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Apr 13, 2017 at 12:19 history edited CommunityBot
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Jan 29, 2015 at 15:18 vote accept Dominic van der Zypen
Jan 29, 2015 at 15:16 answer added Emil Jeřábek timeline score: 4
Jan 29, 2015 at 14:07 comment added Dominic van der Zypen Thank you very much - can you post this as an answer so that I can accept it and we can close this thread?
Jan 29, 2015 at 13:13 comment added Emil Jeřábek The linked argument actually shows that $\chi(G)\le1+\chi(C)$, where $C$ is the subgraph induced by a minimal vertex cover. So, $\tau(G)\ge\chi(G)$ unless $C$ is a clique. For much the same reason, $\tau(G)\ge\chi(G)$ unless one vertex from $V-C$ is connected to everyone in $C$. So, no, there isn’t such a graph.
Jan 29, 2015 at 12:54 history asked Dominic van der Zypen CC BY-SA 3.0