The value of the Ramsey number R(Cn, K4) is 3(n-1)+1 (n≥4)

Abstract

The Ramsey number R(Cn, Km) is the smallest integer p such that any graph G on p vertices either contains a cycle Cn with length n or contains an independent set with order m. In this paper we prove that R(Cn , K 4 ) = 3(n 1) + 1 (n ~ 4). We shall only consider graphs without multiple edges or loops. The Ramsey number R(Cn,Km) is the smallest integer p such that any graph G on p vertices either contains a cycle Cn with length n or contains an independent set with order m. In 1973, J.A. Bondy and P. Erdos proved that following theorem. Theorem 1.1 ([1]). R(Cn , Km) = (n l)(m 1) + 1 for n ~ m 2 2. In 1976, R.H. Schelp and R.J. Faudree in [8] stated the following problem: Problem 1.2 ([8]). Find the range of integers nand m such that R(Cn, Km) = (n 1) (m 1) + 1. In particular, does the equality hold for n ~ m? For this problem, the known results are R(C4 , K 4 ) = 10 (see [2]), R(C5 , K 4 ) = 13, R(C5 ,K5 ) = 17 (see [4], [5]) and R(Cn,K3 ) = 2n 1 (n > 3) (see [3], [6]). However, so far, even for some fixed small m, the problem has not been solved. In the following, we will prove that R(Cn, K 4 ) = 3(n 1) + 1 for n ~ 4. Lemma 1.3. Suppose G is a graph that contains the cycle (VI, V2,'" ,Vn-I) of length n 1 but no cycle of length n. Let X ~ V(G) \ {VI, V2,'" ,Vn-l}. Then (a) No vertex x E X is adjacent to two consecutive vertices on the cycle. (b) If x E X is adjacent to Vi and Vj, then Vi+1 Vj+1 rf E( G). (c) If x E X is adjacent to Vi and Vj, then no vertex x' E X is adjacent to both Vi+1 and Vj+2· * The project supported by NSFC and NSF JS Australasian Journal of Combinatorics 20(1999), pp.205-206 Proof. (a) and (b) were used in [1]. If x is adj acent to Vi and v j and x' is adj acent to Vi+l and Vj+2, then x i= x' and (Vi, x, Vj, Vj-b'" ,Vi+l, x', Vj+2,'" ,vi-d is a cycle of length n in G; this proves (c). 0 Theorem 1.4. For all n ~ 4, R(Cn , K 4 ) = 3(n 1) + 1. Proof. The example G = 3Kn 1 establishes the lower bound R(Cn , K 4 ) ~ 3(n 1) + 1, so it suffices to prove that for n ~ 4 every graph G of order 3(n 1) + 1 contains either Cn or a 4-element independent set. Since the desired result is true for n = 4 and n = 5, we may take n > 5 and assume by induction that R(Cn 1 , K 4 ) = 3(n 2) + 1. Assume that G(V, E) is a graph of order 3(n 1) + 1 that contains neither a Cn nor a 4-element independent set. Using R(Cn , K 3 ) = 2(n 1) + 1 and R(Cn 1 , K 4 ) = 3(n 2) + 1, we find that G contains a 3-element independent set X = {XI,X2,X3}, and, disjoint from X, a cycle (VI,V2,'" ,vn-d of length n 1. Let us refer to (VI, V2, ... ,Vn-l) as simply the cycle. Since G has no 4-element independent set, each vertex on the cycle is adjacent to at least one vertex in X. Since n 1 > 3 at least one vertex in X is adjacent to two or more vertices of the cycle. Thus we may assume that Xl is adjacent to Vi and Vj. By part (b) of Lemma 1.3 Vi+l Vj+l (j; E. Since n > 5 and Xl cannot be adjacent to three or more vertices of the cycle by part (a) and (b) of Lemma 1.3, thus XIVj+2 rJ. E. By part (a) of Lemma 1.3, XIVi+1 (j; E and XIVj+1 (j; E. Since Vj+2 is adjacent to some vertex in X, we may assume that X2Vj+2 E E. By par.t (c) of Lemma 1.3 X2Vi+1 (j; E and by part (a) of Lemma 1.3 X2Vj+1 (j; E. Thus {Xl, X2, Vi+l, Vj+d is a 4-element independent set, a contradiction. 0

Cite this paper

@article{Yang1999TheVO, title={The value of the Ramsey number R(Cn, K4) is 3(n-1)+1 (n≥4)}, author={Jian Sheng Yang and Yi Ru Huang and Kemin Zhang}, journal={Australasian J. Combinatorics}, year={1999}, volume={20}, pages={205-206} }