# Average squared distance in $k$-regular graphs

Let $X=(V,E)$ be a finite, connected, $k$-regular graph. Let $avg(d^2)$ be the averaged square distance between vertices, as defined in Average squared distance vs diameter in vertex-transitive graphs . Is it true that $\sqrt{avg(d^2)}=\Omega(\log(|V|))$? The answer is positive for vertex-transitive graphs. ($\Omega$ is the "Big Omega" Landau notation)

-

## 1 Answer

The number of vertices in the ball of radius $c \log_k(|V|)$ ($c<1$) is small compared to $|V|$, so most pairs of vertices are more than that apart.

-
Thanks! Clearly I should have thought more... – Alain Valette Jun 19 '11 at 7:37