Báo cáo toán học: "The Bipartite Ramsey Numbers b(C2m; K2,2)" pot

The Bipartite Ramsey Numbers b(C 2m ; K 2,2 ) Zhang Rui Sun Yongqi ∗ Sch ool of Computer and Information Technology Beijing Jiaotong University, Beijing, P. R. China yqsun@bjtu.edu.cn Submitted: Nov 26, 2010; Accepted: Feb 13, 2011; Published: Feb 28, 2011 Mathematics Subject Classifications: 05C55, 05C38 Abstract Given bipartite graphs H 1 and H 2 , the bipartite Ramsey number b(H 1 ; H 2 ) is the smallest integer b such that any subgraph G of the complete bipartite graph K b,b , either G contains a copy of H 1 or its complement relative to K b,b contains a copy of H 2 . It is known that b(K 2,2 ; K 2,2 ) = 5, b(K 2,3 ; K 2,3 ) = 9, b(K 2,4 ; K 2,4 ) = 14 and b(K 3,3 ; K 3,3 ) = 17. In this paper we study the case H 1 being even cycles and H 2 being K 2,2 , prove that b(C 6 ; K 2,2 ) = 5 and b(C 2m ; K 2,2 ) = m + 1 for m ≥ 4. Keywords: bipartite graph; Ramsey number; even cycle 1 Introduction We consider only finite undirected gra phs without loops or multiple edges. For a graph G with vertex-set V (G) and edge-set E(G), we denote the order and the size of G by p(G) = |V (G)| and q(G) = |E(G)|. δ(G) and ∆(G) are the minimum degree and the maximum degree o f G respectively. Let K m,n be a complete m by n bipartite graph, that is, K m,n consists of m+n vertices, partitioned into sets of size m and n, and the mn edges between them. P k is a pa th on k vertices, and C k is a cycle of length k. Let H 1 and H 2 be bipartite gra phs, the bipartite Ramsey number b(H 1 ; H 2 ) is the smallest integer b such that given any subgraph G of the complete bipartite graph K b,b , either G contains a copy of H 1 or there exists a copy of H 2 in the complement of G relative to K b,b . Obviously, we have b(H 1 ; H 2 ) = b(H 2 ; H 1 ). Beineke and Schwenk [1] showed that b(K 2,2 ; K 2,2 ) = 5, b(K 2,4 ; K 2,4 ) = 13, b(K 3,3 ; K 3,3 ) = 17. ∗ This resear ch was supported by NSFC(609730 11, 60803034) and SRFDF(20090009120007, 200801081017). the electronic journal of combinatorics 18 (2011), #P51 1 In particular, they proved that b(K 2,n ; K 2,n ) = 4n − 3 for n odd and less than 100 except possibly n = 59 or n = 95. Carnielli and Carmelo [2] proved that b(K 2,n ; K 2,n ) = 4n −3 if 4n−3 is a prime power. They also showed that b(K 2,2 ; K 1,n ) = n+q for q 2 −q+1 ≤ n ≤ q 2 , where q is a prime power. Irving [6] showed that b(K 4,4 ; K 4,4 ) ≤ 48. Hattingh and Henning [4] proved that b(K 2,2 ; K 3,3 ) = 9, b(K 2,2 ; K 4,4 ) = 14. They also determined the values of b(P m ; K 1,n ) in [5]. Faudree and Schelp [3] proved the values of b(H 1 ; H 2 ) when both H 1 and H 2 are two paths. Let G i be t he subgraph of G whose edges are in the i-th color in an r-coloring of the edges of G . If there exists an r-coloring of the edges of G such that H i ⊆ G i for all 1 ≤ i ≤ r, then G is said to b e r-colorable to (H 1 , H 2 , . . . , H r ). The neighborhood of a vertex v ∈ V (G) are denoted by N(v) = {u ∈ V (G)|uv ∈ E(G)}, a nd let d(v) = |N(v)|. G c denotes the complement of G relative to K b,b . GW  denotes the subgraph of G induced by W ⊆ V (G). In this paper we study the case that H 1 being even cycles and H 2 being K 2,2 , prove that b(C 6 ; K 2,2 ) = 5 and b(C 2m ; K 2,2 ) = m + 1 for m ≥ 4. For the sake of convenience, let V (K m,n ) = X ∪ Y , where X = {x i |1 ≤ i ≤ m} and Y = {y j |1 ≤ j ≤ n}, and E(K m,n ) = {x i y j |1 ≤ i ≤ m, 1 ≤ j ≤ n}. 2 The lower bounds of b(C 2m ; K 2,2 ) Theorem 1. b(C 2m ; C 2n ) ≥ m + n − 1. Proof. Let G 1 and G 2 be the subgraphs of K m+n−2,m+n−2 , where G 1 is a complete m − 1 by m + n − 2 bipartite graph, and G 2 is a complete n − 1 by m + n − 2 bipartite graph. And let V (G 1 ) = X 1 ∪ Y, where X 1 = {x i |1 ≤ i ≤ m − 1}, Y = {y i |1 ≤ i ≤ m + n − 2}; V (G 2 ) = X 2 ∪ Y, where X 2 = {x i |m ≤ i ≤ m + n − 2}, Y = {y i |1 ≤ i ≤ m + n − 2}. Then we have E(G 1 ) ∩ E(G 2 ) = ∅ and E(G 1 ) ∪ E(G 2 ) = E(K m+n−2,m+n−2 ). Note that C 2m ⊆ G 1 and C 2n ⊆ G 2 . So K m+n−2,m+n−2 is 2-colorable to (C 2m , C 2n ), that is, b(C 2m ; C 2n ) ≥ m + n − 1. ✷ Setting n = 2 in Theorem 1, we have Corollary 1. b(C 2m ; K 2,2 ) ≥ m + 1. 3 The upper bounds of b(C 2m ; K 2,2 )(m ≥ 3) Lemma 1. b(C 6 ; K 2,2 ) ≤ 5. Proof. We may assume that b(C 6 ; K 2,2 ) > 5, that is, K 5,5 is 2-colorable to (C 6 , K 2,2 ). Since K 2,2  G c and b(K 2,2 ; K 2,2 ) = 5, we have K 2,2 ⊆ G. Without loss of generality, we may assume {x 1 y 1 , y 1 x 2 , x 2 y 2 , y 2 x 1 } ⊆ E(G). the electronic journal of combinatorics 18 (2011), #P51 2 Since K 2,2  G c , there is a t least one edg e between {x 3 , x 4 } and {y 3 , y 4 }, say x 3 y 3 ∈ E(G). Similarly, there is at least one edge between {x 4 , x 5 } and {y 4 , y 5 }, say x 4 y 4 ∈ E(G). And there is at least one edge between {x 1 , x 2 } and {y 3 , y 4 }, say x 1 y 4 ∈ E(G). Since C 6  G, x 4 is nonadjacent to any vertex of {y 1 , y 2 }. Therefore since K 2,2  G c , x 3 has to be adjacent to one vertex of {y 1 , y 2 }, say x 3 y 2 ∈ E(G). x 4 is nonadjacent to any vertex if {y 1 , y 2 , y 3 }, since otherwise we have C 6 ⊆ G. And since K 2,2  G c , x 5 is adjacent to at least two vertices of {y 1 , y 2 , y 3 }. If x 5 is adjacent to both y 1 and y 3 , then we have C 6 ⊆ G, a contradiction. Hence we have x 5 y 1 , x 5 y 2 ∈ E(G) o r x 5 y 2 , x 5 y 3 ∈ E(G). Case 1. Suppose that x 5 y 1 , x 5 y 2 ∈ E(G), see Fig. 1(a). Since C 6  G, x 5 is nonadja cent to y 3 or y 4 . Therefore since K 2,2  G c , x 2 has to be adjacent t o at least one vertex of {y 3 , y 4 }. In any case, we have C 6 ⊆ G, a contradiction. Case 2. Suppose that x 5 y 2 , x 5 y 3 ∈ E(G), see Fig. 1(b). Since C 6  G, x 5 is nonadja cent to y 1 or y 4 . Therefore since K 2,2  G c , x 3 has to be adjacent t o at least one vertex of {y 1 , y 4 }. In any case, we have C 6 ⊆ G, a contradiction too. x 1 x 2 x 3 x 4 x 5 y 1 y 2 y 3 y 4 x 1 x 2 x 3 x 4 x 5 y 1 y 2 y 3 y 4 (a) (b) Fig. 1. The two cases of N(x 5 ) By Case 1 and 2, the assumption that b( C 6 ; K 2,2 ) > 5 does not hold. Then we have the lemma follows. ✷ Lemma 2. Let G be a spanning subgraph of K 5,5 and C 8  G. If K 2,2  G c , then there exists at most one vertex of X(or Y ) whose degrees is at most 2. Proof. Fo r 1 ≤ i, j ≤ 5, if |N(x i ) ∪ N(x j )| ≤ 3(i = j), then t here are at least two vertices of Y are nonadjacent to x i or x j , we have K 2,2 ⊆ G c . Hence we have Claim 1. |N(x i ) ∪ N(x j )| ≥ 4. By way of contradiction, we assume that there exists at least two vertices of X whose degrees are at most 2, say x 1 and x 2 . By Claim 1, we have |N(x 1 ) ∪ N(x 2 )| = 4. We may assume N(x 1 ) = {y 1 , y 2 } and N(x 2 ) = {y 3 , y 4 }. There are two subcases depending on N(y 5 ). Case 1. Suppose that there is at least one vertex of {x 3 , x 4 , x 5 }, say x 3 which is nonadja- cent to y 5 . By Claim 1, we have |N(x 1 ) ∪ N(x 3 )| ≥ 4, x 3 has to be adjacent to both y 3 and y 4 . Similarly we have |N (x 2 ) ∪ N(x 3 )| ≥ 4, x 3 has to be adjacent to both y 1 and y 2 . By Claim 1, we have |N(x 1 ) ∪ N(x 4 )| ≥ 4, x 4 has to be adjacent to a t lease one vertex of {y 3 , y 4 }, say x 4 y 3 ∈ E(G) as shown in F ig. 2(a). Since C 8  G, x 4 is nonadjacent to y 1 the electronic journal of combinatorics 18 (2011), #P51 3 or y 2 . Hence we have |N(x 2 ) ∪ N(x 4 )| ≤ 3, a contradiction to Claim 1. x 1 x 3 x 4 x 2 y 2 y 1 y 4 y 3 x 1 x 5 x 4 x 3 x 2 y 2 y 1 y 5 y 3 y 4 (a) (b) Fig. 2. The two cases of N(y 5 ) Case 2. Suppose that each vertex of {x 3 , x 4 , x 5 } is adjacent to y 5 . By Claim 1, we have |N(x 1 ) ∪ N (x 3 )| ≥ 4, x 3 has to be adjacent to at lease one vertex of {y 3 , y 4 }, say x 3 y 3 ∈ E(G). Similarly, we have |N(x 2 )∪N(x 3 )| ≥ 4, x 3 has to be adjacent to at lease one vertex of {y 1 , y 2 }, say x 3 y 1 ∈ E(G). Since K 2,2  G c , there is at least one edge between {x 4 , x 5 } and {y 2 , y 4 }, say x 4 y 2 ∈ E(G). Since C 8  G, x 4 is nonadjacent to y 4 . By Cla im 1, we have |N(x 1 ) ∪ N(x 4 )| ≥ 4, x 4 has to be adjacent to y 3 as shown in Fig. 2(b). Since C 8  G, x 5 is nonadjacent to y 1 or y 2 . Hence we have |N(x 2 )∪N(x 5 )| ≤ 3, a contradiction to Claim 1. By Case 1 and 2, the assumption does not hold. Then we have the lemma follows. ✷ Lemma 3. b(C 8 ; K 2,2 ) ≤ 5. Proof. We may assume that b(C 8 ; K 2,2 ) > 5, that is, K 5,5 is 2- color able to (C 8 , K 2,2 ), say C 8  G and K 2,2  G c . Since K 2,2  G c and b(C 6 ; K 2,2 ) ≤ 5, we have C 6 ⊆ G. Without loss of generality, we may assume {x 1 y 1 , y 1 x 2 , x 2 y 2 , y 2 x 3 , x 3 y 3 , y 3 x 1 } ⊆ E(G). x 4 y 4 x 5 y 5 x 1 x 2 x 3 y 1 y 2 y 3 Fig. 3. No edge between {x 4 , y 4 } and V (C 6 ) Since K 2,2  G c , there is a t least one edg e between {x 4 , x 5 } and {y 4 , y 5 }, say x 4 y 4 ∈ E(G). Assume that x 4 is nonadjacent to any vertex of {y 1 , y 2 , y 3 } and y 4 is nonadjacent to any vertex of {x 1 , x 2 , x 3 }. Then we have d(x 4 ) ≤ 2 and d(y 4 ) ≤ 2. Since K 2,2  G c , x 5 has to be adjacent to at least two vertices of {y 1 , y 2 , y 3 }, say x 5 y 1 , x 5 y 2 ∈ E(G). Similarly, y 5 has to be adjacent to at least two vertices of { x 1 , x 2 , x 3 }. By symmetry, we may assume that y 5 x 1 , y 5 x 2 ∈ E(G) or y 5 x 1 , y 5 x 3 ∈ E(G). If y 5 x 1 , y 5 x 2 ∈ E(G), then C 8 ⊆ G, a contradiction. Hence we have y 5 x 1 , y 5 x 3 ∈ E(G), as shown in F ig . 3. Since C 8  G, x 2 is the electronic journal of combinatorics 18 (2011), #P51 4 nonadjacent to y 3 or y 5 . So we have d(x 2 ) = 2, a contradiction to Lemma 2. Hence x 4 is adjacent to at least one vertex of {y 1 , y 2 , y 3 } or y 4 is adjacent to at least one vertex of {x 1 , x 2 , x 3 }, say x 4 y 3 ∈ E(G). Since C 8  G, y 4 is nonadjacent to x 1 or x 3 . Therefore since K 2,2  G c , y 5 has to be adjacent to at least one vertex of {x 1 , x 3 }, say y 5 x 1 ∈ E(G). Now we consider the vertex of x 5 , there are three subcases. x 4 y 4 x 5 x 1 x 2 x 3 y 1 y 2 y 3 y 5 Fig. 4. x 5 being nonadjacent to y 4 or y 5 Case 1. Suppose that x 5 is nonadjacent to any vertex of {y 4 , y 5 }. Since C 8  G, y 4 is nonadjacent t o any vertex of {x 1 , x 3 }. Hence we have d(y 4 ) ≤ 2. By Lemma 2, we have d(y 5 ) ≥ 3. Therefore since C 8  G, y 5 has to be adjacent to both x 2 and x 3 as shown in F ig . 4. Since C 8  G, x 4 is nonadjacent to any vertex of {y 1 , y 2 , y 5 }. Hence we have d(x 4 ) = 2. By Lemma 2, we have d(x 5 ) ≥ 3. Hence x 5 has to be adjacent to each vertex of {y 1 , y 2 , y 3 }, we have C 8 ⊆ G, a contradiction. Case 2. Suppose that x 5 is adjacent to just one vertex of {y 4 , y 5 }, that is, x 5 y 4 ∈ E(G) or x 5 y 5 ∈ E(G). Suppose x 5 y 4 ∈ E(G), then x 5 y 5 ∈ E(G). Since C 8  G, we have x 4 y 5 ∈ E(G). Since K 2,2  G c , y 1 is adjacent to at least one vertex of {x 4 , x 5 }. Therefore since C 8  G, y 1 has to be adjacent to x 4 . Similarly, y 2 and y 3 have to be a djacent to x 4 , see Fig. 5(a). Since C 8  G, y 5 is nonadja cent to any vertex of {x 2 , x 3 }. Hence we have d(y 5 ) = 1. By Lemma 2, we have d(y 4 ) ≥ 3. Hence y 4 has to be adjacent to at least one vertex of {x 1 , x 2 , x 3 }. In any case, we have C 8 ⊆ G, a contradiction. Suppose that x 5 y 5 ∈ E(G). Since C 8  G, x 5 is nonadjacent to any vertex of {y 1 , y 3 }. Hence we have d(x 5 ) ≤ 2. By Lemma 2, we have d(x 3 ) ≥ 3. Since C 8  G, x 3 is nonadjacent to y 4 . x 3 has to be adjacent to y 1 , since otherwise K 2,2 ⊆ G c {x 3 , x 5 , y 1 , y 4 }, see Fig. 5(b). By Lemma 2, we have d(x 4 ) ≥ 3. Hence x 4 is adjacent to at least one vertex of {y 1 , y 2 }. In any case, since C 8  G, y 5 is nonadjacent to x 2 or x 3 . Hence we have d(y 5 ) = 2, a contradiction to Lemma 2. Case 3. Suppose that x 5 is adjacent to each vertex of {y 4 , y 5 }, as shown in Fig. 6. Since C 8  G, y 4 is nonadja cent to any vertex o f {x 1 , x 2 , x 3 }. Hence d(y 4 ) = 2. By Lemma 2, we have d(y 2 ) ≥ 3. So y 2 is adjacent to at least one vertex of {x 1 , x 4 , x 5 }. In any case, we have C 8 ⊆ G, a contradiction. By Case 1-3, the assumption does not hold. Then we have the lemma follows. ✷ the electronic journal of combinatorics 18 (2011), #P51 5 x 1 x 2 x 3 y 1 y 2 y 3 x 4 x 5 y 5 y 4 y 5 x 5 x 4 y 4 x 1 x 2 x 3 y 1 y 2 y 3 (a) (b) Fig. 5. x 5 being adjacent to just one of {y 4 , y 5 } y 5 x 5 x 4 y 4 x 1 x 2 x 3 y 1 y 2 y 3 Fig. 6. x 5 being adjacent to y 4 and y 5 Lemma 4. Let G be a spanning subgraph of K k+1,k+1 such that C 2k ⊆ G and x k+1 , y k+1 ∈ V (C 2k ). If x k+1 and y k+1 are adjacent to at least k − 1 vertices of V (C 2k ) respectively, then we have C 2(k+1) ⊆ G. Proof. Without loss of generality, let E(C 2k ) = {x 1 y 1 , y 1 x 2 , . . . , x x y k , y k x 1 }. Then x k+1 is adjacent to at least k − 1 vertices of {y 1 , y 2 , . . . , y k }, say {x k+1 y 1 , x k+1 y 2 , . . . , x k+1 y k−1 } ⊆ E(G). And since y k+1 is adjacent to at least k − 1 vertices o f {x 1 , x 2 , . . . , x k }, y k+1 is nonadjacent to at most one vertex of {x 1 , x k }, say x k y k+1  E(G). Hence we have C 2(k+1) ⊆ G(x 1 y k+1 x 2 y 1 x k+1 y 2 x 3 y 3 , . . . , x k y k x 1 ) as shown in Fig. 7. ✷ Lemma 5. If m ≥ 4, then b(C 2m ; K 2,2 ) ≤ m + 1. Proof. We will prove it by way of induction. (1) For m = 4, by Lemma 3, we have the lemma holds. (2) Suppose that b(C 2k ; K 2,2 ) ≤ k +1 for k ≥ 4. We will show that b(C 2(k+1) ; K 2,2 ) ≤ k +2 as follows. The proof is similar to Lemma 3, however, arbitrary k makes Lemma 2 not applicable, which makes the proof more difficult. By contradiction, we may assume that b(C 2(k+1) ; K 2,2 ) > k + 2, that is, K k+2,k+2 is 2-colorable to (C 2(k+1) , K 2,2 ), say C 2(k+1)  G and K 2,2  G c . By the induction hy- the electronic journal of combinatorics 18 (2011), #P51 6 x 1 x 2 x 3 x k−2 x k−1 x k y 1 y 2 y k−2 y k−1 y k y k+1 x k+1 y 3 y k−3 Fig. 7. The graph with a cycle of length 2(k + 1) pothesis, we have b(C 2k ; K 2,2 ) ≤ k + 1. Therefore since K 2,2  G c , we have C 2k ⊆ G, let E(C 2k ) = {x 1 y 1 , y 1 x 2 , x 2 y 2 , . . . , x k y k , y k x 1 }. Firstly, we consider the four vertices not belong to V (C 2k ), that is, x k+1 , x k+2 , y k+1 and y k+2 . Since K 2,2  G c , there is at least one edge between {x k+1 , x k+2 } and {y k+1 , y k+2 }, say x k+1 y k+1 ∈ E(G). Assume that there is no edge between {x k+1 , y k+1 } and V (C 2k ). Since K 2,2  G c , x k+2 is adjacent to at least k − 1 vertices of {y 1 , y 2 , . . . , y k }, and y k+2 is adjacent to at least k − 1 vertices of {x 1 , x 2 , . . . , x k }. By Lemma 4, we have C 2(k+1) ⊆ G, a contradiction. So there is at least one edge between {x k+1 , y k+1 } and V (C 2k ), say x k+1 y k ∈ E(G). Since C 2(k+1)  G, y k+1 is nonadjacent to x 1 or x k . Therefore since K 2,2  G c , y k+2 has to be adjacent to at least one vertex of {x 1 , x k }, say y k+2 x 1 ∈ E(G) as shown in Fig. 8. Now we consider the edge number between {x k+2 } and {y k+1 , y k+2 }, there are three subcases as follows. y k+2 x k+2 x k+1 y k+1 x 1 x 2 x 3 x k−1 x k y 1 y 2 y k−2 y k−1 y k Fig. 8. The subgraph of G Case 1. Suppose that there is no edge between {x k+2 } and {y k+1 , y k+2 }. Since K 2,2  G c , x k is adjacent to at least one vertex of {y k+1 , y k+2 }. Therefore since C 2(k+1)  G, x k has to be adjacent to y k+2 . Then both x 2 and x k−1 are nonadjacent to y k+1 , since o t herwise C 2(k+1) ⊆ G. If y k+2 is nonadja cent to at least one vertex of {x 2 , x k−1 }, say x 2 , then x 2 , x k+2 , y k+1 and y k+2 would construct a K 2,2 in G c , a contradiction. Hence both x 2 and x k−1 have to be adjacent to y k+2 . x k+1 is nonadjacent to any vertex of {y 1 , y k−1 , y k+2 }, the electronic journal of combinatorics 18 (2011), #P51 7 x 1 x 2 x 3 x k−2 x k−1 x k y 1 y 2 y k−2 y k−1 y k x k+1 x k+2 y k+1 y k+2 Fig. 9. x k+2 being nonadjacent to y k+1 or y k+2 since otherwise C 2(k+1) ⊆ G. If x k+2 is nonadjacent to at least one vertex of {y 1 , y k−1 }, say y 1 , then x k+1 , x k+2 , y 1 and y k+2 would construct a K 2,2 in G c , a contra diction. Hence x k+2 has to be adjacent to both y 1 and y k−1 as shown in Fig. 9. Now we have C 2(k+1) ⊆ G(y 1 x 1 y k x k y k+2 x 2 y 2 , . . . , x k−1 y k−1 x k+2 y 1 ), a contradiction too. Case 2. Suppose that there is just one edge between {x k+2 } and {y k+1 , y k+2 }, namely x k+2 y k+1 ∈ E(G) o r x k+2 y k+2 ∈ E(G). Case 2.1. Suppose that x k+2 y k+1 ∈ E(G), then x k+2 y k+2 ∈ E(G). Since C 2(k+1)  G, we have x k+1 y k+2 , x k+2 y k−1 ∈ E(G). Then y k−1 has to be adjacent to x k+1 , since otherwise x k+1 , x k+2 , y k−1 and y k+2 would construct a K 2,2 in G c . Note that x k+1 together with V (C 2k ) −x k construct a new cycle of length 2k as shown in Fig. 10(a). Since C 2(k+1)  G, y k+2 is nonadjacent to x k or x k+2 . So, the proof is same as Case 1. y k y 1 y 2 y k−2 y k−1 x 1 x 2 x k−2 x k−1 x k+1 x k y k+2 y k+1 x k+2 y k+2 x k+2 x k+1 y k+1 x 1 x 2 x 3 x k−1 x k y 1 y 2 y k−2 y k−1 y k (a) (b) Fig. 10. x k+2 being adjacent to just one of {y k+1 , y k+2 } Case 2.2. Suppose that x k+2 y k+2 ∈ E(G). Then x k+2 y k+1 ∈ E(G). Since C 2(k+1)  G, we have x k y k+1 , x k+2 y 1 ∈ E(G). Then x k has to adjacent to y 1 , since otherwise x k , x k+2 , y 1 and y k+1 would construct a K 2,2 in G c . Note that there exists a path of length 2(k +1) be- tween x k+2 and y k−1 (x k+2 y k+2 x 1 y k x k y 1 x 2 y 2 x 3 y 3 . . . x k−2 y k−2 x k−1 y k−1 ). Hence we have x k+2 is nonadja cent to y k−1 . By symmetry, y k+1 is nonadja cent to x 2 . Therefore since K 2,2  the electronic journal of combinatorics 18 (2011), #P51 8 G c , x 2 has to adjacent to y k−1 . Similarly, there exists a path of length 2(k+1) between x k+2 and y 2 (x k+2 y k+2 x 1 y k x k y 1 x 2 y k−1 x k−1 y k−2 x k−2 . . . y 3 x 3 y 2 ). Hence we have x k+2 is nonadja- cent t o y 2 . By symmetry, y k+1 is nonadjacent to x k−1 . Therefore since K 2,2  G c , y 2 has to adjacent to x k−1 . So for even k, we can have x 3 y k−2 ∈ E(G), y 3 x k−2 ∈ E(G), x 4 y k−3 ∈ E(G), y 4 x k−3 ∈ E(G), . . . , xk 2 −1 yk 2 +2 ∈ E(G), y k 2 −1 xk 2 +2 ∈ E(G), xk 2 yk 2 +1 ∈ E(G) se- quentially. And for odd k, we can have x 3 y k−2 ∈ E(G), y 3 x k−2 ∈ E(G), x 4 y k−3 ∈ E(G), y 4 x k−3 ∈ E(G), . . . , xk−1 2 yk+3 2 ∈ E(G), yk−1 2 xk+3 2 ∈ E(G) sequentially. That is, we will add k − 2 chords on the cycle C 2k as shown in Fig. 10(b). Since C 2(k+1)  G, x k+2 is nonadjacent to any vertex of {y 1 , y 2 , . . . , y k }. Therefore since K 2,2  G c , x k+1 has to be adjacent to at least k − 1 vertices of {y 1 , y 2 , . . . , y k }. By symmetry, we have y k+2 has to be adjacent to at least k − 1 vertices of {x 1 , x 2 , . . . , x k }. By Lemma 4, we have C 2(k+1) ⊆ G, a contradiction. Case 3. Suppose that there are two edges between {x k+2 } and {y k+1 , y k+2 }, namely, x k+2 y k+1 , x k+2 y k+2 ∈ E(G). Since C 2(k+1)  G, we have x k−1 y k , x k+2 y 1 , x k+2 y k ∈ E(G). Then x k−1 has to be adja cent to y 1 , since otherwise x k−1 , x k+2 , y 1 and y k would construct a K 2,2 in G c . By symmetry, y 2 has to be adjacent to x k as shown in Fig. 12. Now we have C 2(k+1) ⊆ G(y k+1 x k+1 y k x k y 2 x 3 . . . y k−2 x k−1 y 1 x 1 y k+2 x k+2 y k+1 ), a contradiction. y k+2 x k+2 x k+1 y k+1 x 1 x 2 x 3 x k−1 x k y 1 y 2 y k−2 y k−1 y k Fig. 12. x k+2 being adjacent to y k+1 and y k+2 By Case 1-3, we have the assumption that b(C 2(k+1) ; K 2,2 ) > k + 2 does not hold. So, we have b(C 2(k+1) ; K 2,2 ) ≤ k + 2 . This completes the induction step, and the proof is finished. ✷ 4 Main results Setting m = 3 in Corollary 1, we have b(C 6 ; K 2,2 ) ≥ 4. Furthermore, we can find that a K 4,4 is a disjoint sum of two subgraphs isomorphic to C 8 . Hence, we have b(C 6 ; K 2,2 ) ≥ 5. By results in [1], Corollary 1, Lemma 1, Lemma 3 and Lemma 5, we obtain the values of b(C 2m ; K 2,2 ) as follows. the electronic journal of combinatorics 18 (2011), #P51 9 Theorem 2. b(C 2m ; K 2,2 ) =  5, m = 2 or 3, m + 1, m ≥ 4. Acknowledgements We would like to thank the referees for their helpful co mments and suggestions which led to the improvement of the present version. References [1] L. W. Beineke, A. J. Schwenk, On a bipartite form of the Ramsey pro blem, Proc. 5th British Combin. Conf. 1975, Congr. Numer. 15 (1975) 17–22. [2] W. A. Carnielli, E. L. Monte Carmelo, K 2,2 − K 1,n and K 2,n − K 2,n bipartite Ramsey numbers, Discrete Math. 223 (2000) 83–92. [3] R. J. Faudree, R. H. Schelp, Path-path Ramsey-type numb ers f or the complete bipartite graphs, J. Combin. Theory Ser. B 19 (1975) 161–173. [4] J. H. Hattingh, M. A. Henning, Bipartite Ramsey numbers, Utilitas Math. 53 (1998) 217–230. [5] J. H. Hattingh, M. A. Henning, Star-path bipartite Ramsey numbers, Discrete Math. 185 (1998) 255–258. [6] R. W. Irving, A bipartite Ramsey problem and the Zarankiewicz numbers, Glasgow Math. 19 (1978) 13-26. the electronic journal of combinatorics 18 (2011), #P51 10 . Henning, Bipartite Ramsey numbers, Utilitas Math. 53 (1998) 217–230. [5] J. H. Hattingh, M. A. Henning, Star-path bipartite Ramsey numbers, Discrete Math. 185 (1998) 255–258. [6] R. W. Irving, A bipartite. length k. Let H 1 and H 2 be bipartite gra phs, the bipartite Ramsey number b(H 1 ; H 2 ) is the smallest integer b such that given any subgraph G of the complete bipartite graph K b,b , either. K 1,n and K 2,n − K 2,n bipartite Ramsey numbers, Discrete Math. 223 (2000) 83–92. [3] R. J. Faudree, R. H. Schelp, Path-path Ramsey- type numb ers f or the complete bipartite graphs, J. Combin. Theory