# Clique number of union.

Let $$\omega(G)$$ denotes clique number of graph $$G$$.

Let $$V(G_{1})=V(G_{2})$$

Is there are a formula for $$\omega(G_{1}\cup G_{2})$$ ?

I thought that it may seem like this:

$$\omega(G_{1}\cup G_{2})=\omega(G_{1})+\omega(G_{2})-\omega(G_{1}\cap G_{2})$$.

But i was wrong.

Any ideas how to solve it?

Regards.

There is no formula for it; at least, nothing more straightforward than computing $$\omega(G_1\cup G_2)$$ separately.
Even restricting our attention to cases where $$G_1 \cap G_2$$ is the empty graph, there are two extreme cases (and everything between them is also possible):
• We could have $$G_1$$ and $$G_2$$ have completely unrelated cliques and get $$\omega(G_1 \cup G_2) = \max\{\omega(G_1), \omega(G_2)\}$$.
• On the other hand, we could take a large clique $$K_n$$ and partition it into graphs $$G_1$$ and $$G_2$$ with $$\omega(G_1), \omega(G_2) = O(\log n)$$, by a lower bound on Ramsey numbers. In this case, $$\omega(G_1 \cup G_2)$$ is exponential in $$\omega(G_1)$$ and $$\omega(G_2)$$.