Average squared distance in $k$-regular graphs - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T00:22:06Zhttp://mathoverflow.net/feeds/question/68199http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/68199/average-squared-distance-in-k-regular-graphsAverage squared distance in $k$-regular graphsAlain Valette2011-06-19T06:34:48Z2011-06-19T06:58:20Z
<p>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 <a href="http://mathoverflow.net/questions/67838/average-squared-distance-vs-diameter-in-vertex-transitive-graphs" rel="nofollow">http://mathoverflow.net/questions/67838/average-squared-distance-vs-diameter-in-vertex-transitive-graphs</a> . 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)</p>
http://mathoverflow.net/questions/68199/average-squared-distance-in-k-regular-graphs/68200#68200Answer by Ori Gurel-Gurevich for Average squared distance in $k$-regular graphsOri Gurel-Gurevich2011-06-19T06:58:20Z2011-06-19T06:58:20Z<p>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.</p>