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Tue, 02 Oct 2007
Van der Waerden's problem
First I'll explain what the programs are about.
Van der Waerden's problemColor each of a row of dots red or blue, so that no three evenly-spaced dots are the same color. (That is, if dots n and n+i are the same color, dot n+2i must be a different color.) How many dots can you do?Well, clearly you can do four: R R B B. And then you can add another red one on the end: R R B B R. And then another that could be either red or blue: R R B B R B. And then the next can be either color, say blue: R R B B R B B. But now you are at the end, because if you make the next dot red, then dots 2, 5, and 8 will all be red (R R B B R B B R), and if you make the next dot blue then dots 6, 7, and 8 will be blue (R R B B R B B B). But maybe we made a mistake somewhere earlier, and if the first seven dots were colored differently, we could have made a row of more than 7 that obeyed the no-three-evenly-spaced-dots requirement. In fact, this is so: R R B B R R B B is an example. But this is the end of the line. Any coloring of a row of 9 dots contains three evenly-spaced dots of the same color. (I don't know a good way to prove this, short of an enumeration of all 512 possible arrangements of dots. Well, of course it is sufficient to enumerate the 256 that begin with R, but that is pretty much the same thing.)
[Addendum 20141208: In this post I give a simple argument that !!V(3,2)\le 9!!.] Van der Waerden's theorem says that for any number of colors, say C, a sufficiently-long row of colored dots will contain n evenly-spaced same-color dots for any n. Or, put another way, if you partition the integers into C disjoint classes, at least one class will contain arbitrarily long arithmetic progressions.The proof of van der Waerden's theorem works by taking C and n and producing a number V such that a row of V dots, colored with C colors, is guaranteed to contain n evenly-spaced dots of a single color. The smallest such V is denoted V(n, C). For example V(3, 2) is 9, because any row of 9 dots of 2 colors is guaranteed to contain 3 evenly-spaced dots of the same color, but this is not true of such row of only 8 dots. Van der Waerden's theorem does not tell you what V(n, C) actually is; it provides only an upper bound. And here's the funny thing about van der Waerden's theorem: the upper bound is incredibly bad. For V(3, 2), the theorem tells you only that V(3, 2) ≤ 325. That is, it tells you that any row of 325 red and blue dots must contain three evenly spaced dots of the same color. This is true, but oh, so sloppy, since the same is true of any row of 9 dots. For V(3, 3), the question is how many red, yellow, and blue dots do you need to guarantee three evenly-spaced same-colored dots. The theorem helpfully suggests that:
$$V(3,3) \leq 7(2\cdot3^7+1)(2\cdot3^{7(2\cdot3^7+1)}+1)$$ This is approximately 5.79·10^{14613}. But what is the actual value of V(3, 3)? It's 27. Urgggh.In fact, there is a rather large cash prize available to be won by the first person who comes up with a general upper bound for V(n, C) that is smaller than a tower of 2's of height n. (That's 2^{22...} with n 2's.) In the rest of this series, a string which does not contain three evenly-spaced equal symbols will be called good, and one which does contain three such symbols will be called bad. Then a special case of Van der Waerden's theorem, with n=3, says that, for any fixed number of symbols, all sufficiently long strings are bad. In college I wanted to investigate this a little more. In particular, I wanted to calculate V(3, 3). These days you can just look it up on Wikipedia, but in those benighted times such information was hard to come by. I also wanted to construct the longest possible good strings, witnesses of length V(3, 3)-1. Although I did not know it at the time, V(3, 3) = 27, so a witness should have length 26. It turns out that there are exactly 48 witnesses of length 26. Here are the 1/6 of them that begin with RB or RRB:
RRBBRRBYBYYRYRRBRBBYRYYBYB RRBBYRRYRYBBYYBBYRYRRYBBRR RRBYBRRYRYBBYYBBYRYRRBYBRR RBRRBRBYYBBYYBRBRRBYYRRYRY RBRBBRRYBBYBYRRYYRRYBYBBYR RBRBBRRYBBYBYRRYYRRYBYBBYB RBRBBYBRRYRYYBYBBRBRYYRRYY RBYYBYBRRBBRRBYBYYBRRYYRYR The rest of the witnesses may be obtained by permuting the colors in these eight. I wrote a series of C programs around 1988 to exhaustively search for good strings. Last month I was in a meeting and I decided to write the program again for some reason. I wrote a much better program. This series of articles will compare the five programs. I will post the first one tomorrow. [ Addendum 20071003: Here is part 1. ] [ Addendum 20071005: Here is part 2. ] [ Addendum 20071005: I made a mistake in the expression I gave for the upper bound on V(3,3) and left out a factor of 7 in the exponent on the last 3. I had said that the upper bound was around 10^{2092}, but actually it is more like the seventh power of this. ] [ Addendum 20071014: Here is part 3. ]
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