1. The independence number $\alpha(G)$ is the maximum size of an independent set of $G$.

  2. A coloring of $G$ is a partition of the vertices into (induced) independent sets (colors), and the chromatic number $\chi(G)$ is the smallest number of colors possible in a coloring of $G$.

What is the relation between $\chi(\bar{G})$ and $\alpha(G)$ for any graph?, where $\bar G$ is complement of $G$.


closed as off-topic by Math1000, Silvia Ghinassi, Chris Godsil, user296602, Harish Chandra Rajpoot Feb 26 '16 at 8:16

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If you take the complement, an independent set becomes a clique. Hence in $\bar{G}$, there is a $K_{\alpha (G)}$ subgraph.

It is known that you can not color a clique of size $k$ with less than $k$ colors, thus $$ \chi(\bar{G}) \geq \alpha (G). $$

  • $\begingroup$ If $G$ is a circle of $n$ points, $\alpha(G) = n-1$ and $\bar{G}$ is a $K_n$ without a circle. You can color $\bar{G}$ with $n-1$ colors since you can choose the same color for the end vertices of an arbitrary edge of $G$. $\endgroup$ – MCsikos Feb 26 '16 at 0:32
  • $\begingroup$ Unfortunately I didn't think about more examples, where the equality holds. For sure it doesn't hold for any graph since if $\bar{G}$ is an odd cycle, $\alpha (G)=2$ but $\chi (\bar{G}) = 3$. Is the proof and this example clear for you? $\endgroup$ – MCsikos Feb 26 '16 at 0:43
  • 1
    $\begingroup$ In fact equality holds if and only if the clique number of $\bar{G}$ is equal to the chromatic number of $\bar{G}$, maybe you can find some tables online which list these numbers for well known graphs. $\endgroup$ – MCsikos Feb 26 '16 at 0:47