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Apr 17, 2016 at 14:25 comment added Ben Barber @LászlóKozma, please do post an answer if you'd like to explain the connection.
Apr 16, 2016 at 12:47 comment added László Kozma It seems that the asymptotics of $f(n)$ is $n^2$, up to a polylog factor, based on the results from "Ordered Ramsey numbers" by David Conlon, Jacob Fox, Choongbum Lee, Benny Sudakov. (hint from Martin Balko) arxiv.org/abs/1410.5292
Feb 1, 2015 at 18:25 comment added László Kozma Very nice - so, together with Ilya Bogdanov's answer the bounds $3n-\epsilon < f(n) \leq n^2$ follow. It seems interesting now to try narrowing the gap.
Jan 31, 2015 at 3:52 comment added The Masked Avenger It breaks down at n=2, with a coloring in which l is at most 1 and k at most 2. Take the main and lower diagonals as 7 elements of one color.
Jan 30, 2015 at 19:37 comment added Gerhard Paseman Try the following induction and see where it breaks down: Every coloring of a 2n by 2n square has all k permutations contained by blacks squares and all l permutations contained by white squares, where k+l >= 2n. Gerhard "It Can't Be That Easy?" Paseman, 2015.01.30
Jan 30, 2015 at 18:59 comment added Tony Huynh Nice. Much better than my answer.
Jan 30, 2015 at 18:09 history answered Ben Barber CC BY-SA 3.0