 34. The appears in the same position in all the two-digit numbers, so it's the digit 1. The other digit of the numeral is , and this must be zero. If it were otherwise, it would appear on its own, as does for example the from or the from .

It is tempting to imagine that is 4. But we can see it's not so. Adding up the rightmost column, we get

+ + + =
+ 11 + + =
(10 + ) + 11 + + = 34,

so that must be an odd number. We know it isn't 1 (because is 1), and it can't be 7 or 9 because appears in the bottom row and there is no 17 or 19. So must be 3 or 5.

Now if were 3, then would be 13, and the third column would be

+ + + =
1 + + 10 + 13 = 34,

and then would be 10, which is too big. So must be 5, and this means that is 4 and is 8. ( appears only a as a single-digit numeral, which is consistent with it being 8.)

The top row has

+ + + =
+ + 1 + 14 =
+ (10 + ) + 1 + 14 = 34

so that + = 9. only appears as a single digit and we already used 8 so must be 7 or 9. But 9 is too big, so it must be 7, and then is 2.

is the only remaining unknown single-digit numeral, and we already know 7 and 8, so is 9. The leftmost column tells us that is 16, and the last two entries, and are easily discovered to be 13 and 3. The decoded square is:


I like that people look at the right-hand column and immediately see 18 + 11 + 4 + 8 but it's actually 14 + 11 + 5 + 4.

This is an extra-special magic square: not only do the ten rows, columns, and diagonals all add up to 34, so do all the four-cell subsquares, so do any four squares arranged symmetrically about the center, and so do all the broken diagonals that you get by wrapping around at the edges.

[ Addendum: It has come to my attention that the digit symbols in the magic square are not too different from the current forms of the digit symbols in the Gujarati script. ]

[ Addendum 20161217: The temple is not very close to Gujarat or to the area in which Gujarati is common, so I guess that the digit symbols in Indian languages have evolved in the past thousand years, with the Gujarati versions remaining closest to the ancient forms, or else perhaps Gujarati was spoken more widely a thousand years ago. I would be interested to hear about this from someone who knows. ]

[ Addendum 20170130: Shreevatsa R. has contributed a detailed discussion of the history of the digit symbols. ]

[Other articles in category /math] permanent link

Fri, 29 Jul 2016

Decomposing a function into its even and odd parts

As I have mentioned before, I am not a sudden-flash-of-insight person. Every once in a while it happens, but usually my thinking style is to minutely examine a large mass of examples and then gradually synthesize some conclusion about them. I am a penetrating but slow thinker. But there have been a few occasions in my life when the solution to a problem struck me suddenly out of the blue.

One such occasion was on the first day of my sophomore honors physics class in 1987. This was one of the best classes I took in my college career. It was given by Professor Stephen Nettel, and it was about resonance phenomena. I love when a course has a single overarching theme and proceeds to examine it in detail; that is all too rare. I deeply regret leaving my copy of the course notes in a restaurant in 1995.

The course was very difficult, But also very satisfying. It was also somewhat hair-raising, because of Professor Nettel's habit of saying, all through the second half “Don't worry if it doesn't seem to make any sense, it will all come together for you during the final exam.” This was not reassuring. But he was right! It did all come together during the final exam.

The exam had two sets of problems. The problems on the left side of the exam paper concerned some mechanical system, I think a rod fixed at one end and free at the other, or something like that. This set of problems asked us to calculate the resonant frequency of the rod, its rate of damping at various driving frequencies, and related matters. The right-hand problems were about an electrical system involving a resistor, capacitor, and inductor. The questions were the same, and the answers were formally identical, differing only in the details: on the left, the answers involved length, mass and stiffness of the rod, and on the right, the resistance, capacitance, and inductance of the electrical components. It was a brilliant exam, and I have never learned so much about a subject during the final exam.

Anyway, I digress. After the first class, we were assigned homework. One of the problems was

Show that every function is the sum of an even function and an odd function.

(Maybe I should explain that an even function is one which is symmetric across the !!y!!-axis; formally it is a function !!f!! for which !!f(x) = f(-x)!! for every !!x!!. For example, the function !!x^2-4!!, shown below left. An odd function is one which is symmetric under a half-turn about the origin; formally it satisfies !!f(x) = -f(-x)!! for all !!x!!. For example !!\frac{x^3}{20}!!, shown below right.)


I found this claim very surprising, and we had no idea how to solve it. Well, not quite no idea: I knew that functions could be expanded in Fourier series, as the sum of a sine series and a cosine series, and the sine part was odd while the cosine part was even. But this seemed like a bigger hammer than was required, particularly since new sophomores were not expected to know about Fourier series.

I had the privilege to be in that class with Ron Buckmire, and I remember we stood outside the class building in the autumn sunshine and discussed the problem. I might have been thinking that perhaps there was some way to replace the negative part of !!f!! with a reflected copy of the positive part to make an even function, and maybe that !!f(x) + f(-x)!! was always even, when I was hit from the blue with the solution:

$$ \begin{align} f_e(x) & = \frac{f(x) + f(-x)}2 \text{ is even},\\ f_o(x) & = \frac{f(x) - f(-x)}2 \text{ is odd, and}\\ f(x) &= f_e(x) + f_o(x) \end{align} $$

So that was that problem solved. I don't remember the other three problems in that day's homework, but I have remembered that one ever since.

But for some reason, it didn't occur to me until today to think about what those functions actually looked like. Of course, if !!f!! itself is even, then !!f_e = f!! and !!f_o = 0!!, and similarly if !!f!! is odd. But most functions are neither even nor odd.

For example, consider the function !!2^x!!, which is neither even nor odd. Then we get

$$ \begin{align} f_e(x) & = \frac{2^x + 2^{-x}}2\\ f_o(x) & = \frac{2^x - 2^{-x}}2 \end{align} $$

The graph is below left. The solid red line is !!2^x!!, and the blue and purple dotted lines are !!f_e!! and !!f_o!!. The red line is the sum of the blue and purple lines. I thought this was very interesting-looking, but a little later I realized that I had already known what these graphs would look like, because !!2^x!! is just like !!e^x!!, and for !!e^x!! the even and odd components are exactly the familiar !!\cosh!! and !!\sinh!! functions. (Below left, !!2^x!!; below right, !!e^x!!.)

I wasn't expecting polynomials to be more interesting, but they were. (Polynomials whose terms are all odd powers of !!x!!, such as !!x^{13} - 4x^5 + x!!, are always odd functions, and similarly polynomials whose terms are all even powers of !!x!! are even functions.) For example, consider !!(x-1)^2!!, which is neither even nor odd. We don't even need the !!f_e!! and !!f_o!! formulas to separate this into even and odd parts: just expand !!(x-1)^2!! as !!x^2 - 2x + 1!! and separate it into odd and even powers, !!-2x!! and !!x^2 + 1!!:

Or we could do !!\frac{(x-1)^3}3!! similarly, expanding it as !!\frac{x^3}3 - x^2 + x -\frac13!! and separating this into !!-x^2 -\frac13!! and !!\frac{x^3}3 + x!!:

I love looking at these and seeing how the even blue line and the odd purple line conspire together to make whatever red line I want.

I kept wanting to try familiar simple functions, like !!\frac1x!!, but many of these are either even or odd, and so are uninteresting for this application. But you can make an even or an odd function into a neither-even-nor-odd function just by translating it horizontally, which you do by replacing !!x!! with !!x-c!!. So the next function I tried was !!\frac1{x+1}!!, which is the translation of !!\frac 1x!!. Here I got a surprise. I knew that !!\frac1{x+1}!! was undefined at !!x=-1!!, so I graphed it only for !!x>-1!!. But the even component is !!\frac12\left(\frac1{1+x}+\frac1{1-x}\right)!!, which is undefined at both !!x=-1!! and at !!x=+1!!. Similarly the odd component is undefined at two points. So the !!f = f_o + f_e!! formula does not work quite correctly, failing to produce the correct value at !!x=1!!, even though !!f!! is defined there. In general, if !!f!! is undefined at some !!x=c!!, then the decomposition into even and odd components fails at !!x=-c!! as well. The limit $$\lim_{x\to -c} f(x) = \lim_{x\to -c} \left(f_o(x) + f_e(x)\right)$$ does hold, however. The graph below shows the decomposition of !!\frac1{x+1}!!.

Vertical translations are uninteresting: they leave !!f_o!! unchanged and translate !!f_e!! by the same amount, as you can verify algebraically or just by thinking about it.

Following the same strategy I tried a cosine wave. The evenness of the cosine function is one of its principal properties, so I translated it and used !!\cos (x+1)!!. The graph below is actually for !!5\cos(x+1)!! to prevent the details from being too compressed:

This reminded me of the time I was fourteen and graphed !!\sin x + \cos x!! and was surprised to see that it was another perfect sinusoid. But I realized that there was a simple way to understand this. I already knew that !!\cos(x + y) = \sin x\cos y + \sin y \cos x!!. If you take !!y=\frac\pi4!! and multiply the whole thing by !!\sqrt 2!!, you get $$\sqrt2\cos\left(x + \frac\pi4\right) = \sqrt2\sin x\cos\frac\pi4 + \sqrt2\cos x\sin\frac\pi4 = \sin x + \cos x$$ so that !!\sin x + \cos x!! is just a shifted, scaled cosine curve. The decomposition of !!\cos(x+1)!! is even simpler because you can work forward instead of backward and find that !!\cos(x+1) = \sin x\cos 1 + \cos x \sin 1!!, and the first term is odd while the second term is even, so that !!\cos(x+1)!! decomposes as a sum of an even and an odd sinusoid as you see in the graph above.

Finally, I tried a Poisson distribution, which is highly asymmetric. The formula for the Poisson distribution is !!\frac{\lambda^xe^\lambda}{x!}!!, for some constant !!\lambda!!. The !!x! !! in the denominator is only defined for non-negative integer !!x!!, but you can extend it to fractional and negative !!x!! in the usual way by using !!\Gamma(x+1)!! instead, where !!\Gamma!! is the Gamma function. The !!\Gamma!! function is undefined at zero and negative integers, but fortunately what we need here is the reciprocal gamma function !!\frac1{\Gamma(x)}!!, which is perfectly well-behaved. The results are spectacular. The graph below has !!\lambda = 0.8!!.

The part of this with !!x\ge 0!! is the most interesting to me, because the Poisson distribution has a very distinctive shape, and once again I like seeing the blue and purple !!\Gamma!! functions working together to make it. I think it's just great how the red line goes gently to zero as !!x!! increases, even though the even and the odd components are going wild. (!!x! !! increases rapidly with !!x!!, so the reciprocal !!\Gamma!! function goes rapidly to zero. But the even and odd components also have a !!\frac1{\Gamma(-x)}!! part, and this is what dominates the blue and purple lines when !!x >4!!.)

On the !!x\lt 0!! side it has no meaning for me, and it's just wiggly lines. It hadn't occurred to me before that you could extend the Poisson distribution function to negative !!x!!, and I still can't imagine what it could mean, but I suppose why not. Probably some statistician could explain to me what the Poisson distribution is about when !!x<0!!.

You can also consider the function !!\sqrt x!!, which breaks down completely, because either !!\sqrt x!! or !!\sqrt{-x}!! is undefined except when !!x=0!!. So the claim that every function is the sum of an even and an odd function fails here too. Except perhaps not! You could probably consider the extension of the square root function to the complex plane, and take one of its branches, and I suppose it works out just fine. The geometric interpretation of evenness and oddness are very different, of course, and you can't really draw the graphs unless you have four-dimensional vision.

I have no particular point to make, except maybe that math is fun, even elementary math (or perhaps especially elementary math) and it's fun to see how it works out.

The beautiful graphs in this article were made with Desmos. I had dreaded having to illustrate my article with graphs from Gnuplot (ugh) or Wolfram|α (double ugh) and was thrilled to find such a handsome alternative.

[ Addendum: I've just discovered that in Desmos you can include a parameter in the functions that it graphs, and attach the parameter to a slider. So for example you can arrange to have it display !!(x+k)^3!! or !!e^{-(x+k)^2}!!, with the value of !!k!! controlled by the slider, and have the graph move left and right on the plane as you adjust the slider, with its even and odd parts changing in real time to match. ]

[ For example, check out travelling Gaussians or varying sinusoid. ]

[Other articles in category /math] permanent link

Tue, 12 Jul 2016

A simple but difficult arithmetic puzzle

Lately my kids have been interested in puzzles of this type: You are given a sequence of four digits, say 1,2,3,4, and your job is to combine them with ordinary arithmetic operations (+, -, ×, and ÷) in any order to make a target number, typically 24. For example, with 1,2,3,4, you can go with $$((1+2)+3)×4 = 24$$ or with $$4×((2×3)×1) = 24.$$

We were stumped trying to make 6,6,5,2 total 24, so I hacked up a solver; then we felt a little foolish when we saw the solutions, because it is not that hard. But in the course of testing the solver, I found the most challenging puzzle of this type that I've ever seen. It is:

Given 6,6,5,2, make 17.

There are no underhanded tricks. For example, you may not concatenate 2 and 5 to make 25; you may not say !!6÷6=1!! and !!5+2=7!! and concatenate 1 and 7 to make !!17!!; you may not interpret the 17 as a base 12 numeral, etc.

I hope to write a longer article about solvers in the next week or so.

[ Addendum 20170305: The next week or so, ha ha. Anyway, here it is. ]

[Other articles in category /math] permanent link

Tue, 19 Apr 2016

The sage and the seven horses

A classic puzzle of mathematics goes like this:

A father dies and his will states that his elder daughter should receive half his horses, the son should receive one-quarter of the horses, and the younger daughter should receive one-eighth of the horses. Unfortunately, there are seven horses. The siblings are arguing about how to divide the seven horses when a passing sage hears them. The siblings beg the sage for help. The sage donates his own horse to the estate, which now has eight. It is now easy to portion out the half, quarter, and eighth shares, and having done so, the sage's horse is unaccounted for. The three heirs return the surplus horse to the sage, who rides off, leaving the matter settled fairly.

(The puzzle is, what just happened?)

It's not hard to come up with variations on this. For example, picking three fractions at random, suppose the will says that the eldest child receives half the horses, the middle child receives one-fifth, and the youngest receives one-seventh. But the estate has only 59 horses and an argument ensues. All that is required for the sage to solve the problem is to lend the estate eleven horses. There are now 70, and after taking out the three bequests, !!70 - 35 - 14 - 10 = 11!! horses remain and the estate settles its debt to the sage.

But here's a variation I've never seen before. This time there are 13 horses and the will says that the three children should receive shares of !!\frac12, \frac13,!! and !!\frac14!!. respectively. Now the problem seems impossible, because !!\frac12 + \frac13 + \frac14 \gt 1!!. But the sage is equal to the challenge! She leaps into the saddle of one of the horses and rides out of sight before the astonished heirs can react. After a day of searching the heirs write off the lost horse and proceed with executing the will. There are now only 12 horses, and the eldest takes half, or six, while the middle sibling takes one-third, or 4. The youngest heir should get three, but only two remain. She has just opened her mouth to complain at her unfair treatment when the sage rides up from nowhere and hands her the reins to her last horse.

[Other articles in category /math] permanent link

Fri, 18 Dec 2015

Math.SE report 2015-08

I only posted three answers in August, but two of them were interesting.

  • In why this !!\sigma\pi\sigma^{-1}!! keeps apearing in my group theory book? (cycle decomposition) the querent asked about the “conjugation” operation that keeps cropping up in group theory. Why is it important? I sympathize with this; it wasn't adequately explained when I took group theory, and I had to figure it out a long time later. Unfortunately I don't think I picked the right example to explain it, so I am going to try again now.

    Consider the eight symmetries of the square. They are of five types:

    1. Rotation clockwise or counterclockwise by 90°.
    2. Rotation by 180°.
    3. Horizontal or vertical reflection
    4. Diagonal reflection
    5. The trivial (identity) symmetry

    What is meant when I say that a horizontal and a vertical reflection are of the same ‘type’? Informally, it is that the horizontal reflection looks just like the vertical reflection, if you turn your head ninety degrees. We can formalize this by observing that if we rotate the square 90°, then give it a horizontal flip, then rotate it back, the effect is exactly to give it a vertical flip. In notation, we might represent the horizontal flip by !!H!!, the vertical flip by !!V!!, the clockwise rotation by !!\rho!!, and the counterclockwise rotation by !!\rho^{-1}!!; then we have

    $$ \rho H \rho^{-1} = V$$

    and similarly

    $$ \rho V \rho^{-1} = H.$$

    Vertical flips do not look like diagonal flips—the diagonal flip leaves two of the corners in the same place, and the vertical flip does not—and indeed there is no analogous formula with !!H!! replaced with one of the diagonal flips. However, if !!D_1!! and !!D_2!! are the two diagonal flips, then we do have

    $$ \rho D_1 \rho^{-1} = D_2.$$

    In general, When !!a!! and !!b!! are two symmetries, and there is some symmetry !!x!! for which

    $$xax^{-1} = b$$

    we say that !!a!! is conjugate to !!b!!. One can show that conjugacy is an equivalence relation, which means that the symmetries of any object can be divided into separate “conjugacy classes” such that two symmetries are conjugate if and only if they are in the same class. For the square, the conjugacy classes are the five I listed earlier.

    This conjugacy thing is important for telling when two symmetries are group-theoretically “the same”, and have the same group-theoretic properties. For example, the fact that the horizontal and vertical flips move all four vertices, while the diagonal flips do not. Another example is that a horizontal flip is self-inverse (if you do it again, it cancels itself out), but a 90° rotation is not (you have to do it four times before it cancels out.) But the horizontal flip shares all its properties with the vertical flip, because it is the same if you just turn your head.

    Identifying this sameness makes certain kinds of arguments much simpler. For example, in counting squares, I wanted to count the number of ways of coloring the faces of a cube, and instead of dealing with the 24 symmetries of the cube, I only needed to deal with their 5 conjugacy classes.

    The example I gave in my answer was maybe less perspicuous. I considered the symmetries of a sphere, and talked about how two rotations of the sphere by 17° are conjugate, regardless of what axis one rotates around. I thought of the square at the end, and threw it in, but I wish I had started with it.

  • How to convert a decimal to a fraction easily? was the month's big winner. OP wanted to know how to take a decimal like !!0.3760683761!! and discover that it can be written as !!\frac{44}{117}!!. The right answer to this is of course to use continued fraction theory, but I did not want to write a long treatise on continued fractions, so I stripped down the theory to obtain an algorithm that is slower, but much easier to understand.

    The algorithm is just binary search, but with a twist. If you are looking for a fraction for !!x!!, and you know !!\frac ab < x < \frac cd!!, then you construct the mediant !!\frac{a+c}{b+d}!! and compare it with !!x!!. This gives you a smaller interval in which to search for !!x!!, and the reason you use the mediant instead of using !!\frac12\left(\frac ab + \frac cd\right)!! as usual is that if you use the mediant you are guaranteed to exactly nail all the best rational approximations of !!x!!. This is the algorithm I described a few years ago in your age as a fraction, again; there the binary search proceeds down the branches of the Stern-Brocot tree to find a fraction close to !!0.368!!.

I did ask a question this month: I was looking for a simpler version of the dogbone space construction. The dogbone space is a very peculiar counterexample of general topology, originally constructed by R.H. Bing. I mentioned it here in 2007, and said, at the time:

[The paper] is on my desk, but I have not read this yet, and I may never.

I did try to read it, but I did not try very hard, and I did not understand it. So my question this month was if there was a simpler example of the same type. I did not receive an answer, just a followup comment that no, there is no such example.

[Other articles in category /math/se] permanent link

Sun, 16 Aug 2015

Math.SE report 2015-07

My overall SE posting volume was down this month, and not only did I post relatively few interesting items, I've already written a whole article about the most interesting one. So this will be a short report.

  • I already wrote up Building a box from smaller boxes on the blog here. But maybe I have a couple of extra remarks. First, the other guy's proposed solution is awful. It's long and complicated, which is forgivable if it had answered the question, but it doesn't. And the key point is “blah blah blah therefore code a solver which visits all configurations of the search space”. Well heck, if this post had just been one sentence that ended with “code a solver which visits all configurations of the search space” I would not have any complaints about that.

    As an undergraduate I once gave a talk on this topic. One of my examples was the problem of packing 31 dominoes into a chessboard from which two squares have been deleted. There is a simple combinatorial argument why this is impossible if the two deleted squares are the same color, say if they are opposite corners: each domino must cover one square of each color. But if you don't take time to think about the combinatorial argument you could waste a lot of time on computer search learning that there is no solution in that case, and completely miss the deeper understanding that it brings you. So this has been on my mind for a long time.

  • I wrote a few posts this month where I thought I gave good hints. In How to scale an unit vector !!u!! in such way that !!a u\cdot u=1!! where !!a!! is a scalar I think I did a good job identifying the original author's confusion; he was conflating his original unit vector !!u!! and the scaled, leading him to write !!au\cdot u=1!!. This is sure to lead to confusion. So I led him to the point of writing !!a(bv)\cdot(bv)=1!! and let him take it from there. The other proposed solution is much more rote and mechanical. (“Divide this by that…”)

    In Find numbers !!\overline{abcd}!! so that !!\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}!! the OP got stuck partway through and I specifically addressed the stuckness; other people solved the problem from the beginning. I think that's the way to go, if the original proposal was never going to work, especially if you stop and say why it was never going to work, but this time OP's original suggestion was perfectly good and she just didn't know how to get to the next step. By the way, the notation !!\overline{abcd}!! here means the number !!1000a+100b+10c+d!!.

    In Help finding the limit of this series !!\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots!! it would have been really easy to say “use the formula” or to analyze the series de novo, but I think I almost hit the nail on the head here: it's just like !!1+\frac12 + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots!!, which I bet OP already knows, except a little different. But I pointed out the wrong difference: I observed that the first sequence is one-fourth the second one (which it is) but it would have been simpler to observe that it's just the second one without the !!1+\frac12!!. I had to review it just now to give the simpler explanation, but I sure wish I'd thought of it at the time. Nobody else pointed it out either. Best of all, would have been to mention both methods. If you can notice both of them you can solve the problem without the advance knowledge of the value of !!1+\frac12+\frac14+\ldots!!, because you have !!4S = 1+\frac12 + S!! and then solve for !!S!!.

    In Visualization of Rhombus made of Radii and Chords it seemed that OP just needed to see a diagram (“I really really don't see how two circles can form a rhombus?”), so I drew one.

[Other articles in category /math/se] permanent link

Tue, 28 Jul 2015

Another ounce of theory

A few months ago I wrote an article here called an ounce of theory is worth a pound of search and I have a nice followup.

When I went looking for that article I couldn't find it, because I thought it was about how an ounce of search is worth a pound of theory, and that I was writing a counterexample. I am quite surprised to discover that that I have several times discussed how a little theory can replace a lot of searching, and not vice versa, but perhaps that is because the search is my default.

Anyway, the question came up on math StackExchange today:

John has 77 boxes each having dimensions 3×3×1. Is it possible for John to build one big box with dimensions 7×9×11?

OP opined no, but had no argument. The first answer that appeared was somewhat elaborate and outlined a computer search strategy which claimed to reduce the search space to only 14,553 items. (I think the analysis is wrong, but I agree that the search space is not too large.)

I almost wrote the search program. I have a program around that is something like what would be needed, although it is optimized to deal with a few oddly-shaped tiles instead of many similar tiles, and would need some work. Fortunately, I paused to think a little before diving in to the programming.

For there is an easy answer. Suppose John solved the problem. Look at just one of the 7×11 faces of the big box. It is a 7×11 rectangle that is completely filled by 1×3 and 3×3 rectangles. But 7×11 is not a multiple of 3. So there can be no solution.

Now how did I think of this? It was a very geometric line of reasoning. I imagined a 7×11×9 carton and imagined putting the small boxes into the carton. There can be no leftover space; every one of the 693 cells must be filled. So in particular, we must fill up the bottom 7×11 layer. I started considering how to pack the bottommost 7×11×1 slice with just the bottom parts of the small boxes and quickly realized it couldn't be done; there is always an empty cell left over somewhere, usually in the corner. The argument about considering just one face of the large box came later; I decided it was clearer than what I actually came up with.

I think this is a nice example of the Pólya strategy “solve a simpler problem” from How to Solve It, but I was not thinking of that specifically when I came up with the solution.

For a more interesting problem of the same sort, suppose you have six 2×2x1 slabs and three extra 1×1×1 cubes. Can you pack the nine pieces into a 3×3x3 box?

[Other articles in category /math] permanent link

Sat, 18 Jul 2015

Math.SE report 2015-04

[ Notice: I originally published this report at the wrong URL. I moved it so that I could publish the June 2015 report at that URL instead. If you're seeing this for the second time, you might want to read the June article instead. ]

A lot of the stuff I've written in the past couple of years has been on Mathematics StackExchange. Some of it is pretty mundane, but some is interesting. I thought I might have a little meta-discussion in the blog and see how that goes. These are the noteworthy posts I made in April 2015.

  • Languages and their relation : help is pretty mundane, but interesting for one reason: OP was confused about a statement in a textbook, and provided a reference, which OPs don't always do. The text used the symbol !!\subset_\ne!!. OP had interpreted it as meaning !!\not\subseteq!!, but I think what was meant was !!\subsetneq!!.

    I dug up a copy of the text and groveled over it looking for the explanation of !!\subset_\ne!!, which is not standard. There was none that I could find. The book even had a section with a glossary of notation, which didn't mention !!\subset_\ne!!. Math professors can be assholes sometimes.

  • Is there an operation that takes !!a^b!! and !!a^c!!, and returns !!a^{bc}!! is more interesting. First off, why is this even a reasonable question? Why should there be such an operation? But note that there is an operation that takes !!a^b!! and !!a^c!! and returns !!a^{b+c}!!, namely, multiplication, so it's plausible that the operation that OP wants might also exist.

    But it's easy to see that there is no operation that takes !!a^b!! and !!a^c!! and returns !!a^{bc}!!: just observe that although !!4^2=2^4!!, the putative operation (call it !!f!!) should take !!f(2^4, 2^4)!! and yield !!2^{4\cdot4} = 2^{16} = 65536!!, but it should also take !!f(4^2, 4^2)!! and yield !!4^{2\cdot2} = 2^4 = 256!!. So the operation is not well-defined. And you can take this even further: !!2^4!! can be written as !!e^{4\log 2}!!, so !!f!! should also take !!f(e^{2\log 4}, e^{2\log 4})!! and yield !!e^{4(\log 4)^2} \approx 2180.37!!.

    They key point is that the representation of a number, or even an integer, in the form !!a^b!! is not unique. (Jargon: "exponentiation is not injective".) You can raise !!a^b!!, but having done so you cannot look at the result and know what !!a!! and !!b!! were, which is what !!f!! needs to do.

    But if !!f!! can't do it, how can multiplication do it when it multiplies !!a^b!! and !!a^c!! and gets !!a^{b+c}!!? Does it somehow know what !!a!! is? No, it turns out that it doesn't need !!a!! in this case. There is something magical going on there, ultimately related to the fact that if some quantity is increasing by a factor of !!x!! every !!t!! units of time, then there is some !!t_2!! for which it is exactly doubling every !!t_2!! units of time. Because of this there is a marvelous group homomophism $$\log : \langle \Bbb R^+, \times\rangle \to \langle \Bbb R ,+\rangle$$ which can change multiplication into addition without knowing what the base numbers are.

    In that thread I had a brief argument with someone who thinks that operators apply to expressions rather than to numbers. Well, you can say this, but it makes the question trivial: you can certainly have an "operator" that takes expressions !!a^b!! and !!a^c!! and yields the expression !!a^{bc}!!. You just can't expect to apply it to numbers, such as !!16!! and !!16!!, because those numbers are not expressions in the form !!a^b!!. I remembered the argument going on longer than it did; I originally ended this paragraph with a lament that I wasted more than two comments on this guy, but looking at the record, it seems that I didn't. Good work, Mr. Dominus.

  • how 1/0.5 is equal to 2? wants a simple explanation. Very likely OP is a primary school student. The question reminds me of a similar question, asking why the long division algorithm is the way it is. Each of these is a failure of education to explain what division is actually doing. The long division answer is that long division is an optimization for repeated subtraction; to divide !!450\div 3!! you want to know how many shares of three cookies each you can get from !!450!! cookies. Long division is simply a notation for keeping track of removing !!100!! shares, leaving !!150!! cookies, then !!5\cdot 10!! further shares, leaving none.

    In this question there was a similar answer. !!1/0.5!! is !!2!! because if you have one cookie, and want to give each kid a share of !!0.5!! cookies, you can get out two shares. Simple enough.

    I like division examples that involve giving cookies to kids, because cookies are easy to focus on, and because the motivation for equal shares is intuitively understood by everyone who has kids, or who has been one.

    There is a general pedagogical principle that an ounce of examples are worth a pound of theory. My answer here is a good example of that. When you explain the theory, you're telling the student how to understand it. When you give an example, though, if it's the right example, the student can't help but understand it, and when they do they'll understand it in their own way, which is better than if you told them how.

  • How to read a cycle graph? is interesting because hapless OP is asking for an explanation of a particularly strange diagram from Wikipedia. I'm familiar with the eccentric Wikipedian who drew this, and I was glad that I was around to say "The other stuff in this diagram is nonstandard stuff that the somewhat eccentric author made up. Don't worry if it's not clear; this author is notorious for that."

  • In Expected number of die tosses to get something less than 5, OP calculated as follows: The first die roll is a winner !!\frac23!! of the time. The second roll is the first winner !!\frac13\cdot\frac23!! of the time. The third roll is the first winner !!\frac13\cdot\frac13\cdot\frac23!! of the time. Summing the series !!\sum_n \frac23\left(\frac13\right)^nn!! we eventually obtain the answer, !!\frac32!!. The accepted answer does it this way also.

    But there's a much easier way to solve this problem. What we really want to know is: how many rolls before we expect to have seen one good one? And the answer is: the expected number of winners per die roll is !!\frac23!!, expectations are additive, so the expected number of winners per !!n!! die rolls is !!\frac23n!!, and so we need !!n=\frac32!! rolls to expect one winner. Problem solved!

    I first discovered this when I was around fifteen, and wrote about it here a few years ago.

    As I've mentioned before, this is one of the best things about mathematics: not that it works, but that you can do it by whatever method that occurs to you and you get the same answer. This is where mathematics pedagogy goes wrong most often: it proscribes that you must get the answer by method X, rather than that you must get the answer by hook or by crook. If the student uses method Y, and it works (and if it is correct) that should be worth full credit.

    Bad instructors always say "Well, we need to test to see if the student knows method X." No, we should be testing to see if the student can solve problem P. If we are testing for method X, that is a failure of the test or of the curriculum. Because if method X is useful, it is useful because for some problems, it is the only method that works. It is the instructor's job to find one of these problems and put it on the test. If there is no such problem, then X is useless and it is the instructor's job to omit it from the curriculum. If Y always works, but X is faster, it is the instructor's job to explain this, and then to assign a problem for the test where Y would take more time than is available.

    I see now I wrote the same thing in 2006. It bears repeating. I also said it again a couple of years ago on itself in reply to a similar comment by Brian Scott:

    If the goal is to teach students how to write proofs by induction, the instructor should damned well come up with problems for which induction is the best approach. And if even then a student comes up with a different approach, the instructor should be pleased. ... The directions should not begin [with "prove by induction"]. I consider it a failure on the part of the instructor if he or she has to specify a technique in order to give students practice in applying it.

[Other articles in category /math/se] permanent link

Fri, 03 Jul 2015

The annoying boxes puzzle: solution
I presented this logic puzzle on Wednesday:

There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "exactly one of the labels is true". The green box is labelled "the treasure is in this box."

Can you figure out which box contains the treasure?

It's not too late to try to solve this before reading on. If you want, you can submit your answer here:

The treasure is in the red box
The treasure is in the green box
There is not enough information to determine the answer
Something else:


There were 506 total responses up to Fri Jul 3 11:09:52 2015 UTC; I kept only the first response from each IP address, leaving 451. I read all the "something else" submissions and where it seemed clear I recoded them as votes for "red", for "not enough information", or as spam. (Several people had the right answer but submitted "other" so they could explain themselves.) There was also one post attempted to attack my (nonexistent) SQL database. Sorry, Charlie; I'm not as stupid as I look.

	 66.52%  300 red
	 25.72   116 not-enough-info
	  3.55    16 green
	  2.00     9 other
	  1.55     7 spam
	  0.44     2 red-with-qualification
	  0.22     1 attack

	100.00   451 TOTAL
One-quarter of respondents got the right answer, that there is not enough information given to solve the problem, Two-thirds of respondents said the treasure was in the red box. This is wrong. The treasure is in the green box.


Let me show you. I stated:

There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "exactly one of the labels is true". The green box is labelled "the treasure is in this box."

The labels are as I said. Everything I told you was literally true.

The treasure is definitely not in the red box.

No, it is actually in the green box.

(It's hard to see, but one of the items in the green box is the gold and diamond ring made in Vienna by my great-grandfather, which is unquestionably a real treasure.)

So if you said the treasure must be in the red box, you were simply mistaken. If you had a logical argument why the treasure had to be in the red box, your argument was fallacious, and you should pause and try to figure out what was wrong with it.

I will discuss it in detail below.


The treasure is undeniably in the green box. However, correct answer to the puzzle is "no, you cannot figure out which box contains the treasure". There is not enough information given. (Notice that the question was not “Where is the treasure?” but “Can you figure out…?”)

(Fallacious) Argument A

Many people erroneously conclude that the treasure is in the red box, using reasoning something like the following:

  1. Suppose the red label is true. Then exactly one label is true, and since the red label is true, the green label is false. Since it says that the treasure is in the green box, the treasure must really be in the red box.
  2. Now suppose that the red label is false. Then the green label must also be false. So again, the treasure is in the red box.
  3. Since both cases lead to the conclusion that the treasure is in the red box, that must be where it is.

What's wrong with argument A?

Here are some responses people commonly have when I tell them that argument A is fallacious:

"If the treasure is in the green box, the red label is lying."

Not quite, but argument A explicitly considers the possibility that the red label was false, so what's the problem?

"If the treasure is in the green box, the red label is inconsistent."

It could be. Nothing in the puzzle statement ruled this out. But actually it's not inconsistent, it's just irrelevant.

"If the treasure is in the green box, the red label is meaningless."

Nonsense. The meaning is plain: it says “exactly one of these labels is true”, and the meaning is that exactly one of the labels is true. Anyone presenting argument A must have understood the label to mean that, and it is incoherent to understand it that way and then to turn around and say that it is meaningless! (I discussed this point in more detail in 2007.)

"But the treasure could have been in the red box."

True! But it is not, as you can see in the pictures. The puzzle does not give enough information to solve the problem. If you said that there was not enough information, then congratulations, you have the right answer. The answer produced by argument A is incontestably wrong, since it asserts that the treasure is in the red box, when it is not.

"The conditions supplied by the puzzle statement are inconsistent."

They certainly are not. Inconsistent systems do not have models, and in particular cannot exist in the real world. The photographs above demonstrate a real-world model that satisfies every condition posed by the puzzle, and so proves that it is consistent.

"But that's not fair! You could have made up any random garbage at all, and then told me afterwards that you had been lying."

Had I done that, it would have been an unfair puzzle. For example, suppose I opened the boxes at the end to reveal that there was no treasure at all. That would have directly contradicted my assertion that "One [box] contains a treasure". That would have been cheating, and I would deserve a kick in the ass.

But I did not do that. As the photograph shows, the boxes, their colors, their labels, and the disposition of the treasure are all exactly as I said. I did not make up a lie to trick you; I described a real situation, and asked whether people they could diagnose the location of the treasure.

(Two respondents accused me of making up lies. One said:

There is no treasure. Both labels are lying. Look at those boxes. Do you really think someone put a treasure in one of them just for this logic puzzle?
What can I say? I did put a treasure in a box just for this logic puzzle. Some of us just have higher standards.)

"But what about the labels?"

Indeed! What about the labels?

The labels are worthless

The labels are red herrings; the provide no information. Consider the following version of the puzzle:

There are two boxes on a table, one red and one green. One contains a treasure.

Which box contains the treasure?

Obviously, the problem cannot be solved from the information given.

Now consider this version:

There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "gcoadd atniy fnck z fbi c rirpx hrfyrom". The green box is labelled "ofurb rz bzbsgtuuocxl ckddwdfiwzjwe ydtd."

Which box contains the treasure?

One is similarly at a loss here.

(By the way, people who said one label was meaningless: this is what a meaningless label looks like.)

There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "exactly one of the labels is true". The green box is labelled "the treasure is in this box."

But then the janitor happens by. "Don't be confused by those labels," he says. "They were stuck on there by the previous owner of the boxes, who was an illiterate shoemaker who only spoke Serbian. I think he cut them out of a magazine because he liked the frilly borders."

Which box contains the treasure?

The point being that in the absence of additional information, there is no reason to believe that the labels give any information about the contents of the boxes, or about labels, or about anything at all. This should not come as a surprise to anyone. It is true not just in annoying puzzles, but in the world in general. A box labeled “fresh figs” might contain fresh figs, or spoiled figs, or angry hornets, or nothing at all.

Why doesn't every logic puzzle fall afoul of this problem?

I said as part of the puzzle conditions that there was a treasure in one box. For a fair puzzle, I am required to tell the truth about the puzzle conditions. Otherwise I'm just being a jerk.

Typically the truth or falsity of the labels is part of the puzzle conditions. Here's a typical example, which I took from Raymond Smullyan's What is the name of this book? (problem 67a):

… She had the following inscriptions put on the caskets:
Portia explained to the suitor that of the three statements, at most one was true.

Which casket should the suitor choose [to find the portrait]?

Notice that the problem condition gives the suitor a certification about the truth of the labels, on which he may rely. In the quotation above, the certification is in boldface.

A well-constructed puzzle will always contain such a certification, something like “one label is true and one is false” or “on this island, each person always lies, or always tells the truth”. I went to What is the Name of this Book? to get the example above, and found more than I had bargained for: problem 70 is exactly the annoying boxes problem! Smullyan says:

Good heavens, I can take any number of caskets that I please and put an object in one of them and then write any inscriptions at all on the lids; these sentences won't convey any information whatsoever.
(Page 65)

Had I known ahead of time that Smullyan had treated the exact same topic with the exact same example, I doubt I would have written this post at all.

But why is this so surprising?

I don't know.

Final notes

16 people correctly said that the treasure was in the green box. This has to be counted as a lucky guess, unacceptable as a solution to a logic puzzle.

One respondent referred me to a similar post on lesswrong.

I did warn you all that the puzzle was annoying.

I started writing this post in October 2007, and then it sat on the shelf until I got around to finding and photographing the boxes. A triumph of procrastination!

[ Addendum 20150911: Steven Mazie has written a blog article about this topic, A Logic Puzzle That Teaches a Life Lesson. ]

[Other articles in category /math/logic] permanent link

Wed, 01 Jul 2015

The annoying boxes puzzle
Here is a logic puzzle. I will present the solution on Friday.

There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "exactly one of the labels is true". The green box is labelled "the treasure is in this box."

Can you figure out which box contains the treasure?

The treasure is in the red box
The treasure is in the green box
There is not enough information to determine the answer
Something else:
Starting on 2015-07-03, the solution will be here.

[Other articles in category /math/logic] permanent link

Thu, 18 Jun 2015

Math.SE report 2015-05

A lot of the stuff I've written in the past couple of years has been on math.StackExchange. Some of it is pretty mundane, but some is interesting. My summary of April's interesting posts was well-received, so here are the noteworthy posts I made in May 2015.

  • What matrix transforms !!(1,0)!! into !!(2,6)!! and tranforms !!(0,1)!! into !!(4,8)!!? was a little funny because the answer is $$\begin{pmatrix}2 & 4 \\ 6 & 8 \end{pmatrix}$$ and yeah, it works exactly like it appears to, there's no trick. But if I just told the guy that, he might feel unnecessarily foolish. I gave him a method for solving the problem and figured that when he saw what answer he came up with, he might learn the thing that the exercise was designed to teach him.

  • Is a “network topology'” a topological space? is interesting because several people showed up right away to say no, it is an abuse of terminology, and that network topology really has nothing to do with mathematical topology. Most of those comments have since been deleted. My answer was essentially: it is topological, because just as in mathematical topology you care about which computers are connected to which, and not about where any of the computers actually are.

    Nobody constructing a token ring network thinks that it has to be a geometrically circular ring. No, it only has to be a topologically circular ring. A square is fine; so is a triangle; topologically they are equivalent, both in networking and in mathematics. The wires can cross, as long as they don't connect at the crossings. But if you use something that isn't topologically a ring, like say a line or a star or a tree, the network doesn't work.

    The term “topological” is a little funny. “Topos” means “place” (like in “topography” or “toponym”) but in topology you don't care about places.

  • Is there a standard term for this generalization of the Euler totient function? was asked by me. I don't include all my answers in these posts, but I think maybe I should have a policy of including all my questions. This one concerned a simple concept from number theory which I was surprised had no name: I wanted !!\phi_k(n)!! to be the number of integers !!m!! that are no larger than !!n!! for which !!\gcd(m,n) = k!!. For !!k=1!! this is the famous Euler totient function, written !!\varphi(n)!!.

    But then I realized that the reason it has no name is that it's simply !!\phi_k(n) = \varphi\left(\frac n k\right)!! so there's no need for a name or a special notation.

    As often happens, I found the answer myself shortly after I asked the question. I wonder if the reason for this is that my time to come up with the answer is Poisson-distributed. Then if I set a time threshold for how long I'll work on the problem before asking about it, I am likely to find the answer to almost any question that exceeds the threshold shortly after I exceed the threshold. But if I set the threshold higher, this would still be true, so there is no way to win this particular game. Good feature of this theory: I am off the hook for asking questions I could have answered myself. Bad feature: no real empirical support.

  • how many ways can you divide 24 people into groups of two? displays a few oddities, and I think I didn't understand what was going on at that time. OP has calculated the first few special cases:

    1:1 2:1 3:3 4:3 5:12 6:15

    which I think means that there is one way to divide 2 people into groups of 2, 3 ways to divide 4 people, and 15 ways to divide 6 people. This is all correct! But what could the 1:1, 3:3, 5:12 terms mean? You simply can't divide 5 people into groups of 2. Well, maybe OP was counting the extra odd person left over as a sort of group on their own? Then odd values would be correct; I didn't appreciate this at the time.

    But having calculated 6 special cases correctly, why can't OP calculate the seventh? Perhaps they were using brute force: the next value is 48, hard to brute-force correctly if you don't have a enough experience with combinatorics.

    I tried to suggest a general strategy: look at special cases, and not by brute force, but try to analyze them so that you can come up with a method for solving them. The method is unnecessary for the small cases, where brute force enumeration suffices, but you can use the brute force enumeration to check that the method is working. And then for the larger cases, where brute force is impractical, you use your method.

    It seems that OP couldn't understand my method, and when they tried to apply it, got wrong answers. Oh well, you can lead a horse to water, etc.

    The other pathology here is:

    I think I did what you said and I got 1.585times 10 to the 21

    for the !!n=24!! case. The correct answer is $$23\cdot21\cdot19\cdot17\cdot15\cdot13\cdot11\cdot9\cdot7\cdot5\cdot3\cdot1 = 316234143225 \approx 3.16\cdot 10^{11}.$$ OP didn't explain how they got !!1.585\cdot10^{21}!! so there's not much hope of correcting their weird error.

    This is someone who probably could have been helped in person, but on the Internet it's hopeless. Their problems are Internet communication problems.

  • Lambda calculus typing isn't especially noteworthy, but I wrote a fairly detailed explanation of the algorithm that Haskell or SML uses to find the type of an expression, and that might be interesting to someone.

  • I think Special representation of a number is the standout post of the month. OP speculates that, among numbers of the form !!pq+rs!! (where !!p,q,r,s!! are prime), the choice of !!p,q,r,s!! is unique. That is, the mapping !!\langle p,q,r,s\rangle \to pq+rs!! is reversible.

    I was able to guess that this was not the case within a couple of minutes, replied pretty much immediately:

    I would bet money against this representation being unique.

    I was sure that a simple computer search would find counterexamples. In fact, the smallest is !!11\cdot13 + 19\cdot 29 = 11\cdot 43 + 13\cdot 17 = 694!! which is small enough that you could find it without the computer if you are patient.

    The obvious lesson to learn from this is that many elementary conjectures of this type can be easily disproved by a trivial computer search, and I frequently wonder why more amateur mathematicians don't learn enough computer programming to investigate this sort of thing. (I wrote recently on the topic of An ounce of theory is worth a pound of search , and this is an interesting counterpoint to that.)

    But the most interesting thing here is how I was able to instantly guess the answer. I explained in some detail in the post. But the basic line of reasoning goes like this.

    Additive properties of the primes are always distributed more or less at random unless there is some obvious reason why they can't be. For example, let !!p!! be prime and consider !!2p+1!!. This must have exactly one of the three forms !!3n-1, 3n,!! or !!3n+1!! for some integer !!n!!. It obviously has the form !!3n+1!! almost never (the only exception is !!p=3!!). But of the other two forms there is no obvious reason to prefer one over the other, and indeed of the primes up to 10,000, 611 are of the type !!3n!! and and 616 are of the type !!3n-1!!.

    So we should expect the value !!pq+rs!! to be distributed more or less randomly over the set of outputs, because there's no obvious reason why it couldn't be, except for simple stuff, like that it's obviously almost always even.

    So we are throwing a bunch of balls at random into bins, and the claim is that no bin should contain more than one ball. For that to happen, there must be vastly more bins than balls. But the bins are numbers, and primes are not at all uncommon among numbers, so the number of bins isn't vastly larger, and there ought to be at least some collisions.

    In fact, a more careful analysis, which I wrote up on the site, shows that the number of balls is vastly larger—to have them be roughly the same, you would need primes to be roughly as common as perfect squares, but they are far more abundant than that—so as you take larger and larger primes, the number of collisions increases enormously and it's easy to find twenty or more quadruples of primes that all map to the same result. But I was able to predict this after a couple of minutes of thought, from completely elementary considerations, so I think it's a good example of Lower Mathematics at work.

    This is an example of a fairly common pathology of questions: OP makes a conjecture that !!X!! never occurs or that there are no examples with property !!X!!, when actually !!X!! almost always occurs or every example has property !!X!!.

    I don't know what causes this. Rik Signes speculates that it's just wishful thinking: OP is doing some project where it would be useful to have !!pq+rs!! be unique, so posts in hope that someone will tell them that it is. But there was nothing more to it than baseless hope. Rik might be right.

[ Addendum 20150619: A previous version of this article included the delightful typo “mathemativicians”. ]

[Other articles in category /math/se] permanent link

Sun, 14 Jun 2015

Math.SE report 2015-06

[ This page originally held the report for April 2015, which has moved. It now contains the report for June 2015. ]

  • Is “smarter than” a transitive relationship? concerns a hypothetical "is smarter than" relation with the following paradoxical-seeming property:

    most X's are smarter than most Y's, but most Y's are such that it is not the case that most X's are smarter than it.

    That is, if !!\mathsf Mx.\Phi(x)!! means that most !!x!! have property !!\Phi!!, then we want both $$\mathsf Mx.\mathsf My.S(x, y)$$ and also $$\mathsf My.\mathsf Mx.\lnot S(x, y).$$

    “Most” is a little funny here: what does it mean? But we can pin it down by supposing that there are an infinite number of !!x!!es and !!y!!s, and agreeing that most !!x!! have property !!P!! if there are only a finite number of exceptions. For example, everyone should agree that most positive integers are larger than 7 and that most prime numbers are odd. The jargon word here is that we are saying that a subset contains “most of” the elements of a larger set if it is cofinite.

    There is a model of this property, and OP reports that they asked the prof if this was because the "smarter than" relation !!S(x,y)!! could be antitransitive, so that one might have !!S(x,y), S(y,z)!! but also !!S(z,x)!!. The prof said no, it's not because of that, but the OP want so argue that it's that anyway. But no, it's not because of that; there is a model that uses a perfectly simple transitive relation, and the nontransitive thing nothing but a distraction. (The model maps the !!x!!es and !!y!!s onto numbers, and says !!x!! is smarter than !!y!! if its number is bigger.) Despite this OP couldn't give up the idea that the model exists because of intransitive relations. It's funny how sometimes people get stuck on one idea and can't let go of it.

  • How to generate a random number between 1 and 10 with a six-sided die? was a lot of fun and attracted several very good answers. Top-scoring is Jack D'Aurizio's, which proposes a completely straightforward method: roll once to generate a bit that selects !!N=0!! or !!N=5!!, and then roll again until you get !!M\ne 6!!, and the result is !!N+M!!.

    But several other answers were suggested, including two by me, one explaining the general technique of arithmetic coding, which I'll probably refer back to in the future when people ask similar questions. Don't miss NovaDenizen's clever simplification of arithmetic coding, which I want to think about more, or D'Aurizio's suggestion that if you threw the die into a V-shaped trough, it would land with one edge pointing up and thus select a random number from 1 to 12 in a single throw.

    Interesting question: Is there an easy-to-remember mapping from edges to numbers from 1–12? Each edge is naturally identified by a pair of distinct integers from 1–6 that do not add to 7.

  • The oddly-phrased Category theory with objects as logical expressions over !!{\vee,\wedge,\neg}!! and morphisms as? asks if there is a standard way to turn logical expressions into a category, which there is: you put an arrow from !!A\to B!! for each proof that !!A!! implies !!B!!; composition of arrows is concatenation of proofs, and identity arrows are empty proofs. The categorial product, coproduct, and exponential then correspond to !!\land, \lor, !! and !!\to!!.

    This got me thinking though. Proofs are properly not lists, they are trees, so it's not entirely clear what the concatenation operation is. For example, suppose proof !!X!! concludes !!A!! at its root and proof !!Y!! assumes !!A!! in more than one leaf. When you concatenate !!X!! and !!Y!! do you join all the !!A!!'s, or what? I really need to study this more. Maybe the Lambek and Scott book talks about it, or maybe the Goldblatt Topoi book, which I actually own. I somehow skipped most of the Cartesian closed category stuff, which is an oversight I ought to correct.

  • In Why is the Ramsey`s theorem a generalization of the Pigeonhole principle I gave what I thought was a terrific answer, showing how Ramsey's graph theorem and the pigeonhole principle are both special cases of Ramsey's hypergraph theorem. This might be my favorite answer of the month. It got several upvotes, but OP preferred a different answer, with fewer details.

    There was a thread a while back about theorems which are generalizations of other theorems in non-obvious ways. I pointed out the Yoneda lemma was a generalization of Cayley's theorem from group theory. I see that nobody mentioned the Ramsey hypergraph theorem being a generalization of the pigeonhole principle, but it's closed now, so it's too late to add it.

  • In Why does the Deduction Theorem use Union? I explained that the English word and actually has multiple meanings. I know I've seen this discussed in elementary logic texts but I don't remember where.

  • Finally, Which is the largest power of natural number that can be evaluated by computers? asks if it's possible for a computer to calculate !!7^{120000000000}!!. The answer is yes, but it's nontrivial and you need to use some tricks. You have to use the multiplying-by-squaring trick, and for the squarings you probably want to do the multiplication with DFT. OP was dissatistifed with the answer, and seemed to have some axe to grind, but I couldn't figure out what it was.

[Other articles in category /math/se] permanent link

Fri, 20 Mar 2015

Rectangles with equal area and perimeter

Wednesday while my 10-year-old daughter Katara was doing her math homework, she observed with pleasure that a !!6×3!! rectangle has a perimeter of 18 units and also an area of 18 square units. I mentioned that there was an infinite family of such rectangles, and, after a small amount of tinkering, that the only other such rectangle with integer sides is a !!4×4!! square, so in a sense she had found the single interesting example. She was very interested in how I knew this, and I promised to show her how to figure it out once she finished her homework. She didn't finish before bedtime, so we came back to it the following evening.

This is just one of many examples of how she has way too much homework, and how it interferes with her education.

She had already remarked that she knew how to write an equation expressing the condition she wanted, so I asked her to do that; she wrote $$(L×W) = ([L+W]×2).$$ I remember being her age and using all different shapes of parentheses too. I suggested that she should solve the equation for !!W!!, getting !!W!! on one side and a bunch of stuff involving !!L!! on the other, but she wasn't sure how to do it, so I offered suggestions while she moved the symbols around, eventually obtaining $$W = 2L\div (L-2).$$ I would have written it as a fraction, but getting the right answer is important, and using the same notation I would use is much less so, so I didn't say anything.

I asked her to plug in !!L=3!! and observe that !!W=6!! popped right out, and then similarly that !!L=6!! yields !!W=3!!, and then I asked her to try the other example she knew. Then I suggested that she see what !!L=5!! did: it gives !!W=\frac{10}3!!, This was new, so she checked it by calculating the area and the perimeter, both !!\frac{50}3!!. She was very excited by this time. As I have mentioned earlier, algebra is magical in its ability to mechanically yield answers to all sorts of questions. Even after thirty years I find it astonishing and delightful. You set up the equations, push the symbols around, and all sorts of stuff pops out like magic. Calculus is somehow much less astonishing; the machinery is all explicit. But how does algebra work? I've been thinking about this on and off for a long time and I'm still not sure.

At that point I took over because I didn't think I would be able to guide her through the next part of the problem without a demonstration; I wanted to graph the function !!W=2L\div(L-2)!! and she does not have much experience with that. She put in the five points we already knew, which already lie on a nice little curve, and then she asked an incisive question: does it level off, or does it keep going down, or what? We discussed what happens when !!L!! gets close to 2; then !!W!! shoots up to infinity. And when !!L!! gets big, say a million, you can see from the algebra that !!W!! is a hair more than 2. So I drew in the asymptotes on the hyperbola.

Katara is not yet familiar with hyperbolas. (She has known about parabolas since she was tiny. I have a very fond memory of visiting Portland with her when she was almost two, and we entered Holladay park, which has fountains that squirt out of the ground. Seeing the water arching up before her, she cried delightedly “parabolas!”)

Once you know how the graph behaves, it is a simple matter to see that there are no integer solutions other than !!\langle 3,6\rangle, \langle 4,4\rangle,!! and !!\langle6,3\rangle!!. We know that !!L=5!! does not work. For !!L>6!! the value of !!W!! is always strictly between !!2!! and !!3!!. For !!L=2!! there is no value of !!W!! that works at all. For !!0\lt L\lt 2!! the formula says that !!W!! is negative, on the other branch of the hyperbola, which is a perfectly good numerical solution (for example, !!L=1, W=-2!!) but makes no sense as the width of a rectangle. So it was a good lesson about how mathematical modeling sometimes introduces solutions that are wrong, and how you have to translate the solutions back to the original problem to see if they make sense.

[ Addendum 20150330: Thanks to Steve Hastings for his plot of the hyperbola, which is in the public domain. ]

[Other articles in category /math] permanent link

Thu, 19 Mar 2015

An ounce of theory is worth a pound of search

The computer is really awesome at doing quick searches for numbers with weird properties, and people with an amateur interest in recreational mathematics would do well to learn some simple programming. People appear on math.stackexchange quite often with questions about tic-tac-toe, but there are only 5,478 total positions, so any question you want to ask can be instantaneously answered by an exhaustive search. An amateur showed up last fall asking “Is it true that no prime larger than 241 can be made by either adding or subtracting 2 coprime numbers made up out of the prime factors 2,3, and 5?” and, once you dig through the jargon, the question is easily answered by the computer, which quickly finds many counterexamples, such as !!162+625=787!! and !!2^{19}+3^4=524369!!.

But sometimes the search appears too large to be practical, and then you need to apply theory. Sometimes you can deploy a lot of theory and solve the problem completely, avoiding the search. But theory is expensive, and not always available. A hybrid approach often works, which uses a tiny amount of theory to restrict the search space to the point where the search is easy.

One of these I wrote up on this blog back in 2006:

A number !!n!! is excellent if it has an even number of digits, and if when you chop it into a front half !!a!! and a back half !!b!!, you have !!b^2 - a^2 = n!!. For example, !!48!! is excellent, because !!8^2 - 4^2 = 48!!, and !!3468!! is excellent, because !!68^2 - 34^2 = 4624 - 1156 = 3468!!.

The programmer who gave me thie problem had tried a brute-force search over all numbers, but to find all 10-digit excellent numbers, this required an infeasible search of 9,000,000,000 candidates. With the application of a tiny amount of algebra, one finds that !!a(10^k+a) = b^2+b!! and it's not hard to quickly test candidates for !!a!! to see if !!a(10^k+a)!! has this form and if so to find the corresponding value of !!b!!. (Details are in the other post.) This reduces the search space for 10-digit excellent numbers from 9,000,000,000 candidates to 90,000, which could be done in under a minute even with last-century technology, and is pretty nearly instantaneous on modern equipment.

But anyway, the real point of this note is to discuss a different problem entirely. A recreational mathematician on stackexchange wanted to find distinct integers !!a,b,c,d!! for which !!a^2+b^2, b^2+c^2, c^2+d^2, !! and !!d^2+a^2!! were all perfect squares. You can search over all possible quadruples of numbers, but this takes a long time. The querent indicated later that he had tried such a search but lost patience before it yielded anything.

Instead, observe that if !!a^2+b^2!! is a perfect square then !!a!! and !!b!! are the legs of a right triangle with integer sides; they are terms in what is known as a Pythagorean triple. The prototypical example is !!3^2 + 4^2 = 5^2!!, and !!\langle 3,4,5\rangle!! is the Pythagorean triple. (The querent was quite aware that he was asking for Pythagorean triples, and mentioned them specifically.)

Here's the key point: It has been known since ancient times that if !!\langle a,b,c\rangle!! is a Pythagorean triple, then there exist integers !!m!! and !!n!! such that: $$\begin{align} \require{align} a & = n^2-m^2 \\ b & = 2mn \\ c & = n^2 + m^2 \end{align}$$

So you don't have to search for Pythagorean triples; you can just generate them with no searching:

    for my $m (1 .. 200) {
      for my $n ($m+1 .. 200) {
        my $a = $n*$n-$m*$m;
        my $b = 2 * $n * $m;
        $trip{$a}{$b} = 1;
        $trip{$b}{$a} = 1;

This builds a hash table, %trip, with two important properties:

  1. $trip{$a} is a sub-table whose keys are all the numbers that can form a triple with !!a!!. For example, $trip{20} is a hash with three keys: 21, 48, and 99, because !!20^2+21^2 = 29^2, 20^2+48^2= 52^2, !! and !!20^2 + 99^2 = 101^2!!, but 20 is not a participant in any other triples.

  2. $trip{$a}{$b} is true if and only if !!a^2+b^2!! is a perfect square, and false otherwise.

The table has only around 40,000 entries. Having constructed it, we now search it:

    for my $a (keys %trip) {
      for my $b (keys %{$trip{$a}}) {
        for my $c (keys %{$trip{$b}}) {
          next if $c == $a;
          for my $d (keys %{$trip{$c}}) {
            next if $d == $b;
            print "$a $b $c $d\n" if $trip{$d}{$a};

The outer loop runs over each !!a!! that is known to be a member of a Pythagorean triple. (Actually the !!m,n!! formulas show that every number bigger than 2 is a member of some triple, but we may as well skip the ones that are only in triples we didn't tabulate.) Then the next loop runs over every !!b!! that can possibly form a triple with !!a!!; that is, every !!b!! for which !!a^2+b^2!! is a perfect square. We don't have to search for them; we have them tabulated ahead of time. Then for each such !!b!! (and there aren't very many) we run over every !!c!! that forms a triple with !!b!!, and again there is no searching and very few candidates. Then then similarly !!d!!, and if the !!d!! we try forms a triple with !!a!!, we have a winner.

The next if $c == $a and next if $d == $b tests are to rule out trivial solutions like !!a=c=3, b=d=4!!, which the querent wasn't interested in anyway. We don't have to test for equality of any of ther other pairs because no number can form a Pythagorean triple with itself (because !!\sqrt2!! is irrational).

This runs in less than a second on so-so hardware and produces 11 solutions:

    3472  7296  10400 2175
    4312  23520 12008 465
    6512  9984  800   6375
    12312 666   1288  8415
    14592 6944  4350  20800
    16830 2576  1332  24624
    19968 13024 12750 1600
    25500 26048 39936 3200
    30192 6175  2400  9856
    41600 29184 13888 8700
    47040 8624  930   24016

Only five of these are really different. For example, the last one is the same as the second, with every element multiplied by 2; the third, seventh, and eighth are similarly the same. In general if !!\langle a,b,c,d\rangle!! is a solution, so is !!\langle ka, kb,kc,kd\rangle!! for any !!k!!. A slightly improved version would require that the four numbers not have any common factor greater than 1; there are few enough solutions that the cost of this test would be completely negligible.

The only other thing wrong with the program is that it produces each solution 8 times; if !!\langle a,b,c,d\rangle!! is a solution, then so are !!\langle b,c,d,a\rangle, \langle d,c,b,a\rangle,!! and so on. This is easily fixed with a little post-filtering; pipe the output through

    perl -nle '$k = join " ",  sort { $a <=> $b } split; print unless $seen{$k}++ '

or something of that sort.

The corresponding run with !!m!! and !!n!! up to 2,000 instead of only 200 takes 5 minutes and finds 445 solutions, of which 101 are distinct, including !!\langle 3614220, 618192, 2080820, 574461\rangle!!. It would take a very long time to find this with a naïve search.

[ For a much larger and more complex example of the same sort of thing, see When do !!n!! and !!2n!! have the same digits?. I took a seemingly-intractable problem and analyzed it mathematically. I used considerably more than an ounce of theory in this case, and while the theory was not enough to solve the problem, it was enough to reduce the pool of candates to the point that a computer search was feasible. ]

[ Addendum 20150728: Another example ]

[Other articles in category /math] permanent link

Sat, 22 Nov 2014

Within this instrument, resides the Universe

When opportunity permits, I have been trying to teach my ten-year-old daughter Katara rudiments of algebra and group theory. Last night I posed this problem:

Mary and Sue are sisters. Today, Mary is three times as old as Sue; in two years, she will be twice as old as Sue. How old are they now?

I have tried to teach Katara that these problems have several phases. In the first phase you translate the problem into algebra, and then in the second phase you manipulate the symbols, almost mechanically, until the answer pops out as if by magic.

There is a third phase, which is pedagogically and practically essential. This is to check that the solution is correct by translating the results back to the context of the original problem. It's surprising how often teachers neglect this step; it is as if a magician who had made a rabbit vanish from behind a screen then forgot to take away the screen to show the audience that the rabbit had vanished.

Katara set up the equations, not as I would have done, but using four unknowns, to represent the two ages today and the two ages in the future:

$$\begin{align} MT & = 3ST \\ MY & = 2SY \\ \end{align} $$

(!!MT!! here is the name of a single variable, not a product of !!M!! and !!T!!; the others should be understood similarly.)

“Good so far,” I said, “but you have four unknowns and only two equations. You need to find two more relationships between the unknowns.” She thought a bit and then wrote down the other two relations:

$$\begin{align} MY & = MT + 2 \\ SY & = ST + 2 \end{align} $$

I would have written two equations in two unknowns:

$$\begin{align} M_T & = 3S_T\\ M_T+2 & = 2(S_T + 2) \end{align} $$

but one of the best things about mathematics is that there are many ways to solve each problem, and no method is privileged above any other except perhaps for reasons of practicality. Katara's translation is different from what I would have done, and it requires more work in phase 2, but it is correct, and I am not going to tell her to do it my way. The method works both ways; this is one of its best features. If the problem can be solved by thinking of it as a problem in two unknowns, then it can also be solved by thinking of it as a problem in four or in eleven unknowns. You need to find more relationships, but they must exist and they can be found.

Katara may eventually want to learn a technically easier way to do it, but to teach that right now would be what programmers call a premature optimization. If her formulation of the problem requires more symbol manipulation than what I would have done, that is all right; she needs practice manipulating the symbols anyway.

She went ahead with the manipulations, reducing the system of four equations to three, then two and then one, solving the one equation to find the value of the single remaining unknown, and then substituting that value back to find the other unknowns. One nice thing about these simple problems is that when the solution is correct you can see it at a glance: Mary is six years old and Sue is two, and in two years they will be eight and four. Katara loves picking values for the unknowns ahead of time, writing down a random set of relations among those values, and then working the method and seeing the correct answer pop out. I remember being endlessly delighted by almost the same thing when I was a little older than her. In The Dying Earth Jack Vance writes of a wizard who travels to an alternate universe to learn from the master “the secret of renewed youth, many spells of the ancients, and a strange abstract lore that Pandelume termed ‘Mathematics.’”

“I find herein a wonderful beauty,” he told Pandelume. “This is no science, this is art, where equations fall away to elements like resolving chords, and where always prevails a symmetry either explicit or multiplex, but always of a crystalline serenity.”

After Katara had solved this problem, I asked if she was game for something a little weird, and she said she was, so I asked her:

Mary and Sue are sisters. Today, Mary is three times as old as Sue; in two years, they will be the same age. How old are they now?

“WHAAAAAT?” she said. She has a good number sense, and immediately saw that this was a strange set of conditions. (If they aren't the same age now, how can they be the same age in two years?) She asked me what would happen. I said (truthfully) that I wasn't sure, and suggested she work through it to find out. So she set up the equations as before and worked out the solution, which is obvious once you see it: Both girls are zero years old today, and zero is three times as old as zero. Katara was thrilled and delighted, and shared her discovery with her mother and her aunt.

There are some powerful lessons here. One is that the method works even when the conditions seem to make no sense; often the results pop out just the same, and can sometimes make sense of problems that seem ill-posed or impossible. Once you have set up the equations, you can just push the symbols around and the answer will emerge, like a familiar building approached through a fog.

But another lesson, only hinted at so far, is that mathematics has its own way of understanding things, and this is not always the way that humans understand them. Goethe famously said that whatever you say to mathematicians, they immediately translate it into their own language and then it is something different; I think this is exactly what he meant.

In this case it is not too much of a stretch to agree that Mary is three times as old as Sue when they are both zero years old. But in the future I plan to give Katara a problem that requires Mary and Sue to have negative ages—say that Mary is twice as old as Sue today, but in three years Sue will be twice as old—to demonstrate that the answer that pops out may not be a reasonable one, or that the original translation into mathematics can lose essential features of the original problem. The solution that says that !!M_T=-2, S_T=-1 !! is mathematically irreproachable, and if the original problem had been posed as “Find two numbers such that…” it would be perfectly correct. But translated back to the original context of a problem that asks about the ages of two sisters, the solution is unacceptable. This is the point of the joke about the spherical cow.

[Other articles in category /math] permanent link

Wed, 23 Jul 2014

When do n and 2n have the same digits?

[This article was published last month on the math.stackexchange blog, which seems to have died young, despite many earnest-sounding promises beforehand from people who claimed they would contribute material. I am repatriating it here.]

A recent question on math.stackexchange asks for the smallest positive integer !!A!! for which the number !!2A!! has the same decimal digits in some other order.

Math geeks may immediately realize that !!142857!! has this property, because it is the first 6 digits of the decimal expansion of !!\frac 17!!, and the cyclic behavior of the decimal expansion of !!\frac n7!! is well-known. But is this the minimal solution? It is not. Brute-force enumeration of the solutions quickly reveals that there are 12 solutions of 6 digits each, all permutations of !!142857!!, and that larger solutions, such as 1025874 and 1257489 seem to follow a similar pattern. What is happening here?

Stuck in Dallas-Fort Worth airport one weekend, I did some work on the problem, and although I wasn't able to solve it completely, I made significant progress. I found a method that allows one to hand-calculate that there is no solution with fewer than six digits, and to enumerate all the solutions with 6 digits, including the minimal one. I found an explanation for the surprising behavior that solutions tend to be permutations of one another. The short form of the explanation is that there are fairly strict conditions on which sets of digits can appear in a solution of the problem. But once the set of digits is chosen, the conditions on that order of the digits in the solution are fairly lax.

So one typically sees, not only in base 10 but in other bases, that the solutions to this problem fall into a few classes that are all permutations of one another; this is exactly what happens in base 10 where all the 6-digit solutions are permutations of !!124578!!. As the number of digits is allowed to increase, the strict first set of conditions relaxes a little, and other digit groups appear as solutions.


The property of interest, !!P_R(A)!!, is that the numbers !!A!! and !!B=2A!! have exactly the same base-!!R!! digits. We would like to find numbers !!A!! having property !!P_R!! for various !!R!!, and we are most interested in !!R=10!!. Suppose !!A!! is an !!n!!-digit numeral having property !!P_R!!; let the (base-!!R!!) digits of !!A!! be !!a_{n-1}\ldots a_1a_0!! and similarly the digits of !!B = 2A!! are !!b_{n-1}\ldots b_1b_0!!. The reader is encouraged to keep in mind the simple example of !!R=8, n=4, A=\mathtt{1042}, B=\mathtt{2104}!! which we will bring up from time to time.

Since the digits of !!B!! and !!A!! are the same, in a different order, we may say that !!b_i = a_{P(i)}!! for some permutation !!P!!. In general !!P!! might have more than one cycle, but we will suppose that !!P!! is a single cycle. All the following discussion of !!P!! will apply to the individual cycles of !!P!! in the case that !!P!! is a product of two or more cycles. For our example of !!a=\mathtt{1042}, b=\mathtt{2104}!!, we have !!P = (0\,1\,2\,3)!! in cycle notation. We won't need to worry about the details of !!P!!, except to note that !!i, P(i), P(P(i)), \ldots, P^{n-1}(i)!! completely exhaust the indices !!0. \ldots n-1!!, and that !!P^n(i) = i!! because !!P!! is an !!n!!-cycle.

Conditions on the set of digits in a solution

For each !!i!! we have $$a_{P(i)} = b_{i} \equiv 2a_{i} + c_i\pmod R $$ where the ‘carry bit’ !!c_i!! is either 0 or 1 and depends on whether there was a carry when doubling !!a_{i-1}!!. (When !!i=0!! we are in the rightmost position and there is never a carry, so !!c_0= 0!!.) We can then write:

$$\begin{align} a_{P(P(i))} &= 2a_{P(i)} + c_{P(i)} \\ &= 2(2a_{i} + c_i) + c_{P(i)} &&= 4a_i + 2c_i + c_{P(i)}\\ a_{P(P(P(i)))} &= 2(4a_i + 2c_i + c_{P(P(i)})) + c_{P(i)} &&= 8a_i + 4c_i + 2c_{P(i)} + c_{P(P(i))}\\ &&&\vdots\\ a_{P^n(i)} &&&= 2^na_i + v \end{align} $$

all equations taken !!\bmod R!!. But since !!P!! is an !!n!!-cycle, !!P^n(i) = i!!, so we have $$a_i \equiv 2^na_i + v\pmod R$$ or equivalently $$\big(2^n-1\big)a_i + v \equiv 0\pmod R\tag{$\star$}$$ where !!v\in\{0,\ldots 2^n-1\}!! depends only on the values of the carry bits !!c_i!!—the !!c_i!! are precisely the binary digits of !!v!!.

Specifying a particular value of !!a_0!! and !!v!! that satisfy this equation completely determines all the !!a_i!!. For example, !!a_0 = 2, v = \color{darkblue}{0010}_2 = 2!! is a solution when !!R=8, n=4!! because !!\bigl(2^4-1\bigr)\cdot2 + 2\equiv 0\pmod 8!!, and this solution allows us to compute

$$\def\db#1{\color{darkblue}{#1}}\begin{align} a_0&&&=2\\ a_{P(0)} &= 2a_0 &+ \db0 &= 4\\ a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 0 \\ a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 1\\ \hline a_{P^4(0)} &= 2a_{P^3(0)} &+ \db0 &= 2\\ \end{align}$$

where the carry bits !!c_i = \langle 0,0,1,0\rangle!! are visible in the third column, and all the sums are taken !!\pmod 8!!. Note that !!a_{P^n(0)} = a_0!! as promised. This derivation of the entire set of !!a_i!! from a single one plus a choice of !!v!! is crucial, so let's see one more example. Let's consider !!R=10, n=3!!. Then we want to choose !!a_0!! and !!v!! so that !!\left(2^3-1\right)a_0 + v \equiv 0\pmod{10}!! where !!v\in\{0\ldots 7\}!!. One possible solution is !!a_0 = 5, v=\color{darkblue}{101}_2 = 5!!. Then we can derive the other !!a_i!! as follows:

$$\begin{align} a_0&&&=5\\ a_{P(0)} &= 2a_0 &+ \db1 &= 1\\ a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 2 \\\hline a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 5\\ \end{align}$$

And again we have !!a_{P^n(0)}= a_0!! as required.

Since the bits of !!v!! are used cyclically, not every pair of !!\langle a_0, v\rangle!! will yield a different solution. Rotating the bits of !!v!! and pairing them with different choices of !!a_0!! will yield the same cycle of digits starting from a different place. In the first example above, we had !!a_0 = 2, v = 0010_2 = 2!!. If we were to take !!a_0 = 4, v = 0100_2 = 4!! (which also solves !!(\star)!!) we would get the same cycle of values of the !! a\_i !! but starting from !!4!! instead of from !!2!!, and similarly if we take !!a_0=0, v = 1000_2 = 8!! or !!a_0 = 1, v = 0001_2!!. So we can narrow down the solution set of !!(\star)!! by considering only the so-called necklaces of !!v!! rather than all !!2^n!! possible values. Two values of !!v!! are considered equivalent as necklaces if one is a rotation of the other. When a set of !!v!!-values are equivalent as necklaces, we need only consider one of them; the others will give the same cyclic sequence of digits, but starting in a different place. For !!n=4!!, for example, the necklaces are !!0000, 0001, 0011, 0101, 0111, !! and !!1111!!; the sequences !!0110, 1100,!! and !!1001!! being equivalent to !!0011!!, and so on.


Let us take !!R=9, n=3!!, so we want to find 3-digit numerals with property !!P_9!!. According to !!(\star)!! we need !!7a_i + v \equiv 0\pmod{9}!! where !!v\in\{0\ldots 7\}!!. There are 9 possible values for !!a_i!!; for each one there is at most one possible value of !!v!! that makes the sum zero:

$$\begin{array}{rrr} a_i & 7a_i & v \\ \hline 0 & 0 & 0 \\ 1 & 7 & 2 \\ 2 & 14 & 4 \\ 3 & 21 & 6 \\ 4 & 28 & \\ 5 & 35 & 1 \\ 6 & 42 & 3 \\ 7 & 49 & 5 \\ 8 & 56 & 7 \\ \end{array} $$

(For !!a_i=4!! there is no solution.) We may disregard the non-necklace values of !!v!!, as these will give us solutions that are the same as those given by necklace values of !!v!!. The necklaces are:

$$\begin{array}{rl} 000 & 0 \\ 001 & 1 \\ 011 & 3 \\ 111 & 7 \end{array}$$

so we may disregard the solutions exacpt when !!v=0,1,3,7!!. Calculating the digit sequences from these four values of !!v!! and the corresponding !!a_i!! we find:

$$\begin{array}{ccl} a_0 & v & \text{digits} \\ \hline 0 & 0 & 000 \\ 5 & 1 & 512 \\ 6 & 3 & 637 \\ 8 & 7 & 888 \ \end{array} $$

(In the second line, for example, we have !!v=1 = 001_2!!, so !!1 = 2\cdot 5 + 0; 2 = 1\cdot 2 + 0;!! and !!5 = 2\cdot 2 + 1!!.)

Any number !!A!! of three digits, for which !!2A!! contains exactly the same three digits, in base 9, must therefore consist of exactly the digits !!000, 125, 367,!! or !!888!!.

A warning

All the foregoing assumes that the permutation !!P!! is a single cycle. In general, it may not be. Suppose we did an analysis like that above for !!R=10, n=5!! and found that there was no possible digit set, other than the trivial set 00000, that satisfied the governing equation !!(2^5-1)a_0 + v\equiv 0\pmod{10}!!. This would not completely rule out a base-10 solution with 5 digits, because the analysis only rules out a cyclic set of digits. There could still be a solution where !!P!! was a product of a !!2!! and a !!3!!-cycle, or a product of still smaller cycles.

Something like this occurs, for example, in the !!n=4, R=8!! case. Solving the governing equation !!(2^5-1)a_0 + v \equiv 0\pmod 8!! yields only four possible digit cycles, namely !!\{0,1,2,4\}, \{1,3,6,4\}, \{2,5,2,5\}!!, and !!\{3,7,6,5\}!!. But there are several additional solutions: !!2500_8\cdot 2 = 5200_8, 2750_8\cdot 2 = 5720_8, !! and !!2775_8\cdot 2 = 5772_8!!. These correspond to permutations !!P!! with more than one cycle. In the case of !!2750_8!!, for example, !!P!! exchanges the !!5!! and the !!2!!, and leaves the !!0!! and the !!7!! fixed.

For this reason we cannot rule out the possibility of an !!n!!-digit solution without first considering all smaller !!n!!.

The Large Equals Odd rule

When !!R!! is even there is a simple condition we can use to rule out certain sets of digits from being single-cycle solutions. Recall that !!A=a_{n-1}\ldots a_0!! and !!B=b_{n-1}\ldots b_0!!. Let us agree that a digit !!d!! is large if !!d\ge \frac R2!! and small otherwise. That is, !!d!! is large if, upon doubling, it causes a carry into the next column to the left.

Since !!b_i =(2a_i + c_i)\bmod R!!, where the !!c_i!! are carry bits, we see that, except for !!b_0!!, the digit !!b_i!! is odd precisely when there is a carry from the next column to the right, which occurs precisely when !!a_{i-1}!! is large. Thus the number of odd digits among !!b_1,\ldots b_{n-1}!! is equal to the number of large digits among !!a_1,\ldots a_{n-2}!!. This leaves the digits !!b_0!! and !!a_{n-1}!! uncounted. But !!b_0!! is never odd, since there is never a carry in the rightmost position, and !!a_{n-1}!! is always small (since otherwise !!B!! would have !!n+1!! digits, which is not allowed). So the number of large digits in !!A!! is exactly equal to the number of odd digits in !!B!!. And since !!A!! and !!B!! have exactly the same digits, the number of large digits in !!A!! is equal to the number of odd digits in !!A!!. Observe that this is the case for our running example !!1042_8!!: there is one odd digit and one large digit (the 4).

When !!R!! is odd the analogous condition is somewhat more complicated, but since the main case of interest is !!R=10!!, we have the useful rule that:

For !!R!! even, the number of odd digits in any solution !!A!! is equal to the number of large digits in !!A!!.

Conditions on the order of digits in a solution

We have determined, using the above method, that the digits !!\{5,1,2\}!! might form a base-9 numeral with property !!P_9!!. Now we would like to arrange them into a base-9 numeral that actually does have that property. Again let us write !!A = a_2a_1a_0!! and !!B=b_2b_1b_0!!, with !!B=2A!!. Note that if !!a_i = 1!!, then !!b_i = 3!! (if there was a carry from the next column to the right) or !!b_i=2!! (if there was no carry), but since !!b_i\in{5,1,2}!!, we must have !!b_i = 2!! and therefore !!a_{i-1}!! must be small, since there is no carry into position !!i!!. But since !!a_{i-1}!! is also one of !!\{5,1,2\}!!, and it cannot also be !!1!!, it must be !!2!!. This shows that the 1, unless it appears in the rightmost position, must be to the left of the !!2!!; it cannot be to the left of the !!5!!. Similarly, if !!a_i = 2!! then !!b_i = 5!!, because !!4!! is impossible, so the !!2!! must be to the left of a large digit, which must be the !!5!!. Similar reasoning produces no constraint on the position of the !!5!!; it could be to the left of a small digit (in which case it doubles to !!1!!) or a large digit (in which case it doubles to !!2!!). We can summarize these findings as follows:

$$\begin{array}{cl} \text{digit} & \text{to the left of} \\ \hline 1 & 1, 2, \text{end} \\ 2 & 5 \\ 5 & 1,2,5,\text{end} \end{array}$$

Here “end” means that the indicated digit could be the rightmost.

Furthermore, the left digit of !!A!! must be small (or else there would be a carry in the leftmost place and !!2A!! would have 4 digits instead of 3) so it must be either 1 or 2. It is not hard to see from this table that the digits must be in the order !!125!! or !!251!!, and indeed, both of those numbers have the required property: !!125_9\cdot 2 = 251_9!!, and !!251_9\cdot 2 = 512_9!!.

This was a simple example, but in more complicated cases it is helpful to draw the order constraints as a graph. Suppose we draw a graph with one vertex for each digit, and one additional vertex to represent the end of the numeral. The graph has an edge from vertex !!v!! to !!v'!! whenever !!v!! can appear to the left of !!v'!!. Then the graph drawn for the table above looks like this:

Graph for 125 base 9

A 3-digit numeral with property !!P_9!! corresponds to a path in this graph that starts at one of the nonzero small digits (marked in blue), ends at the red node marked ‘end’, and visits each node exactly once. Such a path is called hamiltonian. Obviously, self-loops never occur in a hamiltonian path, so we will omit them from future diagrams.

Now we will consider the digit set !!637!!, again base 9. An analysis similar to the foregoing allows us to construct the following graph:

Graph for 367 base 9

Here it is immediately clear that the only hamiltonian path is !!3-7-6-\text{end}!!, and indeed, !!376_9\cdot 2 = 763_9!!.

In general there might be multiple instances of a digit, and so multiple nodes labeled with that digit. Analysis of the !!0,0,0!! case produces a graph with no legal start nodes and so no solutions, unless leading zeroes are allowed, in which case !!000!! is a perfectly valid solution. Analysis of the !!8,8,8!! case produces a graph with no path to the end node and so no solutions. These two trivial patterns appear for all !!R!! and all !!n!!, and we will ignore them from now on.

Returning to our ongoing example, !!1042!! in base 8, we see that !!1!! and !!2!! must double to !!2!! and !!4!!, so must be to the left of small digits, but !!4!! and !!0!! can double to either !!0!! or !!1!! and so could be to the left of anything. Here the constraints are so lax that the graph doesn't help us narrow them down much:

Graph for 1024 base 8

Observing that the only arrow into the 4 is from 0, so that the 4 must follow the 0, and that the entire number must begin with 1 or 2, we can enumerate the solutions:


If leading zeroes are allowed we have also:


All of these are solutions in base 8.

The case of !!R=10!!

Now we turn to our main problem, solutions in base 10.

To find all the solutions of length 6 requires an enumeration of smaller solutions, which, if they existed, might be concatenated into a solution of length 6. This is because our analysis of the digit sets that can appear in a solution assumes that the digits are permuted cyclically; that is, the permutations !!P!! that we considered had only one cycle each.

There are no smaller solutions, but to prove that the length 6 solutions are minimal, we must analyze the cases for smaller !!n!! and rule them out. We now produce a complete analysis of the base 10 case with !!R=10!! and !!n\le 6!!. For !!n=1!! there is only the trivial solution of !!0!!, which we disregard. (The question asked for a positive number anyway.)


For !!n=2!!, we want to find solutions of !!3a_i + v \equiv 0\pmod{10}!! where !!v!! is a two-bit necklace number, one of !!00_2, 01_2, !! or !!11_2!!. Tabulating the values of !!a_i!! and !!v\in\{0,1,3\}!! that solve this equation we get:

$$\begin{array}{ccc} v& a_i \\ \hline 0 & 0 \\ 1& 3 \\ 3& 9 \\ \end{array}$$

We can disregard the !!v=0!! and !!v=3!! solutions because the former yields the trivial solution !!00!! and the latter yields the nonsolution !!99!!. So the only possibility we need to investigate further is !!a_i = 3, v = 1!!, which corresponds to the digit sequence !!36!!: Doubling !!3!! gives us !!6!! and doubling !!6!!, plus a carry, gives us !!3!! again.

But when we tabulate of which digits must be left of which informs us that there is no solution with just !!3!! and !!6!!, because the graph we get, once self-loops are eliminated, looks like this:

graph for 36 base 10

which obviously has no hamiltonian path. Thus there is no solution for !!R=10, n=2!!.


For !!n=3!! we need to solve the equation !!7a_i + v \equiv 0\pmod{10}!! where !!v!! is a necklace number in !!\{0,\ldots 7\}!!, specifically one of !!0,1,3,!! or !!7!!. Since !!7!! and !!10!! are relatively prime, for each !!v!! there is a single !!a_i!! that solves the equation. Tabulating the possible values of !!a_i!! as before, and this time omitting rows with no solution, we have:

$$\begin{array}{rrl} v & a_i & \text{digits}\\ \hline 0& 0 & 000\\ 1& 7 & 748 \\ 3& 1 & 125\\ 7&9 & 999\\ \end{array}$$

The digit sequences !!0,0,0!! and !!9,9,9!! yield trivial solutions or nonsolutions as usual, and we will omit them in the future. The other two lines suggest the digit sets !!1,2,5!! and !!4,7,8!!, both of which fail the “odd equals large” rule.

This analysis rules out the possibility of a digit set with !!a_0 \to a_1 \to a_2 \to a_1!!, but it does not completely rule out a 3-digit solution, since one could be obtained by concatenating a one-digit and a two-digit solution, or three one-digit solutions. However, we know by now that no one- or two-digit solutions exist. Therefore there are no 3-digit solutions in base 10.


For !!n=4!! the governing equation is !!15a_i + v \equiv 0\pmod{10}!! where !!v!! is a 4-bit necklace number, one of !!\{0,1,3,5,7,15\}!!. This is a little more complicated because !!\gcd(15,10)\ne 1!!. Tabulating the possible digit sets, we get:

$$\begin{array}{crrl} a_i & 15a_i& v & \text{digits}\\ \hline 0 & 0 & 0 & 0000\\ 1 & 5 & 5 & 1250\\ 1 & 5 & 15 & 1375\\ 2 & 0 & 0 & 2486\\ 3 & 5 & 5 & 3749\\ 3 & 5 & 15 & 3751\\ 4 & 0 & 0 & 4862\\ 5 & 5 & 5 & 5012\\ 5 & 5 & 5 & 5137\\ 6 & 0 & 0 & 6248\\ 7 & 5 & 5 & 7493\\ 7 & 5 & 5 & 7513\\ 8 & 0 & 0 & 8624 \\ 9 & 5 & 5 & 9874\\ 9 & 5 & 15 & 9999 \\ \end{array}$$

where the second column has been reduced mod !!10!!. Note that even restricting !!v!! to necklace numbers the table still contains duplicate digit sequences; the 15 entries on the right contain only the six basic sequences !!0000, 0125, 1375, 2486, 3749, 4987!!, and !!9999!!. Of these, only !!0000, 9999,!! and !!3749!! obey the odd equals large criterion, and we will disregard !!0000!! and !!9999!! as usual, leaving only !!3749!!. We construct the corresponding graph for this digit set as follows: !!3!! must double to !!7!!, not !!6!!, so must be left of a large number !!7!! or !!9!!. Similarly !!4!! must be left of !!7!! or !!9!!. !!9!! must also double to !!9!!, so must be left of !!7!!. Finally, !!7!! must double to !!4!!, so must be left of !!3,4!! or the end of the numeral. The corresponding graph is:

graph for 3749 base 10

which evidently has no hamiltonian path: whichever of 3 or 4 we start at, we cannot visit the other without passing through 7, and then we cannot reach the end node without passing through 7 a second time. So there is no solution with !!R=10!! and !!n=4!!.


We leave this case as an exercise. There are 8 solutions to the governing equation, all of which are ruled out by the odd equals large rule.


For !!n=6!! the possible solutions are given by the governing equation !!63a_i + v \equiv 0\pmod{10}!! where !!v!! is a 6-bit necklace number, one of !!\{0,1,3,5,7,9,11,13,15,21,23,27,31,63\}!!. Tabulating the possible digit sets, we get:

$$\begin{array}{crrl} v & a_i & \text{digits}\\ \hline 0 & 0 & 000000\\ 1 & 3 & 362486 \\ 3 & 9 & 986249 \\ 5 & 5 & 500012 \\ 7 & 1 & 124875 \\ 9 & 7 & 748748 \\ 11 & 3 & 362501 \\ 13 & 9 & 986374 \\ 15 & 5 & 500137 \\ 21 & 3 & 363636 \\ 23 & 9 & 989899 \\ 27 & 1 & 125125 \\ 31 & 3 & 363751 \\ 63 & 9 & 999999 \\ \end{array}$$

After ignoring !!000000!! and !!999999!! as usual, the large equals odd rule allows us to ignore all the other sequences except !!124875!! and !!363636!!. The latter fails for the same reason that !!36!! did when !!n=2!!. But !!142857!! , the lone survivor, gives us a complicated derived graph containing many hamiltonian paths, every one of which is a solution to the problem:

graph for 124578 base 10

It is not hard to pick out from this graph the minimal solution !!125874!!, for which !!125874\cdot 2 = 251748!!, and also our old friend !!142857!! for which !!142857\cdot 2 = 285714!!.

We see here the reason why all the small numbers with property !!P_{10}!! contain the digits !!124578!!. The constraints on which digits can appear in a solution are quite strict, and rule out all other sequences of six digits and all shorter sequences. But once a set of digits passes these stringent conditions, the constraints on it are much looser, because !!B!! is only required to have the digits of !!A!! in some order, and there are many possible orders, many of which will satisfy the rather loose conditions involving the distribution of the carry bits. This graph is typical: it has a set of small nodes and a set of large nodes, and each node is connected to either all the small nodes or all the large nodes, so that the graph has many edges, and, as in this case, a largish clique of small nodes and a largish clique of large nodes, and as a result many hamiltonian paths.


This analysis is tedious but is simple enough to perform by hand in under an hour. As !!n!! increases further, enumerating the solutions of the governing equation becomes very time-consuming. I wrote a simple computer program to perform the analysis for given !!R!! and !!n!!, and to emit the possible digit sets that satisfied the large equals odd criterion. I had wondered if every base-10 solution contained equal numbers of the digits !!1,2,4,8,5,!! and !!7!!. This is the case for !!n=7!! (where the only admissible digit set is !!\{1,2,4,5,7,8\}\cup\{9\}!!), for !!n=8!! (where the only admissible sets are !!\{1,2,4,5,7,8\}\cup \{3,6\}!! and !!\{1,2,4,5,7,8\}\cup\{9,9\}!!), and for !!n=9!! (where the only admissible sets are !!\{1,2,4,5,7,8\}\cup\{3,6,9\}!! and !!\{1,2,4,5,7,8\}\cup\{9,9,9\}!!). But when we reach !!n=10!! the increasing number of necklaces has loosened up the requirements a little and there are 5 admissible digit sets. I picked two of the promising-seeming ones and quickly found by hand the solutions !!4225561128!! and !!1577438874!!, both of which wreck any theory that the digits !!1,2,4,5,8,7!! must all appear the same number of times.


Thanks to Karl Kronenfeld for corrections and many helpful suggestions.

[Other articles in category /math] permanent link

Fri, 28 Feb 2014

Proof by contradiction

Intuitionistic logic is deeply misunderstood by people who have not studied it closely; such people often seem to think that the intuitionists were just a bunch of lunatics who rejected the law of the excluded middle for no reason. One often hears that intuitionistic logic rejects proof by contradiction. This is only half true. It arises from a typically classical misunderstanding of intuitionistic logic.

Intuitionists are perfectly happy to accept a reductio ad absurdum proof of the following form:

$$(P\to \bot)\to \lnot P$$

Here !!\bot!! means an absurdity or a contradiction; !!P\to \bot!! means that assuming !!P!! leads to absurdity, and !!(P\to \bot)\to \lnot P!! means that if assuming !!P!! leads to absurdity, then you can conclude that !!P!! is false. This is a classic proof by contradiction, and it is intuitionistically valid. In fact, in many formulations of intuitionistic logic, !!\lnot P!! is defined to mean !!P\to \bot!!.

What is rejected by intuitionistic logic is the similar-seeming claim that:

$$(\lnot P\to \bot)\to P$$

This says that if assuming !!\lnot P!! leads to absurdity, you can conclude that !!P!! is true. This is not intuitionistically valid.

This is where people become puzzled if they only know classical logic. “But those are the same thing!” they cry. “You just have to replace !!P!! with !!\lnot P!! in the first one, and you get the second.”

Not quite. If you replace !!P!! with !!\lnot P!! in the first one, you do not get the second one; you get:

$$(\lnot P\to \bot)\to \lnot \lnot P$$

People familiar with classical logic are so used to shuffling the !!\lnot !! signs around and treating !!\lnot \lnot P!! the same as !!P!! that they often don't notice when they are doing it. But in intuitionistic logic, !!P!! and !!\lnot \lnot P!! are not the same. !!\lnot \lnot P!! is weaker than !!P!!, in the sense that from !!P!! one can always conclude !!\lnot \lnot P!!, but not always vice versa. Intuitionistic logic is happy to agree that if !!\lnot P!! leads to absurdity, then !!\lnot \lnot P!!. But it does not agree that this is sufficient to conclude !!P!!.

As is often the case, it may be helpful to try to understand intuitionistic logic as talking about provability instead of truth. In classical logic, !!P!! means that !!P!! is true and !!\lnot P!! means that !!P!! is false. If !!P!! is not false it is true, so !!\lnot \lnot P!! and !!P!! mean the same thing. But in intuitionistic logic !!P!! means that !!P!! is provable, and !!\lnot P!! means that !!P!! is not provable. !!\lnot \lnot P!! means that it is impossible to prove that !!P!! is not provable.

If !!P!! is provable, it is certainly impossible to prove that !!P!! is not provable. So !!P!! implies !!\lnot \lnot P!!. But just because it is impossible to prove that there is no proof of !!P!! does not mean that !!P!! itself is provable, so !!\lnot \lnot P!! does not imply !!P!!.


$$(P\to \bot)\to \lnot P $$

means that if a proof of !!P!! would lead to absurdity, then we may conclude that there cannot be a proof of !!P!!. This is quite valid. But

$$(\lnot P\to \bot)\to P$$

means that if assuming that a proof of !!P!! is impossible leads to absurdity, there must be a proof of !!P!!. But this itself isn't a proof of !!P!!, nor is it enough to prove !!P!!; it only shows that there is no proof that proofs of !!P!! are impossible.

[ Addendum 20141124: This article by Andrej Bauer says much the same thing. ]

[ Addendum 20170508: This article by Robert Harper is another in the same family. ]

[Other articles in category /math] permanent link

Sat, 04 Jan 2014

Cauchy and the continuum

There is a famous mistake of Augustin-Louis Cauchy, in which he is supposed to have "proved" a theorem that is false. I have seen this cited many times, often in very serious scholarly literature, and as often as not Cauchy's purported error is completely misunderstood, and replaced with a different and completely dumbass mistake that nobody could have made.

The claim is often made that Cauchy's Course d'analyse of 1821 contains a "proof" of the following statement: a convergent sequence of continuous functions has a continuous limit. For example, the Wikipedia article on "uniform convergence" claims:

Some historians claim that Augustin Louis Cauchy in 1821 published a false statement, but with a purported proof, that the pointwise limit of a sequence of continuous functions is always continuous…

The nLab article on "Cauchy sum theorem" states:

Non-theorem (attributed to Cauchy, 1821). Let !!f=(f_1,f_2,\ldots)!! be an infinite sequence of continuous functions from the real line to itself. Suppose that, for every real number !!x!!, the sequence !!(f_1(x), f_2(x), \ldots)!! converges to some (necessarily unique) real number !!f_\infty(x)!!, defining a function !!f_\infty!!; in other words, the sequence !!f!! converges pointwise? to !!f_\infty!!. Then !!f_\infty!! is also continuous.

Cauchy never claimed to have proved any such thing, and it beggars belief that Cauchy could have made such a claim, because the counterexamples are so many and so easily located. For example, the sequence !! f_n(x) = x^n!! on the interval !![-1,1]!! is a sequence of continuous functions that converges everywhere on !![0,1]!! to a discontinuous limit. You would have to be a mathematical ignoramus to miss this, and Cauchy wasn't.

Another simple example, one that converges everywhere in !!\mathbb R!!, is any sequence of functions !!f_n!! that are everywhere zero, except that each has a (continuous) bump of height 1 between !!-\frac1n!! and !!\frac1n!!. As !!n\to\infty!!, the width of the bump narrows to zero, and the limit function !!f_\infty!! is everywhere zero except that !!f_\infty(0)=1!!. Anyone can think of this, and certainly Cauchy could have. A concrete example of this type is $$f_n(x) = e^{-x^{2}/n}$$ which converges to 0 everywhere except at !! x=0 !!, where it converges to 1.

Cauchy's controversial theorem is not what Wikipedia or nLab claim. It is that that the pointwise limit of a convergent series of continuous functions is always continuous. Cauchy is not claiming that $$f_\infty(x) = \lim_{i\to\infty} f_i(x)$$ must be continuous if the limit exists and the !!f_i!! are continuous. Rather, he claims that $$S(x) = \sum_{i=1}^\infty f_i(x)$$ must be continuous if the sum converges and the !!f_i!! are continuous. This is a completely different claim. It premise, that the sum converges, is much stronger, and so the claim itself is much weaker, and so much more plausible.

Here the counterexamples are not completely trivial. Probably the best-known counterexample is that a square wave (which has a jump discontinuity where the square part begins and ends) can be represented as a Fourier series.

(Cauchy was aware of this too, but it was new mathematics in 1821. Lakatos and others have argued that the theorem, understood in the way that continuity was understood in 1821, is not actually erroneous, but that the idea of continuity has changed since then. One piece of evidence strongly pointing to this conclusion is that nobody complained about Cauchy's controversial theorem until 1847. But had Cauchy somehow, against all probability, mistakenly claimed that a sequence of continuous functions converges to a continuous limit, you can be sure that it would not have taken the rest of the mathematical world 26 years to think of the counterexample of !!x^n!!.)

The confusion about Cauchy's controversial theorem arises from a perennially confusing piece of mathematical terminology: a convergent sequence is not at all the same as a convergent series. Cauchy claimed that a convergent series of continuous functions has a continuous limit. He did not ever claim that a convergent sequence of continuous functions had a continuous limit. But I have often encountered claims that he did that, even though such such claims are extremely implausible.

The claim that Cauchy thought a sequence of continuous functions converges to a continuous limit is not only false but is manifestly so. Anyone making it has at best made a silly and careless error, and perhaps doesn't really understand what they are talking about, or hasn't thought about it.

[ I had originally planned to write about this controversial theorem in my series of articles about major screwups in mathematics, but the longer and more closely I looked at it the less clear it was that Cauchy had actually made a mistake. ]

[Other articles in category /math] permanent link

Sat, 25 Aug 2012

On the consistency of PA
A monk asked Li Fu, "Master, how do we know that the Peano axioms are consistent?"

Li Fu said, "The axioms are consistent because they have a model."

[Other articles in category /math] permanent link

Fri, 24 Aug 2012

More about ZF's asymmetry between union and intersection
In an article earlier this week, I explored some oddities of defining a toplogy in terms of closed sets rather than open sets, mostly as a result of analogous asymmetry in the ZF set theory axioms.

Let's review those briefly. The relevant axioms concern the operations by which sets can be constructed. There are two that are important. First is the axiom of union, which says that if !!{\mathcal F}!! is a family of sets, then we can form !!\bigcup {\mathcal F}!!, which is the union of all the sets in the family.

The other is actually a family of axioms, the specification axiom schema. It says that for any one-place predicate !!\phi(x)!! and any set !!X!! we can construct the subset of !!X!! for which !!\phi!! holds:

$$\{ x\in X \;|\; \phi(x) \}$$

Both of these are required. The axiom of union is for making bigger sets out of smaller ones, and the specification schema is for extracting smaller sets from bigger ones. (Also important is the axiom of pairing, which says that if !!x!! and !!y!! are sets, then so is the two-element set !!\{x, y\}!!; with pairing and union we can construct all the finite sets. But we won't need it in this article.)

Conspicuously absent is an axiom of intersection. If you have a family !!{\mathcal F}!! of sets, and you want a set of every element that is in some member of !!{\mathcal F}!!, that is easy; it is what the axiom of union gets you. But if you want a set of every element that is in every member of !!{\mathcal F}!!, you have to use specification.

Let's begin by defining this compact notation: $$\bigcap_{(X)} {\mathcal F}$$

for this longer formula: $$\{ x\in X \;|\; \forall f\in {\mathcal F} . x\in f \}$$

This is our intersection of the members of !!{\mathcal F}!!, taken "relative to !!X!!", as we say in the biz. It gives us all the elements of !!X!! that are in every member of !!{\mathcal F}!!. The !!X!! is mandatory in !!\bigcap_{(X)}!!, because ZF makes it mandatory when you construct a set by specification. If you leave it out, you get the Russell paradox.

Most of the time, though, the !!X!! is not very important. When !!{\mathcal F}!! is nonempty, we can choose some element !!f\in {\mathcal F}!!, and consider !!\bigcap_{(f)} {\mathcal F}!!, which is the "normal" intersection of !!{\mathcal F}!!. We can easily show that $$\bigcap_{(X)} {\mathcal F}\subseteq \bigcap_{(f)} {\mathcal F}$$ for any !!X!! whatever, and this immediately implies that $$\bigcap_{(f)} {\mathcal F} = \bigcap_{(f')}{\mathcal F}$$ for any two elements of !!{\mathcal F}!!, so when !!{\mathcal F}!! contains an element !!f!!, we can omit the subscript and just write $$\bigcap {\mathcal F}$$ for the usual intersection of members of !!{\mathcal F}!!.

Even the usually troublesome case of an empty family !!{\mathcal F}!! is no problem. In this case we have no !!f!! to use for !!\bigcap_{(f)} {\mathcal F}!!, but we can still take some other set !!X!! and talk about !!\bigcap_{(X)} \emptyset!!, which is just !!X!!.

Now, let's return to topology. I suggested that we should consider the following definition of a topology, in terms of closed sets, but without an a priori notion of the underlying space:

A co-topology is a family !!{\mathcal F}!! of sets, called "closed" sets, such that:

  1. The union of any two elements of !!{\mathcal F}!! is again in !!{\mathcal F}!!, and
  2. The intersection of any subfamily of !!{\mathcal F}!! is again in !!{\mathcal F}!!.
Item 2 begs the question of which intersection we are talking about here. But now that we have nailed down the concept of intersections, we can say briefly and clearly what we want: It is the intersection relative to !!\bigcup {\mathcal F}!!. This set !!\bigcup {\mathcal F}!! contains anything that is in any of the closed sets, and so !!\bigcup {\mathcal F}!!, which I will henceforth call !!U!!, is effectively a universe of discourse. It is certainly big enough that intersections relative to it will contain everything we want them to; remember that intersections of subfamilies of !!{\mathcal F}!! have a maximum size, so there is no way to make !!U!! too big.

It now immediately follows that !!U!! itself is a closed set, since it is the intersection !!\bigcap_{(U)} \emptyset!! of the empty subfamily of !!{\mathcal F}!!.

If !!{\mathcal F}!! itself is empty, then so is !!U!!, and !!\bigcap_{(U)} {\mathcal F} = \emptyset!!, so that is all right. From here on we will assume that !!{\mathcal F}!! is nonempty, and therefore that !!\bigcap {\mathcal F}!!, with no relativization, is well-defined.

We still cannot prove that the empty set is closed; indeed, it might not be, because even !!M = \bigcap {\mathcal F}!! might not be empty. But as David Turner pointed out to me in email, the elements of !!M!! play a role dual to the extratoplogical points of a topological space that has been defined in terms of open sets. There might be points that are not in any open set anywhere, but we may as well ignore them, because they are topologically featureless, and just consider the space to be the union of the open sets. Analogously and dually, we can ignore the points of !!M!!, which are topologically featureless in the same way. Rather than considering !!{\mathcal F}!!, we should consider !!{\widehat{\mathcal F}}!!, whose members are the members of !!{\mathcal F}!!, but with !!M!! subtracted from each one:

$${\mathcal F}HAT = \{\hat{f}\in 2^U \;|\; \exists f\in {\mathcal F} . \hat{f} = f\setminus M \}$$

So we may as well assume that this has been done behind the scenes and so that !!\bigcap {\mathcal F}!! is empty. If we have done this, then the empty set is closed.

Now we move on to open sets. An open set is defined to be the complement of a closed set, but we have to be a bit careful, because ZF does not have a global notion of the complement !!S^C!! of a set. Instead, it has only relative complements, or differences. !!X\setminus Y!! is defined as: $$X\setminus Y = \{ x\in X \;|\; x\notin Y\} $$

Here we say that the complement of !!Y!! is taken relative to !!X!!.

For the definition of open sets, we will say that the complement is taken relative to the universe of discourse !!U!!, and a set !!G!! is open if it has the form !!U\setminus f!! for some closed set !!f!!.

Anatoly Karp pointed out on Twitter that we know that the empty set is open, because it is the relative complement of !!U!!, which we already know is closed. And if we ensure that !!\bigcap {\mathcal F}!! is empty, as in the previous paragraph, then since the empty set is closed, !!U!! is open, and we have recovered all the original properties of a topology.

But gosh, what a pain it was; in contrast recovering the missing axioms from the corresponding open-set definition of a topology was painless. (John Armstrong said it was bizarre, and probably several other people were thinking that too. But I did not invent this bizarre idea; I got it from the opening paragraph of John L. Kelley's famous book General Topology, which has been in print since 1955.

Here Kelley deals with the empty set and the universe in two sentences, and never worries about them again. In contrast, doing the same thing for closed sets was fraught with technical difficulties, mostly arising from ZF. (The exception was the need to repair the nonemptiness of the minimal closed set !!M!!, which was not ZF's fault.)

I don't think I have much of a conclusion here, except that whatever the advantages of ZF as a millieu for doing set theory, it is overrated as an underlying formalism for actually doing mathematics. (Another view on this is laid out by J.H. Conway in the Appendix to Part Zero of On Numbers and Games (Academic Press, 1976).) None of the problems we encountered were technically illuminating, and nothing was clarified by examining them in detail.

On the other hand, perhaps this conclusion is knocking down a straw man. I think working mathematicians probably don't concern themselves much with whether their stuff works in ZF, much less with what silly contortions are required to make it work in ZF. I think day-to-day mathematical work, to the extent that it needs to deal with set theory at all, handles it in a fairly naïve way, depending on a sort of folk theory in which there is some reasonably but not absurdly big universe of discourse in which one can take complements and intersections, and without worrying about this sort of technical detail.

[ MathJax doesn't work in Atom or RSS syndication feeds, and can't be made to work, so if you are reading a syndicated version of this article, such as you would in Google Reader, or on Planet Haskell or PhillyLinux, you are seeing inlined images provided by the Google Charts API. The MathJax looks much better, and if you would like to compare, please visit my blog's home site. ]

[Other articles in category /math] permanent link

Tue, 21 Aug 2012

The non-duality of open and closed sets
I had long thought that it doesn't matter if we define a topology in terms of open sets or in terms of closed sets, because the two definitions are in every way dual and equivalent. This seems not to be the case: the definition in terms of closed sets seems to be slightly weaker than the definition in terms of open sets.

We can define a topology without reference to the underlying space as follows: A family !!{\mathfrak I}!! of sets is a topology if it is closed under pairwise intersections and arbitrary unions, and we call a set "open" if it is an element of !!{\mathfrak I}!!. From this we can recover the omitted axiom that says that !!\emptyset!! is open: it must be in !!{\mathfrak I}!! because it is the empty union !!\bigcup_{g\in\emptyset} g!!. We can also recover the underlying space of the topology, or at least some such space, because it is the unique maximal open set !!X=\bigcup_{g\in{\mathfrak I}} g!!. The space !!X!! might be embedded in some larger space, but we won't ever have to care, because that larger space is topologically featureless. From a topological point of view, !!X!! is our universe of discourse. We can then say that a set !!C!! is "closed" whenever !!X\setminus C!! is open, and prove all the usual theorems.

If we choose to work with closed sets instead, we run into problems. We can try starting out the same way: A family !!{\mathfrak I}!! of sets is a co-topology if it is closed under pairwise unions and arbitrary intersections, and we call a set "closed" if it is an element of !!{\mathfrak I}!!. But we can no longer prove that !!\emptyset\in{\mathfrak I}!!. We can still recover an underlying space !!X = \bigcup_{c\in{\mathfrak I}} c!!, but we cannot prove that !!X!! is closed, or identify any maximal closed set analogous to the maximal open set of the definition of the previous paragraph. We can construct a minimal closed set !!\bigcap_{c\in{\mathfrak I}} c!!, but we don't know anything useful about it, and in particular we don't know whether it is empty, whereas with the open-sets definition of a topology we can be sure that the empty set is the unique minimal open set.

We can repair part of this asymmetry by changing the "pairwise unions" axiom to "finite unions"; then the empty set is closed because it is a finite union of closed sets. But we still can't recover any maximal closed set. Given a topology, it is easy to identify the unique maximal closed set, but given a co-topology, one can't, and indeed there may not be one. The same thing goes wrong if one tries to define a topology in terms of a Kuratowski closure operator.

We might like to go on and say that complements of closed sets are open, but we can't, because we don't have a universe of discourse in which we can take complements.

None of this may make very much difference in practice, since we usually do have an a priori idea of the universe of discourse, and so we do not care much whether we can define a topology without reference to any underlying space. But it is at least conceivable that we might want to abstract away the underlying space, and if we do, it appears that open and closed sets are not as exactly symmetric as I thought they were.

Having thought about this some more, it seems to me that the ultimate source of the asymmetry here is in our model of set theory. The role of union and intersection in ZF is not as symmetric as one might like. There is an axiom of union, which asserts that the union of the members of some family of sets is again a set, but there is no corresponding axiom of intersection. To get the intersection of a family of sets !!\mathcal S!!, you use a specification axiom. Because of the way specification works, you cannot take an empty intersection, and there is no universal set. If topology were formulated in a set theory with a universal set, such as NF, I imagine the asymmetry would go away.

[ This is my first blog post using MathJax, which I hope will completely replace the ad-hoc patchwork of systems I had been using to insert mathematics. Please email me if you encounter any bugs. ]

[ Addendum 20120823: MathJax depends on executing Javascript, and so it won't render in an RSS or Atom feed or on any page where the blog content is syndicated. So my syndication feed is using the Google Charts service to render formulas for you. If the formulas look funny, try looking at directly. ]

[ Addendum 20120824: There is a followup to this article. ]

[Other articles in category /math] permanent link

Tue, 10 Jan 2012

Elaborations of Russell's paradox
When Katara was five or six, I told her about Russell's paradox in the following form: in a certain library, some books are catalogs that contain lists of other books. For example, there is a catalog of all the books on the second floor, and a catalog of all the books about birds. Some catalogs might include themselves. For example, the catalog of all the books in the library certainly includes itself. Such catalogs have red covers; the other catalogs, which do not include themselves, such as the catalog of all the plays of Shakespeare, have blue covers. Now is there a catalog of all the catalogs with blue covers?

I wasn't sure she would get this, but it succeeded much better than I expected. After I prompted her to consider what color cover it would have, she thought it out, first ruling out one color, and then, when she got to the second color, she just started laughing.

A couple of days ago she asked me if I could think of anything that was like that but with three different colors. Put on the spot, I suggested she consider what would happen if there could be green catalogs that might or might not include themselves. This is somewhat interesting, because you now can have a catalog of all the blue catalogs; it can have a green cover. But I soon thought of a much better extension.

I gave it to Katara like this: say you have a catalog, let's call it X. If X mentions a catalog that mentions X, it has a gold stripe on the spine. Otherwise, it has a silver stripe. Now:

  1. Could there be a red catalog with a gold stripe?
  2. Could there be a red catalog with a silver stripe?
  3. Could there be a blue catalog with a gold stripe?
  4. Could there be a blue catalog with a silver stripe?
And more interesting:

  1. Is there a catalog of all the catalogs with gold stripes?
  2. Is there a catalog of all the catalogs with silver stripes?
I knew that early 20th century logicians, trying to repair the Russell paradox, first tried a very small patch: since comprehension over the predicate XX causes problems, just forbid that predicate. This unfortunately doesn't solve the problem at all; there are an infinite number of equally problematic predicates. (Whitehead and Russell's theory of types is an attempt to fix this; Quine's New Foundations is a different attempt.) One of these predicates is ¬∃Y.X∈Y and Y∈X. You can't construct the set of all X such that ¬∃Y.X∈Y and Y∈X because there is no such set, for reasons similar to the reason why there's no set of all X such that XX, so that's where I got the silver stripe predicate.

Translating this into barber language is left as an exercise for the reader.

[Other articles in category /math] permanent link

Sat, 11 Jun 2011


At a book sale I recently picked up Terence Tao's little book on problem solving for 50¢. One of the exercises (pp. 85–86) is the following little charmer: There are six musicians who will play a series of concerts. At each concert, some of the musicians will be on stage and some will be in the audience. What is the fewest number of concerts that can be played to that each musician gets to see the each of the others play?

Obviously, no more than six concerts are required. (I have a new contribution to the long-debated meaning of the mathematical jargon term "obviously": if my six-year-old daughter Katara could figure out the answer, so can you.) And an easy argument shows that four are necessary: let's say that when a musician views another, that is a "viewing event"; we need to arrange at least 5×6 = 30 viewing events. A concert that has p performers and 6-p in the audience arranges p(6 - p) events, which must be 5, 8, or 9. Three concerts yield no more than 27 events, which is insufficient. So there must be at least 4 concerts, and we may as well suppose that each concert has three musicians in the audience and three onstage, to maximize the number of events at 9·4 = 36. (It turns out there there is no solution otherwise, but that is a digression.)

Each musician must attend at least 2 concerts, or else they would see only 3 other musicians onstage. But 6 musicians attending 2 concerts each takes up all 12 audience spots, so every musician is at exactly 2 concerts. Each musician thus sees exactly six musicians onstage, and since five of them must be different, one is a repeat, and the viewing event is wasted. We knew there would be some waste, since there are 36 viewing avents available and only 30 can be useful, but now we know that each spectator wastes exactly one event.

A happy side effect of splitting the musicians evenly between the stage and the audience in every concert is that we can exploit the symmetry: if we have a solution to the problem, then we can obtain a dual solution by exchanging the performers and the audience in each concert. The conclusion of the previous paragraph is that in any solution, each spectator wastes exactly one event; the duality tells us that each performer is the subject of exactly one wasted event.

Now suppose the same two musicians, say A and B, perform together twice. We know that some spectator must see A twice; this spectator sees B twice also, this wasting two events. But each spectator wastes only one event. So no two musicians can share the stage twice; each two musicians share the stage exactly once. By duality, each two spectators are in the same audience together exactly once.

So we need to find four 3-sets of the elements { A, B, C, D, E, F }, with each element appearing in precisely two sets, and such that each two sets have exactly one element in common.

Or equivalently, we need to find four triangles in K4, none of which share an edge.

The solution is not hard to find:

On stageA B CC D EE F AB D F
In audienceD E FA B FB C DA C E

And in fact this solution is essentially unique.

If you generalize these arguments to 2m musicians, you find that there is a lower bound of $$\left\lceil{4m^2 - 2m \over m^2 }\right\rceil$$ concerts, which is 4. And indeed, even with as few as 4 musicians, you still need four concerts. So it's tempting to wonder if 4 concerts is really sufficient for all even numbers of musicians. Consider 8 musicians, for example. You need 56 viewing events, but a concert with half the musicians onstage and half in the audience provides 16 events, so you might only need as few as 4 concerts to provide the necessary events.

The geometric formulation is that you want to find four disjoint K4s in a K8; or alternatively, you want to find four 4-element subsets of { 1,2,3,4,5,6,7,8 }, such that each element appears in exactly two sets and no two elements are in the same. There seemed to be no immediately obvious reason that this wouldn't work, and I spent a while tinkering around looking for a way to do it and didn't find one. Eventually I did an exhaustive search and discovered that it was impossible.

But the tinkering and the exhaustive search were a waste of time, because there is an obvious reason why it's impossible. As before, each musician must be in exactly two audiences, and can share audiences with each other musician at most once. But there are only 6 ways to be in two audiences, and 8 musicians, so some pair of musicians must be in precisely the same pair of audiences, this wastes too many viewing events, and so there's no solution. Whoops!

It's easy to find solutions for 8 musicians with 5 concerts, though. There is plenty of room to maneuver and you can just write one down off the top of your head. For example:

On stage E F G HB C D HA C D F GA B D E GA B C E F H
In audienceA B C DA E F GB E H C F H D G

Actually I didn't write this one down off the top of my head; I have a method that I'll describe in a future article. But this article has already taken me several weeks to get done, so I'll stop here for now.

[ Addendum: For n = 1…10 musicians, the least number of concerts required is 0, 2, 3, 4, 4, 4, 5, 5, 5, 5; beyond this, I only have bounds. ]

[Other articles in category /math] permanent link

Mon, 15 Nov 2010

A draft of a short introduction to topology
One of my ongoing projects is to figure out how to explain topology briefly. For example, What is Topology?, putatively part 1 of a three-part series that I have not yet written parts 2 or 3 of yet.

CS grad students often have to take classes in category theory. These classes always want to use groups and topological spaces as examples, and my experience is that at this point many of the students shift uncomfortably in their seats since they have not had undergraduate classes in group theory, topology, analysis, or anything else relevant. But you do not have to know much topology to be able to appreciate the example, so I tried to write up the minimal amount necessary. Similarly, if you already understand intuitionistic logic, you do not need to know much topology to understand the way in which topological spaces are natural models for intuitionistic logic—but you do need to know more than zero.

So a couple of years ago I wrote up a short introduction to topology for first-year computer science grad students and other people who similarly might like to know the absolute minimum, and only the absolute minimum, about topology. It came out somewhat longer than I expected, 11 pages, of which 6 are the introduction, and 5 are about typical applications to computer science. But it is a very light, fluffy 11 pages, and I am generally happy with it.

I started writing this shortly after my second daughter was born, and I have not yet had a chance to finish it. It contains many errors. Many, many errors. For example, there is a section at the end about the compactness principle, which can only be taken as a sort of pseudomathematical lorem ipsum. This really is a draft; it is only three-quarters finished.

But I do think it will serve a useful function once it is finished, and that finishing it will not take too long. If you have any interest in this project, I invite you to help.

The current draft is version 0.6 of 2010-11-14. I do not want old erroneous versions wandering around confusing people in my name, so please do not distribute this draft after 2010-12-15. I hope to have an improved draft available here before that.

Please do send me corrections, suggestions, questions, advice, patches, pull requests, or anything else.

[Other articles in category /math] permanent link

Mon, 08 Nov 2010

Semi-boneless ham
The Math Project on Wikipedia is having a discussion about whether or not to have an article about the jargon term "semi-infinite", which I have long considered one of my favorite jargon terms, because it sounds so strange, but makes so much sense. A structure is semi-infinite when it is infinite in one direction but not in the other. For example, the set of positive integers is semi-infinite, since it possesses a least element (1) but no greatest element. Similarly rays in geometry are semi-infinite.

The term is informal, however, and it's not clear just what it should mean in all cases. For example, consider the set S of 1/n for every positive integer n. Is this set semi-infinite? It is bounded in both directions, since it is contained in [0, 1]. But as you move left through the set, you ancounter an infinite number of elements, so it ought to be semi-infinite in the same sense that S ∪ { 1-x : xS } is fully-infinite. Whatever sense that is.

Informal and ill-defined it may be, but the term is widely used; one can easily find mentions in the literature of semi-infinite paths, semi-infinite strips, semi-infinite intervals, semi-infinite cylinders, and even semi-infinite reservoirs and conductors.

The term has spawned an offshoot, the even stranger-sounding "quarter-infinite". This seems to refer to a geometric object that is unbounded in the same way that a quarter-plane is unbounded, where "in the same way" is left rather vague. Consider the set (depicted at left) of all points of the plane for which 0 ≤ |y/x| ≤ √3, for example; is this set quarter-infinite, or only 1/6-infinite? Is the set of points (depicted at right) with xy > 1 and x, y > 0 quarter-infinite? I wouldn't want to say. But the canonical example is simple: the product of two semi-infinite intervals is a quarter-infinite set.

I was going to say that I had never seen an instance of the obvious next step, the eighth-infinite solid, but in researching this article I did run into a few. I can't say it trips off the tongue, however. And if we admit that a half of a quarter-infinite plane segment is also eighth-infinite, we could be getting ourselves into trouble.

(This all reminds me of the complaint of J.H. Conway of the increasing use of the term "biunique". Conway sarcastically asked if he should expect to see "triunique" and soforth, culminating in the idiotic "polyunique".)

Sometimes "semi" really does mean exactly one-half, as in "semimajor axis" (the longest segment from the center of an ellipse to its boundary), "semicubic parabola" (determined by an equation with a term kx3/2), or "semiperimeter" (half the perimeter of a triangle). But just as often, "semi" is one of the dazzling supply of mathematical pejoratives. ("Abnormal, irregular, improper, degenerate, inadmissible, and otherwise undesirable", says Kelley's General Topology.) A semigroup, for example, is not half of a group, but rather an algebraic structure that possesses less structure than a group. Similarly, one has semiregular polyhedra and semidirect products.

I was planning to end with a note that mathematics has so far avoided the "demisemi-" prefix. But alas! Google found this 1971 paper on Demi-semi-primal algebras and Mal'cev-type conditions.

[Other articles in category /math] permanent link

Mon, 14 Dec 2009

Another short explanation of Gödel's theorem
In yesterday's article, I said:

A while back I started writing up an article titled "World's shortest explanation of Gödel's theorem". But I didn't finish it...

I went and had a look to see what was wrong with it, and to my surprise, there seemed to be hardly anything wrong with it. Perhaps I just forgot to post it. So if you disliked yesterday's brief explanation of Gödel's theorem—and many people did—you'll probably dislike this one even more. Enjoy!

A reader wrote to question my characterization of Gödel's theorem in the previous article. But I think I characterized it correctly; I said:

The only systems of mathematical axioms strong enough to prove all true statements of arithmetic, are those that are so strong that they also prove all the false statements of arithmetic.
I'm going to explain how this works.

You start by choosing some system of mathematics that has some machinery in it for making statements about things like numbers and for constructing proofs of theorems like 1+1=2. Many such systems exist. Let's call the one we have chosen M, for "mathematics".

Gödel shows that if M has enough mathematical machinery in it to actually do arithmetic, then it is possible to produce a statement S whose meaning is essentially "Statement S cannot be proved in system M."

It is not at all obvious that this is possible, or how it can be done, and I am not going to get into the details here. Gödel's contribution was seeing that it was possible to do this.

So here's S again:

S: Statement S cannot be proved in system M.

Now there are two possibilities. Either S is in fact provable in system M, or it is not. One of these must hold.

If S is provable in system M, then it is false, and so it is a false statement that can be proved in system M. M therefore proves some false statements of arithmetic.

If S is not provable in system M, then it is true, and so it is a true statement that cannot be proved in system M. M therefore fails to prove some true statements of arithmetic.

So something goes wrong with M: either it fails to prove some true statements, or else it succeeds in proving some false statements.

List of topics I deliberately omitted from this article, that mathematicians should not write to me about with corrections: Presburger arithmetic. Dialetheism. Inexhaustibility. ω-incompleteness. Non-RE sets of axioms.

Well, I see now that left out the step where I go from "M proves a false statement" to "M proves all false statements". Oh well, another topic for another post.

If you liked this post, you may enjoy Torkel Franzén's books Godel's Theorem: An Incomplete Guide to Its Use and Abuse and Inexhaustibility: A Non-Exhaustive Treatment. If you disliked this post, you are even more likely to enjoy them.

Many thanks to Robert Bond for his contribution.

[Other articles in category /math] permanent link

Sun, 13 Dec 2009

World's shortest explanation of Gödel's theorem
A while back I started writing up an article titled "World's shortest explanation of Gödel's theorem". But I didn't finish it, and later I encountered Raymond Smullyan's version, which is much shorter anyway. So here, shamelessly stolen from Smullyan, is the World's shortest explanation of Gödel's theorem.

We have some sort of machine that prints out statements in some sort of language. It needn't be a statement-printing machine exactly; it could be some sort of technique for taking statements and deciding if they are true. But let's think of it as a machine that prints out statements.

In particular, some of the statements that the machine might (or might not) print look like these:

P*x (which means that the machine will print x)
NP*x (which means that the machine will never print x)
PR*x (which means that the machine will print xx)
NPR*x (which means that the machine will never print xx)

For example, NPR*FOO means that the machine will never print FOOFOO. NP*FOOFOO means the same thing. So far, so good.

Now, let's consider the statement NPR*NPR*. This statement asserts that the machine will never print NPR*NPR*.

Either the machine prints NPR*NPR*, or it never prints NPR*NPR*.

If the machine prints NPR*NPR*, it has printed a false statement. But if the machine never prints NPR*NPR*, then NPR*NPR* is a true statement that the machine never prints.

So either the machine sometimes prints false statements, or there are true statements that it never prints.

So any machine that prints only true statements must fail to print some true statements.

Or conversely, any machine that prints every possible true statement must print some false statements too.

The proof of Gödel's theorem shows that there are statements of pure arithmetic that essentially express NPR*NPR*; the trick is to find some way to express NPR*NPR* as a statement about arithmetic, and most of the technical details (and cleverness!) of Gödel's theorem are concerned with this trick. But once the trick is done, the argument can be applied to any machine or other method for producing statements about arithmetic.

The conclusion then translates directly: any machine or method that produces statements about arithmetic either sometimes produces false statements, or else there are true statements about arithmetic that it never produces. Because if it produces something like NPR*NPR* then it is wrong, but if it fails to produce NPR*NPR*, then that is a true statement that it has failed to produce.

So any machine or other method that produces only true statements about arithmetic must fail to produce some true statements.

Hope this helps!

(This explanation appears in Smullyan's book 5000 BC and Other Philosophical Fantasies, chapter 3, section 65, which is where I saw it. He discusses it at considerable length in Chapter 16 of The Lady or the Tiger?, "Machines that Talk About Themselves". It also appears in The Mystery of Scheherezade.)

I gratefully acknowledge Charles Colht for his generous donation to this blog.

[ Addendum 20091214: Another article on the same topic. ]

[ Addendum 20150403: Reddit user cafe_anon has formalized this argument as a Coq proof. ]

[ Addendum 20150406: Reddit user TezlaKoil has formalized this argument as an Agda proof. (Unformatted version) ]

[Other articles in category /math] permanent link

Sun, 21 Jun 2009

Gray code at the pediatrician's office
Last week we took Katara to the pediatrician for a checkup, during which they weighed, measured, and inoculated her. The measuring device, which I later learned is called a stadiometer, had a bracket on a slider that went up and down on a post. Katara stood against the post and the nurse adjusted the bracket to exactly the top of her head. Then she read off Katara's height from an attached display.

How did the bracket know exactly what height to report? This was done in a way I hadn't seen before. It had a photosensor looking at the post, which was printed with this pattern:

(Click to view the other pictures I took of the post.)

The pattern is binary numerals. Each numeral is a certain fraction of a centimeter high, say 1/4 centimeter. If the sensor reads the number 433, that means that the bracket is 433/4 = 108.25 cm off the ground, and so that Katara is 108.25 cm tall.

The patterned strip in the left margin of this article is a straightforward translation of binary numerals to black and white boxes, with black representing 1 and white representing 0:

If you are paying attention, you will notice that although the strip at left is similar to the pattern in the doctor's office, it is not the same. That is because the numbers on the post are Gray-coded.

Gray codes solve the following problem with raw binary numbers. Suppose Katara is close to 104 = 416/4 cm tall, so that the photosensor is in the following region of the post:

0110100001 (417)
0110100000 (416)
0110011111 (415)
0110011110 (414)
But suppose that the sensor (or the post) is slightly mis-aligned, so that instead of properly reading the (416) row, it reads the first half of the (416) row and last half of the (415) row. That makes 0110111111, which is 447 = 111.75 cm, an error of almost 7.5%. (That's three inches, for my American and Burmese readers.) Or the error could go the other way: if the sensor reads the first half of the (415) and the second half of the (416) row, it will see 0110000000 = 384 = 96 cm.

Gray code is a method for encoding numbers in binary so that each numeral differs from the adjacent ones in only one position:

This is the pattern from the post, which you can also see at the right of this article.

Now suppose that the mis-aligned sensor reads part of the (416) line and part of the (417) line. With ordinary binary coding, this could result in an error of up to 7.75 cm. (And worse errors for children of other heights.) But with Gray coding no error results from the misreading:

0101110000 (417)
0101010000 (416)
0101010001 (415)
0101010011 (414)
No matter what parts of 0101110000 and 0101110001 are stitched together, the result is always either 416 or 417.

Converting from Gray code to standard binary is easy: take the binary expansion, and invert every bit that is immediately to the right of a 1 bit. For example, in 1111101000, each red bit is to the right of a 1, and so is inverted to obtain the Gray code 1000011100.

Converting back is also easy: Flip any bit that is anywhere to the right of an odd number of 1 bits, and leave alone the bits that are to the right of an even number of 1 bits. For example, Gray code 1000011100 contains four 1's, and the bits to the right of the first (00001) and third (1), but not the second or fourth, get flipped to 11110 and 0, to give 1111101000.

[ Addendum 20110525: Every so often someone asks why the stadiometer is so sophisticated. Here is the answer. ]

[Other articles in category /math] permanent link

Fri, 22 May 2009

A child is bitten by a dog every 0.07 seconds...
I read in the newspaper today that letter carriers were bitten by dogs 3,000 times last year. (Curiously, this is not a round number; it is exact.) The article then continued: "children ... are 900 times more likely to be bitten than letter carriers."

This is obviously nonsense, because suppose the post office employs half a million letter carriers. (The actual number is actually about half that, but we are doing a back-of-the-envelope estimate of plausibility.) Then the bite rate is six bites per thousand letter carriers per year, and if children are 900 times more likely to be bitten, they are getting bitten at a rate of 5,400 bites per thousand children per year, or 5.4 bites per child. Insert your own joke here, or use the prefabricated joke framework in the title of this article.

I wrote to the reporter, who attributed the claim to the Postal Bulletin 22258 of 7 May 2009. It does indeed appear there. I am trying to track down the ultimate source, but I suspect I will not get any farther. I have discovered that the "900 times" figure appears in the Post Office's annual announcements of Dog Bite Prevention Month as far back as 2004, but not as far back as 2002.

Meantime, what are the correct numbers?

The Centers for Disease Control and Prevention have a superb on-line database of injury data. It immediately delivers the correct numbers for dog bite rate among children:

AgeNumber of
PopulationRate per
0 2,302 4,257,020 54.08
1 7,100 4,182,171 169.77
2 10,049 4,110,458 244.47
3 10,355 4,111,354 251.86
4 9,920 4,063,122 244.15
5 7,915 4,031,709 196.32
6 8,829 4,089,126 215.91
7 6,404 3,935,663 162.72
8 8,464 3,891,755 217.48
9 8,090 3,901,375 207.36
10 7,388 3,927,298 188.11
11 6,501 4,010,171 162.11
12 7,640 4,074,587 187.49
13 5,876 4,108,962 142.99
14 4,720 4,193,291 112.56
15 5,477 4,264,883 128.42
16 4,379 4,334,265 101.03
17 4,459 4,414,523 101.01
Total 133,560 82,361,752 162.16

According to the USPS 2008 Annual Report, in 2008 the USPS employed 211,661 city delivery carriers and 68,900 full-time rural delivery carriers, a total of 280,561. Since these 280,561 carriers received 3,000 dog bites, the rate per 100,000 carriers per year is 1069.29 bites.

So the correct statistic is not that children are 900 times more likely than carriers to be bitten, but rather that carriers are 6.6 times as likely as children to be bitten, 5.6 times if you consider only children under 13. Incidentally, your toddler's chance of being bitten in the course of a year is only about a quarter of a percent, ceteris paribus.

Where did 900 come from? I have no idea.

There are 293 times as many children as there are letter carriers, and they received a total of 44.5 times as many bites. The "900" figure is all over the Internet, despite being utterly wrong. Even with extensive searching, I was not able to find this factoid in the brochures or reports of any other reputable organization, including the American Veterinary Medical Association, the American Academy of Pediatrics, the Centers for Disease Control and Prevention, or the Humane Society of the Uniited States. It appears to be the invention of the USPS.

Also in the same newspaper, the new Indian restaurant on Baltimore avenue was advertising that they "specialize in vegetarian and non-vegetarian food". It's just a cornucopia of stupidity today, isn't it?

[Other articles in category /math] permanent link

Sun, 17 May 2009

Bipartite matching and same-sex marriage
My use of the identifiers husband and wife in Thursday's example code should not be taken as any sort of political statement against same-sex marriage. The function was written as part of a program to solve the stable bipartite matching problem. In this problem, which has historically been presented as concerning "marriage", there are two disjoint equinumerous sets, which we may call "men" and "women". Each man ranks the women in preference order, and each woman ranks the men in preference order. Men are then matched to women. A matching is "stable" if there is no man m and no woman w such that m and w both prefer each other to their current partners. A theorem of Gale and Shapley guarantees the existence of a stable matching and provides an algorithm to construct one.

However, if same-sex marriages are permitted, there may not be a stable matching, so the character of the problem changes significantly.

A minimal counterexample is:

A prefers: B C X
B prefers: C A X
C prefers: A B X
X prefers: A B C

Suppose we match AB, CX. Then since B prefers C to A, and C prefers B to X, B and C divorce their mates and marry each other, yielding BC, AX.

But now C can improve her situation further by divorcing B in favor of A, who is only too glad to dump the miserable X. The marriages are now AC, BX.

B now realizes that his first divorce was a bad idea, since he thought he was trading up from A to C, but has gotten stuck with X instead. So he reconciles with A, who regards the fickle B as superior to her current mate C. The marriages are now AB, CX, and we are back where we started, having gone through every possible matching.

This should not be taken as an argument against same-sex marriage. The model fails to generate the following obvious real-world solution: A, B, and C should all move in together and live in joyous tripartite depravity, and X should jump off a bridge.

[Other articles in category /math] permanent link

Sun, 08 Mar 2009

Happy birthday
Today, my younger daughter Toph is 73 days old, and her elder sister Katara is 123 days old.

Happy birthday, girls.

[Other articles in category /math] permanent link

Thu, 29 Jan 2009

A simple trigonometric identity
A few nights ago I was writing up notes for my category theory reading group, and I wanted to include a commutative diagram on three objects. I was using Paul Taylor's stupendously good diagrams.sty package, which lets you put the vertices of the diagram in the cells of a LaTeX table, and then draw arrows between them. I had drawn the following diagram:

Here I put A at (0,0), B at (4,0), and 1 at (2,2). This is clear enough, but I wished that it were more nearly equilateral.

So that night as I was waiting to fall asleep, I thought about the problem of finding lattice points that are at the vertices of an equilateral triangle. This is a sort of two-dimensional variation on the problem of finding rational approximations to surds, which is a topic that has turned up here many times over the years.

Or rather, I wanted to find lattice points that are almost at the vertices of an equilateral triangle, because I was pretty sure that there were no equilateral lattice triangles. But at the time I could not remember a proof. I started doing some calculations based on the law of cosines, which was a mistake, because nobody but John Von Neumann can do calculations like that in their head as they wait to fall asleep, and I am not John Von Neumann, in case you hadn't noticed.

A simple proof that there are no equilateral lattice triangles has just now occurred to me, though, and I am really pleased with it, so we are about to have a digression.

The area A of an equilateral triangle is s√3/2, where s is the length of the side. And s has the form √t because of the Pythagorean theorem, so A = √(3t)/2, where t is a sum of two squares, because the endpoints of the side are lattice points.

By Pick's theorem, the area of any lattice triangle is a half-integer. So 3t is a perfect square, and thus there are an odd number of threes in t's prime factorization.

But t is a sum of two squares, and by the sum of two squares theorem, its prime factorization must have an even number of threes. We now have a contradiction, so there was no such triangle.

Wasn't that excellent? That is just the sort of thing that I could have thought up while waiting to fall asleep, so it proves even more conclusively that starting with the law of cosines was a mistake.

Okay, end of digression. Back to the law of cosines. We have a triangle with sides a, b, and c, and opposite angles A, B, and C, and you no doubt recall from high school that c2 = a2 + b2 - 2ab cos C. We'll call this "law C".

Before I fell alseep, it occurred to me that you could take the analogous law B, which is b2 = a2 + c2 - 2ac cos B, and substitute the right-hand side for the b2 term in law C. Then a bunch of stuff will cancel out and you should either get something interesting or something tautological. Von Neumann would have known right away which it was, but I needed paper.

So today I got out the paper and did the thing, and came up with the very simple relation that:

c = a cos B + b cos A
Which holds in any triangle. But somehow I had never seen this before, or, if I had, I had completely forgotten it.

The thing is so simple that I thought that it must be wrong, or I would have known it already. But no, it checked out for the easy cases (right triangles, equilateral triangles, trivial triangles) and the geometric proof is easy: Just drop a perpendicular from C. The foot of the perpendicular divides the base c into two segments, which, by the simplest possible trigonometry, have lengths a cos B and b cos A, respectively. QED.

Perhaps that was anticlimactic. Have I mentioned that I have a sign on the door of my office that says "Penn Institute of Lower Mathematics"? This is the kind of thing I'm talking about.

I will let you all know if I come up with anything about the almost-equilateral lattice triangles. Clearly, you can approximate the equilateral triangle as closely as you like by making the lattice coordinates sufficiently large, just as you can approximate √3 as closely as you like with rationals by making the numerator and denominator sufficiently large. Proof: Your computer draws equilateral-seeming triangles on the screen all the time.

I note also that it is important that the lattice is two-dimensional. In three or more dimensions the triangle (1,0,0,0...), (0,1,0,0...), (0,0,1,0...) is a perfectly equilateral lattice triangle with side √2.

[ Addendum 20090130: Vilhelm Sjöberg points out that the area of an equilateral triangle is s2√3/4, not s√3/2. Whoops. This spoils my lovely proof, because the theorem now follows immediately from Pick's: s2 is an integer by Pythagoras, so the area is irrational rather than a half-integer as Pick's theorem requires. ]

[ Addendum 20140403: As a practical matter, one can draw a good lattice approximation to an equilateral triangle by choosing a good rational approximation to !!\sqrt3!!, say !!\frac ab!!, and then drawing the points !!(0,0), (b,a),!! and !!(2b, 0)!!. The rational approximations to !!\sqrt3!! quickly produce triangles that are indistinguishable from equilateral. For example, the rational approximation !!\frac74!! gives the isosceles triangle with vertices !!(0,0), (4,7), (8,0)!! which has one side of length 8 and two sides of length !!\sqrt{65}\approx 8.06!!, an error of less than one percent. The next such approximation, !!\frac{26}{15}!!, gives a triangle that is correct to about 1 part in 1800. (For more about rational approximations to !!\sqrt3!!, see my article on Archimedes and the square root of 3.) ]

[ Addendum 20181126: Even better ways to make 60-degree triangles on lattice points. ]

[Other articles in category /math] permanent link

Tue, 27 Jan 2009

Amusements in Hyperspace
[ Michael Lugo's post on n-spheres today reminded me that I've been wanting for some time to repost this item that I wrote back in 1999. ]

This evening I tried to imagine life in a 1000-dimensional universe. I didn't get too far, but what I did get seemed pretty interesting.

What's it like? Well, it's very dark. Lamps wouldn't work very well, because if the illumination one foot from the source is I, then the illumination two feet from the source is I · 9.3·10-302.

Actually it's even worse than that; there's a double whammy. Suppose you had a cubical room ten feet across. If you thought it was hard to light up the dark corners of a big room in Boston in February, imagine how much worse it is in hyperspace where the corners are 158 feet away.

There are some upsides, however. Rooms won't have to be ten feet on a side because everything will be smaller. You take up about 70,000 cubic centimeters of space; in hyperspace that is just not a lot of room, because a box barely more than a centimeter on a side takes up 70,000 hypercentimeters. In fact, a box barely more than a centimeter on a side can hold as much as you want; an 11 millimeter box already contains 2.5·1041 hypercentimeters.

It's hard to put people in prison in hyperspace, because there are so many directions that you can go to get out. Flatland prison cells have four walls; ours have six, if you count the ceiling and the floor. Hyperspace prison cells have 2000 walls, and each one is very expensive to build.

So that's hyperspace: Big, dark, and easy to get around.

[ Addenda 20120510: An anonymous commenter on Colm Mulcahy's blog observed that "high dimension cubes are qualitatively more like hedgehogs than building blocks". And recently someone asked on stackexchange.math for "What are some examples of a mathematical result being counterintuitive?"; the top-scoring reply concerned the bizarre behavior of high-dimension cubes. ]

[ Addendum 20200325: [More about this]( on the Matrix67 blog. (In Chinese.) ]

[Other articles in category /math] permanent link

Thu, 22 Jan 2009

Archimedes and the square root of 3, revisited
Back in 2006 I discussed Archimedes' calculation of the approximate value of π. In the calculation, he needed rational approximations to several irrational quantities, such as √3, and pulled approximations like 265/153 apparently out of thin air.

I pointed out that although the approximations seem to come out of thin air, a little thought reveals where they probably did come from; it's not very hard. Briefly, you tabulate a2 and 3b2, and look for numbers from one column that are close to numbers from the other; see the previous article for details. But Dr. Chuck Lindsey, the author of a superb explanation of Archimedes' methods, and a professor at Florida Gulf Coast University, seemed mystified by the appearance of the fraction 265/153:

Throughout this proof, Archimedes uses several rational approximations to various square roots. Nowhere does he say how he got those approximations—they are simply stated without any explanation—so how he came up with some of these is anybody's guess.
I left it there for a few years, but just recently I got puzzled email from a gentleman named Peter Nockolds. M. Nockolds was not puzzled by the 265/153. Rather, he wanted to know why so many noted historians of mathematics should be so puzzled by the 265/153.

This was news to me. I did not know anyone else had been puzzled by the 265/153. I had assumed that nearly everyone else saw it the same way that M. Nockolds and I did. But M. Nockolds provided me with a link to an extensive discussion of the matter, which included quotations from several noted mathematicians and historians of mathematics:

It would seem...that [Archimedes] had some (at present unknown) method of extracting the square root of numbers approximately.
W.W Rouse Ball, Short Account of The History of Mathematics, 1908
...the calculation [of π] starts from a greater and lesser limit to the value of √3, which Archimedes assumes without remark as known, namely 265/153 < √3 < 1351/780. How did Archimedes arrive at this particular approximation? No puzzle has exercised more fascination upon writers interested in the history of mathematics... The simplest supposition is certainly [the "Babylonian method"; see Kline below]. Another that the successive solutions in integers of the equations x2-3y2=1 and x2-3y2=-2 may have been a similar way to...the Pythagoreans. The rest of the suggestions amount for the most part to the use of the method of continued fractions more or less disguised.
T. Heath, A History of Greek Mathematics, 1921
Heath said "The simplest supposition is certainly ..." and then followed with the "Babylonian method", which is considerably more complicated than the extremely simple method I suggested in my earlier article. Morris Kline explains the Babylonian method:

He also obtained an excellent approximation to √3, namely 1351/780 > √3 > 265/153, but does not explain how he got this result. Among the many conjectures in the historical literature concerning its derivation the following is very plausible. Given a number A, if one writes it as a2 ± b where a2 is the rational square nearest to A, larger or smaller, and b is the remainder, then a ± b/2a > √A > a ± b/(2a±1). Several applications of this procedure do produce Archimedes' result.
M. Kline, Mathematical Thought From Ancient To Modern Times, 1972
And finally:

Archimedes approximated √3 by the slightly smaller value 265/153... How he managed to extract his square roots with such one of the puzzles that this extraordinary man has bequeathed to us.
P. Beckmann, A History of π, 1977
Nockolds asked me "Have you had any feedback from historians of maths who explain why it wasn't so easy to arrive at 265/153 or even 1351/780? Have you any idea why they make such a big deal out of this?"

No, I'm mystified. Even working with craptastic Greek numerals, it would not take Archimedes very long to tabulate kn2 far enough to discover that 3·7802 = 13512 - 1. Or, if you don't like that theory, try this one: He tabulated n2 and 3n2 far enough to discover the following approximations:

2 / 1
5 / 3
7 / 4
19 / 11
26 / 15
71 / 41
97 / 56
And the pattern is obvious. In the left column, we have 2+5=7, 5+2·7=19, 7+19=26, 19+2·26=71, 26+71=97. In the right column we have 1+3=4, 3+2·4=11, 4+11=15, 11+2·15=41, 15+41=56. It would be trivial to conjecture that the next entries should be 71+2·97 = 265 and 41+2·56 = 153 and then to check 2652 and 3·1532 to see that yes, they are close together. Another couple of iterations will get you to 1351/780, which you can check similarly.

I know someone wants to claim that this is nothing more than the Babylonian method. But this is missing an important point. Although this sort of numeric tinkering might well lead you to discover the Babylonian method, especially if you were Archimedes, it is not the Babylonian method, and it can be done in complete ignorance of the Babylonian method. But it yields the required approximations anyway.

So I will echo Nockolds' puzzlement here. There are a lot of things that Archimedes did that were complex and puzzling, but this is not one of them. You do not need sophisticated algebraic technique to find approximations to surds. You only need to do (at most) a few hours of integer calculation. The puzzle is why people like Rouse Ball and Heath think it is puzzling.

There's an explanation I'm groping for but can't quite articulate, but which goes something like this: Perhaps mathematicians of the late Victorian age lent too much weight to theory and analysis, and not enough to heuristic and simple technique. As a lifelong computer programmer, I have a great appreciation for what can be accomplished by just grinding out the numbers. See my anecdote about the square root algorithm used by the ENIAC, for example. I guessed then that perhaps computer science professors know more about mathematics than I expect, but less about computation. I can imagine the same thing of Victorian mathematicians—but not of Archimedes.

One thing you often hear about pre-19th-century mathematicians is that they were great calculators. I wonder if appreciation of simple arithmetic technique might not have been sometimes lost to the mathematicans from the very end of the pre-computation age, say 1880–1940.

Then again, perhaps I'm not giving them enough credit. Maybe there's something going on that I missed. I haven't checked the original sources to see what they actually say, so who knows? Perhaps Heath discusses the technique I suggested, and then rejects it for some fascinating reason that I, not being an expert in Greek mathematics, can't imagine. If I find out anything else, I will report further.

[Other articles in category /math] permanent link

Tue, 20 Jan 2009

Triples and Closure
Lately I've been reading Lambek and Scott's Introduction to Higher-Order Categorical Logic, which is too advanced for me. (Yoneda Lemma on page 10. Whew!) But you can get some value out of books that are too hard if you pay attention. Last night I learned that monads are analogous to closure operators.

In topology, we have the idea of a "closure" of a set, which is essentially the union of the set with its boundary. For example, consider an open disk D, say the set of all points less than one mile from my house. The boundary of this set is a circle with radius one mile, centered at my house. The closure of D is the union of D with its boundary, and so is closed disk consisting of all points less than or equal to one mile from my house.

Representing the closure of a set S as C(S), we have the obvious theorem that SC(S), because the closure includes everything in S, plus the boundary.

Another easy, but not quite obvious theorem is that C(C(S)) ⊂ C(S). This says that once you take the closure, you have included the boundary, and you do not get any more boundary by taking the closure again. The closure of a set is "closed"; the closure of a "closed" set C is just C.

A third fundamental theorem about closures is that ABC(A) ⊂ C(B).

Now we turn to monads. A monad is first of all a functor, which, if you restrict your attention to programming languages, means that a monad is a type constructor M with an associated function fmap such that for any function f of type α → β, fmap f has type M α → M β.

But a monad is also equipped with two other functions. There is a return function, which has type α → M α, and a join function, which has type M M α → M α.

Haskell provides monads with a "bind" function, written >>=, which is interdefinable with join:

        join x   =  x >>= id
        a >>= b  =  join (fmap b a)
but we are going to forget about >>= for now.

So the monad is equipped with three fundamental operations:

        fmap   ::   (a → b) → (M a → M b)
        join   ::   M M a → M a
        return ::   a → M a
The three basic theorems about topological closures are:

(AB) → (C(A) ⊂ C(B)
C(C(A)) ⊂ C(A)
If we imagine that ⊂ is a special kind of implication, the similarity with the monad laws is clear. And ⊂ is a special kind of implication, since (AB) is just an abbreviation for (xAxB).

If we name the three closure theorems "fmap", "join", and "return", we might guess that "bind" also turns out to be a theorem. And it is, because >>= has the type M a → (aM b) → M b. The corresponding theorem is:

xC(A) → (AC(B)) → xC(B)
If the truth of this is hard to see, it is partly because the implications are in an unnatural order. The theorem is stated in the form PQR, but it would be easier to understand as the equivalent QPR:

AC(B) → xC(A) → xC(B)
Or more briefly:

AC(B) → C(A) ⊂ C(B)
This is quite true. We can prove it from the other three theorems as follows. Suppose AC(B). Then by "fmap", C(A) ⊂ C(C(B)). By "join", C(C(B)) ⊂ C(B). By transitivity of ⊂, C(A) ⊂ C(B). This is what we wanted.

Haskell defines a =<< operator which is the same as >>= except with the arguments forwards instead of backwards:

        =<< ::  (a → M b) → M a → M b
        a =<< b   =   b >>= a
The type of this function is analogous to the bind theorem, and I have seen claims in the literature that the argument order is in some ways more natural. Where the >>= function takes a value first, and then feeds it to a given function, the =<< function makes more sense as a curried function, taking a function of type aM b and yielding the corresponding function of type M aM b.

I think it's also worth noticing that the structure of the proof of the bind theorem (invoke "fmap" and then "join") is exactly the same as the structure of the code that defines "bind".

We can go the other way also, and prove the "join" theorem from the "bind" theorem. The definition of join in terms of >>= is:

        join a = a >>= id
Following the program again, id in the program code corresponds to the theorem that BB for any B. A special case of this theorem is that C(B) ⊂ C(B) for any B. Then in the "bind" theorem:
AC(B) → C(A) ⊂ C(B)
take A = C(B):

C(B) ⊂ C(B) → C(C(B)) ⊂ C(B)
The left side of the implication is satisfied, so we conclude the consequent, C(C(B)) ⊂ C(B), which is what we wanted.

But wait, monad operations are also required to satisfy some monad laws. For example, join (return x) = x. How does this work out in topological closure world?

In programming language world, x here is required to have monad type. Monad types correspond to closed sets, so this is a theorem about closed sets. The theorem says that if X is a closed set, then the closure of X is the same as x. This is true.

The identity between these two things can be found in (surprise) category theory. In category theory, a monad is a (categorial) functor equipped with two natural transformations, the "return" and "join" operations. The categorial version of a closure operator is essentially the same.

Closure operations have a natural opposite. In topology, it is the "interior of" operation. The interior of a set is what you get if you discard the boundary of the set. The interior of a closed disc is an open disc; the interior of an open disc is the same open disc. Interior operations satisfy laws analogous but opposite to those enjoyed by closures:

SC(S) I(T) ⊂ T
C(C(S)) ⊂ C(S) I(T) ⊂ I(I(T))
ABC(A) ⊂ C(B) ABI(A) ⊂ I(B)
Notice that the third theorem does not get turned around. I think this is because it comes from the functor itself, which goes the same way, not from the natural transformations, which go the other way. But I have not finished thinking abhout it carefully yet.

Sooner or later I am going to program in Haskell with comonads, and it gives me a comfortable feeling to know that I am pre-equipped with a way to understand them as interior operations.

I have an idea that the power of mathematics comes principally from the places where it succeeds in understanding two different things as aspects of the same thing. For example, why is group theory so useful? Because it understands transformations of objects (say, rotations of a polyhedron) and algebraic operations as essentially the same thing. If you have a hard problem about one, you can often make it into an easier problem about the other one. Similarly analytic geometry transforms numerical problems into geometric problems and back again. Most often the geometry is harder than the numerical problem, and you use it in that direction, but often you go in the other direction instead.

It is quite possible that this notion is too vague to qualify as an actual theory. But category theory fits the description. Category theory lets you say that types are objects, type constructors are functors, and polymorphic functions are natural transformations. Then you can understand natural transformations as structure-preserving maps of something or other and get some insight into polymorphic functions, or vice-versa.

Category theory is a large agglomeration of such identities. Lambek and Scott's book starts with several slogans about category theory. One of these is that many objects of interest to mathematicians form categories, such as the category of sets. Another is that many objects of interest to mathematicians are categories. (For example, each set is a discrete category.) So one of the reasons category theory is so extremely useful is that it sets up these multiple entities as different aspects of the same thing.

I went to lunch and found more to say on the subject, but it will have to wait until another time.

[Other articles in category /math] permanent link

Mon, 24 Nov 2008

Variations on the Goldbach conjecture

  1. Every prime number is the sum of two even numbers.
  2. Every odd number is the sum of two primes.
  3. Every even number is the product of two primes.

[Other articles in category /math] permanent link

Tue, 11 Nov 2008

Another note about Gabriel's Horn
I forgot to mention in the original article that I think referring to Gabriel's Horn as "paradoxical" is straining at a gnat and swallowing a camel.

Presumably people think it's paradoxical that the thing should have a finite volume but an infinite surface area. But since the horn is infinite in extent, the infinite surface area should be no surprise.

The surprise, if there is one, should be that an infinite object might contain a merely finite volume. But we swallowed that gnat a long time ago, when we noticed that the infinitely wide series of bars below covers only a finite area when they are stacked up as on the right.

The pedigree for that paradox goes at least back to Zeno, so perhaps Gabriel's Horn merely shows that there is still some life in it, even after 2,400 years.

[ Addendum 2014-07-03: I have just learned that this same analogy was also described in this math.stackexchange post of 2010. ]

[Other articles in category /math] permanent link

Mon, 10 Nov 2008

Gabriel's Horn is not so puzzling
Take the curve y = 1/x for x ≥ 1. Revolve it around the x-axis, generating a trumpet-shaped surface, "Gabriel's Horn".

Elementary calculations, with calculus, allow one to show that although the Horn has finite volume, it has an infinite surface area. This is considered paradoxical, because it says that although an infinite amount of paint is required to cover the interior surface of the horn, the entire interior can be filled up with a finite amount of paint.

The calculations themselves do not lend much insight into what is going on here. But I recently read a crystal-clear explanation that I think should be more widely known.

Take out some Play-Doh and roll out a snake. The surface area of the snake (neglecting the two ends, which are small) is the product of the length and the circumference; the circumference is proportional to the diameter. The volume is the product of the length and the cross-sectional area, which is proportional to the square of the diameter.

Now roll the snake with your hands so that it becomes half as thick as it was before. Its diameter decreases by half, so its cross-sectional area decreases to one-fourth. Since the volume must remain the same, the snake is now four times as long as it was before. And the surface area, which is the product of the length and the diameter, has doubled.

As you continue to roll the snake thinner and thinner, the volume stays the same, but the surface area goes to infinity.

Gabriel's Horn does exactly the same thing, except without the rolling, because the parts of the Horn that are far from the origin look exactly the same as very long snakes.

There's nothing going on in the Gabriel's Horn example that isn't also happening in the snake example, except that in the explanation of Gabriel's Horn, the situation is obfuscated by calculus.

I read this explanation in H. Jerome Keisler's caclulus textbook. Keisler's book is an ordinary undergraduate calculus text, except that instead of basing everything on limits and on limiting processes, it is based on nonstandard analysis and explicit infinitesimal quantities. Check it out; it is available online for free. (The discussion of Gabriel's Horn is in chapter 6, page 356.)

[ Addendum 20081110: A bit more about this. ]

[Other articles in category /math] permanent link

Fri, 10 Oct 2008

Representing ordinal numbers in the computer and elsewhere
Lately I have been reading Andreas Abel's paper "A semantic analysis of structural recursion", because it was a referred to by David Turner's 2004 paper on total functional programming.

The Turner paper is a must-read. It's about functional programming in languages where every program is guaranteed to terminate. This is more useful than it sounds at first.

Turner's initial point is that the presence of ⊥ values in languages like Haskell spoils one's ability to reason from the program specification. His basic example is simple:

        loop :: Integer -> Integer
        loop x = 1 + loop x
Taking the function definition as an equation, we subtract (loop x) from both sides and get
0 = 1
which is wrong. The problem is that while subtracting (loop x) from both sides is valid reasoning over the integers, it's not valid over the Haskell Integer type, because Integer contains a ⊥ value for which that law doesn't hold: 1 ≠ 0, but 1 + ⊥ = 0 + ⊥.

Before you can use reasoning as simple and as familiar as subtracting an expression from both sides, you first have to prove that the value of the expression you're subtracting is not ⊥.

By banishing nonterminating functions, one also banishes ⊥ values, and familiar mathematical reasoning is rescued.

You also avoid a lot of confusing language design issues. The whole question of strictness vanishes, because strictness is solely a matter of what a function does when its argument is ⊥, and now there is no ⊥. Lazy evaluation and strict evaluation come to the same thing. You don't have to wonder whether the logical-or operator is strict in its first argument, or its second argument, or both, or neither, because it comes to the same thing regardless.

The drawback, of course, is that if you do this, your language is no longer Turing-complete. But that turns out to be less of a problem in practice than one would expect.

The paper was so interesting that I am following up several of its precursor papers, including Abel's paper, about which the Turner paper says "The problem of writing a decision procedure to recognise structural recursion in a typed lambda calculus with case-expressions and recursive, sum and product types is solved in the thesis of Andreas Abel." And indeed it is.

But none of that is what I was planning to discuss. Rather, Abel introduces a representation for ordinal numbers that I hadn't thought much about before.

I will work up to the ordinals via an intermediate example. Abel introduces a type Nat of natural numbers:

Nat = 1 ⊕ Nat
The "1" here is not the number 1, but rather a base type that contains only one element, like Haskell's () type or ML's unit type. For concreteness, I'll write the single value of this type as '•'.

The ⊕ operator is the disjoint sum operator for types. The elements of the type ST have one of two forms. They are either left(s) where sS or right(t) where tT. So 1⊕1 is a type with exactly two values: left(•) and right(•).

The values of Nat are therefore left(•), and right(n) for any element n of Nat. So left(•), right(left(•)), right(right(left(•))), and so on. One can get a more familiar notation by defining:

0 = left(•)
Succ(n) = right(n)
And then one just considers 3 to be an abbreviation for Succ(Succ(Succ(0))) as usual. (In this explanation, I omitted some technical details about recursive types.)

So much for the natural numbers. Abel then defines a type of ordinal numbers, as:

Ord = (1 ⊕ Ord) ⊕ (NatOrd)
In this scheme, an ordinal is either left(left(•)), which represents 0, or left(right(n)), which represents the successor of the ordinal n, or right(f), which represents the limit ordinal of the range of the function f, whose type is NatOrd.

We can define abbreviations:

Zero = left(left(•))
Succ(n) = left(right(n))
Lim(f) = right(f)
So 0 = Zero, 1 = Succ(0), 2 = Succ(1), and so on. If we define a function id which maps Nat into Ord in the obvious way:

        id :: NatOrd
        id 0       = Zero
        id (n + 1) = Succ(id n)
then ω = Lim(id). Then we easily get ω+1 = Succ(ω), etc., and the limit of this function is 2ω:

        plusomega :: NatOrd
        plusomega 0       = Lim(id)
        plusomega (n + 1) = Succ(plusomega n)
We can define an addition function on ordinals:

        + :: OrdOrdOrd
        ord + Zero    = ord
        ord + Succ(n) = Succ(ord + n)
        ord + Lim(f)  = Limx. ord + f(x))
This gets us another way to make 2ω: 2ω = + ω).

Then this function multiplies a Nat by ω:

        timesomega :: NatOrd
        timesomega 0       = Zero
        timesomega (n + 1) = ω + (timesomega n)
and Lim(timesomega) is ω2. We can go on like this.

But here's what puzzled me. The ordinals are really, really big. Much too big to be a set in most set theories. And even the countable ordinals are really, really big. We often think we have a handle on uncountable sets, because our canonical example is the real numbers, and real numbers are just decimal numbers, which seem simple enough. But the set of countable ordinals is full of weird monsters, enough to convince me that uncountable sets are much harder than most people suppose.

So when I saw that Abel wanted to define an arbitrary ordinals as a limit of a countable sequence of ordinals, I was puzzled. Can you really get every ordinal as the limit of a countable sequence of ordinals? What about Ω, the first uncountable ordinal?

Well, maybe. I can't think of any reason why not. But it still doesn't seem right. It is a very weird sequence, and one that you cannot write down. Because suppose you had a notation for all the ordinals that you would need. But because it is a notation, the set of things it can denote is countable, and so a fortiori the limit of all the ordinals that it can denote is a countable ordinal, not Ω.

And it's all very well to say that the sequence starts out (0, ω, 2ω, ω2, ωω, ε0, ε1, εε0, ...), or whatever, but the beginning of the sequence is totally unimportant; what is important is the end, and we have no way to write the end or to even comprehend what it looks like.

So my question to set theory experts: is every limit ordinal the least upper bound of some countable sequence of ordinals?

I hate uncountable sets, and I have a fantasy that in the mathematics of the 23rd Century, uncountable sets will be looked back upon as a philosophical confusion of earlier times, like Zeno's paradox, or the luminiferous aether.

[ Addendum 20081106: Not every limit ordinal is the least upper bound of some countable sequence of (countable) ordinals, and my guess that Ω is not was correct, but the proof is so simple that I was quite embarrassed to have missed it. More details here. ]

[ Addendum 20160716: In the 8 years since I wrote this article, the link to Turner's paper at Middlesex has expired. Fortunately, Miëtek Bak has taken it upon himself to create an archive containing this paper and a number of papers on related topics. Thank you, M. Bak! ]

[Other articles in category /math] permanent link

Wed, 01 Oct 2008

The Lake Wobegon Distribution
Michael Lugo mentioned a while back that most distributions are normal. He does not, of course, believe any such silly thing, so please do not rush to correct him (or me). But the remark reminded me of how many people do seem to believe that most distributions are normal. More than once on internet mailing lists I have encountered people who ridiculed others for asserting that "nearly all x are above [or below] average". This is a recurring joke on Prairie Home Companion, broadcast from the fictional town of Lake Wobegon, where "all the women are strong, all the men are good looking, and all the children are above average." And indeed, they can't all be above average. But they could nearly all be above average. And this is actually an extremely common situation.

To take my favorite example: nearly everyone has an above-average number of legs. I wish I could remember who first brought this to my attention. James Kushner, perhaps?

But the world abounds with less droll examples. Consider a typical corporation. Probably most of the employees make a below-average salary. Or, more concretely, consider a small company with ten employees. Nine of them are paid $40,000 each, and one is the owner, who is paid $400,000. The average salary is $76,000, and 90% of the employees' salaries are below average.

The situation is familiar to people interested in baseball statistics because, for example, most baseball players are below average. Using Sean Lahman's database, I find that 588 players received at least one at-bat in the 2006 National League. These 588 players collected a total of 23,501 hits in 88,844 at-bats, for a collective batting average of .265. Of these 588, only 182 had an individual batting average higher than 265. 69% of the baseball players in the 2006 National League were below-average hitters. If you throw out the players with fewer than 10 at-bats, you are left with 432 players of whom 279, or 65%, hit worse than their collective average of 23430/88325 = .265. Other statistics, such as earned-run averages, are similarly skewed.

The reason for this is not hard to see. Baseball-hitting talent in the general population is normally distributed, like this:

Here the right side of the graph represents the unusually good hitters, of whom there aren't very many. The left side of the graph represents the unusually bad hitters; there aren't many of those either. Most people are somewhere in the middle, near the average, and there are about as many above-average hitters as below-average hitters in the general population.

But major-league baseball players are not the general population. They are carefully selected, among the best of the best. They are all chosen from the right-hand edge of the normal curve. The people in the middle of the normal curve, people like me, play baseball in Clark Park, not in Quankee Stadium.

Here's the right-hand corner of the curve above, highly magnified:

As you can see here, the shape is not at all like the curve for the general population, which had the vast majority of the population in the middle, around the average. Here, the vast majority of the population is way over on the left side, just barely good enough to play in the majors, hanging on to their jobs by the skin of their teeth, subject at any moment to replacement by some kid up from the triple-A minors. The above-average players are the ones over on the right end, the few of the few.

Actually I didn't present the case strongly enough. There are around 800 regular major-league ballplayers in the USA, drawn from a population of around 300 million, a ratio of one per 375,000. Well, no, the ratio is smaller, since the U.S. leagues also draw the best players from Mexico, Venezuela, Canada, the Dominican Republic, Japan, and elsewhere. The curve above is much too inclusive. The real curve for major-league ballplayers looks more like this:

(Note especially the numbers on the y-axis.)

This has important implications for the analysis of baseball. A player who is "merely" above average is a rare and precious resource, to be cherished; far more players are below average. Skilled analysts know that comparisons with the "average" player are misleading, because baseball is full of useful, effective players who are below average. Instead, analysts compare players to a hypothetical "replacement level", which is effectively the leftmost edge of the curve, the level at which a player can be easily replaced by one of those kids from triple-A ball.

In the Historical Baseball Abstract, Bill James describes some great team, I think one of the Cincinnati Big Red Machine teams of the mid-1970s, as "possibly the only team in history that was above average at every position". That's an important thing to know about the sport, and about team sports in general: you don't need great players to completely clobber the opposition; it suffices to have players that are merely above average. But if you're the coach, you'd better learn to make do with a bunch of players who are below average, because that's what you have, and that's what the other team will beat you with.

The right-skewedness of the right side of a normal distribution has implications that are important outside of baseball. Stephen Jay Gould wrote an essay about how he was diagnosed with cancer and given six months to live. This sounds awful, and it is awful. But six months was the expected lifetime for patients with his type of cancer—the average remaining lifetime, in other words—and in fact, nearly everyone with that sort of cancer lived less than six months, usually much less. The average was only skewed up as high as six months because of a few people who took years to die. Gould realized this, and then set about trying to find out how the few long-lived outliers survived and what he could do to turn himself into one of the long-lived freaks. And he succeeded, and lived for twenty years, dying eventually at age 60.

My heavens, I just realized that what I've written is an article about the "long tail". I had no idea I was being so trendy. Sorry, everyone.

[Other articles in category /math] permanent link

Fri, 26 Sep 2008

Sprague-Grundy theory
I'm on a small mailing list for math geeks, and there's this one guy there, Richard Penn, who knows everything. Whenever I come up with some idle speculation, he has the answer. For example, back in 2003 I asked:

Let N be any positive integer. Does there necessarily exist a positive integer k such that the base-10 representation of kN contains only the digits 0 through 4?
M. Penn was right there with the answer.

Yesterday, M. Penn asked a question to which I happened to know the answer, and I was so pleased that I wrote up the whole theory in appalling detail. Since I haven't posted a math article in a while, and since the mailing list only has about twelve people on it, I thought I would squeeze a little more value out of it by posting it here.

Richard Penn asked:

N dots are placed in a circle. Players alternate moves, where a move consists of crossing out any one of the remaining dots, and the dots on each side of it (if they remain). The winner is the player who crosses out the last dot. What is the optimal strategy with 19 dots? with 20? Can you generalize?
M. Penn observed that there is a simple strategy for the 20-dot circle, but was not able to find one for the 19-dot circle. But solving such problems in general is made easy by the Sprague-Grundy theory, which I will explain in detail.

0. Short Spoilers

Both positions are wins for the second player to move.

The 20-dot case is trivial, since any first-player move leaves a row of 17 dots, from which the second player can leave two disconnected rows of 7 dots each. Then any first-player move in one of these rows can be effectively answered by the second player in the other row.

The 19-dot case is harder. The first player's move leaves a row of 16 dots. The second player can win by removing 3 dots to leave disconnected rows of 6 and 7 dots. After this, the strategy is complicated, but is easily found by the Sprague-Grundy theory. It's at the end of this article if you want to skip ahead.

Sprague-Grundy theory is a complete theory of all finite impartial games, which are games like this one where the two players have exactly the same moves from every position.

The theory says:

  1. Every such game position has a "value", which is a non-negative integer.
  2. A position is a second-player win if and only if its value is zero.
  3. The value of a position can be calculated from the values of the positions to which the players can move, in a simple way.
  4. The value of a collection of disjoint positions (such as two disconnected rows of dots) can be calculated from the values of its component positions in a simple way.
Long details follow. They are also found in "Winning Ways", Vol I, by Berlekamp, Conway, and Guy.

1. Nim

In the game of Nim, one has some piles of beans, and a legal move is to remove some or all of the beans from any one pile. The winner is the player who takes the last bean. Equivalently, the winner is the last player who has a legal move.

Nim is important because every position in every impartial game is somehow equivalent to a position in Nim, as we will see. In fact, every position in every impartial game is equivalent to a Nim position with at most one heap of beans! Since single Nim-heaps are trivially analyzed, one can completely analyze any impartial game position by calculating the Nim-heap to which it is equivalent.

2. Disjoint sums of games

Definition: The "disjoint sum" A # B of two games A and B is a new game whose rules are as follows: a legal move in A # B is either a move in A or a move in B; the winner is the last player with a legal move.

Three easy exercises:

  1. # is commutative.
  2. # is associative.
  3. Let (a,b,c...) represent the Nim position with heaps a, b, c, etc. Then the game (a,b,c,...) is precisely (a) # (b) # (c) # ... .
Consider the trivial game with no legal moves for anyone. This game is called 0, because:

0 # a = a # 0 = a
for all games a. 0 is a win for the previous player: the next player to move has no legal moves, and loses.

We will call the next player to move "P1", and the player who just moved "P2".

Note that a Nim-heap of 0 beans is precisely the 0 game.

3. Sums of Nim-heaps

We usually represent a single Nim-heap with n beans as "∗n". I'll do that from now on.

We observed that ∗0 is a win for the second player. Observe now that when n is positive, ∗n is a win for the first player, by a trivial strategy.

From now on we will use the symbol "=" to mean a weaker relation on games than strict equality. Two games A and B will be equivalent if their outcomes are the same in a rather strong sense:

A = B means that for any game X, A # X is a winning position if and only if B # X is also.
Taking X = 0, the condition A = B implies that both games have the same outcome in isolation: if one is a first-player win, so is the other. But the condition is stronger than that. Both ∗1 and ∗2 are first-player wins, but ∗1 ≠ ∗2, because ∗1 # ∗1 is a second-player win, while ∗2 # ∗1 is a first-player win.

Exercise: ∗x = ∗y if and only if x = y.

It so happens that the disjoint sum of two Nim-heaps is equivalent to a single Nim-heap:

Nim-sum theorem:a # ∗b = ∗(ab), Where ⊕ is the bitwise exclusive-or operation.

I'll omit the proof, which is pretty easy to find. ⊕ is often described as "write a and b in binary, and add, ignoring all carries." For example 1 ⊕ 2 = 3, and 13 ⊕ 7 = 10. This implies that ∗1 # ∗2 = ∗3, and that ∗13 # ∗7 = ∗10.

Although I omitted the proof that # for Nim-heaps is essentially the ⊕ operation in disguise, there are many natural implications of this that you can use to verify that the claim is plausible. For example:

  1. The Nim-sum theorem implies that ∗0 is a neutral element for #, which we already knew.
  2. Since aa = 0, we have:
    a # ∗a = ∗0 for all a
    That is, ∗a # ∗a is a win for P2. And indeed, P2 has an obvious strategy: whatever P1 does in one pile, P2 does in the other pile. P2 never runs out of legal moves until after P1 does, and so must win.

  3. Since aa = 0, we have, more generally:
    a # ∗a # X = X for all a, X
    No matter what X is, its outcome is the same as that of ∗a # ∗a # X. Why?

    Suppose you are the player with a winning strategy for playing X alone. Then it is easy to see that you have a winning strategy in ∗a # ∗a # X, as follows: ignore the ∗a # ∗a component, until your opponent moves in it, when you should copy their move in the other half of that component. Eventually the ∗a # ∗a part will be used up (that is, reduced to ∗0 # ∗0 = 0) and your opponent will be forced to move in X, whereupon you can continue your winning strategy there until you win.

  4. According to the ⊕ operation, ∗1 # ∗2 = ∗3, and so ∗1 # ∗2 # ∗3 = ∗3 # ∗3 = 0, so P2 should have a winning strategy in ∗1 # ∗2 # ∗3. Which he does: If P1 removes any entire heap, P2 can win by equalizing the remaining heaps, leaving ∗1 # ∗1 = 0 or ∗2 # ∗2 = 0, which he wins easily. If P1 equalizes any two heaps, P2 can remove the third heap, winning the same way.

  5. Let's reconsider the game of the previous paragraph, but change the ∗1 to something else. 2 ⊕ 3 ⊕ x > 0 so if ∗x ≠ 1, ∗2 # ∗3 # ∗x = ∗y, where y>0. Since ∗y is a single nonempty Nim-heap, it is obviously a win for P1, and so ∗2 # ∗3 # ∗x should be equivalent, also a win for P1. What is P1's winning strategy in ∗2 # ∗3 # ∗x? It's easy. If x > 1, then P1 can reduce ∗x to ∗1, leaving ∗2 # ∗3 # ∗1, which we saw is a winning position. And if x = 0, then P1 can move to ∗2 # ∗2 and win.

4. The MEX rule

The important thing about disjoint sums is that they abstract away the strategy. If you have some complicated set of Nim-heaps ∗a # ∗b # ... # ∗z, you can ignore them and pretend instead that they are a single heap ∗(ab ⊕ ... ⊕ z). Your best move in the compound heap can be easily worked out from the corresponding best move in the fictitious single heap.

For example, how do you figure out how to play in ∗2 # ∗3 # ∗x? You consider it as (∗2 # ∗3) # ∗x = ∗1 # ∗x. That is, you pretend that the ∗2 and the ∗3 are actually a single heap of size 1. Then your strategy is to win in ∗1 # ∗x, which you obviously do by reducing ∗x to size 1, or, if ∗x is already ∗0, by changing ∗1 to ∗0.

Now, that is very facile, but ∗2 # ∗3 is not the same game as ∗1, because from ∗1 there is just one legal move, which is to ∗0. Whereas from ∗2 # ∗3 there are several moves. It might seem that your opponent could complicate the situation, say by moving from ∗2 # ∗3 to ∗3, which she could not do if it were really ∗1.

But actually this extra option can't possibly help your opponent, because you have an easy response to that move, which is to move right back to ∗1! If pretending that ∗2 # ∗3 was ∗1 was good before, it is certainly good after you make it ∗1 for real.

From ∗2 # ∗3 there are a whole bunch of moves:

Move to ∗3
Move to ∗2
Move to ∗1 # ∗3 = ∗2
Move to ∗2 # ∗1 = ∗3
Move to ∗2 # ∗2 = ∗0
But you can disregard the first four of these, because they are reversible: if some player X has a winning strategy that works by pretending that ∗2 # ∗3 is identical with ∗1, then the extra options of moving to ∗2 and ∗3 won't help X's opponent, because X can reverse those moves and turn the ∗2 # ∗3 component back into ∗1. So we can ignore these options, and say that there's just one move from ∗2 # ∗3 worth considering further, namely to ∗2 # ∗2 = 0. Since this is exactly the same set of moves that is available from ∗1, ∗2 # ∗3 behaves just like ∗1 in all situations, and have just proved that ∗2 # ∗3 = ∗1.

Unlike the other moves, the move from ∗2 # ∗3 to ∗0 is not reversible. Once someone turns ∗2 # ∗3 into ∗0, by equalizing the piles, it cannot then be turned back into ∗1, or anything else.

Considering this in more generality, suppose we have some game position P where the options are to move to one of several possible Nim-heaps, and M is the smallest Nim-heap that is not among the options. Then P = ∗M. Why? Because P has just the same options that ∗M has, namely the options of moving to one of ∗0 ... ∗(M-1). P also has some extra options, but we can ignore these because they're reversible. If you have a winning strategy in X # ∗M, then you have a winning strategy in X # P also, as follows:

  • If your opponent plays in X, then follow your strategy for X # ∗M, since the same move will also be available in X # P.

  • If your opponent makes P into ∗y, with y < M, then they've discarded their extra options, which are now irrelevant; play as you would if they had moved from X # ∗M to X # ∗y.

  • If your opponent makes P into ∗y, with y > M, then just move from ∗y to ∗M, leaving X # ∗M, which you can win.

MEX Theorem: If all the legal moves from a position P are equivalent to Nim-heaps of sizes {s1, ..., sk}, then P itself is equivalent to a nim-heap of size MEX(s1, ..., sk), where the MEX is the "Minimal EXcluded" element of the set: the smallest nonnegative integer that is not in the set.

For example, let's consider what happens if we augment Nim by adding a special token, called ♦. A player may, in lieu of a regular move, replace ♦ by a pile of beans of any positive size. What effect does this have on Nim?

Since the legal moves from ♦ are {∗1, ∗2, ∗3, ...} and the MEX is 0, ♦ should behave like ∗0. That is, adding a ♦ token to any position should leave the outcome unaffected. And indeed it does. If you have a winning strategy in game G, then you have a winning strategy in G # ♦ also, as follows: If your opponent plays in G, reply in G. If your opponent replaces ♦ with a pile of beans, remove it, leaving only G.

Exercise: Let G be a game where all the legal moves are to Nim-heaps. Then G is a win for P1 if and only if one of the legal moves from G is to ∗0, and a win for P2 if and only if none of the legal moves from G is to ∗0.

5. The Sprague-Grundy theory

An "impartial game" is one where both players have the same moves from every position.

Sprague-Grundy theorem: Any finite impartial game is equivalent to some Nim-heap ∗n, which is the "Nim-value" of the game.

Now let's consider Richard Penn's game, which is impartial. A legal move is to cross out any dot, and the adjacent dot or dots, if any.

The Sprague-Grundy theorem says that every row of dots in Penn's game is equivalent to some Nim-heap. Let's tabulate the size of this heap (the Nim-value) for each row of n dots. We'll represent a row of n dots as [οοοοο...ο]. Obviously, [] = ∗0 so the Nim-value of [] is 0. Also obviously, [ο] = ∗1, since they're exactly the same game.

[οο] = ∗1 also, since the only legal move from [οο] is to [] = 0, and the MEX of {0} is 1.

The legal moves from [οοο] are to [] = ∗0 and [ο] = ∗1, so {∗0, ∗1}, and the MEX is 2. So [οοο] = ∗2.

Let's check that this is working. Since the Nim-value of [οοο] is 2, the theory predicts that [οοο] # ∗2 = 0 and so should be a win for P2. P2 should be able to pretend that [οοο] is actually ∗2.

Suppose P1 turns the ∗2 into ∗1, moving to [οοο] # ∗1. Then P2 should turn [οοο] into ∗1 also, which he can do by crossing out an end dot and the adjacent one, leaving [ο] # ∗1, which he easily wins. If P1 turns ∗2 into ∗0, moving to [οοο] # ∗0, then P2 should turn [οοο] into ∗0 also, which he can do by crossing out the middle and adjacent dots, leaving [] # ∗0, which he wins immediately.

If P1 plays in the [οοο] component, she must move to [] or to [ο], each equivalent to some Nim-heap of size x < 2, and P2 can answer by reducing the true Nim-heap ∗2 to contain x beans also.

Continuing our analysis of rows of dots: In Penn's game, the legal moves from [οοοο] are to [οο] and [ο]. Both of these have Nim-value ∗1, so the MEX is 0.

Easy exercise: Since [οοοο] is supposedly equivalent to ∗0, you should be able to show that a player who has a winning strategy in some game G also has a winning strategy in G + [οοοο].

The legal moves from [οοοοο] are to [οοο], [οο], and [ο] # [ο]. The Nim-values of these three games are ∗2, ∗1, and ∗0 respectively, so the MEX is 3 and [οοοοο] = ∗3.

The legal moves from [οοοοοο] are to [οοοο], [οοο], and [ο] # [οο]. The Nim-values of these three games are 0, 2, and 0, so [οοοοοο] = ∗1.

6. Richard Penn's game analyzed

Row of
n dots
0 0  
1 1 []
2 1 []
3 2 []
4 0  
5 3 [ο] # [ο]
6 1 [ο] # [οο]
7 1 [οο] # [οο]
8 0  
9 3 [οοο] # [οοο]
10 3 [οοοοοοοο]
11 2 [οοοο] # [οοοο]
12 2 [οο] # [οοοοοοο]
13 4 [οοοοο] # [οοοοο]
14 0  
15 5 [οοοοοο] # [οοοοοο]
16 2 [ο × 14]
17 2 [οοοοοοο] # [οοοοοοο]
18 3 [οοο] # [ο × 12]
19 3 [οοοοοοοο] # [οοοοοοοο]
20 0  
Continuing in this way, we get the table of Nim-values that you see at left.

The table says that a row of 19 dots should be a win for P1, if she reduces the Nim-value from 3 to 0. And indeed, P1 has an easy winning strategy, which is to cross the 3 dots in the middle of the row, replacing [οοοοοοοοοοοοοοοοοοο] with [οοοοοοοο] # [οοοοοοοο]. But no such easy strategy obtains in a row of 20 dots, which, indeed, is a win for P2.

The original question involved circles of dots, not rows. But from a circle of n dots there is only one legal move, which is to a row of n-3 dots. From a circle of 20 dots, the only legal move is to [ο × 17] = ∗2, which should be a win for P1. P1 should win by changing ∗2 to ∗0, so should look for the move from [ο × 17] to ∗0. This is the obvious solution Richard Penn discovered: move to [οοοοοοο] # [οοοοοοο]. So the circle of 20 dots is an easy win for P2, the second player.

But for the circle of 19 dots the answer is the same, a win for the second player. The first player must move to [ο × 16] = ∗2, and then the second player should win by moving to a 0 position. [ο × 16] must have such a move, because if it didn't, the MEX rule would imply that its Nim-value was 0 instead of 2. So what's the second player's zero move here? There are actually two options. The second player can win by playing to [ο × 14], or by splitting the row into [οοοοοο] # [οοοοοοο].

7. Complete strategy for 19-bean circle

Just for completeness, let's follow one of these purportedly winning moves in detail. I claimed that the second player could win by moving to [οοοοοο] # [οοοοοοο]. But what next?

First recall that any isolated row of four dots, [οοοο], can be disregarded, because any first-player move in such a row can be answered by a second-player move that crosses out the rest of the row. And any pair of isolated rows of one or two dots, [ο] or [οο], can be similarly disregarded, because any move that crosses out one can be answered by a move that crosses out the other. So in what follows, positions like [οο] # [ο] # [οοοο] will be assumed to have been won by the second player, and we will say that the second player "has an easy win" if he has a move to such a position.

  • The first player has three possible moves in the left [οοοοοο] component, as follows:

    1. If the first player moves to [οοοο] # [οοοοοοο], the second player has an easy win by moving to [οοοο] # [οοοο].

    2. If the first player moves to [οοο] # [οοοοοοο] = ∗2 # ∗1, the second player should reduce the left component to ∗1, by moving to [ο] # [οοοοοοο]. Then no matter what the first player does, the second player has an easy win.

    3. If the first player moves to [ο] # [οο] # [οοοοοοο] = ∗1 # ∗1 # ∗1, the second player can disregard the [ο] # [οο] component. The second player instead plays to [ο] # [οο] # [οοοο] and wins.

  • The first player has four moves in the right [οοοοοοο] component, as follows:

    1. If the first player moves to [οοοοοο] # [οοοοο] = ∗1 # ∗3, the second player should move from ∗3 to ∗1. There must be a move in [οοοοο] to a position with Nim-value 1. (If there weren't, [οοοοο] would have Nim-value 1 instead of 3, by the MEX rule.) Indeed, the second player can move to [οοοοοο] # [οο]. Now whatever the first player does the second player has an easy win, either to [οοοο] or to X # X for some row X.

    2. If the first player moves to [οοοοοο] # [οοοο] = ∗1 # ∗0, the second player should move from ∗1 to ∗0. There must be a move in [οοοοοο] to a position with Nim-value 0, and indeed there is: the second player moves to [οοοο] # [οοοο] and wins.

    3. If the first player moves to [οοοοοο] # [ο] # [οοο] = ∗1 # ∗1 # ∗2, the second player can disregard the ∗1 # ∗1 component and should move in the ∗2 component, to ∗0, which he does by eliminating it entirely, leaving the first player with [οοοοοο] # [ο]. After any move by the first player the second player has an easy win.

    4. If the first player moves to [οοοοοο] # [οο] # [οο] = ∗1 # ∗1 # ∗1, the second player has a number of good choices. The simplest thing to do is to disregard the [οο] # [οο] component and move in the [οοοοοο] to some position with Nim-value 0. Moving to [οοοο] # [οο] # [οο] suffices.

So [ο × 17] is indeed a win for the next player to move, and a circle of 20 dots is therefore a win for the previous player, who is the second player.

But the important point here is not the strategy itself, which is hard to remember, and which could have been found by computer search. The important thing to notice is that computing the table of Nim-values for each row of n dots is easy, and once you have done this, the rest of the strategy almost takes care of itself. Do you need to find a good move from [οοοοοοο] # [οοοοοοοοο] # [οοοοοοοοοο]? There's no need to worry, because the table says that this can be viewed as ∗1 # ∗3 # ∗3, and so a good move is to reduce the ∗1 component, the [οοοοοοο], to ∗0, say by changing it to [οοοο] or to [οο] # [οο]. Whatever your opponent does next, calculating your reply will be similarly easy.

[Other articles in category /math] permanent link

Tue, 09 Sep 2008

Factorials are not quite as square as I thought
(This is a followup to yesterday's article.)

Let s(n) be the smallest perfect square larger than n. Then to have n! = a2 - 1 we must have a2 = s(n!), and in particular we must have s(n!) - n! square.

This actually occurs for n in { 4, 5, 6, 7, 8, 9, 10, 11 }, and since 11 was as far as I got on the lunch line yesterday, I had an exaggerated notion of how common it is. had I worked out another example, I would have realized that after n=11 things start going wrong. The value of s(12!) is 218872, but 218872 - 12! = 39169, and 39169 is not a square. (In fact, the n=11 solution is quite remarkable; which I will discuss at the end of this note.)

So while there are (of course) solutions to 12! = a2 - b2, and indeed where b is small compared to a, as I said, the smallest such b takes a big jump between 11 and 12. For 4 ≤ n ≤ 11, the minimal b takes the values 1, 1, 3, 1, 9, 27, 15, 18. But for n = 12, the solution with the smallest b has b = 288.

Calculations with Mathematica by Mitch Harris show that one has n! = s(n!) - b2 only for n in {1, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16}, and then not for any other n under 1,000. The likelihood that I imagine of another solution for n! = a2 - 1, which was already not very high, has just dropped precipitously.

My thanks to M. Harris, and also to Stephen Dranger, who also wrote in with the results of calculations.

Having gotten this far, I then asked OEIS about the sequence 1, 1, 3, 1, 9, 27, 15, 18, and (of course) was delivered a summary of the current state of the art in n! = a2 - 1. Here's my summary of the summary.

The question is known as "Brocard's problem", and was posed by Brocard in 1876. No solutions are known with n > 7, and it is known that if there is a solution, it must have n > 109. According to the Mathworld article on Brocard's problem, it is believed to be "virtually certain" that there are no other solutions.

The calculations for n ≤ 109 are described in this unpublished paper of Berndt and Galway, which I found linked from the Mathworld article. The authors also investigated solutions of n! = a2 - b2 for various fixed b between 2 and 50, and found no solutions with 12 ≤ n ≤ 105 for any of them. The most interesting was the 11! = 63182 - 182 I mentioned already.

[ The original version of this article contained some confusion about whether s(n) was the largest square less than n, or the largest number whose square was less than n. Thanks to Roie Marianer for pointing out the error. ]

[Other articles in category /math] permanent link

Factorials are almost, but not quite, square
This weekend I happened to notice that 7! = 712 - 1. Is this a strange coincidence? Well, not exactly, because it's not hard to see that

$$n! = a^{2} - b^{2}\qquad (*)$$

will always have solutions where b is small compared to a. For example, we have 11! = 63182 - 182.

But to get b=1 might require a lot of luck, perhaps more luck than there is. (Jeremy Kahn once argued that |2x - 3y| = 1 could have no solutions other than the obvious ones, essentially because it would require much more fabulous luck than was available. I sneered at this argument at the time, but I have to admit that there is something to it.)

Anyway, back to the subject at hand. Is there an example of n! = a2 -1 with n > 7? I haven't checked yet.

In related matters, it's rather easy to show that there are no nontrivial examples with b=0.

It would be pretty cool to show that equation (*) implied n = O(f(b)) for some function f, but I would not be surprised to find out that there is no such bound.

This kept me amused for twenty minutes while I was in line for lunch, anyway. Incidentally, on the lunch line I needed to estimate √11. I described in an earlier article how to do this. Once again it was a good trick, the sort you should keep handy if you are the kind of person who needs to know √11 while standing in line on 33rd Street. Here's the short summary: √11 = √(99/9) = √((100-1)/9) = √((100/9)(1 - 1/100) = (10/3)√(1 - 1/100) ≈ (10/3)(1 - 1/200) = (10/3)(199/200) = 199/60.

[ Addendum 20080909: There is a followup article. ]

[Other articles in category /math] permanent link

Sat, 12 Jul 2008

Period three and chaos
In the copious spare time I have around my other major project, I am tinkering with various stuff related to Möbius functions. Like all the best tinkering projects, the Möbius functions are connected to other things, and when you follow the connections you can end up in many faraway places.

A Möbius function is simply a function of the form f : x → (ax + b) / (cx + d) for some constants a, b, c, and d. Möbius functions are of major importance in complex analysis, where they correspond to certain transformations of the Riemann sphere, but I'm mostly looking at the behavior of Möbius functions on the reals, and so restricting a, b, c, and d to be real.

One nice thing about the Möbius functions is that you can identify the Möbius function f : x → (ax + b) / (cx + d) with the matrix ${ a\, b \choose c\,d}$, because then composition of Möbius functions is the same as multiplication of the corresponding matrices, and so the inverse of a Möbius function with matrix M is just the function that corresponds to M-1. Determining whether a set of Möbius functions is closed under composition is the same as determining whether the corresponding matrices form a semigroup; you can figure out what happens when you iterate a Möbius function by looking at the eigenvalues of M, and so on.

The matrices are not quite identical with the Möbius functions, because the matrix ${ 1\, 0 \choose 0\,1}$ and the matrix !!{ 2\, 0 \choose 0\,2}!! are the same Möbius function. So you really need to consider the set of matrices modulo the equivalence relation that makes two matrices equivalent if they are the same up to a scalar factor. If you do this you get a group of matrices called the "projective linear group", PGL(2). This takes us off into classical group theory and Lie groups, which I have been intermittently trying to figure out.

You can also consider various subgroups of PGL(2), such as the subgroup that leaves the set {0, 1, ∞, -1} fixed. The reciprocal function x → 1/x is one such; it leaves 1 and -1 fixed and exchanges 0 and ∞.

In general a Möbius function has three degrees of freedom, since you can choose the four constants a, b, c, and d however you like, but one degree of freedom is removed because of the equivalence relation—or, to look at it another way, you get to pick b/a, c/a, and d/a however you like. So in general you can pick any p, q, and r and find the unique Möbius function m with m(0) = p, m(1) = q, m(-1) = r. These then determine m(∞), which turns out to be (4qr - 2p(q+r))/(q + r - 2p) when that is defined. And sometimes even when it isn't.

You may be worrying about the infinities here, but it's really nothing much to worry about. f(∞) is nothing more than !!\lim_{x\rightarrow\infty} f(x)!!.

If (4qr - 2p(q+r))/(q + r - 2p) in the presence of infinities worries you, try a few examples. For instance, consider m : xx+1. This function has p = m(0) = 1, q = m(1) = 2, r = m(-1) = 0. Plugging into the formula, we get m(∞) = -2pq/(q - 2p) = -4 / (2-2) = -4/0 = ∞, which is just right.

The only other thing you have to remember is that +∞ = -∞, because we're really living on the Riemann sphere. Or rather, we're living on the real part of the Riemann sphere, but either way there's only one ∞. We might call this space the "Riemann circle", but I've never heard it called that. And neither has Google, although it did turn up a bulletin board post in which someone else asked the same question in a similar context. There's a picture of it farther down on the right.

Anyway, most choices of p, q, and r in {0, 1, ∞, -1} do not get you permutations of {0, 1, ∞, -1}, because they end up mapping ∞ outside that set. For example, if you take p = 1, q = -1, r = 0, you get m(∞) = -2/3. But obviously the identity function has the desired property, and if you think about the Riemann circle (excuse me, Riemann sphere) you immediately get the rest: any rigid motion of the Riemann sphere is a Möbius function, and some of those motions permute the four points {0, 1, ∞, -1}. In fact, there are eight such functions, because {0, 1, ∞, -1} are at the vertices of a square, so any rigid motion of the Riemann sphere that permutes {0, 1, ∞, -1} must be a rigid motion of that square, and the square has eight symmetries, namely the elements of the group D4:

D4 element m(0) m(1) m(∞) m(-1) m(x) = ? M
Identity 0 1 -1 x
1 -1 0 (x + 1) / (- x + 1)
Rotate 180° -1 0 1 - (1/x)
-1 0 1 (x - 1) / (x + 1)
0 -1 1 -x
1 0 -1 1/x
diagonally (1)
1 0 -1 (-x + 1) / (x + 1)
diagonally (2)
-1 1 0 (x + 1) / (x - 1)

Here we have eight functions on the reals which make the group D4 under the operation of composition. For example, if f(x) = (x-1)/(x+1), then f(f(f(f(x)))) = x. Isn't that nice?

Anyway, none of that was what I was really planning to talk about. (You knew that was coming, didn't you?)

What I wanted to discuss was the function f : x → 1 / (1 - x). I found this function because I was considering other permutations of {0, 1, ∞, -1}. The f function takes 0 → 1 → ∞ → 0. (It also takes -1 → 1/2, and so is not one of the functions in the D4 table above.) We say that f has a periodic point of order 3 because f(f(f(x))) = x for some x; in this case at least for x ∈ {0, 1, ∞}.

A function with a periodic point of order three is not something you see every day, and I was somewhat surprised that as simple a function as 1/(1-x) had one. But if you do the algebra and calculate f(f(f(x))) explicitly, you find that you do indeed get x, so every point is a periodic point of order 3, or possibly 1.

Or you can do a simpler calculation: since f is the Möbius function that corresponds to the matrix F = !!{ \hphantom{-}0\, 1 \choose -1\,1}!!, just calculate F3. You get !!{ -1\, \hphantom{-}0 \choose \hphantom{-}0\, -1}!!, which is indeed the identity function.

This also gives you a simple matrix M for which M7 = M, if you happened to be looking for such a thing.

I had noticed a couple of years ago that this 1/(1-x) function had period 3, and then forgot about it. Then I noticed it again a few weeks ago, and a nagging question came into my mind, which is reflected in a note I wrote in my notebook at that point: "WHAT ABOUT SARKOVSKY'S THEOREM?"

Well, what about it? Sharkovskii's theorem (I misspelled it in the notebook) is a delightful generalization of the "Period three implies chaos" theorem of Li and Yorke. It says, among other things, that if a continuous function of the reals has a periodic point of order 3, then it also has a periodic point of order n for all positive integers n. In particular, we can take n=1, so the function f, which has a periodic point of order 3 must also have a fixed point. But it's quite easy to see that f has no fixed point on the reals: Just put f(x) = 1/(1-x) = x and solve for x; there are no real solutions.

So what about Sharkovskii's theorem? Oh, it only applies to continuous functions, and f is not, because f(1) = ∞. So that's all right.

The Sharkovskii thing is excellent. The Sharkovskii ordering of the integers is:

3 < 5 < 7 < 9 < ...
  < 6 < 10 < 14 < 18 < ...
  < 12 < 20 < 28 < 36 < ...
... < 16 < 8 < 4 < 2 < 1.

And the theorem says that if a continuous function of the reals has a periodic point of order n, then it also has a periodic point of order m for all m > n in the Sharkovskii ordering. So if the function has a periodic point of order 2, it must also have a fixed point; if it has a periodic point of order 4, it must also have a periodic point of order 2; if it has a periodic point of order 17, it must also have periodic points of all even orders and all odd orders greater than 17, and so on.

The 1/(1-x) function led me to read more about Sharkovskii's theorem and its predecessor, the "period three implies chaos" theorem. Isn't that a great name for a theorem? And Li and Yorke knew it, because that's what they titled their paper. "Chaos" in this context means the following: say that two values a and b are "scrambled" by f if, for any given d and ε, there is some n for which |fn(a) - fn(b)| > d, and some m for which |fm(a) - fm(b)| < ε. That is, a and b are scrambled if repeated application of f drives a and b far apart, then close together, then far apart again, and so on. Then, if f is a continuous function with a periodic point of order 3, there is some uncountable set S of reals such that f scrambles all distinct pairs of values a and b from S. All that was from memory; I hope it got it more or less correct.

(The Li and Yorke paper also includes an example of a continuous function with a periodic point of order 5 but no periodic point of order 3. It's pretty simple.)

Reading about Sharkovskii's theorem and related matters led me to the web pages of James A. Yorke (of Li and Yorke), and then to the book Chaos: An Introduction to Dynamical Systems that he did with Alligood and Sauer, which is very readable.

I was pleased to finally be studying this material, because it was a very early inspiration to me. When I was about fourteen, my cousin Alex, who is an analytic chemist, came to visit, and told me about period-doubling and chaos in the logistic map. (It was all over the news at the time.) The logistic map is just f : x → λx(1-x) for some constant λ. For small λ, the map has a single fixed point, which increases as λ does. But at a certain critical value of λ (λ=3, actually) the function's behavior changes, and it suddenly begins to have a periodic point of order 2. As λ increases further, the behavior changes again, and the periodicity changes from order 2 to order 4. As λ increases, this happens again and again, with the splits occurring at exponentially closer and closer values of λ. Eventually there is a magic value of λ at which the function goes berserk and is chaotic. Chaos continues for a while, and then the function develops a periodic point of order 3, which bifurcates...

(The illustration here, which I copied from Wikipedia, uses r instead of λ.)

I was deeply impressed. For some reason I got the idea that I would need to understand partial differential equations to understand the chaos and the logistic map, so I immediately set out on a program to learn what I thought I would need to know. I enrolled in differential equations courses at Columbia University instead of in something more interesting. The partial differential equations turned out to be a sidetrack, but in those days there were no undergraduate courses in iterated dynamic systems.

I am happy to discover that after only twenty-five years I am finally arriving at the destination.

Cousin Alex also told me to carry a notebook and pen with me wherever I went. That was good advice, and it took me rather less time to learn.

[Other articles in category /math] permanent link

Wed, 23 Apr 2008

Recounting the rationals
I just read a really excellent math paper, Recounting the rationals, by Calkin and Wilf.

Let b(n) be the number of ways of adding up powers of 2 to get n, with each power of 2 used no more than twice. So, for example, b(5) = 2, because there are 2 ways to get 5:
5 = 4 + 1
 = 2 + 2 + 1

And b(10) = 5, because there are 5 ways to get 10:
10 = 8 + 2
  = 8 + 1 + 1
  = 4 + 4 + 2
  = 4 + 4 + 1 + 1
  = 4 + 2 + 2 + 1 + 1

The sequence of values of b(n) begins as follows:

1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 5 4 7 3 8 5 7 2 7 5 8 3 7 4 5 ...
Now consider the sequence b(n) / b(n+1). This is just what you get if you take two copies of the b(n) sequence and place one over the other, with the bottom one shifted left one place, like this:

    1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 5 4 7 3 8 5 7 2 7 5 8 3 7 4 5 ...
    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
    1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 5 4 7 3 8 5 7 2 7 5 8 3 7 4 5 ...
Reading each pair as a rational number, we get the sequence b(n) / b(n+1), which is 1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, 4/3, 3/5, 5/2, ... .

Here is the punchline: This sequence contains each positive rational number exactly once.

If you are just learning to read math papers, or you think you might like to learn to read them, the paper in which this is proved would be a good place to start. It is serious research mathematics, but elementary. It is very short. The result is very elegant. The proofs are straightforward. The techniques used are typical and widely applicable; there is no weird ad-hockery. The discussion in the paper is sure to inspire you to tinker around with it more on your own. All sorts of nice things turn up. The b(n) sequence satisfies a simple recurrence, the fractions organize themselves neatly into a tree structure, and everything is related to everything else. Check it out.

Thanks to Brent Yorgey for bringing this to my attention. I saw it in this old blog article, but then discovered he had written a six-part series about it. I also discovered that M. Yorgey independently came to the same conclusion that I did about the paper: it would be a good first paper to read.

[ Addendum 20080505: Brad Clow agrees that it was a good place to start. ]

[Other articles in category /math] permanent link

Sat, 01 Mar 2008

More rational roots of polynomials
I have a big file of ideas for blog articles, and when I feel like writing but I can't think of a topic, I look over the file. An item from last April was relevant to yesterday's article about finding rational roots of polynomials. It's a trick I saw in the first edition (1768!) of the Encyclopædia Britannica.

Suppose you have a polynomial P(x) = xn + ...+ p = 0. If it has a rational root r, this must be an integer that divides p = P(0). So far so good.

But consider P(x-1). This is a different polynomial, and if r is a root of P(x), then r+1 is a root of P(x-1). So, just as r must divide P(0), r+1 must divide P(-1). And similarly, r-1 must divide P+1.

So we have an extension of the rational root theorem: instead of guessing that some factor r of P(0) is a root, and checking it to see, we first check to see if r+1 is a factor of P(-1), and if r-1 is a factor of P(1), and proceed with the full check only if these two auxiliary tests pass.

My notes conclude with:

Is this really less work than just trying all the divisors of P(0) directly?
Let's find out.

As in the previous article, say P(x) = 3x2 + 6x - 45. The method only works for monic polynomials, so divide everything by 3. (It can be extended to work for non-monic polynomials, but the result is just that you have to divide everything by 3, so it comes to the same thing.) So we consider x2 + 2x - 15 instead. Say r is a rational root of P(x). Then:

r-1 divides P(1) = -12
r divides P(0) = -15
r+1 divides P(-1)= -16

So we need to find three consecutive integers that respectively divide 12, 15, and 16. The Britannica has no specific technique for this; it suggests doing it by eyeball. In this case, 2–3–4 jumps out pretty quickly, giving the root 3, and so does 6–5–4, which is the root -5. But the method also yields a false root: 4–3–2 suggests that -3 might be a root, and it is not.

Let's see how this goes for a harder example. I wrote a little Haskell program that generated the random polynomial x4 - 26x3 + 240 x2 - 918x + 1215.

r-1 divides P(1) = 512 = 29
r divides P(0) = 1215= 35·5
r+1 divides P(-1)= 2400= 25·3·52

That required a fair amount of mental arithmetic, and I screwed up and got 502 instead of 512, which I only noticed because 502 is not composite enough; but had I been doing a non-contrived example, I would not have noticed the error. (Then again, I would have done the addition on paper instead of in my head.) Clearly this example was not hard enough because 2–3–4 and 4–5–6 are obviously solutions, and it will not always be this easy. I increased the range on my random number generator and tried again.

The next time, it came up with the very delightful polynomial x4 - 2735x3 + 2712101 x2 - 1144435245x + 170960860950, and I decided not going to go any farther with it. The table values are easy to calculate, but they will be on the order of 170960860950, and I did not really care to try to factor that.

I decided to try one more example, of intermediate difficulty. The program first gave me x4 - 25x3 + 107 x2 - 143x + 60, which is a lucky fluke, since it has a root at 1. The next example it produced had a root at 3. At that point I changed the program to generate polynomials that had integer roots between 10 and 20, and got x4 - 61x3 + 1364 x2 - 13220x + 46800.

r-1 divides P(1) = 34864 = 22·33·17·19
r divides P(0) = 46800= 24·32·52·13
r+1 divides P(-1)= 61446= 2·3·72·11·19

This is just past my mental arithmetic ability; I got 34884 instead of 34864 in the first row, and balked at factoring 61446 in my head. But going ahead (having used the computer to finish the arithmetic), the 17 and 19 in the first and last rows are suggestive, and there is indeed a 17–18–19 to be found. Following up on the 19 in the first row suggests that we look for 19–20–21, which there is, and following up on the 11 in the last row, hoping for a 9–10–11, finds one of those too. All of these are roots, and I do have to admit that I don't know any better way of discovering that. So perhaps the method does have some value in some cases. But I had to work hard to find examples for which it made sense. I think it may be more reasonable with 18th-century technology than it is with 21st-century technology.

[Other articles in category /math] permanent link

Thu, 28 Feb 2008

Algebra techniques that don't work, except when they do
In Problems I Can't Fix in the Lecture Hall, Rudbeckia Hirta describes the efforts of a student to solve the equation 3x2 + 6x - 45 = 0. She describes "the usual incorrect strategy selected by students who can't do algebra":

3x2 + 6x - 45 = 0
3x2 + 6x = 45
x(3x + 6) = 45

She says "I stopped him before he factored out the x.".

I was a bit surprised by this, because the work so far seemed reasonable to me. I think the only mistake was not dividing the whole thing by 3 in the first step. But it is not too late to do that, and even without it, you can still make progress. x(3x + 6) = 45, so if there are any integer solutions, x must divide 45. So try x = ±1, ±3, ±5, ±9, ±15 in roughly that order. (The "look for the wallet under the lamppost" principle.) x = 3 solves the equation, and then you can get the other root, x=-5, by further application of the same method, or by dividing the original polynomial by x-3, or whatever.

If you get rid of the extra factor of 3 in the first place, the thing is even easier, because you have x(x + 2) = 15, so x = ±1, ±3, or ±5, and it is obviously solved by x=3 and x=-5.

Now obviously, this is not always going to work, but it works often enough that it would have been the first thing I would have tried. It is a lot quicker than calculating b2 - 4ac when c is as big as 45. If anyone hassles you about it, you can get them off your back by pointing out that it is an application of the so-called rational root theorem.

But probably the student did not have enough ingenuity or number sense to correctly carry off this technique (he didn't notice the 3), so that M. Hirta's advice to just use the damn quadratic formula already is probably good.

Still, I wonder if perhaps such students would benefit from exposure to this technique. I can guess M. Hirta's answer to this question: these students will not benefit from exposure to anything.

[ Addendum 20080228: Robert C. Helling points out that I could have factored the 45 in the first place, without any algebraic manipulations. Quite so; I completely botched my explanation of what I was doing. I meant to point out that once you have x(x+2) = 15 and the list [1, 3, 5, 15], the (3,5) pair jumps out at you instantly, since 3+2=5. I spent so much time talking about the unreduced polynomial x(3x+6) that I forgot to mention this effect, which is much less salient in the case of the unreduced polynomial. My apologies for any confusion caused by this omission. ]

[ Addendum 20080301: There is a followup to this article. ]

[Other articles in category /math] permanent link

Fri, 15 Feb 2008

Acta Quandalia
Several readers have emailed me to discuss my recent articles about mathematical screwups, and a few have let drop casual comments that suggest that they think that I invented Acta Quandalia as a joke. I can assure you that no journal is better than Acta Quandalia. Since it is difficult to obtain outside of university libraries, however, I have scanned the cover of one of last year's issues for you to see:

[Other articles in category /math] permanent link

Wed, 13 Feb 2008

The least interesting number
Berry's paradox goes like this: Some natural numbers, like 2, are interesting. Some natural numbers, like 255610679 (I think), are not interesting. Consider the set of uninteresting natural numbers. If this set were nonempty, it would contain a smallest element s. But then s, would have the interesting property of being the smallest uninteresting number. This is a contradiction. So the set of uninteresting natural numbers must be empty.

This reads like a joke, and it is tempting to dismiss it as a trite bit of foolishness. But it has rather interesting and deep connections to other related matters, such as the Grelling-Nelson paradox and Gödel's incompleteness theorem. I plan to write about that someday.

But today my purpose is only to argue that there are demonstrably uninteresting real numbers. I even have an example. Liouville's number L is uninteresting. It is defined as:

$$\sum_{i=1}^\infty {10}^{-i!} = 0.1100010000000000000001000\ldots$$

Why is this number of any concern? In 1844 Joseph Liouville showed that there was an upper bound on how closely an irrational algebraic number could be approximated by rationals. L can be approximated much more closely than that, and so must therefore be transcendental. This was the proof of the existence of transcendental numbers.

The only noteworthy mathematical property possessed by L is its transcendentality. But this is certainly not enough to qualify it as interesting, since nearly all real numbers are transcendental.

Liouville's theorem shows how to construct many transcendental numbers, but the construction generates many similar numbers. For example, you can replace the 10 with a 2, or the n! with floor(en) or any other fast-growing function. It appears that any potentially interesting property possessed by Liouville's number is also possessed by uncountably many other numbers. Its uninterestingness is identical to that of other transcendental numbers constructed by Liouville's method. L was neither the first nor the simplest number so constructed, so Liouville's number is not even of historical interest.

The argument in Berry's paradox fails for the real numbers: since the real numbers are not well-ordered, the set of uninteresting real numbers need have no smallest element, and in fact (by Berry's argument) does not. Liouville's number is not the smallest number of its type, nor the largest, nor anything else of interest.

If someone were to come along and prove that Liouville's number was the most uninteresting real number, that would be rather interesting, but it has not happened, nor is it likely.

[Other articles in category /math] permanent link

Thu, 07 Feb 2008

Trivial theorems
Mathematical folklore contains a story about how Acta Quandalia published a paper proving that all partially uniform k-quandles had the Cosell property, and then a few months later published another paper proving that no partially uniform k-quandles had the Cosell property. And in fact, goes the story, both theorems were quite true, which put a sudden end to the investigation of partially uniform k-quandles.

Except of course it wasn't Acta Quandalia (which would never commit such a silly error) and it didn't concern k-quandles; it was some unspecified journal, and it concerned some property of some sort of topological space, and that was the end of the investigation of those topological spaces.

This would not qualify as a major screwup under my definition in the original article, since the theorems are true, but it certainly would have been rather embarrassing. Journals are not supposed to publish papers about the properties of the empty set.

Hmm, there's a thought. How about a Journal of the Properties of the Empty Set? The editors would never be at a loss for material. And the cover almost designs itself.

Handsome, isn't it? I See A Great Need!

Ahem. Anyway, if the folklore in question is true, I suppose the mathematicians involved might have felt proud rather than ashamed, since they could now boast of having completely solved the problem of partially uniform k-quandles. But on the other hand, suppose you had been granted a doctorate on the strength of your thesis on the properties of objects from some class which was subsequently shown to be empty. Wouldn't you feel at least a bit like a fraud?

Is this story true? Are there any examples? Please help me, gentle readers.

[Other articles in category /math] permanent link

Tue, 05 Feb 2008

Major screwups in mathematics: example 1
Last month I asked for examples of major screwups in mathematics. Specifically, I was looking for cases in which some statement S was considered to be proved, and later turned out to be false. I could not think of any examples myself.

Readers suggested several examples, and I got lucky and turned up one on my own.

Some of the examples were rather obscure technical matters, where Professor Snorfus publishes in Acta Quandalia that all partially uniform k-quandles have the Cosell property, and this goes unchallenged for several years before one of the other three experts in partially uniform quandle theory notices that actually this is only true for Nemontovian k-quandles. I'm not going to report on matters that sounded like that to me, although I realize that I'm running the risk that all the examples that I do report will sound that way to most of the audience. But I'm going to give it a try.

General remarks

I would like to make some general remarks first, but I don't quite know yet what they are. Two readers independently suggested that I should read Proofs and Refutations by Imre Lakatos, and raised a number of interesting points that I'm sure I'd like to expand on, except that I haven't read the book. Both copies are checked out of the Penn library, which is a good sign, and the interlibrary loan copy I ordered won't be here for several days.

Still, I can relate a partial secondhand understanding of the ideas, which seem worth repeating.

Whether a result is "correct" may be largely a matter of definition. Consider Lakatos' principal example, Euler's theorem about polyhedra: Let F, E, and V be the number of faces, edges, and vertices in a polyhedron. Then F - E + V = 2. For example, the cube has (F, E, V) = (6, 12, 8), and 6 - 12 + 8 = 2.

Sometime later, someone observed that Euler's theorem was false for polyhedra with holes in them. For example, consider the object shown at right. It has (F, E, V) = (9, 18, 9), giving F - E + V = 9 - 18 - 9 = 0.

Can we say that Euler was wrong? Not really. The question hinges on the definition of "polyhedron". Euler's theorem is proved for "polyhedra", but we can see from the example above that it only holds for "simply-connected polyhedra". If Euler proved his theorem at a time when "polyhedra" was implicitly meant "simply-connected", and the generally-understood definition changed out from under him, we can't hold that against Euler. In fact, the failure of Euler's theorem for the object above suggests that maybe we shouldn't consider it to be a polyhedron, that it is somehow rather different from a polyhedron in at least one important way. So the theorem drives the definition, instead of the other way around.

Okay, enough introductory remarks. My first example is unquestionably a genuine error, and from a first-class mathematician.

Mathematical background

Some terminology first. A "formula" is just that, for example something like this:

$$\displaylines{ ((\forall a.\lnot R(a,a)) \wedge\cr (\forall b\forall c.R(b,c)\to\lnot R(c,b))\wedge\cr (\forall d\forall e\forall f.(R(d,e)\wedge R(e,f)\to R(d,f))) \to\cr (\forall x\exists y.R(y,x)) }$$

It may contain a bunch of quantified variables (a, b, c, etc.), relations (like R), and logical connectives like ∧. A formula might also include functions and constants (which I didn't) or equality symbols (there are none here).

One can ask whether the formula is true (or, in the jargon, "valid"), which means that it must hold regardless of how one chooses the set S from which the values of the variables will be drawn, and regardless of the meanings assigned to the relation symbols (and to the functions and constants, if there are any). The following formula, although not very interesting, is valid:

$$ \forall a\exists b.(P(a)\wedge P(b))\to P(a) $$

This is true regardless of the meaning we ascribe to P, and regardless of the set from which a and b are required to be drawn.

The longer formula above, which requires that R be a linear order, and then that the linear order R have no minimal element, is not universally valid, but it is valid for some interpretations of R and some sets S from which a...f, x, and y may be drawn. Specifically, it is true if one takes S to be the set of integers and R(x, y) to mean x < y. Such formulas, which are true for some interpretations but not for all, are called "satisfiable". Obviously, valid formulas are satisfiable, because satisfiable formulas are true under some interpretations, but valid formulas are true under all interpretations.

Gödel famously showed that it is an undecidable problem to determine whether a given formula of arithmetic is satisfiable. That is, there is no method which, given any formula, is guaranteed to tell you correctly whether or not there is some interpretation in which the formula is true. But one can limit the form of the allowable formulas to make the problem easier. To take an extreme example, just to illustrate the point, consider the set of formulas of the form:

ab... ((a=0)∨(a=1))∧((b=0)∨(b=1))∧...∧R(a,b,...)

for some number of variables. Since the formula itself requires that a, b, etc. are each either 0 or 1, all one needs to do to decide whether the formula is satisfiable is to try every possible assignment of 0 and 1 to the n variables and see whether R(a,b,...) is true in any of the 2n resulting cases. If so, the formula is satisfiable, if not then not.

Kurt Gödel, 1933

One would like to prove decidability for a larger and more general class of formulas than the rather silly one I just described. How big can the class of formulas be and yet be decidable?

It turns out that one need only consider formulas where all the quantifiers are at the front, because there is a simple method for moving quantifiers to the front of a formula from anywhere inside. So historically, attention has been focused on formulas in this form.

One fascinating result concerns the class of formulas called [∃*2*, all, (0)]. These are the formulas that begin with ∃ab...∃mnpq...∃z, with exactly two ∀ quantifiers, with no intervening ∃s. These formulas may contain arbitrary relations amongst the variables, but no functions or constants, and no equality symbol. [∃*2*, all, (0)] is decidable: there is a method which takes any formula in this form and decides whether it is satisfiable. But if you allow three ∀ quantifiers (or two with an ∃ in between) then the set of formulas is no longer decidable. Isn't that freaky?

The decidability of the class [∃*2*, all, (0)] was shown by none other than Gödel, in 1933. However, in the last sentence of his paper, Gödel added that the same was true even if the formulas were also permitted to include equality:

In conclusion, I would still like to remark that Theorem I can also be proved, by the same method, for formulas that contain the identity sign.


This was believed to be true for more than thirty years, and the result was used by other mathematicians to prove other results. But in the mid-1960s, Stål Aanderaa showed that Gödel's proof would not actually work if the formulas contained equality, and in 1983, Warren D. Goldfarb proved that Gödel had been mistaken, and the satisfiability of formulas in the larger class was not decidable.


Gödel's original 1933 paper is Zum Entscheidungsproblem des logischen Funktionenkalküls (On the decision problem for the functional calculus of logic) which can be found on pages 306–327 of volume I of his Collected Works. (Oxford University Press, 1986.) There is an introductory note by Goldfarb on pages 226–231, of which pages 229–231 address Gödel's error specifically.

I originally heard the story from Val Tannen, and then found it recounted on page 188 of The Classical Decision Problem, by Egon Boerger, Erich Grädel, and Yuri Gurevich. But then blog reader Jeffrey Kegler found the Goldfarb note, of which the Boerger-Grädel-Gurevich account appears to be a summary.

Thanks very much to everyone who contributed, and especially to M. Kegler.

(I remind readers who have temporarily forgotten, that Acta Quandalia is the quarterly journal of the Royal Uzbek Academy of Semi-Integrable Quandle Theory. Professor Snorfus, you will no doubt recall, won the that august institution's prestigious Utkur Prize in 1974.)

[ Addendum 20080206: Another article in this series. ]

[ Addendum 20200206: A serious mistake by Henri Lebesgue. ]

[Other articles in category /math] permanent link

Thu, 31 Jan 2008

Ramanujan's congruences
Let p(n) be the number of partitions of the integer n. For example, p(4) = 5 because there are 5 partitions of the integer 4, namely {4, 3+1, 2+2, 2+1+1, 1+1+1+1}.

Ramanujan's congruences state that:

p(5k+4) =0 (mod 5)
p(7k+5) =0 (mod 7)
p(11k+6) =0 (mod 11)

Looking at this, anyone could conjecture that p(13k+7) = 0 (mod 13), but it isn't so; p(7) = 15 and p(20) = 48·13+3.

But there are other such congruences. For example, according to Partition Congruences and the Andrews-Garvan-Dyson Crank:

$$ p(17\cdot41^4k + 1122838) = 0 \pmod{17} $$

Isn't mathematics awesome?

[Other articles in category /math] permanent link

Fri, 25 Jan 2008

Nonstandard adjectives in mathematics
Ranjit Bhatnagar once propounded the notion of a "nonstandard" adjective. This is best explained by an example. "Red" is not usually a nonstandard adjective, because a red boat is still a boat, a red hat is still a hat, and a red flag is still a flag. But "fake" is typically nonstandard, because a fake diamond is not a diamond, a fake Gucci handbag is not a Gucci handbag.

The property is not really attached to the adjective itself. Red emeralds are not emeralds, so "red" is nonstandard when applied to emeralds. Fake expressions of sympathy are still expressions of sympathy, however insincere. "Toy" often goes both ways: a toy fire engine is not a fire engine, but a toy ball is a ball and a toy dog is a dog.

Adjectives in mathematics are rarely nonstandard. An Abelian group is a group, a second-countable topology is a topology, an odd integer is an integer, a partial derivative is a derivative, a well-founded order is an order, an open set is a set, and a limit ordinal is an ordinal.

When mathematicians want to express that a certain kind of entity is similar to some other kind of entity, but is not actually some other entity, they tend to use compound words. For example, a pseudometric is not (in general) a metric. The phrase "pseudo metric" would be misleading, because a "pseudo metric" sounds like some new kind of metric. But there is no such term.

But there is one glaring exception. A partial function is not (in general) a function. The containment is in the other direction: all functions are partial functions, but not all partial functions are functions. The terminology makes more sense if one imagines that "function" is shorthand for "total function", but that is not usually what people say.

If I were more quixotic, I would propose that partial functions be called "partialfunctions" instead. Or perhaps "pseudofunctions". Or one could go the other way and call them "normal relations", where "normal" can be replaced by whatever adjective you prefer—ejective relations, anyone?

I was about to write "any of these would be preferable to the current confusion", but actually I think it probably doesn't matter very much.

[ Addendum 20080201: Another example, and more discussion of "partial". ]

[ Addendum 20081205: A contravariant functor is not a functor. ]

[ Addendum 20090121: A hom-set is not a set. ]

[ Addendum 20110905: A skew field is not a field. The Wikipedia article about division rings observes that this use of "skew" is counter to the usual behavior of adjectives in mathematics. ]

[ Addendum 20120819: A snub cube is not a cube. Several people have informed me that a quantum group is not a group. ]

[ Addendum 20140708: nLab refers to the red herring principle, that “in mathematics, a ‘red herring’ need not, in general, be either red or a herring”. ]

[ Addendum 20160505: The gaussian integers contain the integers, not vice versa, so a gaussian integer is not in general an integer. ]

[ Addendum 20190503: Timon Salar Gutleb points out that affine spaces were at one time called “affine vector spaces” so that every vector space was an affine vector space, but not vice versa. ]

[Other articles in category /math] permanent link

Wed, 09 Jan 2008

Major screwups in mathematics
I don't remember how I got thinking about this, but for the past week or so I've been trying to think of a major screwup in mathematics. Specifically, I want a statement S such that:

  1. A purported (but erroneous) proof of S was published in the mathematical literature, so that
  2. S was generally accepted as true for a significant period of time, say at least two years, but
  3. S is actually false
I cannot think of an example.

There are many examples of statements that were believed without proof that turned out to be false, such as any number of decidability and completeness (non-)theorems. If it turns out that P=NP, this will be one of those type, but as yet there is no generally accepted proof to the contrary, so it is not an example. Similarly, if would be quite surprising to learn that the Goldbach conjecture was false, but at present mathematicians do not generally believe that it has been proved to be true, so the Goldbach conjecture is not an example of this type, and is unlikely ever to be.

There are a lot of results that could have gone one way or another, such as the three-dimensional kissing number problem. In this case some people believing they could go one way and some the other, and then they found that it was one way, but no proof to the contrary was ever widely accepted.

Then we have results like the independence of the parallel postulate, where people thought for a long time that it should be implied by the rest of Euclidean geometry, and tried to prove it, but couldn't, and eventually it was determined to be independent. But again, there was no generally accepted proof that it was implied by the other postulates. So mathematics got the right answer in this case: the mathematicians tried to prove a false statement, and failed, and then eventually figured it out.

Alfred Kempe is famous for producing an erroneous proof of the four-color map theorem, which was accepted for eleven years before the error was detected. But the four-color map theorem is true. I want an example of a false statement that was believed for years because of an erroneous proof.

If there isn't one, that is an astonishing declaration of success for all of mathematics and for its deductive methods. 2300 years without one major screwup!

It seems too good to be true. Is it?

Glossary for non-mathematicians

  • The "decidability and completeness" results I allude to include the fact that the only systems of mathematical axioms strong enough to prove all true statements of arithmetic, are those that are so strong that they also prove all the false statements of arithmetic. A number of results of this type were big surprises in the early part of the 20th century.

  • If "P=NP" were true, then it would be possible to efficiently find solutions to any problem whose solutions could be efficiently checked for correctness. For example, it is relatively easy to check to see if a proposed conference schedule puts two speakers in the same room at the same time, if it allots the right amount of time for each talk, if it uses no more than the available number of rooms, and so forth. But to generate such schedules seems to be a difficult matter in general. "P=NP" would imply that this problem, and many others that seem equally difficult, was actually easy.

  • The Goldbach conjecture says that every even number is the sum of two prime numbers.

  • The kissing number problem takes a red ping-pong ball and asks how many white ping-pong balls can simultaneously touch it. It is easy to see that there is room for 12 white balls. There is a lot of space left over, and for some time it was an open question whether there was a way to fit in a 13th. The answer turns out to be that there is not.

  • The four-color map theorem asks whether any geographical map (subject to certain restrictions) can be colored with only four colors such that no two adjacent regions are the same color. It is quite easy to see that at least four colors may be necessary (Belgium, France, Germany, and Luxembourg, for example), and not hard to show that five colors are sufficient.

  • Classical Greek geometry contained a number of "postulates", such as "any line can be extended to infinity" and "a circle can be drawn with any radius around any center", but the fifth one, the notorious "parallel postulate", was a complicated and obscure technical matter, which turns out to be equivalent to the statement that, for any line L and point P not on L, there is exactly one line L' through P parallel to L. This in turn is equivalent to the fact that classical geometry is done on a plane, and not on a curved surface.

[ Addendum 20080205: Readers suggested some examples, and I happened upon one myself. For a summary, see this month's addenda. I also wrote a detailed article about a mistake of Kurt Gödel's. ]

[ Addendum 20080206: Another article in this series, asking readers for examples of a different type of screwup. ]

[Other articles in category /math] permanent link

Tue, 11 Dec 2007

More notes on power series
It seems I wasn't done thinking about this. I pointed out in yesterday's article that, having defined the cosine function as:

     coss = zipWith (*) (cycle [1,0,-1,0]) (map ((1/) . fact) [0..])
one has the choice to define the sine function analogously:

     sins = zipWith (*) (cycle [0,1,0,-1]) (map ((1/) . fact) [0..])
or in a totally different way, by reference to cosine:

     sins = (srt . (add one) . neg . sqr) coss
Here is a third way. Sine and cosine are solutions of the differential equation f = -f''. Since I now have enough infrastructure to get Haskell to solve differential equations, I can use this to define sine and cosine:

    solution_of_equation f0 f1 = func
        where func = int f0 (int f1 (neg func))
    sins = solution_of_equation 0 1
    coss = solution_of_equation 1 0
The constants f0 and f1 specify the initial conditions of the differential equation, values for f(0) and f'(0), respectively.

Well, that was fun.

One problem with the power series approach is that the answer you get is not usually in a recognizable form. If what you get out is

then you might recognize it as the cosine function. But last night I couldn't sleep because I was wondering about the equation f·f' = 1, so I got up and put it in, and out came:

Okay, now what? Is this something familiar? I'm wasn't sure. One thing that might help a bit is to get the program to disgorge rational numbers rather than floating-point numbers. But even that won't completely solve the problem.

One thing I was thinking about in the shower is doing Fourier analysis; this should at least identify the functions that are sinusoidal. Suppose that we know (or believe, or hope) that some power series a1x + a3x3 + ... actually has the form c1 sin x + c2 sin 2x + c3 sin 3x + ... . Then we can guess the values of the ci by solving a system of n equations of the form:

$$\sum_{i=1}^n i^kc_i = k!a_k\qquad{\hbox{($k$ from 1 to $n$)}}$$

And one ought to be able to do something analogous, and more general, by including the cosine terms as well. I haven't tried it, but it seems like it might work.

But what about more general cases? I have no idea. If you have the happy inspiration to square the mystery power series above, you get [1, 2, 0, 0, 0, ...], so it is √(2x+1), but what if you're not so lucky? I wasn't; I solved it by a variation of Gareth McCaughan's method of a few days ago: f·f' is the derivative of f2/2, so integrate both sides of f·f' = 1, getting f2/2 = x + C, and so f = √(2x + C). Only after I had solved the equation this way did I try squaring the power series, and see that it was simple.

I'll keep thinking.

[Other articles in category /math] permanent link

Mon, 10 Dec 2007

Lazy square roots of power series return
In an earlier article I talked about wanting to use lazy streams to calculate the power series expansion of the solution of this differential equation:

$$(f(x))^2 + (f'(x))^2 = 1$$

To do that I decided I would need a function to calculate the square root of a power series, which I did figure out; it's in the earlier article. But then I got distracted with other issues, and then folks wrote to me with several ways to solve the differential equation, and I spent a lot of time writing that up, and I didn't get back to the original problem until today, when I had to attend the weekly staff meeting. I get a lot of math work done during that meeting.

At least one person wrote to ask me for the Haskell code for the power series calculations, so here's that first off.

A power series a0 + a1x + a2x2 + a3x3 + ... is represented as a (probably infinite) list of numbers [a0, a1, a2, ...]. If the list is finite, the missing terms are assumed to be all 0.

The following operators perform arithmetic on functions:

	-- add functions a and b
	add [] b = b
	add a [] = a
	add (a:a') (b:b') = (a+b) : add a' b'

	-- multiply functions a and b
	mul [] _ = []
	mul _ [] = []
	mul (a:a') (b:b') = (a*b) : add (add (scale a b')
					     (scale b a'))
					(0 : mul a' b')

	-- termwise multiplication of two series
        mul2 = zipWith (*)

        -- multiply constant a by function b
    	scale a b = mul2 (cycle [a]) b
	neg a = scale (-1) a
And there are a bunch of other useful utilities:

	-- 0, 1, 2, 3, ...
	iota = 0 : zipWith (+) (cycle [1]) iota
	-- 1, 1/2, 1/3, 1/4, ...
	iotaR = map (1/) (tail iota)

	-- derivative of function a
	deriv a = tail (mul2 iota a)

	-- integral of function a
	-- c is the constant of integration
	int c a = c : (mul2 iotaR a)

	-- square of function f
	sqr f = mul f f

	-- constant function
	con c = c : cycle [0]
	one = con 1
The really interesting operators perform division and evolve square roots of functions. I discussed how these work in the earlier article. The reciprocal operation is well-known; it appears in Structure and Interpretation of Computer Programs, Higher-Order Perl, and I presume elsewhere. I haven't seen the square root extractor anywhere else, but I'm sure that's just because I haven't looked.

	-- reciprocal of argument function
	inv (s0:st) = r
	  where r = r0 : scale (negate r0) (mul r st)
		r0 = 1/s0

	-- divide function a by function b
	squot a b = mul a (inv b)

	-- square root of argument function
	srt (s0:s) = r 
	   where r = r0 : (squot s (add [r0] r))
		 r0 = sqrt(s0)

We can define the cosine function as follows:

	coss = zipWith (*) (cycle [1,0,-1,0]) (map ((1/) . fact) [0..])
We could define the sine function analogously, or we can say that sin(x) = √(1 - cos2(x)):

	sins = (srt . (add one) . neg . sqr) coss
This works fine.

Okay, so as usual that is not what I wanted to talk about; I wanted to show how to solve the differential equation. I found I was getting myself confused, so I decided to try to solve a simpler differential equation first. (Pólya says: "Can you solve a simpler problem of the same type?" Pólya is a smart guy. When the voice talking in your head is Pólya's, you better pay attention.) The simplest relevant differential equation seemed to be f = f'. The first thing I tried was observing that for all f, f = f0 : mul2 iotaR f'. This yields the code:

     f = f0 : mul2 iotaR (deriv f)
This holds for any function, and so it's unsolvable. But if you combine it with the differential equation, which says that f = f', you get:

     f = f0 : mul2 iotaR f
       where f0 = 1   -- or whatever the initial conditions dictate
and in fact this works just fine. And then you can observe that this is just the definition of int; replacing the definition with the name, we have:

     f = int f0 f
       where f0 = 1   -- or whatever
This runs too, and calculates the power series for the exponential function, as it should. It's also transparently obvious, and makes me wonder why it took me so long to find. But I was looking for solutions of the form:

     f = deriv f
which Haskell can't figure out. It's funny that it only handles differential equations when they're expressed as integral equations. I need to meditate on that some more.

It occurs to me just now that the f = f0 : mul2 iotaR (deriv f) identity above just says that the integral and derivative operators are inverses. These things are always so simple in hindsight.

Anyway, moving along, back to the original problem, instead of f = f', I want f2 + (f')2 = 1, or equivalently f' = √(1 - f2). So I take the derivative-integral identity as before:

     f = int f0 (deriv f)
and put in √(1 - f2) for deriv f:

     f = int f0 ((srt . (add one) . neg . sqr) f)
       where f0 = sqrt 0.5   -- or whatever
And now I am done; Haskell cheerfully generates the power series expansion for f for any given initial condition. (The parameter f0 is precisely the desired value of f(0).) For example, when f(0) = √(1/2), as above, the calculated terms show the function to be exactly √(1/2)·(sin(x) + cos(x)); when f(0) = 0, the output terms are exactly those of sin(x). When f(0) = 1, the output blows up and comes out as [1, 0, NaN, NaN, ...]. I'm not quite sure why yet, but I suspect it has something to do with there being two different solutions that both have f(0) = 1.

All of this also works just fine in Perl, if you build a suitable lazy-list library; see chapter 6 of HOP for complete details. Sample code is here. For a Scheme implementation, see SICP. For a Java, Common Lisp, Python, Ruby, or SML implementation, do the obvious thing.

But anyway, it does work, and I thought it might be nice to blog about something I actually pursued to completion for a change. Also I was afraid that after my week of posts about Perl syntax, differential equations, electromagnetism, Unix kernel internals, and paint chips in the shape of Austria, the readers of Planet Haskell, where my blog has recently been syndicated, were going to storm my house with torches and pitchforks. This article should mollify them for a time, I hope.

[ Addendum 20071211: Some additional notes about this. ]

[Other articles in category /math] permanent link

Sun, 09 Dec 2007

Four ways to solve a nonlinear differential equation
In a recent article I mentioned the differential equation:

$$(f(x))^2 + \left(df(x)\over dx\right)^2 = 1$$

which I was trying to solve by various methods. The article was actually about calculating square roots of power series; I got sidetracked on this. Before I got back to the original equation, twofour readers of this blog had written in with solutions, all different.

I got interested in this a few weeks ago when I was sitting in on a freshman physics lecture at Penn. I took pretty much the same class when I was a freshman, but I've never felt like I really understood physics. Sitting in freshman physics class again confirms this. Every time I go to a class, I come out with bigger questions than I went in.

The instructor was talking about LC circuits, which are simple circuits with a capacitor (that's the "C") and an inductor (that's the "L", although I don't know why). The physics people claim that in such a circuit the capacitor charges up, and then discharges again, repeatedly. When one plate of the capacitor is full of electrons, the electrons want to come out, and equalize the charge on the plates, and so they make a current flowing from the negative to the positive plate. Without the inductor, the current would fall off exponentially, as the charge on the plates equalized. Eventually the two plates would be equally charged and nothing more would happen.

But the inductor generates an electromotive force that tends to resist any change in the current through it, so the decreasing current in the inductor creates a force that tends to keep the electrons moving anyway, and this means that the (formerly) positive plate of the capacitor gets extra electrons stuffed into it. As the charge on this plate becomes increasingly negative, it tends to oppose the incoming current even more, and the current does eventually come to a halt. But by that time a whole lot of electrons have moved from the negative to the positive plate, so that the positive plate has become negative and the negative plate positive. Then the electrons come out of the newly-negative plate and the whole thing starts over again in reverse.

In practice, of course, all the components offer some resistance to the current, so some of the energy is dissipated as heat, and eventually the electrons stop moving back and forth.

Anyway, the current is nothing more nor less than the motion of the electrons, and so it is proportional to the derivative of the charge in the capacitor. Because to say that current is flowing is exactly the same as saying that the charge in the capacitor is changing. And the magnetic flux in the inductor is proportional to rate of change of the current flowing through it, by Maxwell's laws or something.

The amount of energy in the whole system is the sum of the energy stored in the capacitor and the energy stored in the magnetic field of the inductor. The former turns out to be proportional to the square of the charge in the capacitor, and the latter to the square of the current. The law of conservation of energy says that this sum must be constant. Letting f(t) be the charge at time t, then df/dt is the current, and (adopting suitable units) one has:

$$(f(x))^2 + \left(df(x)\over dx\right)^2 = 1$$

which is the equation I was considering.

Anyway, the reason for this article is mainly that I wanted to talk about the different methods of solution, which were all quite different from each other. Michael Lugo went ahead with the power series approach I was using. Say that:

$$ \halign{\hfil $\displaystyle #$&$\displaystyle= #$\hfil\cr f & \sum_{i=0}^\infty a_{i}x^{i} \cr f' & \sum_{i=0}^\infty (i+1)a_{i+1}x^{i} \cr } $$


$$ \halign{\hfil $\displaystyle #$&$\displaystyle= #$\hfil\cr f^2 & \sum_{i=0}^\infty \sum_{j=0}^{i} a_{i-j} a_j x^{i} \cr (f')^2 & \sum_{i=0}^\infty \sum_{j=0}^{i} (i-j+1)a_{i-j+1}(j+1)a_{j+1} x^{i} \cr } $$

And we want the sum of these two to be equal to 1.

Equating coefficients on both sides of the equation gives us the following equations:

!!a_0^2 + a_1^2!! = 1
!!2a_0a_1 + 4a_1a_2!! = 0
!!2a_0a_2 + a_1^2 + 6a_1a_3 + 4a_2^2!! = 0
!!2a_0a_3 + 2a_1a_2 + 8a_1a_4 + 12a_2a_3!! = 0
!!2a_0a_4 + 2a_1a_3 + a_2^2 + 10a_1a_5 + 16a_2a_4 + 9a_3^2!! = 0
Now here's the thing M. Lugo noticed that I didn't. You can separate the terms involving even subscripts from those involving odd subscripts. Suppose that a0 and a1 are both nonzero. The polynomial from the second line of the table, 2a0a1 + 4a1a2, factors as 2a1(a0 + 2a2), and one of these factors must be zero, so we immediately have a2 = -a0/2.

Now take the next line from the table, 2a0a2 + a12 + 6a1a3 + 4a22. This can be separated into the form 2a2(a0 + 2a2) + a1(a1 + 6a3). The left-hand term is zero, by the previous paragraph, and since the whole thing equals zero, we have a3 = -a1/6.

Continuing in this way, we can conclude that a0 = -2!a2 = 4!a4 = -6!a6 = ..., and that a1 = -3!a3 = 5!a5 = ... . These should look familiar from first-year calculus, and together they imply that f(x) = a0 cos(x) + a1 sin(x), where (according to the first line of the table) a02 + a12 = 1. And that is the complete solution of the equation, except for the case we omitted, when either a0 or a1 is zero; these give the trivial solutions f(x) = ±1.

Okay, that was a lot of algebra grinding, and if you're not as clever as M. Lugo, you might not notice that the even terms of the series depend only on a0 and the odd terms only on a1; I didn't. I thought they were all mixed together, which is why I alluded to "a bunch of not-so-obvious solutions" in the earlier article. Is there a simpler way to get the answer?

Gareth McCaughan wrote to me to point out a really clever trick that solves the equation right off. Take the derivative of both sides of the equation; you immediately get 2ff' + 2f'f'' = 0, or, factoring out f', f'(f + f'') = 0. So there are two solutions: either f'=0 and f is a constant function, or f + f'' = 0, which even the electrical engineers know how to solve.

David Speyer showed a third solution that seems midway between the two in the amount of clever trickery required. He rewrote the equation as:

$${df\over dx} = \sqrt{1 - f^2}$$

$${df\over\sqrt{1 - f^2} } = dx$$

The left side is an old standby of calculus I classes; it's the derivative of the arcsine function. On integrating both sides, we have:

$$\arcsin f = x + C$$

so f = sin(x + C). This is equivalent to the a0 cos(x) + a1 sin(x) form that we got before, by an application of the sum-of-angles formula for the sine function. I think M. McCaughan's solution is slicker, but M. Speyer's is the only one that I feel like I should have noticed myself.

Finally, Walt Mankowski wrote to tell me that he had put the question to Maple, which disgorged the following solution after a few seconds:

  f(x) = 1, f(x) = -1, f(x) = sin(x - _C1), f(x) = -sin(x - _C1).
This is correct, except that the appearance of both sin(x + C) and -sin(x + C) is a bit odd, since -sin(x + C) = sin(x + (C + π)). It seems that Maple wasn't clever enough to notice that. Walt says he will ask around and see if he can find someone who knows what Maple did to get the solution here.

I would like to add a pithy and insightful conclusion to this article, but I've been working on it for more than a week now, and also it's almost lunch time, so I think I'll have to settle for observing that sometimes there are a lot of ways to solve math problems.

Thanks again to everyone who wrote in about this.

[Other articles in category /math] permanent link

Sat, 01 Dec 2007

19th-century elementary arithmetic
In grade school I read a delightful story, by C. A. Stephens, called The Jonah. In the story, which takes place in 1867, Grandma and Grandpa are away for the weekend, leaving the kids alone on the farm. The girls make fried pies for lunch.

They have a tradition that one or two of the pies are "Jonahs": they look the same on the outside, but instead of being filled with fruit, they are filled with something you don't want to eat, in this case a mixture of bran and cayenne pepper. If you get the Jonah pie, you must either eat the whole thing, or crawl under the table to be a footstool for the rest of the meal.

Just as they are about to serve, a stranger knocks at the door. He is an old friend of Grandpa's. They invite him to lunch, of course removing the Jonahs from the platter. But he insists that they be put back, and he gets the Jonah, and crawls under the table, marching it around the dining room on his back. The ice is broken, and the rest of the afternoon is filled with laughter and stories.

Later on, when the grandparents return, the kids learn that the elderly visitor was none other than Hannibal Hamlin, formerly Vice-President of the United States.

A few years ago I tried to track this down, and thanks to the Wonders of the Internet, I was successful. Then this month I had the library get me some other C. A. Stephens stories, and they were equally delightful and amusing.

In one of these, the narrator leaves the pump full of water overnight, and the pipe freezes solid. He then has to carry water for forty head of cattle, in buckets from the kitchen, in sub-freezing weather. He does eventually manage to thaw the pipe. But why did he forget in the first place? Because of fractions:

I had been in a kind of haze all day over two hard examples in complex fractions at school. One of them I still remember distinctly:

$${7\over8} \; {\rm of} \; {60 {5\over10} \over 10 {3\over8}} \; {\rm of} \; {8\over 5} \; \div \; 8{68\over 415} = {\rm What?}$$

At that point I had to stop reading and calculate the answer, and I recommend that you do the same.

I got the answer wrong, by the way. I got 25/64 or 64/25 or something of the sort, which suggests that I flipped over an 8/5 somewhere, because the correct answer is exactly 1. At first I hoped perhaps there was some 19th-century precedence convention I was getting wrong, but no, it was nothing like that. The precedence in this problem is unambiguous. I just screwed up.

Entirely coincidentally (I was investigating the spelling of the word "canceling") I also recently downloaded (from Google Books) an arithmetic text from the same period, The National Arithmetic, on the Inductive System, by Benjamin Greenleaf, 1866. Here are a few typical examples:

  1. If 7/8 of a bushel of corn cost 63 cents, what cost a bushel? What cost 15 bushels?

  2. When 14 7/8 tons of copperas are sold for $500, what is the value of 1 ton? what is the value of 9 11/12 tons?

  3. If a man by laboring 15 hours a day, in 6 days can perform a certain piece of work, how many days would it require to do the same work by laboring 10 hours a day?

  4. Bought 87 3/7 yards of broadcloth for $612; what was the value for 14 7/10 yards?

  5. If a horse eat 19 3/7 bushels of oats in 87 3/7 days, how many will 7 horses eat in 60 days?

Some of these are rather easy, but others are a long slog. For example, #1 and #3 here (actually #1 and #25 in the book) can be solved right off, without paper. But probably very few people have enough skill at mental arithmetic to carry off $612/(83 3/7) * (14 7/10) in their heads.

The "complex fractions" section, which the original problem would have fallen under, had it been from the same book, includes problems like this: "Add 1/9, 2 5/8, 45/(94 7/11), and (47 5/9)/(314 3/5) together." Such exercises have gone out of style, I think.

In addition to the complicated mechanical examples, there is some good theory in the book. For example, pages 227–229 concern continued fraction expansions of rational numbers, as a tool for calculating simple rational approximations of rationals. Pages 417–423 concern radix-n numerals, with special attention given to the duodecimal system. A typical problem is "How many square feet in a floor 48 feet 6 inches long, and 24 feet 3 inches broad?" The remarkable thing here is that the answer is given in the form 1176 sq. feet. 1' 6'', where the 1' 6'' actually means 1/12 + 6/144 square feet— that is, it is a base-12 "decimal".

I often hear people bemoaning the dumbing-down of the primary and secondary school mathematics curricula, and usually I laugh at those people, because (for example) I have read a whole stack of "College Algebra" books from the early 20th century, which deal in material that is usually taken care of in 10th and 11th grades now. But I think these 19th-century arithmetics must form some part of an argument in the other direction.

On the other hand, those same people often complain that students' time is wasted by a lot of "new math" nonsense like base-12 arithmetic, and that we should go back to the tried and true methods of the good old days. I did not have an example in mind when I wrote this paragraph, but two minutes of Google searching turned up the following excellent example:

Most forms of life develop random growths which are best pruned off. In plants they are boles and suckerwood. In humans they are warts and tumors. In the educational system they are fashionable and transient theories of education created by a variety of human called, for example, "Professor Of The Teaching Of Mathematics."

When the Russians launched Sputnik these people came to the rescue of our nation; they leapfrogged the Russians by creating and imposing on our children the "New Math."

They had heard something about digital computers using base 2 arithmetic. They didn't know why, but clearly base 10 was old fashioned and base 2 was in. So they converted a large fraction of children's arithmetic education to learning how to calculate with any base number and to switch from base to base. But why, teacher? Because that is the modern way. No one knows how many potential engineers and scientists were permanently turned away by this inanity.

Fortunately this lunacy has now petered out.

(Smart Machines, by Lawrence J. Kamm; chapter 11, "Smart Machines in Education".)

Pages 417–423 of The National Arithmetic, with their problems on the conversion from base-6 to base-11 numerals, suggest that those people may not know what they are talking about.

[Other articles in category /math] permanent link

Fri, 30 Nov 2007

Lazy square roots of power series
Lately for various reasons I have been investigating the differential equation:

$$(f(x))^2 + (f'(x))^2 = 1$$

where !!f'!! is the derivative of !!f!!. This equation has a couple of obvious solutions (!!f(x) = 1!!; !!f(x) = \sin(x)!!) and a bunch of not-so-obvious ones. Since I couldn't solve the equation symbolically, I decided to fall back on power series. Representing !!f(x)!! as !!a_0 + a_1x + a_2x^2 + \ldots!!, one can manipulate the power series and solve for !!a_0, a_1, a_2!!, etc. In fact, this is exactly the application for which mathematicians first became intersted in power series. The big question is "once you have found !!a_0, a_1!!, etc., do these values correspond to a real function? And for what !!x!! does the power series expression actually make sense?" This question, springing from a desire to solve intractable differential equations, motivates a lot of the theoretical mathematics of the last hundred and fifty years.

I decided to see if I could use the power series methods of chapter 6 of Higher-Order Perl to calculate !!a_0!!, etc. So far, not yet, although I am getting closer. The key is that if $series is the series you want, and if you can calculate at least one term at the front of the series, and then express the rest of $series in terms of $series, you win. For example:

        # Perl
        my $series;
        $series = node(1, promise { scale(2, $series) } );
This is perfectly well-defined code; it runs fine and sets $series to be the series !![1,2,4,8,16...]!!. In Haskell this is standard operating procedure:

        -- Haskell
        series = 1 : scale 2 series
But in Perl it's still a bit outré.

Similarly, the book shows, on page 323, how to calculate the reciprocal of a series !!s!!. Any series can be expressed as the sum of the first term and the rest of the terms:

$$s = s_H + xs_T$$ Now suppose that !!r=\frac1s!!. $$ r = r_H + xr_T$$ We have: $$ \begin{array}{rcl} rs & = & 1 \\ (r_H + xr_T)(s_H+xs_T) & = & 1 \\ r_Hs_H + xr_Hs_T + xr_Ts_H + x^2r_Ts_T & = & 1 \end{array} $$ Equating the constant terms on both sides gives !!r_Hs_H=1!!, so !!r_H = \frac1{s_H}!!. Then equating the non-constant terms gives:

$$\begin{array}{rcl} xr_Hs_T + xr_Ts_H + x^2r_Ts_T & = & 0 \\ x\frac1{s_H}s_T + xr_Ts_H + x^2r_Ts_T & = & 0 \\ \frac1{s_H}s_T + r_Ts_H + xr_Ts_T & = & 0 \\ r_T & = & \frac{-\frac1{s_H}s_T}{s_H+xs_T} \\ r_T & = & \frac{-\frac1{s_H}s_T}s \\ r_T & = & {-\frac1{s_H}s_T}r \end{array} $$ and we win. This same calculation appears on page 323, in a somewhat more confused form. (Also, I considered only the special case where !!s_H = 1!!.) The code works just fine.

To solve the differential equation !!f^2 + (f')^2 = 1!!, I want to do something like this:

$$f = \sqrt{1 - {(f')}^{2}}$$

so I need to be able to take the square root of a power series. This does not appear in the book, and I have not seen it elsewhere. Here it is.

Say we want !!r^2 = s!!, where !!s!! is known. Then write, as usual:

$$\begin{array}{rcl} s &=& \operatorname{head}(s) + x·\operatorname{tail}(s) \\ r &=& \operatorname{head}(r) + x·\operatorname{tail}(r) \\ \end{array} $$ as before, and, since !!r^2 = s!!, we have:

$$ (\operatorname{head}(r))^2 + 2x \operatorname{head}(r) \operatorname{tail}(r) + x^2(\operatorname{tail}(r))^2 = \operatorname{head}(s) + x·\operatorname{tail}(s) $$ so, equating coefficients on both sides, !!(\operatorname{head}(r))^2 = \operatorname{head}(s)!!, and !!\operatorname{head}(r) = \sqrt{\operatorname{head}(s)}!!.

Subtracting the !!\operatorname{head}(s)!! from both sides, and dividing by !!x!!:

$$\begin{array}{rcl} 2\operatorname{head}(r) \operatorname{tail}(r) + x·(\operatorname{tail}(r))^2 &=& \operatorname{tail}(s) \\ \operatorname{tail}(r)·(2·\operatorname{head}(r) + x·\operatorname{tail}(r)) &=& \operatorname{tail}(s) \\ \operatorname{tail}(r)·(\operatorname{head}(r) + r) &=& \operatorname{tail}(s) \\ \operatorname{tail}(r) &=& \frac{\operatorname{tail}(s)}{\operatorname{head}(r) + r} \end{array}$$ and we win. Or rather, we win once we write the code, which would be something like this:

        # Perl
        sub series_sqrt {
          my $s = shift;
          my ($s0, $st) = (head($s), tail($s));
          my $r0 = sqrt($s0);
          my $r;
          $r  = node($r0, 
                      promise {
                               add2(node($r0, undef),
          return $r;
I confess I haven't tried this in Perl yet, but I have high confidence that it will work. I actually did the implementation in Haskell:

        -- Haskell
        series_sqrt (s0:st) = r 
           where r  = r0 : (divide st (add [r0] r))
                 r0 = sqrt(s0)
And when I asked it for the square root of !![1,1,0,0,0,\ldots]!! (that is, of !!1+x!!) it gave me back !![1, 0.5, -0.125, -0.0625, \ldots]!!, which is indeed correct.

The Perl code is skankier than I wish it were. A couple of years ago I said in an interview that "I wish Perl's syntax were less verbose." Some people were surprised by this at the time, since Perl programmers consider Perl's syntax to be quite terse. But comparison of the Perl and Haskell code above demonstrates the sort of thing I mean.

Part of ths issue here, of course, is that the lazy list data structure is built in to Haskell, but I have to do it synthetically in Perl, and so every construction of a lazy list structure in Perl is accompanied by a syntactic marker (such as node(...) or promise { ... }) that is absent, or nearly absent, from the Haskell.

But when I complained about Perl's verbose syntax in 2005, one thing I had specifically in mind was Perl's argument acquisition syntax, here represented by my $s = shift;. Haskell is much terser, with no loss of expressiveness. Haskell gets another win in the automatic destructuring bind: instead of explicitly calling head() and tail() to acquire the values of s0 and st, as in the Perl code, they are implicitly called by the pattern match (s0:st) in the Haskell code, which never mentions s at all. It is quite fair to ascribe this to a failure of Perl's syntax, since there's no reason in principle why Perl couldn't support this, at least for built-in data structures. For example, consider the Perl code:

        sub blah {
          my $href = shift();
          my $a = $href->{this};
          my $tmp = $href->{that};
          my $b = $tmp->[0];
          my $c = $tmp->[2];

          # Now do something with $a, $b, $c
It would be much more convenient to write this as:
        sub blah {
          my { this => $a, that => [$b, undef, $c] } = shift();

          # Now do something with $a, $b, $c
This is a lot easier to understand.

There are a number of interesting user-interface issues to ask about here: What if the assigned value is not in the expected form? Are $a, $b, and $c copied from $href or are they aliases into it? And so on. One easy way to dispense with all of these interesting questions (perhaps not in the best way) is to assert that this notation is just syntactic sugar for the long version.

I talked to Chip Salzenberg about this at one time, and he said he thought it would not be very hard to implement. But even if he was right, what is not very hard for Chip Salzenberg to do can turn out to be nearly impossible for us mortals.

[ Addendum 20071209: There's a followup article that shows several different ways of solving the differential equation, including the power-series method. ]

[ Addendum 20071210: I did figure out how to get Haskell to solve the differential equation. ]

[Other articles in category /math] permanent link

Fri, 12 Oct 2007

The square of the Catalan sequence
Yesterday I went to a talk by Val Tannen about his work on "provenance semirings".

The idea is that when you calculate derived data in a database, such as a view or a selection, you can simultaneously calculate exactly which input tuples contributed to each output tuple's presence in the output. Each input tuple is annotated with an identifier that says who was responsible for putting it there, and the output annotations are polynomials in these identifiers. (The complete paper is here.)

A simple example may make this a bit clearer. Suppose we have the following table R:
a a
a b
a c
b c
c e
d e
We'll write R(p, q) when the tuple (p, q) appears in this table. Now consider the join of R with itself. That is, consider the relation S where S(x, z) is true whenever both R(x, y) and R(y, z) are true:

a a
a b
a c
a e
b e
Now suppose you discover that the R(a, b) information is untrustworthy. What tuples of S are untrustworthy?

If you annotate the tuples of R with identifiers like this:

a a u
a b v
a c w
b c x
c e y
d e z
then the algorithm in the paper calculates polynomials for the tuples of S like this:
a a u2
a b uv
a c uw + xv
a e wy
b e xy
If you decide that R(a, b) is no good, you assign the value 0 to v, which reduces the S table to:

a a u2
a b 0
a c uw
a e wy
b e xy
So we see that tuple S(a, b) is no good any more, but S(a, c) is still okay, because it can be derived from u and w, which we still trust.

This assignment of polynomials generalizes a lot of earlier work on tuple annotation. For example, suppose each tuple in R is annotated with a probability of being correct. You can propagate the probabilities to S just by substituting the appropriate numbers for the variables in the polynomials. Or suppose each tuple in R might appear multiple times and is annotated with the number of times it appears. Then ditto.

If your queries are recursive, then the polynomials might be infinite. For example, suppose you are calculating the transitive closure T of relation R. This is like the previous example, except that instead of having S(x, z) = R(x, y) and R(y, z), we have T(x, z) = R(x, z) or (T(x, y) and R(y, z)). This is a recursive equation, so we need to do a fixpoint solution for it, using certain well-known techniques. The result in this example is:

a a u+
a b u*v
a c u*(vx+w)
a e u*(vx+w)y
b c x
b e xy
d e z
In such a case there might be an infinite number of paths through R to derive the provenance of a certain tuple of T. In this example, R contains a loop, namely R(a, a), so there are an infinite number of derivations of some of the tuples in T, because you can go around the loop as many times as you like. u+ here is an abbreviation for the infinite polynomial u + u2 + u3 + ...; u* here is an abbreviation for 1 + u+.

1 a
2 (a + b)
3 ((a + b) + c)
(a + (b + c))
4 (((a + b) + c) + d)
((a + (b + c)) + d)
((a + b) + (c + d))
(a + ((b + c) + d))
(a + (b + (c + d)))
5 ((((a + b) + c) + d) + e)
(((a + (b + c)) + d) + e)
(((a + b) + (c + d)) + e)
(((a + b) + c) + (d + e))
((a + ((b + c) + d)) + e)
((a + (b + (c + d))) + e)
((a + (b + c)) + (d + e))
((a + b) + ((c + d) + e))
((a + b) + (c + (d + e)))
(a + (((b + c) + d) + e))
(a + ((b + (c + d)) + e))
(a + ((b + c) + (d + e)))
(a + (b + ((c + d) + e)))
(a + (b + (c + (d + e))))
In one example in the paper, the method produces a recursive relation of the form V = s + V2, which can be solved by the same well-known techniques to come up with an (infinite) polynomial for V, namely V = 1 + s + 2s2 + 5s3 + 14s4 + ... . Mathematicians will recognize the sequence 1, 1, 2, 5, 14, ... as the Catalan numbers, which come up almost as often as the better-known Fibonacci numbers. For example, the Catalan numbers count the number of binary trees with n nodes; they also count the number of ways of parenthesizing an expression with n terms, as shown in the table at right.

Anyway, in his talk, Val referred to the sequence as "bizarre", and I had to jump in to point out that it was not at all bizarre, it was the Catalan numbers, which are just what you would expect from a relation like V = s + V2, blah blah, and he cut me off, because of course he knows all about the Catalan numbers. He only called them bizarre as a rhetorical flourish, meant to echo the presumed puzzlement of the undergraduates in the room.

(I never know how much of what kind of math to expect from computer science professors. Sometimes they know things I don't expect at all, and sometimes they don't know things that I expect everyone to know.

(Once I was discussing the algorithm used by ENIAC for computing square roots with a professor, and the professor told me that at the beginning of the program there was a loop, which accumulated a total, each time accumulating the contents of a register that was incremented by 2 each time through the loop, and he did not know what was going on there. I instantly guessed that what was happening was that the register contained the numbers 1, 3, 5, 7, ..., incremented by 2 each time, and so the accumulator contained the totals 1, 4, 9, 16, 25, ..., and so the loop was calculating an initial estimate of the size of the square root. If you count the number of increment-by-2's, then when the accumulator exceeds the radicand, the count contains the integer part of the root.

(This was indeed what was going on, and the professor seemed to think it was a surprising insight. I am not relating this boastfully, because I truly don't think it was a particularly inspired guess.

(Now that I think about it, maybe the answer here is that computer science professors know more about math than I expect, and less about computation.)

Anyway, I digress, and the whole article up to now was not really what I wanted to discuss anyway. What I wanted to discuss was that when I started blathering about Catalan numbers, Val said that if I knew so much about Catalan numbers, I should calculate the coefficient of the x59 term in V2, which also appeared as one of the annotations in his example.

So that's the puzzle, what is the coefficient of the x59 term in V2, where V = 1 + s + 2s2 + 5s3 + 14s4 + ... ?

After I had thought about this for a couple of minutes, I realized that it was going to be much simpler than it first appeared, for two reasons.

The first thing that occurred to me was that the definition of multiplication of polynomials is that the coefficient of the xn term in the product of A and B is Σaibn-i. When A=B, this reduces to Σaian-i. Now, it just so happens that the Catalan numbers obey the relation cn+1 = Σ cicn-i, which is exactly the same form. Since the coefficients of V are the ci, the coefficients of V2 are going to have the form Σcicn-i, which is just the Catalan numbers again, but shifted up by one place.

The next thing I thought was that the Catalan numbers have a pretty simple generating function f(x). This just means that you pretend that the sequence V is a Taylor series, and figure out what function it is the Taylor series of, and use that as a shorthand for the whole series, ignoring all questions of convergence and other such analytic fusspottery. If V is the Taylor series for f(x), then V2 is the Taylor series for f(x)2. And if f has a compact representation, say as sin(x) or something, it might be much easier to square than the original V was. Since I knew in this case that the generating function is simple, this seemed likely to win. In fact the generating function of V is not sin(x) but (1-√(1-4x))/2x. When you square this, you get almost the same thing back, which matches my prediction from the previous paragraph. This would have given me the right answer, but before I actually finished that calculation, I had an "oho" moment.

The generating function is known to satisfy the relation f(x) = 1 + xf(x)2. This relation is where the (1-√(1-4x))/2x thing comes from in the first place; it is the function that satisfies that relation. (You can see this relation prefigured in the equation that Val had, with V = s + V2. There the notation is a bit different, though.) We can just rearrange the terms here, putting the f(x)2 by itself, and get f(x)2 = (f(x)-1)/x.

Now we are pretty much done, because f(x) = V = 1 + x + 2x2 + 5x3 + 14x4 + ... , so f(x)-1 = x + 2x2 + 5x3 + 14x4 + ..., and (f(x)-1)/x = 1 + 2x + 5x2 + 14x3 + ... . Lo and behold, the terms are the Catalan numbers again.

So the answer is that the coefficient of the x59 term is just c(60), calculation of which is left as an exercise for the reader.

I don't know what the point of all that was, but I thought it was fun how the hairy-looking problem seemed likely to be simple when I looked at it a little more carefully, and then how it did turn out to be quite simple.

This blog has had a recurring dialogue between subtle technique and the sawed-off shotgun method, and I often favor the sawed-off shotgun method. Often programmers' big problem is that they are very clever and learned, and so they want to be clever and learned all the time, even when being a knucklehead would work better. But I think this example provides some balance, because it shows a big win for the clever, learned method, which does produce a lot more understanding.

Then again, it really doesn't take long to whip up a program to multiply infinite polynomials. I did it in chapter 6 of Higher-Order Perl, and here it is again in Haskell:

        data Poly a = P [a] deriving Show

        instance (Eq a) => Eq (Poly a) 
                where (P x) == (P y) = (x == y)

        polySum x [] = x
        polySum [] y = y
        polySum (x:xs) (y:ys) = (x+y) : (polySum xs ys)

        polyTimes  [] _ = []
        polyTimes  _ [] = []
        polyTimes  (x:xs) (y:ys) = (x*y) : more
                            more = (polySum (polySum (map (x *) ys) (map (* y) xs))
                                    (0 : (polyTimes xs ys)))

        instance (Num a) => Num (Poly a) 
          where (P x) + (P y) = P (polySum x y)
                (P x) * (P y) = P (polyTimes x y)

[Other articles in category /math] permanent link

Tue, 09 Oct 2007

Relatively prime polynomials over Z2
Last week Wikipedia was having a discussion on whether the subject of "mathematical quilting" was notable enough to deserve an article. I remembered that there had been a mathematical quilt on the cover of some journal I read last year, and I went to the Penn math library to try to find it again. While I was there, I discovered that the June 2007 issue of Mathematics Magazine had a cover story about the probability that two randomly-selected polynomials over Z2 are relatively prime. ("The Probability of Relatively Prime Polynomials", Arthur T. Benjamin and Curtis D Bennett, page 196).

Polynomials over Z2 are one of my favorite subjects, and the answer to the question turned out to be beautiful. So I thought I'd write about it here.

First, what does it mean for two polynomials to be relatively prime? It's analogous to the corresponding definition for integers. For any numbers a and b, there is always some number d such that both a and b are multiples of d. (d = 1 is always a solution.) The greatest such number is called the greatest common divisor or GCD of a and b. The GCD of two numbers might be 1, or it might be some larger number. If it's 1, we say that the two numbers are relatively prime (to each other). For example, the GCD of 100 and 28 is 4, so 100 and 28 are not relatively prime. But the GCD of 100 and 27 is 1, so 100 and 27 are relatively prime. One can prove theorems like these: If p is prime, then either a is a multiple of p, or a is relatively prime to p, but not both. And the equation ap + bq = 1 has a solution (in integers) if and only if p and q are relatively prime.

The definition for polynomials is just the same. Take two polynomials over some variable x, say p and q. There is some polynomial d such that both p and q are multiples of d; d(x) = 1 is one such. When the only solutions are trivial polynomials like 1, we say that the polynomials are relatively prime. For example, consider x2 + 2x + 1 and x2 - 1. Both are multiples of x+1, so they are not relatively prime. But x2 + 2x + 1 is relatively prime to x2 - 2x + 1. And one can prove theorems that are analogous to the ones that work in the integers. The analog of "prime integer" is "irreducible polynomial". If p is irreducible, then either a is a multiple of p, or a is relatively prime to p, but not both. And the equation a(x)p(x) + b(x)q(x) = 1 has a solution for polynomials a and b if and only if p and q are relatively prime.

One uses Euclid's algorithm to calculate the GCD of two integers. Euclid's algorithm is simple: To calculate the GCD of a and b, just subtract the smaller from the larger, repeatedly, until one of the numbers becomes 0. Then the other is the GCD. One can use an entirely analogous algorithm to calculate the GCD of two polynomials. Two polynomials are relatively prime just when their GCD, as calculated by Euclid's algorithm, has degree 0.

Anyway, that was more introduction than I wanted to give. The article in Mathematics Magazine concerned polynomials over Z2, which means that the coefficients are in the field Z2, which is just like the regular integers, except that 1+1=0. As I explained in the earlier article, this implies that a=-a for all a, so there are no negatives and subtraction is the same as addition. I like this field a lot, because subtraction blows. Do you have trouble because you're always dropping minus signs here and there? You'll like Z2; there are no minus signs.

Here is a table that shows which pairs of polynomials over Z2 are relatively prime. If you read this blog through some crappy aggregator, you are really missing out, because the table is awesome, and you can't see it properly. Check out the real thing.

0   [a0]                                                                
1   [a1]                                                                
x   [a2]                                                                
x + 1   [a3]                                                                
x2   [a4]                                                                
x2 + 1   [a5]                                                                
x2 + x   [a6]                                                                
x2 + x + 1   [a7]                                                                
x3   [a8]                                                                
x3 + 1   [a9]                                                                
x3 + x   [b0]                                                                
x3 + x + 1   [b1]                                                                
x3 + x2   [b2]                                                                
x3 + x2 + 1   [b3]                                                                
x3 + x2 + x   [b4]                                                                
x3 + x2 + x + 1   [b5]                                                                
x4   [b6]                                                                
x4 + 1   [b7]                                                                
x4 + x   [b8]                                                                
x4 + x + 1   [b9]                                                                
x4 + x2   [c0]                                                                
x4 + x2 + 1   [c1]                                                                
x4 + x2 + x   [c2]                                                                
x4 + x2 + x + 1   [c3]                                                                
x4 + x3   [c4]                                                                
x4 + x3 + 1   [c5]                                                                
x4 + x3 + x   [c6]                                                                
x4 + x3 + x + 1   [c7]                                                                
x4 + x3 + x2   [c8]                                                                
x4 + x3 + x2 + 1   [c9]                                                                
x4 + x3 + x2 + x   [d0]                                                                
x4 + x3 + x2 + x + 1   [d1]                                                                

A pink square means that the polynomials are relatively prime; a white square means that they are not. Another version of this table appeared on the cover of Mathematics Magazine. It's shown at right.

The thin black lines in the diagram above divide the polynomials of different degrees. Suppose you pick two degrees, say 2 and 2, and look at the corresponding black box in the diagram:

x2   [a4]        
x2 + 1   [a5]        
x2 + x   [a6]        
x2 + x + 1   [a7]        
You will see that each box contains exactly half pink and half white squares. (8 pink and 8 white in that case.) That is, exactly half the possible pairs of degree-2 polynomials are relatively prime. And in general, if you pick a random degree-a polynomial and a random degree-b polynomial, where a and b are not both zero, the polynomials will be relatively prime exactly half the time.

The proof of this is delightful. If you run Euclid's algorithm on two relatively prime polynomials over Z2, you get a series of intermediate results, terminating in the constant 1. Given the intermediate results and the number of steps, you can run the algorithm backward and find the original polynomials. If you run the algorithm backward starting from 0 instead of from 1, for the same number of steps, you get two non-relatively-prime polynomials of the same degrees instead. This establishes a one-to-one correspondence between pairs of relatively prime polynomials and pairs of non-relatively-prime polynomials of the same degrees. End of proof. (See the paper for complete details.)

You can use basically the same proof to show that the probability that two randomly-selected polynomials over Zp is 1-1/p. The argument is the same: Euclid's algorithm could produce a series of intermediate results terminating in 0, in which case the polynomials are not relatively prime, or it could produce the same series of intermediate results terminating in something else, in which case they are relatively prime. The paper comes to an analogous conclusion about monic polynomials over Z.

Some folks I showed the diagram to observed that it looks like a quilt pattern. My wife did actually make a quilt that tabulates the GCD function for integers, which I mentioned in the Wikipedia discussion of the notability of the Mathematical Quilting article. That seems to have brought us back to where the article started, so I'll end here.

[ Puzzle: The (11,12) white squares in the picture are connected to the others via row and column 13, which doesn't appear. Suppose the quilt were extended to cover the entire quarter-infinite plane. Would the white area be connected? ]

[Other articles in category /math] permanent link

Mon, 08 Oct 2007

The Club

Reduces your risk of auto theft by 400%.

[Other articles in category /math] permanent link

Sat, 08 Sep 2007

The missing deltahedron
I recently wrote about the convex deltahedra, which are the eight polyhedra whose faces are all congruent equilateral triangles:

Tetrahedron 464
Triangular dipyramid 695
Octahedron 8126
Pentagonal dipyramid 10157
Snub disphenoid 12188
Triaugmented triangular prism 14219
Gyroelongated square dipyramid 162410
Icosahedron 203012
The names are rather horrible, so I think that from now on I'll just refer to them as D4, D6, D8, D10, D12, D14, D16, and D20.

The number of edges that meet at a vertex is its valence. Vertices in convex deltahedra have valences of 3, 4, or 5. The valence can't be larger than 5 because only six equilateral triangles will fit, and if you fit 6 then they lie flat and the polyhedron is not properly convex.

Let V3, V4, and V5 be the number of vertices of valences 3, 4, and 5, respectively. Then:

D8 6 
D10 52
D12 44
D14 36
D16 28
D20  12
There's a clear pattern here, with V3s turning into V4s two at a time until you reach the octahedron (D8) and then V4s turning into V5s one at a time until you reach the icosahedron (D20). But where is V4=1, V5=10? There's a missing deltahedron. I don't mean it's missing from the table; I mean it's missing from the universe.

Well, this is all oversubtle, I realized later, because you don't need to do the V3V4V5 analysis to see that something is missing. There are convex deltahedra with 4, 6, 8, 10, 12, 14, and 20 faces; what happened to 18?

Still, I did a little work on a more careful analysis that might shed some light on the 18-hedron situation. I'm still in the middle of it, but I'm trying to continue my policy of posting more frequent, partial articles.

Let V be the number of vertices in a convex deltahedron, E be the number of edges, and F be the number of faces.

We then have V = V3 + V4 + V5. We also have E = ½(3V3 + 4V4 + 5V5). And since each face has exactly 3 edges, we have 3F = 2E.

By Euler's formula, F + V = E + 2. Plugging in the stuff from the previous paragraph, we get 3V3 + 2V4 + V5 = 12.

It is very easy to enumerate all possible solutions of this equation. There are 19:

Solutions in green correspond to convex deltahedra. What goes wrong with the other 11 items?

(3,1,1) fails completely because to have V5 > 0 you need V ≥ 6. There isn't even a graph with (V3, V4, V5) = (3,1,1), much less a polyhedron.

There is a graph with (3,0,3), but it is decidedly nonplanar: it contains K3,3, plus an additional triangle. But the graph of any polyhedron must be planar, because you can make a little hole in one of the faces of the polyhedron and flatten it out without the edges crossing.

Another way to think about (3,0,3) is to consider it as a sort of triangular tripyramid. Each of the V5s shares an edge with each of the other five vertices, so the three V5s are all pairwise connected by edges and form a triangle. Each of the three V3s must be connected to each of the three vertices of this triangle. You can add two of the required V3s, by erecting a triangular pyramid on the top and the bottom of the triangle. But then you have nowhere to put the third pyramid.

On Thursday I didn't know what went wrong with (2,2,2); it seemed fine. (I found it a little challenging to embed it in the plane, but I'm not sure if it would still be challenging if it hadn't been the middle of the night.) I decided that when I got into the office on Friday I would try making a model of it with my magnet toy and see what happened.

It turned out that nothing goes wrong with (2,2,2). It makes a perfectly good non-convex deltahedron. It's what you get when you glue together three tetrahedra, face-to-face-to-face. The concavity is on the underside in the picture.

(2,0,6) was a planar graph too, and so the problem had to be geometric, not topological. When I got to the office, I put it together. It also worked fine, but the result is not a polyhedron. The thing you get could be described as a gyroelongated triangular dipyramid. That is, you take an octahedron and glue tetrahedra to two of its opposite faces. But then the faces of the tetrahedra are coplanar with the faces of the octahedron to which they abut, and this is forbidden in polyhedra. When that happens you're supposed to eliminate the intervening edge and consider the two faces to be a single face, a rhombus in this case. The resulting thing is not a polyhedron with 12 triangular faces, but one with six rhombic faces (a rhombohedron), essentially a squashed cube. In fact, it's exactly what you get if you make a cube from the magnet toy and then try to insert another unit-length rod into the diagonal of each of the six faces. You have to squash the cube to do this, of course, since the diagonals had length √2 before and length 1 after.

So there are several ways in which the triples (V3,V4,V5) can fail to determine a convex deltahedron: There is an utter topological failure, as with (3,1,1).

There is a planarity failure, which is also topological, but less severe, as with (3,0,3). (3,0,3) also fails because you can't embed it into R3. (I mean that you cannot embed its 3-skeleton. Of course you can embed its 1-skeleton in R3, but that is not sufficient for the thing to be a polyhedron.) I'm not sure if this is really different from the previous failure; I need to consider more examples. And (3,0,3) fails in yet another way: you can't even embed its 1-skeleton in R3 without violating the constraint that says that the edges must all have unit length. The V5s must lie at the vertices of an equilateral triangle, and then the three unit spheres centered at the V5s intersect at exactly two points of R3. You can put two of the V3s at these points, but this leaves nowhere for the third V3. Again, I'm not sure that this is a fundamentally different failure mode than the other two.

Another failure mode is that the graph might be embeddable into R3, and might satisfy the unit-edge constraint, but in doing so it might determine a concave polyhedron, like (2,2,2) does, or a non-polyhedron, like (2,0,6) does.

I still have six (V3,V4,V5) triples to look into. I wonder if there are any other failure modes?

I should probably think about (0,1,10) first, since the whole point of all this was to figure out what happened to D18. But I'm trying to work up from the simple cases to the harder ones.

I suppose the next step is to look up the proof that there are only eight convex deltahedra and see how it goes.

I suspect that (2,1,4) turns out to be nonplanar, but I haven't looked at it carefully enough to actually find a forbidden minor.

One thing that did occur to me today was that a triple (V3, V4, V5) doesn't necessarily determine a unique graph, and I need to look into that in more detail. I'll be taking a plane trip on Sunday and I plan to take the magnet toy with me and continue my investigations on the plane.

In other news, Katara and I went to my office this evening to drop off some books and pick up some stuff for the trip, including the magnet toy. Katara was very excited when she saw the collection of convex deltahedron models on my desk, each in a different color, and wanted to build models just like them. We got through all of them, except D10, because we ran out of ball bearings. By the end Katara was getting pretty good at building the models, although I think she probably wouldn't be able to do it without directions yet. I thought it was good work, especially for someone who always skips from 14 to 16 when she counts.

On the way home in the car, we were talking about how she was getting older and I rhapsodized about how she was learning to do more things, learning to do the old things better, learning to count higher, and so on. Katara then suggested that when she is older she might remember to include 15.

[Other articles in category /math] permanent link

Thu, 06 Sep 2007

Followup notes about dice and polyhedra
I got a lot of commentary about these geometric articles, and started writing up some followup notes. But halfway through I got stuck in the middle of making certain illustrations, and then I got sick, and then I went to a conference in Vienna. So I decided I'd better publish what I have, and maybe I'll get to the other fascinating points later.

  • Regarding a die whose sides appear with probabilities 1/21 ... 6/21

    • Several people wrote in to cast doubt on my assertion that the probability of an irregular die showing a certain face is proportional to the solid angle subtended by that face from the die's center of gravity. But nobody made the point more clearly than Robert Young, who pointed out that if I were right, a coin would have a 7% chance of landing on its edge. I hereby recant this claim.

    • John Berthels suggested that my analysis might be correct if the die was dropped into an inelastic medium like mud that would prevent it from bouncing.

    • Jack Vickeridge referred me to this web site, which has a fairly extensive discussion of seven-sided dice. The conclusion: if you want a fair die, you have no choice but to use something barrel-shaped.

    • Michael Lugo wrote a detailed followup in which he discusses this and related problems. He says "What makes Mark's problem difficult is the lack of symmetry; each face has to be different." Quite so.

  • Regarding alternate labelings for standard dice

    • Aaron Crane says that these dice (with faces {1,2,2,3,3,4} and {1,3,4,5,6,8}) are sometimes known as "Sicherman dice", after the person who first brought them to the attention of Martin Gardner. Can anyone confirm that this was Col. G.L. Sicherman? I have no reason to believe that it was, except that it would be so very unsurprising if it were true.

    • Addendum 20070905: I now see that the Wikipedia article attributes the dice to "Colonel George Sicherman," which is sufficiently clear that I would feel embarrassed to write to the Colonel to ask if it is indeed he. I also discovered that the Colonel has a Perl program on his web site that will calculate "all pairs of n-sided dice that give the same sums as standard n-sided dice".

    • M. Crane also says that it is an interesting question which set of dice is better for backgammon. Both sets have advantages: the standard set rolls doubles 1/6 of the time, whereas the Sicherman dice only roll doubles 1/9 of the time. (In backgammon, doubles count double, so that whereas a player who rolls ab can move the pieces a total of a+b points, a player who rolls aa can move pieces a total of 4a points.) The standard dice permit movement of 296/36 points per roll, and the Sicherman dice only 274/36 points per roll.

      Ofsetting this disadvantage is the advantage that the Sicherman dice can roll an 8. In backgammon, one's own pieces may not land on a point occupied by more than one opposing piece. If your opponent occupies six conscutive points with two pieces each, they form an impassable barrier. Such a barrier is passable to a player using the Sicherman dice, because of the 8.

    • Doug Orleans points out that in some contexts one might prefer to use a Sicherman variant dice {2,3,3,4,4,5} and {0,2,3,4,5,7}, which retain the property that opposite faces sum to 7, and so that each die shows 3.5 pips on average. Such dice roll doubles as frequently as do standard dice.

    • The Wikipedia article on dice asserts that the {2, 3, 3, 4, 4, 5} die is used in some wargames to express the strength of "regular" troops, and the standard {1, 2, 3, 4, 5, 6} die to express the strength of "irregular" troops. This makes the outcome of battles involving regular forces more predictable than those involving irregular forces.

  • Regarding deltahedra and the snub disphenoid

    • Several people proposed alternative constructions for the snub disphenoid.

      1. Brooks Moses suggested the following construction: Take a square antiprism, squash the top square into a rhombus, and insert a strut along the short diagonal of the rhombus. Then squash and strut the bottom square similarly.

        It seems, when you think about this, that there are two ways to do the squashing. Suppose you squash the bottom square horizontally in all cases. The top square is turned 45° relative to the bottom (because it's an antiprism) and so you can squash it along the -45° diagonal or along the +45° diagonal, obtaining a left- and a right-handed version of the final solid. But if you do this, you find that the two solids are the same, under a 90° rotation.

        This construction, incidentally, is equivalent to the one I described in the previous article: I said you should take two rhombuses and connect corresponding vertices. I had a paragraph that read:

        But this is where I started to get it wrong. The two wings have between them eight edges, and I had imagined that you could glue a rhombic antiprism in between them. . . .

        But no, I was right; you can do exactly this, and you get a snub disphenoid. What fooled me was that when you are looking at the snub disphenoid, it is very difficult to see where the belt of eight triangles from the antiprism got to. It winds around the polyhedron in a strange way. There is a much more obvious belt of triangles around the middle, which is not suitable for an antiprism, being shaped not like a straight line but more like the letter W, if the letter W were written on a cylinder and had its two ends identified. I was focusing on this belt, but the other one is there, if you know how to see it.

        The snub disphenoid has four vertices with valence 4 and four with valence 5. Of its 12 triangular faces, four have two valence-4 vertices and one valence-5 vertex, and eight have one valence-4 vertex and two valence-5 vertices. These latter eight form the belt of the antiprism.

      2. M. Moses also suggested taking a triaugmented triangular prism, which you will recall is a triangular prism with a square pyramid erected on each of its three square faces, removing one of the three pyramids, and then squashing the exposed square face into a rhombus shape, adding a new strut on the diagonal. This one gives me even less intuition about what is going on, and it seems even more strongly that it shou,ld matter whether you put in the extra strut from upper-left to lower-right, or from upper-right to lower-left. But it doesn't matter; you get the same thing either way.

      3. Jacob Fugal pointed out that you can make a snub disphenoid as follows: take a pentagonal dipyramid, and replace one of the equatorial *----*----* figures with a rhombus. This is simple, but unfortunately gives very little intuition for what the disphenoid is like. It is obvious from the construction that there must be pentagons on the front and back, left over from the dipyramid. But it is not at all clear that there are now two new upside-down pentagons on the left and right sides, or that the disphenoid has a vertical symmetry.

    • A few people asked me where John Batzel got they magnet toy that I was using to construct the models. It costs only $5! John gave me his set, and I bought three more, and I now have a beautiful set of convex deltahedra and a stellated dodecahedron on my desk. (Actually, it is not precisely a stellated dodecahedron, since the star faces are not quite planar, but it is very close. If anyone knows the name of this thing, which has 32 vertices, 90 edges, and 60 equilateral triangular faces, I would be pleased to hear about it.) Also I brought my daughter Katara into my office a few weekends ago to show her the stella octangula ("I wanna see the stella octangula, Daddy! Show me the stella octangula!") which she enjoyed; she then stomped on it, and then we built another one together.

    • [ Addendum 20070908: More about deltahedra. ]

[Other articles in category /math] permanent link

Tue, 07 Aug 2007

Different arrangements for standard dice
Gaal Yahas wrote to refer me to an article about a pair of dice that never roll seven. It sounded cool, but but it was too late at night for me to read it, so I put it on the to-do list. But it reminded me of a really nice puzzle, which is to find a nontrivial relabeling of a pair of standard dice that gives the same probability of throwing any sum from 2 to 12. It's a happy (and hardly inevitable) fact that there is a solution.

To understand just what is being asked for here, first observe that a standard pair of dice throws a 2 exactly 1/36 of the time, a 3 exactly 2/36 of the time, and so forth:


The standard dice have faces numbered 1, 2, 3, 4, 5, and 6. It should be clear that if one die had {0,1,2,3,4,5} instead, and the other had {2,3,4,5,6,7}, then the probabilities would be exactly the same. Similarly you could subtract 3.7 from every face of one die, giving it labels {-2.7, -1.7, -0.7, 0.3, 1.3, 2.3}, and if you added the 3.7 to every face of the other die, giving labels {4.7, 5.7, 6.7, 7.7, 8.7, 9.7}, you'd still have the same chance of getting any particular total. For example, there are still exactly 2 ways out of 36 possible rolls to get the total 3: you can roll -2.7 + 5.7, or you can roll -1.7 + 4.7. But the question is to find a nontrivial relabeling.

Like many combinatorial problems, this one is best solved with generating functions. Suppose we represent a die as a polynomial. If the polynomial is Σaixi, it represents a die that has ai chances to produce the value i. A standard die is x6 + x5 + x4 + x3 + x2 + x, with one chance to produce each integer from 1 to 6. (We can deal with probabilities instead of "chances" by requiring that Σai = 1, but it comes to pretty much the same thing.)

The reason it's useful to adopt this representation is that rolling the dice together corresponds to multiplication of the polynomials. Rolling two dice together, we multiply (x6 + x5 + x4 + x3 + x2 + x) by itself and get P(x) = x12 + 2x11 + 3x10 + 4x9 + 5x8 + 6x7 + 5x6 + 4x5 + 3x4 + 2x3 + x2, which gives the chances of getting any particular sum; the coefficient of the x9 term is 4, so there are 4 ways to roll a 9 on two dice.

What we want is a factorization of this 12th-degree polynomial into two polynomials Q(x) and R(x) with non-negative coefficients. We also want Q(1) = R(1) = 6, which forces the corresponding dice to have 6 faces each. Since we already know that P(x) = (x6 + x5 + x4 + x3 + x2 + x)2, it's not hard; we really only have to factor x6 + x5 + x4 + x3 + x2 + x and then see if there's any suitable way of rearranging the factors.

x6 + x5 + x4 + x3 + x2 + x = x(x4 + x2 + 1)(x + 1) = x(x2 + x + 1)(x2 - x + 1)(x + 1). So P(x) has eight factors:

x x2 + x + 1 x2 - x + 1 x + 1
x x2 + x + 1 x2 - x + 1 x + 1

We want to combine these into two products Q(x) and R(x) such that Q(1) = R(1) = 6. If we calculate f(1) for each of these, we get 1, 3 (pink), 1, and 2 (blue). So each of Q and R will require one of the factors that has f(1) = 3 and one that has f(1) = 2; we can distribute the f(1) = 1 factors as needed. For normal dice the way we do this is to assign all the factors in each row to one die. If we want alternative dice, our only real choice is what to do with the x2 - x + 1 and x factors.

Redistributing the lone x factors just corresponds to subtracting 1 from all the faces of one die and adding it back to all the faces of the other, so we can ignore them. The only interesting question is what to do with the x2 - x + 1 factors. The normal distribution assigns one to each die, and the only alternative is to assign both of them to a single die. This gives us the two polynomials:

x(x2 + x + 1)(x + 1)
= x4 + 2x3 + 2x2 + x
x(x2 + x + 1)(x + 1)(x2 - x + 1)2
= x8 + x6 + x5 + x4 + x3 + x

And so the solution is that one die has faces {1,2,2,3,3,4} and the other has faces {1,3,4,5,6,8}:


Counting up entries in the table, we see that there are indeed 6 ways to throw a 7, 4 ways to throw a 9, and so forth.

One could apply similar methods to the problem of making a pair of dice that can't roll 7. Since there are six chances in 36 of rolling 7, we need to say what will happen instead in these 6 cases. We might distribute them equally among some of the other possibilities, say 2, 4, 6, 8, 10, and 12, so that we want the final distribution of results to correspond to the polynomial 2x12 + 2x11 + 4x10 + 4x9 + 6x8 + 6x6 + 4x5 + 4x4 + 2x3 + 2x2. The important thing to notice here is that the coefficient of the x7 term is 0.

Now we want to factor this polynomial and proceed as before. Unfortunately, it is irreducible. (Except for the trivial factor of x2.) Several other possibilities are similarly irreducible. It's tempting to reason from the dice to the algebra, and conjecture that any reducible polynomial that has a zero x7 term must be rather exceptional in other ways, such as by having only even exponents. But I'm not sure it will work, because the polynomials are more general than the dice: the polynomials can have negative coefficients, which are meaningless for the dice. Still, I can fantasize that there might be some result of this type available, and I can even imagine a couple of ways of getting to this result, one combinatorial, another based on Fourier transforms. But I've noticed that I have a tendency to want to leave articles unpublished until I finish exploring all possible aspects of them, and I'd like to change that habit, so I'll stop here, for now.

[ Addendum 20070905: There are some followup notes. ]

[Other articles in category /math] permanent link

Mon, 06 Aug 2007

Standard analytic polyhedra
If you want to consider a cube analytically, you have an easy job. The vertices lie at the points:

And you can see at a glance whether two vertices share an edge (they are the same in two of their three components) or are opposite (they differ in all three components).

Last week I was reading the Wikipedia article about the computer game "Hunt the Wumpus", which I played as a small child. For the Guitar Hero / WoW generation I should explain Wumpus briefly.

The object of "Wumpus" is to kill the Wumpus, which hides in a network of twenty caves arranged in a dodecahedron. Each cave is thus connected to three others. On your turn, you may move to an adjacent cave or shoot a crooked arrow. The arrow can pass through up to five connected caves, and if it enters the room where the Wumpus is, it kills him and you win. Two of the caves contain bottomless pits; to enter these is death. Two of the caves contain giant bats, which will drop you into another cave at random; if it contains a pit, too bad. If you are in a cave adjacent to a pit, you can feel a draft; if you are adjacent to bats, you can hear them. If you are adjacent to the Wumpus, you can smell him. If you enter the Wumpus's cave, he eats you. If you shoot an arrow that fails to kill him, he wakes up and moves to an adjacent cave; if he enters you cave, he eats you. You have five arrows.

I did not learn until much later that the caves are connected in a dodecahedron; indeed, at the time I probably didn't know what a dodecahedron was. The twenty caves were numbered, so that cave 1 was connected to 2, 5, and 8. This necessitated a map, because otherwise it was too hard to remember which room was connected to which.

Or did it? If the map had been a cube, the eight rooms could have been named 000, 001, 010, etc., and then it would have been trivial to remember: 011 is connected to 111, 001, and 010, obviously, and you can see it at a glance. It's even easy to compute all the paths between two vertices: the paths from 011 to 000 are 011–010–000 and 011–001–000; if you want to allow longer paths you can easily come up with 011–111–110–100–000 for example.

And similarly, the Wumpus source code contains a table that records which caves are connected to which, and consults this table in many places. If the caves had been arranged in a cube, no table would have been required. Or if one was wanted, it could have been generated algorithmically.

So I got to wondering last week if there was an analogous nomenclature for the vertices of a dodecahedron that would have obviated the Wumpus map and the table in the source code.

I came up with a very clever proof that there was none, which would have been great, except that the proof also worked for the tetrahedron, and the tetrahedron does have such a convenient notation: you can name the vertices (0,0,0), (0,1,1), (1,0,1), and (1,1,0), where there must be an even number of 1 components. (I mentioned this yesterday in connection with something else and promised to come back to it. Here it is.) So the proof was wrong, which was good, and I kept thinking about it.

The next-simplest case is the octahedron, and I racked my brains trying to come up with a convenient notation for the vertices that would allow one to see at a glance which were connected. When I finally found it, I felt like a complete dunce. The octahedron has six vertices, which are above, below, to the left of, to the right of, in front of, and behind the center. Their coordinates are therefore (1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1) and (0,0,-1). Two vertices are opposite when they have two components the same (necessarily both 0) and one different (necessarily negatives). Otherwise, they are connected by an edge. This is really simple stuff.

Still no luck with the dodecahedron. There are nice canonical representations of the coordinates of the vertices—see the Wikipedia article, for example—but I still haven't looked at it closely enough to decide if there is a simple procedure for taking two vertices and determining their geometric relation at a glance. Obviously, you can check for adjacent vertices by calculating the distance between them and seeing if it's the correct value, but that's not "at a glance"; arithmetic is forbidden.

It's easy to number the vertices in layers, say by calling the top five vertices A1 ... A5, then the five below that B1 ... B5, and so on. Then it's easy to see that A3 will be adjacent to A2, A4, and B3, for example.

But this nomenclature, unlike the good ones above, is not isometric: it has a preferred orientation of the dodecahedron. It's obvious that A1, A2, A3, A4, and A5 form a pentagonal face, but rather harder to see that A2, A3, B2, B3, and C5 do. With the cube, it's easy to see what a rotation or a reflection looks like. For example, rotation of 120° around an axis through a pair of vertices of the cube takes vertex (a, b, c) to (c, a, b); rotation of 90° around an axis through a face takes it to (1-b, a, c). Similarly, rotations and reflections of the tetrahedron correspond to simple permutations of the components of the vertices. Nothing like this exists for the A-B-C-D nomenclature for the dodecahedron.

I'll post if I come up with anything nice.

[Other articles in category /math] permanent link

Sun, 05 Aug 2007

The 123456 die
As a result of my recent article on the snub disphenoid, Paul Keir wrote to me to ask about non-equiprobable dice. Specifically, he wanted a die that, because it was irregular, was twice as likely to land on one face as on any of the others.

That got me thinking about the problem in general. For some reason I've been trying to construct a die whose faces come up with probabilities 1/21, 2/21, 3/21, 4/21, 5/21, and 6/21 respectively.

Unless there is a clever insight I haven't had, I think this will be rather difficult to do explicitly. (Approximation methods will probably work fairly easily though, I think.) I started by trying to make a hexahedron with faces that had areas 1, 2, 3, 4, 5, 6, and even this has so far evaded me. This will not be sufficient to solve the problem, because the probability that the hexahedron will land on face F is not proportional to the area of F, but rather to the solid angle subtended by F from the hexahedron's center of gravity.

Anyway, I got interested in the idea of making a hexahedron whose faces had areas 1..6. First I tried just taking a bunch of simple shapes (right triangles and the like) of the appropriate sizes and fitting them together geometrically; so far that hasn't worked. So then I thought maybe I could get what I wanted by taking a tetrahedron or a disphenoid or some such and truncating a couple of the corners.

As Polya says, if you can't solve the problem, you should try solving a simpler problem of the same sort, so I decided to see if it was possible to take a regular tetrahedron and chop off one vertex so that the resulting pentahedron had faces with areas 1, 2, 3, 4, 5. The regular tetrahedron is quite tractable, geometrically, because you can put its vertices at (0,0,0), (0,1,1), (1,0,1), and (1,1,0), and then a plane that chops off the (0,0,0) vertex cuts the three apical edges at points (0,a,a), (b,0,b), and (c,c,0), for some 0 ≤ a, b, c ≤ 1. The chopped-off areas of the three faces are simply ab√3/4, bc√3/4, ca√3/4, and the un-chopped base has area √3/4, so if we want the three chopped faces to have areas of 2/5, 3/5 and 4/5 times √3/4, respectively, we must have ab = 3/5, bc = 2/5, and ca = 1/5, and we can solve for a, b, c. (We want the new top face to have area 1/5 · √3/4, but that will have to take care of itself, since it is also determined by a, b, and c.) Unfortunately, solving these equations gives b = √6/√5, which is geometrically impossible. We might fantasize that there might be some alternate solution, say with the three chopped faces having areas of 1/5, 2/5 and 4/5 times √3/4, and the top face being 3/5 · √3/4 instead of 1/5 · √3/4, but none of those will work either.

Oh well, it was worth a shot. I do think it's interesting that if you know the areas of the bottom four faces of a truncated regular tetrahedron, that completely determines the apical face. Because you can solve for the lengths of the truncated apical edges, as above, and then that gives you the coordinates of the three apical vertices.

I had a brief idea about truncating a square pyramid to get the hexahedron I wanted in the first place, but that's more difficult, because you can't just pick the lengths of the four apical edges any way you want; their upper endpoints must be coplanar.

The (0,a,a), (b,0,b), (c,c,0) thing has been on my mind anyway, and I hope to write tomorrow's blog article about it. But I've decided that my articles are too long and too intermittent, and I'm going to try to post some shorter, more casual ones more frequently. I recently remembered that in the early days of the blog I made an effort to post every day, and I think I'd like to try to resume that.

[ Addendum 20070905: There are some followup notes. ]

[Other articles in category /math] permanent link

Wed, 01 Aug 2007

The snub disphenoid
The snub disphenoid is pictured at left. I do not know why it is called that, and I ought to know, because I am the principal author (so far) of the Wikipedia article on the disphenoid. Also, I never quite figured out what "snub" means in this context, despite perusing that section of H.S.M. Coxeter's book on polytopes at some length. It has something to do with being halfway between what you get when you cut all the corners off, and what you get when you cut all the corners off again.

Anyway, earlier this week I was visiting John Batzel, who works upstairs from me, and discovered that he had obtained a really cool toy. It was a collection of large steel ball bearings and colored magnetic rods, which could be assembled into various polyhedra and trusses. This is irresistible to me. The pictures at right, taken around 2002, show me modeling a dodecahedron with less suitable materials.

The first thing I tried to make out of John's magnetic sticks and balls was a regular dodecahedron, because it is my favorite polyhedron. (Isn't it everyone's?) This was unsuccessful, because it wasn't rigid enough, and kept collapsing. It's possible that if I had gotten the whole thing together it would have been stable, but holding the 50 separate magnetic parts in the right place long enough to get it together was too taxing, so I tried putting together some other things.

A pentagonal dipyramid worked out well, however. To understand this solid, imagine a regular pyramid, such as the kind that entombs the pharaohs or collects mystical energy. This sort of pyramid is known as a square pyramid, because it has a square base, and thus four triangular sides. Imagine that the base was instead a pentagon, so that there were five triangular sides sides instead of only four. Then it would be a pentagonal pyramid. Now take two such pentagonal pyramids and glue the pentagonal bases together. You now have a pentagonal dipyramid.

The success of the pentagonal dipyramid gave me the idea that rigid triangular lattices were the way to go with this toy, so I built an octahedron (square dipyramid) and an icosahedron to be sure. Even the icosahedron (thirty sticks and twelve balls) held together and supported its own weight. So I had John bring up the Wikipedia article about deltahedra. A deltahedron is just a polyhedron whose faces are all equilateral triangles.

When I was around eight, I was given a wonderful book called Geometric Playthings, by Jean J. Pedersen and Kent Pedersen. The book was in three sections. One section was about Möbius strips, with which I was already familiar; I ignored this section. The second section was about hexaflexagons, with examples to cut out and put together. The third section was about deltahedra, again with cutout models of all eight deltahedra. As an eight-year-old I had cut out and proudly displayed the eight deltahedra, so I knew that there were some reasonably surprising models one would make with John's toy that would be likely to hold together well. Once again, the deltahedra did not disappoint me.

Four of the deltahedra are the tetrahedron (triangular pyramid, with 4 faces), triangular dipyramid (6 faces), octahedron (square dipyramid, 8 faces), and pentagonal dipyramid (10 faces).

Another is the icosahedron. Imagine making a belt of 10 triangles, alternating up and down, and then connect the ends of the belt. The result is a shape called a pentagonal antiprism, shown at left. The edges of the down-pointing triangles form a pentagon on the top of the antiprism, and the edges of the up-pointing triangles form one on the bottom. Attach a pentagonal pyramid to each of these pentagons, and you have an icosahedron, with a total of 20 faces.

The other three deltahedra are less frequently seen. One is the result of taking a triangular prism and appending a square pyramid to each of its three square faces. (Wikipedia calls this a "triaugmented triangular prism"; I don't know how standard that name is.) Since the prism had two triangular faces to begin with, and we have added four more to each of the three square faces of the original prism, the total is 14 faces.

Another deltahedron is the "gyroelongated square dipyramid". You get this by taking two square pyramids, as with the octahedron. But instead of gluing their square bases together directly, you splice a square antiprism in between. The two square faces of the antiprism are not aligned; they are turned at an angle of 45° to each other, so that when you are looking at the top pyramid face-on, you are looking at the bottom pyramid edge-on, and this is the "gyro" in "gyroelongated". (The icosahedron is a gyroelongated pentagonal dipyramid.) I made one of these in John's office, but found it rather straightforward.

The last deltahedron, however, was quite a puzzle. Wikipedia calls it a "snub disphenoid", and as I mentioned before, the name did not help me out at all. It took me several tries to build it correctly. It contains 12 faces and 8 vertices. When I finally had the model I still couldn't figure it out, and spent quite a long time rotating it and examining it. It has a rather strange symmetry. It is front-back and left-right symmetric. And it is almost top-bottom symmetric: If you give it a vertical reflection, you get the same thing back, but rotated 90° around the vertical axis.

When I planned this article I thought I understood it better. Imagine sticking together two equilateral triangles. Call the common edge the "rib". Fold the resulting rhombus along the rib so that the edges go up, down, up, down in a zigzag. Let's call the resulting shape a "wing"; it has a concave side and a convex side. Take two wings. Orient them with the concave sides facing each other, and with the ribs not parallel, but at right angles. So far, so good.

But this is where I started to get it wrong. The two wings have between them eight edges, and I had imagined that you could glue a rhombic antiprism in between them. I'm not convinced that there is such a thing as a rhombic antiprism, but I'll have to do some arithmetic to be sure. Anyway, supposing that there were such a thing, you could glue it in as I said, but if you did the wings would flatten out and what you would get would not be a proper polyhedron because the two triangles in each wing would be coplanar, and polyhedra are not allowed to have abutting coplanar faces. (The putative gyroelongated triangular dipyramid fails for this reason, I believe.)

To make the snub disphenoid, you do stick eight triangles in between the two wings, but the eight triangles do not form a rhombic antiprism. Even supposing that such a thing exists.

I hope to have some nice renderings for you later. I have been doing some fun work in rendering semiregular polyhedra, and I am looking forward to discussing it here. Advance peek: suppose you know how the vertices are connected by edges. How do you figure out where the vertices are located in 3-space?

If you would like to investigate this, the snub disphenoid has 8 vertices, which we can call A, B, ... H. Then:
This vertex:is connected to these:
The two wings here are ABCD and EFGH. We can distinguish three sorts of edges: five inside the top wing, five inside the bottom wing, and eight that go between the two wings.

Here is a list of the eight deltahedra, with links to the corresponding Wikipedia articles:

Tetrahedron 464
Triangular dipyramid 695
Octahedron 8126
Pentagonal dipyramid 10157
Snub disphenoid 12188
Triaugmented triangular prism 14219
Gyroelongated square dipyramid 162410
Icosahedron 203012
[ Addendum 20070905: There are some followup notes. ]

[ Addendum 20070908: More about deltahedra. ]

[Other articles in category /math] permanent link

Thu, 19 Jul 2007

More about fixed points and attractors
A while back I talked about a technique for calculating √2 where you pick a function that has √2 as a fixed point (that is, f(√2) = √2) and then see what happens when you consider the sequence x, f(x), f(f(x)), ..., for various initial values of x. For some such functions the sequence diverges, but often it converges to √2.

I picked a few example functions, some of which worked and some of which didn't.

One glaring omission from the article was that I forgot to mention the so-called "Babylonian method" for calculating square roots. The Babylonian method for calculating √n is simply to iterate the function x → ½(x + n/x). (This is a special case of the Newton-Raphson method for finding the zeroes of a function. In this case the function whose zeroes are being found is is xx2 - n.) The Babylonian method converges quickly for almost all initial values of x. As I was writing the article, at 3 AM, I had the nagging feeling that I was leaving out an important example function, and then later on realized what it was. Oops.

But there's a happy outcome, which is that the Babylonian method points the way to a nice general extension of this general technique. Suppose you've found a function f that has your target value, say √2, as a fixed point, but you find that iterating f doesn't work for some reason. For example, one of the functions I considered in the article was x → 2/x. No matter what initial value you start with (other than √2 and -√2) iterating the function gets you nowhere; the values just hop back and forth between x and 2/x forever.

But as I said in the original article, functions that have √2 as a fixed point are easy to find. Suppose we have such a function, f, which is badly-behaved because the fixed point repels, or because of the hopping-back-and-forth problem. Then we can perturb the function by trying instead x → ½(x + f(x)), which has the same fixed points, but which might be better-behaved. (More generally, x → (ax + bf(x)) / (a + b) has the same fixed points as f for any nonzero a and b, but in this article we'll leave a = b = 1.) Applying this transformation to the function x → 2/x gives us the Babylonian method.

I tried applying this transform to the other example I used in the original article, which was xx2 + x - 2. This has √2 as a fixed point, but the √2 is a repelling fixed point. √2 ± ε → √2 ± (1 + 2√2)ε, so the error gets bigger instead of smaller. I hoped that perturbing this function might improve its behavior, and at first it seemed that it didn't. The transformed version is x → ½(x + x2 + x - 2) = x2/2 + x - 1. That comes to pretty much the same thing. It takes √2 ± ε → √2 + (1 + √2)ε, which has the same problem. So that didn't work; oh well.

But actually things had improved a bit. The original function also has -√2 as a fixed point, and again it's one that repels from both sides, because -√2 ± ε → -√2 ± (1 - 2√2)ε, and |1 - 2√2| > 1. But the transformed function, unlike the original, has -√2 as an attractor, since it takes -√2 ± ε → -√2 ± (1 - √2)ε and |1 - √2| < 1.

So the perturbed function works for calculating √2, in a slightly backwards way; you pick a value close to -√2 and iterate the function, and the iterated values get increasingly close to -√2. Or you can get rid of the minus signs entirely by transforming the function again, and considering -f(-x) instead of f(x). This turns x2/2 + x - 1 into -x2/2 + x + 1. The fixed points change places, so now √2 is the attractor, and -√2 is the repeller, since √2 ± ε → √2 ± (1 - √2)ε. Starting with x = 1, we get:

So that worked out pretty well. One might even make the argument that the method is simpler than the Babylonian method, since the division is a simple x/2 instead of a complex 2/x. I have not yet looked into the convergence properties; I expect it will turn out that the iterated polynomial converges more slowly than the Babylonian method.

I had meant to write about Möbius transformations, but that will have to wait until next week, I think.

[ Addendum 20201018: Another followup article, but I never did get around to discussing the Möbius transformations. ]

[Other articles in category /math] permanent link

Sat, 30 Jun 2007

How to calculate the square root of 2
A few weeks ago I mentioned the following recurrence:

p0 = 1 q0 = 1
pi+1 = pi + 2qi qi+1 = pi + qi
If you carry this out, you get pairs p and q that have p2 - 2q2 = ±1, which means that p/q ≈ √2. The farther you carry the recurrence, the better the approximation is.

I said that this formula comes from consideration of continued fractions. But I was thinking about it a little more, and I realized that there is a way to get such a recurrence for pretty much any algebraic constant you want.

Consider for a while the squaring function s : xx2. This function has two obvious fixed points, namely 0 and 1, by which I mean that s(0) = 0 and s(1) = 1. Actually it has a third fixed point, ∞.

If you consider the behavior on some x in the interval (0, 1), you see that s(x) is also in the same interval. But also, s(x) < x on this interval. Now consider what happens when you iterate s on this interval, calculating the sequence s(x), s(s(x)), and so on. The values must stay in (0, 1), but must always decrease, so that no matter what x you start with, the sequence converges to 0. We say that 0 is an "attracting" fixed point of s, because any starting value x, no matter how far from 0 it is (as long as it's still in (0, 1)), will eventually be attracted to 0. Similarly, 1 is a "repelling" fixed point, because any starting value of x, no matter how close to 1, will be repelled to 0.

Consideration of the interval (1, ∞) is similar. 1 is a repeller and ∞ is an attractor.

Fixed points are not always attractors or repellers. The function x → 1/x has fixed points at ±1, but these points are neither attractors nor repellers.

Also, a fixed point might attract from one side and repel from the other. Consider xx/(x+1). This has a fixed point at 0. It maps the interval (0, ∞) onto (0, 1), which is a contraction, so that 0 attracts values on the right. On the other hand, 0 repels values on the left, because 1/-n goes to 1/(-n+1). -1/4 goes to -1/3 goes to -1/2 goes to -1, at which point the whole thing blows up and goes to -∞.

The idea about the fixed point attractors is suggestive. Suppose we were to pick a function f that had √2 as a fixed point. Then √2 might be an attractor, in which case iterating f will get us increasingly accurate approximations to √2.

So we want to find some function f such that f(√2) = √2. Such functions are very easy to find! For example, take √2. square it, and divide by 2, and add 1, and take the square root, and you have √2 again. So x → √(1+x2/2) is such a function. Or take √2. Take the reciprocal, double it, and you have √2 again. So x → 2/x is another such function. Or take √2. Add 1 and take the reciprocal. Then add 1 again, and you are back to √2. So x → 1 + 1/(x+1) is a function with √2 as a fixed point.

Or we could look for functions of the form ax2 + bx + c. Suppose √2 were a fixed point of this function. Then we would have 2a + b√2 + c = √2. We would like a, b, and c to be simple, since the whole point of this exercise is to calculate √2 easily. So let's take a=b=1, c=-2. The function is now xx2 + x - 2.

Which one to pick? It's an embarrasment of riches.

Let's start with the polynomial, xx2 + x - 2. Well, unfortunately this is the wrong choice. √2 is a fixed point of this function, but repels on both sides: √2 ± ε → √2 ± ε(1 + 2√2), which is getting farther away.

The inverse function of xx2 + x - 2 will have √2 as an attractor on both sides, but it is not so convenient to deal with because it involves taking square roots. Still, it does work; if you iterate ½(-1 + √(9 + 4x)) you do get √2.

Of the example functions I came up with, x → 2/x is pretty simple too, but again the fixed points are not attractors. Iterating the function for any initial value other than the fixed points just gets you in a cycle of length 2, bouncing from one side of √2 to the other forever, and not getting any closer.

But the next function, x → 1 + 1/(x+1), is a winner. (0, ∞) is crushed into (1, 2), with √2 as the fixed point, so √2 attracts from both sides.

Writing x as a/b, the function becomes a/b → 1 + 1/(a/b+1), or, simplifying, a/b → (a + 2b) / (a + b). This is exactly the recurrence I gave at the beginning of the article.

We did get a little lucky, since the fixed point of interest, √2, was the attractor, and the other one, -√2, was the repeller. ((-∞, -1) is mapped onto (-∞, 1), with -√2 as the fixed point; -√2 repels on both sides.) But had it been the other way around we could have exchanged the behaviors of the two fixed points by considering -f(-x) instead. Another way to fix it is to change the attractive behavior into repelling behavior and vice versa by running the function backwards. When we tried this for xx2 + x - 2 it was a pain because of the square roots. But the inverse of x → 1 + 1/(x+1) is simply x → (-x + 2) / (x - 1), which is no harder to deal with.

The continued fraction stuff can come out of the recurrence, instead of the other way around. Let's iterate the function x → 1 + 1/(1+x) formally, repeatedly replacing x with 1 + 1/(1+x). We get:

1 + 1/(1+x)
1 + 1/(1+1 + 1/(1+x))
1 + 1/(1+1 + 1/(1+1 + 1/(1+x)))
So we might expect the fixed point, if there is one, to be 1 + 1/(2 + 1/(2 + 1/(2 + ...))), if this makes sense. Not all such expressions do make sense, but this one is a continued fraction, and continued fractions always make sense. This one is eventually periodic, and a theorem says that such continued fractions always have values that are quadratic surds. The value of this one happens to be √2. I hope you are not too surprised.

In the course of figuring all this out over the last two weeks or so, I investigated many fascinating sidetracks. The x → 1 + 1/(x+1) function is an example of a "Möbius transformation", which has a number of interesing properties that I will probably write about next month. Here's a foretaste: a Möbius transformation is simply a function x → (ax + b) / (cx + d) for some constants a, b, c, and d. If we agree to abbreviate this function as !!{ a\, b \choose c\,d}!!, then the inverse function is also a Möbius transformation, and is in fact !!{a\, b\choose c\,d}^{-1}!!.

[ Addendum 20070719: There is a followup article to this one. ] [ Addendum 20201018: Another followup article. ]

[Other articles in category /math] permanent link

Sun, 17 Jun 2007

Square triangular numbers
A while back I made the erroneous assertion that no numbers are both square and triangular. As I noted in a followup, this is a rather stupid thing to say, since both 0 and 1 are obvious counterexamples. (36 is a nontrivial counterexample.) Also, a few years before I had actually investigated this very question and had determined that the set of such numbers is infinite. Whoops.

I no longer remember how I solved the problem the first time around, but I was tinkering around with it today and came up with an approach that I think is instructive, or at least interesting.

We want to find non-negative integers a and b such that ½(a2 + a) = b2. Or, equivalently, we want a and b such that √(a2 + a) = b√2.

Now, √(a2 + a) is pretty nearly a + ½. So suppose we could find p and q with a + ½ = b·p/q, and p/q a bit larger than √2. a + ½ is a bit too large to be what we want on the left, but p/q is a bit larger than what we want on the right too. Perhaps the fudging on both sides would match up, and we would get √(a2 + a) = b√2 anyway.

If this magic were somehow to occur, then a and b would be the numbers we wanted.

Finding p/q that is a shade over √2 is a well-studied problem, and one of the things I have in my toolbox, because it seems to come up over and over in the solution of other problems, such as this one. It has interesting connections to several other parts of mathematics, and I have written about it here before.

The theoretical part of finding p/q close to √2 is some thing about continued fractions that I don't want to get into today. But the practical part is very simple. The following recurrence generates all the best rational approximations to √2; the farther you carry it, the better the approximation:

p0 = 1 q0 = 1
pi+1 = pi + 2qi qi+1 = pi + qi
This gives us the following examples:

p q p/q
1 1 1.0
3 2 1.5
7 5 1.4
17 12 1.416666666666667
41 29 1.413793103448276
99 70 1.414285714285714
239 169 1.414201183431953
577 408 1.41421568627451
1393 985 1.414213197969543
3363 2378 1.41421362489487
And in all cases p2 - 2q2 = ±1.

Now, we want a + ½ = b·p/q, or equivalently (2a + 1)/2b = p/q. This means we can restrict our attention to the rows of the table that have q even. This is a good thing, because we need p/q a bit larger than √2, and those are precisely the rows with even q. The rows that have q odd have p/q a bit smaller than √2, which is not what we need. So everything is falling into place.

Let's throw away the rows with q odd, put a = (p - 1)/2 and b = q/2, and see what we get:

pqab½(a2+a) = b2
3 2 1 11
17 12 8 636
99 70 49 351225
577 408 288 20441616
3363 2378 1681 11891413721
Lo and behold, our wishful thinking about the fudging on both sides canceling out has come true, and an infinite set of solutions just pops right out.

I have two points to make about this. One is that I have complained in the past about mathematical pedagogy, how the convention is to come up with some magic-seeming guess ahead of time, as when pulling a rabbit from a hat, and then at the end it is revealed to be the right choice, but what really happened was that the author worked out the whole thing, then saw at the end what he would need at the beginning to make it all work, and went back and filled in the details.

That is not what happened here. My apparent luck was real luck. I really didn't know how it was going to come out. I was really just exploring, trying to see if I could get some insight into the answer without necessarily getting all the way there; I thought I might need to go back and do a more careful analysis of the fudge factors, or something. But sometimes when you go exploring you stumble on the destination by accident, and that is what happened this time.

The other point I want to make is that I've written before about how a mixture of equal parts of numerical sloppiness and algebraic tinkering, with a dash of canned theory, can produce useful results, in a sort of alchemical transmutation that turns base metals into gold, or at least silver. Here we see it happen again.

[Other articles in category /math] permanent link

Wed, 13 Jun 2007

How to calculate binomial coefficients, again
Yesterday's article about how to calculate binomial coefficients was well-received. It was posted on Reddit, and to my surprise and gratification, the comments were reasonably intelligent. Usually when a math article of mine shows up on Reddit, all the megacretins come out of the woodwork to say what an idiot I am, and why don't I go back to school and learn basic logic.

A couple of people pointed out that, contrary to what I asserted, the algorithm I described can in fact overflow even when the final result is small enough to fit in a machine word. Consider $8\choose 4$ for example. The algorithm, as I wrote it, calculates intermediate values 8, 8, 56, 28, 168, 56, 280, 70, and 70 is the final answer. If your computer has 7-bit machine integers, the answer (70) will fit, but the calculation will overflow along the way at the 168 and 280 steps.

Perhaps more concretely, !!35\choose11!! is 417,225,900, which is small enough to fit in a 32-bit unsigned integer, but the algorithm I wrote wants to calculate this as !!35{34\choose10}\over11!!, and the numerator here is 4,589,484,900, which does not fit.

One Reddit user suggested that you can get around this as follows: To multiply r by a/b, first check if b divides r. If so, calculate (r/ba; otherwise calculate (r·a)/b. This should avoid both overflow and fractions.

Unfortunately, it does not. A simple example is !!{14\choose4} = {11\over1}{12\over2}{13\over3}{14\over4}!!. After the first three multiplications one has 286. One then wants to multiply by 14/4. 4 does not divide 286, so the suggestion calls for multiplying 286 by 14/4. But 14/4 is 3.5, a non-integer, and the goal was to use integer arithmetic throughout.

Fortunately, this is not hard to fix. Say we want to multiply r by a/b without overflow or fractions. First let g be the greatest common divisor of r and b. Then calculate ((r/g) · a)/(b/g). In the example above, g is 2, and we calculate (286/2) · (14/2) = 143 · 7; this is the best we can do.

I haven't looked, but it is hard to imagine that Volume II of Knuth doesn't discuss this in exhaustive detail, including all the stuff I just said, plus a bunch of considerations that hadn't occurred to any of us.

A few people also pointed out that you can save time when n > m/2 by calculating !!m\choose m-n!! instead of $m \choose n$. For example, instead of calculating !!100\choose98!!, calculate $100\choose 2$. I didn't mention this in the original article because it was irrelevant to the main point, and because I thought it was obvious.

[Other articles in category /math] permanent link

Tue, 12 Jun 2007

How to calculate binomial coefficients
The binomial coefficient !!n\choose k!! is usually defined as:

$${n\choose k} = {n!\over k!(n-k)!}$$

This is a fine definition, brief, closed-form, easy to prove theorems about. But these good qualities seduce people into using it for numerical calculations:

        fact 0 = 1
        fact (n+1) = (n+1) * fact n

        choose n k = (fact n) `div` ((fact k)*(fact (n-k)))
(Is it considered bad form among Haskellites to use the n+k patterns? The Haskell Report is decidedly ambivalent about them.)

Anyway, this is a quite terrible way to calculate binomial coefficients. Consider calculating !!100\choose 2!!, for example. The result is only 4950, but to get there the computer has to calculate 100! and 98! and then divide these two 150-digit numbers. This requires the use of bignums in languages that have bignums, and causes an arithmetic overflow in languages that don't. A straightforward implementation in C, for example, drops dead with an arithmetic exception; using doubles instead, it claims that the value of $100\choose 2$ is -2147483648. This is all quite sad, since the correct answer is small enough to fit in a two-byte integer.

Even in the best case, !!2n\choose n!!, the result is only on the order of 4n, but the algorithm has to divide a numerator of about 4nn2n by a denominator of about n2n to get it.

A much better way to calculate values of $n\choose k$ is to use the following recurrence:

$${n+1\choose k+1} = {n+1\over k+1}{n\choose k}$$

This translates to code as follows:

        choose n 0 = 1
        choose 0 k = 0
        choose (n+1) (k+1) = (choose n k) * (n+1) `div` (k+1)
This calculates !!8\choose 4!! as !!{5\over1}{6\over2}{7\over3}{8\over4} !!. None of the intermediate results are larger than the final answer.

An iterative version is also straightforward:

        unsigned choose(unsigned n, unsigned k) {
          unsigned r = 1;
          unsigned d;
          if (k > n) return 0;
          for (d=1; d <= k; d++) {
            r *= n--;
            r /= d;
          return r;
This is speedy, and it cannot cause an arithmetic overflow unless the final result is too large to be represented.

It's important to multiply by the numerator before dividing by the denominator, since if you do this, all the partial results are integers and you don't have to deal with fractions or floating-point numbers or anything like that. I think I may have mentioned before how much I despise floating-point numbers. They are best avoided.

I ran across this algorithm last year while I was reading the Lilavati, a treatise on arithmetic written about 850 years ago in India. The algorithm also appears in the article on "Algebra" from the first edition of the Encyclopaedia Britannica, published in 1768.

So this algorithm is simple, ancient, efficient, and convenient. And the problems with the other algorithm are obvious, or should be. Why isn't this better known?

[ Addendum 20070613: There is a followup article to this one. ]

[Other articles in category /math] permanent link

Fri, 08 Jun 2007

Counting transitive relations
A relation on a set S is merely a subset of S×S. For example, the relation < on the set {1,2,3} can be identified as {(1,2), (1,3), (2,3)}, the set of all (a, b) with a < b.

A relation is transitive if, whenever it has both (a, b) and (b, c), it also has (a, c).

For the last week I've been trying to find a good way to calculate the