DISC6288 MODC+ ARTICLE IN PRESS. Discrete Mathematics ( ) Note. The Ramsey numbers for disjoint unions of trees

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1 PROD. TYPE: COM PP: - (col.fig.: nil) DISC6288 MODC+ ED: Jaya L PAGN: Padmashini -- SCAN: Global Discrete Mathematics ( ) Note The Ramsey numbers for disjoint unions of trees E.T. Baskoro, Hasmawati 2, H. Assiyatun Combinatorial Mathematics Research Group, Faculty of Mathematics and Natural Sciences, Institut Teknologi Bandung (ITB), Jalan Ganesa 0, Bandung 402, Indonesia Received March 2006; received in revised form 0 May 2006; accepted June Abstract For given graphs G and H, the Ramsey number R(G, H) is the smallest natural number n such that for every graph F of order n: either F contains G or the complement of F contains H. In this paper, we investigate the Ramsey number R( G, H ), where G is a tree and H is a wheel W m or a complete graph K m. We show that if n, then R(kS n,w 4 ) = (k + )n for k 2, even n and R(kS n,w 4 ) = (k + )n for k and odd n. We also show that R( k i= T ni,k m ) = R(T nk,k m ) + k i= n i Published by Elsevier B.V. Keywords: Ramsey number; Star; Wheel; Tree. Introduction For given graphs G and H, the Ramsey number R(G, H) is defined as the smallest positive integer n such that for any graph F of order n, either F contains G or F contains H, where F is the complement of F. In 2, Chvátal and Harary [6] established a useful lower bound for finding the exact Ramsey numbers R(G, H), namely R(G, H) (χ(g) )(c(h ) ) +, where χ(g) is the chromatic number of G and c(h) is the number of vertices of the largest component of H. Since then the Ramsey numbers R(G, H) for many combinations of graphs G and H have been extensively studied by various authors, see a nice survey paper []. Let P n be a path with n vertices and let W m be a wheel of m + vertices that consists of a cycle C m with one additional vertex being adjacent to all vertices of C m.astar S n is the graph on n vertices with one vertex of degree n, called the center, and n vertices of degree. T n is a tree with n vertices and a cocktail-party graph H s is the graph which is obtained by removing s disjoint edges from K 2s. Several results on Ramsey numbers have been obtained for wheels. For instance, Baskoro et al. [] showed that for even m 4 and n (m/2)(m 2), R(P n,w m ) = 2n. They also showed that R(P n,w m ) = n 2 for odd m, and n ((m )/2)(m ). Supported by TWAS Research Grant 04-2 RG-MATHS-AS. 2 Permanent address: Department of Mathematics FMIPA, Universitas Hasanuddin (UNHAS), Jl. Perintis Kemerdekaan Km.0, Makassar 024, Indonesia. addresses: ebaskoro@dns.math.itb.ac.id, ebaskoro@math.itb.ac.id (E.T. Baskoro), hasmawati@math.itb.ac.id (Hasmawati), hilda@math.itb.ac.id (H. Assiyatun) X/$ - see front matter 2006 Published by Elsevier B.V. doi:0.06/j.disc

2 2 E.T. Baskoro et al. / Discrete Mathematics ( ) For a combination of stars with wheels, Surahmat et al. [0] investigated the Ramsey numbers for large stars versus small wheels. Their result is as follows. Theorem (Surahmat and Baskoro [0]). For n, { 2n + if n is even, R(S n,w 4 ) = 2n if n is odd. For odd m, Chen et al. have shown in [4] that R(S n,w m )=n 2 for m and n m. This result was strengthened by Hasmawati et al. in [8], by showing that this Ramsey number remains the same, as given in the following theorem. Theorem 2 (Hasmawati et al. [8]). If m is odd and n ((m + )/2), then R(S n,w m ) = n If n (m + 2)/2, Hasmawati [] gave R(S n,w m ) = n + m 2 for even m and odd n, orr(s n,w m ) = n + m, otherwise. Let G be a graph. The number of vertices in a maximum independent set of G denoted by α 0 (G), and the union of s vertices-disjoint copies of G denoted sg. Burr et al. in [], showed that if the graph G has n vertices and the graph H has n 2 vertices, then n s + n 2 t D R(sG, th) n s + n 2 t D + k, where D = min{sα 0 (G), tα 0 (H )} and k is a constant depending only on G and H. In the following theorem Chvátal gave the Ramsey number for a tree versus a complete graph. Theorem (Chvátal []). For any natural number n and m, R(T n,k m ) = (n )(m ) +. In this paper, we determine the Ramsey numbers R( G, H ) of a disjoint union of a graph G versus a graph H, where G is either a star or a tree, and H is either a wheel or a complete graph. The results are presented in the next three theorems. Theorem 4. If m is odd and n (m + )/2, then R(kS n,w m ) = n 2 + (k )n. Theorem. For n, { (k + )n if n is even and k 2, R(kS n,w 4 ) = (k + )n if n is odd and k. Theorem 6. Let n i n i+ for i =, 2,...,k. If n i (n i n i+ )(m ) for any i, then R( k i= T ni,k m ) = R(T nk,k m ) + k i= n i for an arbitrary m. Before proving the theorems, we present some notations used in this note. Let G(V, E) be a graph. For any vertex v V (G), the neighborhood N(v)is the set of vertices adjacent to v in G. Furthermore, we define N[v]=N(v) {v}. The degree of a vertex v in G is denoted by d G (v). The order of G, G is the number of its vertices, and the minimum (maximum) degree of G is denoted by δ(g)(δ(g)).fors V (G), G[S] represents the subgraph induced by S in G. If G is a graph and H is a subgraph of G, then denote V (G)\V(H)by G\H. Let G =(V,E ) and G 2 =(V 2,E 2 ). The union G=G G 2 has the vertex set V =V V 2 and the edge set E=E E 2. Their join, denoted G + G 2, is the graph with the vertex set V V 2 and the edge set E E 2 {uv : u V,v V 2 }. 2. The proofs of theorems Proof of Theorem 4. Let m be odd and n (m + )/2. We shall use an induction on k. Fork =, we have R(S n,w m )=n 2 (by Theorem 2).Assume the theorem holds for any r<k, namely R(rS n,w m )=(n 2)+(r )n. We will show that R(kS n,w m ) = (n 2) + (k )n.

3 E.T. Baskoro et al. / Discrete Mathematics ( ) V(F ) v v 0 v j A u w Q q 2 X Z q 2 Fig.. The illustration of the proof of R(2S n,w 4 ) n. Let F be a graph with F =n 2 + (k )n. Suppose that F contains no W m. Since F R(rS n,w m ), then F (k )S n. Let A = F \(k )S n and T = F [A]. Thus, T =n 2. Since T contains no W m, then by Theorem 2, T S n. Thus, F contains ks n. Hence, we have R(kS n,w m ) (n 2) + (k )n. On the other hand, it is not difficult to see that F = K kn 2K n contains no ks n and its complement contains no W m. Observe that F has n + (k )n vertices. Therefore, we have R(kS n,w m ) (n 2) + (k )n, and the assertion follows. Proof of Theorem. Let n be even, n 4 and k 2. Consider F = (H (kn 2)/2 + K ) H n/2. Clearly, graph F has (k + )n vertices and contains no ks n. Its complement contains no W 4. Hence, R(kS n,w 4 ) (k + )n. We will prove that R(kS n,w 4 ) = (k + )n for k 2. First we will show that R(2S n,w 4 ) = n. Let F be a graph of order n. Suppose F contains no W 4. By Theorem, we have F S n. Let V(S n ) = {v 0,v,...,v n } with center v 0, A = F \S n and T = F [A]. Thus T =2n. If there exists u T with d T (u) (n ), then T contains S n. Hence F contains 2S n. Therefore we assume that for every vertex u T, d T (u) (n 2). Let u, w T where (u, w) / E(T ). Consider H = N[u] N[w], Q = T \H, Z = N(u) N(w), and X = H \{u, w} (see Fig. ). By contradiction suppose d(u) n. Then 0 Z n, 2 H 2n and 2n 2 Q + Z. Observe that every q Q is adjacent to at least Q 2 other vertices of Q. (Otherwise, there exists q Q which is not adjacent to at least two other vertices of Q, say q and q 2. Then T will contain a W 4 ={q,u,q 2,q,w} with w as a hub, a contradiction). Then, for all q Q, d Q (q) Q 2. Let E(X\Z, Q) ={uv : u X\Z, v Q}. If there exists x X\Z not adjacent to at least two vertices of Q, say q and q 2, then T will contain W 4 ={q,x,q 2,u,w} with w or u as a hub, a contradiction. Hence every x X\Z is adjacent to at least Q vertices in Q. Therefore, we have E(X\Z, Q) X\Z ( Q ). On the other hand, every vertex q Q is incident with at most (n 2) d Q (q) (n 2) ( Q 2) = n Q edges from X\Z. Thus E(X\Z, Q) Q (n Q ). Now, we will show that X\Z ( Q )> Q (n Q ), which leads to a contradiction. Writing X\Z ( Q ) = X\Z Q X\Z and substituting X\Z = 2n 2 Q Z, we obtain X\Z ( Q ) = Q (n Q ) + Q (n 2 Z ) + Q n (n 2 Z ). Noting Q + Z, itcan be verified that Q (n 2 Z ) + Q n (n 2 Z )>0. Thus X\Z ( Q )> Q (n Q ). Hence there is no u T such that d(u) n.

4 4 E.T. Baskoro et al. / Discrete Mathematics ( ) V (F 2 ) 2 S n S k- n Sn v v 2 v (n-) s A' u' w' Q' q q 2 X' Z' Fig. 2. The illustration of the proof of R(kS n,w 4 ) (k + )n. Therefore every vertex u T, d T (u) = n 2. This implies Q =2 + Z. Next, suppose F [Q] is a complete graph. Then for every q Q, d Q (q)= Q. Consequently, every vertex in Q is incident with (n 2) d Q (q)=n Q edges from X\Z. Similarly as in the previous argument, this implies that X\Z ( Q )> Q (n Q ), which is impossible. Hence F [Q] is not a complete graph. Now, choose two vertices in Q which are not adjacent, call q and q 2. Let Y ={q,q 2 } {u, w}, it is clear that Y is an independent set. If there exists a vertex v V(S n ) adjacent to at most one vertex in Y say q, then {v, u, q,q 2,w} will induce a W 4 in F, with a hub w, a contradiction. Therefore, every vertex v V(S n ) is adjacent to at least two vertices in Y. Suppose v 0 and v j in V(S n ) are adjacent to y,y 2 and to y,y 4 in Y, respectively. Note that at least two y i s are distinct. Without loss of generality, assume y = y. Since Y is independent, then we have two new stars, namely S n and S n, where V(S n ) = S n\{v j } {y } with v 0 as the center and V(S n ) = N[y ] {v j } with y as the center (see Fig. ). So, we have F 2S n. Hence, R(2S n,w 4 ) = n. Now, assume the theorem holds for every r<k. We will show that R(kS n,w 4 )=(k +)n. Let F 2 be a graph of order (k +)n. Suppose F 2 contains no W 4. We will show that F 2 ks n. By induction, F 2 (k )S n. Denote the (k )S n as Sn,S2 n,...,sk n with the center v,v 2,...,v k, respectively. Writing A = F 2 \(k )S n and T = F 2 [A ]. Thus T =2n. Similarly, as in the case k=2, every vertex u T, must have degree n 2. Next, let u,w T where (u,w )/ E(T ), H = N[u ] N[w ], Q = T \H, and Y ={q,q 2 } {u,w }, where q,q 2 Q and (q,q 2 )/ E(T ) (see Fig. 2). If the vertex v V ((k )S n ) is adjacent to at most one vertex in Y, say u, then F 2 will contain W 4 = {u,q,v,q 2,w } with w as a hub, a contradiction. Therefore, every vertex v V ((k )S n ) is adjacent to at least two vertices in Y. Suppose v and s in V(S n ) are 2 adjacent to u,q and to u,w, respectively (see Fig. 2). Then, we will alter Sn into S n with V(S n ) = (S n \{s}) {q } and create a new star Sn k where V(Sk n ) = N[u ] {s} with the center u. Hence, we now have k disjoint stars, namely Sn,S2 n,s n,...,sk n and Sn k. Therefore, we have R(kS n,w 4 ) = (k + )n. Let n be odd. Consider F =K kn K n. Clearly, the graph F has order (k +)n 2, without containing ks n and F contains no W 4. Hence, R(kS n,w 4 ) (k + )n. To obtain the Ramsey number we use an induction on k. For k=, we have R(S n,w 4 )=2n. Suppose the theorem holds for every r<k. We show that R(kS n,w 4 )=(k+)n. Let F 4 be a graph of order (k + )n. Suppose F 4 contains no W 4. By the assumption, F 4 contains (k )S n. Let

5 E.T. Baskoro et al. / Discrete Mathematics ( ) B = F 4 \(k )S n and L = F 4 [B]. Thus L =2n. Since L contains no W 4, then by Theorem, L S n. Therefore, F 4 contains ks n. The proof is now complete. 2 2 Proof of Theorem 6. Let n i n i+ and n i (n i n i+ )(m ) for any i. Since F = (m 2)K nk K k i= n i has no k i= T ni and its complement contains no K m, then R( k i= T ni,k m ) (m )(n k ) + k i= n i +. We fix m and apply an induction on k.fork = 2, we show that R(T n T n2,k m ) = (m )(n 2 ) + n +. Let F be a graph with F =(m )(n 2 )++n. Suppose F contains no K m. Since n n 2, then we can write n n 2 =q 0. Substitute n 2 =n q, then we obtain F =(m )(n q )+n +=(m )(n ) q(m )+n + or F =(m )(n ) + +[n (n n 2 )(m )]. Noting n (n n 2 )(m ) 0, it can be verified that F (m )(n ) + i.e. F R(T n,k m ). Hence, F T n.now,leta = F \T n, and H = F [A]. Then H =(m )(n 2 ) +. Since H contains no K m, then by Theorem, H T n2. Therefore, F contains a subgraph T n T n2. Next, assume the theorem holds for all r<k, namely R( r i= T ni,k m )=(m )(n r )+ r i= n i+.we shall show that R( k i= T ni,k m )=(m )(n k )+ k i= n i+.take an arbitrary graph F 2 with order (m )(n k )+ k i= n i+. Suppose F 2 contains no K m. By induction, F 2 contains k i= T n i. Writing B = F 2 \ k i= T n i, and Q = F 2 [B]. Then Q =(m )(n k ) +. Since Q contains no K m, then Q contains T nk. Hence F 2 contains k i= T ni. Therefore, we have R( k i= T ni,k m ) = (m )(n k ) + k i= n i +. The proof is now complete.. Uncited reference [2]. References [] E.T. Baskoro, Surahmat, The Ramsey numbers of path with respect to wheels, Discrete Math. 24 (200). [2] S.A. Burr, On Ramsey numbers for large disjoint unions of graphs, Discrete Math. 0 (88). [] S.A. Burr, P. Erdös, J.H. Spencer, Ramsey theorem for multiple copies of graphs, Trans. Amer. Math. Soc. 20 () 8 8. [4] Y.J. Chen, Y.Q. Zhang, K.M. Zhang, The Ramsey numbers of stars versus wheels, European J. Combin. (2004) [] V. Chvátal, Tree-complete graph Ramsey number, J. Graph Theory (). [6] V. Chvátal, F. Harary, Generalized Ramsey theory for graphs, III: small off-diagonal numbers, Pacific. J. Math. 4 (2) 4. [] Hasmawati, Bilangan Ramsey untuk graf bintang terhadap graf roda, Tesis Magister, Departemen Matematika ITB Indonesia, [8] Hasmawati, E.T. Baskoro, H. Assiyatun, Star-wheel Ramsey numbers, J. Combin. Math. Conbin. Comput. (200) 28. [] S.P. Radziszowski, Small Ramsey numbers, Electron. J. Combin. July (2004) #DS. [0] Surahmat, E.T. Baskoro, On the Ramsey number of a path or a star versus W 4 or W, in: Proceedings of the 2th Australasian Workshop on Combinatorial Algorithms, Bandung, Indonesia, July 4 200, pp. 6 0.