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Neighborhood unions and disjoint chorded cycles in graphs
Discrete Mathematics 312 (2012) 891–897
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Neighborhood unions and disjoint chorded cycles in graphs Shengning Qiao Department of Mathematics, Xidian University, Xi’an, Shaanxi 710071, PR China
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Article history: Received 2 December 2010 Received in revised form 9 October 2011 Accepted 10 October 2011 Available online 28 October 2011
abstract Let s ≥ 0, t ≥ 0 be two integers and G be a graph with at least 3s + 4t vertices. In this paper, we show that if |N (u, G) ∪ N (v, G)| ≥ 3s + 4t + 1 for any two nonadjacent vertices u and v in G, then G contains s + t vertex-disjoint cycles such that t of them are chorded cycles. The condition is sharp. © 2011 Elsevier B.V. All rights reserved.
Keywords: Cycle Chorded cycle Vertex-disjoint
1. Introduction We use Bondy and Murty [3] for terminology and notation not defined here and consider finite simple graphs only. Let G be a graph. A set of subgraphs of G is said to be vertex-disjoint if no two of them have a common vertex in G. Corrádi and Hajnal [5] proved the following well-known result on the existence of vertex-disjoint cycles in graphs. Theorem 1 (Corrádi and Hajnal [5]). Let s ≥ 0 be an integer and G be a graph with at least 3s vertices. If the minimum degree of G is at least 2s, then G contains s vertex-disjoint cycles. Later, this result was generalized by Enomoto [6] and Wang [11], as follows. Theorem 2 (Enomoto [6] and Wang [11]). Let s ≥ 0 be an integer and G be a graph with at least 3s vertices. If the degree sum of any two nonadjacent vertices in G is at least 4s − 1, then G contains s vertex-disjoint cycles. An edge which joins two vertices of a cycle but is not itself an edge of the cycle is a chord of the cycle. A cycle is called a chorded cycle if it has at least one chord. Along a different line, Pósa proved in [10] that any graph with minimum degree at least 3 contains a chorded cycle. Motivated by these results, Finkel [8] proved the following result analogous to Theorem 1. Theorem 3 (Finkel [8]). Let t ≥ 0 be an integer and G be a graph with at least 4t vertices. If the minimum degree of G is at least 3t, then G contains t vertex-disjoint chorded cycles. In fact, in [2], Bialostocki et al. posed the following conjecture. Conjecture 4 (Bialosticki et al. [2]). Let s ≥ 0, t ≥ 0 be two integers and G be a graph with at least 3s + 4t vertices. If the minimum degree of G is at least 2s + 3t, then G contains a vertex-disjoint union of s + t cycles such that t of them are chorded cycles.
E-mail addresses: [email protected], [email protected]. 0012-365X/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2011.10.012
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Very recently, Babu and Diwan [1], Chiba and Fujita [4] and Gao et al. [9] proved that the conjecture is true. Especially, in [1,4], the authors obtained a stronger version of this conjecture for degree sum condition. Theorem 5 (Babu and Diwan [1], Chiba and Fujita [4]). Let s ≥ 0, t ≥ 0 be two integers and G be a graph with at least 3s + 4t vertices. If the degree sum of any two nonadjacent vertices in G is at least 4s + 6t − 1, then G contains a vertex-disjoint union of s + t cycles such that t of them are chorded cycles. Motivated by Theorem 5, in this paper, we consider the neighborhood union condition for the existence of vertex-disjoint cycles and chorded cycles in graphs. In 2005, Faudree and Gould [7] obtained the neighborhood union condition for the existence of vertex-disjoint cycles in graphs, as follows. Theorem 6 (Faudree and Gould [7]). Let s ≥ 0 be an integer. Let G be a graph with at least 3s vertices. If |N (u, G)∪ N (v, G)| ≥ 3s for all nonadjacent pairs of vertices u and v in G, then G contains s vertex-disjoint cycles. Our result is as follows. Theorem 7. Let s ≥ 0, t ≥ 0 be two integers. Let G be a graph with at least 3s + 4t vertices. If |N (u, G)∪ N (v, G)| ≥ 3s + 4t + 1 for all nonadjacent pairs of vertices u and v in G, then G contains s + t vertex-disjoint cycles such that t of them are chorded cycles. The condition is sharp. Let t ≥ 2 be an even. We construct a graph G = K2t +3 ∪ K2t −1 , i.e., the union of K2t +3 and K2t −1 . Clearly, we have |N (u, G)∪ N (v, G)| = 4t for all nonadjacent pairs of vertices u and v in G. But G contains no t vertex-disjoint chorded cycles. Further, we can construct a graph G = K4t +2 ∪ K3s+1 . Let t ≥ 2. Clearly, we cannot apply Theorem 5 to ensure the existence of s + t vertex-disjoint cycles in G such that t of them are chorded cycles. However, by Theorem 7, we can get that G contains s + t vertex-disjoint cycles such that t of them are chorded cycles. We further introduce some notation that will be used later. Let G = (V (G), E (G)) be a graph. By |G| we denote |V (G)|. For a vertex v and a subgraph H of G, we use N (v, H ) and d(v, H ) to denote the set and the number of vertices in H that are adjacent to v , respectively. Thus d(v, G) is the degree of v in G. By G − H we denote G[V (G) \ V (H )]. For a subset S of vertices or edges in G, we use G[S ] to denote the subgraph of G induced by S. In Section 2, we give some lemmas to be used later. In Section 3, we will prove Theorem 7. 2. Some lemmas In the following, we let G be a graph. Lemma 1. Let C be a cycle of G and u, v be two vertices of G − C . If |N (u, C ) ∪ N (v, C )| ≥ 4, then either G[V (C ) ∪ {u, v}] contains a cycle C ′ with |C ′ | < |C | or |C | = 4 and there exists a vertex w ∈ N (u, C ) such that G[(V (C ) ∪ {v}) − {w}] contains a cycle of length 4. Proof. First note that |C | ≥ 4. If u or v is adjacent to two end vertices of an edge of C , then G[V (C ) ∪ {u}] or G[V (C ) ∪ {v}] contains a cycle of length exactly three, respectively. Since |N (u, C ) ∪ N (v, C )| ≥ 4, we have d(u, C ) ≥ 2 or d(v, C ) ≥ 2. By the symmetry, we assume that d(u, C ) ≥ 2. If |C | ≥ 5, then it is not difficult to see that G[V (C ) ∪ {u}] contains a cycle C ′ with |C ′ | < |C |. If |C | = 4, then u and v are adjacent to two nonadjacent vertices on C with N (u, C ) ∩ N (v, C ) = ∅. Then there exists a vertex w ∈ N (u, C ) such that G[(V (C ) ∪ {v}) − {w}] contains a cycle of length 4. Lemma 2. Let C be a chorded cycle of G and u, v be two vertices of G − C . If |N (u, C )∪ N (v, C )| ≥ 5, then either G[V (C )∪{u, v}] ′ ′ contains a chorded cycle C with |C | < |C | or |C | = 6 and there exists a vertex w ∈ N (u, C ) such that G[(V (C ) ∪ {v}) − {w}] contains a chorded cycle of length 6. Proof. Clearly, we have |C | ≥ 5. If d(u, C ) ≥ 4 or d(v, C ) ≥ 4, then we can get that G[V (C ) ∪ {u}] or G[V (C ) ∪ {v}] contains ′ ′ a chorded cycle C with |C | < |C |, respectively. So we may assume that d(u, C ) ≤ 3 and d(v, C ) ≤ 3. Further, we can also obtain that if u or v is adjacent to three consecutive vertices on C , then either G[V (C ) ∪ {u}] or G[V (C ) ∪ {v}] contains a chorded cycle of length 4. Suppose that |C | ≥ 7. By |N (u, C )∪ N (v, C )| ≥ 5, we have d(u, C ) = 3 or d(v, C ) = 3, say d(u, C ) = 3. Then G[V (C )∪{u}] ′ ′ contains a chorded cycle C with |C | < |C |. Suppose that |C | = 6. Without loss of generality, let C = x1 x2 x3 x4 x5 x6 x1 with a chord. First, we consider the case ′ d(v, C ) = 3. If there exists an edge, say x1 x2 ∈ E (C ) with v x1 , v x2 ̸∈ E (G), then C = v x3 x4 x5 x6 v is a chorded cycle of length 5. So, by the symmetry, we assume that v x1 , v x3 , v x5 ∈ E (G). It follows from |N (u, C ) ∪ N (v, C )| ≥ 5 that there exists a vertex w ∈ N (u, C ) such that G[(V (C ) ∪ {v}) − {w}] contains a chorded cycle of length 6. Next, we consider the ′ case d(u, C ) = 3. Similarly, if there exists an edge, say x1 x2 ∈ E (C ) with ux1 , ux2 ̸∈ E (G), then C = ux3 x4 x5 x6 u is a chorded cycle of length 5. So, by the symmetry, we assume that ux1 , ux3 , ux5 ∈ E (G). If |N (u, C ) ∪ N (v, C )| = 6, then d(v, C ) = 3
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because d(u, C ) = 3 and d(v, C ) ≤ 3, and so this case is already done in the previous case. If |N (u, C ) ∪ N (v, C )| = 5, then by the symmetry, we may assume that v x2 , v x4 ∈ E (G). Note that C is a chorded cycle. If x3 is not an end vertex of a chord of C , then we can see that x3 ∈ N (u, C ) and G[(V (C ) ∪ {v}) − {x3 }] contains a chorded cycle of length 6. If the chord of C is x3 x1 or x3 x5 , then ux1 x2 x3 u or ux3 x4 x5 u is a chorded cycle of length 4, respectively. If the chord of C is x3 x6 , then we can see that x1 ∈ N (u, C ) and G[(V (C ) ∪ {v}) − {x1 }] contains a chorded cycle v x2 x3 x6 x5 x4 v of length 6 with the chord x3 x4 . Suppose that |C | = 5. Without loss of generality, let C = x1 x2 x3 x4 x5 x1 with a chord. Then we have N (u, C ) ∪ N (v, C ) = V (C ). We may assume d(u, C ) = 3. If u is adjacent to three consecutive vertices on C , then we are done. So, without loss of generality, suppose ux1 , ux3 , ux5 ∈ E (G). Clearly, v x2 , v x4 ∈ E (G). It is not difficult to deduce that the chord of C is at least one of x1 x3 , x1 x4 , x2 x4 , x2 x5 , x3 x5 . For each of these cases, we can always find a chorded cycle of length 4 in G[V (C ) ∪ {u, v}]. By the symmetry, the case d(v, C ) = 3 can be proved similarly. Lemma 3. Let C be a subgraph of G and P = u1 u2 u3 u4 be a path of length 3 in G − C . Then (i) If C is a cycle and
−
|N (ux , C ) ∪ N (uy , C )| ≥ 10,
(x,y)∈{(1,3),(2,4),(1,4)}
then G[V (C ) ∪ V (P )] contains a cycle C ′ with |C ′ | < |C |; (ii) If C is a chorded cycle and
−
|N (ux , C ) ∪ N (uy , C )| ≥ 13,
(x,y)∈{(1,3),(2,4),(1,4)}
then G[V (C ) ∪ V (P )] contains a chorded cycle C ′ with |C ′ | < |C |. Proof. (i) Since
−
|N (ux , C ) ∪ N (uy , C )| ≥ 10,
(x,y)∈{(1,3),(2,4),(1,4)}
we have |N (u1 , C )∪ N (u3 , C )| ≥ 4, |N (u2 , C )∪ N (u4 , C )| ≥ 4 or |N (u1 , C )∪ N (u4 , C )| ≥ 4. By Lemma 1, we may assume that |C | = 4 and G[V (C ) ∪ V (P )] contains no a cycle of length 3. This implies that ui is not adjacent to two consecutive vertices on C for each i with 1 ≤ i ≤ 4. Further, we have N (u1 , C )∩ N (u2 , C ) = ∅, N (u2 , C )∩ N (u3 , C ) = ∅ and N (u3 , C )∩ N (u4 , C ) = ∅. Then we can obtain that
−
|N (ux , C ) ∪ N (uy , C )| ≤ 8,
(x,y)∈{(1,3),(2,4),(1,4)}
a contradiction. (ii) Since
−
|N (ux , C ) ∪ N (uy , C )| ≥ 13,
(x,y)∈{(1,3),(2,4),(1,4)}
we have |N (u1 , C )∪ N (u3 , C )| ≥ 5, |N (u2 , C )∪ N (u4 , C )| ≥ 5 or |N (u1 , C )∪ N (u4 , C )| ≥ 5. By Lemma 2, we may assume that |C | = 6. If d(ui , C ) ≥ 4 or ui is adjacent to three consecutive vertices on C for some i with 1 ≤ i ≤ 4, then G[V (C ) ∪ V (P )] contains a chorded cycle C ′ with |C ′ | < |C |. So we may assume that d(ui , C ) ≤ 3 and ui is not adjacent to three consecutive vertices on C for all i with 1 ≤ i ≤ 4. Clearly, if d(ui , C ) = 3 and ui is adjacent to two consecutive vertices on C for some i with 1 ≤ i ≤ 4, then G[V (C ) ∪ {ui }] contains a chorded cycle C ′ with |C ′ | < |C |. Suppose that |N (u1 , C ) ∪ N (u3 , C )| ≥ 5. If N (u2 , C ) ∩ (N (u1 , C ) ∪ N (u3 , C )) ̸= ∅ or N (u4 , C ) ∩ N (u3 , C ) ̸= ∅, then G[V (C ) ∪ V (P )] contains a chorded cycle C ′ with |C ′ | < |C |. We may assume that N (u2 , C ) ∩ (N (u1 , C ) ∪ N (u3 , C )) = ∅ and N (u4 , C ) ∩ N (u3 , C ) = ∅. So, if |N (u1 , C ) ∪ N (u3 , C )| = 6, then
−
|N (ux , C ) ∪ N (uy , C )| ≤ 6 + 3 + 3 = 12,
(x,y)∈{(1,3),(2,4),(1,4)}
a contradiction. If |N (u1 , C ) ∪ N (u3 , C )| = 5, then
−
|N (ux , C ) ∪ N (uy , C )| ≤ 5 + 3 + 4 = 12,
(x,y)∈{(1,3),(2,4),(1,4)}
a contradiction. By symmetry, the case |N (u2 , C ) ∪ N (u4 , C )| ≥ 5 can be proved similarly. Suppose that |N (u1 , C )∪N (u4 , C )| ≥ 5. Clearly, if N (u2 , C )∩N (u1 , C ) ̸= ∅ or N (u3 , C )∩N (u4 , C ) ̸= ∅, then G[V (C )∪V (P )] contains a chorded cycle C ′ with |C ′ | < |C |. We may assume that N (u2 , C ) ∩ N (u1 , C ) = ∅ and N (u3 , C ) ∩ N (u4 , C ) = ∅. It is easy to check that
−
|N (ux , C ) ∪ N (uy , C )| ≤ 12,
(x,y)∈{(1,3),(2,4),(1,4)}
a contradiction.
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Lemma 4. Let C be a subgraph of G and C4 = u1 u2 u3 u4 u1 be a cycle of length 4 in G − C . Then (i) If C is a cycle and |N (u1 , C ) ∪ N (u3 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 7, then G[V (C ) ∪ V (C4 )] contains a cycle C ′ with |C ′ | < |C |. (ii) If C is a chorded cycle and |N (u1 , C ) ∪ N (u3 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, then G[V (C ) ∪ V (C4 )] contains a chorded cycle C ′ with |C ′ | < |C |. Proof. (i) Since |N (u1 , C ) ∪ N (u3 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 7, we have d(ui , C ) ≥ 2 for some i with 1 ≤ i ≤ 4. So, either G[V (C ) ∪ V (C4 )] contains a cycle C ′ with |C ′ | < |C | or |C | = 4. If |C | = 4, then it is easy to check that G[V (C ) ∪ V (C4 )] contains a triangle C ′ with |C ′ | < |C |. (ii) By the symmetry, suppose that |N (u1 , C ) ∪ N (u3 , C )| ≥ 5. It follows from Lemma 2 that either G[V (C ) ∪ V (C4 )] contains a chorded cycle C ′ with |C ′ | < |C | or |C | = 6. We may assume |C | = 6. Note that |N (u2 , C ) ∪ N (u4 , C )| ≥ 3. So, N (u2 , C )∩(N (u1 , C )∪ N (u3 , C )) ̸= ∅ or N (u4 , C )∩(N (u1 , C )∪ N (u3 , C )) ̸= ∅. It is not difficult to deduce that G[V (C )∪ V (C4 )] contains a chorded cycle C ′ with |C ′ | < |C |. Lemma 5. Let C be a chorded cycle of G and P = u1 u2 u3 u4 u5 be a path of length 4 in G − C such that u1 u3 , u3 u5 ∈ E (G). If ′ ′ |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u5 , C )| ≥ 9, then G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C |. Proof. Since |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u5 , C )| ≥ 9, by the symmetry, suppose that |N (u1 , C ) ∪ N (u4 , C )| ≥ 5. ′ ′ It follows from Lemma 2 that either G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C | or |C | = 6. We may assume |C | = 6. Clearly |N (u2 , C )∪ N (u5 , C )| ≥ 3. So, N (u2 , C )∩(N (u1 , C )∪ N (u4 , C )) ̸= ∅ or N (u5 , C )∩(N (u1 , C )∪ N (u4 , C )) ̸= ∅. ′
′
It is easy to see that G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C |.
Lemma 6. Let C be a chorded cycle of G and P = u1 u2 u3 u4 be a path of length 3 in G − C such that u1 u3 ∈ E (G). If |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, then at least one of the following holds. ′
′
(i) G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C |; (ii) G[V (C ) ∪ V (P )] contains a cycle of length 4 and a chorded cycle of length 6, all vertex-disjoint. Proof. It is not difficult to see that if N (u1 , C ) ∩ N (u2 , C ) ̸= ∅, N (u1 , C ) ∩ N (u4 , C ) ̸= ∅ or N (u2 , C ) ∩ N (u4 , C ) ̸= ∅, then (i) holds. Therefore, we assume that N (u1 , C ) ∩ N (u2 , C ) = ∅, N (u1 , C ) ∩ N (u4 , C ) = ∅ and N (u2 , C ) ∩ N (u4 , C ) = ∅. Since |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, we have |N (u1 , C ) ∪ N (u4 , C )| ≥ 5 or |N (u2 , C ) ∪ N (u4 , C )| ≥ 5. By the symmetry, we let |N (u1 , C ) ∪ N (u4 , C )| ≥ 5. By Lemma 2, we can obtain that (i) holds or |C | = 6. Now suppose that |C | = 6. Without loss of generality, set C = v1 v2 v3 v4 v5 v6 v1 with a chord. Clearly, if d(ui , C ) ≥ 4 with i ∈ {1, 2, 4}, then (i) holds. In the following we assume that d(ui , C ) ≤ 3. Suppose that |N (u1 , C ) ∪ N (u4 , C )| = 6. This implies that |N (u1 , C )| = 3 and |N (u4 , C )| = 3 with N (u1 , C ) ∩ N (u4 , C ) = ∅. Further, if either u1 or u4 is adjacent to two consecutive vertices on C , then (i) holds. So we may assume that u1 v1 , u1 v3 , u1 v5 ∈ E (G) and u4 v2 , u4 v4 , u4 v6 ∈ E (G). Note that C is a chorded cycle. If the chord of C is v1 v3 , v2 v4 , v3 v5 , v4 v6 , v5 v1 or v6 v2 , then G[V (C ) ∪ V (P )] contains a chorded cycle of length 4, so (i) holds. If the chord of C is v1 v4 , v2 v5 or v3 v6 , by the symmetry, say v1 v4 , then G[V (P ) ∪ {v2 , v3 }] contains a chorded cycle of length 6 and G[V (C ) − {v2 , v3 }] contains a cycle of length 4. This implies that (ii) holds. Suppose that |N (u1 , C ) ∪ N (u4 , C )| = 5. Note that N (u1 , C ) ∩ N (u4 , C ) = ∅. If |N (u1 , C )| = 3 and |N (u4 , C )| = 2, then by |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, we have N (u1 , C ) ∩ N (u2 , C ) ̸= ∅, a contradiction. If |N (u1 , C )| = 2 and |N (u4 , C )| = 3, then we may assume that u4 v1 , u4 v3 , u4 v5 ∈ E (G) and u1 v2 , u1 v4 ∈ E (G). Since |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, we have u2 v6 ∈ E (G). Note that C is a chorded cycle. If the chord of C ′ ′ is v1 v3 , v2 v4 , v3 v5 , v4 v6 , v5 v1 or v6 v2 , then G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C |. So (i) holds. If the chord of C is v1 v4 , v2 v5 or v3 v6 , then we can also get that G[V (C ) ∪ V (P )] contains a cycle of length 4 and a chorded cycle of length 6, all vertex-disjoint, so (ii) holds. Lemma 7. Let P = u1 u2 · · · ul be a path of G with l ≥ 4. Then (A) If u1 u3 , u2 u4 ̸∈ E (G) and |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| ≥ 10, then G[V (P )] contains a chorded cycle. (B) If u1 u3 , u2 u4 , u1 u4 ̸∈ E (G) and |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| + |N (u1 , P ) ∪ N (u4 , P )| ≥ 15, then G[V (P )] contains a chorded cycle. (C) If u1 u3 , u3 u5 ∈ E (G), u1 u4 , u2 u5 ̸∈ E (G) and |N (u1 , P ) ∪ N (u4 , P )| + |N (u2 , P ) ∪ N (u5 , P )| ≥ 10, then G[V (P )] contains a chorded cycle. (D) If u2 u4 ∈ E (G), u1 u4 , u2 u5 , u1 u5 ̸∈ E (G) and |N (u1 , P ) ∪ N (u4 , P )| + |N (u2 , P ) ∪ N (u5 , P )| + |N (u1 , P ) ∪ N (u5 , P )| ≥ 15, then G[V (P )] contains a chorded cycle. (E) If u1 u3 ∈ E (G), u1 u4 , u1 u5 , u3 u5 ̸∈ E (G) and |N (u1 , P ) ∪ N (u4 , P )| + |N (u1 , P ) ∪ N (u5 , P )| + |N (u3 , P ) ∪ N (u5 , P )| ≥ 15, then G[V (P )] contains a chorded cycle.
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Proof. (A). Since |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| ≥ 10, we have |N (u1 , P ) ∪ N (u3 , P )| ≥ 5 or |N (u2 , P ) ∪ N (u4 , P )| ≥ 5. If |N (u1 , P ) ∪ N (u3 , P )| ≥ 5, then u1 or u3 has at least two neighbors in {u5 , . . . , ul }. So (A) holds. If |N (u1 , P ) ∪ N (u3 , P )| ≤ 4, i.e., |N (u2 , P ) ∪ N (u4 , P )| ≥ 6, then u2 or u4 has at least two neighbors in {u6 , . . . , ul }. So (A) also holds. (B). By (A) we know that if |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| ≥ 10, then (B) holds. Thus, we may assume that |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| ≤ 9. Then we have |N (u1 , P ) ∪ N (u4 , P )| ≥ 6. It is not difficult to see that u1 or u4 has at least two neighbors in {u6 , . . . , ul }, so (B) holds. (C). Since |N (u1 , P ) ∪ N (u4 , P )| + |N (u2 , P ) ∪ N (u5 , P )| ≥ 10, we have |N (u1 , P ) ∪ N (u4 , P )| ≥ 5 or |N (u2 , P ) ∪ N (u5 , P )| ≥ 5. If |N (u1 , P ) ∪ N (u4 , P )| ≥ 5, then u1 or u4 has at least one neighbor in {u6 , . . . , ul }, so (C) holds. If |N (u1 , P ) ∪ N (u4 , P )| ≤ 4, i.e., |N (u2 , P ) ∪ N (u5 , P )| ≥ 6, then either u2 has at least one neighbor in {u7 , . . . , ul } or u5 has at least two neighbors in {u7 , . . . , ul }, so (C) also holds. (D). If |N (u1 , P ) ∪ N (u4 , P )| ≥ 5, then either u1 has at least one neighbor in {u6 , . . . , ul } or u4 has at least two neighbors in {u6 , . . . , ul }, so (D) holds. If |N (u2 , P ) ∪ N (u5 , P )| ≥ 6, then either u2 has at least one neighbor in {u7 , . . . , ul } or u5 has at least two neighbors in {u7 , . . . , ul }, so (D) also holds. The case |N (u1 , P ) ∪ N (u5 , P )| ≥ 6 can be proved similarly. (E). Similarly, if |N (u1 , P ) ∪ N (u4 , P )| ≥ 5, |N (u3 , P ) ∪ N (u5 , P )| ≥ 7 or |N (u1 , P ) ∪ N (u5 , P )| ≥ 6, then (E) holds. We may assume that |N (u1 , P ) ∪ N (u4 , P )| = 4, |N (u3 , P ) ∪ N (u5 , P )| = 6 and |N (u1 , P ) ∪ N (u5 , P )| = 5. It is not difficult to deduce that each of u3 , u4 has at least one neighbor in {u6 , . . . , ul }, and u5 has at least one neighbor in {u7 , . . . , ul }. Then G[V (P )] contains a chorded cycle, so (E) holds. 3. Proof of Theorem 7 First we show the following result. Theorem 8. Let G be a graph with at least 4 vertices. If |N (u, G) ∪ N (v, G)| ≥ 5 for all nonadjacent pairs of vertices u and v in G, then G contains a chorded cycle. Proof. Let P = u1 u2 · · · up be a longest path in G. Clearly, p ≥ 2. If u1 up ̸∈ E (G), then by |N (u1 , G) ∪ N (up , G)| ≥ 5 and the choice of P in G, we have d(u1 , P ) ≥ 3 or d(up , P ) ≥ 3. So G[V (P )] contains a chorded cycle. If u1 up ∈ E (G) and p = |G| ≥ 4, then by the neighborhood union condition, G[V (P )] contains a chorded cycle. If u1 up ∈ E (G) and p < |G|, then G[V (P )] contains a chorded cycle, or d(ui , P ) = 2 for all i with 1 ≤ i ≤ p. Further, by the choice of P, we have d(ui , G − P ) = 0. It follows from the neighborhood union condition that the minimum degree of G − P is at least 3. This implies that G − P contains a chorded cycle. Proof of Theorem 7. We prove this theorem by induction on t. If t = 0, then by Theorem 6 the result is true. If s = 0 and t = 1, then by Theorem 8 the result is also true. Now suppose that t ≥ 1 and the result holds for the values less than t. Since
|G| ≥ 3s + 4t > 3s + 4(t − 1) and
|N (u, G) ∪ N (v, G)| ≥ 3s + 4t + 1 > 3s + 4(t − 1) + 1 for all nonadjacent pairs of vertices u and v in G, by induction hypothesis, we can get that G contains s + t − 1 vertexdisjoint cycles such that t − 1 of them are chorded cycles. Without loss of generality, we let C1 , C2 , . . . , Cs be s cycles and C s+1 , C s+2 , . . . , C s+t −1 be t − 1 chorded cycles in G, all vertex-disjoint. Set GC = G −
s
s+t −1
Ci −
i =1
C i.
j=s+1
Let P = u1 u2 · · · up be a longest path in GC . We assume that C1 , C2 , . . . , Cs , C s+1 , C s+2 , . . . , C s+t −1 are chosen in G such that
s
(1)
s+t −1
i =1 C i +
j=s+1 C i is as small as possible. (2) P is as long as possible, subject to (1). (3) If G[V (P )] contains the cycles, then the length of the longest cycles of G[V (P )] is as large as possible, subject to (1) and (2).
Clearly, by induction hypothesis, we can obtain that G − {u, v, x, y} contains s + t − 1 vertex-disjoint cycles such that t − 1 of them are chorded cycles for any four vertices u, v, x, y of G. Together with (1), we have |GC | ≥ 4. We consider two cases as follows. Case 1. p < |GC |. First note that V (GC − P ) ̸= ∅. By the choice of P, we have d(u1 , GC ) = d(u1 , P ). If d(u1 , P ) ≥ 3. Then G[V (P )] contains a chorded cycle. This implies that Theorem 7 holds. So we may assume that d(u1 , GC ) = d(u1 , P ) ≤ 2. Suppose that there exists a vertex v ∈ V (GC − P ) such that d(v, GC ) ≤ 2. Then by u1 v ̸∈ E (G) and |N (u1 , G) ∪ N (v, G)| ≥ 3s + 4t + 1, we have
|N (u1 , G − GC ) ∪ N (v, G − GC )| ≥ 3s + 4(t − 1) + 1.
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Then there exists a cycle Ci with 1 ≤ i ≤ s such that |N (u1 , Ci ) ∪ N (v, Ci )| ≥ 4, or there exists a chorded cycle C j with s + 1 ≤ j ≤ s + t − 1 such that |N (u1 , C j ) ∪ N (v, C j )| ≥ 5. Using Lemmas 1 and 2, together with the choice (1) and (2), we can get a contradiction. Suppose that d(v, GC ) ≥ 3 for each vertex v ∈ V (GC − P ). Let Q = v1 v2 · · · vq be the longest path in GC − P. Clearly q ≥ 3. If d(v1 , Q ) ≥ 3 or d(vq , Q ) ≥ 3, then it is easy to see that GC contains a chorded cycle. So Theorem 7 holds. We may assume that d(v1 , Q ) ≤ 2 and d(vq , Q ) ≤ 2. This implies that d(v1 , P ) ≥ 1 and d(vq , P ) ≥ 1. It is not difficult to deduce that G[V (P ) ∪ V (Q )] contains a chorded cycle. Theorem 7 follows. Case 2. p = |GC |. Note that |GC | ≥ 4, i.e., p ≥ 4. Obviously, if d(u1 , P ) ≥ 3, then GC contains a chorded cycle and so Theorem 7 holds. In the following we assume that GC contains no a chorded cycle. Suppose that d(u1 , P ) = 2. Let ui with 3 ≤ i ≤ p be the vertex in N (u1 , P ). First, we consider the case i ≥ 5. Then u1 u4 , u1 u3 , u2 u4 ̸∈ E (G). By (B) of Lemma 7, we have
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 14.
(x,y)∈{(1,3),(1,4),(2,4)}
By the neighborhood union condition, we can obtain that
−
|N (ux , G − GC ) ∪ N (uy , G − GC )| ≥ 9s + 12(t − 1) + 1.
(x,y)∈{(1,3),(1,4),(2,4)}
Then either there exists a cycle Ck with 1 ≤ k ≤ s such that
−
|N (ux , Ck ) ∪ N (uy , Ck )| ≥ 10,
(I)
(x,y)∈{(1,3),(1,4),(2,4)}
or there exists a chorded cycle C j with s + 1 ≤ j ≤ s + t − 1 such that
−
|N (ux , C j ) ∪ N (uy , C j )| ≥ 13.
(II)
(x,y)∈{(1,3),(1,4),(2,4)}
By Lemma 3 and (1), we can get a contradiction. Next, we consider the case i = 4. Then u1 u3 , u2 u4 ̸∈ E (G). By (A) of Lemma 7, we have
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 9.
(x,y)∈{(1,3),(2,4)}
By the neighborhood union condition, we can obtain that
−
|N (ux , G − GC ) ∪ N (uy , G − GC )| ≥ 6s + 8(t − 1) + 1.
(x,y)∈{(1,3),(2,4)}
Then either there exists a cycle Ck with 1 ≤ k ≤ s such that
−
|N (ux , Ck ) ∪ N (uy , Ck )| ≥ 7,
(III)
(x,y)∈{(1,3),(2,4)}
or there exists a chorded cycle C j with s + 1 ≤ j ≤ s + t − 1 such that
−
|N (ux , C j ) ∪ N (uy , C j )| ≥ 9.
(IV)
(x,y)∈{(1,3),(2,4)}
By Lemma 4 and (1), we can get a contradiction. Finally, we consider the case i = 3. Suppose that p = 4. Then we have u1 u4 , u2 u4 ̸∈ E (G). So we can get that |N (u1 , GC ) ∪ N (u4 , GC )| + |N (u2 , GC ) ∪ N (u4 , GC )| = 4. It follows from the neighborhood union condition that
−
|N (ux , G − GC ) ∪ N (uy , G − GC )| ≥ 6s + 8(t − 1) + 6.
(x,y)∈{(1,4),(2,4)}
Then either (III) or (IV) holds. If the former is true, then by Lemma 1 and (1), we have |Ck | = 4. Note that GC contains a cycle of length 3. This contradicts (1). If the latter is true, then by Lemma 2 and (1), we have |C j | = 6. Further, by Lemma 6, (1) and (3), we can get a contradiction. Suppose that p ≥ 5. So we have u1 u4 , u1 u5 ̸∈ E (G). If u3 u5 ̸∈ E (G), then by (E) of Lemma 7 we have
− (x,y)∈{(1,4),(3,5),(1,5)}
|N (ux , GC ) ∪ N (uy , GC )| ≤ 14.
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Similarly, we can obtain that (I) or (II) holds. By Lemma 3 and (1), we can get a contradiction. If u3 u5 ∈ E (G), then u1 u4 , u2 u5 ̸∈ E (G). It follows from (C) of Lemma 7 that
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 9.
(x,y)∈{(1,4),(2,5)}
Then either (III) or (IV) holds. If the former is true, then by Lemma 1 and (1), we have |Ck | = 4. Note that GC contains a cycle of length 3. This contradicts (1). If the latter is true, then by Lemma 5 and (1), we can also get a contradiction. Suppose that d(u1 , P ) = 1. If u2 u4 ̸∈ E (G), then by (B) of Lemma 7, we have
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 14.
(x,y)∈{(1,4),(2,4),(1,3)}
This means that either (I) or (II) holds. By Lemma 3 and (1), we obtain a contradiction. If u2 u4 ∈ E (G) and p = 4, then
−
|N (ux , GC ) ∪ N (uy , GC )| < 9.
(x,y)∈{(1,3),(1,4)}
So, either (III) or (IV) holds. If the former is true, then by Lemma 1 and (1), we have |Ck | = 4. Note that GC contains a cycle of length 3. This contradicts (1). If the latter is true, then by Lemma 6, (1) and (3), we can also get a contradiction. So we may assume that u2 u4 ∈ E (G) and p ≥ 5. Note that u1 u4 , u1 u5 , u2 u5 ̸∈ E (G). Then by (D) of Lemma 7, we have
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 14.
(x,y)∈{(1,4),(2,5),(1,5)}
This implies that either (I) or (II) holds. By Lemma 3 and (1), we can get a contradiction. The proof of Theorem 7 is complete. 4. Remark In this paper, we give an example for which the bound is tight, when s = 0. In fact, the bound for s > 0 may be 3s + 4t, which is shown by considering a disjoint union of K3s+4t −1 and K2 . This could be proved using the same method, by induction on s. Acknowledgments The author is very grateful to the referees for their valuable suggestions and comments, which helped to improve the presentation of this paper. Supported by NSFC (No. 11001214) and the Fundamental Research Funds for the Central Universities (No. K50510700001). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
C.S. Babu, A.A. Diwan, Disjoint cycles with chords in graphs, J. Graph Theory 60 (2009) 87–98. A. Bialostocki, D. Finkel, A. Gyárfás, Disjoint chorded cycles in graphs, Discrete Math. 308 (2008) 5886–5890. J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, Macmillan Press, New York, 1976. S. Chiba, S. Fujita, Y. Gao, G. Li, On a sharp degree sum condition for disjoint chorded cycles in graphs, Graphs Combin. 26 (2010) 173–186. K. Corrádi, A. Hajnal, On the maximal number of independent circuits in a graph, Acta Math. Acad. Sci. Hungar. 14 (1963) 423–439. H. Enomoto, On the existence of disjoint cycles in a graph, Combinatorica 18 (1998) 487–492. R.J. Faudree, R.J. Gould, A note on neighborhood unions and independent cycles, Ars Combin. 76 (2005) 29–31. D. Finkel, On the number of independent chorded cycle in a graph, Discrete Math. 308 (2008) 5265–5268. Y. Gao, G. Li, Jonathan Z. Sun, J. Yan, Disjoint cycles and chorded cycles in graphs, in: FCS, 2008, pp. 9–13. L. Pósa, On the circuits of finite graphs, Magyar Tud. Akad. Mat. Kutató Int. Közl. 8 (1964) 355–361. H. Wang, On the maximum number of independent cycles in a graph, Discrete Math. 205 (1999) 183–190.
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Neighborhood unions and disjoint chorded cycles in graphs Shengning Qiao Department of Mathematics, Xidian University, Xi’an, Shaanxi 710071, PR China
article
info
Article history: Received 2 December 2010 Received in revised form 9 October 2011 Accepted 10 October 2011 Available online 28 October 2011
abstract Let s ≥ 0, t ≥ 0 be two integers and G be a graph with at least 3s + 4t vertices. In this paper, we show that if |N (u, G) ∪ N (v, G)| ≥ 3s + 4t + 1 for any two nonadjacent vertices u and v in G, then G contains s + t vertex-disjoint cycles such that t of them are chorded cycles. The condition is sharp. © 2011 Elsevier B.V. All rights reserved.
Keywords: Cycle Chorded cycle Vertex-disjoint
1. Introduction We use Bondy and Murty [3] for terminology and notation not defined here and consider finite simple graphs only. Let G be a graph. A set of subgraphs of G is said to be vertex-disjoint if no two of them have a common vertex in G. Corrádi and Hajnal [5] proved the following well-known result on the existence of vertex-disjoint cycles in graphs. Theorem 1 (Corrádi and Hajnal [5]). Let s ≥ 0 be an integer and G be a graph with at least 3s vertices. If the minimum degree of G is at least 2s, then G contains s vertex-disjoint cycles. Later, this result was generalized by Enomoto [6] and Wang [11], as follows. Theorem 2 (Enomoto [6] and Wang [11]). Let s ≥ 0 be an integer and G be a graph with at least 3s vertices. If the degree sum of any two nonadjacent vertices in G is at least 4s − 1, then G contains s vertex-disjoint cycles. An edge which joins two vertices of a cycle but is not itself an edge of the cycle is a chord of the cycle. A cycle is called a chorded cycle if it has at least one chord. Along a different line, Pósa proved in [10] that any graph with minimum degree at least 3 contains a chorded cycle. Motivated by these results, Finkel [8] proved the following result analogous to Theorem 1. Theorem 3 (Finkel [8]). Let t ≥ 0 be an integer and G be a graph with at least 4t vertices. If the minimum degree of G is at least 3t, then G contains t vertex-disjoint chorded cycles. In fact, in [2], Bialostocki et al. posed the following conjecture. Conjecture 4 (Bialosticki et al. [2]). Let s ≥ 0, t ≥ 0 be two integers and G be a graph with at least 3s + 4t vertices. If the minimum degree of G is at least 2s + 3t, then G contains a vertex-disjoint union of s + t cycles such that t of them are chorded cycles.
E-mail addresses: [email protected], [email protected]. 0012-365X/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2011.10.012
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Very recently, Babu and Diwan [1], Chiba and Fujita [4] and Gao et al. [9] proved that the conjecture is true. Especially, in [1,4], the authors obtained a stronger version of this conjecture for degree sum condition. Theorem 5 (Babu and Diwan [1], Chiba and Fujita [4]). Let s ≥ 0, t ≥ 0 be two integers and G be a graph with at least 3s + 4t vertices. If the degree sum of any two nonadjacent vertices in G is at least 4s + 6t − 1, then G contains a vertex-disjoint union of s + t cycles such that t of them are chorded cycles. Motivated by Theorem 5, in this paper, we consider the neighborhood union condition for the existence of vertex-disjoint cycles and chorded cycles in graphs. In 2005, Faudree and Gould [7] obtained the neighborhood union condition for the existence of vertex-disjoint cycles in graphs, as follows. Theorem 6 (Faudree and Gould [7]). Let s ≥ 0 be an integer. Let G be a graph with at least 3s vertices. If |N (u, G)∪ N (v, G)| ≥ 3s for all nonadjacent pairs of vertices u and v in G, then G contains s vertex-disjoint cycles. Our result is as follows. Theorem 7. Let s ≥ 0, t ≥ 0 be two integers. Let G be a graph with at least 3s + 4t vertices. If |N (u, G)∪ N (v, G)| ≥ 3s + 4t + 1 for all nonadjacent pairs of vertices u and v in G, then G contains s + t vertex-disjoint cycles such that t of them are chorded cycles. The condition is sharp. Let t ≥ 2 be an even. We construct a graph G = K2t +3 ∪ K2t −1 , i.e., the union of K2t +3 and K2t −1 . Clearly, we have |N (u, G)∪ N (v, G)| = 4t for all nonadjacent pairs of vertices u and v in G. But G contains no t vertex-disjoint chorded cycles. Further, we can construct a graph G = K4t +2 ∪ K3s+1 . Let t ≥ 2. Clearly, we cannot apply Theorem 5 to ensure the existence of s + t vertex-disjoint cycles in G such that t of them are chorded cycles. However, by Theorem 7, we can get that G contains s + t vertex-disjoint cycles such that t of them are chorded cycles. We further introduce some notation that will be used later. Let G = (V (G), E (G)) be a graph. By |G| we denote |V (G)|. For a vertex v and a subgraph H of G, we use N (v, H ) and d(v, H ) to denote the set and the number of vertices in H that are adjacent to v , respectively. Thus d(v, G) is the degree of v in G. By G − H we denote G[V (G) \ V (H )]. For a subset S of vertices or edges in G, we use G[S ] to denote the subgraph of G induced by S. In Section 2, we give some lemmas to be used later. In Section 3, we will prove Theorem 7. 2. Some lemmas In the following, we let G be a graph. Lemma 1. Let C be a cycle of G and u, v be two vertices of G − C . If |N (u, C ) ∪ N (v, C )| ≥ 4, then either G[V (C ) ∪ {u, v}] contains a cycle C ′ with |C ′ | < |C | or |C | = 4 and there exists a vertex w ∈ N (u, C ) such that G[(V (C ) ∪ {v}) − {w}] contains a cycle of length 4. Proof. First note that |C | ≥ 4. If u or v is adjacent to two end vertices of an edge of C , then G[V (C ) ∪ {u}] or G[V (C ) ∪ {v}] contains a cycle of length exactly three, respectively. Since |N (u, C ) ∪ N (v, C )| ≥ 4, we have d(u, C ) ≥ 2 or d(v, C ) ≥ 2. By the symmetry, we assume that d(u, C ) ≥ 2. If |C | ≥ 5, then it is not difficult to see that G[V (C ) ∪ {u}] contains a cycle C ′ with |C ′ | < |C |. If |C | = 4, then u and v are adjacent to two nonadjacent vertices on C with N (u, C ) ∩ N (v, C ) = ∅. Then there exists a vertex w ∈ N (u, C ) such that G[(V (C ) ∪ {v}) − {w}] contains a cycle of length 4. Lemma 2. Let C be a chorded cycle of G and u, v be two vertices of G − C . If |N (u, C )∪ N (v, C )| ≥ 5, then either G[V (C )∪{u, v}] ′ ′ contains a chorded cycle C with |C | < |C | or |C | = 6 and there exists a vertex w ∈ N (u, C ) such that G[(V (C ) ∪ {v}) − {w}] contains a chorded cycle of length 6. Proof. Clearly, we have |C | ≥ 5. If d(u, C ) ≥ 4 or d(v, C ) ≥ 4, then we can get that G[V (C ) ∪ {u}] or G[V (C ) ∪ {v}] contains ′ ′ a chorded cycle C with |C | < |C |, respectively. So we may assume that d(u, C ) ≤ 3 and d(v, C ) ≤ 3. Further, we can also obtain that if u or v is adjacent to three consecutive vertices on C , then either G[V (C ) ∪ {u}] or G[V (C ) ∪ {v}] contains a chorded cycle of length 4. Suppose that |C | ≥ 7. By |N (u, C )∪ N (v, C )| ≥ 5, we have d(u, C ) = 3 or d(v, C ) = 3, say d(u, C ) = 3. Then G[V (C )∪{u}] ′ ′ contains a chorded cycle C with |C | < |C |. Suppose that |C | = 6. Without loss of generality, let C = x1 x2 x3 x4 x5 x6 x1 with a chord. First, we consider the case ′ d(v, C ) = 3. If there exists an edge, say x1 x2 ∈ E (C ) with v x1 , v x2 ̸∈ E (G), then C = v x3 x4 x5 x6 v is a chorded cycle of length 5. So, by the symmetry, we assume that v x1 , v x3 , v x5 ∈ E (G). It follows from |N (u, C ) ∪ N (v, C )| ≥ 5 that there exists a vertex w ∈ N (u, C ) such that G[(V (C ) ∪ {v}) − {w}] contains a chorded cycle of length 6. Next, we consider the ′ case d(u, C ) = 3. Similarly, if there exists an edge, say x1 x2 ∈ E (C ) with ux1 , ux2 ̸∈ E (G), then C = ux3 x4 x5 x6 u is a chorded cycle of length 5. So, by the symmetry, we assume that ux1 , ux3 , ux5 ∈ E (G). If |N (u, C ) ∪ N (v, C )| = 6, then d(v, C ) = 3
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because d(u, C ) = 3 and d(v, C ) ≤ 3, and so this case is already done in the previous case. If |N (u, C ) ∪ N (v, C )| = 5, then by the symmetry, we may assume that v x2 , v x4 ∈ E (G). Note that C is a chorded cycle. If x3 is not an end vertex of a chord of C , then we can see that x3 ∈ N (u, C ) and G[(V (C ) ∪ {v}) − {x3 }] contains a chorded cycle of length 6. If the chord of C is x3 x1 or x3 x5 , then ux1 x2 x3 u or ux3 x4 x5 u is a chorded cycle of length 4, respectively. If the chord of C is x3 x6 , then we can see that x1 ∈ N (u, C ) and G[(V (C ) ∪ {v}) − {x1 }] contains a chorded cycle v x2 x3 x6 x5 x4 v of length 6 with the chord x3 x4 . Suppose that |C | = 5. Without loss of generality, let C = x1 x2 x3 x4 x5 x1 with a chord. Then we have N (u, C ) ∪ N (v, C ) = V (C ). We may assume d(u, C ) = 3. If u is adjacent to three consecutive vertices on C , then we are done. So, without loss of generality, suppose ux1 , ux3 , ux5 ∈ E (G). Clearly, v x2 , v x4 ∈ E (G). It is not difficult to deduce that the chord of C is at least one of x1 x3 , x1 x4 , x2 x4 , x2 x5 , x3 x5 . For each of these cases, we can always find a chorded cycle of length 4 in G[V (C ) ∪ {u, v}]. By the symmetry, the case d(v, C ) = 3 can be proved similarly. Lemma 3. Let C be a subgraph of G and P = u1 u2 u3 u4 be a path of length 3 in G − C . Then (i) If C is a cycle and
−
|N (ux , C ) ∪ N (uy , C )| ≥ 10,
(x,y)∈{(1,3),(2,4),(1,4)}
then G[V (C ) ∪ V (P )] contains a cycle C ′ with |C ′ | < |C |; (ii) If C is a chorded cycle and
−
|N (ux , C ) ∪ N (uy , C )| ≥ 13,
(x,y)∈{(1,3),(2,4),(1,4)}
then G[V (C ) ∪ V (P )] contains a chorded cycle C ′ with |C ′ | < |C |. Proof. (i) Since
−
|N (ux , C ) ∪ N (uy , C )| ≥ 10,
(x,y)∈{(1,3),(2,4),(1,4)}
we have |N (u1 , C )∪ N (u3 , C )| ≥ 4, |N (u2 , C )∪ N (u4 , C )| ≥ 4 or |N (u1 , C )∪ N (u4 , C )| ≥ 4. By Lemma 1, we may assume that |C | = 4 and G[V (C ) ∪ V (P )] contains no a cycle of length 3. This implies that ui is not adjacent to two consecutive vertices on C for each i with 1 ≤ i ≤ 4. Further, we have N (u1 , C )∩ N (u2 , C ) = ∅, N (u2 , C )∩ N (u3 , C ) = ∅ and N (u3 , C )∩ N (u4 , C ) = ∅. Then we can obtain that
−
|N (ux , C ) ∪ N (uy , C )| ≤ 8,
(x,y)∈{(1,3),(2,4),(1,4)}
a contradiction. (ii) Since
−
|N (ux , C ) ∪ N (uy , C )| ≥ 13,
(x,y)∈{(1,3),(2,4),(1,4)}
we have |N (u1 , C )∪ N (u3 , C )| ≥ 5, |N (u2 , C )∪ N (u4 , C )| ≥ 5 or |N (u1 , C )∪ N (u4 , C )| ≥ 5. By Lemma 2, we may assume that |C | = 6. If d(ui , C ) ≥ 4 or ui is adjacent to three consecutive vertices on C for some i with 1 ≤ i ≤ 4, then G[V (C ) ∪ V (P )] contains a chorded cycle C ′ with |C ′ | < |C |. So we may assume that d(ui , C ) ≤ 3 and ui is not adjacent to three consecutive vertices on C for all i with 1 ≤ i ≤ 4. Clearly, if d(ui , C ) = 3 and ui is adjacent to two consecutive vertices on C for some i with 1 ≤ i ≤ 4, then G[V (C ) ∪ {ui }] contains a chorded cycle C ′ with |C ′ | < |C |. Suppose that |N (u1 , C ) ∪ N (u3 , C )| ≥ 5. If N (u2 , C ) ∩ (N (u1 , C ) ∪ N (u3 , C )) ̸= ∅ or N (u4 , C ) ∩ N (u3 , C ) ̸= ∅, then G[V (C ) ∪ V (P )] contains a chorded cycle C ′ with |C ′ | < |C |. We may assume that N (u2 , C ) ∩ (N (u1 , C ) ∪ N (u3 , C )) = ∅ and N (u4 , C ) ∩ N (u3 , C ) = ∅. So, if |N (u1 , C ) ∪ N (u3 , C )| = 6, then
−
|N (ux , C ) ∪ N (uy , C )| ≤ 6 + 3 + 3 = 12,
(x,y)∈{(1,3),(2,4),(1,4)}
a contradiction. If |N (u1 , C ) ∪ N (u3 , C )| = 5, then
−
|N (ux , C ) ∪ N (uy , C )| ≤ 5 + 3 + 4 = 12,
(x,y)∈{(1,3),(2,4),(1,4)}
a contradiction. By symmetry, the case |N (u2 , C ) ∪ N (u4 , C )| ≥ 5 can be proved similarly. Suppose that |N (u1 , C )∪N (u4 , C )| ≥ 5. Clearly, if N (u2 , C )∩N (u1 , C ) ̸= ∅ or N (u3 , C )∩N (u4 , C ) ̸= ∅, then G[V (C )∪V (P )] contains a chorded cycle C ′ with |C ′ | < |C |. We may assume that N (u2 , C ) ∩ N (u1 , C ) = ∅ and N (u3 , C ) ∩ N (u4 , C ) = ∅. It is easy to check that
−
|N (ux , C ) ∪ N (uy , C )| ≤ 12,
(x,y)∈{(1,3),(2,4),(1,4)}
a contradiction.
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Lemma 4. Let C be a subgraph of G and C4 = u1 u2 u3 u4 u1 be a cycle of length 4 in G − C . Then (i) If C is a cycle and |N (u1 , C ) ∪ N (u3 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 7, then G[V (C ) ∪ V (C4 )] contains a cycle C ′ with |C ′ | < |C |. (ii) If C is a chorded cycle and |N (u1 , C ) ∪ N (u3 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, then G[V (C ) ∪ V (C4 )] contains a chorded cycle C ′ with |C ′ | < |C |. Proof. (i) Since |N (u1 , C ) ∪ N (u3 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 7, we have d(ui , C ) ≥ 2 for some i with 1 ≤ i ≤ 4. So, either G[V (C ) ∪ V (C4 )] contains a cycle C ′ with |C ′ | < |C | or |C | = 4. If |C | = 4, then it is easy to check that G[V (C ) ∪ V (C4 )] contains a triangle C ′ with |C ′ | < |C |. (ii) By the symmetry, suppose that |N (u1 , C ) ∪ N (u3 , C )| ≥ 5. It follows from Lemma 2 that either G[V (C ) ∪ V (C4 )] contains a chorded cycle C ′ with |C ′ | < |C | or |C | = 6. We may assume |C | = 6. Note that |N (u2 , C ) ∪ N (u4 , C )| ≥ 3. So, N (u2 , C )∩(N (u1 , C )∪ N (u3 , C )) ̸= ∅ or N (u4 , C )∩(N (u1 , C )∪ N (u3 , C )) ̸= ∅. It is not difficult to deduce that G[V (C )∪ V (C4 )] contains a chorded cycle C ′ with |C ′ | < |C |. Lemma 5. Let C be a chorded cycle of G and P = u1 u2 u3 u4 u5 be a path of length 4 in G − C such that u1 u3 , u3 u5 ∈ E (G). If ′ ′ |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u5 , C )| ≥ 9, then G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C |. Proof. Since |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u5 , C )| ≥ 9, by the symmetry, suppose that |N (u1 , C ) ∪ N (u4 , C )| ≥ 5. ′ ′ It follows from Lemma 2 that either G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C | or |C | = 6. We may assume |C | = 6. Clearly |N (u2 , C )∪ N (u5 , C )| ≥ 3. So, N (u2 , C )∩(N (u1 , C )∪ N (u4 , C )) ̸= ∅ or N (u5 , C )∩(N (u1 , C )∪ N (u4 , C )) ̸= ∅. ′
′
It is easy to see that G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C |.
Lemma 6. Let C be a chorded cycle of G and P = u1 u2 u3 u4 be a path of length 3 in G − C such that u1 u3 ∈ E (G). If |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, then at least one of the following holds. ′
′
(i) G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C |; (ii) G[V (C ) ∪ V (P )] contains a cycle of length 4 and a chorded cycle of length 6, all vertex-disjoint. Proof. It is not difficult to see that if N (u1 , C ) ∩ N (u2 , C ) ̸= ∅, N (u1 , C ) ∩ N (u4 , C ) ̸= ∅ or N (u2 , C ) ∩ N (u4 , C ) ̸= ∅, then (i) holds. Therefore, we assume that N (u1 , C ) ∩ N (u2 , C ) = ∅, N (u1 , C ) ∩ N (u4 , C ) = ∅ and N (u2 , C ) ∩ N (u4 , C ) = ∅. Since |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, we have |N (u1 , C ) ∪ N (u4 , C )| ≥ 5 or |N (u2 , C ) ∪ N (u4 , C )| ≥ 5. By the symmetry, we let |N (u1 , C ) ∪ N (u4 , C )| ≥ 5. By Lemma 2, we can obtain that (i) holds or |C | = 6. Now suppose that |C | = 6. Without loss of generality, set C = v1 v2 v3 v4 v5 v6 v1 with a chord. Clearly, if d(ui , C ) ≥ 4 with i ∈ {1, 2, 4}, then (i) holds. In the following we assume that d(ui , C ) ≤ 3. Suppose that |N (u1 , C ) ∪ N (u4 , C )| = 6. This implies that |N (u1 , C )| = 3 and |N (u4 , C )| = 3 with N (u1 , C ) ∩ N (u4 , C ) = ∅. Further, if either u1 or u4 is adjacent to two consecutive vertices on C , then (i) holds. So we may assume that u1 v1 , u1 v3 , u1 v5 ∈ E (G) and u4 v2 , u4 v4 , u4 v6 ∈ E (G). Note that C is a chorded cycle. If the chord of C is v1 v3 , v2 v4 , v3 v5 , v4 v6 , v5 v1 or v6 v2 , then G[V (C ) ∪ V (P )] contains a chorded cycle of length 4, so (i) holds. If the chord of C is v1 v4 , v2 v5 or v3 v6 , by the symmetry, say v1 v4 , then G[V (P ) ∪ {v2 , v3 }] contains a chorded cycle of length 6 and G[V (C ) − {v2 , v3 }] contains a cycle of length 4. This implies that (ii) holds. Suppose that |N (u1 , C ) ∪ N (u4 , C )| = 5. Note that N (u1 , C ) ∩ N (u4 , C ) = ∅. If |N (u1 , C )| = 3 and |N (u4 , C )| = 2, then by |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, we have N (u1 , C ) ∩ N (u2 , C ) ̸= ∅, a contradiction. If |N (u1 , C )| = 2 and |N (u4 , C )| = 3, then we may assume that u4 v1 , u4 v3 , u4 v5 ∈ E (G) and u1 v2 , u1 v4 ∈ E (G). Since |N (u1 , C ) ∪ N (u4 , C )| + |N (u2 , C ) ∪ N (u4 , C )| ≥ 9, we have u2 v6 ∈ E (G). Note that C is a chorded cycle. If the chord of C ′ ′ is v1 v3 , v2 v4 , v3 v5 , v4 v6 , v5 v1 or v6 v2 , then G[V (C ) ∪ V (P )] contains a chorded cycle C with |C | < |C |. So (i) holds. If the chord of C is v1 v4 , v2 v5 or v3 v6 , then we can also get that G[V (C ) ∪ V (P )] contains a cycle of length 4 and a chorded cycle of length 6, all vertex-disjoint, so (ii) holds. Lemma 7. Let P = u1 u2 · · · ul be a path of G with l ≥ 4. Then (A) If u1 u3 , u2 u4 ̸∈ E (G) and |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| ≥ 10, then G[V (P )] contains a chorded cycle. (B) If u1 u3 , u2 u4 , u1 u4 ̸∈ E (G) and |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| + |N (u1 , P ) ∪ N (u4 , P )| ≥ 15, then G[V (P )] contains a chorded cycle. (C) If u1 u3 , u3 u5 ∈ E (G), u1 u4 , u2 u5 ̸∈ E (G) and |N (u1 , P ) ∪ N (u4 , P )| + |N (u2 , P ) ∪ N (u5 , P )| ≥ 10, then G[V (P )] contains a chorded cycle. (D) If u2 u4 ∈ E (G), u1 u4 , u2 u5 , u1 u5 ̸∈ E (G) and |N (u1 , P ) ∪ N (u4 , P )| + |N (u2 , P ) ∪ N (u5 , P )| + |N (u1 , P ) ∪ N (u5 , P )| ≥ 15, then G[V (P )] contains a chorded cycle. (E) If u1 u3 ∈ E (G), u1 u4 , u1 u5 , u3 u5 ̸∈ E (G) and |N (u1 , P ) ∪ N (u4 , P )| + |N (u1 , P ) ∪ N (u5 , P )| + |N (u3 , P ) ∪ N (u5 , P )| ≥ 15, then G[V (P )] contains a chorded cycle.
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Proof. (A). Since |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| ≥ 10, we have |N (u1 , P ) ∪ N (u3 , P )| ≥ 5 or |N (u2 , P ) ∪ N (u4 , P )| ≥ 5. If |N (u1 , P ) ∪ N (u3 , P )| ≥ 5, then u1 or u3 has at least two neighbors in {u5 , . . . , ul }. So (A) holds. If |N (u1 , P ) ∪ N (u3 , P )| ≤ 4, i.e., |N (u2 , P ) ∪ N (u4 , P )| ≥ 6, then u2 or u4 has at least two neighbors in {u6 , . . . , ul }. So (A) also holds. (B). By (A) we know that if |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| ≥ 10, then (B) holds. Thus, we may assume that |N (u1 , P ) ∪ N (u3 , P )| + |N (u2 , P ) ∪ N (u4 , P )| ≤ 9. Then we have |N (u1 , P ) ∪ N (u4 , P )| ≥ 6. It is not difficult to see that u1 or u4 has at least two neighbors in {u6 , . . . , ul }, so (B) holds. (C). Since |N (u1 , P ) ∪ N (u4 , P )| + |N (u2 , P ) ∪ N (u5 , P )| ≥ 10, we have |N (u1 , P ) ∪ N (u4 , P )| ≥ 5 or |N (u2 , P ) ∪ N (u5 , P )| ≥ 5. If |N (u1 , P ) ∪ N (u4 , P )| ≥ 5, then u1 or u4 has at least one neighbor in {u6 , . . . , ul }, so (C) holds. If |N (u1 , P ) ∪ N (u4 , P )| ≤ 4, i.e., |N (u2 , P ) ∪ N (u5 , P )| ≥ 6, then either u2 has at least one neighbor in {u7 , . . . , ul } or u5 has at least two neighbors in {u7 , . . . , ul }, so (C) also holds. (D). If |N (u1 , P ) ∪ N (u4 , P )| ≥ 5, then either u1 has at least one neighbor in {u6 , . . . , ul } or u4 has at least two neighbors in {u6 , . . . , ul }, so (D) holds. If |N (u2 , P ) ∪ N (u5 , P )| ≥ 6, then either u2 has at least one neighbor in {u7 , . . . , ul } or u5 has at least two neighbors in {u7 , . . . , ul }, so (D) also holds. The case |N (u1 , P ) ∪ N (u5 , P )| ≥ 6 can be proved similarly. (E). Similarly, if |N (u1 , P ) ∪ N (u4 , P )| ≥ 5, |N (u3 , P ) ∪ N (u5 , P )| ≥ 7 or |N (u1 , P ) ∪ N (u5 , P )| ≥ 6, then (E) holds. We may assume that |N (u1 , P ) ∪ N (u4 , P )| = 4, |N (u3 , P ) ∪ N (u5 , P )| = 6 and |N (u1 , P ) ∪ N (u5 , P )| = 5. It is not difficult to deduce that each of u3 , u4 has at least one neighbor in {u6 , . . . , ul }, and u5 has at least one neighbor in {u7 , . . . , ul }. Then G[V (P )] contains a chorded cycle, so (E) holds. 3. Proof of Theorem 7 First we show the following result. Theorem 8. Let G be a graph with at least 4 vertices. If |N (u, G) ∪ N (v, G)| ≥ 5 for all nonadjacent pairs of vertices u and v in G, then G contains a chorded cycle. Proof. Let P = u1 u2 · · · up be a longest path in G. Clearly, p ≥ 2. If u1 up ̸∈ E (G), then by |N (u1 , G) ∪ N (up , G)| ≥ 5 and the choice of P in G, we have d(u1 , P ) ≥ 3 or d(up , P ) ≥ 3. So G[V (P )] contains a chorded cycle. If u1 up ∈ E (G) and p = |G| ≥ 4, then by the neighborhood union condition, G[V (P )] contains a chorded cycle. If u1 up ∈ E (G) and p < |G|, then G[V (P )] contains a chorded cycle, or d(ui , P ) = 2 for all i with 1 ≤ i ≤ p. Further, by the choice of P, we have d(ui , G − P ) = 0. It follows from the neighborhood union condition that the minimum degree of G − P is at least 3. This implies that G − P contains a chorded cycle. Proof of Theorem 7. We prove this theorem by induction on t. If t = 0, then by Theorem 6 the result is true. If s = 0 and t = 1, then by Theorem 8 the result is also true. Now suppose that t ≥ 1 and the result holds for the values less than t. Since
|G| ≥ 3s + 4t > 3s + 4(t − 1) and
|N (u, G) ∪ N (v, G)| ≥ 3s + 4t + 1 > 3s + 4(t − 1) + 1 for all nonadjacent pairs of vertices u and v in G, by induction hypothesis, we can get that G contains s + t − 1 vertexdisjoint cycles such that t − 1 of them are chorded cycles. Without loss of generality, we let C1 , C2 , . . . , Cs be s cycles and C s+1 , C s+2 , . . . , C s+t −1 be t − 1 chorded cycles in G, all vertex-disjoint. Set GC = G −
s
s+t −1
Ci −
i =1
C i.
j=s+1
Let P = u1 u2 · · · up be a longest path in GC . We assume that C1 , C2 , . . . , Cs , C s+1 , C s+2 , . . . , C s+t −1 are chosen in G such that
s
(1)
s+t −1
i =1 C i +
j=s+1 C i is as small as possible. (2) P is as long as possible, subject to (1). (3) If G[V (P )] contains the cycles, then the length of the longest cycles of G[V (P )] is as large as possible, subject to (1) and (2).
Clearly, by induction hypothesis, we can obtain that G − {u, v, x, y} contains s + t − 1 vertex-disjoint cycles such that t − 1 of them are chorded cycles for any four vertices u, v, x, y of G. Together with (1), we have |GC | ≥ 4. We consider two cases as follows. Case 1. p < |GC |. First note that V (GC − P ) ̸= ∅. By the choice of P, we have d(u1 , GC ) = d(u1 , P ). If d(u1 , P ) ≥ 3. Then G[V (P )] contains a chorded cycle. This implies that Theorem 7 holds. So we may assume that d(u1 , GC ) = d(u1 , P ) ≤ 2. Suppose that there exists a vertex v ∈ V (GC − P ) such that d(v, GC ) ≤ 2. Then by u1 v ̸∈ E (G) and |N (u1 , G) ∪ N (v, G)| ≥ 3s + 4t + 1, we have
|N (u1 , G − GC ) ∪ N (v, G − GC )| ≥ 3s + 4(t − 1) + 1.
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Then there exists a cycle Ci with 1 ≤ i ≤ s such that |N (u1 , Ci ) ∪ N (v, Ci )| ≥ 4, or there exists a chorded cycle C j with s + 1 ≤ j ≤ s + t − 1 such that |N (u1 , C j ) ∪ N (v, C j )| ≥ 5. Using Lemmas 1 and 2, together with the choice (1) and (2), we can get a contradiction. Suppose that d(v, GC ) ≥ 3 for each vertex v ∈ V (GC − P ). Let Q = v1 v2 · · · vq be the longest path in GC − P. Clearly q ≥ 3. If d(v1 , Q ) ≥ 3 or d(vq , Q ) ≥ 3, then it is easy to see that GC contains a chorded cycle. So Theorem 7 holds. We may assume that d(v1 , Q ) ≤ 2 and d(vq , Q ) ≤ 2. This implies that d(v1 , P ) ≥ 1 and d(vq , P ) ≥ 1. It is not difficult to deduce that G[V (P ) ∪ V (Q )] contains a chorded cycle. Theorem 7 follows. Case 2. p = |GC |. Note that |GC | ≥ 4, i.e., p ≥ 4. Obviously, if d(u1 , P ) ≥ 3, then GC contains a chorded cycle and so Theorem 7 holds. In the following we assume that GC contains no a chorded cycle. Suppose that d(u1 , P ) = 2. Let ui with 3 ≤ i ≤ p be the vertex in N (u1 , P ). First, we consider the case i ≥ 5. Then u1 u4 , u1 u3 , u2 u4 ̸∈ E (G). By (B) of Lemma 7, we have
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 14.
(x,y)∈{(1,3),(1,4),(2,4)}
By the neighborhood union condition, we can obtain that
−
|N (ux , G − GC ) ∪ N (uy , G − GC )| ≥ 9s + 12(t − 1) + 1.
(x,y)∈{(1,3),(1,4),(2,4)}
Then either there exists a cycle Ck with 1 ≤ k ≤ s such that
−
|N (ux , Ck ) ∪ N (uy , Ck )| ≥ 10,
(I)
(x,y)∈{(1,3),(1,4),(2,4)}
or there exists a chorded cycle C j with s + 1 ≤ j ≤ s + t − 1 such that
−
|N (ux , C j ) ∪ N (uy , C j )| ≥ 13.
(II)
(x,y)∈{(1,3),(1,4),(2,4)}
By Lemma 3 and (1), we can get a contradiction. Next, we consider the case i = 4. Then u1 u3 , u2 u4 ̸∈ E (G). By (A) of Lemma 7, we have
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 9.
(x,y)∈{(1,3),(2,4)}
By the neighborhood union condition, we can obtain that
−
|N (ux , G − GC ) ∪ N (uy , G − GC )| ≥ 6s + 8(t − 1) + 1.
(x,y)∈{(1,3),(2,4)}
Then either there exists a cycle Ck with 1 ≤ k ≤ s such that
−
|N (ux , Ck ) ∪ N (uy , Ck )| ≥ 7,
(III)
(x,y)∈{(1,3),(2,4)}
or there exists a chorded cycle C j with s + 1 ≤ j ≤ s + t − 1 such that
−
|N (ux , C j ) ∪ N (uy , C j )| ≥ 9.
(IV)
(x,y)∈{(1,3),(2,4)}
By Lemma 4 and (1), we can get a contradiction. Finally, we consider the case i = 3. Suppose that p = 4. Then we have u1 u4 , u2 u4 ̸∈ E (G). So we can get that |N (u1 , GC ) ∪ N (u4 , GC )| + |N (u2 , GC ) ∪ N (u4 , GC )| = 4. It follows from the neighborhood union condition that
−
|N (ux , G − GC ) ∪ N (uy , G − GC )| ≥ 6s + 8(t − 1) + 6.
(x,y)∈{(1,4),(2,4)}
Then either (III) or (IV) holds. If the former is true, then by Lemma 1 and (1), we have |Ck | = 4. Note that GC contains a cycle of length 3. This contradicts (1). If the latter is true, then by Lemma 2 and (1), we have |C j | = 6. Further, by Lemma 6, (1) and (3), we can get a contradiction. Suppose that p ≥ 5. So we have u1 u4 , u1 u5 ̸∈ E (G). If u3 u5 ̸∈ E (G), then by (E) of Lemma 7 we have
− (x,y)∈{(1,4),(3,5),(1,5)}
|N (ux , GC ) ∪ N (uy , GC )| ≤ 14.
S. Qiao / Discrete Mathematics 312 (2012) 891–897
897
Similarly, we can obtain that (I) or (II) holds. By Lemma 3 and (1), we can get a contradiction. If u3 u5 ∈ E (G), then u1 u4 , u2 u5 ̸∈ E (G). It follows from (C) of Lemma 7 that
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 9.
(x,y)∈{(1,4),(2,5)}
Then either (III) or (IV) holds. If the former is true, then by Lemma 1 and (1), we have |Ck | = 4. Note that GC contains a cycle of length 3. This contradicts (1). If the latter is true, then by Lemma 5 and (1), we can also get a contradiction. Suppose that d(u1 , P ) = 1. If u2 u4 ̸∈ E (G), then by (B) of Lemma 7, we have
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 14.
(x,y)∈{(1,4),(2,4),(1,3)}
This means that either (I) or (II) holds. By Lemma 3 and (1), we obtain a contradiction. If u2 u4 ∈ E (G) and p = 4, then
−
|N (ux , GC ) ∪ N (uy , GC )| < 9.
(x,y)∈{(1,3),(1,4)}
So, either (III) or (IV) holds. If the former is true, then by Lemma 1 and (1), we have |Ck | = 4. Note that GC contains a cycle of length 3. This contradicts (1). If the latter is true, then by Lemma 6, (1) and (3), we can also get a contradiction. So we may assume that u2 u4 ∈ E (G) and p ≥ 5. Note that u1 u4 , u1 u5 , u2 u5 ̸∈ E (G). Then by (D) of Lemma 7, we have
−
|N (ux , GC ) ∪ N (uy , GC )| ≤ 14.
(x,y)∈{(1,4),(2,5),(1,5)}
This implies that either (I) or (II) holds. By Lemma 3 and (1), we can get a contradiction. The proof of Theorem 7 is complete. 4. Remark In this paper, we give an example for which the bound is tight, when s = 0. In fact, the bound for s > 0 may be 3s + 4t, which is shown by considering a disjoint union of K3s+4t −1 and K2 . This could be proved using the same method, by induction on s. Acknowledgments The author is very grateful to the referees for their valuable suggestions and comments, which helped to improve the presentation of this paper. Supported by NSFC (No. 11001214) and the Fundamental Research Funds for the Central Universities (No. K50510700001). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
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