Comments on: DHJ: Writing the second paper III. http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/ Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao Fri, 12 Mar 2010 20:58:09 +0000 http://wordpress.com/ hourly 1 By: Kevin O'Bryant http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-42713 Kevin O'Bryant Sun, 06 Dec 2009 03:43:30 +0000 http://terrytao.wordpress.com/?p=2398#comment-42713 According to this site (http://www.math.uni.wroc.pl/~jwr/non-ave.htm), Rodolfo Niborski has proved that $latex r_3(194)=41$. The witness is 1 2 5 7 11 16 18 19 24 26 38 39 42 44 48 53 55 56 61 63 112 127 128 133 135 136 140 146 148 149 151 155 172 173 178 179 181 187 188 192 194 with the optimality coming by exhaustive search. According to this site (http://www.math.uni.wroc.pl/~jwr/non-ave.htm), Rodolfo Niborski has proved that r_3(194)=41. The witness is

1 2 5 7 11 16 18 19 24 26 38 39 42 44 48 53 55 56 61 63 112 127 128 133 135 136 140 146 148 149 151 155 172 173 178 179 181 187 188 192 194

with the optimality coming by exhaustive search.

]]> By: Klas Markström http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-42313 Klas Markström Sun, 08 Nov 2009 09:23:57 +0000 http://terrytao.wordpress.com/?p=2398#comment-42313 You might be thinking of this set of results. http://www.st.ewi.tudelft.nl/sat/waerden.php n=200 sounds much larger than anything I can recall having seen so if there really is such a result the methods behind it would be very interesting to see. You might be thinking of this set of results.
http://www.st.ewi.tudelft.nl/sat/waerden.php

n=200 sounds much larger than anything I can recall having seen so if there really is such a result the methods behind it would be very interesting to see.

]]> By: Kevin O'Bryant http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-42307 Kevin O'Bryant Sun, 08 Nov 2009 01:09:22 +0000 http://terrytao.wordpress.com/?p=2398#comment-42307 The preprint I found was definitely concerning $latex r_3(n)$, and it was reporting the outcome of a large parallel computation. The number 200 is from my memory, and could well have been ``almost 200''. While 200 is too small for the good stuff, some elementary arguments are still powerful. The original 1930s Erdos-Turan paper uses the bound $latex r_3(m+n) \leq r_3(m)+r_3(n)$, for example, and the greedy algorithm to compute $latex r_3(41)=16$ (by hand!). The preprint I found was definitely concerning r_3(n), and it was reporting the outcome of a large parallel computation. The number 200 is from my memory, and could well have been “almost 200”.

While 200 is too small for the good stuff, some elementary arguments are still powerful. The original 1930s Erdos-Turan paper uses the bound r_3(m+n) \leq r_3(m)+r_3(n), for example, and the greedy algorithm to compute r_3(41)=16 (by hand!).

]]> By: Kristal Cantwell http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-42302 Kristal Cantwell Sat, 07 Nov 2009 18:04:45 +0000 http://terrytao.wordpress.com/?p=2398#comment-42302 I can look for the van der Warden for 5 colors and length note it is greater than 170 from the table here: http://en.wikipedia.org/wiki/Van_der_Waerden_number Then divide by 5 and get a set of 34 points out of the 200 without an arithmetic progression of length 3 so the maximum must be 34 or more. I can look for the van der Warden for 5 colors and length note it is greater than 170 from the table here:

http://en.wikipedia.org/wiki/Van_der_Waerden_number

Then divide by 5 and get a set of 34 points out of the 200 without an arithmetic progression of length 3 so the maximum must be 34 or more.

]]> By: Terence Tao http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-42295 Terence Tao Sat, 07 Nov 2009 00:52:32 +0000 http://terrytao.wordpress.com/?p=2398#comment-42295 That sounds somewhat large to me, given that the Fourier-analytic upper bounds are quite lossy, and the Behrend examples are also not very strong at this scale, and there are too many combinations for a brute force approach. Perhaps it was van der Waerden numbers instead? That sounds somewhat large to me, given that the Fourier-analytic upper bounds are quite lossy, and the Behrend examples are also not very strong at this scale, and there are too many combinations for a brute force approach. Perhaps it was van der Waerden numbers instead?

]]> By: Kevin O'Bryant http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-42268 Kevin O'Bryant Fri, 06 Nov 2009 07:31:24 +0000 http://terrytao.wordpress.com/?p=2398#comment-42268 This is slightly off-topic, but does anyone know of explicit computations of $latex r_k(n)$, the maximum size of a subset of $latex \{1,2,\ldots,n\}$ that does not have $latex k$ terms in AP? I recall seeing $latex r_3(200)$ somewhere, but I can't find it now. This is slightly off-topic, but does anyone know of explicit computations of r_k(n), the maximum size of a subset of \{1,2,\ldots,n\} that does not have k terms in AP?

I recall seeing r_3(200) somewhere, but I can’t find it now.

]]> By: Terence Tao http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-40727 Terence Tao Tue, 28 Jul 2009 04:18:06 +0000 http://terrytao.wordpress.com/?p=2398#comment-40727 Huh, that's somewhat strange symmetry (it doesn't have a geometric interpretation, despite the fact that the lines are geometric), but I guess you're right. So for instance, for k=4, one could swap all 1s with 4s in a Moser set while keeping 2s and 3s unchanged, or instead swap 1s with 2s and 3s with 4s, boosting the group of symmetries by an order of 4. Every little bit helps, I suppose... Huh, that’s somewhat strange symmetry (it doesn’t have a geometric interpretation, despite the fact that the lines are geometric), but I guess you’re right. So for instance, for k=4, one could swap all 1s with 4s in a Moser set while keeping 2s and 3s unchanged, or instead swap 1s with 2s and 3s with 4s, boosting the group of symmetries by an order of 4. Every little bit helps, I suppose…

]]> By: Michael Peake http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-40726 Michael Peake Tue, 28 Jul 2009 02:57:58 +0000 http://terrytao.wordpress.com/?p=2398#comment-40726 Just a remark on the symmetry group in the Moser problem, mentioned in the introduction. The coordinates $latex 1\ldots k$ naturally pair up into $latex (1,k), (2,k-1)$ and so on. When $latex k>3$, these pairs can be swapped around, which increases the symmetry group by a factor $latex \lfloor k/2\rfloor !$. Each pair can be flipped, which gives an extra factor $latex 2^{\lfloor k/2\rfloor}/2$. (The symmetry of flipping every pair is the same as flipping every coordinate dimension, so it is already included in $latex 2^n n!$.) Just a remark on the symmetry group in the Moser problem, mentioned in the introduction. The coordinates 1\ldots k naturally pair up into (1,k), (2,k-1) and so on. When k>3, these pairs can be swapped around, which increases the symmetry group by a factor \lfloor k/2\rfloor !. Each pair can be flipped, which gives an extra factor 2^{\lfloor k/2\rfloor}/2. (The symmetry of flipping every pair is the same as flipping every coordinate dimension, so it is already included in 2^n n!.)

]]> By: Kristal Cantwell http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-40725 Kristal Cantwell Mon, 27 Jul 2009 22:06:56 +0000 http://terrytao.wordpress.com/?p=2398#comment-40725 With regard to my above post I am not able to prove this result. With regard to my above post I am not able to prove this result.

]]> By: Kristal Cantwell http://terrytao.wordpress.com/2009/07/09/dhj-writing-the-second-paper-iii/#comment-40721 Kristal Cantwell Mon, 27 Jul 2009 18:49:09 +0000 http://terrytao.wordpress.com/?p=2398#comment-40721 I think I can find lower bounds to a couple more coloring problems. They involve the cube when viewed as a torus. This allows more lines. I think these lines are sets of n points with fixed coordinates and if we are looking at a k-cube k types of moving coordinates they move up by one but can start at any of the k-values. Let us refer to these as n=T(k,r) is the smallest integer to force a monochromatic line. If we allow lines to move sown as well as up we have 2n types of lines and we can define n=D(k,r). By using arguments similar to the arguments relating colorings without combinatorial lines and coloring without geometric lines. I can get lower bounds of roughly 2^n/3 and 2^n/6 for these number if the number of colors is two. Two colors is important because it related to the drawing postions in two player games. These two toric variants are interesting because their hypergraph has a property that makes a variant of the two player game easy to analyze. I think I can find lower bounds to a couple more coloring problems. They involve the cube when viewed as a torus. This allows more lines. I think these lines are sets of n points with fixed coordinates and if we are looking at a k-cube k types of moving coordinates they move up by one but can start at any of the k-values. Let us refer to these as n=T(k,r) is the smallest integer to force a monochromatic line. If we allow lines to move sown as well as up we have 2n types of lines and we can define n=D(k,r). By using arguments similar to the arguments relating colorings without combinatorial lines and coloring without geometric lines. I can get lower bounds of roughly 2^n/3 and 2^n/6 for these number if the number of colors is two. Two colors is important because it related to the drawing postions in two player games. These two toric variants are interesting because their hypergraph has a property that makes a variant of the two player game easy to analyze.

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