Given a graph $H$, let $R_k(H)$ be the smallest integer $N$ such that in every $k$-coloring of the edges $K_N$ there is a monochromatic copy of $H$ (in other words, $R_k(H)$ is the ordinary $k$-color Ramsey number of $H$).

Does there exist a sequence of graphs $(H_n)_{n\in \mathbb{N}}$ (where $H_n$ has $n$ vertices) such that $R_2(H_n)=O(n)$, but $R_3(H_n)=\omega(n)$?


1 Answer 1


There is an almost example, 'almost' in three senses: it's for $3$-uniform hypergraphs rather than graphs; for $4$ colours rather than $3$; and the $2$-colour case isn't quite known to be linear (though it is conjectured to be).

To be more specific, the example is the $3$-uniform hypergraph known as the hedgehog. This is the $3$-uniform hypergraph $H_t$ with vertex set $[t + \binom{t}{2}]$ such that for every pair $(i, j)$ with $1 \leq i < j \leq t$ there is a unique vertex $k > t$ for which $ijk$ is an edge. It is known that $R_2(H_t) = O(t^2 \log t)$, which is almost linear in the number of vertices, while $R_4(H_t) \geq 2^{\Omega(t)}$.

For more information, see https://arxiv.org/pdf/1511.00563.pdf and https://arxiv.org/pdf/1902.10221.pdf.

  • $\begingroup$ I just read the fantastically titled "Hedgehogs are not colour blind" paper. This is a very nice example with such a short, simple proof. Based on what you say in the abstract, there won't be a result in the literature which already answers my question (and thus it seems unlikely that my question has an easy answer). $\endgroup$
    – Louis D
    Mar 4, 2021 at 11:55

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