The Independence Number of Dense Graphs with Large Odd Girth

Let G be a graph with n vertices and odd girth 2k+3. Let the degree of a vertex v of G be d1(v). Let (G) be the independence number of G. Then we show (G) 2 ( k 1 k ) "X v2G d1(v) 1 k 1 #(k 1)=k . This improves and simpli es results proven by Denley [1]. AMS Subject Classi cation. 05C35 Let G be a graph with n vertices and odd girth 2k + 3. Let di(v) be the number of points of degree i from a vertex v. Let (G) be the independence number of G. We will prove lower bounds for (G) which improve and simplify the results proven by Denley [1]. We will consider rst the case k = 1. We need the following lemma. Lemma 1: Let G be a triangle-free graph. Then (G) X v2G d1(v)=[1 + d1(v) + d2(v)]: Proof. Randomly label the vertices of G with a permutation of the integers from 1 to n. Let A be the set of vertices v such that the minimum label on vertices at distance 0; 1 or 2 from v is on a vertex at distance 1. Clearly the probability that A contains a vertex v is d1(v)=[1+d1(v)+d2(v)]. Hence the expected size of A is X v2G d1(v)=[1+d1(v)+d2(v)]. Furthermore, A must be an independent set since if A contains an edge it is easy to see that it must lie in a triangle of G a contradiction. The result follows at once. Typeset by AMS-TEX 1 the electronic journal of combinatorics 2 (1995), #R5 2 We can now prove the following theorem. Theorem 1. Suppose G contains no 3 or 5 cycles. Let d be the average degree of vertices of G. Then (G) q n d=2: Proof. Since G contains no 3 or 5 cycles, we have (G) d1(v) (consider the neighbors of v) and (G) 1 + d2(v) (consider v and the points at distance 2 from v) for any vertex v of G. Hence (G) X v2G d1(v)=[1+d1(v)+d2(v)] X v2G d1(v)=2 (G) (by lemma 1 and the preceding remark). Therefore (G) n d=2 or (G) p n d2 as claimed. This improves Denley's Theorems 1 and 2. It is sharp for the regular complete bipartite graphs Kaa. The above results are readily extended to graphs of larger odd girth. Lemma 2: Let G have odd girth 2k + 1 or greater (k 2). Then