For a hypergraph let $\chi$ be the least number of colours needed to colour the vertices, so that in each edge, each colour is used at most once (i.e., the strong chromatic number). Let $\Delta$ be the maximum number of hyperedges containing any vertex. Let $\omega$ be the maximum size of a clique, meaning a vertex set such that for every pair of vertices in the clique, some edge contains both.

Question: is there $\epsilon>0$ so that $\chi \le \Delta \omega / (1+\epsilon)$ in all hypergraphs?

Motive: let $R$ be the maximum edge size. A simple greedy algorithm for colouring can be used to establish that $\chi \le 1 + \Delta(R-1)$, and this bound cannot be improved in general. Note $\chi \ge \omega \ge R$; so I am essentially asking if $\omega$ approximates $\chi$ more closely than $R$.


1 Answer 1


Very nice question, Dave. I had a look at it with Matej Stehlik, here in Grenoble. We found a way to give a positive answer to your question, although it uses more heavy machinery than we would like. You would expect there is an easy argument, but if it exists, we haven't found it yet.

Given a hypergraph $H$, form a graph $G$ by replacing all hyperedges by cliques. Than $\chi(G)=\chi$, $\omega(G)=\omega$ and $\Delta(G)\le\Delta(R-1)$. Bruce Reed, in his paper "$\omega$, $\Delta$ and $\chi$", J. Graph Theory 27 (1998) 177-212, proves that if $a=1/140000000$, then for $\Delta(G)$ large enough, we have $\chi(G)\le a\omega(G)+(1-a)(\Delta(G)+1)$. So for the hypergraph we have $$\chi\le a\omega+(1-a)(\Delta(R-1)+1)\le a\omega+(1-a)\Delta R\le a\omega+(1-a)\Delta\omega=(1-a+a/\Delta)\cdot\Delta\omega.$$ So if $\Delta\ge2$ (and $\Delta(R-1)$ large enough to allow Bruce's theorem), we get $\chi\le(1-a/2)\Delta\omega$.

I would be interested to hear about better bounds or more elementary approaches (Bruce's proof is not for the faint of heart).

  • $\begingroup$ That's great! That totally answers the original question as far as I can see, assuming $\Delta, R \ge 2$ (and I notice now my conjecture was false for $\Delta = 1$), since for bounded $\Delta(R-1)$ we must have that $\Delta$ and $R$ are bounded, in which case $\Delta(R-1)+1$ is indeed at most $\Delta\omega/(1+\epsilon)$. $\endgroup$ Jan 24, 2011 at 19:10
  • $\begingroup$ Just for the record, Bruce and I now have an easier proof of this theorem posted on arXiv, but it still uses a combination of structural and probabilistic arguments. $\endgroup$ Jan 15, 2013 at 13:09

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