Báo cáo toán học: "The Ramsey number r(K5 − P3, K5)" pptx

The Ramsey number r(K 5 − P 3 , K 5 ) Luis Boza ∗ Departamento de Matem´atica Aplicada I Universidad de Sevilla, Seville, Sp ain boza@us.es Submitted: Dec 2, 2010; Accepted: Apr 4, 2011; Published: Apr 14, 2011 Mathematics Subject Classification: 05D10 Abstract For two given graphs G 1 and G 2 , the Ramsey number r(G 1 , G 2 ) is the smallest integer n such that for any graph G of order n, either G contains G 1 or the comp le- ment of G contains G 2 . Let K m denote a complete graph of order m and K n − P 3 a complete graph of order n without two incident edges. In this paper, we prove that r(K 5 − P 3 , K 5 ) = 25 without help of computer algorithms. 1 Introduction All graphs considered in this paper are simple graphs without loops. For two given graphs G 1 and G 2 and a given integer n, let R(G 1 , G 2 ; n) denote the set of all graphs G of order n, such that G does not contain G 1 and G does not contain G 2 , where G is the complement of G. The Ramsey number r(G 1 , G 2 ) is the smallest integer n such that R(G 1 , G 2 ; n) is empty. The values of r(G 1 , G 2 ) f or all graphs G 1 and G 2 of order at most five up to the three cases that G 1 is one of the graphs K 5 − P 3 , K 5 − e and K 5 and G 2 = K 5 are found in [1, 3, 5, 6, 7, 8, 9, 10, 11, 12, 15, 16, 17, 18]. Kalbfleisch [13] proved that r(K 5 − P 3 , K 5 ) ≥ 25 and McKay and Radziszowski [15] found 350904 graphs belonging to R(K 4 , K 5 ; 24) ⊆ R(K 5 −P 3 , K 5 ; 24), but there might be more of them. Recently, Black, Leven and Radziszowski [2] proved that r(K 5 − P 3 , K 5 ) ≤ 26 and Clavert, Schuster a nd Ra dziszowski [4] computed the main result of the present paper. ∗ This research is supported by the Andalusian Government under project P06-FQM-01649. the electronic journal of combinatorics 18 (2011), #P90 1 2 Main result In this paper we find the value of r(K 5 − P 3 , K 5 ) without help of computer algorithms. The main result is the following: Theorem 2.1 r(K 5 − P 3 , K 5 ) = 25. In order to prove Theorem 2.1, we proceed by reduction to the absurd. Suppose that there exists a graph G ∈ R(K 5 − P 3 , K 5 ; 25). Since r(K 4 , K 5 ) = 25 [15] we have G contains K 4 . Let K be the set of cliques of G of order 4, let K ∈ K be such that  v∈V (K) d(v) = max{  v∈V (X) d(v) : X ∈ K } and let v 1 , v 2 , v 3 and v 4 be the vertices of K. We may suppose without loss of generality that d(v 1 ) ≥ d(v 2 ) ≥ d(v 3 ) ≥ d(v 4 ). Let |A| denote the cardinality of the set A. If F is a graph then V (F ) denotes its vertex set. The neighborhood N F (v) of a vertex v is the set of vertices adjacent to v in the graph F . If G 1 is isomorphic to a subgraph of G 2 then we use G 1 ⊆ G 2 to denote it. If A is a subset of V (F ), then F [A] is the subgraph induced by A. If v ∈ V (F ), d F (v) is the degree of v in F . The maximum and minimum degree of F are denoted by ∆(F ) and δ(F ), respectively. Let d, V and N denote d G , V (G) and N G , respectively. If k is a positive integer, F ∈ R(K m − P 3 , K n ; k) and v ∈ V (F ) then F [N F (v)] ∈ R(K m−1 −P 3 , K n ; d F (v)) and F[N F (v)] ∈ R(K m −P 3 , K n−1 ; k−1 −d F (v)). Thus ∆(F ) ≤ r(K m−1 − P 3 , K n ) − 1 and δ(F ) ≥ k − r(K m − P 3 , K n−1 ). Since r(K 5 − P 3 , K 4 ) = 18 [7] we have δ(G) ≥ 7. In the rest of the paper, i and j are two different integers with 1 ≤ i, j ≤ 4. Let A i = N(v i ) − V (K) and D = V (G) − V (K) −  4 k=1 A k . A i ∩ A j = ∅, because in otherwise G should contain K 5 − P 3 . Hence {V (K), A 1 , A 2 , A 3 , A 4 , D} is a partition of V . Obviously, |A i | = d(v i ) − 3 ≥ 4. If u ∈ A i or u ∈ D then |N(u) ∩ A i |, the number of vertices belonging to A j adjacent to u, is denoted by e j (u). Let H i denote the graph G[V (G) − (A i ∪ V (K))] = G[N G (v i )]. Clearly H i ∈ R(K 5 − P 3 , K 4 ; 21 − |A i |). If u ∈ V − V (K) − D, g(u) will represent the integer k for which u ∈ A k . Also, if u ∈ D, we define g(u) = 0. In order to prove Theorem 2.1, we need the following r esults: Lemma 2.1 Let u ∈ A i and w ∈ D. Then d(u) − 9 ≤ e j (u) ≤ d(u) + |A j | − 13 and d(w) − 8 ≤ e i (w) ≤ d(w) + |A i | − 12. Proof. Since H j ∈ R(K 5 −P 3 , K 4 ; 21−|A j |), r(K 4 −P 3 , K 4 ) = 9 [6] and r(K 5 −P 3 , K 3 ) = 9 [7], we have d H j (u), d H j (w) ≤ ∆(H j ) ≤ r(K 4 − P 3 , K 4 ) − 1 = 8 and d H j (u), d H j (w) ≥ δ(H j ) ≥ 21 − |A j | − r(K 5 − P 3 , K 3 ) = 12 − |A j |. The results follow on noting t hat d(u) = d H j (u) + e j (u) + 1 and d(w) = d H i (w) + e i (w). Corollary 2.1 Let k be, with 1 ≤ k ≤ 4 and k = i, and let u ∈ A i . Th en e j (u)+4−|A j | ≤ e k (u). the electronic journal of combinatorics 18 (2011), #P90 2 Proof. The result is obtained from e j (u) ≤ d(u) + |A j | − 13 and d(u) − 9 ≤ e k (u). Corollary 2.2 The degree in G of every vertex of D is 1 0 . Proof. Let w ∈ D. Since d(w) = d D (w) +  4 k=1 e k (w), by Lemma 2.1, we have d D (w) + 4(d(w) − 8) ≤ d(w) ≤ d D (w) +  4 k=1 (d(w) + |A k | − 12). Thus 3d(w) ≥ 48 −  4 k=1 |A k | − d D (w) ≥ 48 −  4 k=1 |A k |−(|D|−1) = 28 and 3d(w) ≤ 32−d D (w) ≤ 32. As 28 ≤ 3 d(w) ≤ 32, the result follows. Lemma 2.2 The vertices of degree 7 or 8 in G belong to K. Proof. Let u ∈ V (G) − V (K). On the one hand, if u ∈ D then, by Corollary 2.2, d(u) = 10, thus d(u) ≥ 9. On the other hand, if u ∈ A i then, by Lemma 2.1, d(u) = 1+d G[A i ] (u)+|D∩N(u)|+  4 k=1,k=i e k (u) ≤ 1+(|A i |−1)+|D|+  4 k=1,k=i (d(u)+|A k |−13) = 3d(u) +  4 k=1 |A k | + |D| − 39 = 3d(u) + 21 − 39. Therefore d(u) ≥ 9. Corollary 2.3 G has exactly one subgraph isomorph i c to K 4 . Proof. Suppose, to the contrary, that there exists K ′ ∈ K − {K}. Since K 5 − P 3  G, we have |V (K) ∩ V (K ′ )| ≤ 1 and, by Lemma 2.2, there are at least three vertices in K ′ with degree in G at least 9. Thus  v∈V (K ′ ) d(v) ≥ 3 · 9 + 1 · 7 = 34. As 21 ≥  4 k=1 |A k | =  4 k=1 (d(v k ) − 3), we have  4 k=1 d(v k ) ≤ 33, contradicting the definition of K. Corollary 2.4 Let u ∈ V − V (K), then K 3  G[N( u)]. Proof. If there is a clique of order 3 in G[N(u)], let u 1 , u 2 and u 3 be its vertices. G[{u, u 1 , u 2 , u 3 }] is a subgraph of G different of K isomorphic to K 4 , contradicting Lemma 2.3. Corollary 2.5 G[A i ] ∈ R(K 3 , K 4 ; |A i |). Proof. If K 3 ⊆ G[A i ], then let u 1 , u 2 and u 3 be the vertices of a clique of order 3 of G[A i ]. G[{u 1 , u 2 , u 3 , v i }] is a subgraph of G isomorphic to K 4 different from K, contradicting Corollary 2.3. If K 4 ⊆ G[A i ], then let u 1 , u 2 , u 3 and u 4 be the four vertices of a clique of o rder 4 of G[A i ]. G[{u 1 , u 2 , u 3 , u 4 , v j }] is a subgraph of G isomorphic to K 5 , a contradiction. Thus G[A i ] ∈ R(K 3 , K 4 ; |A i |). Corollary 2.6 H i ∈ R(K 4 , K 4 ; 21 − |A i |). Proof. K is not a subgraph of H i , thus, by Corollary 2.3, K 4  H i . Since H i ∈ R(K 5 − P 3 , K 4 ; 21 − |A i |) we have K 4  H i , concluding the proof. Lemma 2.3 D = ∅. the electronic journal of combinatorics 18 (2011), #P90 3 Proof. Suppose, to the contrary, that there exists w ∈ D. Let X = V (G) − V (K) − N(w) − {w}. Since N(w) ∩ V (K) = ∅, and, by Corollary 2.2, |N(w)| = 1 0, we have |X| = 25 − 4 − 10 − 1 = 10. As K is not a subgraph of G[X], by Corollary 2.3, K 4  G[X] = G[X]. r(K 3 , K 4 ) = 9 [10], thus R(K 3 , K 4 ; 10) = ∅ and G[X] /∈ R(K 3 , K 4 ; 10), hence K 3 ⊆ G[X]. Let u 1 , u 2 and u 3 be the vertices of a clique of order 3 of G[X] and let k ∈ {1, 2, 3, 4} − {g(u 1 ), g(u 2 ), g(u 3 )}. Then G[{w, v k , u 1 , u 2 , u 3 }] ⊆ G is isomorphic to K 5 , a contradiction. Lemma 2.4 If G[A i ] contains a clique of order 3, with ve rtices w 1 , w 2 and w 3 , then |(N(w 1 ) − N(w 2 ) − N(w 3 )) ∩ V (H i )| ≤ 2. Proof. Let Y = (N(w 1 ) − N(w 2 ) − N(w 3 )) ∩ V (H i ). By Corollary 2.4, K 3  G [Y ] ⊆ G[N(w 1 )]. If K 2 ⊆ G[Y ] then let u 1 u 2 be an edge of G[Y ] and let k ∈ {1, 2, 3, 4} − {i, g(u 1 ), g(u 2 )}. G[{w 1 , w 2 , u 1 , u 2 , v k }] is isomorphic to K 5 , a contradiction. Thus K 2  G[Y ] and G[Y ] ∈ R(K 3 , K 2 ; |Y |). Therefore |(N(w 1 ) − N(w 2 ) − N(w 3 )) ∩ V (H i )| = |Y | ≤ r(K 3 , K 2 ) − 1 = 2. Lemma 2.5 If G[A i ] contains two adjacent vertices w 1 and w 2 then |V (H i ) ∩ N(w 1 ) ∩ N(w 2 )| ≤ 3. Proof. Let Y = V (H i ) ∩ N(w 1 ) ∩ N(w 2 ). If K 2 ⊆ G[Y ] then let u 1 and u 2 be the vertices of an edge of G[Y ]. G[{w 1 , w 2 , u 1 , u 2 }] is a subgraph of G isomorphic to K 4 and different from K, contradicting Corollary 2.3. Thus K 2  G[Y ]. By Corollary 2.6, K 4  G[Y ] ⊆ H i . Thus G[Y ] ∈ R(K 2 , K 4 ; |Y |) and |V (H i )∩N(w 1 )∩ N(w 2 )| = |Y | ≤ r( K 2 , K 4 ) − 1 = 3. Lemma 2.6 Let u ∈ A i , then the following statements are verified: d G[A i ] (u) ≤ 3,  4 k=1,k=i e k (u) ≤ 8 and d(u) ≤ 11. Proof. Suppose, to the contrary, that d G[A i ] (u) ≥ 4. Let u 1 , u 2 , u 3 and u 4 be four different vertices belonging to N G[A i ] (u). If K 2 ⊆ G[{u 1 , u 2 , u 3 , u 4 }], then let u p , u q ∈ {u 1 , u 2 , u 3 , u 4 } be two adjacent vertices. G[{u p , u q , u, v i }] is a subgraph of G isomorphic to K 4 different from K, contradicting Corollary 2.3. Thus K 2  G[{u 1 , u 2 , u 3 , u 4 }] and G[{u 1 , u 2 , u 3 , u 4 }] ⊆ H i is isomorphic to K 4 , contradicting Corollary 2.6. Hence d G[A i ] (u) ≤ 3. By Coro llar y 2.4, K 3  G[V (H i ) ∩ N(u)] ⊆ G[N(u)] and, by Corollary 2.6, K 4  G[V (H i ) ∩ N(u)] ⊆ H i . Therefore G[V (H i ) ∩ N(u)] ∈ R(K 3 , K 4 ; |V (H i ) ∩ N(u)|) and 8 = r(K 3 , K 4 ) − 1 ≥ |V (H i ) ∩ N(u)| =  4 k=1,k=i |V (A k ) ∩ N(u)| =  4 k=1,k=i e k (u), completing the second part of the proof. Finally, by Lemma 2.1, d(u) − 9 ≤ e j (u), thus 3d(u) − 27 ≤  4 k=1,k=i e k (u) ≤ 8 and d(u) ≤ 11. Lemma 2.7 Let u ∈ A i , then |A i | ≤ 2d(u) + d G[A i ] (u) − 17. the electronic journal of combinatorics 18 (2011), #P90 4 Proof. By Lemma 2.1, e j (u) ≤ d(u) − 13 + |A j |. Therefore d(u) − d G[A i ] (u) − 1 =  4 k=1,k=i e k (u) ≤ 3d(u) − 39 +  4 k=1,k=i |A k | = 3d(u) − 39 + (21 − |A i |) and |A i | ≤ 2d(u) + d G[A i ] (u) − 17. Corollary 2.7 Let u ∈ A i , then d G[A i ] (u) ≥ 1 an d if d G[A i ] (u) = 1 then |A i | = 4. Proof. By Lemma 2.6, d(u) = 1+  4 k=1,k=i e k (u)+d G[A i ] (u) ≤ 9+d G[A i ] (u). If d G[A i ] (u) = 0 then d(u) ≤ 9 and, by Lemma 2.7, |A i | ≤ 1, contradicting |A i | ≥ 4. If d G[A i ] (u) = 1 then d(u) ≤ 10 and, by Lemma 2.7, |A i | ≤ 4, thus |A i | = 4. Corollary 2.8 If |A i | = 7 and u ∈ A i then d(u) = 11. Proof. By Lemmas 2.6 and 2.7, 7 = |A i | ≤ 2d(u) + d G[A i ] (u) − 17 ≤ 2d(u) + 3 − 17, thus 2d(u) ≥ 21 and d(u) ≥ 11. Since d(u) ≤ 1 1, we conclude the proof. In the rest of the paper if we assign the name W to an ordered set of vertices, {u 1 , . . . , u p } ⊆ A i , with p ≥ 3, then |(N(u k )−  p t=1,t=k N(u t ))∩ V (H i )| will be denoted by a k , |(N(u h ) ∩ N(u k ) −  p t=1,t=h,k N(u t )) ∩ V (H i )| by b h,k , and |(N(u h ) ∩ N(u k ) ∩ N(u l ) −  p t=1,t=h,k,l N(u t )) ∩ V (H i )| by c h,k,l . Lemma 2.8 If G[A i ] contains a clique of order 3, with vertices u 1 , u 2 and u 3 , then 39 − 2|A i | ≤ d(u 1 ) + d(u 2 ) + d(u 3 ) − d G[A i ] (u 1 ) − d G[A i ] (u 2 ) − d G[A i ] (u 3 ). Proof. Let W = {u 1 , u 2 , u 3 }, let w ∈ V (H i ), and let k ∈ {1, 2, 3, 4} − { i, g(w)}. Since G[{u 1 , u 2 , u 3 , w, v k }] is not isomorphic to K 5 , w is adjacent to at least a vertex of W and, therefore, every vertex o f V (H i ) is adjacent to at least a vertex of W . Hence: a 1 + a 2 + a 3 + b 1,2 + b 1,3 + b 2,3 + c 1,2,3 = |V (H i )| = 21 − |A i | (1) On the one hand, since u 1 is adjacent to d(u 1 ) − d G[A i ] (u 1 ) − 1 vertices of H i we have: a 1 + b 1,2 + b 1,3 + c 1,2,3 = d(u 1 ) − d G[A i ] (u 1 ) − 1 (2) Analogously, u 2 and u 3 are adjacent to d(u 2 ) − d G[A i ] (u 2 ) − 1 and d(u 3 ) − d G[A i ] (u 3 ) − 1 vertices of H i respectively, thus: a 2 + b 1,2 + b 2,3 + c 1,2,3 = d(u 2 ) − d G[A i ] (u 2 ) − 1 (3) a 3 + b 1,3 + b 2,3 + c 1,2,3 = d(u 3 ) − d G[A i ] (u 3 ) − 1 (4) On the other hand, by Lemma 2.4: a 1 = |(N(u 1 ) − N(u 2 ) − N(u 3 )) ∩ V (H i )| ≤ 2 (5) a 2 = |(N(u 2 ) − N(u 1 ) − N(u 3 )) ∩ V (H i )| ≤ 2 (6) the electronic journal of combinatorics 18 (2011), #P90 5 a 3 = |(N(u 3 ) − N(u 1 ) − N(u 2 )) ∩ V (H i )| ≤ 2 (7) Finally, from (2) + ( 3) + (4 ) + (5) + (6) + (7) − 2(1) we have: c 1,2,3 ≤ d(u 1 ) + d(u 2 ) + d(u 3 ) − d G[A i ] (u 1 ) − d G[A i ] (u 2 ) − d G[A i ] (u 3 ) − 39 + 2|A i |. We obtain the result noting that c 1,2,3 is non-negative. Corollary 2.9 If G[A i ] contains a clique of orde r 3, then |A i | ≥ 6 and if |A i | = 6 then for any vertex u of the clique, d G[A i ] (u) = 2 and d(u) = 1 1 . Proof. Let u 1 , u 2 , and u 3 be the three vertices of the clique. By Corollary 2.7, d G[A i ] (u 1 ), d G[A i ] (u 2 ), d G[A i ] (u 3 ) ≥ 1 and, by Lemma 2.6, d(u 1 ), d(u 2 ), d(u 3 ) ≤ 11. Therefore, by Lemma 2.8, 39 − 2|A i | ≤ d(u 1 ) + d(u 2 ) + d(u 3 ) − d G[A i ] (u 1 ) − d G[A i ] (u 2 ) − d G[A i ] (u 3 ) ≤ 11 + 11 + 11 − 1 − 1 − 1 = 30 and |A i | ≥ 5. Hence, by Corollary 2.7, d G[A i ] (u 1 ), d G[A i ] (u 2 ), d G[A i ] (u 3 ) ≥ 2, and 39 − 2|A i | ≤ d(u 1 ) + d(u 2 ) + d(u 3 ) − d G[A i ] (u 1 ) − d G[A i ] (u 2 ) − d G[A i ] (u 3 ) ≤ 11 + 11 + 11 − 2 − 2 − 2 = 27, thus |A i | ≥ 6. If |A i | = 6, on the one hand, 27 = 39 − 2|A i | ≤ d(u 1 ) + d(u 2 ) + d(u 3 ) − d G[A i ] (u 1 ) − d G[A i ] (u 2 ) − d G[A i ] (u 3 ) ≤ d(u) + 11 + 11 − 2 − 2 − 2 = d(u) + 16, hence d(u) = 11. On the other hand, 27 = 39 − 2|A i | ≤ d(u 1 ) + d(u 2 ) + d(u 3 ) − d G[A i ] (u 1 ) − d G[A i ] (u 2 ) − d G[A i ] (u 3 ) ≤ 11 + 11 + 11 − d G[A i ] (u) − 2 − 2 = 29 − d G[A i ] (u), therefore d G[A i ] (u) = 2. Corollary 2.10 Let u ∈ A i , then d(u) ≥ 10. Proof. By Lemma 2.6, d G[A i ] (u) ≤ 3. If d(u) ≤ 9 then by Lemma 2.2, d(u) = 9 and, by Lemma 2.7, |A i | ≤ 2d(u) + d G[A i ] (u) − 17 ≤ 18 + 3 − 17 = 4, thus |A i | = 4. By Lemma 2.1, 8 − d G[A i ] (u) = d(u) − 1 − d G[A i ] (u) =  4 k=1,k=i e k (u) ≤  4 k=1,k=i (|A k | − 13 + d(u)) =  4 k=1,k=i |A k | − 12 = (21 − 4) −12 = 5. Hence d G[A i ] (u) ≥ 3 and u is adjacent to the three vertices of A i − {u}. By Corollary 2.9, K 3  G[A i ] and at least two of the three vertices of A i − {u} are adjacent. Let w 1 and w 2 denote them. Then u, w 1 and w 2 are the vertices of a clique of G[A i ], contradicting Corollary 2.5 and completing the proof. From Corollaries 2.5 and 2.9 and Lemma 2.6, it is easy to check the next result: Corollary 2.11 1. If |A i | = 4 then G[A i ] is isomorph i c to 2K 2 , P 4 or C 4 . 2. If |A i | = 5 then G[A i ] is isomorph i c to C 5 . 3. If |A i | = 6 then G[A i ] is isomorphic to C 6 or SK 2,3 (the g raph obtained subdividing one edge of K 2,3 ). Now, we prove that G[A i ] is isomorphic neither to C 5 nor to SK 2,3 . Lemma 2.9 G[A i ] is not isomorphi c to C 5 . the electronic journal of combinatorics 18 (2011), #P90 6 Proof. Suppose, to the contrary, that G[A i ] is isomorphic to C 5 . Let W = {u 1 , . . . , u 5 } denote its vertices, with its edges being u 1 u 2 , u 2 u 3 , u 3 u 4 , u 4 u 5 and u 1 u 5 . Let w ∈ V (H i ). By Lemmas 2.1 and 2.6 and Corollary 2.10, 1 ≤ d(w) − 9 ≤ e i (w) ≤ d(w) − 13 + |A i | = d(w) − 8 ≤ 3, thus every vertex of H i is adjacent to 1, 2 or 3 vertices of G[A i ] and we have: 5  k=1 a k +  1≤k<m≤5 b k,m +  1≤k<m<n≤5 c k,m,n = |V (H i )| = 16 (8) On the one hand, by Lemma 2.5: b 1,2 + c 1,2,3 + c 1,2,4 + c 1,2,5 = |V (H i ) ∩ N(u 1 ) ∩ N(u 2 )| ≤ 3 (9) b 2,3 + c 1,2,3 + c 2,3,4 + c 2,3,5 = |V (H i ) ∩ N(u 2 ) ∩ N(u 3 )| ≤ 3 (10) b 3,4 + c 1,3,4 + c 2,3,4 + c 3,4,5 = |V (H i ) ∩ N(u 3 ) ∩ N(u 4 )| ≤ 3 (11) b 4,5 + c 1,4,5 + c 2,4,5 + c 3,4,5 = |V (H i ) ∩ N(u 4 ) ∩ N(u 5 )| ≤ 3 (12) b 1,5 + c 1,2,5 + c 1,3,5 + c 1,4,5 = |V (H i ) ∩ N(u 1 ) ∩ N(u 5 )| ≤ 3 (13) On the other hand, let Y = V (H i ) − N(u 1 ) − N(u 3 ). By Corollary 2.6, K 4  G[Y ] ⊆ H i . If K 2 ⊆ G [Y ] then let w 1 and w 2 be the vertices of an edge of G[Y ] and let k ∈ {1, 2, 3, 4} − {i, g(w 1 ), g(w 2 )}. G[{u 1 , u 3 , w 1 , w 2 , v k }] is isomorphic to K 5 , a contradiction. Thus K 2  G[Y ], G[Y ] ∈ R(K 4 , K 2 ; |Y |) and: a 2 + a 4 + a 5 + b 2,4 + b 2,5 + b 4,5 + c 2,4,5 = |V (H i ) − N(u 1 ) − N(u 3 )| = = |Y | ≤ r(K 4 , K 2 ) − 1 = 3 (14) Similarly G[V (H i ) − N(u 1 ) − N(u 4 )] ∈ R(K 4 , K 2 ; |V (H i ) − N(u 1 ) − N(u 4 )|) and: a 2 + a 3 + a 5 + b 2,3 + b 2,5 + b 3,5 + c 2,3,5 ≤ 3 (15) G[V (H i ) − N(u 2 ) − N(u 4 )] ∈ R(K 4 , K 2 ; |V (H i ) − N(u 2 ) − N(u 4 )|) and: a 1 + a 3 + a 5 + b 1,3 + b 1,5 + b 3,5 + c 1,3,5 ≤ 3 (16) G[V (H i ) − N(u 2 ) − N(u 5 )] ∈ R(K 4 , K 2 ; |V (H i ) − N(u 2 ) − N(u 5 )|) and: a 1 + a 3 + a 4 + b 1,3 + b 1,4 + b 3,4 + c 1,3,4 ≤ 3 (17) the electronic journal of combinatorics 18 (2011), #P90 7 G[V (H i ) − N(u 3 ) − N(u 5 )] ∈ R(K 4 , K 2 ; |V (H i ) − N(u 3 ) − N(u 5 )|) and: a 1 + a 2 + a 4 + b 1,2 + b 1,4 + b 2,4 + c 1,2,4 ≤ 3 (18) From (9) + · · · + (18 ) we have: 3 5  k=1 a k + 2  1≤k<m≤5 b k,m + 2  1≤k<m<n≤5 c k,m,n ≤ 30 (19) Finally, from (19) − 2(8) we obtain  5 k=1 a k ≤ −2. This contradiction completes the proof. Lemma 2.10 G[A i ] is not isomorphi c to SK 2,3 . Proof. Suppose, to the contrary, that G[A i ] is isomorphic to SK 2,3 . Let W = {u 1 , . . . , u 4 } be the set of vertices of A i of degree 2 in G[A i ], with its only edge being u 1 u 2 . Every vertex of W belongs to a clique of order 3 contained in G[A i ], thus, by Corollary 2.9, each vertex of W has degree 11 in G. Let h = |V (H i ) ∩  4 n=1 N(u n )|. Since d(u 1 ) = 11 and d G[A i ] (u 1 ) = 2 we have that, |N(u)∩V (H i )|, the number of edges incident to u 1 and a vertex of H i is: a 1 + b 1,2 + b 1,3 + b 1,4 + c 1,2,3 + c 1,2,4 + c 1,3,4 + h = d(u 1 ) − d G[A i ] (u 1 ) − 1 = 8 (20) On the one hand, by Lemma 2.4: a 1 + b 1,2 = |(N(u 1 ) − N(u 3 ) − N(u 4 )) ∩ V (H i )| ≤ 2 (21) On the other hand, let Y = V (H i ) ∩ N(u 1 ) ∩ N(u 3 ). By Corollary 2 .4 , K 3  G[Y ] ⊆ G[N(u 1 )]. If K 2 ⊆ G[Y ] then let w 1 and w 2 be the vertices of an edge of G[Y ] and let k ∈ {1, 2, 3, 4} − {i, g(w 1 ), g(w 2 )}. G[{u 1 , u 3 , w 1 , w 2 , v k }] is isomorphic to K 5 , a contradiction. Therefore K 2  G[Y ], G[Y ] ∈ R(K 3 , K 2 ; |Y |) and: b 1,3 + c 1,2,3 + c 1,3,4 + h = |V (H i ) ∩ N(u 1 ) ∩ N(u 3 )| = |Y | ≤ r( K 3 , K 2 ) − 1 = 2 (22) Similarly G[V (H i ) ∩ N(u 1 ) ∩ N(u 4 )] ∈ R(K 3 , K 2 ; |V (H i ) ∩ N(u 1 ) ∩ N(u 4 )|) and: b 1,4 + c 1,2,4 + c 1,3,4 + h ≤ 2 (23) From (21) + (22) + (23) − (20) we obtain c 1,3,4 + h ≤ −2, a contra diction. Finally, we prove Theorem 2.1: Proof. By Corollaries 2.8 and 2.11 and Lemmas 2.9 and 2.10, |A 1 | = 7, |A 2 | = 6, |A 3 | = |A 4 | = 4, G[A 2 ] is isomorphic to C 6 and all vertices of A 1 ∪ A 2 have degree 11 in G. Let s denote the number of edges of G[A 3 ]. the electronic journal of combinatorics 18 (2011), #P90 8 By Corollary 2.6, H 4 ∈ R(K 4 , K 4 ; 17). Kalbfleisch [14] proved t hat there is exactly one graph in the set R(K 4 , K 4 ; 17) and every vertex of this graph has degree 8, thus, for any w ∈ V (H 4 ), d H 4 (w) = 8. If u ∈ A 1 ∪ A 2 , by Lemma 2.1, 2 = d(u) − 9 ≤ e 3 (u) ≤ d(u) − 13 + |A 3 | = 2, thus e 3 (u) = 2 and the number of edges of H 4 with a vertex belonging to A 1 ∪ A 2 and another vertex belonging to A 3 is  w∈A 3 (e 1 (w) + e 2 (w)) =  u∈A 1 ∪A 2 e 3 (u) = 2|A 1 ∪ A 2 | = 26. If w ∈ A 3 , then 8 = d H 4 (w) = d G[A 3 ] (w) + e 1 (w) + e 2 (w). Thus 32 =  w∈A 3 8 =  w∈A 3 d G[A 3 ] (w) +  w∈A 3 (e 1 (w) + e 2 (w)) = 2s + 26 and s = 3. Hence, by Corollary 2.11, G[A 3 ] is isomorphic to P 4 . In the following, we assume that i = 3. Let W = {u 1 , . . . , u 4 } be the vertices of A 3 , with the edges of G[A 3 ] being u 1 u 2 , u 2 u 3 and u 3 u 4 . Let w ∈ V (H 3 ). By Lemmas 2.1 and 2.6 and Corollary 2.10, 1 ≤ d(w) − 9 ≤ e 3 (w) ≤ d(w) − 13 + |A 3 | = d(w) − 9 ≤ 2, thus every vertex of H 3 is adjacent to 1 or 2 vertices of A 3 and |N(u 2 ) ∩ V (H 3 )| is: a 2 + b 1,2 + b 2,3 + b 2,4 = d(u 2 ) − d G[A 3 ] (u 2 ) − 1 ≥ 10 − 2 − 1 = 7 (24) Let Y 1 = V (H 3 ) ∩ N(u 2 ) − N(u 1 ) − N(u 3 ). By Corollary 2.4, K 3  G[Y 1 ] ⊆ G[N(u 2 )]. If K 2 ⊆ G[Y 1 ] then let w 1 and w 2 be the vertices of an edge of G[Y 1 ] and let k ∈ {1, 2, 4} − {g(w 1 ), g(w 2 )}. G[{u 1 , u 3 , w 1 , w 2 , v k }] is isomorphic to K 5 , a contradiction. Hence K 2  G[Y 1 ], G[Y 1 ] ∈ R(K 3 , K 2 ; |Y 1 |) and: a 2 + b 2,4 = |V (H 3 ) ∩ N(u 2 ) − N(u 1 ) − N(u 3 )| = |Y 1 | ≤ r(K 3 , K 2 ) − 1 = 2 (25) On the one hand, by Lemma 2.5: b 1,2 = |V (H 3 ) ∩ N(u 1 ) ∩ N(u 2 )| ≤ 3 (26) On the other hand, let Y 2 = V (H 3 )∩ N(u 2 )∩ N(u 3 ). If K 2 ⊆ G[Y 2 ] then let w 1 and w 2 be the vertices of an edge of G[Y 2 ]. G[{w 1 , w 2 , u 2 , u 3 }] is a subgraph of G isomor phic to K 4 and different from K, contradicting Corollary 2.3, thus K 2  G[Y 2 ]. If K 2 ⊆ G[Y 2 ] then let w 1 and w 2 be the vertices of an edge of G[Y 2 ] and let k ∈ {1, 2, 4} − {g(w 1 ), g(w 2 )}. G[{u 1 , u 4 , w 1 , w 2 , v k 3 }] is isomorphic to K 5 , a contradiction. Hence K 2  G[Y 2 ], G[Y 2 ] ∈ R(K 2 , K 2 ; |Y 2 |) and: b 2,3 = |V (H 3 ) ∩ N(u 2 ) ∩ N(u 3 )| = |Y 2 | ≤ r(K 2 , K 2 ) − 1 = 1 (27) From (25)+(26) + ( 27) we obtain a 2 + b 1,2 + b 2,3 + b 2,4 ≤ 6, contradicting (24) a nd completing the proof of Theorem 2.1. the electronic journal of combinatorics 18 (2011), #P90 9 References [1] A. Babak, S.P. Radziszowski and Kung-Kuen Tse. Computation of the Ramsey Number R(B 3 , K 5 ). Bulletin of the Institute of Combinatorics and its Applications, 41:71–76, 2004. [2] K. Black, D. Leven and S.P. Radziszowski. New Bounds on Some Ramsey Num- bers. 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Stinehour and Kung-Kuen Tse. Computation of the Ramsey Number R(W 5 , K 5 ). Bulletin of the Ins titute of Combin atoric s and its Applications, 47:53–57, 2006. [18] Y. Yuansheng and G.R.T. Hendry. The Ramsey Number r(K 1 +C 4 , K 5 −e). Journal of Graph Theory, 19:13–15, 1995. the electronic journal of combinatorics 18 (2011), #P90 10 . R(K m − P 3 , K n ; k) and v ∈ V (F ) then F [N F (v)] ∈ R(K m−1 −P 3 , K n ; d F (v)) and F[N F (v)] ∈ R(K m −P 3 , K n−1 ; k−1 −d F (v)). Thus ∆(F ) ≤ r(K m−1 − P 3 , K n ) − 1 and δ(F ) ≥ k −. D. Then d(u) − 9 ≤ e j (u) ≤ d(u) + |A j | − 13 and d(w) − 8 ≤ e i (w) ≤ d(w) + |A i | − 12. Proof. Since H j ∈ R(K 5 −P 3 , K 4 ; 2 1−| A j |), r(K 4 −P 3 , K 4 ) = 9 [6] and r(K 5 −P 3 , K 3 ). have d D (w) + 4(d(w) − 8) ≤ d(w) ≤ d D (w) +  4 k=1 (d(w) + |A k | − 12). Thus 3d(w) ≥ 48 −  4 k=1 |A k | − d D (w) ≥ 48 −  4 k=1 |A k |−( |D |−1 ) = 28 and 3d(w) ≤ 32−d D (w) ≤ 32. As 28 ≤