The number of independent sets in a graph with small maximum degree

Details The number of independent sets in a graph with small maximum degree The number of independent sets in a graph with small maximum degree

2010-07-27

arxiv.org/abs/1007.4803v1

Mathematics

Combinatorics

Article will appear in {\em Graphs and Combinatorics}

Scientific paper

Let ${\rm ind}(G)$ be the number of independent sets in a graph $G$. We show that if $G$ has maximum degree at most $5$ then $$ {\rm ind}(G) \leq 2^{{\rm iso}(G)} \prod_{uv \in E(G)} {\rm ind}(K_{d(u),d(v)})^{\frac{1}{d(u)d(v)}} $$ (where $d(\cdot)$ is vertex degree, ${\rm iso}(G)$ is the number of isolated vertices in $G$ and $K_{a,b}$ is the complete bipartite graph with $a$ vertices in one partition class and $b$ in the other), with equality if and only if each connected component of $G$ is either a complete bipartite graph or a single vertex. This bound (for all $G$) was conjectured by Kahn. A corollary of our result is that if $G$ is $d$-regular with $1 \leq d \leq 5$ then $$ {\rm ind}(G) \leq \left(2^{d+1}-1\right)^\frac{|V(G)|}{2d}, $$ with equality if and only if $G$ is a disjoint union of $V(G)/2d$ copies of $K_{d,d}$. This bound (for all $d$) was conjectured by Alon and Kahn and recently proved for all $d$ by the second author, without the characterization of the extreme cases. Our proof involves a reduction to a finite search. For graphs with maximum degree at most $3$ the search could be done by hand, but for the case of maximum degree $4$ or $5$, a computer is needed.

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