Extremal Graphs without Topological Complete Subgraphs

c 2004 Society for Industrial and Applied Mathematics
Downloaded 01/22/16 to 150.214.182.82. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php SIAM J. DISCRETE MATH. Vol. 18, No. 2, pp. 388–396 EXTREMAL GRAPHS WITHOUT TOPOLOGICAL COMPLETE SUBGRAPHS∗ M. CERA† , A. DIÁNEZ‡ , AND A. MÁRQUEZ§ ⌉ ≤ p < n Abstract. The exact values of the function ex(n; T Kp ) are known for ⌈ 2n+5 3 (see [Cera, Diánez, and Márquez, SIAM J. Discrete Math., 13 (2000), pp. 295–301]), where ex(n; T Kp ) is the maximum number of edges of a graph of order n not containing a subgraph homeomorphic to the complete graph of order p. In this paper, for ⌈ 2n+6 ⌉ ≤ p < n − 3, we characterize the family of 3 extremal graphs EX(n; T Kp ), i.e., the family of graphs with n vertices and ex(n; T Kp ) edges not containing a subgraph homeomorphic to the complete graph of order p. Key words. extremal graph theory, topological complete subgraphs AMS subject classifications. 05C35, 05C70 DOI. 10.1137/S0895480100378677 1. Introduction. The study of the function ex(n; T Kp )—i.e., the maximum number of edges of a graph of order n not containing a subgraph homeomorphic to Kp , where Kp is the complete graph with p vertices—is one of the most general extremal problems, as pointed out by Bollobas in [1]. Exact values for this function are known only in some cases, as can be seen in Table 1.1. Table 1.1 Exact values of the function ex(n; T Kp ).
2n+5  3 p 3 4 5 ex(n; T Kp ) n−1 2n − 3 3n − 6 [3] [4], [8], [9] . . . . . . . . . 4 4  n 2  − (5n − 6p + 3) [2] ≤p<n  n 2  − (2n − 2p + 1) [2] ≤p<
3n+2  Reference
3n+2  The aim of this work is to characterize a family of extremal graphs EX(n; T Kp ) for appropriate values of n and p, i.e., the set of graphs of order n, with ex(n; T Kp ) ∗ Received by the editors September 28, 2000; accepted for publication (in revised form) July 1, 2004; published electronically December 9, 2004. This research was partially supported by the Ministry of Science and Technology, Spain, Research Project BMF2001-2474. http://www.siam.org/journals/sidma/18-2/37867.html † E.U.I.T. Agrı́cola, Universidad de Sevilla, Ctra. Utrera s/n, 41013-Sevilla, Spain ([email protected]). ‡ E.T.S. Arquitectura, Universidad de Sevilla, Reina Mercedes 2, 41012-Sevilla, Spain (anadianez@ us.es). § E.T.S.I. Informática, Universidad de Sevilla, Reina Mercedes s/n, 41012-Sevilla, Spain (almar@ us.es). 388 389 Downloaded 01/22/16 to 150.214.182.82. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php EXTREMAL TOPOLOGICAL SUBGRAPHS edges and not containing any subgraph homeomorphic to Kp . Actually, we characterize the family EX(n; T Kp ) for ⌈ 2n+6 3 ⌉ ≤ p < n − 3: EX(n; T Kp ) = ⎧ ⎨ (3n − 4p + 2)K3 + (6p − 4n − 3)K2 ⎩ K4p−3n−2 + (2n − 2p + 1)K2 for for
2n+6  3
3n+2  4 ≤p<
3n+2  4 , ≤ p < n − 3. 2. Definitions and notation. Given a graph H and a set {v1 , . . . , vq } of vertices of H, we denote by H0 = H and by Hk for k = 1, . . . , q the induced subgraph in H by the set of vertices V (H) − {v1 , . . . , vk }. We denote by ∆(H) the maximum degree of the graph H and by δH (v) the degree of the vertex v in the graph H. The complement graph of H will be denoted by H. Let q and s be a pair of nonnegative integers; Cqs denotes the set of graphs H such that there exists a set {v1 , . . . , vq } of vertices of H verifying the following: (1) δHj−1 (vj ) ≥ δHj (vj+1 ) for j = 1, . . . , q − 1. (2) For each positive integer h, if there exists k ∈ {1, . . . , q} and v ∈ Hk such that δHk (v) ≥ h, then δHj (vj+1 ) ≥ h for all j = 1, . . . , k. (3) Hq has at most s edges (i.e., |E(Hq )| ≤ s). The next results show different conditions to guarantee that a graph belongs to the family described above (see [2]). Lemma 2.1 (see [2]). Let H be a graph with n vertices. Then, for any q ≤ n, there exists s such that H is in Cqs . When s = q, we know sufficient conditions for the edges of a graph to belong to the class Cqq . Lemma 2.2 (see [2]). Let n and q be two positive integers, with q < n. If H is a graph with n vertices and 2q edges, then 1. H ∈ Cqq , 2. δHq (v) ≤ 1 for v ∈ V (Hq ). Lemma 2.3 (see [2]). Let q and k be two positive integers with k ≤ q − 2. Let H be a graph with 4q − k + 1 vertices and 2q + k + 1 edges. Then H ∈ Cqq . Notation and terminology not given here can be found in [1] and [2]. 3. The family of extremal graphs. In this section, we will characterize the family EX(n; T Kp ) for ⌈ 2n+6 3 ⌉ ≤ p < n − 3. This problem is equivalent to characterizing EX(n; T Kn−q ) for n ≥ 4q + 2 with q ≥ 4 (case ⌈ 3n+2 4 ⌉ ≤ p < n − 3) and ⌉ ≤ p < ⌈ 3n+2 n = 4q − k + 1 with q ≥ 5, 0 ≤ k ≤ q − 5 (the case ⌈ 2n+6 3 4 ⌉). In order to avoid excessive repetition, we define the graphs H(n; T Kn−q ): H(n; T Kn−q ) = ⎧ ⎨ Kn−(4q+2) + (2q + 1)K2 ⎩ (k + 1)K3 + (2(q − k) − 1)K2 for n ≥ 4q + 2, for n = 4q − k + 1, 0 ≤ k ≤ q − 5. For n ≥ 4q + 2, a graph G belongs to the family {H(n; T Kn−q )} if G has n vertices and G is formed by 2q + 1 nonadjacent edges (see Figure 3.1). For n = 4q − k + 1 with q ≥ 5 and 0 ≤ k ≤ q − 5, a graph G belongs to the family {H(n; T Kn−q )} if it has 4q − k + 1 vertices and G is formed by k + 1 nonadjacent triangles and 2(q − k) − 1 nonadjacent edges, as Figure 3.2 shows. In the next two sections, we will prove the following theorem. Theorem 3.1. EX(n; T Kp ) = {H(n; T Kp )} for ⌈ 2n+6 3 ⌉ ≤ p < n − 3. Downloaded 01/22/16 to 150.214.182.82. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php 390 M. CERA, A. DIÁNEZ, AND A. MÁRQUEZ 2q + 1 nonadjacent edges t t t t t q q q t n − (4q + 2) isolated vertices q q q t t t Fig. 3.1. Structure of G for n ≥ 4q + 2. t t t k + 1 triangles t t q t t t t t q t t 2(q − k) − 1 nonadjacent edges t t t t q q t t t t t q q t t Fig. 3.2. Structure of G for n = 4q − k + 1. ⌉ ≤ p < n − 3. The aim of this section is to prove Theorem 3.1 4. Case ⌈ 3n+2 4 when n and p are related by the expression ⌈ 3n+2 4 ⌉ ≤ p < n − 3. Proposition 4.1. Let n and p be two positive integers such that ⌈ 3n+2 4 ⌉ ≤ p < n − 3. It is verified that EX(n; T Kp ) = {H(n; T Kp )}. In order to provide this proposition, we need some previous results. First, we recall the following results about the function ex(n; T Kn−q ) (see [2]). Theorem 4.2 (see [2]). Let n and q be two positive integers. If n ≥ 4q + 2, then  n ex(n; T Kn−q ) = − (2q + 1). 2 Also, we recall that, given a graph H and v ∈ H, the set of vertices adjacent to v in H is denoted by Γ(v) (see [1]). Given a bipartite graph B whose classes are X and Y with |X| ≤ |Y |, we say that B has a complete matching if there exists a set of nonadjacent edges in B with cardinality |X|. If we need to show the existence of a complete matching in a bipartite graph, then we can use Hall’s condition. Theorem 4.3 (see [5]). Given a bipartite graph with classes X and Y , if |Γ(A)| ≥ |A| for all A ⊆ X, where Γ(A) = v∈A Γ(v), then there exists a complete matching. The next result asserts that for any graph G ∈ EX(n; T Kn−q ) its complement graph G is extremal for Cqq+1 in the sense that G ∈ Cqq+1 and G ∈ Cqq . Lemma 4.4. Let n and q be two nonnegative integers with q ≥ 4 and n ≥ 4q + 2. For every graph G from the family of graphs EX(n; T Kn−q ), we have G ∈ Cqq+1 − Cqq . Proof. Let G be a graph such that G ∈ EX(n; T Kn−q ). The graph G does not contain a subgraph homeomorphic to Kn−q , so by Theorem 4.2, we know that  n |E(G)| = − (2q + 1). 2 Downloaded 01/22/16 to 150.214.182.82. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php EXTREMAL TOPOLOGICAL SUBGRAPHS 391 Hence, |E(H)| = 2q + 1, where H = G. By Lemma 2.1, there exists an integer s such that H ∈ Cqs . This means that there exists a subset {v1 , . . . , vq } of vertices of G verifying |E(Hq )| ≤ s, where Hq = H − {v1 , . . . , vq }. If s ≤ q + 1, then H ∈ Cqq+1 . Otherwise (s > q + 1), let H ∗ be the graph obtained from H by removing one of the edges of the subgraph Hq . The graph H ∗ has n ≥ 4q + 2 vertices and 2q edges, and applying Lemma 2.2 results in H ∗ ∈ Cqq . Furthermore, by the construction of the graph H ∗ , the set of vertices chosen to prove that H ∗ belongs to the class of graphs Cqq is the same as the one we chose previously in H; thus |E(Hq )| ≤ q + 1 and H ∈ Cqq+1 . Now we will prove that the number of edges of Hq may not be equal to or less than q, i.e., H ∈ Cqq . Suppose that H ∈ Cqq . This means there exists a set of vertices {v1 , . . . , vq } guaranteeing this assertion. Let e1 = (a1 , b1 ), . . . , es = (as , bs ) be the edges of Hq with 1 ≤ s ≤ q. We consider the bipartite graph B whose classes are X = {e1 , . . . , es } and Y = {v1 , . . . , vq } such that ei is adjacent to vj in B if the path ai vj bi exists in G. We note that if there exists a complete matching in B, then we have that G contains a subgraph homeomorphic to Kn−q . Now Hall’s condition implies the existence of a complete matching. Thus, we will prove that |Γ(A)| ≥ |A| for each A ⊆ X. Let A = {ei } be a subset of X with |A| = 1 for i ∈ {1, . . . , s}. If |Γ(A)| = 0, then ei is nonadjacent to any vertex of the set {vq−2 , vq−1 , vq } in B. Hence, no vertex v ∈ {vq−2 , vq−1 , vq } is adjacent to both ai and bi in G. Consequently, δHq−1 (ai ) ≥ 2 or δHq−1 (bi ) ≥ 2 and, furthermore, δHq−3 (ai ) ≥ 3 or δHq−3 (bi ) ≥ 3. Thus, using property (2) of the definition of Cqq , we obtain that δHj−1 (vj ) ≥ 3 for j = 1, . . . , q − 2 and δHj−1 (vj ) ≥ 2 for j = q − 1, q. Therefore, since s ≥ 1 we have that |E(H)| ≥ 3(q − 2) + 2 · 2 + s ≥ 2q + 2 for q ≥ 3. But this is not possible since |E(H)| = 2q + 1. We consider A = {ei , ej } ⊆ X for i, j ∈ {1, . . . , s} with i = j, and we suppose |Γ(A)| ≤ 1. This means that at least three vertices of the set {vq−3 , vq−2 , vq−1 , vq } are nonadjacent to ei and to ej in B. Taking into account property (2) of the definition of Cqq , we have that δHj−1 (vj ) ≥ 3 for j = 1, . . . , q − 3, δHj−1 (vj ) ≥ 2 for j = q − 2, q − 1 and δHq−1 (vq ) ≥ 1 (see Figure 4.1). Hence, |E(H)| ≥ 3(q − 3) + 2 · 2 + 1 + s ≥ 2q + 2 for q ≥ 4, and this is a contradiction, as in the previous case. Let m be an integer with 3 ≤ m ≤ s. Let A be the set of vertices {ei1 , . . . , eim } ⊆ {e1 , . . . , es } with i1 < i2 < · · · < im . If |Γ(A)| ≤ m − 1, then there v1 t q vq−3 vq−2 vq−1 vq q q t t t t ☎☎❚❚ ✔✔❉❉ ✔✔❉❉ ☎ ❚✔ ❉ ✔ ❉ ☎ ✔ ❚ ❉✔ ❉ ❚✔ ❉ ❉ ☎ ✔ ☎✔ ✔❚❉ ❉ ✔ ☎t t✔ ❚❉t ❉t aj ai bj bi Fig. 4.1. Possible structure of H for the most unfavorable case for A = {ei , ej }. Downloaded 01/22/16 to 150.214.182.82. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php 392 M. CERA, A. DIÁNEZ, AND A. MÁRQUEZ v1 t q q t ei1 vq−m vq−(m−1) vj vq q t t q q q t q q q t ☞❭ ☞ ❭ ☞ ❭ ☞ ❭ ☞ ❭ ☞ t t☞ t q q q❭❭t t eim ei2 Fig. 4.2. Possible structure of H for the most unfavorable case for 3 ≤ m ≤ s. exists i ∈ {q − (m − 1), . . . , q} in such a way that vi is not adjacent to any vertex of the set A in the graph B. By applying condition (2) of the definition of Cqq , we obtain that δHq−m (vq−(m−1) ) ≥ m and, therefore, δHj−1 (vj ) ≥ m for 1 ≤ j ≤ q − (m − 1) (see Figure 4.2). Furthermore, δHj−1 (vj ) ≥ 1 for q −(m−2) ≤ j ≤ q and |E(Hq )| = s ≥ m. Consequently, |E(H)| ≥ m(q − (m − 1)) + m − 1 + s ≥ mq − m2 + 3m − 1. Since E(H) = 2q + 1, we have that 2q + 1 ≥ mq − m2 + 3m − 1 and, therefore, 2 −3m+2 q ≤ m m−2 ≤ m − 1 < m ≤ s, but this is not possible. Therefore, |Γ(A)| ≥ |A| for each A ⊆ X. Thus, by Hall’s condition, there exists a complete matching in B and, thereby, the graph G contains a subgraph homeomorphic to Kn−q . This is not possible, and the result follows. Now we can prove Proposition 4.1. Proof of Proposition 4.1. It is equivalent to prove that EX(n; T Kn−q ) = {H(n; T Kn−q )} for q ≥ 4 and n ≥ 4q + 2. Let G be a graph belonging to {H(n; T Kn−q )} with n ≥ 4q + 2. It is easy to check that G does not contain a subgraph homeomorphic to Kn−q . Furthermore, by denoting |E(G)| as the number of edges of G, we have that  n |E(G)| = ex(n; T Kn−q ) = − (2q + 1). 2 Thus, by Theorem 4.2, G is maximal on edges and {H(n; T Kn−q )} ⊆ EX(n; T Kn−q ). In order to prove that EX(n; T Kn−q ) ⊆ {H(n; T Kn−q )}, let G be a graph belonging to EX(n; T Kn−q ), and we set H = G. By Theorem 4.2 we have that |E(H)| = 2q + 1. By Lemma 2.1, we know there exists s such that H ∈ Cqs . Let {v1 , . . . , vq } be a set of q vertices guaranteeing this property. We know that there exists a vertex v ∈ Hq such that δHq (v) ≥ 1, because otherwise Hq is empty and H ∈ Cqq . But this is not possible because, by Lemma 4.4, we know that H ∈ Cqq . If δ(v1 ) ≥ 2, then |E(Hq )| ≤ 2q + 1 − (2 + q − 1) = q and therefore H ∈ Cqq , a contradiction. Therefore, δ(v1 ) ≤ 1. Thus, as v1 is the vertex of maximum degree in H, we have that δ(v) ≤ 1 for all v ∈ H, and then the graph H is formed by 2q + 1 nonadjacent edges. Therefore, the result follows. Downloaded 01/22/16 to 150.214.182.82. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php EXTREMAL TOPOLOGICAL SUBGRAPHS 393 5. Case ⌈ 2n+6 ⌉ ≤ p < ⌈ 3n+2 ⌉. In this section, we will characterize the family 3 4 of extremal graphs EX(n; T Kn−q ) for n = 4q − k + 1 with 0 ≤ k ≤ q − 5 in such a way that we will show that EX(n; T Kn−q ) = {H(n; T Kn−q )}, applying techniques based on the same ideas as in the previous section. 3n+2 Theorem 5.1. Let n and p be two positive integers with ⌈ 2n+6 3 ⌉ ≤ p < ⌈ 4 ⌉. Then EX(n; T Kp ) = {H(n; T Kp )}. In order to prove this result, we also need to recall some results about the function ex(n; T Kn−q ) (see [2]). Lemma 5.2 (see [2]). Let k be a nonnegative integer and H be a graph with maximum degree 2 and at least 3k + 1 vertices of maximum degree. Then there exist at least k + 1 nonadjacent vertices with degree 2. Theorem 5.3 (see [2]). Let n, k, and q be three nonnegative integers with 0 ≤ k ≤ q − 4 and n = 4q − k + 1. It is verified that  n ex(n; T Kn−q ) = − (2q + k + 2). 2 Now we will show, as in Lemma 4.4, that if G ∈ EX(n; T Kn−q ) with n = 4q−k+1, then G ∈ Cqq+1 but G ∈ Cqq . Lemma 5.4. Let k, n, and q be three nonnegative integers such that q ≥ 5, 0 ≤ k ≤ q − 5, and n = 4q − k + 1. If G ∈ EX(n; T Kn−q ), then G ∈ Cqq+1 − Cqq . Proof. Let G be a graph belonging to EX(n; T Kn−q ). This graph does not contain a graph homeomorphic to Kn−q , and by Theorem 5.3 we know that  n |E(G)| = − (2q + k + 2). 2 Thus, H = G has 2q + k + 2 edges. Let H ∗ be the graph obtained from H by removing one edge, similar to what we have done in Lemma 4.4. Since H ∗ is a graph formed by 4q − k + 1 vertices and 2q + k + 1 edges, then applying Lemma 2.3 yields H ∗ ∈ Cqq , and then H ∈ Cqq+1 . Now we will show that H ∈ Cqq . To the contrary, suppose H ∈ Cqq and let {v1 , . . . , vq } be a set of vertices of H guaranteeing that H ∈ Cqq . Let e1 = (a1 , b1 ), . . . , es = (as , bs ) be the edges of Hq with s ≤ q. We consider the bipartite graph B constructed as in Lemma 4.4, i.e., the graph whose classes are X = {e1 , . . . , es } and Y = {v1 , . . . , vq } in such a way that ei is adjacent to vj if the path ai vj bi exists in the graph G. In this case, if we show the existence of a complete matching in B, then we would have that G contains a subgraph homeomorphic to Kn−q . Therefore, we will show that |Γ(A)| ≥ |A| for each A ⊆ X. If |A| = m = 1, by reasoning as in the proof of Lemma 4.4, we have that |E(H)| ≥ 3(q − 2) + 4 + s = 3q + s − 2 ≥ 3q − 1. Downloaded 01/22/16 to 150.214.182.82. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php 394 M. CERA, A. DIÁNEZ, AND A. MÁRQUEZ Since k ≤ q − 4, it is verified that 3q − 1 ≥ 2q + k + 4 − 1 > 2q + k + 2, but this is not possible. For m = 2, by considering as done previously, we have that |E(H)| ≥ 3(q − 3) + 4 + 1 + s = 3q − 4 + s ≥ 3q − 2. Taking into account that k ≤ q − 5, it is verified that |E(H)| > 2q + k + 2, and this is a contradiction. We consider m = 3. Let A = {ei1 , ei2 , ei3 } be a subset of vertices of X with 1 ≤ i1 < i2 < i3 ≤ s. If |Γ(A)| ≤ 2, then there exists i ∈ {q − 2, . . . , q} in such a way that vi is not adjacent to any vertex of the set A in the graph B. Hence, by applying property (2) of the definition of Cqq , we have that δHq−3 (vq−2 ) ≥ 3. Thus, |E(H)| ≥ 3(q − 2) + 2 + s ≥ 3q − 1 > 2q + k + 2 since k ≤ q − 4. In general, if 4 ≤ m ≤ s, then we consider A as the set of vertices {ei1 , . . . , eim } ⊆ {e1 , . . . , es } with i1 < i2 < · · · < im . If |Γ(A)| ≤ m − 1, then there exists i ∈ {q − (m − 1), . . . , q} in such a way that vi is not adjacent to any vertex of the set A in the graph B. Hence, as in the proof of Lemma 4.4, we have that δHq−m (vq−(m−1) ) ≥ m and, therefore, |E(H)| ≥ m(q − (m − 1)) + m − 1 + s ≥ mq − m2 + 3m − 1. But |E(H)| = 2q + k + 2 ≤ 3q − 3 for k ≤ q − 5. Thus, 3q − 3 ≥ mq − m2 + 3m − 1 2 and, thereby, q ≤ m − m−3 < m, but this is not possible. Thus, using Hall’s condition, there exists a complete matching in B, and consequently, G contains a subgraph homeomorphic to Kn−q , but this is not possible. Hence, H ∈ Cqq and the result follows. The next result is devoted to proving the existence of nonadjacent triangles in graphs with maximum degree 2 and the prescribed number of vertices of maximum degree. Lemma 5.5. Let r be a nonnegative integer, and let H be a graph with maximum degree 2. If H has 3r + 3 vertices of degree 2 and r + 1 of them form an independent set, then H contains r + 1 nonadjacent triangles. Proof. We apply induction on r. For r = 0 the result is obvious, because the triangle is the unique graph formed by 3 vertices of degree 2 and all of them are adjacent among themselves. Now suppose that r + 1 ≥ 2 and the result holds for r. Let H be a graph with 3(r + 1) + 3 = 3(r + 2) vertices of degree 2, and let w1 , . . . , wr+2 be r + 2 nonadjacent vertices of H. If there exist i, j ∈ {1, . . . , r + 2} with i = j such that Γ(wi ) ∩ Γ(wj ) = ∅, then r+2 | k=1 {Γ(wk ) ∪ wk }| < 3(r + 2). Thus, there exists w ∈ H with degree 2 nonadjacent to wi for all i. Hence, {w, w1 , . . . , wr+2 } is a set of r + 3 nonadjacent vertices of degree 2, but this is a contradiction. Therefore, Γ(wi ) ∩ Γ(wj ) = ∅ for all i = j. Furthermore, if w ∈ H is adjacent to any wi for i ∈ {1, . . . , r + 2}, then w has degree 2; otherwise, since the number of vertices of degree 2 is 3(r + 2), there exists v ∈ H with degree 2 nonadjacent to wi for all i, and we have seen above that this is not possible. Now, let a and b be the vertices adjacent to wr+2 . If the edge (a, b) does not belong to H, we have that {w1 , . . . , wr+1 , a, b} is a set of r + 3 nonadjacent vertices of degree 2. Thus, the vertices w1 , a, and b form a triangle. Downloaded 01/22/16 to 150.214.182.82. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php EXTREMAL TOPOLOGICAL SUBGRAPHS 395 Denote by H ∗ the graph obtained from H, removing the previous triangle. Therefore, H ∗ is a graph with 3r + 3 vertices of degree 2, and r + 1 of them are nonadjacent; by induction hypothesis, H ∗ contains r + 1 nonadjacent triangles. Thus, H contains r + 2 nonadjacent triangles. To finish this section, we give the proof of Theorem 5.1, using the previous results. Proof of Theorem 5.1. It is equivalent to show that EX(n; T Kn−q ) = {H(n; T Kn−q )} for n = 4q − k + 1 with q ≥ 5, 0 ≤ k ≤ q − 5. Let G be a graph belonging to the set {H(n; T Kn−q )}. By checking the structure of this graph G, it is easy to prove that G does not contain a subgraph homeomorphic to Kn−q . Since |E(G)| = ex(n; T Kn−q ) = ( n2 ) − (2q + k + 2), we have that G ∈ EX(n; T Kn−q ). In order to show that EX(n; T Kn−q ) ⊆ {H(n; T Kn−q )}, let G be a graph belonging to EX(n; T Kn−q ). We denote by H = G. By Theorem 5.3, |E(H)| = 2q + k + 2. First, we will prove that ∆(H) ≤ 2. Suppose the contrary, that ∆(H) ≥ 3. By applying Lemma 5.4, we have H ∈ Cqq+1 − Cqq . Hence, there exists a subset of vertices {v1 , . . . , vq } of H guaranteeing this property. Furthermore, |E(Hq )| = q + 1. We claim there exists j ∈ {1, . . . , q} such that ∆(Hj−1 ) ≥ 3 and ∆(Hj ) ≤ 2, because otherwise we have δHi−1 (vi ) ≥ 3 for each 1 ≤ i ≤ q, and |E(H)| ≥ 3q + (q + 1) > 2q + k + 2, but this is not possible. Now we distinguish the cases j ≥ k + 1 and j ≤ k. For j ≥ k + 1, we consider the fact that ∆(Hj−1 ) ≥ 3 and ∆(Hj ) ≤ 2. Taking into account property (2) of the definition of Cqq+1 and |E(Hq )| > 0, we have δHi−1 (vi ) ≥ 3 for 1 ≤ i ≤ j and δHi−1 (vi ) ≥ 1 for j + 1 ≤ i ≤ q. Hence, |E(Hq )| ≤ 2q + k + 2 − (3j + (q − j)) ≤ q − j + 1 ≤ q. But this is not possible since |E(Hq )| = q + 1. For j ≤ k, we have that δHi−1 (vi ) ≥ 3 for 1 ≤ i ≤ j. If ∆(Hk ) ≤ 1, then 2|E(Hk )| ≤ |V (Hk )| and 4q − 2k + 1 = |V (Hk )| ≥ 2|E(Hk )| ≥ 2(q − k + q + 1) = 4q − 2k + 2, and this is a contradiction. Thus, ∆(Hk ) = 2 and δHi−1 (vi ) ≥ 2 for j + 1 ≤ i ≤ k. Hence, |E(Hq )| ≤ 2q + k + 2 − (3j + 2(k − j + 1) + (q − k + 1)) = q − j + 1 ≤ q, and this not possible. Thus, ∆(H) ≤ 2. Since 2|E(H)| > |V (H)|, we have ∆(H) ≥ 2 and, consequently, ∆(H) = 2. Next we are going to study the structure of H. On the one hand, if H has at least 3(k + 1) + 1 vertices of degree 2, then by Lemma 5.2 we have that k + 2 of those vertices {w1 , . . . , wk+2 } are nonadjacent. Let wk+3 , . . . , wq be q − (k + 2) vertices of H such that the set {w1 , . . . , wk+2 , wk+3 , . . . , wq } verifies properties (1) and (2) of the definition of Cqs . For this set of vertices, we have that |E(Hq )| ≤ 2q + k + 2 − (2(k + 2) + q − (k + 2)) = q, Downloaded 01/22/16 to 150.214.182.82. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php 396 M. CERA, A. DIÁNEZ, AND A. MÁRQUEZ and therefore, H ∈ Cqq , a contradiction. Thus, H has at most 3k + 3 vertices of degree 2. On the other hand, if we denote by ni the number of vertices of degree i in H, we have that 2n2 + n1 = 2(2q + k + 2) n2 + n1 + n0 = 4q − k + 1 . Thus, n2 = 3k + 3 + n0 ≥ 3k + 3 and the number of vertices of degree 2 in H is n2 = 3k + 3. Furthermore, as we have shown previously, H may not have k + 2 nonadjacent vertices of degree 2. Since H has 3k +3 ≥ 3k +1 vertices of degree 2, by Lemma 5.2 we have that H has at least k + 1 nonadjacent vertices. Hence, H has maximum degree 2 and 3k + 3 vertices of degree 2, and k + 1 of them are nonadjacent. Therefore, by applying Lemma 5.5, H contains k + 1 nonadjacent triangles. Additionally, n0 = 0, n1 = 4q − 4k − 2, and the result follows. Acknowledgment. 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