# Inferring Shannon Capacity of pentagon

Can we infer from the fact that the number of independent sets in product of $5$-cycle two times is $5$ and in product of $5$-cycle four times is $25$, that the capacity of pentagon is $\sqrt{5}$? That is does $\alpha(C_5 \boxtimes C_5)=5$ and $\alpha(C_5 \boxtimes C_5\boxtimes C_5 \boxtimes C_5)=25$ imply $\theta(C_5)=\sqrt{5}$? Why? Or why not then?

Here $C_5$ is pengtagon, $\alpha(G)$ is independence number of graph $G$, $\boxtimes$ is strong product and $\theta(G)$ is Shannon capacity of graph $G$.

From the givens you can only deduce that the Shannon capacity is at least $\sqrt{5}$. For example, it could be that $\alpha(C_5^8)^{1/8} > \sqrt{5}$, which doesn't contradict your data.
As an aside, the Shannon capacity is usually denoted by $\Theta$, whereas $\theta$ is the Lovász function, which is an upper bound on the Shannon capacity. From the givens you can only deduce that $\theta(C_5) \geq \sqrt{5}$.
• So you are saying even though $\alpha(C_n^{m_1m_2})=\alpha(C_n^{m_1})^{m_2}$ seems to stabilize, the higher powers may satisfy $\alpha(C_n^{m_1m_2m_3})>\alpha(C_n^{m_1})^{m_2m_3}$? – Turbo Nov 9 '14 at 6:18
• We know that the capacity of the pentagon is $\sqrt{5}$, since we know that $\theta(C_5) = \sqrt{5}$ (here $\theta$ is the Lovász function). Regarding the behavior of powers, I recall someone mentioning that it could stabilize and then grow at an arbitrary power, but that is a separate question you might want to ask. – Yuval Filmus Nov 9 '14 at 16:30