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A new sufficient condition for pancyclability of graphs
Discrete Applied Mathematics 162 (2014) 142–148
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Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
A new sufficient condition for pancyclability of graphs Junqing Cai a,∗ , Hao Li b,c a
School of Management, Qufu Normal University, Rizhao, 276826, PR China
b
L.R.I, UMR 8623, CNRS and Université Paris-Sud 11, F-91405 Orsay, France
c
Institute for Interdisciplinary Research, Jianghan University, Wuhan, PR China
article
info
Article history: Received 9 March 2011 Received in revised form 14 August 2013 Accepted 19 August 2013 Available online 12 September 2013 Keywords: Cyclable S-vertex S-length S-pancyclable
abstract Let G be a graph of order n and S be a set of s vertices. We call G, S-pancyclable, if for every i with 3 ≤ i ≤ s, there exists a cycle C in G such that |V (C ) ∩ S | = i. For any two nonadjacent vertices u, v of S, we say that u, v are of distance 2 in S, denoted by dS (u, v) = 2, if there is a path P in G connecting u and v such that |V (P ) ∩ S | ≤ 3. In this paper, we will prove that: Let G be a 2-connected graph of order n and S be a subset of V (G) with |S | ≥ 3. If max{d(u), d(v)} ≥ n/2 for all pairs of vertices u, v of S with dS (u, v) = 2, then G is S-pancyclable or else |S | = 4r and G[S ] is a spanning subgraph of F4r , or else |S | = n is even and G is the complete bipartite graph Kn/2,n/2 , or else |S | = n ≥ 6 is even and G is Kn′ /2,n/2 , or else G[S ] = K2,2 and the structure of G is well characterized. This generalizes a result of Benhocine and Wojda for the case when S = V (G). [A. Benhocine, A.P. Wojda, The Geng-Hua Fan conditions for pancyclic or Hamilton-connected graph, J. Combin. Theory B 42 (1987) 167–180]. © 2013 Elsevier B.V. All rights reserved.
1. Introduction In this paper, we consider only finite undirected and simple graphs. Given a graph G, we denote by V (G), E (G), respectively, the sets of vertices and edges of G. For A ⊆ V (G), G[A] is the subgraph of G induced by A. If H and S are subsets of V (G) or subgraphs of G, we denote by NH (S ) = {u ∈ H: there exists a vertex v ∈ S such that uv ∈ E (G)}, and set dH (S ) = |NH (S )|. In particular, if H = G and S = {u}, then set N (u) = NG (u) and d(u) = dG (u). Paths and cycles in a graph G are considered as subgraphs of G. Let S be a subset of V (G). Then v is called an S-vertex if v ∈ S. Following [4,7], the set S is called cyclable in G if all vertices of S belong to a common cycle in G. Following [5], the S-length of a cycle in G is defined as the number of the S-vertices that it contains, and the graph G is said to be S-pancyclable, if it contains cycles of all S-lengths from 3 to |S |. Following [6], for any two nonadjacent vertices u, v of S, we say that u, v are of distance 2 in S, denoted by dS (u, v) = 2, if there is a path P in G connecting u and v such that |V (P ) ∩ S | ≤ 3. Other notations not defined in this paper can be found in [3]. For a cycle (or a path) C in G with a given orientation and a vertex a in C , a+ and a− denote the successor and the predecessor of a in C , respectively. For a subset X of V (C ), X + and X − denote the set of the successors and the predecessors of the vertices of X in C , respectively. For two vertices a and b in C , we define C [a, b] (C [a, b), C (a, b), respectively) to be the subpath of C from a to b (from a to b− , from a+ to b− , respectively). We use C [b, a] for the path from b to a in the reversed direction of C . In order to give the results of this paper, we define some special graphs. Kn′ ,m is the complete bipartite graph Kn,m minus one edge. Given an integer r ≥ 2, F4r is the graph with 4r vertices containing a complete graph K2r , a set of r independent edges, denoted by Er and a matching between the sets of the vertices of K2r and Er .
∗
Corresponding author. Tel.: +86 15163380587. E-mail addresses: [email protected] (J. Cai), [email protected] (H. Li).
0166-218X/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.dam.2013.08.026
J. Cai, H. Li / Discrete Applied Mathematics 162 (2014) 142–148
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From the definitions, we see that cyclability and S-pancyclability are generalizations of Hamiltonicity and pancyclability of the whole graph (set S = V (G)), respectively. In recent years, people have given different definitions and results containing certain subsets of vertices, and some related papers can be found in [4,5,7,9,10]. Let us cite for example the following ones. Theorem 1 (Benhocine and Wojda [1]). Let G be a 2-connected graph of order n ≥ 3. If max{d(u); d(v)} ≥ n/2 for any pair of vertices u and v in G with dG (u, v) = 2, then G is pancyclic unless n = 4r , r > 2, and G is F4r , or n is even and G = Kn/2,n/2 or else n ≥ 6 is even and G = Kn′ /2,n/2 . Theorem 2 (Flandrin, Li and Wei [6]). Let G be a 2-connected graph of order n and S be a subset of V (G) with |S | ≥ 3. If max{d(u), d(v)} ≥ n/2 holds for all pairs u, v in S with dS (u, v) = 2, then S is cyclable in G. Theorem 3 (Flandrin, Li and Wei [6]). Let G be a 2-connected graph of order n and S be a subset of V (G) with |S | ≥ 3. If max{d(u), d(v)} ≥ (n + 1)/2 holds for all pairs u, v in S with dS (u, v) = 2, then G is S-pancyclable unless |S | = 4r and G[S ] is a spanning subgraph of F4r . Theorem 4 (Schmeichel and Hakimi [8]). Let G be a graph containing a Hamiltonian cycle C = x1 x2 x3 · · · xn x1 with n ≥ 3. If d(x1 ) + d(xn ) ≥ n, then G is either (i) pancyclic, or (ii) bipartite, or (iii) missing only (n − 1)-cycles. Moreover, if (iii) holds, then d(xn−2 ), d(xn−1 ), d(x2 ), d(x3 ) < n/2. In [2], Bondy suggested the meta-conjecture that almost any nontrivial condition on graphs which implies that the graph is Hamiltonian also implies that the graph is pancyclic (except maybe for a special family of graphs). Many results have been obtained in this problem. We follow, in this paper, the idea of some analogy between Hamiltonicity and pancyclicity, and prove the Conjecture mentioned in [6]. Its proof is related to the method in [6]. Theorem 5. Let G be a 2-connected graph of order n and S be a subset of V (G) with |S | ≥ 3. If max{d(u), d(v)} ≥ n/2 for all pairs of vertices u, v of S with dS (u, v) = 2, then G is S-pancyclable or else |S | = 4r and G[S ] is a spanning subgraph of F4r , or else |S | = n is even and G is the complete bipartite graph Kn/2,n/2 , or else |S | = n ≥ 6 is even and G is Kn′ /2,n/2 , or else G[S ] = K2,2 := x1 x2 x3 x4 x1 and the structure of G is as follows: V (G) is partitioned into S ∪ V1 ∪ V2 ∪ V3 ∪ V4 ; for any i, 1 ≤ i ≤ 4, G[Vi ] is any graph on |Vi | vertices, and each vertex xi is adjacent to all the vertices of Vi+1 and Vi , where the index i is taken as modulo 4. 2. The proof of Theorem 5 By Theorem 2, there exists a cycle containing all the vertices of S in G. Choose such a cycle C such that |C | is as small as possible and give C an arbitrary orientation. If |S | = 3, then Theorem 5 holds. Thus, we may assume that |S | ≥ 4. Set R = G − C and |S | = s. Let x1 , x2 , . . . , xs be the vertices of V (C ) ∩ S, the order 1, 2, . . . , s following the orientation of C , and consider the subscripts modulo s. Two S-vertices xi and xi+1 are said to be S-consecutive. And the segment C [xi , xi+1 ) is denoted by Si , 1 ≤ i ≤ s. We use Cl to denote a cycle of S-length l in G. In [5], it was proved: Theorem 6. Let G be a graph, S be a subset of V (G) such that S is cyclable in G, and let C be a shortest cycle through all the vertices of S. If dC (x) + dC (y) ≥ |C | + 1 for some pair of S-consecutive vertices x and y in C , then G is S-pancyclable. Theorem 7. Let G be a graph, S be a subset of V (G) such that S is cyclable in G, and let C be a shortest cycle through all the vertices of S. If dC (x) + dC (y) ≥ |C | for some pair of S-consecutive vertices x and y in C , then G is S-pancyclable or S = C and G[S ] is bipartite or there is just no Cs−1 . By using a similar method as that used in the proof of Theorems 6 and 7 in [5], we can get the following lemma. Lemma 1. Let G be a graph, S be a subset of V (G) such that S is cyclable in G and let C be a shortest cycle through all the vertices |C |+1 of S. If there exists some 1 ≤ i ≤ s such that xi−1 xi+1 ∈ E (G) and dC (xi ) ≥ 2 , then G is S-pancyclable. Proof. Suppose G is not S-pancyclable, and G misses Cl for some l, 4 ≤ l ≤ s − 2. Without loss of generality, suppose i = 1. Let fl (k) =
s + k + 1 − l, k − l + 2,
if 2 ≤ k ≤ l − 1; if l ≤ k ≤ s − 2.
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Then N (x1 ) ∩ Si ̸= ∅, implies N (x1 ) ∩ Sf+(i) = ∅. (Otherwise, we can get a Cl in G.) By the choice of C , dS1 (x1 ) + dS + (x1 ) + l
1
dSs (x1 ) + dS + (x1 ) ≤ |S1 | + |Ss |. For any i, 2 ≤ i ≤ s − 1, s
dSi (x1 ) + dS + (x1 ) ≤
i
if dSi (x1 ) = 0 or dS + (x1 ) = 0;
|Si |, |Si | + 1,
otherwise.
i
Let A1 = {i : N (x1 )∩ Si ̸= ∅, N (x1 )∩ Si+ = ∅ and 2 ≤ i ≤ s − 1}, A2 = {i : N (x1 )∩ Si = ∅, N (x1 )∩ Si+ ̸= ∅ and 2 ≤ i ≤ s − 1} and A3 = {i : N (x1 ) ∩ Si ̸= ∅, N (x1 ) ∩ Si+ ̸= ∅ and 2 ≤ i ≤ s − 1}. Then i ∈ A1 ∪ A3 , implies fl (i) ̸∈ A2 ∪ A3 . Thus, |A1 ∪ A3 | ≤ s − 2 − |A2 ∪ A3 |, i.e. |A3 | ≤ s − 2 − |A1 ∪ A2 ∪ A3 |. Therefore, s−1 |Si | + (|Si+ | + 1) (dSi (x1 ) + dS + (x1 )) ≤ i=2
i
i∈A1 ∪A2
=
i∈A3
|Si | + |A3 |
i∈A1 ∪A2 ∪A3
≤
|Si | + (s − 2 − |A1 ∪ A2 ∪ A3 |)
i∈A1 ∪A2 ∪A3
≤
s−1
|Si |.
i=2
Then |C | + 1 = 2dC (x1 ) =
s−1 i =1
(dSi (x1 ) + dS + (x1 )) ≤ |C |, a contradiction. Then G is S-pancyclable. i
Now, let T = {v ∈ S : d(v) ≥ n/2}. Notice that for any 1 ≤ i ≤ s, xi xi+2 ∈ E (G) when {xi , xi+2 } ⊆ S − T and xi xj ∈ E (G) for any j ̸= i when N (xi ) ∩ N (xj ) ̸= ∅ and {xi , xj } ⊆ S − T . It is easy to see the following: Remark 1. If there is no pair of S-consecutive vertices x, y in C [xi , xj ](i ̸= j) such that {x, y} ⊆ T , then G[V (C [xi , xj ])∩(S − T )] is a clique of G. Lemma 2. If there exists at most one pair of S-consecutive vertices which are both in T , then Theorem 5 holds. Proof. If |S | = 4, since G is not S-pancyclable, G[S ] must be isomorphic to an induced cycle x1 x2 x3 x4 x1 . Thus, Lemma 2 is true. Next we assume that |S | ≥ 5. If there is no pair of S-consecutive vertices which are both in T , then G[V (C ) ∩ (S − T )] is a clique by Remark 1. Thus, G is S-pancyclable. If there is exactly one pair of S-consecutive vertices, say {xs , x1 } ⊆ T , then G[V (C [x2 , xs−1 ]) ∩ (S − T )] is a clique, especially, x2 xs−1 ∈ E (G). Thus, we can easily check that G is S-pancyclable. In fact, suppose to the contrary that there is no Cl for some 3 ≤ l ≤ s − 1. Then x2 xl+1 ̸∈ E (G) and xs−1 xl ̸∈ E (G). Thus, d(xl + 1) ≥ n/2 and d(xl ) ≥ n/2 by the conditions of Theorem 5, a contradiction. So Lemma 2 is true. Next, we will show four structural lemmas for some special paths containing vertices of S. These lemmas will play very important roles in the proof of Theorem 5. Lemma 3. Let P = u1 u2 . . . up be a path connecting u1 and up in G. If dP (u1 )+ dP (up ) ≥ |V (P )|− 1 and {u1 u3 , xp xp−2 }∩ E (G) = ∅, then there exists a cycle C in G such that V (C ) = V (P ); or p is odd and NP (u1 ) = NP (up ) = {u2 , u4 , u6 , . . . , up−1 }. Proof. Suppose that there is no cycle C in G such that V (C ) = V (P ). Then u1 up ̸∈ E (G). We can easily check that if u1 ui+1 ∈ E (G), then up ui ̸∈ E (G) for any i with 2 ≤ i ≤ p − 2. We can claim that u1 ui ∈ E (G) and u1 ui+1 ∈ E (G) cannot hold at the same time for any i with 4 ≤ i ≤ p − 2. Otherwise, suppose that there exists some i such that u1 ui ∈ E (G) and u1 ui+1 ∈ E (G). Then u2 up ̸∈ E (G). (Since, otherwise, C = P [u2 , ui ]u1 P [ui+1 , up ]u2 is a cycle in G with V (C ) = V (P ), a contradiction.) Therefore, dP (u1 ) + dP (up ) ≤ dP (u1 ) + (|V (P )| − 1 − dP (u1 )) − 1 = |V (P )| − 2, a contradiction. By symmetry, up uj ∈ E (G) and up uj+1 ∈ E (G) cannot hold at the same time for any j with 2 ≤ j ≤ p − 4. Then dP (u1 ) ≤ If p is even, then dP (u1 ) ≤
|V (P )|
|V (P )|
|V (P )|−1 2
and dP (up ) ≤
|V (P )|−1 2
.
− 1 and dP (up ) ≤ 2 − 1. So dP (u1 ) + dP (up ) ≤ |V (P )| − 2, a contradiction. So p is 2 |V (P )|−1 odd. Since dP (u1 ) + dP (up ) ≥ |V (P )| − 1, dP (u1 ) = dP (up ) = . Moreover, NP (u1 ) = NP (up ) = {u2 , u4 , u6 , . . . , up−1 }. 2
Hence, Lemma 3 holds.
Lemma 4. Suppose P = u1 . . . u2 . . . up−1 . . . up is a path in G[V (C )] such that |V (P ) ∩ S | = l + 1 ≥ 4, {u1 , u2 , up−1 , up } ⊆ T , {u1 , u2 }, {up−1 , up } are two pairs of S-consecutive vertices on C and (V (P (u1 , u2 )) ∪ V (P (up−1 , up ))) ∩ S = ∅. Then there exists a Cl in G or dC (u1 ) + dC (u2 ) = dC (up−1 ) + dC (up ) = |C |. Proof. Suppose there exists no Cl in G. Recall that R = G − C . If NR (u1 ) ∩ NR (up−1 ) ̸= ∅ or NR (u2 ) ∩ NR (up ) ̸= ∅, then there exists a Cl in G, a contradiction. So NR (u1 ) ∩ NR (up−1 ) = ∅ and NR (u2 ) ∩ NR (up ) = ∅. Noting that {u1 , u2 , up−1 , up } ⊆ T , we have dC (u1 ) + dC (u2 ) + dC (up−1 ) + dC (up ) ≥ 2|C |. Thus either dC (u1 ) + dC (u2 ) ≥ |C | or dC (up−1 ) + dC (up ) ≥ |C |. If dC (u1 ) + dC (u2 ) ≥ |C | + 1 or dC (up−1 ) + dC (up ) ≥ |C | + 1, then G is S-pancyclable by Theorem 6, a contradiction. Hence, dC (u1 ) + dC (u2 ) = dC (up−1 ) + dC (up ) = |C |. Now the proof of Lemma 4 is complete.
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Lemma 5. Let P = u1 u2 . . . up be a path in G such that |V (P ) ∩ S | = l ≥ 3. If {u1 , up } ⊆ T and there is no Cl in G, then (i) |(N (u1 ) ∩ N (up ) − V (P )) ∩ (V (G) − S )| = 0; (ii) |N (u1 ) ∩ N (up ) ∩ S ∩ (V (G) − V (P ))| ≥ 1; and there exists a Cl+1 which contains P as its subpath; (iii) when P = C [xi , xj ] for some j = l + i − 1(3 ≤ l ≤ s − 2) with {xi , xj } ⊆ T , then there exists a pair of S-consecutive vertices y and z in V (C (xj , xi )) such that y ∈ N (xi ) ∩ N (xj ) and z ∈ N (xj ) (or z ∈ N (xi )), and there exists a Cl+1 and a Cl+2 which contain C [xi , xj ] as a subpath. Moreover V (Cl+1 ) ∩ S ⊆ V (Cl+2 ). Proof. Since there is no Cl in G, (i) is obvious and |N (u1 ) ∩ V (P )| + |N (up ) ∩ V (P )| ≤ |V (P )| − 1 by Lemma 3. As d(u1 ) + d(up ) ≥ n, by (i), it is easy to check that (ii) holds. (iii) As d(xi ) + d(xj ) ≥ n and S ∩ R = ∅, Lemmas 3 and 5(i) imply |N (xi ) ∩ V (C (xj , xi )) ∩ S | + |N (xj ) ∩ V (C (xj , xi )) ∩ S | ≥ |V (C (xj , xi )) ∩ S | + 1. Thus (iii) holds. Lemma 6. Let P = u1 u2 . . . up be a path in G[V (C )] such that V (P ) ∩ S = {v1 , v2 , . . . , vl }, where v1 = u1 , vl = up and the order 1, 2, . . . , l follows the orientation of P from u1 to up . Suppose that l ≥ 5 and there is no Cl in G. If there exists a Cl+m and a Cl+m+1 in G(m ∈ {1, 2}), both of which contain P as a subpath and V (Cl+m ) ∩ S ⊆ V (Cl+m+1 ), then for any 1 ≤ i ≤ l − m − 2, (i) vi vi+m+1 ̸∈ E (G) and vi vi+m+2 ̸∈ E (G); (ii) {vi , vi+m+2 } ∩ (S − T ) ̸= ∅. Proof. Let C ′ = Cl+m+1 and C ∗ = Cl+m . Since P is a subpath of both C ′ and C ∗ , C ′ [v1 , vl ] = C ∗ [v1 , vl ] = P [v1 , vl ]. (i) If vi vi+m+1 ∈ E (G) or vi vi+m+2 ∈ E (G) for some 1 ≤ i ≤ l − m − 2, then replace C ∗ [vi , vi+m+1 ] or C ′ [vi , vi+m+2 ] with the edge vi vi+m+1 or vi vi+m+2 . We can get a Cl in G, a contradiction. (ii) Since there is no Cl in G, NR (vi ) ∩ NR (vi+m+2 ) ∩ (V (G) − V (C ′ )) = ∅, (N (vi ) ∩ V (C (vi+2 , vi+m+2 ))) ∪ (N (vi+m+2 ) ∩ V (C (vi , vi+1 ))) = ∅ and vi+2 ̸∈ N (vi ) ∩ N (vi+m+2 ). These imply that |N (vi ) ∩ V (C (vi , vi+m+2 ))| + |N (vi+m+2 ) ∩ V (C (vi , vi+m+2 ))| ≤ |V (C (vi , vi+m+2 ))|. Since P ′ = C ′ [vi+m+2 , vi ] is a path with |VP ′ ∩ S | = l, we have vi vi+m+1 ̸∈ E (G) and dC ′ (vi ) + dC ′ (vi+m+2 ) < |C ′ | be Lemma 3. If {vi , vi+m+2 } ⊆ T , then there exists at least one vertex, say x, in N (vi ) ∩ N (vi+m+2 ) ∩ (V (G) − V (C ′ )). If x ̸∈ S, then there is a Cl which contains V (C ′ [vi+m+2 , vi ]) ∪ {x}, a contradiction. If x ∈ S, since V (Cl+m ) ∩ S ⊆ V (Cl+m+1 ), x ̸∈ V (C ∗ ). Then we can get a Cl containing V (C ∗ [vi+m+2 , vi ]) ∪ {x}, a contradiction. Hence, {vi , vi+m+2 } ∩ (S − T ) ̸= ∅, and (ii) holds. From Lemma 2, we may assume that |T | ≥ 3 and there exist at least two pairs of S-consecutive vertices which are all in T . Without loss of generality, let {xs , x1 } ⊆ T such that |NR (x1 ) ∩ NR (xs )| = min{|NR (x) ∩ NR (y)| : {x, y} ⊆ T and x, y are S-consecutive}. If dC (x1 ) + dC (xs ) ≥ |C |, then by Theorem 7, G is S-pancyclable or S = C and G[S ] is bipartite or there is just no Cs−1 . Thus in the rest of the proof, we assume that dC (x1 ) + dC (xs ) ≤ |C | − 1 and let M = NR (x1 ) ∩ NR (xs ). Lemma 7. If there exists some 2 ≤ i ≤ s − 2 such that {xi , xi+1 } ⊆ T and dC (xi ) + dC (xi+1 ) ≤ |C | − 1, then (i) |(NR (x1 ) ∪ NR (xs )) ∩ NR (xi ) ∩ NR (xi+1 )| ≥ 1; (ii) there exists a cycle C3 and a cycle C4 in G. Proof. (i) By the choice of x1 and xs , we have |M | ≤ |NR (xi ) ∩ NR (xi+1 )|. Thus,
|R| + 1 ≤ ≤ = ≤
|NR (x1 ) ∪ NR (xs )| + |M | |NR (x1 ) ∪ NR (xs )| + |NR (xi ) ∩ NR (xi+1 )| |(NR (x1 ) ∪ NR (xs )) ∪ (NR (xi ) ∩ NR (xi+1 ))| + |(NR (x1 ) ∪ NR (xs )) ∩ (NR (xi ) ∩ NR (xi+1 ))| |R| + |(NR (x1 ) ∪ NR (xs )) ∩ (NR (xi ) ∩ NR (xi+1 ))|.
From the above inequalities, we can easily check that (i) holds. (ii) Since (NR (x1 ) ∪ NR (xs )) ∩ NR (xi ) ∩ NR (xi+1 ) ̸= ∅, without loss of generality, we may choose a vertex, say v , in (NR (x1 ) ∪ NR (xs )) ∩ NR (xi+1 ). Notice that {xs , x1 , xi , xi+1 } ⊆ T . Assume that there is no C3 in G. Applying Lemma 5(ii) to the path P = C [xi , xi+1 ]v xs , we can getN (xi ) ∩ N (xs ) ∩ (S − V (P )) ̸= ∅. This implies that there is a C3 as v ̸∈ S, a contradiction. Thus, there is a C3 in G. Similarly, applying Lemma 5(ii) to the path P ′ = C [xi , xi+1 ]v C [xs , x1 ], we can get a C4 in G. Hence, (ii) holds. Lemma 8. If there is no Cl in G for some integer l ≥ 3, then S = C and l = s − 1 or G[S ] is bipartite. Proof. Suppose 3 ≤ l ≤ s − 2 and G[S ] is not bipartite. By Theorem 7, dC (x) + dC (y) ≤ |C | − 1 for any pair of S-consecutive vertices x and y in C . Thus, by the assumption, M ̸= ∅ as dC (x1 ) + dC (xs ) ≤ |C | − 1 and |{xl−1 , xl } ∩ T | ≤ 1 by applying Lemma 4 to C [xs , xl ]. Case 1. xl ∈ T . Then xl−1 ̸∈ T . Case 1.1. xl+1 ̸∈ T .
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Then xl−1 xl+1 ∈ E (G) and there exists a C3 in G. So l ≥ 4. By Lemma 1, dC (xl ) ≤ 2 implies dR (xl ) ≥ 2 . Since NR (xl ) ∩ NR (x1 ) = ∅ and dR (x1 ) + dR (xs ) ≥ |R| + 1, 2|R| + |NR (xs ) ∩ NR (xl )| ≥ |NR (x1 ) ∪ NR (xl )| + |NR (xs ) ∪ NR (xl )| + |NR (xs ) ∩ NR (xl )| ≥ dR (x1 ) + dR (xs ) + 2dR (xl ) ≥ 2|R| + 1. This implies |NR (xs ) ∩ NR (xl )| ≥ 1 and there exists a Cl+1 = v C [xs , xl ]v and a Cl+2 = v C [xs , xl−1 ]xl+1 C (xl+1 , xl ]v for some v ∈ NR (xs ) ∩ NR (xl ). Both Cl+1 and Cl+2 contain C [xs , xl−1 ] as a subpath and V (Cl+1 ) ∩ S ⊆ V (Cl+2 ). When l = 4, then x4 ∈ T . Since there is no C4 in G, we have x2 x5 ̸∈ E (G), x2 x4 ̸∈ E (G). Since x3 x5 ∈ E (G) and x5 ̸∈ T , we have dS (x2 , x5 ) = 2, NR (x2 )∩ NR (x4 ) = ∅ and x2 ∈ T . If (N (x2 )∩ N (x4 )− V (C [x2 , x4 ]))∩(V (G)− S ) ̸= ∅, we can get a C4 which contains V (C [xs , x2 ]) ∪ {u, v, x4 } when u ∈ V (C ) − V (C [xs , x2 ]) or V (C [x4 , x5 ]) ∪ V (C [x2 , x3 ]) ∪ {u} when u ∈ V (C [xs , x2 ]), a contradiction. Suppose (N (x2 ) ∩ N (x4 ) − V (C [x2 , x4 ])) ∩ (V (G) − S ) = ∅. Since there is no C4 in G, we can obtain that N (x2 ) ∩ V (C (x3 , x4 ]) = ∅ and N (x4 ) ∩ V (C [x2 , x3 )) = ∅. Also by the minimality of |C |, we have |N (x2 ) ∩ V (C [x2 , x3 ])| = 1 and |N (x4 ) ∩ V (C [x3 , x4 ])| = 1. So dC [x2 ,x4 ] (x2 ) + dC [x2 ,x4 ] (x4 ) ≤ |C [x2 , x4 ]| − 1. Since {x2 , x4 } ⊆ T (that is d(x2 ) + d(x4 ) ≥ n), there exists a vertex, say w , in N (x2 ) ∩ N (x4 ) ∩ S ∩ (V (G) − V (C [x2 , x4 ])). Then we can get a C4 containing V (C [x2 , x4 ]) ∪ {w}, a contradiction. When l ≥ 5, since {xs , x1 } ⊆ T , by applying Lemma 6 with m = 1, we have x2 x4 ̸∈ E (G) and {x3 , x4 } ⊆ S − T . Then x2 ∈ T . If l ≥ 6, then x5 ∈ S − T by Lemma 6(ii), which implies x3 x5 ∈ E (G), contrary to Lemma 6(i). So l = 5, that is x5 ∈ T . Since d(x2 ) + d(x5 ) ≥ n and x2 x5 ̸∈ E (G) as x4 x6 ∈ E (G), |N (x2 ) ∩ N (x5 )| ≥ 2. If (N (x2 ) ∩ N (x5 )) − V (C [x2 , x6 ]) ̸= ∅, then, let u ∈ (N (x2 ) ∩ N (x5 )) − V (C [x2 , x6 ]). Since x4 x6 ∈ E (G), we can get a C5 containing V (C [x2 , x6 ] − C (x4 , x5 )) ∪ {u} whenever u ̸∈ S or V (C [x2 , x5 ]) ∪ {u} whenever u ∈ S, a contradiction. So (N (x2 ) ∩ N (x5 )) − V (C [x2 , x6 ]) = ∅. Since there is no C5 in G, N (x2 ) ∩ V (C (x3 , x5 ]) = ∅ and x3 x5 ̸∈ E (G). Also by the minimality of |C |, we have |N (x2 ) ∩ V (C [x2 , x3 ])| = 1 and |N (x5 ) ∩ V (C [x5 , x6 ])| = 1. Similarly as l = 4, we can get a C5 , a contradiction. |C |
|R|+1
Case 1.2. xl+1 ∈ T . Since there is no Cl in G, NR (x1 ) ∩ NR (xl ) = ∅ and dC (xl ) + dC (xl+1 ) ≤ |C | − 1. Thus, there is a vertex, say x, in NR (xs )∩NR (xl )∩NR (xl+1 ) and l ≥ 5 by Lemma 7. Hence, there exists a Cl+1 and a Cl+2 , which contain C [xs , xl ]∪{x} as a subpath. Since l ≥ 5 and {xs , x1 } ⊆ T , by Lemma 6 with m = 1, we obtain {x3 , x4 } ⊆ S − T . By applying Lemma 4 to C [x1 , xl+1 ] and the fact that dC (x) + dC (y) ≤ |C | − 1 for any pair of S-consecutive vertices x and y, we have x2 ̸∈ T . Then x2 x4 ∈ E (G). We can get a Cl = xV (C [xs , x2 ])x4 V (C (x4 , xl ])x, a contradiction. Case 2. xl−1 ∈ T . Then xl ̸∈ T . Case 2.1. xl−2 ̸∈ T . Then xl−2 xl ∈ E (G) and (NR (x1 ) ∪ NR (xs )) ∩ NR (xl−1 ) = ∅ since there is no Cl . Since 2|NR (x1 ) ∪ NR (xs )| ≥ dR (x1 ) + dR (xs ) ≥ 1 . Then G is S-pancyclable by Lemma 1, a contradiction. |R| + 1, dC (xl−1 ) ≥ |C |+ 2 Case 2.2. xl−2 ∈ T . If l ̸= 3, noting that NR (xs ) ∩ NR (xl−1 ) = ∅, |NR (x1 ) ∩ NR (xl−1 ) ∩ NR (xl−2 )| ≥ 1 and l ≥ 5 by Lemma 7. Since {xs , xl−1 } ⊆ T , there exists a Cl+1 and a Cl+2 containing C [xs , xl−1 ] as a subpath and V (Cl+1 ) ∩ S ⊆ V (Cl+2 ) by Lemma 5(iii). Thus, by Lemma 6(ii) with m = 1, we can get {x3 , x4 } ⊆ S − T . Hence, l ≥ 7 and {x2 , x5 } ∩ (S − T ) ̸= ∅. This implies x2 x4 ∈ E (G) or x3 x5 ∈ E (G), contrary to Lemma 6(i). So l = 3. Then {xs , x1 , x2 } ⊆ T , xs x2 ̸∈ E (G) and NR (xs ) ∩ NR (x2 ) = ∅. (Otherwise, there exists a C3 .) By the minimality of |C |, |N (xs ) ∩ V (C [xs , x1 ])| = 1 and |N (x2 ) ∩ V (C [x1 , x2 ])| = 1. And by Lemma 3, |N (xs ) ∩ V (C [xs , x2 ])| + |N (x2 ) ∩ V (C [xs , x2 ])| ≤ |V (C [xs , x2 ])| − 1. Thus, |N (xs ) ∩ V (C (x2 , xs ))| + |N (x2 ) ∩ V (C (x2 , xs ))| ≥ |V (C (x2 , xs ))| + 1. If there is some i with 2 ≤ i ≤ s − 1 such that either {xi , xi+1 } ⊆ N (xs ) or {xi , xi+1 } ⊆ N (x2 ), then we can get a C3 containing V (C [xi , xi+1 ]) ∪ {xs } or V (C [xi , xi+1 ]) ∪ {x2 }, a contradiction. Hence, there exists some i with 2 ≤ i ≤ s − 1 such that N (xs ) ∩ N (x2 ) ∩ V (C (xi , xi+1 )) ̸= ∅. Then we can find a C3 containing {xs , x1 , x2 }, a contradiction. Case 3. xl ̸∈ T and xl−1 ̸∈ T , that is {xl , xl−1 } ∩ T = ∅. Case 3.1. There is no pair of S-consecutive vertices x and y in V (C [xl+1 , xs−1 ]) such that {x, y} ⊆ T . Then G[V (C [xl−1 , xs−1 ]) ∩ (S − T )] is a clique by Remark 1. Since l ≤ s − 2, |V (C [xl−1 , xs−1 ]) ∩ S | ≥ 3. Case 3.1.1. xs−1 ̸∈ T . Then xl−1 xs−1 ∈ E (G) and xl xs−1 ∈ E (G). Thus, there exists a C3 , a Cl+1 and a Cl+2 . And both Cl+1 and Cl+2 contain C [xs−1 , xl−1 ] as a subpath and V (Cl+1 ) ∩ S ⊆ V (Cl+2 ). Then l ≥ 4 and by Lemma 6(i) {xl−2 , xl−3 } ⊆ T . By applying Lemma 6 to C [xs , xl−1 ] with m = 1, we have {x3 , x4 } ⊆ S − T . This implies that x2 ∈ T and l − 1 ≥ 7 or l − 1 = 3 as {xl−2 , xl−3 } ⊆ T . If l ≥ 8, by Lemma 6(ii) again, x5 ∈ S − T as x2 ∈ T . Then x3 x5 ∈ E (G), contrary to Lemma 6(i). Hence, l = 4. Since there is no C4 and {x3 , x4 } ⊆ N (xs−1 ), we have N (xs ) ∩ V (C (x1 , x3 ]) = ∅, N (x2 ) ∩ V (C [xs−1 , x1 )) = ∅ and (N (x2 ) ∩ N (xs ) − V (C [xs , x2 ])) ∩ (V (G) − S ) = ∅. Since |N (xs ) ∩ V (C [xs , x2 ])| + |N (x2 ) ∩ V (C [xs , x2 ])| ≤ |V (C [xs , x2 ])| − 1 and d(xs ) + d(x1 ) ≥ n, there exists a vertex, say w , in N (x2 )∩ N (xs )∩ S ∩(V (G)− V (C [xs , x2 ])). Then we can get a C4 containing V (C [xs , x2 ])∪{w}, a contradiction. Case 3.1.2. xs−1 ∈ T . Then xs−2 ∈ S − T . Since xl−1 ∈ S − T , xl−1 xs−2 ∈ E (G). Thus, there exists a Cl+2 in G which contains C [xs−2 , xl−1 ] as a subpath. If l ≤ s − 3, then xl xs−2 ∈ E (G) and there exists a C3 . If xl−2 ̸∈ T , then xl−2 xs−2 ∈ E (G) and there exists a C4 and a Cl+1 containing C [xs−2 , xl−2 ] as a subpath. If xl−2 ∈ T , since xs−1 ∈ T , by Lemma 5(iii) there exists a pair of S-consecutive vertices a and b in V (C (xl−2 , xs−1 )) such that a ∈ N (xl−2 ) ∩ N (xs−1 ) and b ∈ N (xl−2 ) (or b ∈ N (xs−1 )).
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Fig. 1.
Then there exists a Cl+1 and a Cl+2 containing C [xs−1 , xl−2 ] ∪ {y} as a subpath. Since {xs−1 , x1 } ⊆ T and there exists no Cl , |N (xs−1 ) ∩ V (C [xs−1 , x1 ])| + |N (x1 ) ∩ V (C [xs−1 , x1 ])| ≤ |V (C [xs−1 , x1 ])| − 1. We can get a C4 . Thus, in both cases, l ≥ 5 and V (Cl+1 ) ∩ S ⊆ V (Cl+2 ). By using Lemma 6 with m = 1 and the facts that xl−1 ∈ S − T and {xs−1 , xs , x1 } ⊆ T , we can get {x2 , x3 , x4 } ⊆ S − T and x2 x4 ∈ E (G). If l ≥ 6, then replacing Cl+1 [x2 , x4 ] with the edge x2 x4 , we can get a Cl in G, a contradiction. If l = 5, then we can get a C5 = x2 C [x4 , x5 ]xs−2 C [x3 , x2 ] (see Fig. 1), a contradiction. If l = s − 2 and xl−2 ̸∈ T , then xl−2 xl ∈ E (G). This implies there exists a C3 and a Cl+1 containing C [xs−1 , xl−2 ] as a subpath. If l ≥ 5, by Lemma 6(ii) and {xs−1 , xs , x1 } ⊆ T , we can get {x2 , x3 , x4 } ∩ T = ∅ and x2 x4 ∈ E (G). Then we can get a Cl , a contradiction. So l = 4 and s = 6. Since there exists no C4 and x2 x4 ∈ E (G), dC (x1 ) = 2 and dC (x5 ) = 2 by the minimality of |C |. Since {x1 , x5 } ⊆ T , NR (x1 ) ∩ NR (x5 ) ̸= ∅. Then we can get a C4 containing V (C [x1 , x2 ]) ∪ V (C [x4 , x5 ]) ∪ {w} where w ∈ NR (x1 ) ∩ NR (x5 ), a contradiction. Hence, l = s − 2 and xl−2 ∈ T . By applying Lemma 5(ii) to C [xs−1 , xl−2 ], we have {xl−1 , xl } ∩ N (xs−1 ) ∩ N (xl−2 ) ̸= ∅. If {xl−1 , xl } ⊆ N (xs−1 ) ∩ N (xl−2 ), then there exists a C3 , a C4 and a Cl+1 in G which contains C [xs−1 , xl−2 ] as a subpath. If {xl−1 , xl } ̸⊆ N (xs−1 ) ∩ N (xl−2 ), without loss of generality, we assume xl−1 ̸∈ N (xs−1 ) ∩ N (xl−2 ) and xl ∈ N (xs−1 ) ∩ N (xl−2 ). This implies that there exists a C3 and a Cl+1 . In both cases, we have l ≥ 4. If l ≥ 5, we can derive a contradiction as before by Lemma 6. If l = 4, similarly as l = s − 2 and xl−2 ̸∈ T , we can derive a contradiction. Case 3.2. There exists a pair of S-consecutive vertices x and y in V (C [xl+1 , xs−1 ]) such that {x, y} ⊆ T . Choose l + 1 ≤ t < s − 1 such that {xt , xt +1 } ⊆ T and t is as small as possible. Then by Remark 1, we have G[V (C [xl−1 , xt )) ∩ (S − T )] is a clique of G. This implies xt −1 xl−1 ∈ E (G). Let P = C [x1 , xl−1 ]xt −1 . By Lemma 7, |(NR (x1 ) ∪ NR (xs )) ∩ NR (xt ) ∩ NR (xt +1 )| ≥ 1 and l ≥ 5. We distinguish the following subcases: Case 3.2.1. |NR (x1 ) ∩ NR (xt ) ∩ NR (xt +1 )| ≥ 1. Then we can get a Cl+1 and a Cl+2 in G, both of which contain P as a subpath. Since l ≥ 5, {x3 , x4 } ⊆ S − T by Lemma 6. This implies x2 ∈ T . Using Lemma 4 for the path P ′ = C [x2 , xl−1 ]C [xt −1 , xt +1 ], we have xj ∈ S − T , where j = 5 when l ≥ 6 or j = t − 1 when l = 5. This implies x3 xj ∈ E (G), contrary to Lemma 6(i). Case 3.2.2. |NR (x1 ) ∩ NR (xt ) ∩ NR (xt +1 )| = 0. By Lemma 7(i), there is a vertex, say w , in NR (xs ) ∩ NR (xt ) ∩ NR (xt +1 ). Thus, there exists a Cl+2 = w C [xs , xl−1 ]C [xt −1 , xt ]w and a Cl+3 = w C [xs , xl−1 ]C [xt −1 , xt +1 ]w which contain P = C [x1 , xl−1 ]xt −1 as a subpath. Since l ≥ 5 and {xs , x1 } ⊆ T , by applying Lemma 6(ii) with m = 2, we have {x4 , xj } ⊆ S − T , where j = 5 when l ≥ 6 and j = t − 1 when l = 5. If x2 ̸∈ T , then x2 x5 ∈ E (G) or x2 xt −1 ∈ E (G), contrary to Lemma 6(i). Thus x2 ∈ T . Since xs ∈ T , x3 ̸∈ T by Lemma 4 and x3 xj ∈ E (G). Since {x2 , xt } ⊆ T , applying Lemma 5(ii) to the path P ∗ = C [x2 , xl−1 ]C [xt −1 xt ], we can get a Cl+1 containing P ∗ as a subpath. Since x3 xj ∈ E (G), we can get a Cl in G, a contradiction. Now, we complete the proof of Lemma 8. Now, we turn to prove Theorem 5. By Theorem 7 and Lemma 8, if there exists a Cs−1 , then G is S-pancyclable. Thus, in the rest of the proof, we assume that there is no Cs−1 in G. This implies for any 1 ≤ i ≤ s, xi−1 xi+1 ̸∈ E (G) and NR (xi−1 )∩ NR (xi+1 ) = ∅. Consequently, |{xi−1 , xi+1 }∩ T | ≥ 1 as |V (C [xi−1 , xi+1 ])∩ S | = 3. If there exists some i with 1 ≤ i ≤ s such that {xi−1 , xi+1 } ⊆ T , then dC (xi−1 ) + dC (xi+1 ) ≥ |C |. Since N (xi−1 ) ∩ V (C (xi , xi+1 ]) = ∅ and N (xi+1 ) ∩ V (C [xi−1 , xi )) = ∅, dP (xi−1 ) + dP (xi+1 ) ≥ |V (P )| − 1 for P = C [xi+1 , xi−1 ]. Since xi+1 xi+3 ̸∈ E (G) and xi−1 xi−3 ̸∈ E (G), by Lemma 3 and the choice of C , we can get a Cs−1 in G or else |S | = |V (C )| = s is even and NC (xi−1 ) = NC (xi+1 ) = {x1 , x3 , . . . , xs−1 } when i is odd or NC (xi−1 ) = NC (xi+1 ) = {x2 , x4 , . . . , xs } when i is even. Without loss of generality, we assume i is odd. Thus, by the assumption, s is even and NC (xi−1 ) = NC (xi+1 ) = {x1 , x3 , . . . , xs−1 }. If xi ̸∈ T , since there is no Cs−1 , {xi−2 , xi+2 } ⊆ T . By applying Lemma 4 to C [xi+1 , xi−1 ], we can get dC (xi+1 ) + dC (xi+2 ) = dC (xi−1 ) + dC (xi−2 ) = |C |. Since xi−2 xi+1 ∈ E (G) and xi−1 xi+2 ∈ E (G), we can get that G is S-pancyclable or G[S ] is bipartite by Theorem 4. Since G is not S-pancyclable, we conclude
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that G[S ] is a bipartite graph and dC (xi+1 ) = dC (xi+2 ) = dC (xi−1 ) = dC (xi−2 ) = |C |/2. Since xi ̸∈ T , NC (xi+1 ) \ {xi } ⊆ T and dC (x) = |C |/2 for any vertex x ∈ NC (xi+1 ) \ {xi }. Moreover, G[S ] is the bipartite graph K|S |/2,|S |/2 or else |S | ≥ 6 and G[S ] is isomorphic to K|′S |/2,|S |/2 . Next, we assume xi ∈ T . If xi−2 ∈ T or xi+2 ∈ T , without loss of generality, we assume xi+2 ∈ T . Since there exists no Cs−1 , we can get that dC (xi ) + dC (xi+2 ) = |C |. Recalling that dC (xi−1 ) + dC (xi+1 ) = |C |, we can get dC (xi−1 ) + dC (xi ) ≥ |C | or dC (xi+1 ) + dC (xi+2 ) ≥ |C |. By Theorem 7 and the assumption, similarly to the above, we can get that G[S ] is a bipartite graph K|S |/2,|S |/2 or K|′S |/2,|S |/2 . If {xi−2 , xi+2 }∩ T = ∅, then xi−2 xi+2 ∈ E (G). Thus, we can get Cs−1 = C [xi−1 , xi+1 ]xi−2 C [xi+2 , xi−4 ]xi−1 , a contradiction. Hence, by the above discussion, if S = V (G), then G is isomorphic to Kn/2,n/2 or else n ≥ 6 and G is isomorphic to Kn′ /2,n/2 . If S ̸= V (G), then any two nonadjacent S-vertices belonging to the same class of K|S |/2,|S |/2 or K|′S |/2,|S |/2 and any two nonadjacent S-vertices both in T have degree sum in R equal to at least |R|. Moreover, if |S | ≥ 6, there must be two nonadjacent S-vertices u and v with a common neighbor w in R. (Since G[S ] is isomorphic to K|S |/2,|S |/2 or K|′S |/2,|S |/2 and any two nonadjacent vertices have degree sum at least n.) Clearly G[S ] contains cycles of all even lengths between 4 and |S |. Since G[S ] contains all paths of end-vertices u and v of all odd lengths between 3 and |S | − 1 of S-vertices, then these paths together with the vertex w give cycles containing exactly all odd lengths between 3 and |S | − 1 of S-vertices. Thus G is S-pancyclable, which contradicts the assumption. Therefore |S | = 4 and G[S ] is isomorphic to an induced cycle x1 x2 x3 x4 x1 . The vertices x1 and x3 , and similarly x2 and x4 , have no common neighbors since otherwise G would contain cycles C3 and would be S-pancyclable. It is then easy to see that G has the structure described in Theorem 5. Thus we may assume that for any 1 ≤ i ≤ s, |{xi−1 , xi+1 } ∩ T | = 1. Noting that {xs , x1 } ⊆ T , we obtain s = 4r , {x4p , x4p+1 } ⊆ T and {x4p+2 , x4p+3 } ⊆ S − T for any 1 ≤ p ≤ r. Since C is the shortest cycle containing S in G, V (C (x4p+2 , x4p+3 )) = ∅ for any 1 ≤ p ≤ r. In order to show that G[S ] has the exceptional structure described in the statement of Theorem 5, we need to show that N (x4p+2 ) ∩ S ⊆ {x4p+1 , x4p+3 } and N (x4p+3 ) ∩ S ⊆ {x4p+2 , x4p+4 } for any 1 ≤ p ≤ r. Since there is no Cs−1 , x4p+1 x4p+3 ̸∈ E (G) and x4p+2 x4p+4 ̸∈ E (G). Assume that (N (x4p+2 ) ∪ N (x4p+3 )) ∩ {x4t +1 , x4t +2 , x4t +3 , x4t +4 } ̸= ∅ for some p and t with 1 ≤ p ̸= t ≤ r. Then G[{x4p+2 , x4p+3 , x4t +2 , x4t +3 }] is a clique since {x4p+2 , x4p+3 , x4t +2 , x4t +3 } ⊆ S − T . Let P = C [x4p+4 , x4t +2 ]x4p+2 x4t +3 C (x4t +3 , x4p+1 ]. Then |V (P ) ∩ S | = s − 1. If dP (x4p+1 ) + dP (x4p+4 ) ≥ |V (P )| − 1, then there exists a Cs−1 in G by Lemma 3, a contradiction. Thus, dP (x4p+1 )+ dP (x4p+4 ) ≤ |V (P )|− 2. Since {x4p+1 , x4p+4 } ⊆ T , there is a vertex, say w , in NG−P (x4p+1 ) ∩ NG−P (x4p+4 ) \ {x4p+3 }. Then we can get a Cs−1 containing V (P ) ∪ {w} in G, a contradiction. Hence, N (x4p+2 ) ∩ S ⊆ {x4p+1 , x4p+3 } and N (x4p+3 ) ∩ S ⊆ {x4p+2 , x4p+4 } for any 1 ≤ p ≤ r. Consequently, we can derive that G[S ] is a spanning subgraph of F4r . Therefore, the proof of Theorem 5 is complete. Acknowledgments The authors are very grateful to the anonymous referees whose helpful comments and suggestions have led to a substantial improvement of the paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
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A new sufficient condition for pancyclability of graphs Junqing Cai a,∗ , Hao Li b,c a
School of Management, Qufu Normal University, Rizhao, 276826, PR China
b
L.R.I, UMR 8623, CNRS and Université Paris-Sud 11, F-91405 Orsay, France
c
Institute for Interdisciplinary Research, Jianghan University, Wuhan, PR China
article
info
Article history: Received 9 March 2011 Received in revised form 14 August 2013 Accepted 19 August 2013 Available online 12 September 2013 Keywords: Cyclable S-vertex S-length S-pancyclable
abstract Let G be a graph of order n and S be a set of s vertices. We call G, S-pancyclable, if for every i with 3 ≤ i ≤ s, there exists a cycle C in G such that |V (C ) ∩ S | = i. For any two nonadjacent vertices u, v of S, we say that u, v are of distance 2 in S, denoted by dS (u, v) = 2, if there is a path P in G connecting u and v such that |V (P ) ∩ S | ≤ 3. In this paper, we will prove that: Let G be a 2-connected graph of order n and S be a subset of V (G) with |S | ≥ 3. If max{d(u), d(v)} ≥ n/2 for all pairs of vertices u, v of S with dS (u, v) = 2, then G is S-pancyclable or else |S | = 4r and G[S ] is a spanning subgraph of F4r , or else |S | = n is even and G is the complete bipartite graph Kn/2,n/2 , or else |S | = n ≥ 6 is even and G is Kn′ /2,n/2 , or else G[S ] = K2,2 and the structure of G is well characterized. This generalizes a result of Benhocine and Wojda for the case when S = V (G). [A. Benhocine, A.P. Wojda, The Geng-Hua Fan conditions for pancyclic or Hamilton-connected graph, J. Combin. Theory B 42 (1987) 167–180]. © 2013 Elsevier B.V. All rights reserved.
1. Introduction In this paper, we consider only finite undirected and simple graphs. Given a graph G, we denote by V (G), E (G), respectively, the sets of vertices and edges of G. For A ⊆ V (G), G[A] is the subgraph of G induced by A. If H and S are subsets of V (G) or subgraphs of G, we denote by NH (S ) = {u ∈ H: there exists a vertex v ∈ S such that uv ∈ E (G)}, and set dH (S ) = |NH (S )|. In particular, if H = G and S = {u}, then set N (u) = NG (u) and d(u) = dG (u). Paths and cycles in a graph G are considered as subgraphs of G. Let S be a subset of V (G). Then v is called an S-vertex if v ∈ S. Following [4,7], the set S is called cyclable in G if all vertices of S belong to a common cycle in G. Following [5], the S-length of a cycle in G is defined as the number of the S-vertices that it contains, and the graph G is said to be S-pancyclable, if it contains cycles of all S-lengths from 3 to |S |. Following [6], for any two nonadjacent vertices u, v of S, we say that u, v are of distance 2 in S, denoted by dS (u, v) = 2, if there is a path P in G connecting u and v such that |V (P ) ∩ S | ≤ 3. Other notations not defined in this paper can be found in [3]. For a cycle (or a path) C in G with a given orientation and a vertex a in C , a+ and a− denote the successor and the predecessor of a in C , respectively. For a subset X of V (C ), X + and X − denote the set of the successors and the predecessors of the vertices of X in C , respectively. For two vertices a and b in C , we define C [a, b] (C [a, b), C (a, b), respectively) to be the subpath of C from a to b (from a to b− , from a+ to b− , respectively). We use C [b, a] for the path from b to a in the reversed direction of C . In order to give the results of this paper, we define some special graphs. Kn′ ,m is the complete bipartite graph Kn,m minus one edge. Given an integer r ≥ 2, F4r is the graph with 4r vertices containing a complete graph K2r , a set of r independent edges, denoted by Er and a matching between the sets of the vertices of K2r and Er .
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Corresponding author. Tel.: +86 15163380587. E-mail addresses: [email protected] (J. Cai), [email protected] (H. Li).
0166-218X/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.dam.2013.08.026
J. Cai, H. Li / Discrete Applied Mathematics 162 (2014) 142–148
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From the definitions, we see that cyclability and S-pancyclability are generalizations of Hamiltonicity and pancyclability of the whole graph (set S = V (G)), respectively. In recent years, people have given different definitions and results containing certain subsets of vertices, and some related papers can be found in [4,5,7,9,10]. Let us cite for example the following ones. Theorem 1 (Benhocine and Wojda [1]). Let G be a 2-connected graph of order n ≥ 3. If max{d(u); d(v)} ≥ n/2 for any pair of vertices u and v in G with dG (u, v) = 2, then G is pancyclic unless n = 4r , r > 2, and G is F4r , or n is even and G = Kn/2,n/2 or else n ≥ 6 is even and G = Kn′ /2,n/2 . Theorem 2 (Flandrin, Li and Wei [6]). Let G be a 2-connected graph of order n and S be a subset of V (G) with |S | ≥ 3. If max{d(u), d(v)} ≥ n/2 holds for all pairs u, v in S with dS (u, v) = 2, then S is cyclable in G. Theorem 3 (Flandrin, Li and Wei [6]). Let G be a 2-connected graph of order n and S be a subset of V (G) with |S | ≥ 3. If max{d(u), d(v)} ≥ (n + 1)/2 holds for all pairs u, v in S with dS (u, v) = 2, then G is S-pancyclable unless |S | = 4r and G[S ] is a spanning subgraph of F4r . Theorem 4 (Schmeichel and Hakimi [8]). Let G be a graph containing a Hamiltonian cycle C = x1 x2 x3 · · · xn x1 with n ≥ 3. If d(x1 ) + d(xn ) ≥ n, then G is either (i) pancyclic, or (ii) bipartite, or (iii) missing only (n − 1)-cycles. Moreover, if (iii) holds, then d(xn−2 ), d(xn−1 ), d(x2 ), d(x3 ) < n/2. In [2], Bondy suggested the meta-conjecture that almost any nontrivial condition on graphs which implies that the graph is Hamiltonian also implies that the graph is pancyclic (except maybe for a special family of graphs). Many results have been obtained in this problem. We follow, in this paper, the idea of some analogy between Hamiltonicity and pancyclicity, and prove the Conjecture mentioned in [6]. Its proof is related to the method in [6]. Theorem 5. Let G be a 2-connected graph of order n and S be a subset of V (G) with |S | ≥ 3. If max{d(u), d(v)} ≥ n/2 for all pairs of vertices u, v of S with dS (u, v) = 2, then G is S-pancyclable or else |S | = 4r and G[S ] is a spanning subgraph of F4r , or else |S | = n is even and G is the complete bipartite graph Kn/2,n/2 , or else |S | = n ≥ 6 is even and G is Kn′ /2,n/2 , or else G[S ] = K2,2 := x1 x2 x3 x4 x1 and the structure of G is as follows: V (G) is partitioned into S ∪ V1 ∪ V2 ∪ V3 ∪ V4 ; for any i, 1 ≤ i ≤ 4, G[Vi ] is any graph on |Vi | vertices, and each vertex xi is adjacent to all the vertices of Vi+1 and Vi , where the index i is taken as modulo 4. 2. The proof of Theorem 5 By Theorem 2, there exists a cycle containing all the vertices of S in G. Choose such a cycle C such that |C | is as small as possible and give C an arbitrary orientation. If |S | = 3, then Theorem 5 holds. Thus, we may assume that |S | ≥ 4. Set R = G − C and |S | = s. Let x1 , x2 , . . . , xs be the vertices of V (C ) ∩ S, the order 1, 2, . . . , s following the orientation of C , and consider the subscripts modulo s. Two S-vertices xi and xi+1 are said to be S-consecutive. And the segment C [xi , xi+1 ) is denoted by Si , 1 ≤ i ≤ s. We use Cl to denote a cycle of S-length l in G. In [5], it was proved: Theorem 6. Let G be a graph, S be a subset of V (G) such that S is cyclable in G, and let C be a shortest cycle through all the vertices of S. If dC (x) + dC (y) ≥ |C | + 1 for some pair of S-consecutive vertices x and y in C , then G is S-pancyclable. Theorem 7. Let G be a graph, S be a subset of V (G) such that S is cyclable in G, and let C be a shortest cycle through all the vertices of S. If dC (x) + dC (y) ≥ |C | for some pair of S-consecutive vertices x and y in C , then G is S-pancyclable or S = C and G[S ] is bipartite or there is just no Cs−1 . By using a similar method as that used in the proof of Theorems 6 and 7 in [5], we can get the following lemma. Lemma 1. Let G be a graph, S be a subset of V (G) such that S is cyclable in G and let C be a shortest cycle through all the vertices |C |+1 of S. If there exists some 1 ≤ i ≤ s such that xi−1 xi+1 ∈ E (G) and dC (xi ) ≥ 2 , then G is S-pancyclable. Proof. Suppose G is not S-pancyclable, and G misses Cl for some l, 4 ≤ l ≤ s − 2. Without loss of generality, suppose i = 1. Let fl (k) =
s + k + 1 − l, k − l + 2,
if 2 ≤ k ≤ l − 1; if l ≤ k ≤ s − 2.
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Then N (x1 ) ∩ Si ̸= ∅, implies N (x1 ) ∩ Sf+(i) = ∅. (Otherwise, we can get a Cl in G.) By the choice of C , dS1 (x1 ) + dS + (x1 ) + l
1
dSs (x1 ) + dS + (x1 ) ≤ |S1 | + |Ss |. For any i, 2 ≤ i ≤ s − 1, s
dSi (x1 ) + dS + (x1 ) ≤
i
if dSi (x1 ) = 0 or dS + (x1 ) = 0;
|Si |, |Si | + 1,
otherwise.
i
Let A1 = {i : N (x1 )∩ Si ̸= ∅, N (x1 )∩ Si+ = ∅ and 2 ≤ i ≤ s − 1}, A2 = {i : N (x1 )∩ Si = ∅, N (x1 )∩ Si+ ̸= ∅ and 2 ≤ i ≤ s − 1} and A3 = {i : N (x1 ) ∩ Si ̸= ∅, N (x1 ) ∩ Si+ ̸= ∅ and 2 ≤ i ≤ s − 1}. Then i ∈ A1 ∪ A3 , implies fl (i) ̸∈ A2 ∪ A3 . Thus, |A1 ∪ A3 | ≤ s − 2 − |A2 ∪ A3 |, i.e. |A3 | ≤ s − 2 − |A1 ∪ A2 ∪ A3 |. Therefore, s−1 |Si | + (|Si+ | + 1) (dSi (x1 ) + dS + (x1 )) ≤ i=2
i
i∈A1 ∪A2
=
i∈A3
|Si | + |A3 |
i∈A1 ∪A2 ∪A3
≤
|Si | + (s − 2 − |A1 ∪ A2 ∪ A3 |)
i∈A1 ∪A2 ∪A3
≤
s−1
|Si |.
i=2
Then |C | + 1 = 2dC (x1 ) =
s−1 i =1
(dSi (x1 ) + dS + (x1 )) ≤ |C |, a contradiction. Then G is S-pancyclable. i
Now, let T = {v ∈ S : d(v) ≥ n/2}. Notice that for any 1 ≤ i ≤ s, xi xi+2 ∈ E (G) when {xi , xi+2 } ⊆ S − T and xi xj ∈ E (G) for any j ̸= i when N (xi ) ∩ N (xj ) ̸= ∅ and {xi , xj } ⊆ S − T . It is easy to see the following: Remark 1. If there is no pair of S-consecutive vertices x, y in C [xi , xj ](i ̸= j) such that {x, y} ⊆ T , then G[V (C [xi , xj ])∩(S − T )] is a clique of G. Lemma 2. If there exists at most one pair of S-consecutive vertices which are both in T , then Theorem 5 holds. Proof. If |S | = 4, since G is not S-pancyclable, G[S ] must be isomorphic to an induced cycle x1 x2 x3 x4 x1 . Thus, Lemma 2 is true. Next we assume that |S | ≥ 5. If there is no pair of S-consecutive vertices which are both in T , then G[V (C ) ∩ (S − T )] is a clique by Remark 1. Thus, G is S-pancyclable. If there is exactly one pair of S-consecutive vertices, say {xs , x1 } ⊆ T , then G[V (C [x2 , xs−1 ]) ∩ (S − T )] is a clique, especially, x2 xs−1 ∈ E (G). Thus, we can easily check that G is S-pancyclable. In fact, suppose to the contrary that there is no Cl for some 3 ≤ l ≤ s − 1. Then x2 xl+1 ̸∈ E (G) and xs−1 xl ̸∈ E (G). Thus, d(xl + 1) ≥ n/2 and d(xl ) ≥ n/2 by the conditions of Theorem 5, a contradiction. So Lemma 2 is true. Next, we will show four structural lemmas for some special paths containing vertices of S. These lemmas will play very important roles in the proof of Theorem 5. Lemma 3. Let P = u1 u2 . . . up be a path connecting u1 and up in G. If dP (u1 )+ dP (up ) ≥ |V (P )|− 1 and {u1 u3 , xp xp−2 }∩ E (G) = ∅, then there exists a cycle C in G such that V (C ) = V (P ); or p is odd and NP (u1 ) = NP (up ) = {u2 , u4 , u6 , . . . , up−1 }. Proof. Suppose that there is no cycle C in G such that V (C ) = V (P ). Then u1 up ̸∈ E (G). We can easily check that if u1 ui+1 ∈ E (G), then up ui ̸∈ E (G) for any i with 2 ≤ i ≤ p − 2. We can claim that u1 ui ∈ E (G) and u1 ui+1 ∈ E (G) cannot hold at the same time for any i with 4 ≤ i ≤ p − 2. Otherwise, suppose that there exists some i such that u1 ui ∈ E (G) and u1 ui+1 ∈ E (G). Then u2 up ̸∈ E (G). (Since, otherwise, C = P [u2 , ui ]u1 P [ui+1 , up ]u2 is a cycle in G with V (C ) = V (P ), a contradiction.) Therefore, dP (u1 ) + dP (up ) ≤ dP (u1 ) + (|V (P )| − 1 − dP (u1 )) − 1 = |V (P )| − 2, a contradiction. By symmetry, up uj ∈ E (G) and up uj+1 ∈ E (G) cannot hold at the same time for any j with 2 ≤ j ≤ p − 4. Then dP (u1 ) ≤ If p is even, then dP (u1 ) ≤
|V (P )|
|V (P )|
|V (P )|−1 2
and dP (up ) ≤
|V (P )|−1 2
.
− 1 and dP (up ) ≤ 2 − 1. So dP (u1 ) + dP (up ) ≤ |V (P )| − 2, a contradiction. So p is 2 |V (P )|−1 odd. Since dP (u1 ) + dP (up ) ≥ |V (P )| − 1, dP (u1 ) = dP (up ) = . Moreover, NP (u1 ) = NP (up ) = {u2 , u4 , u6 , . . . , up−1 }. 2
Hence, Lemma 3 holds.
Lemma 4. Suppose P = u1 . . . u2 . . . up−1 . . . up is a path in G[V (C )] such that |V (P ) ∩ S | = l + 1 ≥ 4, {u1 , u2 , up−1 , up } ⊆ T , {u1 , u2 }, {up−1 , up } are two pairs of S-consecutive vertices on C and (V (P (u1 , u2 )) ∪ V (P (up−1 , up ))) ∩ S = ∅. Then there exists a Cl in G or dC (u1 ) + dC (u2 ) = dC (up−1 ) + dC (up ) = |C |. Proof. Suppose there exists no Cl in G. Recall that R = G − C . If NR (u1 ) ∩ NR (up−1 ) ̸= ∅ or NR (u2 ) ∩ NR (up ) ̸= ∅, then there exists a Cl in G, a contradiction. So NR (u1 ) ∩ NR (up−1 ) = ∅ and NR (u2 ) ∩ NR (up ) = ∅. Noting that {u1 , u2 , up−1 , up } ⊆ T , we have dC (u1 ) + dC (u2 ) + dC (up−1 ) + dC (up ) ≥ 2|C |. Thus either dC (u1 ) + dC (u2 ) ≥ |C | or dC (up−1 ) + dC (up ) ≥ |C |. If dC (u1 ) + dC (u2 ) ≥ |C | + 1 or dC (up−1 ) + dC (up ) ≥ |C | + 1, then G is S-pancyclable by Theorem 6, a contradiction. Hence, dC (u1 ) + dC (u2 ) = dC (up−1 ) + dC (up ) = |C |. Now the proof of Lemma 4 is complete.
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Lemma 5. Let P = u1 u2 . . . up be a path in G such that |V (P ) ∩ S | = l ≥ 3. If {u1 , up } ⊆ T and there is no Cl in G, then (i) |(N (u1 ) ∩ N (up ) − V (P )) ∩ (V (G) − S )| = 0; (ii) |N (u1 ) ∩ N (up ) ∩ S ∩ (V (G) − V (P ))| ≥ 1; and there exists a Cl+1 which contains P as its subpath; (iii) when P = C [xi , xj ] for some j = l + i − 1(3 ≤ l ≤ s − 2) with {xi , xj } ⊆ T , then there exists a pair of S-consecutive vertices y and z in V (C (xj , xi )) such that y ∈ N (xi ) ∩ N (xj ) and z ∈ N (xj ) (or z ∈ N (xi )), and there exists a Cl+1 and a Cl+2 which contain C [xi , xj ] as a subpath. Moreover V (Cl+1 ) ∩ S ⊆ V (Cl+2 ). Proof. Since there is no Cl in G, (i) is obvious and |N (u1 ) ∩ V (P )| + |N (up ) ∩ V (P )| ≤ |V (P )| − 1 by Lemma 3. As d(u1 ) + d(up ) ≥ n, by (i), it is easy to check that (ii) holds. (iii) As d(xi ) + d(xj ) ≥ n and S ∩ R = ∅, Lemmas 3 and 5(i) imply |N (xi ) ∩ V (C (xj , xi )) ∩ S | + |N (xj ) ∩ V (C (xj , xi )) ∩ S | ≥ |V (C (xj , xi )) ∩ S | + 1. Thus (iii) holds. Lemma 6. Let P = u1 u2 . . . up be a path in G[V (C )] such that V (P ) ∩ S = {v1 , v2 , . . . , vl }, where v1 = u1 , vl = up and the order 1, 2, . . . , l follows the orientation of P from u1 to up . Suppose that l ≥ 5 and there is no Cl in G. If there exists a Cl+m and a Cl+m+1 in G(m ∈ {1, 2}), both of which contain P as a subpath and V (Cl+m ) ∩ S ⊆ V (Cl+m+1 ), then for any 1 ≤ i ≤ l − m − 2, (i) vi vi+m+1 ̸∈ E (G) and vi vi+m+2 ̸∈ E (G); (ii) {vi , vi+m+2 } ∩ (S − T ) ̸= ∅. Proof. Let C ′ = Cl+m+1 and C ∗ = Cl+m . Since P is a subpath of both C ′ and C ∗ , C ′ [v1 , vl ] = C ∗ [v1 , vl ] = P [v1 , vl ]. (i) If vi vi+m+1 ∈ E (G) or vi vi+m+2 ∈ E (G) for some 1 ≤ i ≤ l − m − 2, then replace C ∗ [vi , vi+m+1 ] or C ′ [vi , vi+m+2 ] with the edge vi vi+m+1 or vi vi+m+2 . We can get a Cl in G, a contradiction. (ii) Since there is no Cl in G, NR (vi ) ∩ NR (vi+m+2 ) ∩ (V (G) − V (C ′ )) = ∅, (N (vi ) ∩ V (C (vi+2 , vi+m+2 ))) ∪ (N (vi+m+2 ) ∩ V (C (vi , vi+1 ))) = ∅ and vi+2 ̸∈ N (vi ) ∩ N (vi+m+2 ). These imply that |N (vi ) ∩ V (C (vi , vi+m+2 ))| + |N (vi+m+2 ) ∩ V (C (vi , vi+m+2 ))| ≤ |V (C (vi , vi+m+2 ))|. Since P ′ = C ′ [vi+m+2 , vi ] is a path with |VP ′ ∩ S | = l, we have vi vi+m+1 ̸∈ E (G) and dC ′ (vi ) + dC ′ (vi+m+2 ) < |C ′ | be Lemma 3. If {vi , vi+m+2 } ⊆ T , then there exists at least one vertex, say x, in N (vi ) ∩ N (vi+m+2 ) ∩ (V (G) − V (C ′ )). If x ̸∈ S, then there is a Cl which contains V (C ′ [vi+m+2 , vi ]) ∪ {x}, a contradiction. If x ∈ S, since V (Cl+m ) ∩ S ⊆ V (Cl+m+1 ), x ̸∈ V (C ∗ ). Then we can get a Cl containing V (C ∗ [vi+m+2 , vi ]) ∪ {x}, a contradiction. Hence, {vi , vi+m+2 } ∩ (S − T ) ̸= ∅, and (ii) holds. From Lemma 2, we may assume that |T | ≥ 3 and there exist at least two pairs of S-consecutive vertices which are all in T . Without loss of generality, let {xs , x1 } ⊆ T such that |NR (x1 ) ∩ NR (xs )| = min{|NR (x) ∩ NR (y)| : {x, y} ⊆ T and x, y are S-consecutive}. If dC (x1 ) + dC (xs ) ≥ |C |, then by Theorem 7, G is S-pancyclable or S = C and G[S ] is bipartite or there is just no Cs−1 . Thus in the rest of the proof, we assume that dC (x1 ) + dC (xs ) ≤ |C | − 1 and let M = NR (x1 ) ∩ NR (xs ). Lemma 7. If there exists some 2 ≤ i ≤ s − 2 such that {xi , xi+1 } ⊆ T and dC (xi ) + dC (xi+1 ) ≤ |C | − 1, then (i) |(NR (x1 ) ∪ NR (xs )) ∩ NR (xi ) ∩ NR (xi+1 )| ≥ 1; (ii) there exists a cycle C3 and a cycle C4 in G. Proof. (i) By the choice of x1 and xs , we have |M | ≤ |NR (xi ) ∩ NR (xi+1 )|. Thus,
|R| + 1 ≤ ≤ = ≤
|NR (x1 ) ∪ NR (xs )| + |M | |NR (x1 ) ∪ NR (xs )| + |NR (xi ) ∩ NR (xi+1 )| |(NR (x1 ) ∪ NR (xs )) ∪ (NR (xi ) ∩ NR (xi+1 ))| + |(NR (x1 ) ∪ NR (xs )) ∩ (NR (xi ) ∩ NR (xi+1 ))| |R| + |(NR (x1 ) ∪ NR (xs )) ∩ (NR (xi ) ∩ NR (xi+1 ))|.
From the above inequalities, we can easily check that (i) holds. (ii) Since (NR (x1 ) ∪ NR (xs )) ∩ NR (xi ) ∩ NR (xi+1 ) ̸= ∅, without loss of generality, we may choose a vertex, say v , in (NR (x1 ) ∪ NR (xs )) ∩ NR (xi+1 ). Notice that {xs , x1 , xi , xi+1 } ⊆ T . Assume that there is no C3 in G. Applying Lemma 5(ii) to the path P = C [xi , xi+1 ]v xs , we can getN (xi ) ∩ N (xs ) ∩ (S − V (P )) ̸= ∅. This implies that there is a C3 as v ̸∈ S, a contradiction. Thus, there is a C3 in G. Similarly, applying Lemma 5(ii) to the path P ′ = C [xi , xi+1 ]v C [xs , x1 ], we can get a C4 in G. Hence, (ii) holds. Lemma 8. If there is no Cl in G for some integer l ≥ 3, then S = C and l = s − 1 or G[S ] is bipartite. Proof. Suppose 3 ≤ l ≤ s − 2 and G[S ] is not bipartite. By Theorem 7, dC (x) + dC (y) ≤ |C | − 1 for any pair of S-consecutive vertices x and y in C . Thus, by the assumption, M ̸= ∅ as dC (x1 ) + dC (xs ) ≤ |C | − 1 and |{xl−1 , xl } ∩ T | ≤ 1 by applying Lemma 4 to C [xs , xl ]. Case 1. xl ∈ T . Then xl−1 ̸∈ T . Case 1.1. xl+1 ̸∈ T .
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Then xl−1 xl+1 ∈ E (G) and there exists a C3 in G. So l ≥ 4. By Lemma 1, dC (xl ) ≤ 2 implies dR (xl ) ≥ 2 . Since NR (xl ) ∩ NR (x1 ) = ∅ and dR (x1 ) + dR (xs ) ≥ |R| + 1, 2|R| + |NR (xs ) ∩ NR (xl )| ≥ |NR (x1 ) ∪ NR (xl )| + |NR (xs ) ∪ NR (xl )| + |NR (xs ) ∩ NR (xl )| ≥ dR (x1 ) + dR (xs ) + 2dR (xl ) ≥ 2|R| + 1. This implies |NR (xs ) ∩ NR (xl )| ≥ 1 and there exists a Cl+1 = v C [xs , xl ]v and a Cl+2 = v C [xs , xl−1 ]xl+1 C (xl+1 , xl ]v for some v ∈ NR (xs ) ∩ NR (xl ). Both Cl+1 and Cl+2 contain C [xs , xl−1 ] as a subpath and V (Cl+1 ) ∩ S ⊆ V (Cl+2 ). When l = 4, then x4 ∈ T . Since there is no C4 in G, we have x2 x5 ̸∈ E (G), x2 x4 ̸∈ E (G). Since x3 x5 ∈ E (G) and x5 ̸∈ T , we have dS (x2 , x5 ) = 2, NR (x2 )∩ NR (x4 ) = ∅ and x2 ∈ T . If (N (x2 )∩ N (x4 )− V (C [x2 , x4 ]))∩(V (G)− S ) ̸= ∅, we can get a C4 which contains V (C [xs , x2 ]) ∪ {u, v, x4 } when u ∈ V (C ) − V (C [xs , x2 ]) or V (C [x4 , x5 ]) ∪ V (C [x2 , x3 ]) ∪ {u} when u ∈ V (C [xs , x2 ]), a contradiction. Suppose (N (x2 ) ∩ N (x4 ) − V (C [x2 , x4 ])) ∩ (V (G) − S ) = ∅. Since there is no C4 in G, we can obtain that N (x2 ) ∩ V (C (x3 , x4 ]) = ∅ and N (x4 ) ∩ V (C [x2 , x3 )) = ∅. Also by the minimality of |C |, we have |N (x2 ) ∩ V (C [x2 , x3 ])| = 1 and |N (x4 ) ∩ V (C [x3 , x4 ])| = 1. So dC [x2 ,x4 ] (x2 ) + dC [x2 ,x4 ] (x4 ) ≤ |C [x2 , x4 ]| − 1. Since {x2 , x4 } ⊆ T (that is d(x2 ) + d(x4 ) ≥ n), there exists a vertex, say w , in N (x2 ) ∩ N (x4 ) ∩ S ∩ (V (G) − V (C [x2 , x4 ])). Then we can get a C4 containing V (C [x2 , x4 ]) ∪ {w}, a contradiction. When l ≥ 5, since {xs , x1 } ⊆ T , by applying Lemma 6 with m = 1, we have x2 x4 ̸∈ E (G) and {x3 , x4 } ⊆ S − T . Then x2 ∈ T . If l ≥ 6, then x5 ∈ S − T by Lemma 6(ii), which implies x3 x5 ∈ E (G), contrary to Lemma 6(i). So l = 5, that is x5 ∈ T . Since d(x2 ) + d(x5 ) ≥ n and x2 x5 ̸∈ E (G) as x4 x6 ∈ E (G), |N (x2 ) ∩ N (x5 )| ≥ 2. If (N (x2 ) ∩ N (x5 )) − V (C [x2 , x6 ]) ̸= ∅, then, let u ∈ (N (x2 ) ∩ N (x5 )) − V (C [x2 , x6 ]). Since x4 x6 ∈ E (G), we can get a C5 containing V (C [x2 , x6 ] − C (x4 , x5 )) ∪ {u} whenever u ̸∈ S or V (C [x2 , x5 ]) ∪ {u} whenever u ∈ S, a contradiction. So (N (x2 ) ∩ N (x5 )) − V (C [x2 , x6 ]) = ∅. Since there is no C5 in G, N (x2 ) ∩ V (C (x3 , x5 ]) = ∅ and x3 x5 ̸∈ E (G). Also by the minimality of |C |, we have |N (x2 ) ∩ V (C [x2 , x3 ])| = 1 and |N (x5 ) ∩ V (C [x5 , x6 ])| = 1. Similarly as l = 4, we can get a C5 , a contradiction. |C |
|R|+1
Case 1.2. xl+1 ∈ T . Since there is no Cl in G, NR (x1 ) ∩ NR (xl ) = ∅ and dC (xl ) + dC (xl+1 ) ≤ |C | − 1. Thus, there is a vertex, say x, in NR (xs )∩NR (xl )∩NR (xl+1 ) and l ≥ 5 by Lemma 7. Hence, there exists a Cl+1 and a Cl+2 , which contain C [xs , xl ]∪{x} as a subpath. Since l ≥ 5 and {xs , x1 } ⊆ T , by Lemma 6 with m = 1, we obtain {x3 , x4 } ⊆ S − T . By applying Lemma 4 to C [x1 , xl+1 ] and the fact that dC (x) + dC (y) ≤ |C | − 1 for any pair of S-consecutive vertices x and y, we have x2 ̸∈ T . Then x2 x4 ∈ E (G). We can get a Cl = xV (C [xs , x2 ])x4 V (C (x4 , xl ])x, a contradiction. Case 2. xl−1 ∈ T . Then xl ̸∈ T . Case 2.1. xl−2 ̸∈ T . Then xl−2 xl ∈ E (G) and (NR (x1 ) ∪ NR (xs )) ∩ NR (xl−1 ) = ∅ since there is no Cl . Since 2|NR (x1 ) ∪ NR (xs )| ≥ dR (x1 ) + dR (xs ) ≥ 1 . Then G is S-pancyclable by Lemma 1, a contradiction. |R| + 1, dC (xl−1 ) ≥ |C |+ 2 Case 2.2. xl−2 ∈ T . If l ̸= 3, noting that NR (xs ) ∩ NR (xl−1 ) = ∅, |NR (x1 ) ∩ NR (xl−1 ) ∩ NR (xl−2 )| ≥ 1 and l ≥ 5 by Lemma 7. Since {xs , xl−1 } ⊆ T , there exists a Cl+1 and a Cl+2 containing C [xs , xl−1 ] as a subpath and V (Cl+1 ) ∩ S ⊆ V (Cl+2 ) by Lemma 5(iii). Thus, by Lemma 6(ii) with m = 1, we can get {x3 , x4 } ⊆ S − T . Hence, l ≥ 7 and {x2 , x5 } ∩ (S − T ) ̸= ∅. This implies x2 x4 ∈ E (G) or x3 x5 ∈ E (G), contrary to Lemma 6(i). So l = 3. Then {xs , x1 , x2 } ⊆ T , xs x2 ̸∈ E (G) and NR (xs ) ∩ NR (x2 ) = ∅. (Otherwise, there exists a C3 .) By the minimality of |C |, |N (xs ) ∩ V (C [xs , x1 ])| = 1 and |N (x2 ) ∩ V (C [x1 , x2 ])| = 1. And by Lemma 3, |N (xs ) ∩ V (C [xs , x2 ])| + |N (x2 ) ∩ V (C [xs , x2 ])| ≤ |V (C [xs , x2 ])| − 1. Thus, |N (xs ) ∩ V (C (x2 , xs ))| + |N (x2 ) ∩ V (C (x2 , xs ))| ≥ |V (C (x2 , xs ))| + 1. If there is some i with 2 ≤ i ≤ s − 1 such that either {xi , xi+1 } ⊆ N (xs ) or {xi , xi+1 } ⊆ N (x2 ), then we can get a C3 containing V (C [xi , xi+1 ]) ∪ {xs } or V (C [xi , xi+1 ]) ∪ {x2 }, a contradiction. Hence, there exists some i with 2 ≤ i ≤ s − 1 such that N (xs ) ∩ N (x2 ) ∩ V (C (xi , xi+1 )) ̸= ∅. Then we can find a C3 containing {xs , x1 , x2 }, a contradiction. Case 3. xl ̸∈ T and xl−1 ̸∈ T , that is {xl , xl−1 } ∩ T = ∅. Case 3.1. There is no pair of S-consecutive vertices x and y in V (C [xl+1 , xs−1 ]) such that {x, y} ⊆ T . Then G[V (C [xl−1 , xs−1 ]) ∩ (S − T )] is a clique by Remark 1. Since l ≤ s − 2, |V (C [xl−1 , xs−1 ]) ∩ S | ≥ 3. Case 3.1.1. xs−1 ̸∈ T . Then xl−1 xs−1 ∈ E (G) and xl xs−1 ∈ E (G). Thus, there exists a C3 , a Cl+1 and a Cl+2 . And both Cl+1 and Cl+2 contain C [xs−1 , xl−1 ] as a subpath and V (Cl+1 ) ∩ S ⊆ V (Cl+2 ). Then l ≥ 4 and by Lemma 6(i) {xl−2 , xl−3 } ⊆ T . By applying Lemma 6 to C [xs , xl−1 ] with m = 1, we have {x3 , x4 } ⊆ S − T . This implies that x2 ∈ T and l − 1 ≥ 7 or l − 1 = 3 as {xl−2 , xl−3 } ⊆ T . If l ≥ 8, by Lemma 6(ii) again, x5 ∈ S − T as x2 ∈ T . Then x3 x5 ∈ E (G), contrary to Lemma 6(i). Hence, l = 4. Since there is no C4 and {x3 , x4 } ⊆ N (xs−1 ), we have N (xs ) ∩ V (C (x1 , x3 ]) = ∅, N (x2 ) ∩ V (C [xs−1 , x1 )) = ∅ and (N (x2 ) ∩ N (xs ) − V (C [xs , x2 ])) ∩ (V (G) − S ) = ∅. Since |N (xs ) ∩ V (C [xs , x2 ])| + |N (x2 ) ∩ V (C [xs , x2 ])| ≤ |V (C [xs , x2 ])| − 1 and d(xs ) + d(x1 ) ≥ n, there exists a vertex, say w , in N (x2 )∩ N (xs )∩ S ∩(V (G)− V (C [xs , x2 ])). Then we can get a C4 containing V (C [xs , x2 ])∪{w}, a contradiction. Case 3.1.2. xs−1 ∈ T . Then xs−2 ∈ S − T . Since xl−1 ∈ S − T , xl−1 xs−2 ∈ E (G). Thus, there exists a Cl+2 in G which contains C [xs−2 , xl−1 ] as a subpath. If l ≤ s − 3, then xl xs−2 ∈ E (G) and there exists a C3 . If xl−2 ̸∈ T , then xl−2 xs−2 ∈ E (G) and there exists a C4 and a Cl+1 containing C [xs−2 , xl−2 ] as a subpath. If xl−2 ∈ T , since xs−1 ∈ T , by Lemma 5(iii) there exists a pair of S-consecutive vertices a and b in V (C (xl−2 , xs−1 )) such that a ∈ N (xl−2 ) ∩ N (xs−1 ) and b ∈ N (xl−2 ) (or b ∈ N (xs−1 )).
J. Cai, H. Li / Discrete Applied Mathematics 162 (2014) 142–148
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Fig. 1.
Then there exists a Cl+1 and a Cl+2 containing C [xs−1 , xl−2 ] ∪ {y} as a subpath. Since {xs−1 , x1 } ⊆ T and there exists no Cl , |N (xs−1 ) ∩ V (C [xs−1 , x1 ])| + |N (x1 ) ∩ V (C [xs−1 , x1 ])| ≤ |V (C [xs−1 , x1 ])| − 1. We can get a C4 . Thus, in both cases, l ≥ 5 and V (Cl+1 ) ∩ S ⊆ V (Cl+2 ). By using Lemma 6 with m = 1 and the facts that xl−1 ∈ S − T and {xs−1 , xs , x1 } ⊆ T , we can get {x2 , x3 , x4 } ⊆ S − T and x2 x4 ∈ E (G). If l ≥ 6, then replacing Cl+1 [x2 , x4 ] with the edge x2 x4 , we can get a Cl in G, a contradiction. If l = 5, then we can get a C5 = x2 C [x4 , x5 ]xs−2 C [x3 , x2 ] (see Fig. 1), a contradiction. If l = s − 2 and xl−2 ̸∈ T , then xl−2 xl ∈ E (G). This implies there exists a C3 and a Cl+1 containing C [xs−1 , xl−2 ] as a subpath. If l ≥ 5, by Lemma 6(ii) and {xs−1 , xs , x1 } ⊆ T , we can get {x2 , x3 , x4 } ∩ T = ∅ and x2 x4 ∈ E (G). Then we can get a Cl , a contradiction. So l = 4 and s = 6. Since there exists no C4 and x2 x4 ∈ E (G), dC (x1 ) = 2 and dC (x5 ) = 2 by the minimality of |C |. Since {x1 , x5 } ⊆ T , NR (x1 ) ∩ NR (x5 ) ̸= ∅. Then we can get a C4 containing V (C [x1 , x2 ]) ∪ V (C [x4 , x5 ]) ∪ {w} where w ∈ NR (x1 ) ∩ NR (x5 ), a contradiction. Hence, l = s − 2 and xl−2 ∈ T . By applying Lemma 5(ii) to C [xs−1 , xl−2 ], we have {xl−1 , xl } ∩ N (xs−1 ) ∩ N (xl−2 ) ̸= ∅. If {xl−1 , xl } ⊆ N (xs−1 ) ∩ N (xl−2 ), then there exists a C3 , a C4 and a Cl+1 in G which contains C [xs−1 , xl−2 ] as a subpath. If {xl−1 , xl } ̸⊆ N (xs−1 ) ∩ N (xl−2 ), without loss of generality, we assume xl−1 ̸∈ N (xs−1 ) ∩ N (xl−2 ) and xl ∈ N (xs−1 ) ∩ N (xl−2 ). This implies that there exists a C3 and a Cl+1 . In both cases, we have l ≥ 4. If l ≥ 5, we can derive a contradiction as before by Lemma 6. If l = 4, similarly as l = s − 2 and xl−2 ̸∈ T , we can derive a contradiction. Case 3.2. There exists a pair of S-consecutive vertices x and y in V (C [xl+1 , xs−1 ]) such that {x, y} ⊆ T . Choose l + 1 ≤ t < s − 1 such that {xt , xt +1 } ⊆ T and t is as small as possible. Then by Remark 1, we have G[V (C [xl−1 , xt )) ∩ (S − T )] is a clique of G. This implies xt −1 xl−1 ∈ E (G). Let P = C [x1 , xl−1 ]xt −1 . By Lemma 7, |(NR (x1 ) ∪ NR (xs )) ∩ NR (xt ) ∩ NR (xt +1 )| ≥ 1 and l ≥ 5. We distinguish the following subcases: Case 3.2.1. |NR (x1 ) ∩ NR (xt ) ∩ NR (xt +1 )| ≥ 1. Then we can get a Cl+1 and a Cl+2 in G, both of which contain P as a subpath. Since l ≥ 5, {x3 , x4 } ⊆ S − T by Lemma 6. This implies x2 ∈ T . Using Lemma 4 for the path P ′ = C [x2 , xl−1 ]C [xt −1 , xt +1 ], we have xj ∈ S − T , where j = 5 when l ≥ 6 or j = t − 1 when l = 5. This implies x3 xj ∈ E (G), contrary to Lemma 6(i). Case 3.2.2. |NR (x1 ) ∩ NR (xt ) ∩ NR (xt +1 )| = 0. By Lemma 7(i), there is a vertex, say w , in NR (xs ) ∩ NR (xt ) ∩ NR (xt +1 ). Thus, there exists a Cl+2 = w C [xs , xl−1 ]C [xt −1 , xt ]w and a Cl+3 = w C [xs , xl−1 ]C [xt −1 , xt +1 ]w which contain P = C [x1 , xl−1 ]xt −1 as a subpath. Since l ≥ 5 and {xs , x1 } ⊆ T , by applying Lemma 6(ii) with m = 2, we have {x4 , xj } ⊆ S − T , where j = 5 when l ≥ 6 and j = t − 1 when l = 5. If x2 ̸∈ T , then x2 x5 ∈ E (G) or x2 xt −1 ∈ E (G), contrary to Lemma 6(i). Thus x2 ∈ T . Since xs ∈ T , x3 ̸∈ T by Lemma 4 and x3 xj ∈ E (G). Since {x2 , xt } ⊆ T , applying Lemma 5(ii) to the path P ∗ = C [x2 , xl−1 ]C [xt −1 xt ], we can get a Cl+1 containing P ∗ as a subpath. Since x3 xj ∈ E (G), we can get a Cl in G, a contradiction. Now, we complete the proof of Lemma 8. Now, we turn to prove Theorem 5. By Theorem 7 and Lemma 8, if there exists a Cs−1 , then G is S-pancyclable. Thus, in the rest of the proof, we assume that there is no Cs−1 in G. This implies for any 1 ≤ i ≤ s, xi−1 xi+1 ̸∈ E (G) and NR (xi−1 )∩ NR (xi+1 ) = ∅. Consequently, |{xi−1 , xi+1 }∩ T | ≥ 1 as |V (C [xi−1 , xi+1 ])∩ S | = 3. If there exists some i with 1 ≤ i ≤ s such that {xi−1 , xi+1 } ⊆ T , then dC (xi−1 ) + dC (xi+1 ) ≥ |C |. Since N (xi−1 ) ∩ V (C (xi , xi+1 ]) = ∅ and N (xi+1 ) ∩ V (C [xi−1 , xi )) = ∅, dP (xi−1 ) + dP (xi+1 ) ≥ |V (P )| − 1 for P = C [xi+1 , xi−1 ]. Since xi+1 xi+3 ̸∈ E (G) and xi−1 xi−3 ̸∈ E (G), by Lemma 3 and the choice of C , we can get a Cs−1 in G or else |S | = |V (C )| = s is even and NC (xi−1 ) = NC (xi+1 ) = {x1 , x3 , . . . , xs−1 } when i is odd or NC (xi−1 ) = NC (xi+1 ) = {x2 , x4 , . . . , xs } when i is even. Without loss of generality, we assume i is odd. Thus, by the assumption, s is even and NC (xi−1 ) = NC (xi+1 ) = {x1 , x3 , . . . , xs−1 }. If xi ̸∈ T , since there is no Cs−1 , {xi−2 , xi+2 } ⊆ T . By applying Lemma 4 to C [xi+1 , xi−1 ], we can get dC (xi+1 ) + dC (xi+2 ) = dC (xi−1 ) + dC (xi−2 ) = |C |. Since xi−2 xi+1 ∈ E (G) and xi−1 xi+2 ∈ E (G), we can get that G is S-pancyclable or G[S ] is bipartite by Theorem 4. Since G is not S-pancyclable, we conclude
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that G[S ] is a bipartite graph and dC (xi+1 ) = dC (xi+2 ) = dC (xi−1 ) = dC (xi−2 ) = |C |/2. Since xi ̸∈ T , NC (xi+1 ) \ {xi } ⊆ T and dC (x) = |C |/2 for any vertex x ∈ NC (xi+1 ) \ {xi }. Moreover, G[S ] is the bipartite graph K|S |/2,|S |/2 or else |S | ≥ 6 and G[S ] is isomorphic to K|′S |/2,|S |/2 . Next, we assume xi ∈ T . If xi−2 ∈ T or xi+2 ∈ T , without loss of generality, we assume xi+2 ∈ T . Since there exists no Cs−1 , we can get that dC (xi ) + dC (xi+2 ) = |C |. Recalling that dC (xi−1 ) + dC (xi+1 ) = |C |, we can get dC (xi−1 ) + dC (xi ) ≥ |C | or dC (xi+1 ) + dC (xi+2 ) ≥ |C |. By Theorem 7 and the assumption, similarly to the above, we can get that G[S ] is a bipartite graph K|S |/2,|S |/2 or K|′S |/2,|S |/2 . If {xi−2 , xi+2 }∩ T = ∅, then xi−2 xi+2 ∈ E (G). Thus, we can get Cs−1 = C [xi−1 , xi+1 ]xi−2 C [xi+2 , xi−4 ]xi−1 , a contradiction. Hence, by the above discussion, if S = V (G), then G is isomorphic to Kn/2,n/2 or else n ≥ 6 and G is isomorphic to Kn′ /2,n/2 . If S ̸= V (G), then any two nonadjacent S-vertices belonging to the same class of K|S |/2,|S |/2 or K|′S |/2,|S |/2 and any two nonadjacent S-vertices both in T have degree sum in R equal to at least |R|. Moreover, if |S | ≥ 6, there must be two nonadjacent S-vertices u and v with a common neighbor w in R. (Since G[S ] is isomorphic to K|S |/2,|S |/2 or K|′S |/2,|S |/2 and any two nonadjacent vertices have degree sum at least n.) Clearly G[S ] contains cycles of all even lengths between 4 and |S |. Since G[S ] contains all paths of end-vertices u and v of all odd lengths between 3 and |S | − 1 of S-vertices, then these paths together with the vertex w give cycles containing exactly all odd lengths between 3 and |S | − 1 of S-vertices. Thus G is S-pancyclable, which contradicts the assumption. Therefore |S | = 4 and G[S ] is isomorphic to an induced cycle x1 x2 x3 x4 x1 . The vertices x1 and x3 , and similarly x2 and x4 , have no common neighbors since otherwise G would contain cycles C3 and would be S-pancyclable. It is then easy to see that G has the structure described in Theorem 5. Thus we may assume that for any 1 ≤ i ≤ s, |{xi−1 , xi+1 } ∩ T | = 1. Noting that {xs , x1 } ⊆ T , we obtain s = 4r , {x4p , x4p+1 } ⊆ T and {x4p+2 , x4p+3 } ⊆ S − T for any 1 ≤ p ≤ r. Since C is the shortest cycle containing S in G, V (C (x4p+2 , x4p+3 )) = ∅ for any 1 ≤ p ≤ r. In order to show that G[S ] has the exceptional structure described in the statement of Theorem 5, we need to show that N (x4p+2 ) ∩ S ⊆ {x4p+1 , x4p+3 } and N (x4p+3 ) ∩ S ⊆ {x4p+2 , x4p+4 } for any 1 ≤ p ≤ r. Since there is no Cs−1 , x4p+1 x4p+3 ̸∈ E (G) and x4p+2 x4p+4 ̸∈ E (G). Assume that (N (x4p+2 ) ∪ N (x4p+3 )) ∩ {x4t +1 , x4t +2 , x4t +3 , x4t +4 } ̸= ∅ for some p and t with 1 ≤ p ̸= t ≤ r. Then G[{x4p+2 , x4p+3 , x4t +2 , x4t +3 }] is a clique since {x4p+2 , x4p+3 , x4t +2 , x4t +3 } ⊆ S − T . Let P = C [x4p+4 , x4t +2 ]x4p+2 x4t +3 C (x4t +3 , x4p+1 ]. Then |V (P ) ∩ S | = s − 1. If dP (x4p+1 ) + dP (x4p+4 ) ≥ |V (P )| − 1, then there exists a Cs−1 in G by Lemma 3, a contradiction. Thus, dP (x4p+1 )+ dP (x4p+4 ) ≤ |V (P )|− 2. Since {x4p+1 , x4p+4 } ⊆ T , there is a vertex, say w , in NG−P (x4p+1 ) ∩ NG−P (x4p+4 ) \ {x4p+3 }. Then we can get a Cs−1 containing V (P ) ∪ {w} in G, a contradiction. Hence, N (x4p+2 ) ∩ S ⊆ {x4p+1 , x4p+3 } and N (x4p+3 ) ∩ S ⊆ {x4p+2 , x4p+4 } for any 1 ≤ p ≤ r. Consequently, we can derive that G[S ] is a spanning subgraph of F4r . Therefore, the proof of Theorem 5 is complete. Acknowledgments The authors are very grateful to the anonymous referees whose helpful comments and suggestions have led to a substantial improvement of the paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
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