All Ramsey numbers for brooms in graphs

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1 All Ramsey numbers for brooms in graphs Pei Yu Department of Mathematics Tongji University Shanghai, China Yusheng Li Department of Mathematics Tongji University Shanghai, China li Submitted: Jul 17, 015; Accepted: Aug, 016; Published: Aug 19, 016 Mathematics Subject Classifications: 05D10 Abstract For k, l 1, a broom B k,l is a tree on n = k + l vertices obtained by connecting the central vertex of a star K 1,k with an end-vertex of a path on l 1 vertices. As B n, is a star and B 1,n 1 is a path, their Ramsey number have been determined among rarely known R(T n ) of trees T n of order n. Erdős, Faudree, Rousseau and Schelp determined the value of R(B k,l ) for l. We shall determine all other R(B k,l ) in this paper, which says that, for fixed n, R(B n l,l ) decreases first on 1 l n/ from n or n to 4n 1, and then it increases on n/ < l n from 4n 1 to n 1. Hence R(B n l,l) may attain the maximum and minimum values of R(T n ) as l varies. Keywords: Ramsey number; Tree; Broom 1 Introduction Given a graph G, the Ramsey number R(G) is the smallest integer N such that every redblue coloring of the edges of K N contains a monochromatic G. Let T n be a tree of order n. Finding R(T n ) for an arbitrary T n is a difficult unsolved problem in Ramsey theory. Most works focus on improving the known bounds, see [10]. Erdős and Sós conjectured that if a graph G has average degree greater than n 1, then G contains every tree of n edges, which implies that R(T n ) n for n. A result of Erdős, Faudree, Rousseau and Schelp in [4] yields 4n r(t n ) 1, (1) under () by minimizing the lower bound with b = a, and the lower bound can be attained by some brooms. For k, l 1, a broom B k,l is a tree on k + l vertices obtained by connecting the central vertex of a star K 1,k with an end-vertex of a path on l 1 the electronic journal of combinatorics () (016), #P.9 1

2 vertices. Thus B k,1 = K 1,k, B k, = K 1,k+1 and B 1,l = P l+1, where P l+1 is a path of order l + 1. They obtained the following result. Theorem 1. ([4]) Let k and l be integers with l and n = k + l. Then l R(B k,l ) = n + 1. Thus R(B k,l ) = 4n 1 for l {, +1, +} and n = k+l, which attain the lower bound in (1). In this paper, we shall determine the values of R(B k,l ) for 1 l 1. Note that when k is fixed and l is sufficient large, B k,l is similar to a path P n ; when l is fixed and k is sufficient large, B k,l is similar to a star K 1,n 1. Among few known results of R(T n ), R(P n ) and R(K 1,n 1 ) have been determined completely. For n, the exact value of R(P n ) was determined in [7] as n R(P n ) = 1, and R(K 1,n 1 ) was determined in [] as { n if n is odd, R(K 1,n 1 ) = n otherwise. As B 1,l = P l+1, B k,1 = K 1,k and B k, = K 1,k+1, their Ramsey numbers can be determined by the above results. It was proved that R(B k, ) = R(K 1,k+1 ) in []. Thus we shall consider the case l 4 and k. Theorem. Let k and l be integers with k and n = k + l. Then { n + l 1 if l 1, R(B k,l ) = n l 1 if 4 l. Remark. Roughly speaking, for fixed n, R(B n l,l ) decreases first on l n 1 from n or n to 4n n 1 1, and then increases on < l n from 4n 1 to n 1. Hence R(B n l,l) may attain the maximum and minimum values of R(T n ) when l varies, as it is believed that R(K 1,n 1 ) is the maximum value of R(T n ). Proofs For any red-blue edge-coloring of K N, denote R and B be the induced red and blue subgraph, respectively, and N R (x) and N B (x) be the red neighborhood and blue neighborhood of x, respectively. Let N R [x] = N R (x) {x}, N B [x] = N B (x) {x}, deg R (x) = N R (x), and deg B = N B (x). For a graph G and disjoint subset A and D, denote by G(A) the subgraph of G induced by A, and G(A, D) the bipartite subgraph of G induced by A and D. If G is the red-blue edge-colored K N, we write G R (A) = G(A) R and G R (A, D) = G(A, D) R. Notation not specifically mentioned will follow from [1]. We do not distinguish the vertex set and the graph when there is no danger of confusion. the electronic journal of combinatorics () (016), #P.9

3 Consider a tree T n as a bipartite graph with two parts of size a and b, respectively, where a b, a + b = n. Observing that a red-blue edge-colored K a+b with R = K a 1 K a+b 1 contains no monochromatic T n, and a red-blue edge-colored K b with R = K b 1 K b 1 contains no monochromatic T n. We see that { } R(T n ) max a + b 1, b 1, () where a and b are determined by T n. Note that B k,l is a bipartite graphs on parts of sizes a = l and b = k + l, then { R(B k,l ) max k + l l } 1, + 1. () We shall prove the cases l = 1 and l = 4 in Theorem via the following two lemmas. Lemma. Let k be an integer. Then Lemma 4. Let k be an integer. Then R(B k, 1 ) = 4k. R(B k,4 ) = + In order to prove Lemma, we need two results from [9] and [6], respectively. Lemma 5. ([9]) Let G(A, D) be a bipartite graph on parts A and D with A = k and D = such that min{d(x) : x A} k. Then G(A, D) contains a cycle C. Lemma 6. ([6]) R(C ) = k 1 for integer k Proof of Lemma. It is easy to see that R(B, ) = 6, we assume that k. As the lower bound () implies R(B k, 1 ) 4k, it suffices to show the opposite inequality. Let G be a red-blue edge-colored K 4k. We shall show that G contains a monochromatic B k, 1. By lemma 6, G contains a monochromatic cycle C. Without loss of generality, we assume that this C is blue and denote it by C (B). Let D = G \ C(B). Then D =. If there exists a vertex x C (B) such that N B (x) D k 1, then G contains a blue B k, 1. We then assume that N B (x) D k for each x C (B). The fact that N B (x) D + N R (x) D = D = implies that N R (x) D k for each x C (B), hence the number of red edges between and D is at least. So there exists a vertex u D such that C (B) N R (u) C (B) D = k + 1. the electronic journal of combinatorics () (016), #P.9

4 Let {u 0, u 1,..., u k } N R (u) C (B), and A = C(B) \ {u 1, u,..., u k }. Then A = k. Consider the bipartite graph G R (A, D). By Lemma 5, there is a red cycle C between A and D. Denote by C (R) the red C. Since C (R) contains A hence u 0, so the graph on {u 1, u,..., u k }, u, u 0, C (R) induced by red edges contains a red B k, 1. This completes the proof of Lemma. Proof of Lemma 4. The lower bound () implies R(B k,4 ) +, and we shall show R(B k,4 ) +. Let G be a red-blue edge-colored K +. Assume that G contains no monochromatic B k,4. If k =, + = 7. As R(P 5 ) = 6 < 7, we suppose that G contain a red P 5. Label the vertices of the path in order as {x 1, x,..., x 5 }, denote another two vertices as y 1, y. Since there is no red B,4, all the edges between {x, x 4 } and {y 1, y } are red. If there is red edge between {x 1, x 5 } and {y 1, y }, say edge x 5 y is red. Then edges x 5 y 1, x y 1,x y are blue. Now {x, x }, y, x 4, y 1, x 5 contain a blue B,4, a contradiction. If all the edges between {x 1, x 5 } and {y 1, y } are blue, then {x 1, x }, y, x 4, y 1, x 5 contain a blue B,4, a contradiction. Now we consider the case k. As R(K 1,k+1 ) < +, we suppose that there is a blue star K 1,k+1, which is denoted by K (B) 1,k+1 and D = G \ K (B) 1,k+1. Then A = D = k Let x be the center of K(B) 1,k+1, A = K(B) 1,k+1 \ {x} Claim. D induces a red K k+1. Proof. Suppose to the contrary, there is a blue edge uv in D. Since G contains no blue B k,4, the edges between {u, v} and A are all red, and thus all the edges between {u, v} and D \ {u, v} are blue from the assumption that G contains no red B k,4. Now consider the blue edges between {u, v} and A. With a similar analysis, we get that D induces a blue K k+1 and all edges between D and A are red. Now, consider the adjacency between x and a vertex of D, say xu, no matter what the color of xu is, we have a monochromatic B k,4, leading to a contradiction and the claim is proved. Now D is a red K k+1. If there exists a red edge xw with w D, then D {x} induces a red K 1,k+1 with center w. As A = V (G) \ (D {x}), a similar analysis for the above claim tells us that A is a blue K k+1. If the number of blue edges between A and D is at least k +, then there exists a vertex y A such that N B (y) D. Now choose two vertices {y 1, y } N B (y) D and two vertices {a 1, a } A \ y, then (A\{a 1, a, y}) {y 1, y }, y, a 1, a, x contains a blue B k,4, a contradiction. Thus assume to the contrary, there exists a vertex z D such that N R (z) A (k+1) (k+1) = k. If D w, z are the same vertex, we can choose two vertices {z 1, z } N R (z) A and three vertices {d 1, d, d } D \ z for D = k Then (D \ {d 1, d, d, z}) {z 1, z, x}, z, d 1, d, d contain a red B k,4. If w, z are different, choose a vertex d 1 D \ {z, w}, then (D \ {z, w, d 1 }) {z 1, z }, z, d 1, w, x contain a red B k,4, a contradiction. Finally, assume that x is adjacent to D completely blue. Choose any set F A D such that F = k + 1 and denote M = V (G) \ (F x). A similar analysis for the claim says that M is a red K k+1. The choice of F tells us that A D is a red K +, hence G the electronic journal of combinatorics () (016), #P.9 4

5 contains a red B k,4, which is a contradiction too. This completes the proof of Lemma 4. Lemma 7. For integers k, l, N with 5 l and N + l 1, let the edges of K N be colored by two colors i 0, 1 (mod ). Suppose i is a color and x is a vertex such that deg i (x) = max max{deg i (v), deg i+1 (v)}. v If there exist vertices y, z N i+1 (x), not necessarily distinct, satisfying 1. N i+1 (y) N i (x) k, and. deg i+1 (z) N l then G contains a monochromatic B k,l. Proof of Lemma 7. Since N i+1 (y) N i (x) k, we can choose a subset A in N i+1 (y) N i (x) such that A = k. Let H = G \ A, then H = N k. Since R(C t ) = t 1 for t, H contains a monochromatic C t in color j, denoted by C (j) t, where N k + 1 k + l/ t l for 5 l. The choice of vertex x implies deg i (x) deg i+1 (z) N l, and thus N i [x] \ A + C (j) H + 1, N i+1 [z] \ A + C (j) H + 1, t which implies that both N i [x] \ A and N i+1 [z] \ A contain a vertex of C (j) t. Case 1. y = z. If j = i, namely, C (j) t = C (i) t is in color i, there is a monochromatic B k,l in color i in A {x} C (i) t, and otherwise j = i + 1, there exists a monochromatic B k,l in color i + 1 in A {y} C (i+1) t. Case. y z. Similarly, we can find a monochromatic B k,l either in A {x} C (j) t or in A {y, x, z} C (j). This completes the proof of Lemma 7. The next two lemmas are results about the extremal edges in graph that contains no path P t. Lemma 8. ([5]) Let t be an integer, and G a graph of order N that contains no P t. Let e(g) be the number of edges of G, then e(g) (t )N. Lemma 9. ([8]) Let G(X B, X R ) be a bipartite graph on parts X B and X R with X R X B. If G(X B, X R ) contains no P t with (t 1) X R, then [ ] e(g(x B, X R )) (t 1) X B + X R (t 1). t the electronic journal of combinatorics () (016), #P.9 5

6 Proof of Theorem. We may assume that 5 l from Theorem 1, Lemma and Lemma 4. Set N = n l 1 = + l 1. Let G be a red-blue edge-colored K N, and let R and B be the induced red and blue subgraph, respectively. Without loss of generality, we may assume that the maximal monochromatic degree of G is the maximum blue degree and x is a vertex such that deg B (x) = max max{deg B (v), deg R (v)}. v To simplify the notation, we write X B = N B (x), X R = N R (x) and t = l + k X B 1. The choice of x implies N B (u) X B for each vertex u X R as otherwise N B [x] N R (u) and thus deg R (u) > deg B (x), which is impossible. We shall separate the proof into three cases depending on X B. Case 1. X B < k + l 1, and either G(X R ) contains a blue P t, denoted by P (B) t, or G(X B, X R ) contains a blue P t, denoted P (B) t. In G(X B X R ), let P (B) be the longest blue path extended from P (B) t such that one of its end-vertices is in X B if G(X R ) contains a blue P t, or that from P (B) t otherwise. If P (B) l 1, then there exists a blue B k,l. Thus we assume that P (B) l, then P (B) fails to contain at least X B (l t) = k + 1 vertices of X B. Let y be the other end-vertex of P (B). Then N R (y) X B k + 1. The maximality of P (B) implies N R (y) N 1 (l ) = N l + 1, which and Lemma 7 imply that G contains a monochromatic B k,l. Case. X B k + l 1. Let P (B) be the longest blue path in G(X B X R ) that has an end-vertex in X B. A similar analysis in Case 1 implies that G contains a monochromatic B k,l. Case. X B < k + l 1, and neither G(X R ) contains a blue P t nor G(X B, X R ) contains a blue P t. As t = l + k X B 1, then X B l + k and X R = N 1 X B k. Since G(X R ) contains no blue P t, Lemma 8 implies e(g B (X R )) (t ) X R /. The choice of x implies that the min{deg R (v), deg B (v)} X R for each vertex v of G, and thus e(g B (X B, X R )) X R X R (t ) X R = ( X R t + ) X R. Since G(X B, X R ) contains no blue P t, Lemma 9 yields e(g B (X B, X R )) M B, where [ ] M B = (t 1) X B + X R (t 1). Claim for Case. G(X B, X R ) has at most X R ( X B k ) 1 blue edges. the electronic journal of combinatorics () (016), #P.9 6

7 where Proof. Suppose opposite, then e(g B (X B, X R )) X R max{ X R t +, X B k} m B. m B = k( X R t + ) + ( X R k)( X B k). Note that M B and m B are upper and lower bound of number of blue edges in G(X B, X R ), respectively, and thus M B m B. Case.1. l is even. In this subcase, X B + X R = + l and we have t = l + k X B 1 = X R k + 1, m B = k(k + 1) + ( X R k)( X B k), [ ] M B = ( X R k) X B + k ( X R k) m B M B = k ( X R k) + ( X R k) + k = ( X R ) + k > 0, which is a contradiction. Case.. l is odd. In this subcase, X R + X B = + l and We have t = l + k X B 1 = X R k +, m B = k + ( X R k)( X B k), [ ] M B = ( X R k + 1) X B + k ( X R k). m B M B = ( X R ) + X R X B 4k + = ( X R )( X R 4) + l + 5. For l,is odd, we get X R N 1. If X R 4, m B M B l + 5 > 0; if l =, m B M B = l + > 0, a contradiction, hence the claim holds. and We now have ( ) XR e(g R (X R )) (t ) X R (k 1) X R. e((g R (X B, X R )) X R X B [ X R ( X B k) 1 ] = k X R + 1, Recall X R = N R (x), and thus v X R N R (v) e(g R (X B, X R )) + e(g R (X R )) + X R = X R + 1. Therefore, there exist y, z X R = N R (x),not necessarily distinct, such that N R (y) N B (x) k + 1 and N R (z) + 1 N l. Then, Lemma 7 implies that G contains a monochromatic B k,l. This completes the proof of Theorem. the electronic journal of combinatorics () (016), #P.9 7

8 Acknowledgements The authors are grateful to the editors and referees for their invaluable comments and suggestions, particularly, their careful reading and detailed marking on the manuscript. References [1] B. Bollobás. Modern Graph Theory. Springer-Verlag, New York, [] P. Bahls and T. Spencer. On the Ramsey numbers of trees with small diameter. Graphs Combin., 9:9 44, 01. [] V. Chvátal and F. Harary. Generalized Ramsey theory for graphs II, Small diagonal numbers. Proc. Amer. Math. Soc., :89 94, 197. [4] P. Erdős, R. Faudree, C. Rousseau and R. Schelp. Ramsey numbers for brooms. Congr. Numer., 5:8 9, 198. [5] P. Erdős and T. Gallai. On maximal paths and circuits graphs. Acta Math. Acad. Sci. Hung., 10: 7 56, [6] R. Faudree and R. Schelp. All Ramsey numbers for cycles in graphs. Discrete Math., 8:5 5,1974. [7] L. Geréncser and A. Gyárfas. On Ramsey-type problems. Ann. Univ. Sci. Budapest, Eötvös Sect. Math., 10: , [8] A. Gyárfás, C. Rousseau and R. Schelp. An extremal problem for paths in bipartite graphs, J. Graph Theory., 8:8 95, [9] B. Jackson. Cycles in bipartite graphs. J. Combin. Theory Ser.B., 0: 4, [10] S. Radziszowski. Small Ramsey numbers. Electronic J. Combin., #DS the electronic journal of combinatorics () (016), #P.9 8