The Universe of Discourse

Mon, 01 Dec 2014

People are frequently surprised that my book, Higher-Order Perl, is available as a free download from my web site. They ask if it spoiled my sales, or if it was hard to convince the publisher. No and no.

I sent the HOP proposal to five publishers, expecting that two or three would turn it down, and that I would pick from the remaining two or three, but somewhat to my dismay, all five offered to publish it, and I had to decide who.

One of the five publishers was Morgan Kaufmann. I had never heard of Morgan Kaufmann, but one day around 2002 I was reading the web site of Philip Greenspun. Greenspun was incredibly grouchy. He found fault with everything. But he had nothing but praise for Morgan Kaufmann. I thought that if Morgan Kaufmann had pleased Greenspun, who was nearly impossible to please, then they must be really good, so I sent them the proposal. (They eventually published the book, and did a superb job; I have never regretted choosing them.)

But not only Morgan Kaufmann but four other publishers had offered to publish the book. So I asked a number of people for advice. I happened to be in London one week and Greenspun was giving a talk there, which I went to see. After the talk I introduced myself and asked for his advice about picking the publisher.

Greenspun reiterated his support for Morgan Kaufmann, but added that the publisher was not important. Instead, he said, I should make sure to negotiate permission to make the book available for free on my web site. He told me that compared with the effort that you put into the book, the money you get back is insignificant. So if you write a book it should not be because you want to make a lot of money from it but because you have an idea that you want to present to the world. And as an author, you owe it to yourself to get your idea in front of as many people as possible. By putting the book in your web site, you make it available to many people who would not otherwise have access to it: poor people, high school students, people in developing countries, and so on.

I thought that Greenspun's idea made sense; I wanted my ideas about programming to get to as many people as possible. Also, demanding that I make the book available on my web site for free seemed like a good way to narrow down the five publishers to two or three.

The first part of that plan worked out well. The second part not so well: all five publishers agreed. Some agreed reluctantly and some agreed willingly, but they all agreed. Eventually I had the book published by Morgan Kaufmann, and after a delay that seemed long at the time but in retrospect seems not so long, I put the book on my web site. It has been downloaded many times. (It's hard to say how many, since browsers often download just the portion of the PDF file that they need to display.)

Would the book have made more money if it were not available as a free download? We can't know for sure, but I don't think so. The book has always sold well, and has made a significant amount of money for me and for Morgan Kaufmann. The amount I made is small compared to the amount of work I had to put in, just as Greenspun said, but it was nothing to sneeze at either. Even now, ten years later, it is still selling and I still get a royalty check every six months. For my book to have lasted ten years is extremely rare. Most computer books disappear without a trace after six months.

Part of this is that it's an unusually good book. But I think the longevity is partly because it is available as a free download. Imagine that person A asks a question on an Internet forum, and person B says that HOP has a section that could help with the question. If B wants to follow up, they now must find a copy of HOP. If the book is out of print, this can be difficult. It may not be in the library; it almost certainly isn't in the bookstore. Used copies may be available, but you have to order them and have them shipped, and if you don't like it once it arrives, you are stuck with it. The barrier is just too high to be convenient. But since HOP is available on my web site, A can include a link, or B can find it with an easy web search. The barrier is gone! And now I have another reader who might mention it to someone else, and they might even buy a copy. Instead of drifting away into obscurity, HOP is a book that people can recommend over and over.

So my conclusion is, Greenspun's advice was exactly correct. As an author, you owe it to yourself to make your book available to as many people as possible. And the publisher may agree, so be sure to ask.

[ Addendum: Some people are just getting the news, but the book was published in 2005, and has been available as a free download since 2008. ]

Sun, 30 Nov 2014

I don't have impostor syndrome about programming, advanced mathematics, or public speaking. I cheerfully stand up in rooms full of professional programmers and authoritatively tell them what I think they should do.

However, when I put up shelves in the bathroom back in May, I was a psychological mess. For every little thing that went wrong—and there were quite a lot—I got all stressed out and wondered why I dared to perform this task. The outcome was good, but I had a lot of stress getting there.

I put in one plexiglass shelf, for which I had bought heavy-duty wall anchors in case the kids leaned on it, and two metal shelves higher up, which came with their own screws and anchors.

Here's a partial list of things that worried me:

1. The two upper shelves came with a paper template that I held up to the wall to mark where the holes should be drilled. What if the two shelves were slightly different and their templates were different and I needed to use both templates on the wall instead of using the same template twice?
2. When I putting the heavy-duty wall anchors into the drywall, big divots of plaster fell out of the wall around the anchors.
3. Then I filled in the holes with filler, and got filler in the screw holes in the wall anchors, and stressed about this. What if the filler in the sockets somehow prevented the screws from going into the anchors or caused some other unforeseeable problem?
4. The filler looked sloppy and I worried that it would look absurdly ugly to everyone who came into the bathroom. (The shelf would have hidden the ugly screw job from a normal view, except that it was made of plexiglass, so the filled holes were visible through it.)
5. I didn't know how big to drill the holes for the smaller wall anchors and stressed about it, examining the wall anchor packaging for some hint. There was none.
6. I wanted to insert the wall anchors into the holes with my rubber mallet. Where the hell is it? Then I stressed about using a claw hammer instead and maybe squishing the anchors, and spent a while looking for a piece of wood or something to soften the hammer blows. Eventually I gave up looking, wondering if I was dooming the project.
7. I guessed how big to make the hole for the anchor, and was wrong; my hole was too small. I didn't realize this until I had the first anchor halfway in. Then I stressed that I might ruin it when I pulled it back out of the wall.
8. Then I stressed about the size of the holes again when I drilled larger holes. What if I make the hole too big, and then have to fill all the holes and re-measure and re-drill the whole thing?
9. The anchors didn't go into two of the holes. I needed to yank them back out, then redrill the holes, with the outer end a little messy, or the anchors wouldn't go all the way into the holes. Again I worried about spoiling the anchors.
10. When I drilled the holes, sometimes the drill suddenly went all the way into the wall and the rotating chuck left a circular scar on the paint.
11. Also, two of the holes didn't drill easily; I had to lean on the drill really hard to get it to go through. For a while I was concerned that there was some undrillable metal thing in the wall just where I wanted my hole, and I would have to fill in all the holes and remeasure and redrill the whole thing.
12. Even though I had marked the wall for the lower shelf by holding the shelf against the wall and then poking a pencil through the actual holes, when time came to put the bolts in place, I found that the two holes were slightly too far apart. Somehow this worked itself out.

On review, I see that several of these worries could have been completely avoided if I had had a supply of extra wall anchors.

Stuff that could have worried me but (rightly or wrongly) didn't:

1. I knew enough to go to the store to buy wall anchors and screws for the bottom shelf, which did not come with its own hardware. There are a lot of different kinds of anchors, and I did not worry too much that I was getting the wrong thing.

2. I was concerned (although not worried) that the screws holding the bottom shelf to the wall might stress the plastic too much and cause it to crack, either immediately or over time. Obvious solution: insert washers between the screw heads and the shelf. I went to the hardware store to get nylon washers; they didn't have any. So I got thin metal washers instead. I did not worry about this; I was sure (perhaps wrongly) that metal washers would do the job.

3. When I asked the hardware people for plastic washers, they looked totally blank. “Plastic... washers?” they asked, as if this were a heretofore unimaginable combination. I could have felt like an idiot, but instead I felt, correctly I think, that they were idiots.

4. For some reason, I was not even slightly worried about properly leveling the marks for the holes. I used a spirit level, which I consider pretty fancy.

5. I was careful not to over-tighten the screws holding the plexiglass shelf in place, so as to avoid cracking them, but I was at no time afraid that I would somehow crack them anyway.

[Added in July: I have reread this article for the first time. I can report that all the worries I had about whether the shelves would look good have come to nothing; they all look just fine and I had forgotten all the things I was afraid would look bad. But I really do need to buy a couple of boxes of plastic wall anchors so I can stop worrying about spoiling the four I have.]

[The shelves look crooked in the picture, but that is because I am holding the camera crooked; in real life they look great.]

[ A later visit to a better hardware store confirmed that plastic washers do exist, and I did not hallucinate them. The rubber mallet still has not come to light.]

Sat, 22 Nov 2014

When opportunity permits, I have been trying to teach my ten-year-old daughter rudiments of algebra and group theory. Last night I posed this problem:

Mary and Sue are sisters. Today, Mary is three times as old as Sue; in two years, she will be twice as old as Sue. How old are they now?

I have tried to teach Ms. 10 that these problems have several phases. In the first phase you translate the problem into algebra, and then in the second phase you manipulate the symbols, almost mechanically, until the answer pops out as if by magic.

There is a third phase, which is pedagogically and practically essential. This is to check that the solution is correct by translating the results back to the context of the original problem. It's surprising how often teachers neglect this step; it is as if a magician who had made a rabbit vanish from behind a screen then forgot to take away the screen to show the audience that the rabbit had vanished.

Ms. 10 set up the equations, not as I would have done, but using four unknowns, to represent the two ages today and the two ages in the future:

\begin{align} MT & = 3ST \\ MY & = 2SY \\ \end{align}

(!!MT!! here is the name of a single variable, not a product of !!M!! and !!T!!; the others should be understood similarly.)

“Good so far,” I said, “but you have four unknowns and only two equations. You need to find two more relationships between the unknowns.” She thought a bit and then wrote down the other two relations:

\begin{align} MY & = MT + 2 \\ SY & = ST + 2 \end{align}

I would have written two equations in two unknowns:

\begin{align} M_T & = 3S_T\\ M_T+2 & = 2(S_T + 2) \end{align}

but one of the best things about mathematics is that there are many ways to solve each problem, and no method is privileged above any other except perhaps for reasons of practicality. Ms. 10's translation is different from what I would have done, and it requires more work in phase 2, but it is correct, and I am not going to tell her to do it my way. The method works both ways; this is one of its best features. If the problem can be solved by thinking of it as a problem in two unknowns, then it can also be solved by thinking of it as a problem in four or in eleven unknowns. You need to find more relationships, but they must exist and they can be found.

Ms. 10 may eventually want to learn a technically easier way to do it, but to teach that right now would be what programmers call a premature optimization. If her formulation of the problem requires more symbol manipulation than what I would have done, that is all right; she needs practice manipulating the symbols anyway.

She went ahead with the manipulations, reducing the system of four equations to three, then two and then one, solving the one equation to find the value of the single remaining unknown, and then substituting that value back to find the other unknowns. One nice thing about these simple problems is that when the solution is correct you can see it at a glance: Mary is six years old and Sue is two, and in two years they will be eight and four. Ms. 10 loves picking values for the unknowns ahead of time, writing down a random set of relations among those values, and then working the method and seeing the correct answer pop out. I remember being endlessly delighted by almost the same thing when I was a little older than her. In The Dying Earth Jack Vance writes of a wizard who travels to an alternate universe to learn from the master “the secret of renewed youth, many spells of the ancients, and a strange abstract lore that Pandelume termed ‘Mathematics.’”

“I find herein a wonderful beauty,” he told Pandelume. “This is no science, this is art, where equations fall away to elements like resolving chords, and where always prevails a symmetry either explicit or multiplex, but always of a crystalline serenity.”

After Ms. 10 had solved this problem, I asked if she was game for something a little weird, and she said she was, so I asked her:

Mary and Sue are sisters. Today, Mary is three times as old as Sue; in two years, they will be the same age. How old are they now?

“WHAAAAAT?” she said. She has a good number sense, and immediately saw that this was a strange set of conditions. (If they aren't the same age now, how can they be the same age in two years?) She asked me what would happen. I said (truthfully) that I wasn't sure, and suggested she work through it to find out. So she set up the equations as before and worked out the solution, which is obvious once you see it: Both girls are zero years old today, and zero is three times as old as zero. Ms. 10 was thrilled and delighted, and shared her discovery with her mother and her aunt.

There are some powerful lessons here. One is that the method works even when the conditions seem to make no sense; often the results pop out just the same, and can sometimes make sense of problems that seem ill-posed or impossible. Once you have set up the equations, you can just push the symbols around and the answer will emerge, like a familiar building approached through a fog.

But another lesson, only hinted at so far, is that mathematics has its own way of understanding things, and this is not always the way that humans understand them. Goethe famously said that whatever you say to mathematicians, they immediately translate it into their own language and then it is something different; I think this is exactly what he meant.

In this case it is not too much of a stretch to agree that Mary is three times as old as Sue when they are both zero years old. But in the future I plan to give Ms. 10 a problem that requires Mary and Sue to have negative ages—say that Mary is twice as old as Sue today, but in three years Sue will be twice as old—to demonstrate that the answer that pops out may not be a reasonable one, or that the original translation into mathematics can lose essential features of the original problem. The solution that says that !!M_T=-2, S_T=-1 !! is mathematically irreproachable, and if the original problem had been posed as “Find two numbers such that…” it would be perfectly correct. But translated back to the original context of a problem that asks about the ages of two sisters, the solution is unacceptable. This is the point of the joke about the spherical cow.

Wed, 23 Jul 2014

[This article was published last month on the math.stackexchange blog, which seems to have died young, despite many earnest-sounding promises beforehand from people who claimed they would contribute material. I am repatriating it here.]

A recent question on math.stackexchange asks for the smallest positive integer !!A!! for which the number !!2A!! has the same decimal digits in some other order.

Math geeks may immediately realize that !!142857!! has this property, because it is the first 6 digits of the decimal expansion of !!\frac 17!!, and the cyclic behavior of the decimal expansion of !!\frac n7!! is well-known. But is this the minimal solution? It is not. Brute-force enumeration of the solutions quickly reveals that there are 12 solutions of 6 digits each, all permutations of !!142857!!, and that larger solutions, such as 1025874 and 1257489 seem to follow a similar pattern. What is happening here?

Stuck in Dallas-Fort Worth airport one weekend, I did some work on the problem, and although I wasn't able to solve it completely, I made significant progress. I found a method that allows one to hand-calculate that there is no solution with fewer than six digits, and to enumerate all the solutions with 6 digits, including the minimal one. I found an explanation for the surprising behavior that solutions tend to be permutations of one another. The short form of the explanation is that there are fairly strict conditions on which sets of digits can appear in a solution of the problem. But once the set of digits is chosen, the conditions on that order of the digits in the solution are fairly lax.

So one typically sees, not only in base 10 but in other bases, that the solutions to this problem fall into a few classes that are all permutations of one another; this is exactly what happens in base 10 where all the 6-digit solutions are permutations of !!124578!!. As the number of digits is allowed to increase, the strict first set of conditions relaxes a little, and other digit groups appear as solutions.

### Notation

The property of interest, !!P_R(A)!!, is that the numbers !!A!! and !!B=2A!! have exactly the same base-!!R!! digits. We would like to find numbers !!A!! having property !!P_R!! for various !!R!!, and we are most interested in !!R=10!!. Suppose !!A!! is an !!n!!-digit numeral having property !!P_R!!; let the (base-!!R!!) digits of !!A!! be !!a_{n-1}\ldots a_1a_0!! and similarly the digits of !!B = 2A!! are !!b_{n-1}\ldots b_1b_0!!. The reader is encouraged to keep in mind the simple example of !!R=8, n=4, A=\mathtt{1042}, B=\mathtt{2104}!! which we will bring up from time to time.

Since the digits of !!B!! and !!A!! are the same, in a different order, we may say that !!b_i = a_{P(i)}!! for some permutation !!P!!. In general !!P!! might have more than one cycle, but we will suppose that !!P!! is a single cycle. All the following discussion of !!P!! will apply to the individual cycles of !!P!! in the case that !!P!! is a product of two or more cycles. For our example of !!a=\mathtt{1042}, b=\mathtt{2104}!!, we have !!P = (0\,1\,2\,3)!! in cycle notation. We won't need to worry about the details of !!P!!, except to note that !!i, P(i), P(P(i)), \ldots, P^{n-1}(i)!! completely exhaust the indices !!0. \ldots n-1!!, and that !!P^n(i) = i!! because !!P!! is an !!n!!-cycle.

### Conditions on the set of digits in a solution

For each !!i!! we have $$a_{P(i)} = b_{i} \equiv 2a_{i} + c_i\pmod R$$ where the ‘carry bit’ !!c_i!! is either 0 or 1 and depends on whether there was a carry when doubling !!a_{i-1}!!. (When !!i=0!! we are in the rightmost position and there is never a carry, so !!c_0= 0!!.) We can then write:

\begin{align} a_{P(P(i))} &= 2a_{P(i)} + c_{P(i)} \\ &= 2(2a_{i} + c_i) + c_{P(i)} &&= 4a_i + 2c_i + c_{P(i)}\\ a_{P(P(P(i)))} &= 2(4a_i + 2c_i + c_{P(P(i)})) + c_{P(i)} &&= 8a_i + 4c_i + 2c_{P(i)} + c_{P(P(i))}\\ &&&\vdots\\ a_{P^n(i)} &&&= 2^na_i + v \end{align}

all equations taken !!\bmod R!!. But since !!P!! is an !!n!!-cycle, !!P^n(i) = i!!, so we have $$a_i \equiv 2^na_i + v\pmod R$$ or equivalently $$\big(2^n-1\big)a_i + v \equiv 0\pmod R\tag{\star}$$ where !!v\in{0,\ldots 2^n-1}!! depends only on the values of the carry bits !!c_i!!—the !!c_i!! are precisely the binary digits of !!v!!.

Specifying a particular value of !!a_0!! and !!v!! that satisfy this equation completely determines all the !!a_i!!. For example, !!a_0 = 2, v = \color{darkblue}{0010}_2 = 2!! is a solution when !!R=8, n=4!! because !!\bigl(2^4-1\bigr)\cdot2 + 2\equiv 0\pmod 8!!, and this solution allows us to compute

\def\db#1{\color{darkblue}{#1}}\begin{align} a_0&&&=2\\ a_{P(0)} &= 2a_0 &+ \db0 &= 4\\ a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 0 \\ a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 1\\ \hline a_{P^4(0)} &= 2a_{P^3(0)} &+ \db0 &= 2\\ \end{align}

where the carry bits !!c_i = \langle 0,0,1,0\rangle!! are visible in the third column, and all the sums are taken !!\pmod 8!!. Note that !!a_{P^n(0)} = a_0!! as promised. This derivation of the entire set of !!a_i!! from a single one plus a choice of !!v!! is crucial, so let's see one more example. Let's consider !!R=10, n=3!!. Then we want to choose !!a_0!! and !!v!! so that !!\left(2^3-1\right)a_0 + v \equiv 0\pmod{10}!! where !!v\in{0\ldots 7}!!. One possible solution is !!a_0 = 5, v=\color{darkblue}{101}_2 = 5!!. Then we can derive the other !!a_i!! as follows:

\begin{align} a_0&&&=5\\ a_{P(0)} &= 2a_0 &+ \db1 &= 1\\ a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 2 \\\hline a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 5\\ \end{align}

And again we have !!a_{P^n(0)}= a_0!! as required.

Since the bits of !!v!! are used cyclically, not every pair of !!\langle a_0, v\rangle!! will yield a different solution. Rotating the bits of !!v!! and pairing them with different choices of !!a_0!! will yield the same cycle of digits starting from a different place. In the first example above, we had !!a_0 = 2, v = 0010_2 = 2!!. If we were to take !!a_0 = 4, v = 0100_2 = 4!! (which also solves !!(\star)!!) we would get the same cycle of values of the !!a_i!! but starting from !!4!! instead of from !!2!!, and similarly if we take !!a_0=0, v = 1000_2 = 8!! or !!a_0 = 1, v = 0001_2!!. So we can narrow down the solution set of !!(\star)!! by considering only the so-called bracelets of !!v!! rather than all !!2^n!! possible values. Two values of !!v!! are considered equivalent as bracelets if one is a rotation of the other. When a set of !!v!!-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 !!n=4!!, for example, the bracelets are !!0000, 0001, 0011, 0101, 0111, !! and !!1111!!; the sequences !!0110, 1100,!! and !!1001!! being equivalent to !!0011!!, and so on.

#### Example

Let us take !!R=9, n=3!!, so we want to find 3-digit numerals with property !!P_9!!. According to !!(\star)!! we need !!7a_i + v \equiv 0\pmod{9}!! where !!v\in{0\ldots 7}!!. There are 9 possible values for !!a_i!!; for each one there is at most one possible value of !!v!! that makes the sum zero:

$$\pi \approx 3$$

$$\begin{array}{rrr} a_i & 7a_i & v \\ \hline 0 & 0 & 0 \\ 1 & 7 & 2 \\ 2 & 14 & 4 \\ 3 & 21 & 6 \\ 4 & 28 & \\ 5 & 35 & 1 \\ 6 & 42 & 3 \\ 7 & 49 & 5 \\ 8 & 56 & 7 \\ \end{array}$$

(For !!a_i=4!! there is no solution.) We may disregard the non-bracelet values of !!v!!, as these will give us solutions that are the same as those given by bracelet values of !!v!!. The bracelets are:

$$\begin{array}{rl} 000 & 0 \\ 001 & 1 \\ 011 & 3 \\ 111 & 7 \end{array}$$

so we may disregard the solutions exacpt when !!v=0,1,3,7!!. Calculating the digit sequences from these four values of !!v!! and the corresponding !!a_i!! we find:

$$\begin{array}{ccl} a_0 & v & \text{digits} \\ \hline 0 & 0 & 000 \\ 5 & 1 & 512 \\ 6 & 3 & 637 \\ 8 & 7 & 888 \ \end{array}$$

(In the second line, for example, we have !!v=1 = 001_2!!, so !!1 = 2\cdot 5 + 0; 2 = 1\cdot 2 + 0;!! and !!5 = 2\cdot 2 + 1!!.)

Any number !!A!! of three digits, for which !!2A!! contains exactly the same three digits, in base 9, must therefore consist of exactly the digits !!000, 125, 367,!! or !!888!!.

#### A warning

All the foregoing assumes that the permutation !!P!! is a single cycle. In general, it may not be. Suppose we did an analysis like that above for !!R=10, n=5!! and found that there was no possible digit set, other than the trivial set 00000, that satisfied the governing equation !!(2^5-1)a_0 + v\equiv 0\pmod{10}!!. This would not completely rule out a base-10 solution with 5 digits, because the analysis only rules out a cyclic set of digits. There could still be a solution where !!P!! was a product of a !!2!! and a !!3!!-cycle, or a product of still smaller cycles.

Something like this occurs, for example, in the !!n=4, R=8!! case. Solving the governing equation !!(2^5-1)a_0 + v \equiv 0\pmod 8!! yields only four possible digit cycles, namely !!{0,1,2,4}, {1,3,6,4}, {2,5,2,5}!!, and !!{3,7,6,5}!!. But there are several additional solutions: !!2500_8\cdot 2 = 5200_8, 2750_8\cdot 2 = 5720_8, !! and !!2775_8\cdot 2 = 5772_8!!. These correspond to permutations !!P!! with more than one cycle. In the case of !!5720_8!!, for example, !!P!! exchanges the !!5!! and the !!2!!, and leaves the !!0!! and the !!7!! fixed.

For this reason we cannot rule out the possibility of an !!n!!-digit solution without first considering all smaller !!n!!.

#### The Large Equals Odd rule

When !!R!! is even there is a simple condition we can use to rule out certain sets of digits from being single-cycle solutions. Recall that !!A=a_{n-1}\ldots a_0!! and !!B=b_{n-1}\ldots b_0!!. Let us agree that a digit !!d!! is large if !!d\ge \frac R2!! and small otherwise. That is, !!d!! is large if, upon doubling, it causes a carry into the next column to the left.

Since !!b_i =(2a_i + c_i)\bmod R!!, where the !!c_i!! are carry bits, we see that, except for !!b_0!!, the digit !!b_i!! is odd precisely when there is a carry from the next column to the right, which occurs precisely when !!a_{i-1}!! is large. Thus the number of odd digits among !!b_1,\ldots b_{n-1}!! is equal to the number of large digits among !!a_1,\ldots a_{n-2}!!. This leaves the digits !!b_0!! and !!a_{n-1}!! uncounted. But !!b_0!! is never odd, since there is never a carry in the rightmost position, and !!a_{n-1}!! is always small (since otherwise !!B!! would have !!n+1!! digits, which is not allowed). So the number of large digits in !!A!! is exactly equal to the number of odd digits in !!B!!. And since !!A!! and !!B!! have exactly the same digits, the number of large digits in !!A!! is equal to the number of odd digits in !!A!!. Observe that this is the case for our running example !!1042_8!!: there is one odd digit and one large digit (the 4).

When !!R!! is odd the analogous condition is somewhat more complicated, but since the main case of interest is !!R=10!!, we have the useful rule that:

For !!R!! even, the number of odd digits in any solution !!A!! is equal to the number of large digits in !!A!!.

# Conditions on the order of digits in a solution

We have determined, using the above method, that the digits !!{5,1,2}!! might form a base-9 numeral with property !!P_9!!. Now we would like to arrange them into a base-9 numeral that actually does have that property. Again let us write !!A = a_2a_1a_0!! and !!B=b_2b_1b_0!!, with !!B=2A!!. Note that if !!a_i = 1!!, then !!b_i = 3!! (if there was a carry from the next column to the right) or !!2!! (if there was no carry), but since !!b_i=3!! is impossible, we must have !!a_i = 2!! and therefore !!a_{i-1}!! must be small, since there is no carry into position !!i!!. But since !!a_{i-1}!! is also one of !!{5,1,2}!!, and it cannot also be !!1!!, it must be !!2!!. This shows that the 1, unless it appears in the rightmost position, must be to the left of the !!2!!; it cannot be to the left of the !!5!!. Similarly, if !!a_i = 2!! then !!b_i = 5!!, because !!4!! is impossible, so the !!2!! must be to the left of a large digit, which must be the !!5!!. Similar reasoning produces no constraint on the position of the !!5!!; it could be to the left of a small digit (in which case it doubles to !!1!!) or a large digit (in which case it doubles to !!2!!). We can summarize these findings as follows:

$$\begin{array}{cl} \text{digit} & \text{to the left of} \\ \hline 1 & 1, 2, \text{end} \\ 2 & 5 \\ 5 & 1,2,5,\text{end} \end{array}$$

Here “end” means that the indicated digit could be the rightmost.

Furthermore, the left digit of !!A!! must be small (or else there would be a carry in the leftmost place and !!2A!! would have 4 digits instead of 3) so it must be either 1 or 2. It is not hard to see from this table that the digits must be in the order !!125!! or !!251!!, and indeed, both of those numbers have the required property: !!125_9\cdot 2 = 251_9!!, and !!251_9\cdot 2 = 512_9!!.

This was a simple example, but in more complicated cases it is helpful to draw the order constraints as a graph. Suppose we draw a graph with one vertex for each digit, and one additional vertex to represent the end of the numeral. The graph has an edge from vertex !!v!! to !!v'!! whenever !!v!! can appear to the left of !!v'!!. Then the graph drawn for the table above looks like this:

A 3-digit numeral with property !!P_9!! corresponds to a path in this graph that starts at one of the nonzero small digits (marked in blue), ends at the red node marked ‘end’, and visits each node exactly once. Such a path is called hamiltonian. Obviously, self-loops never occur in a hamiltonian path, so we will omit them from future diagrams.

Now we will consider the digit set !!637!!, again base 9. An analysis similar to the foregoing allows us to construct the following graph:

Here it is immediately clear that the only hamiltonian path is !!3-7-6-\text{end}!!, and indeed, !!376_9\cdot 2 = 763_9!!.

In general there might be multiple instances of a digit, and so multiple nodes labeled with that digit. Analysis of the !!0,0,0!! case produces a graph with no legal start nodes and so no solutions, unless leading zeroes are allowed, in which case !!000!! is a perfectly valid solution. Analysis of the !!8,8,8!! case produces a graph with no path to the end node and so no solutions. These two trivial patterns appear for all !!R!! and all !!n!!, and we will ignore them from now on.

Returning to our ongoing example, !!1042!! in base 8, we see that !!1!! and !!2!! must double to !!2!! and !!4!!, so must be to the left of small digits, but !!4!! and !!0!! can double to either !!0!! or !!1!! and so could be to the left of anything. Here the constraints are so lax that the graph doesn't help us narrow them down much:

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:

1042
1204
2041
2104

If leading zeroes are allowed we have also:

0412
0421

All of these are solutions in base 8.

### The case of !!R=10!!

Now we turn to our main problem, solutions in base 10.

To find all the solutions of length 6 requires an enumeration of smaller solutions, which, if they existed, might be concatenated into a solution of length 6. This is because our analysis of the digit sets that can appear in a solution assumes that the digits are permuted cyclically; that is, the permutations !!P!! that we considered had only one cycle each.

There are no smaller solutions, but to prove that the length 6 solutions are minimal, we must analyze the cases for smaller !!n!! and rule them out. We now produce a complete analysis of the base 10 case with !!R=10!! and !!n\le 6!!. For !!n=1!! there is only the trivial solution of !!0!!, which we disregard. (The question asked for a positive number anyway.)

#### !!n=2!!

For !!n=2!!, we want to find solutions of !!3a_i + v \equiv 0\pmod{10}!! where !!v!! is a two-bit bracelet number, one of !!00_2, 01_2, !! or !!11_2!!. Tabulating the values of !!a_i!! and !!v\in{0,1,3}!! that solve this equation we get:

$$\begin{array}{ccc} v& a_i \\ \hline 0 & 0 \\ 1& 3 \\ 3& 9 \\ \end{array}$$

We can disregard the !!v=0!! and !!v=3!! solutions because the former yields the trivial solution !!00!! and the latter yields the nonsolution !!99!!. So the only possibility we need to investigate further is !!a_i = 3, v = 1!!, which corresponds to the digit sequence !!36!!: Doubling !!3!! gives us !!6!! and doubling !!6!!, plus a carry, gives us !!3!! again.

But when we tabulate of which digits must be left of which informs us that there is no solution with just !!3!! and !!6!!, because the graph we get, once self-loops are eliminated, looks like this:

which obviously has no hamiltonian path. Thus there is no solution for !!R=10, n=2!!.

#### !!n=3!!

For !!n=3!! we need to solve the equation !!7a_i + v \equiv 0\pmod{10}!! where !!v!! is a bracelet number in !!{0,\ldots 7}!!, specifically one of !!0,1,3,!! or !!7!!. Since !!7!! and !!10!! are relatively prime, for each !!v!! there is a single !!a_i!! that solves the equation. Tabulating the possible values of !!a_i!! as before, and this time omitting rows with no solution, we have:

$$\begin{array}{rrl} v & a_i & \text{digits}\\ \hline 0& 0 & 000\\ 1& 7 & 748 \\ 3& 1 & 125\\ 7&9 & 999\\ \end{array}$$

The digit sequences !!0,0,0!! and !!9,9,9!! yield trivial solutions or nonsolutions as usual, and we will omit them in the future. The other two lines suggest the digit sets !!1,2,5!! and !!4,7,8!!, both of which fails the “odd equals large” rule.

This analysis rules out the possibility of a digit set with !!a_0 \to a_1 \to a_2 \to a_1!!, but it does not completely rule out a 3-digit solution, since one could be obtained by concatenating a one-digit and a two-digit solution, or three one-digit solutions. However, we know by now that no one- or two-digit solutions exist. Therefore there are no 3-digit solutions in base 10.

#### !!n=4!!

For !!n=4!! the governing equation is !!15a_i + v \equiv 0\pmod{10}!! where !!v!! is a 4-bit bracelet number, one of !!{0,1,3,5,7,15}!!. This is a little more complicated because !!\gcd(15,10)\ne 1!!. Tabulating the possible digit sets, we get:

$$\begin{array}{crrl} a_i & 15a_i& v & \text{digits}\\ \hline 0 & 0 & 0 & 0000\\ 1 & 5 & 5 & 1250\\ 1 & 5 & 15 & 1375\\ 2 & 0 & 0 & 2486\\ 3 & 5 & 5 & 3749\\ 3 & 5 & 15 & 3751\\ 4 & 0 & 0 & 4862\\ 5 & 5 & 5 & 5012\\ 5 & 5 & 5 & 5137\\ 6 & 0 & 0 & 6248\\ 7 & 5 & 5 & 7493\\ 7 & 5 & 5 & 7513\\ 8 & 0 & 0 & 8624 \\ 9 & 5 & 5 & 9874\\ 9 & 5 & 15 & 9999 \\ \end{array}$$

where the second column has been reduced mod !!10!!. Note that even restricting !!v!! to bracelet numbers the table still contains duplicate digit sequences; the 15 entries on the right contain only the six basic sequences !!0000, 0125, 1375, 2486, 3749, 4987!!, and !!9999!!. Of these, only !!0000, 9999,!! and !!3749!! obey the odd equals large criterion, and we will disregard !!0000!! and !!9999!! as usual, leaving only !!3749!!. We construct the corresponding graph for this digit set as follows: !!3!! must double to !!7!!, not !!6!!, so must be left of a large number !!7!! or !!9!!. Similarly !!4!! must be left of !!7!! or !!9!!. !!9!! must also double to !!9!!, so must be left of !!7!!. Finally, !!7!! must double to !!4!!, so must be left of !!3,4!! or the end of the numeral. The corresponding graph is:

which evidently has no hamiltonian path: whichever of 3 or 4 we start at, we cannot visit the other without passing through 7, and then we cannot reach the end node without passing through 7 a second time. So there is no solution with !!R=10!! and !!n=4!!.

#### !!n=5!!

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.

#### !!n=6!!

For !!n=6!! the possible solutions are given by the governing equation !!63a_i + v \equiv 0\pmod{10}!! where !!v!! is a 6-bit bracelet number, one of !!{0,1,3,5,7,9,11,13,15,21,23,27,31,63}!!. Tabulating the possible digit sets, we get:

$$\begin{array}{crrl} v & a_i & \text{digits}\\ \hline 0 & 0 & 000000\\ 1 & 3 & 362486 \\ 3 & 9 & 986249 \\ 5 & 5 & 500012 \\ 7 & 1 & 124875 \\ 9 & 7 & 748748 \\ 11 & 3 & 362501 \\ 13 & 9 & 986374 \\ 15 & 5 & 500137 \\ 21 & 3 & 363636 \\ 23 & 9 & 989899 \\ 27 & 1 & 125125 \\ 31 & 3 & 363751 \\ 63 & 9 & 999999 \\ \end{array}$$

After ignoring !!000000!! and !!999999!! as usual, the large equals odd rule allows us to ignore all the other sequences except !!124875!! and !!363636!!. The latter fails for the same reason that !!36!! did when !!n=2!!. But !!142857!! , the lone survivor, gives us a complicated derived graph containing many hamiltonian paths, every one of which is a solution to the problem:

It is not hard to pick out from this graph the minimal solution !!125874!!, for which !!125874\cdot 2 = 251748!!, and also our old friend !!142857!! for which !!142857\cdot 2 = 285714!!.

We see here the reason why all the small numbers with property !!P_{10}!! contain the digits !!124578!!. The constraints on which digits can appear in a solution are quite strict, and rule out all other sequences of six digits and all shorter sequences. But once a set of digits passes these stringent conditions, the constraints on it are much looser, because !!B!! is only required to have the digits of !!A!! in some order, and there are many possible orders, many of which will satisfy the rather loose conditions involving the distribution of the carry bits. This graph is typical: it has a set of small nodes and a set of large nodes, and each node is connected to either all the small nodes or all the large nodes, so that the graph has many edges, and, as in this case, a largish clique of small nodes and a largish clique of large nodes, and as a result many hamiltonian paths.

### Onward

This analysis is tedious but is simple enough to perform by hand in under an hour. As !!n!! increases further, enumerating the solutions of the governing equation becomes very time-consuming. I wrote a simple computer program to perform the analysis for given !!R!! and !!n!!, and to emit the possible digit sets that satisfied the large equals odd criterion. I had wondered if every base-10 solution contained equal numbers of the digits !!1,2,4,8,5,!! and !!7!!. This is the case for !!n=7!! (where the only admissible digit set is !!\{1,2,4,5,7,8\}\cup\{9\}!!), for !!n=8!! (where the only admissible sets are !!\{1,2,4,5,7,8\}\cup \{3,6\}!! and !!\{1,2,4,5,7,8\}\cup\{9,9\}!!), and for !!n=9!! (where the only admissible sets are !!\{1,2,4,5,7,8\}\cup\{3,6,9\}!! and !!\{1,2,4,5,7,8\}\cup\{9,9,9\}!!). But when we reach !!n=10!! the increasing number of 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 !!4225561128!! and !!1577438874!!, both of which wreck any theory that the digits !!1,2,4,5,8,7!! must all appear the same number of times.

### Acknowledgments

Thanks to Karl Kronenfeld for corrections and many helpful suggestions.

Sun, 02 Mar 2014

This week there has been an article floating around about “What happens when placeholder text doesn't get replaced. This reminds me of the time I made this mistake myself.

In 1996 I was programming a web site for a large company which sold cosmetics and skin care products in hundreds of department stores and malls around the country. The technology to actually buy the stuff online wasn't really mature yet, and the web was new enough that the company was worried that selling online would anger the retail channels. They wanted a web page where you would put in your location and it would tell you where the nearby stores were.

The application was simple; it accepted a city and state, looked them up in an on-disk hash table, and then returned a status code to the page generator. The status code was for internal use only. For example, if you didn't fill in the form completely, the program would return the status code MISSING, which would trigger the templating engine to build a page with a suitable complaint message.

If the form was filled out correctly, but there was no match in the database, the program would return a status code that the front end translated to a suitably apologetic message. The status code I selected for this was BACKWATER.

Which was all very jolly, until one day there was a program bug and some user in Podunk, Iowa submitted the form and got back a page with BACKWATER in giant letters.

Anyone could have seen that coming; I have no excuse.

Sat, 01 Mar 2014

Intuitionistic logic is deeply misunderstood by people who have not studied it closely; such people often seem to think that the intuitionists were just a bunch of lunatics who rejected the law of the excluded middle for no reason. One often hears that intuitionistic logic rejects proof by contradiction. This is only half true. It arises from a typically classical misunderstanding of intuitionistic logic.

Intuitionists are perfectly happy to accept a reductio ad absurdum proof of the following form:

$$(P\to \bot)\to \lnot P$$

Here !!\bot!! means an absurdity or a contradiction; !!P\to \bot!! means that assuming !!P!! leads to absurdity, and !!(P\to \bot)\to \lnot P!! means that if assuming !!P!! leads to absurdity, then you can conclude that !!P!! is false. This is a classic proof by contradiction, and it is intuitionistically valid. In fact, in many formulations of intuitionistic logic, !!\lnot P!! is defined to mean !!P\to \bot!!.

What is rejected by intuitionistic logic is the similar-seeming claim that:

$$(\lnot P\to \bot)\to P$$

This says that if assuming !!\lnot P!! leads to absurdity, you can conclude that !!P!! is true. This is not intuitionistically valid.

This is where people become puzzled if they only know classical logic. “But those are the same thing!” they cry. “You just have to replace !!P!! with !!\lnot P!! in the first one, and you get the second.”

Not quite. If you replace !!P!! with !!\lnot P!! in the first one, you do not get the second one; you get:

$$(\lnot P\to \bot)\to \lnot \lnot P$$

People familiar with classical logic are so used to shuffling the !!\lnot !! signs around and treating !!\lnot \lnot P!! the same as !!P!! that they often don't notice when they are doing it. But in intuitionistic logic, !!P!! and !!\lnot \lnot P!! are not the same. !!\lnot \lnot P!! is weaker than !!P!!, in the sense that from !!P!! one can always conclude !!\lnot \lnot P!!, but not always vice versa. Intuitionistic logic is happy to agree that if !!\lnot P!! leads to absurdity, then !!\lnot \lnot P!!. But it does not agree that this is sufficient to conclude !!P!!.

As is often the case, it may be helpful to try to understand intuitionistic logic as talking about provability instead of truth. In classical logic, !!P!! means that !!P!! is true and !!\lnot P!! means that !!P!! is false. If !!P!! is not false it is true, so !!\lnot \lnot P!! and !!P!! mean the same thing. But in intuitionistic logic !!P!! means that !!P!! is provable, and !!\lnot P!! means that !!P!! is not provable. !!\lnot \lnot P!! means that it is impossible to prove that !!P!! is not provable.

If !!P!! is provable, it is certainly impossible to prove that !!P!! is not provable. So !!P!! implies !!\lnot \lnot P!!. But just because it is impossible to prove that there is no proof of !!P!! does not mean that !!P!! itself is provable, so !!\lnot \lnot P!! does not imply !!P!!.

Similarly,

$$(P\to \bot)\to \lnot P$$

means that if a proof of !!P!! would lead to absurdity, then we may conclude that there cannot be a proof of !!P!!. This is quite valid. But

$$(\lnot P\to \bot)\to P$$

means that if assuming that a proof of !!P!! is impossible leads to absurdity, there must be a proof of !!P!!. But this itself isn't a proof of !!P!!, nor is it enough to prove !!P!!; it only shows that there is no proof that proofs of !!P!! are impossible.

Thu, 27 Feb 2014

I was able to release a pretty nice piece of software today, courtesy of my employer, ZipRecruiter. If you have a family of web pages, and whenever you are looking at one you want to know when someone else is looking at the same page, you can use my package. The package is called 2banner, because it pops up a banner on a page whenever two people are looking at it. With permission from ZipRecruiter, I have put it on github, and you can download and use it for free.

You can similarly trick out the menu page itself, to hide the menu items that someone is already looking out.

I wanted to use someone else's package for this, but I was not able to find one, so I wrote one myself. It was not as hard as I had expected. The system comprises three components:

• The back-end database for recording who started looking at which pages and when. I assumed a SQL database and wrote a component that uses Perl's DBIx::Class module to access it, but it would be pretty easy throw this away and use something else instead.

• An API server that can propagate notifications like “user X is now looking at page Y” and “user X is no longer looking at page Y” into the database, and which can answer the question “who else is looking at page Y right now?”. I used Perl's Catalyst framework for this, because our web app already runs under it. It would not be too hard to throw this away and use something else instead. You could even write a standalone server using HTTP::Server, and borrow most of the existing code, and probably have it working in under an hour.

• A JavaScript thingy that lives in the web page, sends the appropriate notifications to the API server, and pops up the banner when necessary. I used jQuery for this. Probably there is something else you could use instead, but I have no idea what it is, because I know next to nothing about front-end web programming. I was happy to have the chance to learn a little about jQuery for this project.

Often a project seems easy but the more I think about it the harder it seems. This project was the opposite. I thought it sounded hard, and I didn't want to do it. It had been an outstanding request of the CS people for some time, but I guess everyone else thought it sounded hard too, because nobody did anything about it. But the longer I let it percolate around in my head, the simpler it seemed. As I considered it, one difficulty after another evaporated.

Other than the jQuery stuff, which I had never touched before, the hardest part was deciding what to do about the API server. I could easily have written a standalone, special-purpose API server, but I felt like it was the wrong thing, and anyway, I didn't want to administer one. But eventually I remembered that our main web site is driven by Catalyst, which is itself a system for replying to HTTP requests, which already has access to our database, and which already knows which user is making each request.

So it was natural to say that the API was to send HTTP requests to certain URLs on our web site, and easy-peasy to add a couple of handlers to the existing Catalyst application to handle the API requests, query the database, and send the right replies.

I don't know why it took me so long to think of doing the API server with Catalyst. If it had been someone else's suggestion I would probably feel dumb for not having thought of it myself, because in hindsight it is so obvious. Happily, I did think of it, because it is clearly the right solution for us.

It was not too hard to debug. The three components are largely independent of one another. The draft version of the API server responded to GET requests, which are easier to generate from the browser than the POST requests that it should be getting. Since the responses are in JSON, it was easy to see if the API server was replying correctly.

I had to invent techniques for debugging the jQuery stuff. I didn't know the right way to get diagnostic messages out of jQuery, so I put a big text area on the web page, and had the jQuery routines write diagnostic messages into it. I don't know if that's what other people do, but I thought it worked pretty well. JavaScript is not my ideal language, but I program in Perl all day, so I am not about to complain. Programming the front end in JavaScript and watching stuff happen on the page is fun! I like writing programs that make things happen.

The package is in ZipRecruiter's GitHub repository, and is available under a very lenient free license. Check it out.

(By the way, I really like working for ZipRecruiter, and we are hiring Perl and Python developers. Please send me email if you want to ask what it is like to work for them.)

Sun, 09 Feb 2014

You sometimes hear people claim that there is no perfectly efficient machine, that every machine wastes some of its input energy in noise or friction.

However, there is a counterexample. An electric space heater is perfectly efficient. Its purpose is to heat the space around it, and 100% of the input energy is applied to this purpose. Even the electrical energy lost to resistance in the cord you use to plug it into the wall is converted to heat.

Wait, you say, the space heater does waste some of its energy. The coils heat up, and they emit not only heat, but also light, which is useless, being a dull orange color. Ah! But what happens when that light hits the wall? Most of it is absorbed, and heats up the wall. Some is reflected, and heats up a different wall instead.

Similarly, a small fraction of the energy is wasted in making a quiet humming noise—until the sound waves are absorbed by the objects in the room, heating them slightly.

Now it's true that some heat is lost when it's radiated from the outside of the walls and ceiling. But some is also lost whenever you open a window or a door, and you can't blame the space heater for your lousy insulation. It heated the room as much as possible under the circumstances.

So remember this when you hear someone complain that incandescent light bulbs are wasteful of energy. They're only wasteful in warm weather. In cold weather, they're free.

Sat, 08 Feb 2014

I wrote some time ago about Moonpig's use of GUIDs: every significant object was given a unique ID. I said that this was a useful strategy I had only learned from Rik, and I was surprised to see how many previously tricky programming problems became simpler once the GUIDs were available. Some of these tricky problems are artifacts of Perl's somewhat limited implementation of hashes; hash keys must be strings, and the GUID gives you an instantaneous answer to any question about what the keys should be.

But it reminds me of a similar maxim which I was thinking about just yesterday: Every table in a relational database should have a record ID field. It often happens that I am designing some table and there is no obvious need for such a field. I now always put one in anyway, having long ago learned that I will inevitably want it for something.

Most recently I was building a table to record which web pages were being currently visited by which users. A record in the table is naturally identified by the pair of user ID and page URL; it is not clear that it needs any further keys.

But I put in a record ID anyway, because my practice is to always put in a record ID, and sure enough, within a few hours I was glad it was there. The program I was writing has not yet needed to use the record IDs. But to test the program I needed to insert and manipulate some test records, and it was much easier to write this:

update table set ... where record_id = 113;

than this:

update table set ... where user_id = 97531 and url = 'http://hostname:port/long/path/that/is/hard/to/type';

If you ever end up with two objects in the program that represesent record sets and you need to merge or intersect them synthetically, having the record ID numbers automatically attached to the records makes this quite trivial, whereas if you don't have them it is a pain in the butt. You should never be in such a situation, perhaps, but stranger things have happened. Just yesterday I found myself writing

function relativize (pathPat) {
var dummyA = document.createElement('a');
dummyA.href = document.URL;
return "http://" + dummyA.host + pathPat;
}

which nobody should have to do either, and yet there I was. Sometimes programming can be a dirty business.

During the bootstrapping of the user-url table project some records with bad URLs were inserted by buggy code, and I needed to remove them. The URLs all ended in % signs, and there's probably some easy way to delete all the records where the URL ends in a % sign. But I couldn't remember the syntax offhand, and looking up the escape sequence for LIKE clauses would have taken a lot longer than what I did do, which was:

delete from table where record_id in (43, 47, 49)

So the rule is: giving things ID numbers should be the default, because they are generally useful, like handles you can use to pick things up with. You need a good reason to omit them.

Sat, 01 Feb 2014

My current employer uses an online quiz to pre-screen applicants for open positions. The first question on the quiz is a triviality, just to let the candidate get familiar with the submission and testing system. The question is to write a program that copies standard input to standard output. Candidates are allowed to answer the questions using whatever language they prefer.

Sometimes we get candidates who get a zero score on the test. When I see the report that they failed to answer even the trivial question, my first thought is that this should not reflect badly on the candidate. Clearly, the testing system itself is so hard to use that the candidate was unable to submit even a trivial program, and this is a failure of the testing system and not the candidate.

But it has happened more than once that when I look at the candidate's incomplete submissions I see that the problem, at least this time, is not necessarily in the testing system. There is another possible problem that had not even occurred to me. The candidate failed the trivial question because they tried to write the answer in Java.

I am reminded of Dijkstra's remark that the teaching of BASIC should be rated as a criminal offense. Seeing the hapless candidate get bowled over by a question that should be a mere formality makes me wonder if the same might be said of Java.

I'm not sure. It's possible that this is still a failure of the quiz. It's possible that the Java programmers have valuable skills that we could use, despite their inability to produce even a trivial working program in a short amount of time. I could be persuaded, but right now I have a doubtful feeling.

When you learn Perl, Python, Ruby, or Javascript, one of the things you learn is a body of technique for solving problems using hashes, which are an integral part of the language. When you learn Haskell, you similarly learn a body of technique for solving problems with lazy lists and monads. These kinds of powerful general-purpose tools are at the forefront of the language.

But when you learn Java, there aren't any powerful language features you can use to solve many problems. Instead, you spend your time learning a body of technique for solving problems in the language. Java has hashes, but if you are aware of them at all, they are just another piece of the immense Collections library, lost among the many other sorts of collections, and you have no particular reason to know about them or think about them. A good course of Java instruction might emphasize the more useful parts of the Collections, but since they're just another part of the library it may not be obvious that hashes are any more or less useful than, say, AbstractAction or zipOutputStream.

### I really like Java

I am glad to have had the experience of programming in Java. I liked programming in Java mainly because I found it very relaxing. With a bad language, like say Fortran or csh, you struggle to do anything at all, and the language fights with you every step of the way forward. With a good language there is a different kind of struggle, to take advantage of the language's strengths, to get the maximum amount of functionality, and to achieve the clearest possible expression.

Java is neither a good nor a bad language. It is a mediocre language, and there is no struggle. In Haskell or even in Perl you are always worrying about whether you are doing something in the cleanest and the best way. In Java, you can forget about doing it in the cleanest or the best way, because that is impossible. Whatever you do, however hard you try, the code will come out mediocre, verbose, redundant, and bloated, and the only thing you can do is relax and keep turning the crank until the necessary amount of code has come out of the spout. If it takes ten times as much code as it would to program in Haskell, that is all right, because the IDE will generate half of it for you, and you are still being paid to write the other half.

So you turn the crank, draw your paycheck, and you don't have to worry about the fact that it takes at least twice as long and the design is awful. You can't solve any really hard design problems, but there is a book you can use to solve some of the medium-hard ones, and solving those involves cranking out a lot more Java code, for which you will also be paid. You are a coder, your job is to write code, and you write a lot of code, so you are doing your job and everyone is happy.

You will not produce anything really brilliant, but you will probably not produce anything too terrible either. The project might fail, but if it does you can probably put the blame somewhere else. After all, you produced 576 classes that contain 10,000 lines of Java code, all of it seemingly essential, so you were doing your job. And nobody can glare at you and demand to know why you used 576 classes when you should have used 50, because in Java doing it with only 50 classes is probably impossible.

(Different languages have different failure modes. With Perl, the project might fail because you designed and implemented a pile of shit, but there is a clever workaround for any problem, so you might be able to keep it going long enough to hand it off to someone else, and then when it fails it will be their fault, not yours. With Haskell someone probably should have been fired in the first month for choosing to do it in Haskell.)

So yes, I enjoyed programming in Java, and being relieved of the responsibility for producing a quality product. It was pleasant to not have to worry about whether I was doing a good job, or whether I might be writing something hard to understand or to maintain. The code was ridiculously verbose, of course, but that was not my fault. It was all out of my hands.

So I like Java. But it is not a language I would choose for answering test questions, unless maybe the grade was proportional to the number of lines of code written. On the test, you need to finish quickly, so you need to optimize for brevity and expressiveness. Java is many things, but it is neither brief nor expressive.

When I see that some hapless job candidate struggled for 15 minutes and 14 seconds to write a Java program for copying standard input to standard output, and finally gave up, without even getting to the real questions, it makes me sad that their education, which was probably expensive, has not equipped them with with better tools or to do something other than grind out Java code.

Thu, 30 Jan 2014

(Like everything else in this section, these are notes for a project that was never completed.)

### Introduction

These incomplete notes from 1997-2001 are grappling with the problem of transforming data structures in a language like Perl, Python, Java, Javascript, or even Haskell. A typical problem is to take an input of this type:

[
[Chi, Ill],
[NY, NY],
[Alb, NY],
[Spr, Ill],
[Tr, NJ],
[Ev, Ill],
]

and to transform it to an output of this type:

{ Ill => [Chi, Ev, Spr],
NY  => [Alb, NY],
NJ  => [Tr],
}

One frequently writes code of this sort, and it should be possible to specify the transformation with some sort of high-level declarative syntax that is easier to read and write than the following gibberish:

my $out; for my$pair (@$in) { push @{$out->{$pair->[0]}},$pair->[1];
}
for my $k (keys %$out) {
@{$out->{$k}} = sort @{$out->{$k}};
}

This is especially horrible in Perl, but it is bad in any language. Here it is in a hypothetical language with a much less crusty syntax:

for pair (in.items) :
out[pair[0]].append(pair[1])
for list (out.values) :
list.sort

You still can't see what it really going on without executing the code in your head. It is hard for a beginner to write, and hard to anyone to understand.

### Original undated notes from around 1997–1998

Consider this data structure DS1:

[
[Chi, Ill],
[NY, NY],
[Alb, NY],
[Spr, Ill],                DS1
[Tr, NJ],
[Ev, Ill],
]

This could be transformed several ways:

{
Chi => Ill,
NY => NY,
Alb => NY,
Spr => Ill,                DS2
Tr => NJ,
Ev => Ill,
}

{ Ill => [Chi, Spr, Ev],
NY  => [NY, Alb],        DS3
NJ  => Tr,
}
{ Ill => 3,
NY  => 2,
NJ  => 1,
}

[ Chi, Ill, NY, NY, Alb, NY, Spr, Ill, Tr, NJ, Ev, Ill]    DS4

Basic idea: Transform original structure of nesting depth N into an N-dimensional table. If Nth nest is a hash, index table ranks by hash keys; if an array, index by numbers. So for example, DS1 becomes

1        2
1   Chi Ill
2   NY  NY
3   Alb NY
4   Spr Ill
5   Tr  NJ
6   Ev  Ill

Or maybe hashes should be handled a little differently? The original basic idea was more about DS2 and transformed it into

Ill        NY        NJ
Chi     X
NY              X
Alb             X
Spr     X
Tr                      X
Ev      X

Maybe the rule is: For hashes, use a boolean table indexed by keys and values; for arrays, use a string table index by integers.

Notation idea: Assign names to the dimensions of the table, say X and Y. Then denote transformations by:

[([X, Y])]        (DS1)
{(X => Y)}        (DS2)
{X => [Y]}        (DS3)
[(X, Y)]        (DS4)

The (...) are supposed to incdicate a chaining of elements within the larger structure. But maybe this isn't right.

At the bottom: How do we say whether

X=>Y, X=>Z

turns into

[ X => Y, X => Z ]                (chaining)

or [ X => [Y, Z] ] (accumulation)

Consider

A B C
D . . .
E . .
F .   .

<...> means ITERATE over the thing inside and make a list of the results. It's basically map'.

Note that:

<X,Y>     |=   (D,A,D,B,D,C,E,A,E,B,F,A,F,C)
<X,[<Y>]> |=   (D,[A,B,C],E,[A,B],F,[A,C])

Brackets and braces just mean brackets and braces. Variables at the same level of nesting imply a loop over the cartesian join. Variables subnested imply a nested loop. So:

<X,Y> means

for x in X
for y in Y
push @result, (x,y) if present(x,y);

But

<X,<Y>> means
for x in X
for y in Y
push @yresult, (y) if present(x,y);
push @result, @yresult

Hmmm. Maybe there's a better syntax for this.

Well, with this plan:

DS1: [ <[X,Y]> ]
DS2: { <X=>Y>  }
DS3: { <X => [<Y>]> }
DS4: [ <X, Y> ]

It seems pretty flexible. You could just as easily write

{ <X => max(<Y>) }

and you'd get

{ D => C, E => B, F => C }

If there's a count' function, you can get

{ D => 3, E => 2, F => 2 }

or maybe we'll just overload scalar' to meancount'.

Question: How to invert this process? That's important so that you can ask it to convert one data structure to another. Also, then you could write something like

[ <city, state> ]  |= { <state => [<city>] > }

and omit the X's and Y's.

Real example: From proddir. Given

ID / NAME / SHADE / PALETTE / DESC

For example:

A / AAA / red     / pink   / Aaa
B / BBB / yellow  / tawny  / Bbb
A / AAA / green   / nude   / Aaa
B / BBB / blue    / violet / Bbb
C / CCC / black   / nude   / Ccc

Turn this into

{ A => [ AAA, [ [red, pink], [green, nude] ], Aaa],
B => [ BBB, [ [yellow, tawny], [blue, violet] ], Bbb],
C => [ CCC, [ [black, nude] ], CCC]
}

{ < ID => [
name,
desc
]>
}

Something interesting happened here. Suppose we have

[ [A, B]
[A, B]
]

And we ask for <A, B>. Do we get (A, B, A, B), or just (A, B)? Does it remove duplicate items for us or not? We might want either.

In the example above, why didn't we get

{ A => [ AAA, [ [red, pink], [green, nude] ], Aaa],
A => [ AAA, [ [red, pink], [green, nude] ], Aaa],
B => [ BBB, [ [yellow, tawny], [blue, violet] ], Bbb],
B => [ BBB, [ [yellow, tawny], [blue, violet] ], Bbb],
C => [ CCC, [ [black, nude] ], CCC]
}

If the outer iteration was supposed to be over all id-name-desc triples? Maybe we need

<...>                all triples
<!...!>                unique triples only

Then you could say

<X>  |= <!X!>

to indicate that you want to uniq a list.

But maybe the old notation already allowed this:

<X> |= keys %{< X => 1 >}

It's still unclear how to write the example above, which has unique key-triples. But it's in a hash, so it gets uniqed on ID anyway; maybe that's all we need.

### 1999-10-23

Rather than defining some bizarre metalanguage to describe the transformation, it might be easier all around if the user just enters a sample input, a sample desired output, and lets the twingler figure out what to do. Certainly the parser and internal representation will be simpler.

For example:

[ [ A, B ],
[ C, B ],
[ D, E ] ]
--------------
{ B => [A, C],
E => [D],
}

should be enough for it to figure out that the code is:

for my $a1 (@$input) {
my ($e1,$e2) = @$a1; push @{$output{$e2}},$e1;
}

Advantage: After generating the code, it can run it on the sample input to make sure that the output is correct; otherwise it has a bug.

Input grammar:

%token ELEMENT
expr: array | hash ;
array: '[' csl ']' ;
csl: ELEMENT | ELEMENT ',' csl | /* empty */ ;
hash: '{' cspl '}' ;
cspl: pair | pair ',' cspl | /* empty */ ;
pair: ELEMENT '=>' ELEMENT;

Simple enough. Note that (...) lines are not allowed. They are only useful at the top level. A later version can allow them. It can replace the outer (...) with [...] or {...] as appropirate when it sees the first top-level separator. (If there is a => at the top level, it is a hash, otherwise an array.)

Idea for code generation: Generate pseudocode first. Then translate to Perl. Then you can insert a peephole optimizer later. For example

foreachkey k (somehash) {
push somearray, \$somehash{k}
}

could be optimized to

somearray = values somehash;

How do we analyze something like:

[ [ A, B ],
[ C, B ],
[ D, E ] ]
--------------
{ B => [A, C],
E => [D],
}

Idea: Analyze structure of input. Analyze structure of output and figure out an expression to deposit each kind of output item. Iterate over input items. Collect all input items into variables. Deposit items into output in appropriate places.

For an input array, tag the items with index numbers. See where the indices go in the output. Try to discern a pattern. The above example:

Try #1:
A: 1
B: 2
C: 1
B: 2  -- consistent with B above
D: 1
E: 2

Output:  2 => [1, 1]
2 => [1]

OK—2s are keys, 1s are array elements.

A different try fails:

A: 1
B: 1
C: 2
B: 2 -- inconsistent, give up on this.

Now consider:

[ [ A, B ],
[ C, B ],
[ D, E ] ]
--------------
{ A => B,
C => B,
D => E,
}

A,C,D get 1; B,E get 2. this works again. 1s are keys, 2s are values.

I need a way of describing an element of a nested data structure as a simple descriptor so that I can figure out the mappings between descriptors. For arrays and nested arrays, it's pretty easy: Use the sequence of numeric indices. What about hashes? Just K/V? Or does V need to be qualified with the key perhaps?

Example above:

IN:        A:11  B:12 22 C:21 D:31 E:32
OUT:    A:K   B:V     C:K  D:K  E:V

Now try to find a mapping from the top set of labels to the bottom. x1 => K, x2 => V works.

Problem with this:

[ [ A, B ],
[ B, C ],
]
------------
{ A => B,
B => C,
}

is unresolvable. Still, maybe this works well enough in most common cases.

Let's consider:

[[ A , AAA , red     , pink   , Aaa],
[ B , BBB , yellow  , tawny  , Bbb],
[ A , AAA , green   , nude   , Aaa],
[ B , BBB , blue    , violet , Bbb],
[ C , CCC , black   , nude   , Ccc],
]
-------------------------------------------------------------
{ A => [ AAA, [ [red, pink], [green, nude] ], Aaa],
B => [ BBB, [ [yellow, tawny], [blue, violet] ], Bbb],
C => [ CCC, [ [black, nude] ], CCC]
}

A:        00,20 => K
AAA:        01,21 => V0
red:        02    => V100
pink:        03    => V101
Aaa:    04    => V2
B:      10,30 => K
C:      40    => K

etc.

Conclusion: x0 => K; x1 => V0; x2 => V100; x3 => V101; x4 => V2

How to reverse?

Simpler reverse example:

{ A => [ B, C ],
E => [ D ],
}
---------------------
[ [ A, B ],
[ A, C ],
[ E, D ],
]

A: K        => 00, 10
B: V0   => 01
C: V1   => 11
D: V0   => 21
E: K    => 20

Conclusion: K => x0; V => x1 This isn't enough information. Really, V => k1, where k is whatever the key was!

What if V items have the associated key too?

A: K        => 00, 10
B: V{A}0=> 01
C: V{A}1=> 11
D: V{E}0=> 21
E: K    => 20

Now there's enough information to realize that B amd C stay with the A, if we're smart enough to figure out how to use it.

### 2001-07-28

Sent to Nyk Cowham

### 2001-08-24

Sent to Timur Shtatland

### 2001-10-28

Here's a great example. The output from HTML::LinkExtor is a list like

([img, src, URL1, losrc, URL2],
[a, href, URL3],
...
)

we want to transform this into

(URL1 => undef,
URL2 => undef,
URL3 => undef,
...
)

Sun, 19 Jan 2014

(This post inaugurates a new section on my blog, for incomplete notes. It often happens that I have some idea, usually for software, and I write up a bunch of miscellaneous notes about it, and then never work on it. I'll use this section to post some of those notes, mainly just because I think they might be interesting, but also in the faint hope that someone might get interested and do something with it.)

### Why are simple SQL queries so verbose?

For example:

UPDATE batches b
join products p using (product_id)
join clients c using (client_id)
SET b.scheduled_date = NOW()
WHERE b.scheduled_date > NOW()
and b.batch_processor = 'batchco'

(This is 208 characters.)

I guess about two-thirds of this is unavoidable, but those join-using clauses ought to be omittable, or inferrable, or abbreviatable, or something.

b.batch_processor should be abbreviated to at least batch_processsor, since that's the only field in those three tables with that name. In fact it could probably be inferred from b_p. Similarly c.login_name -> login_name -> log or l_n.

update batches set sch_d = NOW()
where sch_d > NOW()
and bp = 'batchco'
and cl.ln = 'mjd'

(Only 94 characters.)

cl.ln is inferrable: Only two tables begin with cl. None of the field names in the client_transaction_view table look like ln. So cl.ln unambiguously means client.login_name.

Then the question arises of how to join the batches to the clients. This is the only really interesting part of this project, and the basic rule is that it shouldn't do anything really clever. There is a graph, which the program can figure out from looking at the foreign key constraints. And the graph should clearly have a short path from batches through products to clients.

bp might be globally ambiguous, but it can be disambiguated by assuming it's in one of the three tables involved.

If something is truly ambiguous, we can issue an intelligent request for clarification:

"bp" is ambiguous. Did you mean:
1. batches.batch_processor
2. batches.bun_predictor
0. None of the above
which? _

### Overview

1. Debreviate table names
2. Figure out required joins and join fields
3. Debreviate field names

Can 1 and 2 really be separated? They can in the example above, but maybe not in general.

I think separating 3 and putting it at the end is a good idea: don't try to use field name abbreviations to disambiguate and debreviate table names. Only go the other way. But this means that we can't debreviate cl, since it might be client_transaction_view.

What if something like cl were left as ambiguous after stage 1, and disambiguated only in stage 3? Then information would be unavailable to the join resolution, which is the part that I really want to work.

Abbreviations for batch_processor:

bp
b_p
ba_pr
batch_p

There is a tradeoff here: the more different kinds of abbreviations you accept, the more likely there are to be ambiguities.

There could also be a preferences file that lists precedences for tables and fields: if it lists clients, then anything that could debreviate to clients or to client_transaction_view automatically debreviates to clients. The first iteration could just be a list of table names.

Short join paths are preferred to long join paths.

If it takes a long time to generate the join graph, cache it. Build it automatically on the first run, and then rebuild it on request later on.

### More examples

(this section blank)

### Implementation notes

Maybe convert the input to a SQL::Abstract first, then walk the resulting structure, first debreviating names, then inserting joins, then debreviating the rest of the names. Then you can output the text version of the result if you want to.

Note that this requires that the input be valid SQL. Your original idea for the abbreviated SQL began with

update set batches.sch_d = NOW()

rather than

update batches set sch_d = NOW()

but the original version would probably be ruled out by this implementation. In this case that is not a big deal, but this choice of implementation might rule out more desirable abbreviations in the future.

Correcting dumb mistakes in the SQL language design might be in Quirky's purview. For example, suppose you do

select * from table where (something)

### Application notes

RJBS said he would be reluctant to use the abbreviated version of a query in a program. I agree: it would be very foolish to do so, because adding a table or a field might change the meaning of an abbreviated SQL query that was written into a program ten years ago and has worked ever since. This project was never intended to abbreviate queries in program source code.

Quirky is mainly intended for one-off queries. I picture it going into an improved replacement for the MySQL command-line client. It might also find use in throwaway programs. I also picture a command-line utility that reads your abbreviated query and prints the debreviated version for inserting into your program.

### Miscellaneous notes

(In the original document this section was blank. I have added here some notes I made in pen on a printout of the foregoing, on an unknown date.)

Maybe also abbreviate update => u, where => w, and => &. This cuts the abbreviated query from 94 to 75 characters.

Since debreviation is easier [than join inference] do it first!

Another idea: "id" always means the main table's primary key field.