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Dec 29, 2015 at 21:48 comment added Johnny Cage I do not get this. This same reason would give that for $C_4$ the same is trivial, and it is not at all (the upper and lower bound are of order $n^{3/2}$...)
Dec 29, 2015 at 16:54 answer added Fedor Petrov timeline score: 3
Dec 29, 2015 at 16:08 comment added joro By $P_4$ do you really mean path with 4 edges or with 4 vertices? If you don't mean induced subgraph, then $K_{n,n}$ is solution for all $P_k$, no matter which way you define $P_4$.
Dec 29, 2015 at 15:40 history edited Johnny Cage CC BY-SA 3.0
added 20 characters in body
Dec 29, 2015 at 15:32 history asked Johnny Cage CC BY-SA 3.0