On Selkow's Bound on the Independence Number of Graphs

TL;DR

This paper provides a new probabilistic proof for Selkow's lower bound on the independence number of graphs, correcting the original proof which was found to be incorrect.

Contribution

It introduces a valid probabilistic proof for Selkow's bound, enhancing the theoretical understanding of graph independence numbers.

Findings

Established a correct probabilistic proof of Selkow's bound

Confirmed the validity of the lower bound on independence number

Clarified the conditions under which the bound holds

Abstract

For a graph $G$ with vertex set $V(G)$ and independence number $\alpha(G)$, S. M. Selkow (Discrete Mathematics, 132(1994)363--365) established the famous lower bound $\sum\limits_{v\in V(G)}\frac{1}{d(v)+1}(1+\max\{\frac{d(v)}{d(v)+1}-\sum\limits_{u\in N(v)}\frac{1}{d(u)+1},0 \})$ on $\alpha(G)$, where $N(v)$ and $d(v)=|N(v)|$ denote the neighborhood and the degree of a vertex $v\in V(G)$, respectively. However, Selkow's original proof of this result is incorrect. We give a new probabilistic proof of Selkow's bound here.

Full text

(Discrete Mathematics, 132(1994)363–365) established the famous lower bound

$\sum\limits_{v\in V(G)}\frac{1}{d(v)+1}(1+\max\{\frac{d(v)}{d(v)+1}-\sum\limits_{u\in N(v)}\frac{1}{d(u)+1},0\})$ on $\alpha(G)$, where $N(v)$ and

$d(v)=|N(v)|$ denote the neighborhood and the degree of a vertex $v\in V(G)$, respectively. However, Selkow’s original proof of this result is incorrect. We give a new probabilistic proof of Selkow’s bound here.

On Selkow’s Bound on the Independence Number of Graphs

Jochen Harant, Samuel Mohr

Ilmenau University of Technology, Department of Mathematics, Germany

Keywords: Graph, Independence Number

We consider a finite, simple, and undirected graph $G$ with vertex set $V(G)$. Let $N_{G}(v)$ and $d_{G}(v)=|N_{G}(v)|$ denote the neighborhood and the degree of $v\in V(G)$, respectively. A set of vertices $I\subseteq V(G)$ is independent if no two vertices in $I$ are adjacent. The independence number $\alpha(G)$ of $G$ is the maximum cardinality of an independent set of $G$.

The independence number is one of the most fundamental and well-studied graph parameters. In view of its computational hardness, various bounds on the independence number have been proposed. The classical lower bound $CW(G)=\sum\limits_{v\in V(G)}\frac{1}{d_{G}(v)+1}$ on $\alpha(G)$ is due to Y. Caro [2] and V. K. Wei [4]. It is natural to ask whether improvements of $\alpha(G)\geq CW(G)$ are possible if more information about $G$ is known than just its degrees. The following result takes not only the degree of every vertex but also the degree distribution in its neighborhood into account.

Unfortunately, Selkow’s original proof of Theorem 1 is not correct. To our best knowledge, this has not been discovered earlier, and we are not aware of an alternative, correct proof. In order to extract the problematic part of Selkow’s argument, let us repeat his proof:

For an event $A$ and a random variable $X$ let $P(A)$ and $E(X)$ denote be the probability of $A$ and the expectation of $X$, respectively.

First, a uniformly chosen ordering $<$ of $V(G)$ is considered.

Obviously, the set $I_{1}=\{v\in V(G)~{}|~{}u\in N_{G}(v)\Rightarrow v<u\}$ is independent and it is easy to show that $E(|I_{1}|)=CW(G)$ (e. g. see [1]).

Next, let the graph $H$ (depending on the ordering $<$) be obtained from $G$ by removing $I_{1}\cup\bigcup\limits_{x\in I_{1}}N_{G}(x)$ and consider the set $I_{2}=\{v\in V(H)~{}|~{}u\in N_{H}(v)\Rightarrow v<u\}$.

It follows that $\alpha(G)\geq E(|I_{1}|+|I_{2}|)=E(|I_{1}|)+E(|I_{2}|)=CW(G)+\sum\limits_{v\in V(G)}P(v\in I_{2})$, since $I_{1}\cap I_{2}=\emptyset$ and $I_{1}\cup I_{2}$ is an independent set of $G$. To finish the proof of Theorem 1 in [3], the inequality $P(v\in I_{2})\geq\frac{1}{d_{G}(v)+1}\max\{\frac{d_{G}(v)}{d_{G}(v)+1}-\sum\limits_{u\in N_{G}(v)}\frac{1}{d_{G}(u)+1},0\}$ for all $v\in V(G)$ ([3], page 364, lines 14–17) is used. This turns out to be false in the following general sense.

*For every $\varepsilon>0$, there is a graph $G$ and a vertex $v\in V(G)$, such that

$0<P(v\in I_{2})<\varepsilon\cdot\frac{1}{d_{G}(v)+1}(\frac{d_{G}(v)}{d_{G}(v)+1}-\sum\limits_{u\in N_{G}(v)}\frac{1}{d_{G}(u)+1})$.

*To see this, let $n$ be a large positive integer. Consider an arbitrary graph $F$ on $n-3$ vertices and let the graph $G$ on $n$ vertices be obtained by adding three new vertices $v,w,x$ and the edges $vw$, $wx$, and $xy$ for all $y\in V(F)$. For an arbitrary ordering $<$ of $V(G)$, $v\in I_{2}$ if and only if $x\in I_{1}$, $x<w<v$, and $x<y$ for all $y\in V(F)$. It is easy to see that there are exactly ${n-1\choose 2}(n-3)!$ such orderings with the property $v\in I_{2}$, thus, $P(v\in I_{2})=\frac{{n-1\choose 2}(n-3)!}{n!}=\frac{1}{2n}$, however, $\frac{1}{d_{G}(v)+1}(\frac{d_{G}(v)}{d_{G}(v)+1}-\sum\limits_{u\in N_{G}(v)}\frac{1}{d_{G}(u)+1})=\frac{1}{2}(\frac{1}{2}-\frac{1}{3})=\frac{1}{12}$.

Eventually, we present a new probabilistic proof of Theorem 1.

Proof of Theorem 1. As in Selkow’s proof, consider a uniformly chosen ordering $<$ of $V(G)$, the set $I_{1}=\{v\in V(G)~{}|~{}u\in N_{G}(v)\Rightarrow v<u\}$, and the graph $H$ induced by $V(G)\setminus(I_{1}\cup\bigcup\limits_{x\in I_{1}}N_{G}(x))$.

With $f(v)=0$ if $v\notin V(H)$ and $f(v)=\frac{1}{d_{G}(v)+1}$ if $v\in V(H)$, it follows

$\alpha(G)\geq E(|I_{1}|+\alpha(H))=E(|I_{1}|)+E(\alpha(H))\geq E(|I_{1}|)+E(CW(H))=CW(G)+E(\sum\limits_{v\in V(H)}\frac{1}{d_{H}(v)+1})$

$\geq CW(G)+E(\sum\limits_{v\in V(H)}\frac{1}{d_{G}(v)+1})=CW(G)+E(\sum\limits_{v\in V(G)}f(v))$

$=CW(G)+\sum\limits_{v\in V(G)}\frac{1}{d_{G}(v)+1}P(v\in V(H))$.

Using $P(v\in V(H))\geq 0$ and $P(v\notin V(H))=P(v\in I_{1}\vee(\bigvee\limits_{u\in N(v)}u\in I_{1}))$

$\leq P(v\in I_{1})+\sum\limits_{u\in N(v)}P(u\in I_{1})=\frac{1}{d_{G}(v)+1}+\sum\limits_{u\in N(v)}\frac{1}{d_{G}(u)+1}$, Theorem 1 is proved. $\Box$