This is a continuation of the previous thread here in the polymath1 project, which is now full. Ostensibly, the purpose of this thread is to continue writing up the paper containing many of the things achieved during this side of the project, though we have also been spending time on chasing down more results, in particular using new computer data to narrow down the range of the maximal size of 6D Moser sets (currently we can pin this down to between 353 and 355). At some point we have to decide what results to put in in full detail in the paper, what results to summarise only (with links to the wiki), and what results to defer to perhaps a subsequent paper, but these decisions can be taken at a leisurely pace.
I guess we’ve abandoned the numbering system now, but I suppose that if necessary we can use timestamps or URLs to link to previous comments.
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9 July, 2009 at 8:28 am
DHJ: Writing the second paper III. « What’s new
[…] July, 2009 in math.CO | Tags: polymath1 | by Terence Tao This is a continuation of the preceding two threads here on the polymath1 project, which are now full. We are closing on on the computation […]
9 July, 2009 at 8:14 pm
Michael Peake
Mostly, it was a depth-first search, with trimming every few layers to check I hadn’t completed any rows.
For 5,12,12,4,1, make use of the central (2,2,2,2) point.
Half the c-points remain, one from each symmetric pair.
It is easy to see that Gamma(2,2,0) is excluded, so Gamma(0,2,2) is included. The remaining six c points are one from each pair of Gamma(1,2,1).
It turned out, by brute force, that Gamma(3,1,0) and Gamma(0,1,3) were excluded, and the twelve b points were then one from each of the twelve remaining pairs.
The a points are one each from five of the eight symmetric pairs.
For 5,12,18,4,0, the c-points are determined by the d-points, as Kristal found in 15 June 12:11. Then the a-points turned out to be five of the six from the 6,12,18,4,0 solution. Then I just did a brute-force search through the combin(32,12) possible sets of b points.
For 6,8,12,8,0, the c-points are non-adjacent pairs from cxx22 and the other c-sets. By the cube’s symmetry, I can assume they are 1122,3322;1212,3232;1221,3223, and three other pairs.
Then place the a-points, and I think it turned out that the only solution so far is that the c-points are Gamma(2,2,0) and Gamma(0,2,2), and the a-points are Gamma(2,0,2). Then brute-force through the b-points.