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Feb 28, 2017 at 17:44 comment added Will Orrick …seem skeptical, both that such things might be true and there there's any hope of attacking such questions with current methods.
Feb 28, 2017 at 17:43 comment added Will Orrick @Gerry Unfortunately not much. The writeup mentioned in my comment above is arxiv.org/abs/1112.4160. It is not published yet due mostly to inattention on my part and partly to lack of enthusiasm for this kind of work from the referees. I did have some ideas a few years back for how to adapt the method for the $11\times 11$ and possibly $13\times 13$ cases, at some increase in complexity of the code and decrease in its efficiency, but have not seriously pursued it. I believe there's been no progress related to any of the conjectures in my answer below. Experts I've spoken with...
Feb 25, 2017 at 22:09 comment added Gerry Myerson @Will, any update?
S Feb 25, 2017 at 15:52 history suggested Rodrigo de Azevedo
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Feb 25, 2017 at 15:40 review Suggested edits
S Feb 25, 2017 at 15:52
Dec 31, 2016 at 14:41 history edited Qfwfq CC BY-SA 3.0
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Apr 15, 2010 at 22:25 answer added Will Orrick timeline score: 4
Apr 15, 2010 at 3:27 comment added Will Orrick The 12 x 12 case has recently been settled by a calculation of Richard Brent, Judy-anne Osborn, Paul Zimmermann, and myself: the conjecture at indiana.edu/~maxdet/spectrum13.html is correct. (The link is to the n=13 case of {-1,1} matrices, which corresponds to the 12 x 12 case of {0,1} matrices.) We hope to write up these and other results soon. The answer for 11 x 11 is still unknown, as our code works only for odd-sized {-1,1} matrices.
Mar 19, 2010 at 15:52 vote accept Ross Snider
Mar 18, 2010 at 18:36 comment added Ross Snider Ahh, yes there seems to be a small problem with the determinant calculating code. I grabbed a determinant function that works with libboost online. Should have checked it for accuracy in all cases.
Mar 18, 2010 at 14:51 comment added Gerhard Paseman I also did the smaller cases by hand. For 4x4 I got 7, for 5x5 I got 11 and 6x6 I got 19. You might check your computer program. Gerhard "Ask Me About System Design" Paseman, 2010.03.18
Mar 18, 2010 at 7:27 comment added Gerhard Paseman I remember doing the 7x7 spectrum by hand years ago after some computer searching with a single desktop computer of 1990's vintage. It feels good to be compared to a computing cluster. Gerhard "Ask Me About System Design" Paseman, 2010.03.18
Mar 18, 2010 at 7:22 answer added Gerhard Paseman timeline score: 4
Mar 18, 2010 at 6:52 comment added Cam McLeman Just a comment that you can certainly improve the efficiency of your brute force approach. For example, once you have one determinant calculated, you may as well cross off all matrices which are row or column swaps. While we're at it, the 6x6 determinants properly include into the 7x7 determinants (multiple times!), so you could skip over these if you've already done them. And I'm sure you could be much more clever than these...this was just a first thought.
Mar 18, 2010 at 6:40 answer added Gerry Myerson timeline score: 3
Mar 18, 2010 at 6:30 comment added Gerry Myerson I'm sorry, I was a bit careless above; the 10 x 10 number is 269, but the 11 x 11 is unknown. See A013588 in the Online Encyclopedia of Integer Sequences, oeis.org/A013588
Mar 18, 2010 at 6:27 comment added Gerry Myerson I think this is a hard problem. As evidence, I note that the smallest positive integer which is not the determinant of a 9 x 9 binary matrix was only recently found to be 103 (arXiv.org/pdf/math/0511636v1); the smallest positive integer not the determinant of a 10 x 10 binary matrix seems to be unknown.
Mar 18, 2010 at 6:26 comment added Qiaochu Yuan Singular matrices certainly do have a determinant; it's zero...
Mar 18, 2010 at 6:02 history asked Ross Snider CC BY-SA 2.5