Is it NP complete to decide whether a graph has bigger clique or bigger independence number?

Question

Is it $\mathsf{NP}$ complete to decide whether a graph $G$ has bigger clique $\omega(G)$ or bigger independence number $\alpha(G)$?

Answer 1

Note that there is a slight technicality with regards to whether you consider the graph where the clique is larger a "yes"-instance, or a graph where the independent set is larger a "yes"-instance (i.e. the complement problem) and what you do in the event of a tie. I have picked one particular way of settling ties, but this proof can easily be adapted to the other way of settling ties (i.e. by fudging with a $\pm 1$).

If $NP\not = coNP$ then neither this problem nor its complement is in $NP$, so neither problem is likely to be $NP$-complete.

It is $NP$-complete to decide whether a graph has an independent set of size $k$. We show that, if there exist polynomial certificates for your problem then there exist polynomial certificates for the complement of the independent set problem. We note that IS remains $NP$-complete even when restricted to maximum degree $3$ graphs and $k>4$.

Given a graph $G$ that does not have an independent set of size $k$, we create a new graph $G'$ as the union of $G$ and a clique of size $k$ (note that this increases the independence number by one).

$G'$ is a "yes"-instance of the problem "Is the Clique greater or equal to the IS?". A polynomial certificate for this fact would give a polynomial certificate testifying to the fact that $G$ does not have an IS of size $k$, which is impossible unless $NP=coNP$. (The largest clique in $G$ is of size $4$ since $G$ has maximum degree $3$, so the largest clique in $G'$ is of size $k$. The certificate for $G'$ thus has to certify that $G'$ has no IS of size $k+1$ and equivalently, that $G$ has no IS of size $k$).

If we instead consider the complement problem "Is the Clique strictly smaller than the IS?", we create $G'$ by adding a clique of size $k+1$ and then taking the complement graph $\bar{G'}$. Since a Clique becomes an IS in the complement graph and vice-versa, the largest IS in $\bar{G'}$ is of size $k+1$ while the largest Clique is of size at most $k$. $\bar{G'}$ is thus a "yes"-instance of the complement problem, but a certificate for this fact would give a certificate for $G$ not having an IS of size $k$ which is impossible unless $NP=coNP$.

However, the problem "Is the Clique strictly smaller than the IS?" is $NP$-hard. We reduce from Independent Set on a maximum degree $3$ graphs $G$ where $k>4$. We create $G'$ by adding a clique of size $k$ to $G$. If $G$ has an IS of size $k$ then $G'$ has one of size $k+1$ (obtained by adding any of the clique's vertices to the IS). The largest clique in $G'$ has size $k$, so $G'$ is a "yes"-instance if and only if $G$ has an Independent Set of size $k$.