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Given a $k$-colorable graph $G$ and vertices $u$ and $v$ of $G$, what is the complexity of deciding if every $k$-coloring of $G$ must assign the same color to both $u$ and $v$?

It does not seem obvious to me how to use the above problem as an oracle to construct a coloring, so it is conceivable that this problem is efficiently decidable. On the other hand, minor variants of NP-hard problems, such as this one, typically remain NP-hard. But most of all, I do not know what problem is a good choice to reduce from.

Any thoughts?

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    $\begingroup$ Add the edge $(u,v)$. Your property holds if and only if $G$ is still $k$-colorable. $\endgroup$ – Yuval Filmus Apr 29 '13 at 23:55
  • $\begingroup$ Ah, very good. Will you make this an answer? $\endgroup$ – Tyson Williams Apr 30 '13 at 3:02
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    $\begingroup$ You should specify that they should take the same color in every k-coloring. Otherwise, the problem is not interesting. $\endgroup$ – Austin Buchanan May 1 '13 at 3:03
  • $\begingroup$ @Austin Yes, this is what I meant. Thanks for the correction. $\endgroup$ – Tyson Williams May 1 '13 at 13:49
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Add the edge $(u,v)$. Your property holds if and only if $G$ is no longer $k$-colorable.

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    $\begingroup$ Let $G$ be obtained from the complete graph with $n$ vertices by deleting the edge $uv$. Then $G$ is $(n-1)$-colourable and every $(n-1)$-colouring of $G$ must assign the same colour to both $u$ and $v$. But $G$ plus the edge $(u,v)$ is not $(n-1)$-colourable. $\endgroup$ – user13136 Apr 30 '13 at 9:45
  • $\begingroup$ There was a mistake in the original answer. The current answer should work. $\endgroup$ – Yuval Filmus May 1 '13 at 20:59
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If you can solve this problem, it implies a coloring algorithm for Uniquely colorable graphs.

Though, I'm not sure the problem of coloring uniquely colorable graphs is NP-hard.

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