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Sep 24, 2023 at 6:14 history edited Tony Huynh
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Jun 3, 2016 at 16:30 answer added László Kozma timeline score: 3
Feb 1, 2015 at 20:56 comment added David Eppstein Re "Was this problem considered before?": it's close to, but not the same thing as, the concept of a superpattern — see en.wikipedia.org/wiki/Superpattern
Jan 31, 2015 at 8:58 answer added Ilya Bogdanov timeline score: 11
Jan 30, 2015 at 18:09 answer added Ben Barber timeline score: 9
Jan 30, 2015 at 17:53 comment added László Kozma added clarification on relative ordering.
Jan 30, 2015 at 17:53 history edited László Kozma CC BY-SA 3.0
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Jan 30, 2015 at 17:40 comment added Włodzimierz Holsztyński Phrase "$x_i$ have the same relative ordering as $\sigma_i$" didn't seem defined.
Jan 30, 2015 at 17:31 answer added Tony Huynh timeline score: 1
Jan 30, 2015 at 16:52 history edited László Kozma CC BY-SA 3.0
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Jan 30, 2015 at 16:49 comment added László Kozma For the first observation, color the top-right square of size $(n/2+1) \times (n/2+1)$ white. The remaining B has almost $3/4$ of the points, yet it does not include the identity permutation.
Jan 30, 2015 at 16:45 comment added The Masked Avenger Also I don't see the first easy observation: can you provide an example where B has more than half the nodes and not all the permutations of n/2?
Jan 30, 2015 at 16:37 comment added The Masked Avenger I don't know if this precise form is in the literature. Looking at permanents, binary matrices, and enumerating matrices avoiding a certain pattern may get you literature which gets close to the form above.
Jan 30, 2015 at 15:28 comment added László Kozma No, it seemed like it should be true for some obvious reason (that I am missing), but it will be worth checking if a proof will not be found.
Jan 30, 2015 at 15:14 comment added Per Alexandersson Have you run any computer experiments on this?
Jan 30, 2015 at 15:03 history asked László Kozma CC BY-SA 3.0