The Universe of Discourse
http://blog.plover.com
The Universe of Discourse (Mark Dominus Blog)enRectangles with equal area and perimeter
http://blog.plover.com/2015/03/20#rectangles
<p>Wednesday while my 10-year-old daughter Katara was doing her math
homework, she observed with pleasure that a <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%246%c3%973%24"> 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%c3%974%24"> 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.</p>
<p>This is just one of many examples of how she has way too much
homework, and how it interferes with her education.</p>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%24">, getting <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%24"> on one side and a bunch of
stuff involving <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%24"> 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.</p>
<p>I asked her to plug in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%3d3%24"> and observe that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%3d6%24"> popped right
out, and then similarly that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%3d6%24"> yields <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%3d3%24">, and then I asked
her to try the other example she knew. Then I suggested that she see
what <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%3d5%24"> did: it gives <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%3d%5cfrac%7b10%7d3%24">, This was new, so she
checked it by calculating the area and the perimeter, both
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac%7b50%7d3%24">. She was very excited by this time. <a href="http://blog.plover.com/math/algebra.html">As I have
mentioned earlier</a>, 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.</p>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%3d2L%5cdiv%28L%2d2%29%24"> 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%24"> gets close to
2; then <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%24"> shoots up to infinity. And when <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%24"> gets big, say a
million, you can see from the algebra that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%24"> is a hair more than
2. So I drew in the asymptotes on the hyperbola. </p>
<p align=center><a href="http://pic.blog.plover.com/math/rectangles/plot.png"><img border=0 src="http://pic.blog.plover.com/math/rectangles/plot-th.png"></a></p>
<p><a href="http://pic.blog.plover.com/math/rectangles/holladay1.jpg"><img border=0 align=right src="http://pic.blog.plover.com/math/rectangles/holladay1-th.jpg"></a></p>
<p>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!”)</p>
<p><br clear=right></p>
<p>Once you know how the graph behaves, it is a simple matter to see that
there are no integer solutions other than <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle%203%2c6%5crangle%2c%0a%5clangle%204%2c4%5crangle%2c%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle6%2c3%5crangle%24">. We know that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%3d5%24">
does not work. For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%3e6%24"> the value of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%24"> is always strictly
between <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24">. For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%3d2%24"> there is no value of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%24"> that works at
all. For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%5clt%20L%5clt%202%24"> the formula says that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24W%24"> is negative, on the
other branch of the hyperbola, which is a perfectly good <em>numerical</em>
solution (for example, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24L%3d1%2c%20W%3d%2d2%24">) 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.</p>
<p>[ Addendum 20150330: Thanks to Steve Hastings for his plot of the
hyperbola, which is in the public domain. ]</p>
An ounce of theory is worth a pound of search
http://blog.plover.com/2015/03/19#ounce-of-theory
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24162%2b625%3d787%24"> and
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%5e%7b19%7d%2b3%5e4%3d524369%24">.</p>
<p>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.</p>
<p>One of these <a href="http://blog.plover.com/math/excellent.html">I wrote up on this blog back in 2006</a>:</p>
<blockquote>
<p>A number
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> is excellent if it has an even number of digits, and if when
you chop it into a front half <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24"> and a back half <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%24">, you have
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%5e2%20%2d%20a%5e2%20%3d%20n%24">. For example, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2448%24"> is excellent, because <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%248%5e2%20%2d%0a%3e%204%5e2%20%3d%2048%24">, and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243468%24"> is excellent, because <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2468%5e2%20%2d%2034%5e2%20%3d%204624%0a%3e%20%2d%201156%20%3d%203468%24">.</p>
</blockquote>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%2810%5ek%2ba%29%20%3d%20b%5e2%2bb%24"> and it's not
hard to quickly test candidates for <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24"> to see if <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%2810%5ek%2ba%29%24"> has
this form and if so to find the corresponding value of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%24">.
(Details are in <a href="http://blog.plover.com/math/excellent.html">the other post</a>.) 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.</p>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%2cb%2cc%2cd%24"> for which <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%5e2%2bb%5e2%2c%0ab%5e2%2bc%5e2%2c%20c%5e2%2bd%5e2%2c%20%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24d%5e2%2ba%5e2%24"> 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.</p>
<p>Instead, observe that if <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%5e2%2bb%5e2%24"> is a perfect square then <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24">
and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%24"> are the legs of a right triangle with integer sides; they
are terms in what is known as a <a href="https://en.wikipedia.org/wiki/Pythagorean_triple">Pythagorean
triple</a>. The
prototypical example is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%5e2%20%2b%204%5e2%20%3d%205%5e2%24">, and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle%0a3%2c4%2c5%5crangle%24"> is the Pythagorean triple. (The querent was quite
aware that he was asking for Pythagorean triples, and mentioned them
specifically.)</p>
<p>Here's the key point: It has been known since ancient times that if
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle%20a%2cb%2cc%5crangle%24"> is a Pythagorean triple, then there exist
integers <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24m%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> such that: $$\begin{align}
\require{align}
a & = n^2-m^2 \\
b & = 2mn \\
c & = n^2 + m^2
\end{align}$$</p>
<p>So you don't have to search for Pythagorean triples; you can just
generate them with no searching:</p>
<pre><code> 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}{$a} = 1;
}
}
</code></pre>
<p>This builds a hash table, <code>%trip</code>, with two important properties:</p>
<ol>
<li><p><code>$trip{$a}</code> is a sub-table whose keys are all the numbers that can
form a triple with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24">. For example, <code>$trip{20}</code> is a hash with
three keys: 21, 48, and 99, because <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2420%5e2%2b21%5e2%20%3d%2029%5e2%2c%2020%5e2%2b48%5e2%3d%0a%20%20%2052%5e2%2c%20%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2420%5e2%20%2b%2099%5e2%20%3d%20101%5e2%24">, but 20 is not a participant
in any other triples.</p></li>
<li><p><code>$trip{$a}{$b}</code> is true if and only if <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%5e2%2bb%5e2%24"> is a perfect
square, and false otherwise.</p></li>
</ol>
<p>The table has only around 40,000 entries. Having constructed it, we
now search it:</p>
<pre><code> 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};
}
}
}
}
</code></pre>
<p>The outer loop runs over each <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24"> that is known to be a member of a
Pythagorean triple. (Actually the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24m%2cn%24"> formulas show that
every number bigger than 2 is a member of <em>some</em>
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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%24"> that can
possibly form a triple with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24">; that is, every <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%24"> for which
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%5e2%2bb%5e2%24"> is a perfect square. We don't have to search for them; we
have them tabulated ahead of time. Then for each such <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%24"> (and
there aren't very many) we run over every <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24c%24"> that forms a triple
with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%24">, and again there is no searching and very few candidates.
Then then similarly <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24d%24">, and if the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24d%24"> we try forms a triple with
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24">, we have a winner.</p>
<p>The <code>next if $c == $a</code> and <code>next if $d == $b</code> tests are to rule out
trivial solutions like <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%3dc%3d3%2c%20b%3dd%3d4%24">, 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5csqrt2%24"> is irrational).</p>
<p>This runs in less than a second on so-so hardware and produces 11
solutions:</p>
<pre><code> 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
</code></pre>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle%0aa%2cb%2cc%2cd%5crangle%24"> is a solution, so is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle%20ka%2c%20kb%2ckc%2ckd%5crangle%24">
for any <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%24">. 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.</p>
<p>The only other thing wrong with the program is that it produces each
solution 8 times; if <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle%20a%2cb%2cc%2cd%5crangle%24"> is a solution, then
so are <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle%20b%2cc%2cd%2ca%5crangle%2c%20%5clangle%20d%2cc%2cb%2ca%5crangle%2c%24"> and so on.
This is easily fixed with a little post-filtering; pipe the output
through</p>
<pre><code> perl -nle '$k = join " ", sort { $a <=> $b } split; print unless $seen{$k}++ '
</code></pre>
<p>or something of that sort.</p>
<p>The corresponding run with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24m%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> up to 2,000 instead of
only 200 takes 5 minutes and finds 445 solutions, of which 101 are distinct, including <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle%0a3614220%2c%20618192%2c%202080820%2c%20574461%5crangle%24">. It would take a very long
time to find this with a naïve search.</p>
<p>[ For a much larger and more complex example of the same sort of
thing, see <a href="http://blog.plover.com/math/dd.html">When do <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242n%24"> have the same
digits?</a>. 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. ]<p></p>
Within this instrument, resides the Universe
http://blog.plover.com/2014/11/22#algebra
<p>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:</p>
<blockquote>
<p>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?</p>
</blockquote>
<p>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.</p>
<p>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.</p>
<p>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:</p>
<p>$$\begin{align}
MT & = 3ST \\
MY & = 2SY \\
\end{align}
$$</p>
<p>(<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24MT%24"> here is the name of a single variable, not a product of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24M%24">
and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24T%24">; the others should be understood similarly.)</p>
<p>“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:</p>
<p>$$\begin{align}
MY & = MT + 2 \\
SY & = ST + 2
\end{align}
$$</p>
<p>I would have written two equations in two unknowns:</p>
<p>$$\begin{align}
M_T & = 3S_T\\
M_T+2 & = 2(S_T + 2)
\end{align}
$$</p>
<p>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 <em>both</em> 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 <em>also</em> 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.</p>
<p>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.</p>
<p>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 <em>The Dying Earth</em> 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.’”</p>
<blockquote>
<p>“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.”</p>
</blockquote>
<p>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:</p>
<blockquote>
<p>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?</p>
</blockquote>
<p>“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
<em>once you see it</em>: 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.</p>
<p>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.</p>
<p>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.</p>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24M_T%3d%2d2%2c%20S_T%3d%2d1%20%24"> 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.</p>
When do n and 2n have the same digits?
http://blog.plover.com/2014/07/23#dd
<p>[This article was published last month on <a href="http://math.blogoverflow.com/">the math.stackexchange
blog</a>, 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.]</p>
<p>A <a href="http://math.stackexchange.com/questions/782334/interview-question-asked-in-yahoo?">recent question on math.stackexchange</a> asks for the smallest positive integer <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> for which the number <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242A%24"> has the same
decimal digits in some other order.</p>
<p>Math geeks may immediately realize that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24142857%24"> has this property, because it is the first 6 digits of the decimal expansion of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac%2017%24">, and the cyclic behavior of the decimal expansion of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac%20n7%24"> is well-known. But is this the <em>minimal</em> solution? It is not. Brute-force enumeration of the solutions quickly reveals that there are 12 solutions of 6 digits each, all permutations of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24142857%24">, and that larger solutions, such as 1025874 and 1257489 seem to follow a similar pattern. What is happening here?</p>
<p>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 <em>sets</em> of digits can appear in a solution of the problem. But once the set of digits is chosen, the conditions on that <em>order</em> of the digits in the solution are fairly lax.</p>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24124578%24">. 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.</p>
<h3>Notation</h3>
<p>The property of interest, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P_R%28A%29%24">, is that the numbers <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%3d2A%24"> have exactly the same base-<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> digits. We would like to find numbers <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> having property <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P_R%24"> for various <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24">, and we are most interested in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d10%24">. Suppose <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> is an <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24">-digit numeral having property <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P_R%24">; let the (base-<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24">) digits of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> be <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_%7bn%2d1%7d%5cldots%20a_1a_0%24"> and similarly the digits of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%20%3d%202A%24"> are <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_%7bn%2d1%7d%5cldots%20b_1b_0%24">. The reader is encouraged to keep in mind the simple example of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d8%2c%20n%3d4%2c%20A%3d%5cmathtt%7b1042%7d%2c%20B%3d%5cmathtt%7b2104%7d%24"> which we will bring up from time to time.</p>
<p>Since the digits of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> are the same, in a different order, we may say that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_i%20%3d%20a_%7bP%28i%29%7d%24"> for some permutation <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">. In general <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> might have more than one cycle, but we will suppose that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is a single cycle. All the following discussion of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> will apply to the individual cycles of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> in the case that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is a product of two or more cycles. For our example of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%3d%5cmathtt%7b1042%7d%2c%20b%3d%5cmathtt%7b2104%7d%24">, we have <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%20%3d%20%280%5c%2c1%5c%2c2%5c%2c3%29%24"> in cycle notation. We won't need to worry about the details of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">, except to note that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24i%2c%20P%28i%29%2c%20P%28P%28i%29%29%2c%20%5cldots%2c%20P%5e%7bn%2d1%7d%28i%29%24"> completely exhaust the indices <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%2e%20%5cldots%20n%2d1%24">, and that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%5en%28i%29%20%3d%20i%24"> because <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is an <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24">-cycle.</p>
<h3>Conditions on the set of digits in a solution</h3>
<p>For each <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24i%24"> we have $$a_{P(i)} = b_{i} \equiv 2a_{i} + c_i\pmod R
$$ where the ‘carry bit’ <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24c_i%24"> is either 0 or 1 and depends on whether there was a carry when doubling <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_%7bi%2d1%7d%24">. (When <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24i%3d0%24"> we are in the rightmost position and there is never a carry, so <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24c_0%3d%200%24">.) We can then write:</p>
<p>$$\begin{align}
a_{P(P(i))} &= 2a_{P(i)} + c_{P(i)} \\
&= 2(2a_{i} + c_i) + c_{P(i)} &&= 4a_i + 2c_i + c_{P(i)}\\
a_{P(P(P(i)))} &= 2(4a_i + 2c_i + c_{P(P(i)})) + c_{P(i)} &&= 8a_i + 4c_i + 2c_{P(i)} + c_{P(P(i))}\\
&&&\vdots\\
a_{P^n(i)} &&&= 2^na_i + v
\end{align}
$$</p>
<p>all equations taken <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbmod%20R%24">. But since <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is an <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24">-cycle, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%5en%28i%29%20%3d%20i%24">, 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%5cin%5c%5c%7b0%2c%5cldots%202%5en%2d1%5c%5c%7d%24"> depends only on the values of the carry bits <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24c_i%24">—the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24c_i%24"> are precisely the binary digits of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24">.</p>
<p>Specifying a particular value of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_0%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> that satisfy this equation completely determines all the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%24">. For example, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_0%20%3d%202%2c%20v%20%3d%20%5ccolor%7bdarkblue%7d%7b0010%7d_2%20%3d%202%24"> is a solution when <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d8%2c%20n%3d4%24"> because <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigl%282%5e4%2d1%5cbigr%29%5ccdot2%20%2b%202%5cequiv%200%5cpmod%208%24">, and this solution allows us to compute</p>
<p>$$\def\db#1{\color{darkblue}{#1}}\begin{align}
a_0&&&=2\\
a_{P(0)} &= 2a_0 &+ \db0 &= 4\\
a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 0 \\
a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 1\\ \hline
a_{P^4(0)} &= 2a_{P^3(0)} &+ \db0 &= 2\\
\end{align}$$</p>
<p>where the carry bits <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24c_i%20%3d%20%5clangle%200%2c0%2c1%2c0%5crangle%24"> are visible in the third column, and all the sums are taken <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cpmod%208%24">. Note that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_%7bP%5en%280%29%7d%20%3d%20a_0%24"> as promised. This derivation of the entire set of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%24"> from a single one plus a choice of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> is crucial, so let's see one more example. Let's consider <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d10%2c%20n%3d3%24">. Then we want to choose <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_0%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> so that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cleft%282%5e3%2d1%5cright%29a_0%20%2b%20v%20%5cequiv%200%5cpmod%7b10%7d%24"> where <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%5cin%5c%5c%7b0%5cldots%207%5c%5c%7d%24">. One possible solution is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%5c_0%20%3d%205%2c%20v%3d%5ccolor%7bdarkblue%7d%7b101%7d%5c_2%20%3d%205%24">. Then we can derive the other <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%24"> as follows:</p>
<p>$$\begin{align}
a_0&&&=5\\
a_{P(0)} &= 2a_0 &+ \db1 &= 1\\
a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 2 \\\hline
a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 5\\
\end{align}$$</p>
<p>And again we have <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_%7bP%5en%280%29%7d%3d%20a_0%24"> as required.</p>
<p>Since the bits of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> are used cyclically, not every pair of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clangle%20a_0%2c%20v%5crangle%24"> will yield a different solution. Rotating the bits of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> and pairing them with different choices of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_0%24"> will yield the same cycle of digits starting from a different place. In the first example above, we had <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_0%20%3d%202%2c%20v%20%3d%200010_2%20%3d%202%24">. If we were to take <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_0%20%3d%204%2c%20v%20%3d%200100_2%20%3d%204%24"> (which also solves <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28%5cstar%29%24">) we would get the same cycle of values of the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%24"> but starting from <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%24"> instead of from <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24">, and similarly if we take <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_0%3d0%2c%20v%20%3d%201000_2%20%3d%208%24"> or <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_0%20%3d%201%2c%20v%20%3d%200001_2%24">. So we can narrow down the solution set of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28%5cstar%29%24"> by considering only the so-called <a href="https://en.wikipedia.org/wiki/Bracelet_%28combinatorics%29">bracelets</a> of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> rather than all <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%5en%24"> possible values. Two values of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> are considered equivalent as bracelets if one is a rotation of the other. When a set of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24">-values are equivalent as bracelets, we need only consider one of them; the others will give the same cyclic sequence of digits, but starting in a different place. For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d4%24">, for example, the bracelets are <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240000%2c%200001%2c%200011%2c%200101%2c%200111%2c%20%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241111%24">; the sequences <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240110%2c%201100%2c%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241001%24"> being equivalent to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240011%24">, and so on.</p>
<h4>Example</h4>
<p>Let us take <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d9%2c%20n%3d3%24">, so we want to find 3-digit numerals with property <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P_9%24">. According to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28%5cstar%29%24"> we need <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247a_i%20%2b%20v%20%5cequiv%200%5cpmod%7b9%7d%24"> where <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%5cin%5c%5c%7b0%5cldots%0a7%5c%5c%7d%24">. There are 9 possible values for <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%24">; for each one there is at most one possible value of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> that makes the sum zero:</p>
<p>$$\begin{array}{rrr}
a_i & 7a_i & v \\ \hline
0 & 0 & 0 \\
1 & 7 & 2 \\
2 & 14 & 4 \\
3 & 21 & 6 \\
4 & 28 & \\
5 & 35 & 1 \\
6 & 42 & 3 \\
7 & 49 & 5 \\
8 & 56 & 7 \\
\end{array}
$$</p>
<p>(For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%3d4%24"> there is no solution.) We may disregard the non-bracelet values of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24">, as these will give us solutions that are the same as those given by bracelet values of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24">. The bracelets are:</p>
<p>$$\begin{array}{rl} 000 & 0 \\ 001 & 1 \\ 011 & 3 \\ 111 & 7 \end{array}$$</p>
<p>so we may disregard the solutions exacpt when <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%3d0%2c1%2c3%2c7%24">. Calculating the digit sequences from these four values of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> and the corresponding <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%24"> we find:</p>
<p>$$\begin{array}{ccl}
a_0 & v & \text{digits} \\ \hline
0 & 0 & 000 \\
5 & 1 & 512 \\
6 & 3 & 637 \\
8 & 7 & 888 \
\end{array}
$$</p>
<p>(In the second line, for example, we have <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%3d1%20%3d%20001_2%24">, so <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%20%3d%202%5ccdot%205%20%2b%200%3b%202%20%3d%201%5ccdot%202%20%2b%200%3b%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%245%20%3d%202%5ccdot%202%20%2b%201%24">.)</p>
<p>Any number <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> of three digits, for which <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242A%24"> contains exactly the same three digits, in base 9, must therefore consist of exactly the digits <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24000%2c%20125%2c%20367%2c%24"> or <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24888%24">.</p>
<h4>A warning</h4>
<p>All the foregoing assumes that the permutation <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is <em>a single cycle</em>. In general, it may not be. Suppose we did an analysis like that above for <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d10%2c%20n%3d5%24"> and found that there was no possible digit set, other than the trivial set <code>00000</code>, that satisfied the governing equation <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%282%5e5%2d1%29a_0%20%2b%20v%5cequiv%200%5cpmod%7b10%7d%24">. This would <em>not</em> completely rule out a base-10 solution with 5 digits, because the analysis only rules out a <em>cyclic</em> set of digits. There could still be a solution where <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> was a product of a <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24"> and a <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24">-cycle, or a product of still smaller cycles.</p>
<p>Something like this occurs, for example, in the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d4%2c%20R%3d8%24"> case. Solving the governing equation <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%282%5e5%2d1%29a_0%20%2b%20v%20%5cequiv%200%5cpmod%208%24"> yields only four possible digit cycles, namely <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b0%2c1%2c2%2c4%5c%5c%7d%2c%20%5c%5c%7b1%2c3%2c6%2c4%5c%5c%7d%2c%20%5c%5c%7b2%2c5%2c2%2c5%5c%5c%7d%24">, and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b3%2c7%2c6%2c5%5c%5c%7d%24">. But there are several additional solutions: <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242500_8%5ccdot%202%20%3d%205200_8%2c%202750_8%5ccdot%202%20%3d%20%20%205720_8%2c%20%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242775_8%5ccdot%202%20%3d%205772_8%24">. These correspond to permutations <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> with more than one cycle. In the case of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242750_8%24">, for example, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> exchanges the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%245%24"> and the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24">, and leaves the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24"> and the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24"> fixed.</p>
<p>For this reason we cannot rule out the possibility of an <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24">-digit solution without first considering all smaller <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24">.</p>
<h4>The Large Equals Odd rule</h4>
<p>When <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> is even there is a simple condition we can use to rule out certain sets of digits from being single-cycle solutions. Recall that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%3da_%7bn%2d1%7d%5cldots%20a_0%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%3db_%7bn%2d1%7d%5cldots%20b_0%24">. Let us agree that a digit <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24d%24"> is <em>large</em> if <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24d%5cge%20%5cfrac%20R2%24"> and <em>small</em> otherwise. That is, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24d%24"> is large if, upon doubling, it causes a carry into the next column to the left.</p>
<p>Since <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_i%20%3d%282a_i%20%2b%20c_i%29%5cbmod%20R%24">, where the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24c_i%24"> are carry bits, we see that, except for <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_0%24">, the digit <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_i%24"> is odd precisely when there is a carry from the next column to the right, which occurs precisely when <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_%7bi%2d1%7d%24"> is large. Thus the number of odd digits among <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_1%2c%5cldots%20b_%7bn%2d1%7d%24"> is equal to the number of large digits among <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_1%2c%5cldots%20a_%7bn%2d2%7d%24">.
This leaves the digits <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_0%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_%7bn%2d1%7d%24"> uncounted. But <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_0%24"> is never odd, since there is never a carry in the rightmost position, and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_%7bn%2d1%7d%24"> is always small (since otherwise <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24"> would have <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%2b1%24"> digits, which is not allowed). So the number of large digits in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> is exactly equal to the number of odd digits in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24">. And since <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24"> have exactly the same digits, the number of large digits in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> is equal to the number of odd digits in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24">. Observe that this is the case for our running example <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241042_8%24">: there is one odd digit and one large digit (the 4).</p>
<p>When <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> is odd the analogous condition is somewhat more complicated, but since the main case of interest is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d10%24">, we have the useful rule that:</p>
<blockquote>
For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> even, the number of odd digits in any solution <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> is equal to the number of large digits in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24">.
</blockquote>
<h1>Conditions on the order of digits in a solution</h1>
<p>We have determined, using the above method, that the digits <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b5%2c1%2c2%5c%5c%7d%24"> might form a base-9 numeral with property <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P_9%24">. Now we would like to arrange them into a base-9 numeral that actually does have that property. Again let us write <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%20%3d%20a_2a_1a_0%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%3db_2b_1b_0%24">, with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%3d2A%24">. Note that if <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%20%3d%201%24">, then <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_i%20%3d%203%24"> (if there was a carry from the next column to the right) or <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24"> (if there was no carry), but since <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_i%3d3%24"> is impossible, we must have <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%20%3d%202%24"> and therefore <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_%7bi%2d1%7d%24"> must be small, since there is no carry into position <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24i%24">. But since <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_%7bi%2d1%7d%24"> is also one of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b5%2c1%2c2%5c%5c%7d%24">, and it cannot also be <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24">, it must be <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24">. This shows that the 1, unless it appears in the rightmost position, must be to the left of the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24">; it cannot be to the left of the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%245%24">. Similarly, if <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%20%3d%202%24"> then <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b_i%20%3d%205%24">, because <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%24"> is impossible, so the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24"> must be to the left of a large digit, which must be the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%245%24">. Similar reasoning produces no constraint on the position of the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%245%24">; it could be to the left of a small digit (in which case it doubles to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24">) or a large digit (in which case it doubles to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24">). We can summarize these findings as follows:</p>
<p>$$\begin{array}{cl}
\text{digit} & \text{to the left of} \\ \hline
1 & 1, 2, \text{end} \\
2 & 5 \\
5 & 1,2,5,\text{end}
\end{array}$$</p>
<p>Here “end” means that the indicated digit could be the rightmost.</p>
<p>Furthermore, the left digit of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> must be small (or else there would be a carry in the leftmost place and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242A%24"> 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24125%24"> or <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24251%24">, and indeed, both of those numbers have the required property: <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24125_9%5ccdot%202%20%3d%20251_9%24">, and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24251_9%5ccdot%202%20%3d%20512_9%24">.</p>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%27%24"> whenever <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> can appear to the left of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%27%24">. Then the graph drawn for the table above looks like this:</p>
<p><img src="http://pic.plover.com/math-se-blog/double-digits/huhooa.png" alt="Graph for 125 base 9" /></p>
<p>A 3-digit numeral with property <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P_9%24"> 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 <em><a href="https://en.wikipedia.org/wiki/Hamiltonian_path">hamiltonian</a></em>. Obviously, self-loops never occur in a hamiltonian path, so we will omit them from future diagrams.</p>
<p>Now we will consider the digit set <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24637%24">, again base 9. An analysis similar to the foregoing allows us to construct the following graph:</p>
<p><img src="http://pic.plover.com/math-se-blog/double-digits/ovecmp.png" alt="Graph for 367 base 9" /></p>
<p>Here it is immediately clear that the only hamiltonian path is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%2d7%2d6%2d%5ctext%7bend%7d%24">, and indeed, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24376_9%5ccdot%202%20%3d%20763_9%24">.</p>
<p>In general there might be multiple instances of a digit, and so multiple nodes labeled with that digit. Analysis of the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%2c0%2c0%24"> case produces a graph with no legal start nodes and so no solutions, unless leading zeroes are allowed, in which case <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24000%24"> is a perfectly valid solution. Analysis of the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%248%2c8%2c8%24"> case produces a graph with no path to the end node and so no solutions. These two trivial patterns appear for all <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> and all <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24">, and we will ignore them from now on.</p>
<p>Returning to our ongoing example, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241042%24"> in base 8, we see that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24"> must double to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%24">, so must be to the left of small digits, but <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24"> can double to either <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24"> or <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24"> 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:</p>
<p><img src="http://pic.plover.com/math-se-blog/double-digits/mludlo.png" alt="Graph for 1024 base 8" /></p>
<p>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:</p>
<pre>
1042
1204
2041
2104
</pre>
<p>If leading zeroes are allowed we have also:</p>
<pre>
0412
0421
</pre>
<p>All of these are solutions in base 8.</p>
<h3>The case of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d10%24"></h3>
<p>Now we turn to our main problem, solutions in base 10.</p>
<p>To find <em>all</em> 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 <em>cyclically</em>; that is, the permutations <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> that we considered had only one cycle each.</p>
<p>There are no smaller solutions, but to prove that the length 6 solutions are minimal, we must analyze the cases for smaller <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> and rule them out. We now produce a complete analysis of the base 10 case with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d10%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%5cle%206%24">. For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d1%24"> there is only the trivial solution of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24">, which we disregard. (The question asked for a positive number anyway.)</p>
<h4><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d2%24"></h4>
<p>For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d2%24">, we want to find solutions of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243a_i%20%2b%20v%20%5cequiv%200%5cpmod%7b10%7d%24"> where <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> is a two-bit bracelet number, one of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2400_2%2c%2001_2%2c%20%24"> or <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2411_2%24">. Tabulating the values of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%5cin%5c%5c%7b0%2c1%2c3%5c%5c%7d%24"> that solve this equation we get:</p>
<p>$$\begin{array}{ccc}
v& a_i \\ \hline
0 & 0 \\
1& 3 \\
3& 9 \\
\end{array}$$</p>
<p>We can disregard the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%3d0%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%3d3%24"> solutions because the former yields the trivial solution <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2400%24"> and the latter yields the nonsolution <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2499%24">. So the only possibility we need to investigate further is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%20%3d%203%2c%20v%20%3d%201%24">, which corresponds to the digit sequence <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2436%24">: Doubling <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24"> gives us <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%246%24"> and doubling <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%246%24">, plus a carry, gives us <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24"> again.</p>
<p>But when we tabulate of which digits must be left of which informs us that there is no solution with just <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%246%24">, because the graph we get, once self-loops are eliminated, looks like this:</p>
<p><img src="http://pic.plover.com/math-se-blog/double-digits/xthkmc.png" alt="graph for 36 base 10" /></p>
<p>which obviously has no hamiltonian path. Thus there is no solution for <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d10%2c%20n%3d2%24">.</p>
<h4><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d3%24"></h4>
<p>For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d3%24"> we need to solve the equation <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247a_i%20%2b%20v%20%5cequiv%200%5cpmod%7b10%7d%24"> where <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> is a bracelet number in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b0%2c%5cldots%207%5c%5c%7d%24">, specifically one of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%2c1%2c3%2c%24"> or <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24">. Since <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2410%24"> are relatively prime, for each <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> there is a single <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%24"> that solves the equation.
Tabulating the possible values of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_i%24"> as before, and this time omitting rows with no solution, we have:</p>
<p>$$\begin{array}{rrl}
v & a_i & \text{digits}\\ \hline
0& 0 & 000\\
1& 7 & 748 \\
3& 1 & 125\\
7&9 & 999\\
\end{array}$$</p>
<p>The digit sequences <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%2c0%2c0%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%249%2c9%2c9%24"> yield trivial solutions or nonsolutions as usual, and we will omit them in the future. The other two lines suggest the digit sets <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%2c2%2c5%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%2c7%2c8%24">, both of which fails the “odd equals large” rule.</p>
<p>This analysis rules out the possibility of a digit set with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a_0%20%5cto%20a_1%20%5cto%20a_2%20%5cto%20a_1%24">, but it does not <em>completely</em> 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.</p>
<h4><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d4%24"></h4>
<p>For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d4%24"> the governing equation is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2415a_i%20%2b%20v%20%5cequiv%200%5cpmod%7b10%7d%24"> where <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> is a 4-bit bracelet number, one of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b0%2c1%2c3%2c5%2c7%2c15%5c%5c%7d%24">. This is a little more complicated because <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cgcd%2815%2c10%29%5cne%201%24">. Tabulating the possible digit sets, we get:</p>
<p>$$\begin{array}{crrl}
a_i & 15a_i& v & \text{digits}\\ \hline
0 & 0 & 0 & 0000\\
1 & 5 & 5 & 1250\\
1 & 5 & 15 & 1375\\
2 & 0 & 0 & 2486\\
3 & 5 & 5 & 3749\\
3 & 5 & 15 & 3751\\
4 & 0 & 0 & 4862\\
5 & 5 & 5 & 5012\\
5 & 5 & 5 & 5137\\
6 & 0 & 0 & 6248\\
7 & 5 & 5 & 7493\\
7 & 5 & 5 & 7513\\
8 & 0 & 0 & 8624 \\
9 & 5 & 5 & 9874\\
9 & 5 & 15 & 9999 \\
\end{array}$$</p>
<p>where the second column has been reduced mod <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2410%24">. Note that even restricting <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> to bracelet numbers the table still contains duplicate digit sequences; the 15 entries on the right contain only the six basic sequences <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240000%2c%200125%2c%201375%2c%202486%2c%203749%2c%204987%24">, and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%249999%24">. Of these, only <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240000%2c%209999%2c%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243749%24"> obey the odd equals large criterion, and we will disregard <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%240000%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%249999%24"> as usual, leaving only <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243749%24">. We construct the corresponding graph for this digit set as follows: <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24"> must double to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24">, not <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%246%24">, so must be left of a large number <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24"> or <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%249%24">. Similarly <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%24"> must be left of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24"> or <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%249%24">. <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%249%24"> must also double to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%249%24">, so must be left of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24">. Finally, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24"> must double to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%24">, so must be left of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%2c4%24"> or the end of the numeral. The corresponding graph is:</p>
<p><img src="http://pic.plover.com/math-se-blog/double-digits/wmelxc.png" alt="graph for 3749 base 10" /></p>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%3d10%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d4%24">.</p>
<h4><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d5%24"></h4>
<p>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.</p>
<h4><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d6%24"></h4>
<p>For <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d6%24"> the possible solutions are given by the governing equation <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2463a_i%20%2b%20v%20%5cequiv%200%5cpmod%7b10%7d%24"> where <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24v%24"> is a 6-bit bracelet number, one of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b0%2c1%2c3%2c5%2c7%2c9%2c11%2c13%2c15%2c21%2c23%2c27%2c31%2c63%5c%5c%7d%24">. Tabulating the possible digit sets, we get:</p>
<p>$$\begin{array}{crrl}
v & a_i & \text{digits}\\ \hline
0 & 0 & 000000\\
1 & 3 & 362486 \\
3 & 9 & 986249 \\
5 & 5 & 500012 \\
7 & 1 & 124875 \\
9 & 7 & 748748 \\
11 & 3 & 362501 \\
13 & 9 & 986374 \\
15 & 5 & 500137 \\
21 & 3 & 363636 \\
23 & 9 & 989899 \\
27 & 1 & 125125 \\
31 & 3 & 363751 \\
63 & 9 & 999999 \\
\end{array}$$</p>
<p>After ignoring <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24000000%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24999999%24"> as usual, the large equals odd rule allows us to ignore all the other sequences except <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24124875%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24363636%24">. The latter fails for the same reason that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%2436%24"> did when <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d2%24">. But <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24142857%24"> , the lone survivor, gives us a complicated derived graph containing many hamiltonian paths, every one of which is a solution to the problem:</p>
<p><img src="http://pic.plover.com/math-se-blog/double-digits/uswkfz.png" alt="graph for 124578 base 10" /></p>
<p>It is not hard to pick out from this graph the minimal solution <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24125874%24">, for which <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24125874%5ccdot%202%20%3d%20251748%24">, and also our old friend <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24142857%24"> for which <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24142857%5ccdot%202%20%3d%20285714%24">.</p>
<p>We see here the reason why all the small numbers with property <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P_%7b10%7d%24"> contain the digits <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24124578%24">. The constraints on <em>which</em> 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24"> is only required to have the digits of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> in <em>some</em> 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 <em>all</em> the small nodes or <em>all</em> 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.</p>
<h3>Onward</h3>
<p>This analysis is tedious but is simple enough to perform by hand in under an hour. As <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24">, and to emit the possible digit sets that satisfied the large equals odd criterion. I had wondered if <em>every</em> base-10 solution contained equal numbers of the digits <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%2c2%2c4%2c8%2c5%2c%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24">. This is the case for <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d7%24"> (where the only admissible digit set is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b1%2c2%2c4%2c5%2c7%2c8%5c%5c%7d%5ccup%5c%5c%7b9%5c%5c%7d%24">), for <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d8%24"> (where the only admissible sets are <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b1%2c2%2c4%2c5%2c7%2c8%5c%5c%7d%5ccup%20%5c%5c%7b3%2c6%5c%5c%7d%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b1%2c2%2c4%2c5%2c7%2c8%5c%5c%7d%5ccup%5c%5c%7b9%2c9%5c%5c%7d%24">), and for <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d9%24"> (where the only admissible sets are <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b1%2c2%2c4%2c5%2c7%2c8%5c%5c%7d%5ccup%5c%5c%7b3%2c6%2c9%5c%5c%7d%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%5c%7b1%2c2%2c4%2c5%2c7%2c8%5c%5c%7d%5ccup%5c%5c%7b9%2c9%2c9%5c%5c%7d%24">). But when we reach <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d10%24"> the increasing number of bracelets 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%244225561128%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241577438874%24">, both of which wreck any theory that the digits <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%2c2%2c4%2c5%2c8%2c7%24"> must all appear the same number of times.</p>
<h3>Acknowledgments</h3>
<p>Thanks to <a href="http://math.stackexchange.com/users/67848/karl-kronenfeld">Karl
Kronenfeld</a>
for corrections and many helpful suggestions.</p>
Proof by contradiction
http://blog.plover.com/2014/03/01#IL-contradiction
<p>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.</p>
<p>Intuitionists are perfectly happy to accept a reductio ad absurdum
proof of the following form:</p>
<p>$$(P\to \bot)\to \lnot P$$</p>
<p>Here <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbot%24"> means an absurdity or a contradiction; <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%5cto%20%5cbot%24"> means that assuming
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> leads to absurdity, and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28P%5cto%20%5cbot%29%5cto%20%20%5clnot%20P%24"> means that if assuming <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">
leads to absurdity, then you can conclude that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is false. This
is a classic proof by contradiction, and it is intuitionistically
valid. In fact, in many formulations of intuitionistic logic, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20P%24"> is
<em>defined</em> to mean <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%5cto%20%5cbot%24">.</p>
<p>What is rejected by intuitionistic logic is the similar-seeming claim
that:</p>
<p>$$(\lnot P\to \bot)\to P$$</p>
<p>This says that if assuming <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20P%24"> leads to absurdity, you can conclude
that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is true. This is <em>not</em> intuitionistically valid.</p>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20P%24"> in the first one, and you get the second.”</p>
<p>Not quite. If you replace <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20P%24"> in the first one, you do not get the second
one; you get:</p>
<p>$$(\lnot P\to \bot)\to \lnot \lnot P$$</p>
<p>People familiar with classical logic are so used to shuffling the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%24">
signs around and treating <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%5clnot%20P%24"> the same as <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> that they often don't
notice when they are doing it. But in intuitionistic logic, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%5clnot%20P%24">
are not the same. <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%5clnot%20P%24"> is weaker than <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">, in the sense that
from <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> one can always conclude <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%5clnot%20P%24">, but not always vice versa. Intuitionistic logic
is happy to agree that if <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20P%24"> leads to absurdity, then <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%5clnot%20P%24">. But it
does not agree that this is sufficient to conclude <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">.</p>
<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, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> means that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is true and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20P%24">
means that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is false. If <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is not false it is true, so <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%5clnot%20P%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">
mean the same thing. But in intuitionistic logic <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> means that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is
<em>provable</em>, and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20P%24"> means that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is not provable. <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%5clnot%20P%24"> means that it is
impossible to prove that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is not provable.</p>
<p>If <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is provable, it is certainly impossible to prove that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is not
provable. So <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> implies <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%5clnot%20P%24">. But just because it is impossible to
prove that there is no proof of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> does not mean that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> itself is
provable, so <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%20%5clnot%20P%24"> does not imply <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">.</p>
<p>Similarly,</p>
<p>$$(P\to \bot)\to \lnot P $$</p>
<p>means that if a proof of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> would lead to absurdity, then we may conclude
that there cannot be a proof of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">. This is quite valid. But</p>
<p>$$(\lnot P\to \bot)\to P$$</p>
<p>means that if assuming that a proof of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is impossible leads to
absurdity, there must be a proof of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">. But this itself isn't a
proof of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">, nor is it enough to prove <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">; it only shows that
there is no proof that proofs of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> are impossible.</p>
<p>[ Addendum 20141124: <a href="http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/">This article by Andrej Bauer</a> says much the same thing. ]</p>
Cauchy and the continuum
http://blog.plover.com/2014/01/04#Cauchy
<p>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.</p>
<p>The claim is often made that Cauchy's <em>Course d'analyse</em> of 1821
contains a "proof" of the following statement: a convergent sequence
of continuous functions has a continuous limit. For example, <a href="https://en.wikipedia.org/w/index.php?title=Uniform_convergence&oldid=583507328#History">the
Wikipedia article on "uniform
convergence"</a>
claims:</p>
<blockquote>
<p>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…</p>
</blockquote>
<p><a href="http://ncatlab.org/nlab/show/Cauchy+sum+theorem#statements">The nLab article on "Cauchy sum theorem"</a> states:</p>
<blockquote>
<p>Non-theorem (attributed to Cauchy, 1821). Let <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%3d%28f_1%2cf_2%2c%5cldots%29%24">
be an infinite sequence of continuous functions from the real line
to itself. Suppose that, for every real number <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24x%24">, the sequence
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28f_1%28x%29%2c%20f_2%28x%29%2c%20%5cldots%29%24"> converges to some (necessarily unique)
real number <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f_%5cinfty%28x%29%24">, defining a function <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f_%5cinfty%24">; in
other words, the sequence <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> converges pointwise? to
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f_%5cinfty%24">. Then <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f_%5cinfty%24"> is also continuous.</p>
</blockquote>
<p>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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%20f_n%28x%29%20%3d%20x%5en%24"> on the interval <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5b%2d1%2c1%5d%24"> is a sequence of
continuous functions that converges everywhere on <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5b0%2c1%5d%24"> to a
discontinuous limit. You would have to be a mathematical ignoramus to
miss this, and Cauchy wasn't. </p>
<p>Another simple example, one that converges everywhere in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cmathbb%0aR%24">, is any sequence of functions <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f_n%24"> that are everywhere zero,
except that each has a (continuous) bump of height 1 between
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2d%5cfrac1n%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac1n%24">. As <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%5cto%5cinfty%24">, the width of the
bump narrows to zero, and the limit function <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f_%5cinfty%24"> is
everywhere zero except that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f_%5cinfty%280%29%3d1%24">. Anyone can think of
this, and certainly Cauchy could have. A concrete example of this type
is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=f_n%28x%29%20%3d%20e%5e%7b%2dx%5e%7b2%7d%2fn%7d"> which converges to 0
everywhere except at <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%20x%3d0%20%24">, where it converges to 1.</p>
<p>Cauchy's controversial theorem is not what Wikipedia or nLab claim.
It is that that the pointwise limit of a convergent <strong>series</strong> of
continuous functions is always continuous. Cauchy is not claiming that
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=f_%5cinfty%28x%29%20%3d%20%5clim_%7bi%5cto%5cinfty%7d%20f_i%28x%29"> must be continuous if the
limit exists and the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f_i%24"> are continuous. Rather, he claims that
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=S%28x%29%20%3d%20%5csum_%7bi%3d1%7d%5e%5cinfty%20f_i%28x%29"> must be continuous if
the sum converges and the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f_i%24"> 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. </p>
<p>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.</p>
<p>(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
<em>sequence</em> 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
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24x%5en%24">.)</p>
<p>The confusion about Cauchy's controversial theorem arises from a
perennially confusing piece of mathematical terminology: a convergent
<em>sequence</em> is not at all the same as a convergent <em>series</em>. Cauchy
claimed that a convergent <em>series</em> of continuous functions has a
continuous limit. He did not ever claim that a convergent <em>sequence</em>
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.</p>
<p>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.</p>
<p>[ I had originally planned to write about this controversial theorem
in <a href="http://blog.plover.com/math/major-screwups.html">my series of articles about major screwups in
mathematics</a>, but the
longer and more closely I looked at it the less clear it was that
Cauchy had actually made a mistake. ]</p>
On the consistency of PA
http://blog.plover.com/2012/08/25#koan
A monk asked Li Fu, "Master, how do we know that the Peano axioms are
consistent?"<p>
Li Fu said, "The axioms are consistent because they have a model."<p>More about ZF's asymmetry between union and intersection
http://blog.plover.com/2012/08/24#topology-closed-2
In <a href="http://blog.plover.com/math/topology-closed.html">an article earlier this
week</a>, 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 <a
href="http://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory">ZF
set theory axioms</a>.<p>
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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> is a family
of sets, then we can form <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcup%20%7b%5cmathcal%20F%7d%24">, which is the union of all
the sets in the family.<p>
The other is actually a family of axioms, the
specification axiom schema. It says that for any one-place predicate
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cphi%28x%29%24"> and any set <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24"> we can construct the subset of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24"> for
which <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cphi%24"> holds:<p>
<p align=center><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%0a%24%24%5c%7b%20x%5cin%20X%20%5c%3b%7c%5c%3b%20%5cphi%28x%29%20%5c%7d%24%24%0a"></p><p>
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
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24x%24"> and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24y%24"> are sets, then so is the two-element set <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%7bx%2c%20y%5c%7d%24">;
with pairing and union we can construct all the finite sets. But we
won't need it in this article.)<p>
Conspicuously absent is an axiom of intersection. If you have a
family <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> of sets, and you want a set of every element that is in
<i>some</i> member of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">, that is easy; it is what the axiom of union gets
you. But if you want a set of every element that is in <i>every</i>
member of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">, you have to use specification. <p>
Let's begin by defining this compact notation:
<p align=center><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%0a%24%24%5cbigcap_%7b%28X%29%7d%20%7b%5cmathcal%20F%7d%24%24%0a"></p><p>
for this longer formula:
<p align=center><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%0a%24%24%5c%7b%20x%5cin%20X%20%5c%3b%7c%5c%3b%20%5cforall%20f%5cin%20%7b%5cmathcal%20F%7d%20%2e%20x%5cin%20f%20%5c%7d%24%24%0a"></p><p>
This is our intersection of the members of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">, taken "relative to
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24">", as we say in the biz. It gives us all the elements of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24">
that are in every member of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">. The <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24"> is mandatory in
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap_%7b%28X%29%7d%24">, because ZF makes it mandatory when you construct a
set by specification. If you leave it out, you get the Russell paradox.<p>
Most of the time, though, the <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24"> is not very important. When
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> is nonempty, we can choose some element <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%5cin%20%7b%5cmathcal%20F%7d%24">, and
consider <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap_%7b%28f%29%7d%20%7b%5cmathcal%20F%7d%24">, which is the "normal" intersection of
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">. We can easily show that
<p align=center><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%0a%24%24%5cbigcap_%7b%28X%29%7d%20%7b%5cmathcal%20F%7d%5csubseteq%20%5cbigcap_%7b%28f%29%7d%20%7b%5cmathcal%20F%7d%24%24%0a"></p>
for any <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24"> whatever, and this immediately implies that
<p align=center><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%0a%24%24%5cbigcap_%7b%28f%29%7d%20%7b%5cmathcal%20F%7d%20%3d%20%5cbigcap_%7b%28f%27%29%7d%7b%5cmathcal%20F%7d%24%24%0a"></p>
for any two elements
of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">, so when <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> contains an element <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24">, we can omit the
subscript and just write
<p align=center><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%0a%24%24%5cbigcap%20%7b%5cmathcal%20F%7d%24%24%0a"></p>
for the usual intersection of members of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">.<p>
Even the usually troublesome case of an
empty family <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> is no problem. In this case we have no <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> to
use for <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap_%7b%28f%29%7d%20%7b%5cmathcal%20F%7d%24">, but we can still take some other set
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24"> and talk about <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap_%7b%28X%29%7d%20%5cemptyset%24">, which is just
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24">.<p>
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:<p>
A <b>co-topology</b> is a family <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> of sets, called "closed"
sets, such that:<p>
<ol><li>The union of any two elements of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> is again in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">, and
<li>The intersection of any subfamily of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> is again in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">.
</ol>
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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcup%20%7b%5cmathcal%20F%7d%24">. This set <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcup%20%7b%5cmathcal%20F%7d%24"> contains
anything that is in any of the closed sets, and so <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcup%20%7b%5cmathcal%20F%7d%24">,
which I will henceforth call <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24U%24">, 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> have a maximum size, so there
is no way to make <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24U%24"> too big. <p>
It now immediately follows that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24U%24"> itself is a closed set, since it
is the intersection <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap_%7b%28U%29%7d%20%5cemptyset%24"> of
the empty subfamily of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">.<p>
If <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> itself is empty, then so is <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24U%24">, and <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap_%7b%28U%29%7d%20%7b%5cmathcal%20F%7d%0a%3d%20%5cemptyset%24">, so that is all right. From here on we will assume that
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24"> is nonempty, and therefore that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap%20%7b%5cmathcal%20F%7d%24">, with no
relativization, is well-defined.<p>
We still cannot prove that the empty set is closed; indeed, it might
not be, because even <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24M%20%3d%20%5cbigcap%20%7b%5cmathcal%20F%7d%24"> might not be empty. But as David
Turner pointed out to me in email, the elements of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24M%24"> 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24M%24">, which are topologically
featureless in the same way. Rather than considering <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">, we
should consider <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cwidehat%7b%5cmathcal%20F%7d%7d%24">, whose members are the members of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathcal%20F%7d%24">,
but with <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24M%24"> subtracted from each one:<p>
<p align=center><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%0a%24%24%7b%5cwidehat%7b%5cmathcal%20F%7d%7d%20%3d%20%5c%7b%5chat%7bf%7d%5cin%202%5eU%20%5c%3b%7c%5c%3b%20%5cexists%20f%5cin%20%7b%5cmathcal%20F%7d%20%2e%20%5chat%7bf%7d%20%3d%20f%5csetminus%20M%20%5c%7d%24%24%0a"></p><p>
So we may as well assume that this has been done behind the scenes and
so that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap%20%7b%5cmathcal%20F%7d%24"> is empty. If we have done this, then the
empty set is closed.<p>
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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24S%5eC%24"> of a set. Instead, it has only
relative complements, or differences. <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%5csetminus%20Y%24"> is defined as:
<p align=center><img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%0a%24%24X%5csetminus%20Y%20%3d%20%5c%7b%20x%5cin%20X%20%5c%3b%7c%5c%3b%20x%5cnotin%20Y%5c%7d%20%24%24%0a"></p><p>
Here we say that the complement of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Y%24"> is taken relative to <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24">.<p>
For the definition of open sets, we will say that the complement is
taken relative to the universe of discourse <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24U%24">, and a set <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24G%24"> is
open if it has the form <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24U%5csetminus%20f%24"> for some closed set <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24">.<p>
Anatoly Karp pointed out on Twitter that we know that the empty set is
open, because it is the relative complement of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24U%24">, which we already
know is closed. And if we ensure that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap%20%7b%5cmathcal%20F%7d%24"> is empty, as in
the previous paragraph, then since the empty set is closed, <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24U%24"> is
open, and we have recovered all the original properties of a
topology.<p>
<div class="bookbox"><table align=right width="14%" bgcolor="#ffffdd" border=1><tr><td align=center>
<font size="-1">Order</font><br>
<cite><font size="-1">General Topology</font></cite><br>
<A HREF="http://www.powells.com/partner/29575/biblio/0387901256"><IMG SRC="http://www.powells.com/cgi-bin/imageDB.cgi?isbn=0387901256" BORDER="0" ALIGN="center" ALT="General Topology" ></a><BR>
<A HREF="http://www.powells.com/partner/29575/biblio/0387901256"><font size="-1">with kickback</font></a><br>
<A HREF="http://www.powells.com/biblio/0387901256"><font size="-1">no kickback</font></a>
</td></tr></table></div>
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 <a
href="https://secure.wikimedia.org/wikipedia/en/wiki/John_L._Kelley">John L. Kelley</a>'s
famous book <cite>General Topology</cite>, which has been in print
since 1955. <p>
<p align=center><a href="http://pic.blog.plover.com/math/topology-closed-2/kelley.png"><img align=center
src="http://pic.blog.plover.com/math/topology-closed-2/kelley-th.png"></a><p></p>
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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24M%24">, which
was not ZF's fault.)<p>
<br clear="right"><div class="bookbox"><table align=right width="14%" bgcolor="#ffffdd" border=1><tr><td align=center>
<font size="-1">Order</font><br>
<cite><font size="-1">On Numbers and Games</font></cite><br>
<A HREF="http://www.powells.com/partner/29575/biblio/9781568811277"><IMG SRC="http://www.powells.com/cgi-bin/imageDB.cgi?isbn=9781568811277" BORDER="0" ALIGN="center" ALT="On Numbers and Games" ></a><BR>
<A HREF="http://www.powells.com/partner/29575/biblio/9781568811277"><font size="-1">with kickback</font></a><br>
<A HREF="http://www.powells.com/biblio/9781568811277"><font size="-1">no kickback</font></a>
</td></tr></table></div>
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 <cite>On Numbers and Games</cite> (Academic
Press, 1976).) None of the problems we encountered were technically
illuminating, and nothing was clarified by examining them in
detail.<p>
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.<p>
[ MathJax doesn't work in Atom or RS<!-- -->S 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 <a href="http://blog.plover.com/math/topology-closed-2.html">my
blog's home site</a>. ]<p>The non-duality of open and closed sets
http://blog.plover.com/2012/08/22#topology-closed
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.<p>
We can define a topology without reference to
the underlying space as follows: A family <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathfrak%20I%7d%24"> 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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathfrak%20I%7d%24">. From this we can
recover the omitted axiom that says that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cemptyset%24"> is open: it must
be in <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathfrak%20I%7d%24"> because it is the empty union <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcup_%7bg%5cin%5cemptyset%7d%0ag%24">. We can also recover the underlying space of the topology, or at
least <i>some</i> such space, because it is the unique maximal open set
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%3d%5cbigcup_%7bg%5cin%7b%5cmathfrak%20I%7d%7d%20g%24">. The space <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24"> 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,
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24"> is our universe of discourse. We can then say that a set <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24C%24"> is
"closed" whenever <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%5csetminus%20C%24"> is open, and prove all the usual
theorems.<p>
If we choose to work with closed sets instead, we run into problems.
We can try starting out the same way: A family <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathfrak%20I%7d%24"> 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
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%7b%5cmathfrak%20I%7d%24">. But we can no longer prove that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cemptyset%5cin%7b%5cmathfrak%20I%7d%24">. We can
still recover an underlying space <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%20%3d%20%5cbigcup_%7bc%5cin%7b%5cmathfrak%20I%7d%7d%20c%24">, but we
cannot prove that <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24X%24"> 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 <i>minimal</i> closed set
<img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigcap_%7bc%5cin%7b%5cmathfrak%20I%7d%7d%20c%24">, 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.<p>
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.<p>
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.<p>
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.<p>
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 <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cmathcal%20S%24">, 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.<p>
[ This is my first blog post using <a
href="http://www.mathjax.org">MathJax</a>, 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. ]<p>
[ 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 <a
href="http://blog.plover.com/">http://blog.plover.com/</a>
directly. ]<p>
[ Addendum 20120824: <a
href="http://blog.plover.com/math/topology-closed-2.html">There is a followup to this
article</a>. ]<p>Elaborations of Russell's paradox
http://blog.plover.com/2012/01/10#russell
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?<p>
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.<p>
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 <i>can</i> have a catalog of
all the blue catalogs; it can have a green cover. But I soon thought
of a much better extension.<p>
I gave it to Katara like this: say you have a catalog, let's call it
<i>X</i>. If <i>X</i> mentions a catalog that mentions <i>X</i>, it
has a gold stripe on the spine. Otherwise, it has a silver stripe.
Now:<p>
<ol>
<li>Could there be a red catalog with a gold stripe?
<li>Could there be a red catalog with a silver stripe?
<li>Could there be a blue catalog with a gold stripe?
<li>Could there be a blue catalog with a silver stripe?
</ol>
And more interesting:<p>
<ol start=5>
<li>Is there a catalog of all the catalogs with gold stripes?
<li>Is there a catalog of all the catalogs with silver stripes?
</ol>
I knew that early 20th century logicians, trying to repair the Russell paradox,
first tried a very small patch: since comprehension over the predicate
<i>X</i>∉<i>X</i> 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
<i>X</i> 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
<i>X</i> such that <i>X</i>∉<i>X</i>, so that's where I got the
silver stripe predicate.<p>
Translating this into barber language is left as an exercise for the
reader.<p>Concerts
http://blog.plover.com/2011/06/11#concerts
<div class="bookbox"><table align=right width="14%" bgcolor="#ffffdd" border=1><tr><td align=center>
<font size="-1">Order</font><br>
<cite><font size="-1">Unknown book with tag 'Tao problems'</font></cite><br>
<A HREF="http://www.powells.com/partner/29575/biblio/0000000000"></a><BR>
<A HREF="http://www.powells.com/partner/29575/biblio/0000000000"><font size="-1">with kickback</font></a><br>
<A HREF="http://www.powells.com/biblio/0000000000"><font size="-1">no kickback</font></a>
</td></tr></table></div>
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?<p>
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 <i>p</i> performers and 6-<i>p</i>
in the audience arranges <i>p</i>(6 - <i>p</i>) 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.)<p>
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.<p>
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.<p>
Now suppose the same two musicians, say <i>A</i> and <i>B</i>, perform
together twice. We know that some spectator must see <i>A</i> twice;
this spectator sees <i>B</i> 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.<p>
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.<p>
Or equivalently, we need to find four triangles in <i>K</i><sub>4</sub>, none of which
share an edge.<p>
The solution is not hard to find:<p>
<p align=center>
<table cellpadding=2>
<tr><td > <td bgcolor="pink" align="center">1<td bgcolor="white" align="center">2<td bgcolor="pink" align="center">3<td bgcolor="white" align="center">4
<tr><td align="center">On stage<td bgcolor="pink" align="center">A B C<td bgcolor="white" align="center">C D E<td bgcolor="pink" align="center">E F A<td bgcolor="white" align="center">B D F
<tr><td align="center">In audience<td bgcolor="pink" align="center">D E F<td bgcolor="white" align="center">A B F<td bgcolor="pink" align="center">B C D<td bgcolor="white" align="center">A C E
</table>
</p>
And in fact this solution is essentially unique.<p>
If you generalize these arguments to 2<i>m</i> musicians, you find
that there is a lower bound of <img src="https://chart.googleapis.com/chart?chf=bg,s,00000000&cht=tx&chl=%5cleft%5clceil%7b4m%5e2%20%2d%202m%20%5cover%20m%5e2%20%7d%5cright%5crceil"> 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.<p>
The geometric formulation is that you want to find four
disjoint <i>K</i><sub>4</sub>s in a <i>K</i><sub>4</sub>; 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.<p>
But the tinkering and the exhaustive search were a waste of time,
because there <i>is</i> 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!<p>
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:<p>
<p align=center>
<table cellpadding=2 >
<tr><td align="center"> <td bgcolor="pink" align="center">1<td bgcolor="white" align="center">2<td bgcolor="pink" align="center">3<td bgcolor="white" align="center">4<td bgcolor="pink" align="center">5
<tr><td " align="center">On stage <td bgcolor="pink" align="center">E F G H<td bgcolor="white" align="center">B C D H<td bgcolor="pink" align="center">A C D F G<td bgcolor="white" align="center">A B D E G<td bgcolor="pink" align="center">A B C E F H
<tr><td align="center">In audience<td bgcolor="pink" align="center">A B C D<td bgcolor="white" align="center">A E F G<td bgcolor="pink" align="center">B E H <td bgcolor="white" align="center">C F H <td bgcolor="pink" align="center">D G
</table>
</p>
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.<p>
[ Addendum: For <i>n</i> = 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. ]<p>A draft of a short introduction to topology
http://blog.plover.com/2010/11/15#topology-doc
One of my ongoing projects is to figure out how to explain topology
briefly. For example, <a href="http://blog.plover.com/math/metric.html">What is
Topology?</a>, putatively part 1 of a three-part series that I have
not yet written parts 2 or 3 of yet.<p>
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.<p>
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.<p>
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 <a
href="http://en.wikipedia.org/wiki/Compactness_theorem">compactness
principle</a>, which can only be taken as a sort of pseudomathematical
lorem ipsum. This really is a draft; it is only three-quarters
finished.<p>
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.<p>
The current draft is version <b>0.6</b> of <b>2010-11-14</b>. I do
not want old erroneous versions wandering around confusing people in
my name, so please do not distribute this draft after
<b>2010-12-15</b>. I hope to have an improved draft available here
before that.<p>
Please do send me corrections, suggestions, questions, advice,
patches, pull requests, or anything else.<p>
<ul>
<li><a href="http://pic.blog.plover.com/math/topology-doc/topology-draft-0.6.pdf">PDF</a><br>
<li><a href="http://pic.blog.plover.com/math/topology-doc/topology-draft-0.6.tex">LaTeX source</a>
<li><a href="https://github.com/mjdominus/topology-doc">Github git
repository</a> (<tt>git://github.com/mjdominus/topology-doc.git</tt>)
</ul>