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Let $G = (V, E)$ be a graph and $G^*$ its edge complement (that is, $G^* = (V, E^*)$, where an edge $\{u, v\} \in E^* \Leftrightarrow \{u, v\} \not \in E$).

What is the relationship between a coloring in $G$ and a coloring in $G^*$ ?

I was expecting something like

"If $G$ accepts a $k$-coloring, than $G^*$ accepts a $(n - k)$-coloring"

but I can't prove that.

(Of course, I am dealing with proper coloring)

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    $\begingroup$ The claim is false. Take a complete graph of $5$ nodes and remove any edge, and call that $G$. Then $G$ is $4$-colorable, but certainly $G^*$ is not 1-colorable. $\endgroup$ Sep 13, 2016 at 11:46
  • $\begingroup$ Thanks! What about the chromatic numbers? For the complete graph $K_5$, it is $5$, and for its complement, it is $1$. Maybe there is some relation of the type $X(G) = k \Leftrightarrow X(G^*) = n - k + 1$... What do you think? $\endgroup$ Sep 13, 2016 at 13:20
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    $\begingroup$ @Vitor Take the 4-cycle, which has chromatic number 2. Its complement is a pair of disjoint edges, which has chromatic number $2\neq 4-2+1$. $\endgroup$ Sep 13, 2016 at 13:25

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Sorry, this is false:

For example, take $G$ as a size-$10$ independent set. It has a $1$-coloring, or even a $3$-coloring... its complement is a clique, which admits neither a $9$- nor $7$-coloring.

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  • $\begingroup$ However, I think you might get something like $n-k-2$ where $k$ is the smallest integer such that $G$ is $k$-colorable... I don't have a proof though $\endgroup$
    – tarulen
    Sep 13, 2016 at 11:49
  • $\begingroup$ The 5-cycle has chromatic number 3. Its complement is still a 5-cycle, which does not have a proper colouring with 5-3-2=0 colours. $\endgroup$ Sep 13, 2016 at 13:21
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    $\begingroup$ Well, at least I didn't prove a false statement (: $\endgroup$ Sep 13, 2016 at 13:21
  • $\begingroup$ Right, my conjecture was $n-k+2$, I misstyped there. I don't find a counter-example but I didn't try too much though. $\endgroup$
    – tarulen
    Sep 13, 2016 at 15:13

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