vertex pancyclicity and new sufficient conditions

Proc. Indian Acad. Sci. (Math. Sci.) Vol. 122, No. 3, August 2012, pp. 319–328.c© Indian Academy of Sciences

Vertex pancyclicity and new sufficient conditions

KEWEN ZHAO and YUE LIN

Department of Mathematics, Qiongzhou University, Sanya, Hainan 572022,People’s Republic of ChinaE-mail: [email protected]

MS received 23 November 2010; revised 24 May 2012

Abstract. For a graph G, δ(G) denotes the minimum degree of G. In 1971, Bondyproved that, if G is a 2-connected graph of order n and d(x) + d(y) ≥ n for each pairof non-adjacent vertices x, y in G, then G is pancyclic or G = Kn/2,n/2. In 2001, Xuproved that, if G is a 2-connected graph of order n ≥ 6 and |N (x) ∪ N (y)| + δ(G) ≥ nfor each pair of non-adjacent vertices x, y in G, then G is pancyclic or G = Kn/2,n/2.In this paper, we introduce a new sufficient condition of generalizing degree sum andneighborhood union and prove that, if G is a 2-connected graph of order n ≥ 6 and|N (x) ∪ N (y)| + d(w) ≥ n for any three vertices x, y, w of d(x, y) = 2 and wxor wy �∈ E(G) in G, then G is 4-vertex pancyclic or G belongs to two classes ofwell-structured exceptional graphs. This result also generalizes the above results.

Keywords. Hamiltonian graphs; vertex pancyclic; degree sum; neighborhood union;sufficient conditions.

1. Introduction

We generalize two well-known degree sum and neighborhood union for characterizingHamiltonian graphs, in particular for vertex-pancyclic. We first give a few definitions andsome notation. We consider only finite undirected graphs with no loops or multiples. Wedenote by δ(G) the minimum degree of G, and Kh the complete graph of order h. If uis a vertex and H is a subgraph of G, then let NH (u) = {v ∈ V (H) : uv ∈ E(G)}be the vertex set of H that are adjacent to vertex u, and if S is a vertex set or subgraphof G, then set NH (S) = ∪u∈S NH (u) and set N [u] = N (u) ∪ {u}. Let G − H andG[S] denote the subgraphs of G induced by V (G) − V (H) and S, respectively. If Cm =x1x2 · · · xm x1 is a cycle of order m, let N+

Cm(u) = {xi+1 : xi ∈ NCm (u)}, N−

Cm(u) =

{xi−1 : xi ∈ NCm (u)}, and N±Cm

(u) = N+Cm

(u) ∪ N−Cm

(u), where subscripts are takenmodulo m. For a graph G of order n, Ore [7] introduced degree sum condition d(u) +d(v) ≥ n for Hamiltonian graphs; Faudree et al. [3] introduced neighborhood unionNC = min{|N (x) ∪ N (y)| : x, y ∈ V (G), xy �∈ E(G)}; and Lindquester [6] introducedneighborhood union of each pair of vertices at distance 2 as follows: NC2 = min{|N (x)∪N (y)| : x, y ∈ V (G), d(x, y) = 2}. In this paper, we introduce new and sufficientcondition for generalizing degree sum and neighborhood union as follows: DNC2 =min{|N (x) ∪ N (y)| + d(w) : x, y, w ∈ V (G), d(x, y) = 2, wx or wy �∈ E(G)}. For

319

320 Kewen Zhao and Yue Lin

graphs A and B the joining operator A ∨ B of A and B is the graph constructed from Aand B by adding all the edges joining the vertices of A and B.

If no ambiguity can arise we sometimes write N (u) instead of NG(u), δ instead ofδ(G), etc.

If a graph G has a Hamiltonian cycle (a cycle containing every vertex of G), then G iscalled Hamiltonian. A graph G is said to be pancyclic if G contains cycles of every lengthk, 3 ≤ k ≤ n. For each vertex u of G, if u is contained in cycles of G of every length k,r ≤ k ≤ n, then graph G is called r -vertex pancyclic. Terminologies and notations can befound in [2].

Ore [7] obtained the following well-known Hamiltonian result on sum condition.

Theorem 1.1 [7]. If G is a graph of order n and d(x) + d(y) ≥ n for each pair ofnon-adjacent vertices x, y in G, then G is Hamiltonian.

Bondy [1] considered Ore’s condition for pancyclic graphs.

Theorem 1.2 [1]. If G is a 2-connected graph of order n and d(x) + d(y) ≥ n for eachpair of non-adjacent vertices x, y in G, then G is pancyclic or G = Kn/2,n/2.

Faudree et al. [3] proved the following result on neighborhood union.

Theorem 1.3 [3]. If G is a 2-connected graph of order n and |N (x) ∪ N (y)| + δ ≥ n foreach pair of non-adjacent vertices x, y in G, then G is Hamiltonian.

Xu [8] proved the following pancyclic result.

Theorem 1.4 [8]. If G is a 2-connected graph of order n ≥ 6 and |N (x)∪ N (y)|+δ ≥ nfor each pair of non-adjacent vertices x, y in G, then G is pancyclic or G = Kn/2,n/2.

Lin and Song [5] also proved the following vertex pancyclic result on neighborhoodunion.

Theorem 1.5 [5]. If G is a 2-connected graph of order n ≥ 6 and |N (x)∪ N (y)|+δ ≥ nfor each pair of non-adjacent vertices x, y of d(x, y) = 2 in G, then G is 4-vertex-pancyclic or G = Kn/2,n/2.

Recently, some sufficient conditions for the characterizing of Hamiltonian and pan-cyclic graphs which motivate the work of this paper was shown in [4,9].

In this paper, we introduce a new and sufficient condition and prove the following resultwhich generalizes the above results.

Theorem 1.6. If G is a 2-connected graph of order n ≥ 6 and |N (x)∪ N (y)|+d(w) ≥ nfor any three vertices x, y, w in G of d(x, y) = 2 and wx or wy �∈ E(G), then G is a4-vertex pancyclic or G ∈ {Kn/2,n/2, Gh : (K1 ∪ Kn−h−1)}.

Here Gh is some graph of order h, h = 2, 3. Gh : (K1 ∪ Kn−h−1) is the graphobtained from each vertex of graph Gh adjacent to some vertices of two disjoint com-plete graphs K1 and Kn−h−1. More detailed descriptions can be found in the proof ofLemma 2.1.

Note. If the condition of Theorem 1.2 holds, i.e., each pair of nonadjacent vertices of agraph G satisfy the condition of Theorem 1.2, then any three vertices x, y, w with xw �∈E(G) or yw �∈ E(G) satisfy |N (x)∪N (y)|+d(w) ≥ n. Since graphs Gh : (K1∪Kn−h−1)

Vertex pancyclicity and new sufficient conditions 321

for h = 2, 3 are pancyclic and clearly G satisfies condition of Theorem 1.2 which has3-cycle, Theorem 1.6 implies Theorem 1.2. Also, if each pair of nonadjacent vertices ofa graph G satisfy the condition of Theorem 1.4 or Theorem 1.5, then any three verticesx, y, w of d(x, y) = 2 satisfy |N (x) ∪ N (y)| + d(w) ≥ n, and thus Theorem 1.6 impliesTheorems 1.4 and 1.5 too.

2. Proof of the main result

Obviously, Theorem 1.6 can be obtained immediately by Lemmas 2.1 and 2.6, so we onlyneed to prove Lemmas 2.1 and 2.6.

Lemma 2.1. If G is a 2-connected graph of order n ≥ 6 and DNC2 ≥ n, then for eachvertex u in G, G has C3, C4 or C4, C5 all containing u or G ∈ {G2 : (K1 ∪ Kn−3), G3 :(K1 ∪ Kn−4), Kn/2,n/2}, where G2 is a subgraph of order 2, G2 : (K1 ∪ Kn−3) is thegraph obtained from each vertex of G2 adjacent to some vertices of two disjoint completegraphs K1 and Kn−3. G3 is a subgraph of order 3 with |E(G3)| = 1. The description ofG3 : (K1 ∪ Kn−4) can be found in the following proof.

Proof. Let u be any vertex of G. Then we consider the following two cases:

Case 1. d(u) ≥ 3. Let x, y, z ∈ N (u). Consider the following:

Subcase 1.1. There exist two adjacent vertices in N (u). In this case, G has C3 containingu. Then, we consider the following two subcases:

Subcase 1.1.1. The induced subgraph G[N (u)] contains path of order 3. In this case, Ghas C4 containing u, so the proof of Lemma 2.1 is complete.

Subcase 1.1.2. G[N (u)] does not contain any path of order 3. If N (u) has two verticeswho have common neighbor in G − N [u], then clearly, G has C4 containing u, and so theproof is complete.

Consider the case where each pair of vertices of N (u) have no common neighbor inG − N [u]. Without loss of generality, assume x, y, z ∈ N (u) satisfying xy �∈ E(G),together without a path of order 3 in G[N (u)] such that zx �∈ E(G) or zy �∈ E(G).Without loss of generality, assume zy /∈ E(G). (1). If zx /∈ E(G), then we easily checkthat |N (x)∪ N (y)|+|N (z)| ≤ n−1, a contradiction. (2). If zx ∈ E(G), clearly, |N (u)| =3. Otherwise, if |N (u)| ≥ 4, we can check that |N (x) ∪ N (y)| + |N (z)| ≤ n − 1, acontradiction. Then, we have that G − N [u] is a complete subgraph. Otherwise, if thereexist two vertices w, t ∈ V (G − N [u]) with d(w, t) = 2, then, we have |N (w)∪ N (t)| +|N (z)| ≤ n − 1, a contradiction. We denote the graph without C4 of containing u byG3 : (K1 ∪ Kn−4), where V (G3) = {x, y, z}, V (K1) = {u}. Each vertex of Kn−4 isadjacent and only adjacent to one of V (G3).

Subcase 1.2. N (u) does not have two adjacent vertices. In this case, by the condition ofLemma 2.1 that |N (x) ∪ N (y)| + |N (z)| ≥ n, we can check that N (u) has two verticeswho have common neighbor in G − N [u], so G has C4 containing u. If G has no C5containing u, then we consider the following:


Let v,w, z ∈ N (u), since d(w, z) = 2, by condition of Lemma 2.1 that |N (w) ∪N (z)| + d(v) ≥ n, we can check |V (G − N [u])| ≥ |N (u)| − 1. Consider the followingtwo cases:

(1) When |V (G−N [u])| = |N (u)|−1. In this case, by |N (w)∪N (z)|+d(v) ≥ n for anythree vertices w, z, v in N (u), each vertex of N (u) must be adjacent to every vertexof G − N (u) for example, if v ∈ N (u) is not adjacent to some vertex of G − N (u),let w, z ∈ N (u) \ {v}, and then we can check |N (w) ∪ N (z)| + d(v) ≤ n − 1, acontradiction, so G ∈ N (u)∨(G − N (u)). Since G has no C5 containing u, G − N (u)

is an empty subgraph, and this implies that G = Kn/2,n/2.(2) When |V (G − N [u])| ≥ |N (u)|.

(2)–(1) There exists some v of N (u) that is adjacent to two x, y of G − N [u] withxy /∈ E(G). In this case, since G is 2-connected, G − N [u] is not an empty sub-graph. Also, since G has no C5 containing u, so G − N [u] is not a complete subgraph.Thus, there exist two vertices of distance 2 in V (G − N [u]). Let x, y be two vertices ofdistance 2 in V (G − N [u]). By |N (x) ∪ N (y)| + d(u) ≥ n, we can check that thereare at most |N (u)| − 1 vertices of G − N [u] that are not adjacent to x and y, so thereare at least |V (G − N [u])| − (|N (u)| − 1) vertices of G − N [u] that are adjacent tox or y.

Since G has no C5 containing u, so clearly, vertex w is not adjacent to any of N (x) ∪N (y) for each w ∈ N (u) \ {u, v}. Here v is the notation represented in case (2)–(1).

By the condition of Lemma 2.1, we have that there exists at least a vertex of N (u) \ {v}that is adjacent to at least two vertices of V (G − N [u]). Otherwise, let w, z ∈ N (u) \ {v}.We can check that |N (w) ∪ N (v)| + d(z) ≤ n − 1, a contradiction. So without loss ofgenerality, let w ∈ N (u) \ {v} is adjacent to at least two vertices q, r of V (G − N [u]).Consider the following:

(i) When qr ∈ E(G). Since G has no C5 containing u, so both q, r are not adjacent toboth v, z. Here z ∈ N (u) \ {v,w}, so we have d(w) ≤ |V (G − N [u])| − |N (x) ∪N (y)|+ |{u}| ≤ |V (G − N [u])|− (|V (G − N [u])|− (|N (u)|−1))+|{u}| ≤ |N (u)|.And we have |N (t)∪ N (z)|+d(w) ≤ (|V (G − N [u])|− |{q, r}|)+|{u}|+ |N (u)| ≤|V (G − N (u) ∪ {u})| + |N (u)| ≤ n − 1, a contradiction.

(ii) When qr /∈ E(G). Here also, we can check d(w) ≤ |V (G − N [u])| − |N (x) ∪N (y)| + |{u}| ≤ |V (G − N [u])| − (|V (G − N [u])| − (|N (u)| − 1)) + |{u}| ≤|N (u)|. Then, by the condition of Lemma 2.1, |N (q) ∪ N (r)| + d(u) ≥ n and by|V (G − N [u])| ≥ |N (u)|, we can see that q or r must be adjacent to at least avertex of V (G − N [u]), i.e., there exist two adjacent vertices h, k of G − N [u] withh or k adjacent to w. Without loss of generality, assume k is adjacent to w. Then,since G has no C5 containing u, both h is not adjacent to both v, z. Then we have|N (v) ∪ N (z)| ≤ (|V (G − N [u])| − |{h}|) + |{u}|, together with d(w) ≤ |N (u)|.So we have |N (v) ∪ N (z)| + d(w) ≤ (|V (G − N [u])| − 1) + |{u}| + |N (u)| ≤|V (G − N (u) ∪ {u})| + |N (u)| ≤ n − 1, a contradiction.

(2)–(2) xy ∈ E(G) for any v of N (u) that is adjacent to two x, y of G − N [u]. In thiscase, since G has no C5 containing u, v of N (u) is adjacent to x, y of G − N [u], and w isnot adjacent to both x and y for each w ∈ N (u) \ {v}. Thus, let w, z, v ∈ N (u). We cancheck |N (w) ∪ N (z)| + d(v) ≤ n − 1, a contradiction.


Case 2. d(u) = 2. In this case, if G − N (u) ∪ {u} contains two nonadjacent vertices,then we can choose two vertices x, y in G − N (u) ∪ {u} of distance 2, but we check|N (x) ∪ N (y)| + |N (u)| ≤ n − 1. This contradicts the condition of the lemma. Thiscontradiction shows that G − N (u) ∪ {u} is a complete subgraph Kn−3, and this impliesthat G ∈ G2 : (K1 ∪ Kn−3). �

To prove Lemma 2.6, we need the following lemmas.

Lemma 2.2. If G is a 2-connected graph of order n ≥ 6, and DNC2 ≥ n, Cm is a cycleof order m, u is a vertex of G − Cm, |NCm (u)| ≥ 2, then the following two hold:(1) If xi+1, x j+1 ∈ N+

Cm(u) and d(xi+1, x j+1) = 2 and xi+1, x j+1 are not adjacent to

any of N [u]\(Cm), then there exists xk ∈ NCm (u) satisfying xi+1xk+1 or x j+1xk+1 ∈E(G).

(2) If there exist xi+1, x j+1 ∈ N+Cm

(u) satisfying d(xi+1, x j+1) ≥ 3 and {xi+1, xi+2, . . .,x j−1}∩ NCm (u) = ∅ and x j+1 is not adjacent to any of N [u] \ (Cm), then there existsat least a vertex xk in P = x j+1x j+2 · · · xi such that xk ∈ N (x j+1) with xk−1xi+1 orxk−1u ∈ E(G).

Clearly, (1) and (2) imply that V (Cm) ∪ {u} structures a Cm+1.

Proof. Suppose Lemma 2.2(1) is false, i.e., for any x ∈ N (u) ∪ {u}, when x �∈ V (Cm), xis not adjacent to xi+1, x j+1. When x = xk ∈ V (Cm), xk+1 is not adjacent to xi+1, x j+1.This implies that |N (xi+1) ∪ N (x j+1)| ≤ n − |N (u) ∪ {u}|, so |N (xi+1) ∪ N (x j+1)| +d(u) ≤ n − 1, a contradiction. �

Therefore, (1) holds, and we can construct cycle Cm+1 = xi uxk xk−1 · · · xi+1xk+1xk+2· · · xi if xk+1xi+1 ∈ E(G) for some xk ∈ NCm (u) or cycle Cm+1 = x j uxk xk−1 · · ·x j+1xk+1xk+2 · · · x j if xk+1x j+1 ∈ E(G) for some xk ∈ NCm (u), the two cyclesconsisting of u and Cm .

If (2) is false, i.e., for any x ∈ N (x j+1), when x �∈ V (Cm), x is not adjacent tovertex u and since d(xi+1, x j+1) ≥ 3, x is also not adjacent to xi+1. When x = xk

in path P = x j+1x j+2 · · · xi then xk−1xi+1, xk−1u �∈ E(G), i.e., none of N−P (x j+1)

is adjacent to xi+1, u. When x in path R = xi+1xi+2 · · · x j−1, d(xi+1, x j+1) ≥ 3 and{xi+1, xi+2, · · · , x j−1} ∩ NCm (u) = ∅, so x is not adjacent to u, xi+1, i.e., none ofN+

R (x j+1) is adjacent to xi+1, u. Also xi+1, u are not adjacent to xi+1, u. Clearly in thethree sets N−

P (x j+1), N+R (x j+1) and {u, xi+1}, there are no two sets that have any com-

mon vertex, and |N+R (x j+1)| + |N−

P (x j+1)| = |NCm(x j+1)| − 1. Hence we can check|N (u) ∪ N (xi+1)| ≤ n − |N+

R (x j+1)| − |N−P (x j+1)| − |{xi+1, u}| ≤ n − d(x j+1) − 1.

This implies that |N (u) ∪ N (xi+1)| + d(x j+1) ≤ n − 1, a contradiction.Therefore, (2) holds, and we can construct cycle Cm+1 =: xi ux j x j−1 · · · xi+1xk−1xk−2

· · · x j+1xk xk+1 · · · xi or x j uxk−1xk−2 · · · x j+1xk xk+1 · · · x j , respectively.Similarly, it is also easy to obtain the following result by considering the reverse

direction on Cm from Lemma 2.2.

Lemma 2.3. If G is a 2-connected graph of order n ≥ 6, and DNC2 ≥ n, Cm is a cycleof order m, u is a vertex of G − Cm , |NCm (u)| ≥ 2, then the following two hold:(1) If xi−1, x j−1 ∈ N−

Cm(u) and d(xi−1, x j−1) = 2 and xi−1, x j−1 are not adjacent to

any of N [u]\(Cm), then there exists xk ∈ NCm (u) satisfying xi−1xk−1 or x j−1xk−1 ∈E(G).


(2) If there exist xi−1, x j−1 ∈ N−Cm

(u) satisfying d(xi−1, x j−1) ≥ 3 and {x j+1, x j+2,

. . . , xi−1} ∩ NCm (u) = ∅ and x j−1 is not adjacent to any of N [u] \ (Cm), thenthere exists at least a vertex xk in P = x j+1x j+2 · · · xi such that xk ∈ N (x j+1) withxk+1xi+1 or xk+1u ∈ E(G).

To prove Lemma 2.6, we also need the following two propositions:

PROPOSITION 2.4

Let Cm+1 =: y1 y2 · · · ym+1 y1 be the cycle of order m + 1 obtained from (1) or (2) ofLemma 2.2, if v ∈ V (G − Cm+1) is adjacent to some yh in V (Cm+1) ∩ V (Cm), whenyh ∈ {xi , xi+1, x j , x j+1, xk−1, xk, xk+1} as described in Lemma 2.2. Then clearly yh+1or yh−1 ∈ N±

Cm(v). When yh �∈ {xi , xi+1, x j , x j+1, xk, xk+1}, then clearly yh+1, yh−1 ∈

N±Cm

(v).

PROPOSITION 2.5

Let Cm+1 =: y1 y2 · · · ym+1 y1 be the cycle of order m + 1 obtained from (1) or (2) ofLemma 2.2 consisting of u and Cm, where u, y ∈ V (G−Cm) and the condition of Lemma2.2 holds. If x j ∈ NCm (y) satisfying this path x j−1x j x j+1 of Cm is also a path of Cm+1,and if yh ∈ NCm+1(y) \ {x j } satisfies yh+1 or yh−1 ∈ N±

Cm(y), then we can obtain Cm+2

consisting of y and Cm+1.

That is, under hypothesis of Proposition 2.4, there must exist yk, yh ∈ NCm+1(y) sat-isfying both yk+1, yh+1 or both yk−1, yh−1 which are not adjacent to any NG−Cm+1(y).(For example, let x j−1x j x j+1 = yi−1 yi yi+1, if yh+1 ∈ N±

Cm(y), then yi+1, yh+1 are not

adjacent to any N (y) \ V (Cm+1); if yh−1 ∈ N±Cm

(y), then yi−1, yh−1 are not adjacent toany N (y) \ V (Cm+1).) Thus, the cycle Cm+1 and graph G must satisfy the condition ofLemma 2.2, so we can obtain Cm+2 consisting of y and Cm+1 immediately by Lemma 2.2.

Lemma 2.6. If G is a 2-connected graph of order n ≥ 6 and u is a vertex of G, ifDNC2 ≥ n and Cm, Cm+1 are two cycles of containing u, then G has Cm+2 containing u.

Proof. Assume on the contrary, that G has no Cm+2 containing u. Under the hypothesis,clearly, for each vertex x of G−Cm , none of N±

Cm(x) are adjacent to any of N (x)\V (Cm).

(Otherwise, if some vertex of N±Cm

(x) is adjacent to some vertex of N (x) \ V (Cm), for

example, if xi+1 ∈ N+Cm

(u) or xi−1 ∈ N−Cm

(u) is adjacent to y ∈ N (x) \ V (Cm), then weobtain cycle xi xyxi+1xi+2 · · · xi or cycle xi xyxi−1xi−2 · · · xi , respectively, that are twoCm+2 consisting of V (Cm) ∪ {x, y}, so this Cm+2 contains u, a contradiction.) Then weconsider the following cases:

Case 1. There exist two distinct vertices x, y in G − Cm such that |NCm (x)| ≥ 2 and|NCm (y)| ≥ 2.

Subcase 1.1. NCm (x) = NCm (y) = {xi , x j }. In this case, we have xy ∈ E(G). Otherwise,if xy �∈ E(G), let xk ∈ V (Cm) \ {xi , x j } satisfying xk ∈ {xi+1, x j+1}. Then we can check|N (x) ∪ N (y)| ≤ n − |N [xk] \ {xi , x j }| − |{x, y}| = n − |N [xk]| = n − d(xk) − 1. Thisimplies that |N (x) ∪ N (y)| + d(xk) ≤ n − 1, a contradiction.

Thus, xy ∈ E(G) holds. Then by Lemma 2.2(1) or Lemma 2.2(2), we can constructCm+1 consisting of V (Cm) ∪ {x}. Clearly Cm+1 contains u and it contains edge xxi oredge xx j . Since y is adjacent to x, xi and x j , we have Cm+2 containing u.


Subcase 1.2. NCm (x) �= NCm (y) or max{|NCm (x)|, |NCm (y)|} ≥ 3.

Subcase 1.2.1. xy ∈ E(G).

Subcase 1.2.1.1. NCm (x) ∩ NCm (y) �= ∅. In this case, let x j ∈ NCm (x) ∩ NCm (y). Thenwe choose xi ∈ NCm (x) of satisfying {xi+1, xi+2, · · · , x j−1} ∩ NCm (x) = ∅.

(I) If d(xi+1, x j+1) ≥ 3, by Lemma 2.2(2), there exists xk ∈ NCm (x j+1) satisfy-ing xi+1xk−1 or xxk−1 ∈ E(G). When xi+1xk−1, then Cm+2 = xi xyx j x j−1 · · ·xi+1xk−1xk−2 · · · x j+1xk xk+1 · · · xi is a cycle of order m + 2 containing u, a contra-diction. When xxk−1, then Cm+2 = x j yxxk−1xk−2 · · · x j+1xk xk+1 · · · x j contains u.

(II) If d(xi+1, x j+1) = 2, by Lemma 2.2(1), there exists xk ∈ NCm (x) satisfy-ing x j+1xk+1 or xi+1xk+1 ∈ E(G). When x j+1xk+1 ∈ E(G), then Cm+2 =x j yxxk xk−1 · · · x j+1xk+1xk+2 · · · x j contains u. When xi+1xk+1 ∈ E(G), we firstconstruct Cm+1 = y1 y2 · · · ym+1 y1 consisting of x and Cm via Lemma 2.2(1).Clearly, since x j ∈ NCm (x) ∩ NCm (y) \ {xi }, the path x j−1x j x j+1 of Cm is also apath of Cm+1, together with Proposition 2.5 and since |NCm (y)| ≥ 2, there must existvertex yr ∈ NCm+1(y) \ {x j } together with x j Without loss of generality, assume x j

of Cm to be yh of Cm+1 satisfying {yh+1, yh+2, . . . , yr−1}∩ NCm+1(y) = ∅ with bothyr+1, yh+1 or both yr−1, yh−1 are not adjacent to any N (y) \ V (Cm+1), by Lemma2.2(1) or Lemma 2.2(2) and we obtain Cm+2 containing u.

Subcase 1.2.1.2. NCm (x) ∩ NCm (y) = ∅.

Subcase 1.2.1.2.1. There exist consecutive xi , xi+1 on Cm satisfying xi , xi+1 ∈ NCm (x)

or NCm (y).Without loss of generality, assume xi , xi+1 ∈ NCm (x), so we obtain Cm+1 =

xi xxi+1xi+2 . . . xi . Then for each x j ∈ NCm (y) path x j−1x j x j+1 of Cm is also a path ofCm+1, similar to the subcase 1.2.1.1. By Lemma 2.2(1) or Lemma 2.2(2), we can obtainCm+2 containing u.

Subcase 1.2.1.2.2. There are no consecutive xi , xi+1 on Cm such that xi , xi+1 ∈ NCm (x)

or NCm (y).In this case using x and Cm we first construct Cm+1.By the hypothesis of G that has no Cm+2 of containing u and it is no subcase 1.2.1.2.1,

together with xy as an edge, for any xi ∈ NCm (x), y is not adjacent to xi−1, xi , xi+1. Since|NCm (y)| ≥ 2, there exists at least a vertex x j ∈ NCm (y) satisfying path x j−1x j x j+1of Cm which is also a path of Cm+1. By (1) or (2) of Lemma 2.2, we can obtain Cm+2containing u.

Case 1.2.2. xy �∈ E(G).

Subcase 1.2.2.1. There exist consecutive xi , xi+1 on Cm satisfying xi , xi+1 ∈ NCm (x) orNCm (y).

Without loss of generality, assume xi , xi+1 ∈ NCm (x), so we have Cm+1 =xi xxi+1xi+2 · · · xi . Then if y is adjacent to two consecutive vertices of Cm+1, we haveCm+2 containing u. Otherwise, since this is not Case 1.1, there exists at least a vertexx j ∈ NCm (y) satisfying path x j−1x j x j+1 of Cm which is also the path of Cm+1. ByProposition 2.4 and Lemma 2.2, we have Cm+2 containing u.


Subcase 1.2.2.2. There are no consecutive xi , xi+1 on Cm such that xi , xi+1 ∈ NCm (x)

or NCm (y). By Lemma 2.2, we first construct Cm+1 by x and Cm containing u. Then (1)if y is adjacent to some consecutive two vertices on Cm+1, we have Cm+2 containing u.(2) If y is not adjacent to any consecutive two vertices on Cm+1, then we consider thefollowing:

(2–1) When xi ∈ NCm (x)∩ NCm (y) �= ∅, since this is not Case 1.2.2.1, so xxi+1, yxi+1 �∈E(G), and by |N (x) ∪ N (y)| + d(xi+1) ≥ n, since any one of {x, y} and xi+1 do nothave common neighbor in G − Cm (otherwise, we have Cm+2 containing u immedi-ately), so x, y are at least adjacent to three vertices of Cm , together with |NCm (x)| ≥ 2and |NCm (y)| ≥ 2, so there must exist xh, xk ∈ V (Cm) \ {xi } that are adjacent to x, y,respectively. By |N (x) ∪ N (y)| + d(xi+1) ≥ n, we have xi+1xh+1 ∈ E(G). Since yis not adjacent to any consecutive two vertices on Cm+1, so clearly, there exists at leastx j ∈ NCm (y) satisfying path x j−1x j x j+1 of Cm which is also the path of Cm+1. ByProposition 2.5 and Lemma 2.2, we have Cm+2 containing u.(2–2) When NCm (x) ∩ NCm (y) = ∅, there must exist xh, xk ∈ NCm (x) satis-fying {xh+1, xh+2, . . . , xk−1} ∩ NCm (x) = ∅ and x j ∈ NCm (y) satisfying x j ∈{xh+1, xh+2, . . . , xk−1} \ {xh+1}. Then by Lemma 2.2 we first construct Cm+1 consistingof V (Cm) ∪ {x} containing u. Then clearly path x j−1x j x j+1 of Cm is also the path ofCm+1. By Proposition 2.5 and Lemma 2.2, we have Cm+2 containing u.

Case 2. There exists a vertex x in G − Cm such that |NCm (x)| ≥ 2.In this case, under the assumption that G does not have any Cm+2 containing u, we have

claim (I) that if y ∈ V (G − Cm) and NCm (y) = {xi }, then induced subgraph G[Cm − xi ]is a complete subgraph of order m − 1.

That is, since G is 2-connected, so there must at least exist a vertex y ∈ V (G−Cm) suchthat |NCm (y)| = 1. Let NCm (y) = {xi }, then we first prove xi−1xi+1 ∈ E(G), Otherwise,if xi−1xi+1 �∈ E(G), since there does not exist Cm+2 containing u, so if w ∈ N [y] \ {xi }then w is not adjacent to xi+1, xi−1. Hence we can check |N (xi+1) ∪ N (xi−1)| ≤ n −|N [y] \ {xi }| − |{xi+1, xi−1}|, a contradiction.

Similarly, we have xi−1xi+2 ∈ E(G), if xi−1xi+2 �∈ E(G). Since G has no Cm+2containing u, w ∈ N [y]\{xi } and w is not adjacent to xi+2, xi−1 (for example, if wxi+2 ∈E(G), we have Cm+2 = xi−1xi+1xi ywxi+2xi+3 · · · xi−1 containing u). Hence we have|N (xi+1) ∪ N (xi−1)| ≤ n − |N [y] \ {xi }| − |{xi+1, xi−1}|, a contradiction.

Now, we use induction for the following proof. That is, under the assumptionxi−1xi+r ∈ E(G), we have xi−1xi+r+1 ∈ E(G).

Thus, xi+1, xi+2, xi+3, . . . , xi−3, xi−2 are all adjacent to xi−1. Then, clearly for eachpair xh, xk in Cm \ {xi }, d(xh, xk) ≤ 2, if xh xk �∈ E(G). If w ∈ N [y] \ {xi } then w is notadjacent to xh, xk (for example, wxh ∈ E(G), together with xi−1xi+h−1 ∈ E(G), so wehave Cm+2 = xi−1xi+h−1xi+h−2 · · · xi ywxi+h xi+h+1 · · · xi−1 containing u). Hence wehave |N (xh) ∪ N (xk)| ≤ n − |N [y] \ {xi }| − |{xh, xk}|, a contradiction.

Therefore, induced subgraph G[V (Cm) \ {xi }] is a complete subgraph of orderm − 1.

Let P = y1 y2 · · · yk be a path of G − Cm whose two end-vertices y1, yk adjacent totwo vertices xi , x j on Cm with order k of path P is as small as possible. Without loss ofgenerality, assume NCm (y1) = {xi }, so G[Cm −xi ] is a complete subgraph of order m −1.Then we consider the following cases on order k of path P .


Case 2.1. k = 2. In this case, if | j − i | = 1 or m − 1, then we obtain Cm+2 =xi y1 yk xi+1xi+2 · · · xi containing u, a contradiction. If 2 ≤ | j − i | ≤ m − 2, by Claim (I),we have xi+1x j+1 ∈ E(G) so we obtain Cm+2 = xi y1 yk x j x j−1 · · · xi+1x j+1x j+2 · · · xi

containing u, a contradiction.

Case 2.2. k = 3.

Subcase 2.2.1. u is at least adjacent to one of y1, yk .

Note. u is the vertex that will contain Cm+2.

Since u ∈ V (Cm) it is equal to u ∈ {xi , x j }. When m = 3, Cm+2 = C5 = xi y1 y2 y3x j xi

containing u, a contradiction. When m ≥ 4 | j − i | = 1 without loss of generality, assumej ≥ i . By Claim (I), xi−1xi−3 ∈ E(G), so we have Cm+2 = xi y1 y2 y3x j x j+1 · · · xi−3xi−1xi

not containing xi−2, and containing u, a contradiction. If | j − i | ≥ 2, clearly xi+1x j+2 ∈E(G), so we have Cm+2 = xi y1 y2 y3x j x j−1 · · · xi+1x j+2x j+3 · · · xi not containing x j+1,and containing u, a contradiction.

Subcase 2.2.2. u is not adjacent to y1, yk . In this case we consider the following cases onlength m of Cm .

When m = 3, Cm = C3 = xi x j uxi , so for any w ∈ N [u] \ {xi , x j }, w is not adjacentto y1, y3 (for example, if wy1 ∈ E(G), then C5 = ux j xi y1wu is a cycle of order 5 withcontaining u). Hence we have |N (y1) ∪ N (y3)| ≤ n − |N [u] \ {xi , x j }| − |{y1, y3}|, acontradiction.

When m ≥ 4. u is adjacent to xi , together with G[Cm − xi ] which is a completesubgraph. Then it is easy to obtain Cm+2 containing u, if u is not adjacent to xi . Whenm = 4, we have |N (y1) ∪ N (y3)| ≤ n − |N (u) \ {x j }| − |{y1, y3}|, a contradiction. Whenm ≥ 5, since G[Cm − xi ] is complete subgraph, it is easy to obtain Cm+2 containing u, acontradiction.

Case 2.3. k ≥ 4. Let xi , x j in Cm be adjacent to y1, yk , respectively, and |NCm (y1)| =|{xi }|. Let xt ∈ V (Cm) \ {xi , x j }. So for any w ∈ N [xt ] \ {xi , y4}, w �∈ V (Cm), sincek ≥ 4. Then w is not adjacent to y1, y3. (Otherwise, if w is adjacent to y1, in this case wehave path P = wy1 of order 2 of G − Cm whose two end-vertices w, y1 are adjacent totwo vertices of Cm , respectively. This contradicts our choice that k is as small as possible.When w is adjacent to y3, then wy3 y4 · · · yk is a path of order less than k of G−Cm whosetwo end-vertices w, yk are adjacent to two vertices of Cm , respectively, a contradiction.)When w ∈ V (Cm)\{xi }, then w is not adjacent to y1, y3. (This is because NCm (y1) = {xi },so wy1 �∈ E(G). If wy3 ∈ E(G), then y1 y2 y3 is a path of order less than k of G − Cm

whose two end-vertices y1, y3 are adjacent to two vertices xi , w of Cm , respectively, acontradiction.) Hence we have |N (y1) ∪ N (y3)| ≤ n − |N [xt ] \ {xi , y4}| − |{y1, y3}|, acontradiction.

This completes the proof of Lemma 2.6. �

Proof of Theorem 1.6. Under the condition of Theorem 1.6, by Lemma 2.1, G con-tains cycles C3, C4 or C4, C5 all containing u for each vertex u in G or G ∈ {G2 :(K1 ∪ Kn−3), G3 : (K1 ∪ Kn−4), Kn/2,n/2}. By Lemma 2.6, if G has cycles Cm, Cm+1containing u, then G has Cm+2 containing u. Thus, if G ∈ {G2 : (K1 ∪ Kn−3), G3 :(K1 ∪ Kn−4), Kn/2,n/2}, then G has cycles C4, C5, . . . , Cn all containing u, i.e., G is4-vertex pancyclic. This completes the proof of Theorem 1.6.


3. Conclusion

Edge-pancyclic graphs have also been studied as a subject in graph theory for the pastfew years. However, the condition DNC2 ≥ n of Theorem 1.6 has not been used forstudying edge-pancyclic graph. Clearly, condition DNC2 ≥ n that has been used foredge-pancyclic graphs is more difficult than for vertex-pancyclic graphs. Moreover, it isalso interesting to generalize the condition DNC2 ≥ n of Theorem 1.6 for the researchof vertex-pancyclic graphs. For example, what is the result if we consider the condition|N (x)∪ N (y)|+d(w) ≥ n for any three vertices x, y, w in G of d(x, y) = d(y, w) = 2?

Acknowledgements

The authors are very grateful to the anonymous referee for helpful remarks and comments.The authors are also very grateful to Professor Ping Zhang, co-author of their publication[9] from which some methods were used for the research of this paper.

References

[1] Bondy J A, Pancyclic graphs I, J. Combin. Theory B11 (1971) 80–84[2] Bondy J A and Murty U S R, Graph Theory with Applications (London: Macmillan and

New York: Elsevier) (1976)[3] Faudree R J, Gould R J, Jacobson M S and Lesniak L, Neighborhood unions and highly

Hamilton graphs, Ars Combin. 31 (1991) 139–148[4] Gould R J and Zhao Kewen, A new sufficient condition for Hamiltonian graphs, Arkiv.

Mate. 44(2) (2006) 299–308[5] Lin W S and Song Z M, Neighborhood union condition with distance for vertex-

pancyclicity, Ars Combin. 61 (2001) 119–127[6] Lindquester T E, The effects of distance and neighborhood unions conditions on hamilto-

nian properties in graphs, J. Graph Theory 13 (1989) 335–352[7] Ore O, Note on Hamilton circuits, Amer. Math. Monthly 67 (1960) 55[8] Xu J, A sufficient condition for pancyclism of Hamiltonian graphs, Acta Math. Appl.

Sinica 24 (2001) 310–313[9] Zhao Kewen, Lin Yue and Zhang Ping, A sufficient condition for pancyclic graphs, Inform.

Proc. Lett. 109(16) (2009) 991–996

vertex pancyclicity and new sufficient conditions

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