On the girth of extremal graphs without shortest cycles

Discrete Mathematics 308 (2008) 5682–5690 www.elsevier.com/locate/disc On the girth of extremal graphs without shortest cycles✩ C. Balbuena a,∗ , M. Cera b , A. Diánez b , P. Garcı́a-Vázquez b a Departament de Matemàtica Aplicada III, Universitat Politècnica de Catalunya, Campus Nord, Edifici C2, C/Jordi Girona 1 i 3, E-08034 Barcelona, Spain b Departamento de Matemática Aplicada I, Universidad de Sevilla, Avda. Reina Mercedes 2, E-41012 Sevilla, Spain Received 13 June 2006; received in revised form 16 October 2007; accepted 23 October 2007 Abstract Let E X (ν; {C3 , . . . , Cn }) denote the set of graphs G of order ν that contain no cycles of length less than or equal to n which have maximum number of edges. In this paper we consider a problem posed by several authors: does G contain an n + 1 cycle? We prove that the diameter of G is at most n − 1, and present several results concerning the above question: the girth of G is g = n + 1 if (i) ν ≥ n + 5, diameter equal to n − 1 and minimum degree at least 3; (ii) ν ≥ 12, ν 6∈ {15, 80, 170} and n = 6. Moreover, if ν = 15 we find an extremal graph of girth 8 obtained from a 3-regular complete bipartite graph subdividing its edges. (iii) We prove that if ν ≥ 2n − 3 and n ≥ 7 the girth is at most 2n − 5. We also show that the answer to the question is negative for ν ≤ n + 1 + ⌊(n − 2)/2⌋. c 2007 Elsevier B.V. All rights reserved. Keywords: Extremal graphs; Girth; Forbidden cycles; Cages 1. Introduction Throughout this paper, only undirected simple graphs without loops or multiple edges are considered. Unless otherwise stated, we follow [4] for terminology and definitions. Let V (G) and E(G) denote the set of vertices and the set of edges of a graph G, respectively. The order of G is denoted by |V (G)| = ν(G) and the size by |E(G)| = e(G). The minimum length of a cycle contained in G is the girth g(G) of G. If G does not contain a cycle we set g(G) = ∞. By Cs we will denote the cycle of length s, s ≥ 3. The distance dG (x, y) in G between two vertices x, y is the length of a shortest x − y path in G. The greatest distance between any two vertices in G is the diameter D(G) of G. Diameter and girth are related by g(G) ≤ 2D(G) + 1. Let F be a family of graphs. The extremal number ex(ν, F ) is the maximum number of edges in a graph of order ν that does not contain any graph of F as a subgraph. The graphs of order ν and size ex(ν, F)not containing any ✩ Research supported by the Ministry of Education and Science, Spain, and the European Regional Development Fund (ERDF) under project MTM2005-08990-C02-02; and under the Andalusian Government project P06-FQM-01649. ∗ Corresponding author. Tel.: +34 934016917; fax: +34 934011825. E-mail addresses: m.camino.balbuena@upc.edu (C. Balbuena), mcera@us.es (M. Cera), anadianez@us.es (A. Diánez), pgvazquez@us.es (P. Garcı́a-Vázquez). 0012-365X/$ - see front matter c 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2007.10.037 C. Balbuena et al. / Discrete Mathematics 308 (2008) 5682–5690 5683 F ∈ F as a subgraph are the extremal graphs and the set of all such graphs is denoted by E X (ν, F). We refer to graphs from E X (ν, F) as extremal F-free graphs of order ν, or just extremal. Let ex(ν; {C3 , . . . , Cn }) denote the maximum number of edges in a graph of order ν and girth at least n + 1, and E X (ν; {C3 , . . . , Cn }) denote the set of all graphs of order ν, girth at least n + 1, and with ex(ν; {C3 , . . . , Cn }) edges. Erdös and Sachs [5] showed that an r -regular graph of girth g at least n + 1 with the least possible number of vertices has girth g = n + 1. (A proof of this result can be found in Lovász [12], pp. 66, 384, 385, see also the references therein.) These graphs are called (r ; g)-cages. In this paper we consider a similar problem: What is the girth of an extremal {C3 , . . . , Cn }-free graph of order ν? Is it always n + 1 or can it be greater? This problem has been studied in [2,8,9,13]. Some of the results obtained in these references are listed below. Theorem 1.1. Let G ∈ E X (ν; {C3 , . . . , Cn }). The following assertions hold: (i) the girth is n + 1 provided that • n = 4 and ν ≥ 7 [8,9]; • n = 5 and ν ≥ 8 [13]; • n = 6 and ν ≥ 171 [2]; • n ≥ 7 and ν > (2(n − 2)n−2 + n − 5)/(n − 3) [2]; • the maximum degree ∆ of G is ∆ ≥ n [13]; (ii) the girth is at most n + 2 provided that the minimum degree is δ(G) ≥ 2 and • the maximum degree ∆ of G is ∆ ≥ ⌈(n + 1)/2⌉ [2]. • n ≥ 7 and ν > (2(t − 2)n−2 + t − 5)/(t − 3), where t = ⌈(n + 1)/2⌉ [2]; (iii) if ν ≥ 2n − 2, then g ≤ 2n − 4 [2]. The same kind of structural properties as contained in points (i), (ii) of the above theorem for bipartite graphs are stated in [3]. This paper contributes to the study of this problem by proving the following results for a graph G belonging to the extremal family E X (ν, {C3 , . . . , Cn }): (i) If ν is such that n + 2 ≤ ν ≤ n + 1 + ⌊(n − 2)/2⌋, then there exist extremal graphs of girth g ≥ n + 2. (ii) The diameter D(G) is at most n − 1. Furthermore, if D(G) = n − 1 and the minimum degree is δ(G) ≥ 3, then the girth is g = n + 1, and for n = 6, 7 the requirement on the minimum degree is not needed. (iii) If n = 6, ν ≥ 12 and ν 6∈ {15, 80, 170}, then the girth is g = 7. For ν = 15 there exists an extremal graph of girth 8, which is a subdivision of a complete 3-regular bipartite graph. (iv) If n ≥ 7 and ν ≥ 2n − 3, then the girth g satisfies g ≤ 2n − 5. (v) If n = 7 and ν ≥ 11, then the girth g satisfies 8 ≤ g ≤ 9. Furthermore, as an immediate consequence of our results we obtain that the girth of an extremal graph is n + 1 for n = 3, 4, 5, which are the mentioned known results [8,9,13]. 2. Results Throughout the paper n ≥ 3 is an integer. We begin with a theorem providing a negative answer to the question pointed in the introduction for small values of ν. Theorem 2.1. Let n ≥ 4 and ν be integers such that ν ≥ n + 1. Let G ∈ E X (ν; {C3 , . . . , Cn }). Then G contains just one cycle if and only if n + 1 ≤ ν ≤ n + 1 + ⌊(n − 2)/2⌋. Hence for this range of ν we have ex(ν; {C3 , . . . , Cn }) = ν, and if ν > n + 1 there exist graphs G ∈ E X (ν; {C3 , . . . , Cn }) with girth g(G) ≥ n + 2. Proof. Let G ∈ E X (ν; {C3 , . . . , Cn }) be with ν ≥ n + 1. First, assume that ν ≤ n + 1 + ⌊(n − 2)/2⌋. Then G contains a cycle Cs of length s, and notice that s ≥ g(G) ≥ n + 1. Moreover, we may assume that s ≤ ν − 1, because otherwise G = Cs and we are done. If Cs is the only cycle contained in G, then G consists of Cs joined to one or several paths or trees, which implies that e(G) = ν and we have finished. Then suppose that G contains two cycles, Cs and C ′ and we will arrive at a contradiction. As g(G) ≥ n + 1 and ν < 2n + 2, it follows that Cs and C ′ must share a path Π , hence n + 1 ≤ |V (C ′ )| ≤ |V (Π )| + ν − s. 5684 C. Balbuena et al. / Discrete Mathematics 308 (2008) 5682–5690 Then |V (Π )| ≥ n + 1 + s − ν ≥ s − ⌊(n − 2)/2⌋. Let x, y be the end vertices of the path Π . Then the cycle induced by (C ′ − Π ) ∪ (Cs − Π ) ∪ {x, y} has length at most ν − (|V (Π )| − 2) ≤ n + 1 + ⌊(n − 2)/2⌋ − s + ⌊(n − 2)/2⌋ + 2 ≤ 2(1 + ⌊(n − 2)/2⌋) ≤ n, which is a contradiction. Therefore G contains just one cycle. Conversely, an extremal graph G with ν ≥ n + 2 + ⌊(n − 2)/2⌋ contains two cycles. It suffices to take a cycle of length n + 1 and two vertices x, y of the cycle at distance ⌊(n + 1)/2⌋ apart; and take a vertex-disjoint path with the cycle starting in x and finishing in y of length 1+⌊(n −2)/2⌋. In this way we obtain that G contains at least two cycles and g(G) = n + 1. In order to finish the proof it is enough to observe that any graph resulting from joining a cycle of length at least n + 2 and a vertex-disjoint path of order at most ⌊(n − 2)/2⌋ − 1 is an extremal {C3 , . . . , Cn }-free graph with girth g(G) ≥ n + 2.  In what follows the set of neighbors of u ∈ V (G) is denoted by N G (u). The number of neighbors of u is the degree dG (u) of u in G, or briefly d(u) when it is clear which graph is meant. Let G − u denote the graph formed from G by deleting u. Next, we will make use of the following property. Definition 2.2. Let G be a connected graph with diameter D. We say that G has the property Q if the following assertion holds: For each pair of vertices u, v ∈ V (G) such that dG (u, v) = D, we have dG (u, x) + dG (x, v) = D, for all x ∈ V (G). Theorem 2.1 allows us to deduce that there exist extremal {C3 , . . . , Cn }-free graphs with and without the property Q. For instance, if n is odd we may consider the graph G resulting from joining a cycle of length n + 1 and a path of (n − 1)/2 vertices, both intersecting in one unique vertex. By Theorem 2.1 it is clear that G has diameter D(G) = n − 1 and it is an extremal {C3 , . . . , Cn }-free graph satisfying the property Q. Analogously, if n is even an extremal {C3 , . . . , Cn }-free graph of diameter D(G) = n − 1 without property Q results from joining a cycle of length n + 1 and a path of n/2 vertices both intersecting in one unique vertex. Furthermore, notice that any extremal graph of order ν, n + 2 ≤ ν ≤ n + 1 + ⌊(n − 2)/2⌋ having girth g ≥ n + 2 must have diameter D ≤ n − 1 if n is even, and D(G) ≤ n − 2 if n is odd. As the next theorem indicates, the diameter of an extremal {C3 , . . . , Cn }-free graph G is at most n − 1, which will be very helpful to study the girth of G. Theorem 2.3. Every graph G ∈ E X (ν; {C3 , . . . , Cn }) satisfies: (i) The diameter is D(G) ≤ n − 1. (ii) If there exists u ∈ V (G) with d(u) = 1, then D(G −u) ≤ n−2, this inequality being an equality if D(G) = n−1. (iii) If D(G) = n − 1 and G does not satisfy the property Q, then the girth is g(G) = n + 1. Proof. (i) Let D = D(G) be the diameter of G ∈ E X (n; {C3 , . . . , Cn }), and let us take two vertices u, v at distance dG (u, v) = D. Then D ≤ n − 1 because otherwise adding the edge uv to G, we would obtain a graph G ′ of order ν having girth g(G ′ ) ≥ n + 1 and more edges than G which contradicts the maximality of G. (ii) Assume that there exists u ∈ V (G) with degree 1. Let G ′ = G − u and v, w ∈ V (G ′ ) be two vertices such that dG ′ (v, w) = D(G ′ ). We consider the new graph G ∗ obtained by adding to G ′ a new vertex u ∗ and a new edge u ∗ v. Clearly, D(G ∗ ) = 1 + D(G ′ ) and G ∗ ∈ E X (ν; {C3 , . . . , Cn }), hence from (i) it follows that D(G ∗ ) ≤ n − 1, which implies that D(G ′ ) ≤ n − 2. As D(G ′ ) ≥ D(G) − 1, then D(G ′ ) = n − 2 if D(G) = n − 1. (iii) Assume by way of contradiction that g(G) ≥ n + 2. Since G has no property Q, then there exist u, v ∈ V (G) such that dG (u, v) = n − 1 and there exists x ∈ V (G) \ {u, v} for which dG (u, x) + dG (x, v) ≥ n. This means that all the possible paths passing through x that connect u with v have length at least n. Take any vertex y ∈ N G (x) and consider the graph G ′ resulting by contracting the edge x y in G. The girth of this new graph is g(G ′ ) ≥ n + 1 and the diameter is D(G ′ ) = n − 1. So let u ′ , v ′ ∈ V (G ′ ) be such that dG ′ (u ′ , v ′ ) = n − 1, and denote by G ∗ the graph obtained from G ′ by adding a new vertex u ∗ and the edges u ′ u ∗ and u ∗ v ′ . Clearly, ν(G ∗ ) = ν(G ′ ) + 1 = ν and g(G ∗ ) = n + 1, but e(G ∗ ) = e(G ′ ) + 2 = e(G) + 1. This contradicts the extremality of G and therefore g(G) = n + 1.  Clearly, every extremal graph G ∈ E X (ν; {C3 , . . . , Cn }) must be connected. As a consequence of the preceding theorem we may compute the connectivities of an extremal graph. C. Balbuena et al. / Discrete Mathematics 308 (2008) 5682–5690 5685 Recall that a graph G is called connected if every pair of vertices is joined by a path; that is, if D(G) < ∞. If S ⊂ V and G − S is not connected, then S is said to be a cut set. A (non-complete) connected graph is called k-connected if every cut set has cardinality at least k. The connectivity κ(G) of a (non-complete) connected graph G is defined as the maximum integer k such that G is k-connected. The connectivity of a complete graph K δ+1 on δ + 1 vertices is defined as κ(K δ+1 ) = δ. Connectivity has an edge-analogue. An edge-cut in a graph G is a set W of edges of G such that G − W is non-connected. The edge-connectivity λ(G) of a graph G is the minimum cardinality of an edge-cut of G. A classic result due to Whitney is that κ(G) ≤ λ(G) ≤ δ(G) for every graph G of minimum degree δ(G). A graph is maximally connected if κ(G) = δ(G), and maximally edge-connected if λ(G) = δ(G). Sufficient conditions for a graph G of minimum degree δ(G) to be maximally connected have been given in terms of its diameter and its girth. In this regard, the following result is contained in [6,7,11]: λ(G) = δ(G) if D(G) ≤ 2⌊(g(G) − 1)/2⌋; κ(G) = δ(G) if D(G) ≤ 2⌊(g(G) − 1)/2⌋ − 1. (1) Corollary 1. Every graph G ∈ E X (ν; {C3 , . . . , Cn }) has λ(G) = δ(G). Furthermore, if D(G) ≤ n − 2, then κ(G) = δ(G). Proof. By Theorem 2.3, we have D(G) ≤ n − 1 ≤ g(G) − 2, because g(G) ≥ n + 1. Therefore, from (1) it follows  that λ(G) = δ(G). Moreover, if D(G) ≤ n − 2, then D(G) ≤ g − 3, and hence κ(G) = δ(G) by (1). The following lemmas describe the structure of a graph satisfying the property Q under certain hypothesis. Lemma 1. Let G be a graph satisfying the property Q. Then each vertex u ∈ V (G) has at most one vertex v at maximum distance. Moreover, G has at most two vertices of degree one in G, which must be at maximum distance. Proof. Let us denote by D the diameter of G and take three vertices u, v, w such that dG (u, v) = D and dG (u, w) = D. Property Q implies that dG (u, w) + dG (w, v) = D, which gives dG (w, v) = 0 or in other words, w = v. Therefore u has at most one vertex at maximum distance. Now take two vertices u, v such that dG (u, v) = D. Suppose that there exists u ′ ∈ V (G) \ {u, v} of degree one, and consider the vertex u ′1 ∈ N G (u ′ ). Then dG (u, u ′ ) + dG (u ′ , v) = D and dG (u, u ′1 ) + dG (u ′1 , v) = D, because of the property Q. But taking into account that dG (u, u ′ ) = 1 + dG (u, u ′1 ) and dG (u ′ , v) = 1 + dG (u ′1 , v) we obtain that D = D + 2, which is a contradiction. Therefore the only vertices which may have degree one are u and v.  Lemma 2. Let G be a connected graph with diameter D, girth g and satisfying the property Q. Let u, v ∈ V (G) be two vertices such that dG (u, v) = D, and let Π : u = u 0 , u 1 , . . . , u D = v be any shortest path joining u with v. Then (i) if g ≥ D + 2, then d(u i ) = 2 for any i = ⌊D/2⌋, ⌈D/2⌉; (ii) if g ≥ 2D − 3, then d(u i ) = 2 for any i = 2, . . . , D − 2; (iii) if δ(G) ≥ 2 and g ≥ max{2D − 3, D + 3}, then G is a cycle of length 2D. Proof. (i) Suppose that d(u i ) ≥ 3 for some i ∈ {⌊D/2⌋, ⌈D/2⌉} and let y ∈ N G (u i ) \ {u i−1 , u i+1 }. Notice that dG (u, u i ) = i and dG (u i , v) = D − i. Then we have dG (u, y) ≤ dG (u, u i ) + 1 = i + 1. If dG (u, y) = i + 1, then dG (y, v) = D − i − 1 because G satisfies the property Q. As y, u i , u i+1 , . . . , v is a path of length D − i + 1, it follows that G contains a cycle of length 2D − 2i ≤ 2D − 2⌊D/2⌋ ≤ D + 1 < g, which is impossible. Therefore dG (u, y) ≤ i, which means that there is a cycle in G of length at most dG (u, y) + i + 1 ≤ 2i + 1 ≤ 2⌈D/2⌉ + 1 ≤ 2⌈(g − 2)/2⌉ + 1. (2) This is only possible if g is odd in which case all the inequalities in (2) become equalities; that is, dG (u, y) = i = ⌈D/2⌉, and g = D + 2 with D odd (because g must be odd). Since G satisfies the property Q, then dG (y, v) = D − dG (u, y) = D − (D + 1)/2 = (D − 1)/2. However, y, u i , u i+1 , . . . , v is a path of length D − i + 1 = (D + 1)/2. Hence, there is a cycle in G of length at most (D − 1)/2 + (D + 1)/2 = D < g which is a contradiction. (ii) Notice that if D ≤ 4, then the result is true by (i). Thus, assume that D ≥ 5 and suppose that d(u i ) ≥ 3 for some i ∈ {2, . . . , D − 2} and let y ∈ N G (u i ) \ {u i−1 , u i+1 }. As in the above case we have dG (u, y) ≤ i + 1. If 5686 C. Balbuena et al. / Discrete Mathematics 308 (2008) 5682–5690 Fig. 1. An extremal graph for n = 7, ν = 12, e = 13, g = 8 satisfying property Q. dG (u, y) = i + 1 then dG (y, v) = D − i − 1 because of the property Q. As y, u i , u i+1 , . . . , v is a path of length D − i + 1, then there is a cycle in G of length at most 2D − 2i ≤ 2D − 4 < g which is impossible. Thus dG (u, y) ≤ i which implies that there is a cycle in G of length at most dG (u, y) + i + 1 ≤ 2i + 1 ≤ 2D − 3. (3) Since g ≥ 2D − 3 we have that all the above inequalities of (3) are equalities; that is, g = 2D − 3, dG (u, y) = i = D − 2. As G satisfies the property Q, then dG (y, v) = 2, say v, x, y the shortest path. But v, x, y, u D−2 , u D−1 , v is a cycle of length 5, which implies that g = 2D − 3 ≤ 5, hence D ≤ 4 against the assumption D ≥ 5. (iii) By hypothesis D ≤ g − 3 ≤ 2⌊(g − 1)/2⌋ − 1, which means by (1) that G is 2-connected because δ(G) ≥ 2. Then by Menger’s Theorem, there exist two vertex-disjoint paths between any two vertices. Thus we find a path Px : u, x1 , x2 , . . . , xr , v that joins u with v, verifying V (Π )∩V (Px ) = {u, v}. Moreover, Px must be of length D since G satisfies the property Q. Hence these two paths form a cycle C of length 2D. By (ii) we obtain d(u i ) = d(xi ) = 2 for i ∈ {2, . . . , D − 2}. Moreover, dG (u 1 , x D−1 ) ≥ D − 2, and dG (x1 , u D−1 ) ≥ D − 2, since dG (u, v) = D. All these facts imply that dG (u 2 , x D−2 ) = min{1 + dG (u 1 , x D−1 ) + 1, D − 2 + dG (u D−1 , x D−1 )} = D and analogously, dG (x2 , u D−2 ) = D. Thus, (ii) applied to the path u 2 , u 3 , . . . , u D−1 , v, x D−1 , x D−2 gives d(u D−1 ) = d(v) = 2. Considering the path u 2 , u 1 , u 0 , x1 , . . . , x D−2 it follows that d(u 0 ) = d(x1 ) = 2; and finally considering the path x2 , . . . , x D−1 , v, u D−1 , u D−2 and the path x2 , x1 , u, u 1 , . . . , u D−2 we obtain that d(x D−1 ) = d(u 1 ) = 2. Therefore, G must be a cycle of length 2D.  An immediate consequence of Lemma 2 is that every graph satisfying the property Q has minimum degree at most two. Next we obtain a theorem in which any extremal {C3 , . . . , Cn }-free graph satisfying the property Q for small values of n and with g ≥ n + 2 is shown to have diameter at most n − 2 or to be C6 or a C8 joined to a pendant edge. In [1] some extremal numbers have been calculated. Among other results, it is proved that ex(ν; {C3 , . . . , Cn }) = ν + 1 provided that ⌈(3n + 1)/2⌉ ≤ ν ≤ 2n − 1. (4) For instance, an extremal graph for n = 7 and ν = 12 is obtained from a bipartite complete K 2,3 by subdividing its edges and adding a pendant edge to a vertex of degree 3, see Fig. 1. Notice that this graph is an example of extremal graph satisfying property Q of order ν ≥ n + 2 + ⌊(n − 2)/2⌋ for n = 7. Theorem 2.4. Let n, ν be two integers such that 3 ≤ n ≤ 7 and ν ≥ n + 2. Let G ∈ E X (ν; {C3 , . . . , Cn }) satisfying the property Q and suppose that g ≥ n + 2. Then n ≥ 4 and either the diameter D ≤ n − 2, or n = 4 and the graph is G = C6 , or n = 6 and the graph consists of a C8 joined to a pendant edge. Proof. Let G be a graph satisfying the hypothesis of the theorem. From Lemma 2, it follows that δ(G) ≤ 2, and from Theorem 2.3 we have D ≤ n − 1. Notice that if n = 3 then D = 2. Assume that D = n − 1, otherwise the theorem is true. Hence we have g ≥ n + 2 = D + 3 ≥ 2D − 3 because n ≤ 7. Thus if δ(G) = 2 from Lemma 2, it follows that G must be a cycle of length 2D = 2n − 2. This implies that n = 3 is impossible. For n = 4 the graph is C6 . For n = 5 the possible graph is C7 , which does not satisfies the property Q. For n = 6, 7, the graph G = C2n−2 is not an extremal graph, because the graph G ∗ resulting by gluing two cycles of length n + 1 by a common path of length 3, has 2n − 2 vertices and one more edge than G (see Theorem 2.1.) Hence if δ(G) = 2 the only possible graph is C6 corresponding to n = 4. Therefore assume that δ(G) = 1 and let u be a vertex with d(u) = 1. By Lemma 1 there is only one vertex v such that dG (u, v) = D, and d(x) ≥ 2 for every vertex x different from u and v. Let us consider a path Π : u = u 0 , u 1 , . . . , u D−1 , u D = v of length D. By Lemma 2 we have d(u i ) = 2 for i = 2, . . . , D − 2 because g ≥ 2D − 3. As D = n − 1 and ν ≥ n + 1, d(u j ) ≥ 3 for some j ∈ {1, D − 1, D}. Let us see that d(u 1 ) ≥ 3. Otherwise d(u 1 ) = 2, which implies that d(v) = 1, because if x ∈ N G (v) − u D−1 , C. Balbuena et al. / Discrete Mathematics 308 (2008) 5682–5690 5687 Fig. 2. A graph of ν = 12, e = 14, g = 7. Fig. 3. A subdivision of K 3,3 and a derived graph. then dG (u, x) = D − 1 + dG (u D−1 , x) ≥ D + 1, which is impossible. Therefore d(u D−1 ) ≥ 3, and for all y ∈ N G (u D−1 ) \ {u D−2 , v}, dG (u, y) = dG (u, u D−1 ) + 1 = D, contradicting Lemma 1 because y 6= v. Hence d(u 1 ) ≥ 3 as claimed. Take x ∈ N G (u 1 ) \ {u 2 , u}. As G satisfies the property Q we have dG (x, v) = D − 2. Notice that d(v) = d(u 1 ) − 1 because if d(v) 6= d(u 1 ) − 1, then dG (x, u D−1 ) = D − 3, and this shortest path together with the path x, u 1 , u 2 , . . . , u D−1 of length D − 1 form a cycle of length 2D − 4 which contradicts g ≥ 2D − 3. Therefore we have obtained that G has a cycle of length 2D − 2, that is, n + 2 ≤ g ≤ 2D − 2 = 2n − 4 which is only possible if n = 6, 7, thus item (i) is valid. Observe that for n = 6, 7 the obtained graph G has ν = (n − 3)(d(u 1 ) − 1) + 3 and e = (n − 2)(d(u 1 ) − 1) + 1. From Theorem 1.1(i) we may suppose that ∆ = ∆(G) ≤ n − 1, where ∆ is the maximum degree of G, because by hypothesis g ≥ n + 2. Then we have the following cases: (a) For n = 6 we obtain (ν, e) = {(9, 9), (12, 13), (15, 17)}. A graph of order ν = 12, e = 14 and girth g = 7 is shown in Fig. 2. Then the pair (12, 13) cannot be the parameters of an extremal graph G ∈ E X (ν; {C3 , . . . , C6 }). Also, the graph resulting from the subdivision of the complete bipartite K 3,3 has ν = 15, e = 18, and girth g = 8, see graph G of Fig. 3. Thus, the pair (15, 17) cannot be the parameters of an extremal graph. Therefore we conclude that when n = 6 the graph G is formed by a cycle of length 8 joined to a pendant edge which has ν = e = 9. (b) For n = 7 we obtain (ν, e) = {(11, 11), (15, 16), (19, 21), (23, 26)}. The pair (11, 11) cannot be the parameter of an extremal graph, because the subdivision of K 2,3 has ν = 11 and e = 12. See also (4). We already has shown that the pair (15, 16) cannot be the parameter of an extremal graph. And (19, 21) and (23, 26) cannot be the parameters of an extremal graph, because the subdivision of K 3,4 has ν = 19 and e = 24, and the subdivision of K 3,5 has ν = 23 and e = 30. Therefore we conclude that when n = 7 there is no extremal graph of degree 1 with diameter n − 1 satisfying property Q.  Combining both Theorems 2.3 and 2.4 we obtain the following result, which contains the known results for the cases n = 3, 4, 5, see [8,9,13]. Theorem 2.5. Let G ∈ E X (ν; {C3 , . . . , Cn }). Then g(G) = n + 1 if one of the following assertions holds: (i) The minimum degree is δ(G) ≥ 3 and the diameter is D(G) = n − 1. (ii) n = 3, 4, 5, 6, 7, ν ≥ n + 4 and the diameter is D(G) = n − 1. (iii) n = 3, 4, 5 and ν ≥ n + 4. Proof. (i) Since δ(G) ≥ 3, then G does not satisfy the property Q by Lemma 2. Therefore from Theorem 2.3(iii), it follows that g = n + 1. 5688 C. Balbuena et al. / Discrete Mathematics 308 (2008) 5682–5690 (ii) Since ν ≥ n + 4, n ≤ 7 and D(G) = n − 1, then from Theorem 2.4, it follows that g = n + 1 and the result is true, or G does not satisfy the property Q. In this latter case we also obtain that g = n + 1 by applying again Theorem 2.3(iii). (iii) For n = 3, 4 the result follows from item (ii), because otherwise we would have ⌊(n + 2)/2⌋ ≤ D ≤ n − 2 which is impossible. Moreover, item (iii) of Theorem 1.1 implies that g = 6 if n = 5. Hence the result is valid.  Theorem 2.5(ii) guarantees that every graph G ∈ E X (ν; {C3 , . . . , Cn }) has girth g(G) = n + 1 for n = 6, 7 whenever ν ≥ n + 4 and D(G) = n − 1. In what follows the objective is to prove that the girth is g(G) = n + 1 for the cases n = 6, 7 assuming D(G) ≤ n − 2. Let e be an edge of G. As usual we will denote by G/e the graph obtained from G by contracting the edge e. We begin by proving a technical lemma. Lemma 3. Let G ∈ E X (ν; {C3 , . . . , Cn }) be with ν ≥ 2n − 3 and n ≥ 6. If G contains a cycle of length 2n − 3, then the girth is g ≤ 2n − 5. Proof. Assume by way of contradiction that g ≥ 2n − 4. Then the diameter is D ≥ ⌊g/2⌋ ≥ n − 2. Let C2n−3 = u 0 , u 1 , . . . , u n−2 , vn−2 , . . . , v1 , u 0 be a cycle of length 2n − 3 contained in G. We may assume that there is x ∈ N G (u 0 ) − C2n−3 because otherwise G = C2n−3 which is not an extremal graph by Theorem 2.1. Taking into account that in a graph with girth g the paths of length at most ⌊g/2⌋ are shortest paths, we deduce that dG (u 0 , u n−2 ) = dG (u 0 , vn−2 ) = n − 2. Let us see that every cycle C in G containing both edges u 0 x and u n−2 vn−2 must have length at least 2n − 3. Indeed, we have dG (x, u n−2 ) ≥ n − 3 and dG (x, vn−2 ) ≥ n − 3. Furthermore, if dG (x, u n−2 ) = n − 3 then dG (x, vn−2 ) ≥ n − 2 because g ≥ 2n − 4. Thus every cycle C in G containing both edges u 0 x and u n−2 vn−2 must have length at least 2n − 3. Let us consider the graph G ′ = G/{u 0 x, u n−2 vn−2 }, i.e., G ′ is the resulting graph from G by contracting the edges u 0 x and u n−2 vn−2 . Clearly, g(G ′ ) ≥ 2n − 5 ≥ n + 1, because n ≥ 6. Moreover, this operation transforms the cycle C2n−3 of G in a cycle C ′ in G ′ of length 2n − 4, and hence D(G ′ ) ≥ n − 2. Observe that ν(G ′ ) = ν(G) − 2 and e(G ′ ) = e(G) − 2. Let u ′ , v ′ be two vertices of G ′ such that dG ′ (u ′ , v ′ ) = D(G ′ ) and consider the graph G ∗ obtained by adding to G ′ the path u ′ , x1 , x2 , v ′ , where x1 , x2 6∈ V (G ′ ). Clearly, g(G ∗ ) = D(G ′ ) + 3 ≥ n + 1, ν(G ∗ ) = ν(G ′ ) + 2 = ν(G) and e(G ∗ ) = e(G ′ ) + 3 = e(G) + 1. This is a contradiction because G is extremal. Therefore g ≤ 2n − 5.  The incidence graph of an incidence structure I ⊂ P × L is a bipartite graph G = (P, L) in which two vertices a, b are adjacent if and only if they are incident, i.e. (a, b) ∈ I (See Chapter 5 of the book by Godsil and Royle [10]). The vertices of P are called points and the vertices of L are called lines. Two points being incident to a common line are called collinear and two lines being incident to a common point are called concurrent. A partial linear space is an incidence structure in which any two points are incident with at most one line. This implies that any two lines are incident with at most one point. A generalized quadrangle is a partial linear space containing non-collinear points and non-concurrent lines satisfying that given any line L and a point p not on L there is a unique point p ′ on L such that p and p ′ are collinear. Each of the known (r ; 8)-cages for r odd prime is the incidence graph of a generalized quadrangle, see the survey by Wong [15] or the book [10]. Lemma 4. Let G be a graph of diameter 4 and girth 8 free of cycles of length 9. Then G is the incidence graph of a generalized quadrangle. Proof. First let us see that G is a bipartite graph. Otherwise suppose that G contains odd cycles and let l ≥ 11 be chosen as the minimum number such Cl is an odd cycle of G. Then Cl has two vertices x and y joined by two disjoint paths, the shortest one of length ⌊l/2⌋ ≥ 5 denoted by Cl1 and the other one of length ⌈l/2⌉ ≥ 6 denoted by Cl2 . As the diameter is 4, we have a shortest Px y path of length dG (x, y) ≤ 4 between x and y different from Cli , i = 1, 2. Thus these two semi-cycles together with the path Px y form two cycles of length at most ⌊l/2⌋ + dG (x, y) < l and ⌈l/2⌉ + dG (x, y) < l, respectively. By the way of l has been chosen both numbers ⌊l/2⌋ + dG (x, y) and ⌈l/2⌉ + dG (x, y) must be even, which is a contradiction because ⌈l/2⌉ = ⌊l/2⌋ + 1. Therefore, G is a bipartite graph with classes P and L. Let us call the vertices of P points and the vertices of L lines. Consider a line L and point p at distance 3. Since the girth is 8, there is a unique path L, p ′ , L ′ , p from L to p. This provides the unique point p ′ satisfying the condition defining a generalized quadrangle.  C. Balbuena et al. / Discrete Mathematics 308 (2008) 5682–5690 5689 Fig. 4. Case ν = 35 and e(G) = 50. As a consequence of Theorems 2.3 and 2.5 and Lemmas 3 and 4 we obtain the following theorem. Theorem 2.6. Let G ∈ E X (ν; {C3 , C4 , C5 , C6 }) be with ν ≥ 12 and ν 6∈ {15, 80, 170}. Then the girth is g(G) = 7. Proof. By Theorem 2.3 it follows that D(G) ≤ 5. If D(G) = 5 then g(G) = 7 by applying Theorem 2.5(ii). Moreover, if D(G) = 3 then g(G) = 7. So assume that D(G) = 4 and proceed by contradiction supposing that g(G) ≥ 8. By Lemma 3 we may suppose that G has no cycles of length 9, which leads to a graph G with g(G) = 8. As D(G) = 4, by Lemma 4 we may conclude that G must be the incidence graph of a generalized quadrangle and thus all the vertices of one class have degree α ≥ 2 and all the vertices of the other class have degree β ≥ 2 (see [10] pp. 235). Moreover, from Theorem 1.1(iii) we may suppose that ∆ = ∆(G) ≤ 5, where ∆ is the maximum degree of G, hence 2 ≤ α, β ≤ 5. By algebraic methods it has been proved that (α, β) ∈ {(3, 5), (4, 5)} cannot exist (see [10] pp. 236–238). Consequently if 3 ≤ α, β ≤ 5 we obtain (ν, e) = {(30, 45), (72, 135), (80, 160), (170, 425)}; or G is a subdivision of a complete bipartite graph K α,β with 2 ≤ α, β ≤ 5 giving the following pairs of values: (ν, e) = {(14, 16), (15, 18), (17, 20), (19, 24), (23, 30), (24, 32), (29, 40), (35, 50)}. Let us study all these values. Fig. 3 shows a graph G of girth 8, which is a subdivision of a 3-regular complete bipartite graph on ν = 15 and e = 18. Contracting an edge (for instance ab in Fig. 3) we obtain a graph on ν = 14 and e = 17 of girth g = 7, thus the pair (ν, e) = (14, 16) cannot be the parameters of an extremal graph. Moreover, adding to G two new vertices x, y and the path vx yw where dG (v, w) = 4 we obtain a graph H (see Fig. 3) on ν = 17 and e = 21 of girth g = 7, thus the pair (ν, e) = (17, 20) cannot be the parameters of an extremal graph. Furthermore, these graphs are extremal because in [1], it has been proved that ex(14; {C3 , . . . , C6 }) = 17 and ex(17; {C3 , . . . , C6 }) = 22. Let Γ be the (3, 7)-cage which has ν(Γ ) = 24 and e(Γ ) = 36, hence ex(24; {C3 , . . . , C6 }) ≥ 36. Then the pairs (ν, e) = (23, 30), (24, 32) are not valid. Moreover, removing from Γ a set of vertices formed by a vertex u, its neighbors u 1 , u 2 , u 3 and one neighbor of u 1 we obtain a new graph Γ ′ on ν = 19 and e = 25, hence the pair (ν, e) = (19, 24) is not valid. Let Γ be the (3, 8)-cage which has ν(Γ ) = 30 and e(Γ ) = 45, hence the pair (29, 40) is not valid. Further, in [14] has been constructed a graph on 30 vertices and 47 edges having g = 7. Then the (3, 8)-cage is not an extremal graph for n = 6. Observe that the (3, 8)-cage has three vertices u, v, w such that d(u, v) = d(u, w) = d(v, w) = 4. Let Γ ′ be the graph obtained from Γ by adding two new vertex-disjoint paths, P1 of length 3 connecting vertex u to vertex v and P2 of length 4 connecting u and w, see Fig. 4. In this way we obtain a graph Γ ′ on 35 vertices, 52 edges and girth 7. So the pair (ν, e) = (35, 50) is not valid. Let Γ be the (4, 8)-cage which has ν(Γ ) = 80 and e(Γ ) = 160. Removing from Γ a set of vertices formed by a vertex u, its neighbors u 1 , u 2 , u 3 , u 4 and all the neighbors of u 1 , and adding an additional edge u 41 u 42 between two neighbors of u 4 we obtain a new graph Γ ′ on ν = 72 and e = 160 − 25 + 1 = 136 of girth 7, hence the pair (ν, e) = (72, 135) is not valid.  5690 C. Balbuena et al. / Discrete Mathematics 308 (2008) 5682–5690 In [1], it has been proved that ex(15; {C3 , C4 , C5 , C6 }) = 18. For ν = 15 Fig. 3 shows an extremal {C3 , C4 , C5 , C6 }-free graph of girth 8, which is a subdivision of a 3-regular complete bipartite graph. The problem of knowing if for n = 6 the (4; 8)-cage and (5; 8)-cage of order 80, 170, respectively, are extremal graphs is still opened. As mentioned in the introduction, in [2] was proved that every graph G ∈ E X (ν; {C3 , . . . , Cn }) of minimum degree δ(G) ≥ 2 and maximum degree ∆ ≥ ⌈(n + 1)/2⌉ has girth g ≤ n + 2. As a consequence it was derived in the same paper that the girth of an extremal graph G ∈ E X (ν; {C3 , . . . , C7 }) of minimum degree δ(G) ≥ 2 is g(G) ≤ 9 if ν ≥ 64. Next, we present a theorem from which we can derive an improvement of this result. Theorem 2.7. Let G ∈ E X (ν; {C3 , . . . , Cn }) be with ν ≥ 2n − 3 for n ≥ 7. Then the girth is g(G) ≤ 2n − 5. Proof. We reason by contradiction assuming that g = g(G) ≥ 2n − 4. Hence, from Lemma 3, it follows that G does not contain a cycle of length 2n − 3. Let C : u 0 , u 1 , . . . , u g−1 , u 0 be a shortest cycle in G, and consider the graph G ′ = G/{u 0 u 1 , u 1 u 2 }, which has g(G ′ ) ≥ 2n − 6 ≥ n + 1 because n ≥ 7. Hence the diameter D(G ′ ) ≥ ⌊g(G ′ )/2⌋ ≥ (2n − 6)/2 = n − 3. Assume first that D(G ′ ) ≥ n − 2 and let u ′ , v ′ be two vertices of G ′ such that dG ′ (u ′ , v ′ ) = D(G ′ ). Let us consider the graph G ∗ obtained from G ′ by adding two new vertices x1 , x2 and the three edges u ′ x1 , x1 x2 , x2 v ′ . We have g(G ∗ ) = min{g(G ′ ), D(G ′ ) + 3} ≥ n + 1; |V (G ∗ )| = |V (G ′ )| + 2 = ν, and e(G ∗ ) = e(G)+1, which contradicts the maximality of G. Consequently, D(G ′ ) = ⌊g(G ′ )/2⌋ = (2n −6)/2 = n −3, which implies that g(G ′ ) = 2n − 6, as otherwise g(G ′ ) = 2n − 5, yielding g(G) = 2n − 3, contradicting the fact that G does not contain cycles of length 2n − 3. Therefore, every path in G of length n − 2 is a shortest one because g = g(G) = 2n − 4. Let x ∈ N G (u 0 ) \ {u 1 , u g−1 }, and let û 012 denote the new vertex in G ′ resulting from the contraction. Clearly we have dG (x, u n−1 ) = n − 2 because x, u 0 , u 2n−5 , . . . , u n−1 is a shortest path; and the paths passing through the edges u 0 u 1 or u 1 u 2 have length at least n. Then dG ′ (x, u n−1 ) = 1 + dG ′ (û 012 , u n−1 ) = 1 + n − 3 = n − 2 which is a contradiction with the fact that D(G ′ ) = n − 3. Therefore the girth is g ≤ 2n − 5.  Corollary 2. The girth of every graph G ∈ E X (ν; {C3 , C4 , C5 , C6 , C7 }) with ν ≥ 11 is 8 ≤ g(G) ≤ 9. Acknowledgments We would like to express our thanks to the referees for their helpful comments and suggestions. References [1] E. Abajo, A. Diánez, Extremal graphs with fixed excess (submitted for publication). [2] C. Balbuena, P. Garcı́a-Vázquez, On the minimum order of extremal graphs to have a prescribed girth, SIAM J. Discrete Math. 21 (1) (2007) 251–257. [3] C. Balbuena, P. Garcı́a-Vázquez, X. Marcote, J.C. 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