# Domination Number equals Independence Number?

Let $\gamma(G)$ and $\alpha(G)$ be the domination number and independence number for an Graph $G$. Further, let $i(G)$ be the minimum-size Independent Dominating Set. Then, it is known that $\forall G, \gamma(G) \leq i(G) \leq \alpha(G)$.

According to this paper, it is also known that $\gamma(G) = i(G)$ for Claw-free Graphs and $i(G) = \alpha(G)$ for Block Graphs.

What is the most general family of graphs for which $\gamma(G)=i(G)=\alpha(G)$? Further, what is the most general family of graphs for which $\frac{\alpha(G)}{c}\leq\gamma(G)\leq \alpha(G)$ for a constant c >1?

• What do you mean by "the most general family"? A characterisation like this: $G$ has $\gamma(G) = i(G)$ if and only if $G$ has some property $\Phi$? – Dominic van der Zypen Oct 25 '17 at 9:20
• yes, that is correct. – Ainesh Bakshi Oct 26 '17 at 3:54

Since $$\gamma(G)\leq i(G)\leq \alpha(G)$$ for all graphs $$G$$, the problem of characterizing all graphs $$G$$ for which $$\gamma(G)=i(G)=\alpha(G)$$ is equivalent to the characterizing all graphs $$G$$ for which $$\gamma(G)=\alpha(G)$$. The theorem "$$\gamma(G)=i(G)$$ holds for all claw-free graphs $$G$$" was proved by Bollobas and Cockayne in
However, the problem of finding the family of all graphs $$G$$ for which $$\gamma(G)=\alpha(G)$$ is still open.