Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G = (V, E)$ is a connected graph $\Rightarrow$ $\alpha(G) \le \frac{|E|}{\delta(G)}$, where $\alpha(G)$ is the vertex independence number of $G$ and $\delta(G)$ − minimum degree of all vertices.

Give some clue please!

Thanks anyway!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

This is a rather easy application of the hand-shaking lemma.

Separate the graph $G$ into independent vertex sets. Suppose $S$ is the largest such set. Then since $G$ is connected, each vertex of $S$ must be connected to another, but no vertices of $S$ are adjacent to each other since the set is by assumption independent. Consider the set $T = S\cup N(S)$, the set of vertices of $S$ with all their connected neighbors $N(S)$. If we apply the handshaking lemma to $T$ $$2|E(G)|\ge2|E(T)| = \sum_{v\in V(T)}\deg(v)=2\sum_{v\in V(S)}\deg(v) \ge 2\alpha\delta$$ The second equality is given as follows. For each degree we introduce in the vertices of $S$, we necessarily introduce one degree in a neighbor of $S$ since the vertices of $S$ are non-adjacent. Therefore the net degree of $S$ and $N(S)$ must be equal.

share|improve this answer
    
Simply and clean. Thank you for your help. I learn from you a lot new. –  jofisher Oct 4 '12 at 17:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.