Digital Object Identifier (DOI) 10.1007/s00373-005-0622-3Graphs and Combinatorics (2005) 21:489–501

Graphs andCombinatorics© Springer-Verlag 2005

Fan-Type Theorem for Long Cycles Containing a SpecifiedEdge

Mei Lu1,�, Huiqing Liu2 and Feng Tian3,��

1 Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China.e-mail: mlu@math.tsinghua.edu.cn2 School of Mathematical Sciences and LPMC, Nankai University, Tianjin 300071, China.e-mail: hqliu@nankai.edu.cn3 Institute of Systems Science, Academy of Mathematics and Systems Sciences, Chinese Acad-emy of Sciences, Beijing 100080, China.e-mail: ftian@mail.iss.ac.cn

Abstract. In this paper, we prove that if G is 3-connected noncomplete graph of order nsatisfying min{max{d(u), d(v)} : d(u, v) = 2} = µ, then for each edge e, G has a cyclecontaining e of length at least min{n, 2µ}, unless G is a spanning subgraph of Kµ + Kc

n−µ orK3 + (lKµ−2 ∪ Ks), where n = l(µ − 2) + s + 3, 1 ≤ s ≤ µ − 2.

Key words. Cycle, Specified edge, Complete subgraph

1. Introduction

We use Bondy and Murty [1] for terminology and notation not defined here andconsider simple undirected graphs only. In particular, we use Kk to denote a completegraph with k vertices and Kc

k its complement graph. Denote H1∪H2 and H1+H2 theunion graph and the join graph of H1 and H2, respectively, and lH the union graphof l copies of H . Let G be a graph. If S ⊆ V (G), then N(S) denotes the neighborsof S. For a subgraph H of G and S ⊆ V (G) \ V (H), let NH (S) = N(S) ∩ V (H).When H = G, we write N(S) instead of NG(S). If S = {x}, then we will denoteN({x}) = N(x) and d(x) = |N(x)|. For convenience, we let N [x] = N(x) ∪ {x}.For ∅ �= X ⊂ V (G), 〈X〉G is the subgraph of G induced by X and G − X is thesubgraph of G induced by V (G) \X. For ∅ �= E0 ⊆ E(G), G[E0] is the subgraph ofG induced by E0. Let A, B ⊆ V (G), denote E(A, B) = {uv ∈ E(G) |u ∈ A, v ∈ B}.The distance between x and y in G is denoted by d(x, y). For two distinct verticesx and y in G, we will use G − xy to denote the graph that arises from G by deletingthe edge xy ∈ E(G). Similarly, G + xy is a graph that arises from G by adding theedge xy (G + xy = G if xy ∈ E(G)).

� Partially supported by NNSFC(No. 60172005);�� Partially supported by NNSFC(No. 10431020);

490 M. Lu et al.

Let P = v0v1 · · · vl be a path of G. We say that the vertex vi precedes vertex vj onP if i < j , and indicate this relationship by vi ≺ vj and we use vi � vj to denote therelationship vi ≺ vj or vi = vj . Let X be a cycle or a path of G with a given orienta-tion. For u ∈ V (X), we use u+ to denote the successor of u and u− its predecessor.If u, v ∈ V (X), we use X[u, v] or uXv to denote the subpath uu+ · · · v−v of X andthe same subpath, in reverse order, is denoted by X[v, u] or vXu. And for conve-nience, we consider them as both paths and vertex sets. Set X(u, v] = X[u, v] \ {u},X[u, v) = X[u, v] \ {v}, etc., and consider them as both paths and vertex sets. Thelength of X, denoted by l(X), is defined as the number of edges of X. For a pathX of G, let Int (X) = {v ∈ V (X) : v is not any end vertex of X}. For e ∈ E(G), weuse ce(G) to denote the maximum length of cycles in G which pass through e. Thesymmetric difference of two sets A and B is defined by A B = (A \ B) ∪ (B \ A).

For a connected noncomplete graph G, let

σ = min{d(u) + d(v) : uv /∈ E(G), u �= v},τ = min{d(u) + d(v) : d(u, v) = 2},µ = min{max{d(u), d(v)} : d(u, v) = 2}.

There are many results about the existence of the longest cycle in a graph.

Theorem 1 [2]. Each 2-connected noncomplete graph of order n contains a cycle oflength at least min{n, σ }.

Enomoto et al. discussed the existence of large cycles containing a specified edgein 3-connected graph and gave the following two results.

Theorem 2 [3]. Suppose G is a 3-connected graph of order n satisfying σ ≥ m. Thenthrough each edge of G there passes a cycle of length ≥ min{n, m − 1}. Furthermore,suppose n > m−1 and some edge is contained in no cycle of length > m−1. Then either

(K2 ∪ Kcm2 −2) + Kc

n− m2

⊆ G ⊆ Km2

+ Kcn− m

2

or

(K2 ∪ K1) + lKm2 −2 ⊆ G ⊆ K3 + lKm

2 −2,

where l = 2(n − 3)/(m − 4).

Theorem 3 [4]. SupposeG is a 3-connected noncomplete graph of ordern. Then througheach edge of G there passes a cycle of length ≥ min{n, τ − 1}.

In [8], a result similar to Theorem 2 was given.

Theorem 4 [8]. Suppose that G is a 3-connected noncomplete graph of order n. Then,for any edge e of G, there exists a cycle containing e of length at least min{n, τ },unless G belongs to the two classes of the exceptional graphs defined in Theorem 2(with m = τ).

In 2002, Hu et al. gave the following result which improved Theorem 3.

Fan-Type Theorem for Long Cycles Containing a Specified Edge 491

Theorem 5 [5]. Suppose G is a (s + 2)-connected noncomplete graph, where s ≥ 1is an integer. Then through each path of length s of G there passes a cycle of length≥ min{|V (G)|, 2µ − s}.

Motivated by Theorems 2 and 4, we prove the following theorem which improvesTheorems 2 and 4.

Theorem 6. Suppose G is a 3-connected noncomplete graph of order n. Then througheach edge of G there passes a cycle of length ≥ min{n, 2µ}, unless G is a spanning sub-graph of Kµ+Kc

n−µ or K3 +(lKµ−2 ∪Ks), where n = l(µ−2)+s+3, 1 ≤ s ≤ µ−2.

2. Lemmas

In this section, we give some lemmas which will be used in Section 3.

Lemma 1. Let G be a connected graph. Suppose A ⊆ V (G) with |A| ≤ µ + 1 (µ =min{max{d(u), d(v)} : d(u, v) = 2}) and B ⊆ A with N(B) ⊆ A. Then 〈B〉G is acomplete subgraph.

Proof. If 〈B〉G is not a complete subgraph, then existv, v′ ∈ B such thatd(v, v′) = 2.But max{d(v), d(v′)} ≤ |A| − 2 ≤ µ − 1, a contradiction. �

Lemma 2 [5]. Let G be a connected graph, e ∈ E(G) and let P be a path in G con-taining e with |V (P )| maximum, if ce(G) = |V (P )|, then G has a Hamilton cyclecontaining e.

Lemma 3 [5]. Let G be a 3-connected noncomplete graph and e ∈ E(G). SupposeP = P [x, y] is a path of G such that

(i) e ∈ E(P );(ii) Subject to (i), |V (P )| is maximum;

(iii) Subject to (i) and (ii), d(x) + d(y) is maximum.

Then either G has a Hamilton cycle passing through e or min{d(x), d(y)} ≥ µ.

A subvine (given in [5]) on a path P = P [x, y] in a graph G is a set P = {Pi [xi, yi ] :1 ≤ i ≤ m} of internally-disjoint paths such that

V (Pi) ∩ V (P ) = {xi, yi}, 1 ≤ i ≤ m,

x = x1 ≺ x2 ≺ y1 � x3 ≺ y2 � x4 ≺ · · · � xm ≺ ym−1 ≺ ym on P.

Here, ym is called the end vertex of P ; if ym = y, then P is called a vine on P inG. Clearly, if P is a subvine on P = P [x, y] in G with the end vertex v, then P is avine on P [x, v] in G.

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Let P = {Pi [xi, yi ] : 1 ≤ i ≤ m} be a subvine on P = P [x, y]. Then we setC(P, P) = G[E(C1) E(C2) · · · E(Cm)], where Ci = xiPyiPixi, 1 ≤ i ≤ m.Note that C(P, P) is a cycle of G with the edge sets E(C(P ,P)) = ⋃m

i=1(E(Pi) ∪E(P [yi−1, xi+1])), where y0 = x1 = x and xm+1 = ym.

Let P = P [x, y] be a path of G and e an edge of P . For v ∈ V (P ), setP(G; P [x, v], e) = {P : P is a subvine on P in G with the end vertex v such thate ∈ E(C(P ,P)) ∩ E(P [x, v])}.

LetGbe a connected graph andx,y be distinct vertices ofG. DefineD(G; x, y) ={(u, v)|u, v ∈ V (G) \ {x, y}, d(u, v) = 2}. Set

m(G; x, y) ={

min(u,v)∈D(G;x,y)

max{d(u), d(v)} if D(G; x, y) �= ∅;|V (G)| − 1 if D(G; x, y) = ∅.

Note that m(G; x, y) ≤ |V (G)|−1 and the equality holds if and only if D(G; x, y) =∅.

Lemma 4 [7]. Let G be a block. Then each pair of vertices x and y of G are joined bya path of length at least m(G; x, y).

Lemma 5 [5]. Let P = P [x, y] be a path of a graph G and let e be an edge of P . If G

is 2-connected, then there exists a vertex v of P such that P(G; P [x, v], e) �= ∅.

Lemma 6. Let P = P [x, y] be a path that satisfies the condition of Lemma 3. IfP(G; P [x, y], e) �= ∅, then G has a cycle C of length ≥ min{|V (G)|, 2µ} passingthrough e, unless G is a spanning subgraph of Kµ + Kc

n−µ or K3 + (lKµ−2 ∪ Ks),

where n = l(µ − 2) + s + 3, 1 ≤ s ≤ µ − 2.

Proof. Let P = {Pi [xi, yi ], 1 ≤ i ≤ m} be a vine on P with x1 = x, ym = y.Consider the cycle C = C(P, P) and assume that P is chosen so that

(i) P ∈ P(G; P [x, y], e), i.e., e ∈ E(C) ∩ E(P );(ii) Subject to (i), E(C) ∩ E(P ) is maximal.

Obviously, N(x) ∪ N(y) ⊆ V (P ). By Lemma 2, we can assume that xy /∈ E(G)

and m ≥ 2. Set P = v0v1 · · · vt where v0 = x, vr = y1, vkvk+1 = e and vt = y, anddefine A = {vi : xvi+1 ∈ E(G)} and B = {vi : viy ∈ E(G)}. From the proof ofLemma 13 in [5], we have A ∩ B ⊆ {vk}, N(x) ∩ P(xl+1, yl ] ⊆ {y1}, 1 ≤ l ≤ m − 1and N(y) ∩ P [xl+1, yl) ⊆ {xm}, 1 ≤ l ≤ m − 1. Hence (A \ {vr−1}) ∪ B ⊆ V (C) ifx2 �= vr−1 or A ∪ B ⊆ V (C) if x2 = vr−1.

Note that y ∈ V (C)\(A∪B) and |A∩B| ≤ 1. If m ≥ 3, then x2 ∈ V (C)\(A∪B)

when x2 �= vr−1. Hence |V (C)| ≥ |(A\{vr−1})∪B∪{y}∪{x2}| ≥ |A|+|B|−|A∩B|+1 ≥ d(x) + d(y) if x2 �= vr−1 or |V (C)| ≥ |A ∪ B ∪ {y}| ≥ |A| + |B| − |A ∩ B| + 1 ≥d(x) + d(y) if x2 = vr−1. If there exists i (1 ≤ i ≤ m) such that l(Pi) > 1, then|V (C)| ≥ |(A \ {vr−1}) ∪ B ∪ {y}| + 1 ≥ d(x) + d(y). By Lemma 3, d(x) ≥ µ andd(y) ≥ µ. Thus |V (C)| ≥ 2µ.

Fan-Type Theorem for Long Cycles Containing a Specified Edge 493

Therefore, in the next proof, we assume that m = 2, x2 �= vr−1 and l(Pi) = 1,

i = 1, 2. Then we have

|V (C)| ≥ |A| + |B| − |A ∩ B| ≥ d(x) + d(y) − 1 ≥ 2µ − 1. (*)

Suppose ce(G) < 2µ. Then |V (C)| = 2µ − 1. This implies that all the equalities in(∗) hold. Thus we can made the following assumptions.

Assumption 1. |N [x]| = |N [y]| = µ + 1, A ∩ B = {vk}, N(x) ∩ P(x2, y1] ={y1}, N(y) ∩ P [x2, y1) = {x2}.

Assumption 2. Denote C = xPx2yPy1x. Then V (C) = (A \ {vr−1}) ∪ B ∪ {y}.

Note 1.

(1) If there exist vi, vj ∈ N(x) such that N(x) ∩ N(y) ∩ P [vi, vj ) = ∅, P [vi, vj ] ⊆V (C) and e /∈ E(P [vi, vj ]), then P [vi, vj ] ⊆ N(x).

(2) If there exist vi, vj ∈ N(y) such that N(x) ∩ N(y) ∩ P(vi, vj ] = ∅, P [vi, vj ] ⊆V (C) and e /∈ E(P [vi, vj ]), then P [vi, vj ] ⊆ N(y).

Proof. We show only (1) holds. Otherwise, we may choose vq, i < q < j such thatvq ∈ N(x) and vq+1 /∈ N(x). Since N(x) ∩ N(y) ∩ P [vi, vj ) = ∅, vq /∈ N(y), i.e.vq /∈ B. By Assumption 2 we have vq ∈ A, and hence vq+1 ∈ N(x), which is acontradiction. �

Assume, without loss of generality, that e = vkvk+1 ∈ E(P [vr , vt ]), where vr =y1, vt = y. Then xvk+1, yvk ∈ E(G) and

(N)

|N(x) ∩ N(y) ∩ P(x, x2]| ≥ 1,

|N(x) ∩ N(y) ∩ P [vr , vk]| ≥ 1,

|N(x) ∩ N(y) ∩ P [vk+1, y)| ≥ 1.

Note 2. If there exists a vertex v ∈ P(x, y) such that v /∈ N(x)∪N(y) and d(v) ≥ µ,then G is a spanning subgraph of Kµ + Kc

n−µ.

Proof. Since A ∩ B = {vk}, v �= vk, vk+1. By Assumption 2, v+ ∈ N(x), v− ∈N(y). Then N(v) ⊆ V (P ), and we have a cycle C′ = xPv−yPv+x of length atleast 2µ − 1. So v is not adjacent to consecutive vertices on P except {vk, vk+1},(otherwise, we can get a longer cycle than C by inserting v into C′). Thus µ ≤d(v) ≤ 1

2 (|V (C)| + 1) = µ, and hence d(v) = µ, t = 2µ − 1, (vt = y) and N(v) ={v1, v3, . . . , vk} ∪ {vk+1, vk+3, . . . , vt−1}. Since ce(G) < 2µ and d(x) = d(y) = µ,neither x nor y is adjacent to consecutive vertices on P except {vk, vk+1}. HenceN(x) = N(y) = N(v). Thus we have

(i) V (P )\N(x) is an independent set and for each u ∈ V (P )\N(x), N(u) ⊆ V (P )

by the maximality of P .

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Let H be any component of G \ V (P ). We shall prove that |H | = 1. Other-wise, by the connectedness of G, we get u1vq, u1u2 ∈ E(G), where u1, u2 ∈V (H), vq ∈ V (P ). Thus the path

P ′ =

u2u1v1Py, if q = 1,

u2u1vqPyv1Pvq−1, if q �= k + 1, 1,

u2u1vk+1Pv1yPvk+2, if q = k + 1

would be longer than P , a contradiction. Thus(ii) V (G) \ V (P ) is an independent set.

By (i), (ii), we get that V (G) \ N(x) is an independent set of order n − µ. ThusG is a spanning subgraph of Kµ + Kc

n−µ with V (Kcn−µ) = V (G) \ N(x). �

Now, from (N), we prove Lemma 6 by considering three cases separately. Denotex2 by vs .

Case 1. |N(x) ∩ N(y) ∩ P(x, x2]| ≥ 2.

Since |N(x) ∩ N(y) ∩ P(x, x2]| ≥ 2, there exists vi ∈ P(x, x2) such that viy,

xvi+2 ∈ E(G) and vi+1 /∈ N(x)∪N(y) by Assumption 2. Hence |P [vs+1, vr−1]| = 1,vi+1vs+1 /∈ E(G) and t = 2µ − 1(vt = y).

By Note 2, we may assume that d(vi+1) < µ and d(vs+1) < µ, then d(vi+1,

vs+1) > 2 by vi+1vs+1 /∈ E(G). Thus vk+1 /∈ N(vi+1) ∩ N(vs+1). Since e ∈E(P [vr , vt ]), we can assume, without loss of generality, that vk+1vi+1 �∈ E(G).By G being 3-connected, d(vi+1) ≥ 3. Let v ∈ N(vi+1)\{vi, vi+2}. Then v− �= vk byv �= vk+1. Also vi+1 is not adjacent to consecutive vertices on P except {vk, vk+1},hence d(v−, vi+1) = 2 and d(v+, vi+1) = 2. We will show that v+ /∈ N(x) ∪ N(y)

or v− /∈ N(x) ∪ N(y).

(a) v ∈ P [x, vi−1]. Then the cycles xPvvi+1PyviP v+x and xPvvi+1Pv+yPvi+2x,which are longer than C, show that v+ /∈ N(x) ∪ N(y);

(b) v ∈ P [vi+3, y]. Then the cycles xPviyPvvi+1Pv−x and xPvi+1vPyv−Pvi+2x,which are longer than C, show that v− /∈ N(x) ∪ N(y).

By (a) and (b), there exists w ∈ P(x, y) such that w /∈ N(x) ∪ N(y) andd(w, vi+1) = 2. By d(vi+1) < µ, we get d(w) ≥ µ. Then we derive that G is aspanning subgraph of Kµ + Kc

n−µ by Note 2.

Case 2. |N(x) ∩ N(y) ∩ P(x, x2]| = 1 and |N(x) ∩ N(y) ∩ P [y1, y)| ≥ 3.

Similar to Case 1, there exists vi ∈ P [y1, y), where i �= k, such that viy, xvi+2 ∈E(G) but vi+1 /∈ N(x) ∪ N(y). Hence s = r − 2, vs+1vi+1 /∈ E(G) and t =2µ − 1(vt = y). Also N(vi+1) ∪ N(vs+1) ⊆ V (P ) and neither vi+1 nor vs+1 isadjacent to consecutive vertices on P except {vk, vk+1}. Hence d(vi+1) ≤ µ andd(vs+1) ≤ µ.

By Note 2, we can assume that d(vi+1) < µ and d(vs+1) < µ. Since vi+1vs+1 /∈E(G), we have that d(vi+1, vs+1) > 2 and hance vk+1 /∈ N(vi+1) ∩ N(vs+1).

Fan-Type Theorem for Long Cycles Containing a Specified Edge 495

(1) If e ∈ E(P [vi+2, vt ]), then we can assume, without loss of generality, thatvk+1vi+1 /∈ E(G). By the same argument as (a) and (b) of Case 1, there existsw ∈ P(x, y) such that w /∈ N(x) ∪ N(y) but d(w) ≥ µ and thus we can derive thatG is a spanning subgraph of Kµ + Kc

n−µ by Note 2.By (1), we assume that e ∈ E(P [vs+2, vi ]).(2) If d(vi+1) ≥ 4 or d(vs+1) ≥ 4, then there exists v ∈ N(vi+1) \ {vi, vi+2, vk}

or v ∈ N(vs+1) \ {vs, vs+2, vk+1}. By the same argument as (a) and (b) of Case 1,we can show that v+ /∈ N(x) ∪ N(y) or v− /∈ N(x) ∪ N(y) with degree at least µ,and then G is a spanning subgraph of Kµ + Kc

n−µ by Note 2.(3) If vs+1vk+1 /∈ E(G) or vi+1vk /∈ E(G), say vi+1vk /∈ E(G), then there is

a vertex v ∈ N(vi+1) \ {vi, vi+2, vk}, and we can show that v+ /∈ N(x) ∪ N(y) orv− /∈ N(x) ∪ N(y) with degree at least µ by the same argument as (a) and (b) ofCase 1, and then G is a spanning subgraph of Kµ + Kc

n−µ by Note 2.By (1), (2) and (3), we can make the following assumption.

Assumption 3. d(vi+1) = 3 < µ, d(vs+1) = 3 < µ, e ∈ E(P [vs+2, vi ]), vs+1vk+1 ∈E(G) and vi+1vk ∈ E(G).

(4) If there exists v ∈ V (P ) \ {vi+1, vs+1} such that v /∈ N(x) ∪ N(y), then wecan assume that d(v) < µ by Note 2. By Assumption 3 and d(v) < µ, we have thatvvk, vvk+1 /∈ E(G). Thus we can derive that G is a spanning subgraph of Kµ+Kc

n−µ

by the same argument as (a) and (b) of Case 1.By (4), we can make the following Assumption.

Assumption 4. vs+1 and vi+1 are the only two vertices on P that are adjacent neitherto x nor to y, i.e. |N(x) ∩ N(y) ∩ P [y1, y)| = 3.

By Assumption 4 and the assumption of Case 2, |N(x)∩N(y)| = 4 and V (P ) =N [x] ∪N [y] ∪ {vi+1, vs+1}. Set N(x)∩N(y) = {vi1 , vi2 , vi3 , vi4} with vi1 ∈ P [v1, vs ],vi2 ∈ P [vs+2, vk], vi3 ∈ P [vk+1, vi ] and vi4 ∈ P [vi+2, vt−1]. By Note 1, P [v1, vi1 ] ∪P [vs+2, vi2 ]∪P [vk+1, vi3 ]∪P [vi+2, vi4 ] = N(x)andP [vi1 , vs ]∪P [vi2 , vk]∪P [vi3 , vi ]∪P [vi4 , vt−1] = N(y).

By Assumption 3, we have d(vi+1, vk−1) = 2, d(vs+1, vk) = 2, d(vs+1, vi+1) > 2,and hence vk+1 �= vi and vk �= vs+2.

If xvk ∈ E(G), then vi2 = vk by Assumption 4, N(vk−1) ⊆ V (P ) by the choiceof P and d(vk−1) ≥ µ by d(vi+1, vk−1) = 2 and d(vi+1) < µ. If there existsv ∈ (N [y] ∪ {vi1−1}) \ {vs, vi2 , vi3 , vi4} such that vvk−1 ∈ E(G), then the cycle

C′ ={

xPvvk−1Pv+yPvkx if v ∈ P [vi1−1, vs−1],xPvk−1vPyv−Pvkx if v ∈ P [vi3+1, vi ] ∪ P [vi4+1, y]

have length 2µ, a contradiction. Hence N(vk−1) �⊆ (N [y]∪{vi1−1})\{vs, vi2 , vi3 , vi4}.Since V (P ) = N [x]∪N [y]∪{vi+1, vs+1}, N(vk−1) ⊆ (N [x]∪{vs})\{vi1−1, vi1 , vk−1}.Thus d(vk−1) < µ, a contradiction.

If xvk /∈ E(G), then yvk−1 ∈ E(G) by Assumption 2 and hence N(vk) ⊆ V (P ).Since d(vs+1, vk) = 2 and d(vs+1) < µ, d(vk) ≥ µ. If there exists v ∈ (N [x] ∪{vi4+1}) \ {vi1 , vi2 , vk+1, vi+2} such that vvk ∈ E(G), then the cycle

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C′ ={

xPvvkPyvk−1Pv+x if v ∈ P [x, vi1−1] ∪ P [vs+2, vi2−1],xPvk−1yPvvkPv−x if v ∈ P(vk+1, vi3 ] ∪ P(vi+2, vi4+1]

have length 2µ, a contradiction. Thus N(vk) �⊆ (N [x]∪{vi4+1})\{vi1 , vi2 , vk+1, vi+2}.Hence N(vk) ⊆ (N [y] ∪ {vk+1, vi+2}) \ {vk, vi3 , vi4 , vi4+1}. Thus d(vk) < µ, a con-tradiction.

Case 3. |N(x) ∩ N(y) ∩ P(x, x2]| = 1 and |N(x) ∩ N(y) ∩ P [y1, y)| = 2.Set vi1 ∈ N(x) ∩ N(y) ∩ P(x, x2], vi2 , vi3 ∈ N(x) ∩ N(y) ∩ P [y1, y) with i2 < i3.

Then by (N), i2 ≤ k, i3 ≥ k + 1. By Assumption 2, N [x] ⊆ P [x, vi3 ] and N [y] ⊆P [vi1 , y].

Claim 1. |P [vs+1, vr−1]| ≤ µ − 2.

Proof. If |P [vs+1, vr−1]| ≥ µ − 1, then the cycle xPvi3x would have length ≥ µ −1 + µ + 1 = 2µ, a contradiction. �

Claim 2. (1) N(u) ⊆ V (P ) for u ∈ (N [x] ∪ N [y]) \ {vi1 , vi2 , vi3};(2) E(P [x, vs−1], P [vs+1, vr−1]) = ∅ and E(P [vr+1, y] \ {vk+1}, P [vs+1,

vr−1]) = ∅;(3) 〈N [x] \ {vi1 , vi2 , vi3}〉G and 〈N [y] \ {vi1 , vi2 , vi3 , vs}〉G are complete subgraphs.

Proof. By Note 1 and N(x) ∪ N(y) ⊆ V (C), N(x) = P(x, vi1 ] ∪ P [vr , vi2 ] ∪P [vk+1, vi3 ] and N(y) = P [vi1 , vs ] ∪ P [vi2 , vk] ∪ P [vi3 , y). Thus, (1) holds by thechoice of P .

If there exist u ∈ N [x] \ {vk+1, vr} and v ∈ P [vs+1, vr−1] such that uv ∈ E(G),then the cycle

C′ =

xPuvPyvsPu+x if u ∈ P [x, vi1),

xPvi1vPvi1+1yPvrx if u = vi1 and i1 �= s,

xPvsyPuvPu−x if u ∈ P [vr+1, vi2 ] ∪ P(vk+1, vi3 ]

would have length ≥ |V (C)|+|{v}| = 2µ, a contradiction. Hence E(N [x]\{vk+1, vr},P [vs+1, vr−1]) = ∅ and similarly, E(N [y] \ {vs}, P [vs+1, vr−1]) = ∅. Thus, (2) holds.It is easy to check that E(N [x] \ {vi1 , vi2 , vi3}, N [y] \ {vi1 , vi2 , vi3 , vs}) \ {e} = ∅ andN(vs) ∩ N [x] ⊆ {vi1 , vi2 , vi3}. Since d(x) = d(y) = µ, (3) holds by Lemma 1. �

Claim 3. If d(vr−1) ≥ µ or d(vs+1) ≥ µ, then N(P [vs+1, vr−1]) ⊆ P [vs, vr ]∪{vk+1},|P [vs+1, vr−1]| = µ − 2 and 〈P [vs+1, vr−1]〉G is a complete subgraph.

Proof. Assumed(vr−1) ≥ µ. Obviously,N(vr−1) ⊆ V (P ). By Claim 2(2),N [vr−1] ⊆P [vs, vr ] ∪ {vk+1}. Thus µ ≤ d(vr−1) ≤ |P [vs, vr−2] ∪ {vr , vk+1}| ≤ µ by Claim 1.Hence we have that d(vr−1) = µ, P [vs, vr ]∪{vk+1} = N [vr−1] and |P [vs+1, vr−1]| =µ−2. For each u ∈ P [vs+2, vr−2], the path uPvr−1u

−PxvrPy and Claim 2(2) showthat N [u] ⊆ P [vs, vr ] ∪ {vk+1}. So N(P [vs+1, vr−1]) ⊆ P [vs, vr ] ∪ {vk+1} and byLemma 1, 〈P [vs+1, vr−1]〉G is a complete subgraph. �

Fan-Type Theorem for Long Cycles Containing a Specified Edge 497

Claim 4. i1 = s, i2 = k = r, i3 = k + 1.

Proof. Assume i2 �= k. Since 〈N [y] \ {vi1 , vi2 , vi3 , vs}〉G is a complete subgraphby Claim 2(3), the cycle xPvi2yPvi3+1vi2+1Pvi3x would be a contradiction. Hencei2 = k. Similarly, we have s − 1 ≤ i1 ≤ s, k − 1 ≤ r ≤ k and k + 1 ≤ i3 ≤ k + 2.

We first show that Claim 4 holds if vi1 = x+ or vi3 = y−. Assume, withoutloss of generality, that vi3 = y−. By Note 1, i2 = k and vi3 = y−, we haveN(y) = P [vi1 , vs ] ∪ {vi2 , vi3}. If i1 < s, then the cycle xPvs−1yvsPy−x wouldbe longer than C, a contradiction. Thus i1 = s and N(y) = {vi1 , vi2 , vi3} whichimplies µ = 3. By Assumption 1, d(x) = µ = 3. By Assumptions 1 and 2, wehave {vi1 , vr , vi2 , vi3 , vk+1} ⊆ N(x). Since d(x) = 3, k − 1 ≤ r ≤ k = i2 andk + 1 ≤ i3 ≤ k + 2, there must be i1 = s, i2 = r = k and i3 = k + 1, that is, Claim 4holds.

Hence we can assume that vi1 �= x+ and vi3 �= y−.In the following proof, we always use u, v to denote arbitrary vertices of P(x, vi1)

and P(vi3 , y), respectively. By Claim 2(1), N(u) ∪ N(v) ⊆ V (P ). Denote P1 =xPu−vi1−1Pu and P2 = vPvi3+1v

+Py.(1) Suppose i1 = s − 1 and r = k − 1. By Assumption 2 and Claim 2(2),

N [u] ⊆ P [x, vs ] ∪ P [vk−1, vi3 ] and N [v] ⊆ P [vk, y] ∪ {vs−1, vs}. Now we show thatuvs, uvk−1, uvk, vvs−1, vvs /∈ E(G), and when i3 = k + 2, uvk+1 /∈ E(G).

We consider the following cycles:

C1 = xP1uvsPyvs−1x, if uvs ∈ E(G);C2 = xP1uvk−1Pvs−1yPvkx, if uvk−1 ∈ E(G);C3 = xP1uvkPyvs−1Pvk−1x, if uvk ∈ E(G);C4 = xP1uvk+1Pvs−1yPvk+2x, if i3 = k + 2 and uvk+1 ∈ E(G);C5 = xPvs−1vP2yvsP vi3x, if vvs−1 ∈ E(G);C6 = xPvs−1yP2vvsP vi3x, if vvs ∈ E(G).

It is easy to see that the length of cycle Ci(1 ≤ i ≤ 6) is longer than C, a contra-diction. Hence N [u] ⊆ P [x, vs−1] ∪ {vi3} ⊆ N [x] \ {vk−1, vk} and N [v] ⊆ N [y] \{vs−1, vs}. Thus d(u) < µ, d(v) < µ and then uvi3 /∈ E(G) by vvi3 ∈ E(G), wherechoose v = vi3+1. But G − {x, vs−1} is not connected, a contradiction.

(2) Suppose i1 = s and r = k−1. The cycles C2, C3 and C4 in (1) (replacing vs−1by vs) show that N [u] ⊆ P [x, vs ]∪{vi3}. Thus d(u) < µ. Since G is 3-connected, thereexists u0 ∈ P(x, vi1) such that u0vi3 ∈ E(G). The cycle xP1u0vi3PvkvP2yvsP vk−1x

shows vvk /∈ E(G). Thus N [v] ⊆ N [y] \ {vk} and hance d(v) < µ. By the arbitrari-ness of u and v, d(u0) < µ and d(vi3+1) < µ, a contradiction with d(u0, vi3+1) = 2.

(3) Suppose i1 = s − 1 and r = k. The cycles C5 and C6 in (1) show that N [v] ⊆P [vk, y]. If i3 = k+2, then vvk+1 /∈ E(G) (otherwise, the cycle xPvk+1vPyv−Pvi3x

would be longer than C, a contradiction). Thus N [v] ⊆ N [y] \ {vs, vs−1}. Henced(v) < µ. Choose v = v+

i3. Since v+

i3y ∈ E(G), d(vs, v

+i3

) = 2 and then d(vs) ≥ µ.It is easy to check N [vs ] ⊆ P [vs−1, vk+1] ∪ {y}. Thus |P [vs−1, vk+1]| ≥ µ. If|P [vs−1, vk+1]| ≥ µ + 1, then the cycle xPvs−1yvsP vi3x would have length ≥ 2µ,a contradiction. Hence |P [vs−1, vk+1]| = µ and N [vs ] = P [vs−1, vk+1] ∪ {y}. Letw be arbitrary vertex of P [vs+1, vk−1]. By P [vs−1, vk+1] ⊆ N(vs) and P being thelongest path, N(w) ⊆ V (P ). By Claim 2(2), N [w] ⊆ P [vs, vk+1]. Thus d(w) < µ.

498 M. Lu et al.

Since v+i3

vi3 ∈ E(G) and d(v+i3

) < µ, wvi3 �∈ E(G). If i3 = k +2 and wvk+1 ∈ E(G),

then the cycle xPwvk+1vkyPvi3x is longer than C, a contradiction, and hencewvk+1 /∈ E(G) if i3 = k + 2. Hence N [w] ⊆ P [vs, vk]. Thus G − {vs, vk} is notconnected, a contradiction.

By (1), (2) and (3), i1 = s and r = k.(4) Suppose i3 = k+2. Then cycles xP1uvk+2PyvsPvk+1x and C4 in (1) (replac-

ing vs−1 by vs) show that N [u] ⊆ P [x, vs ]∪{vk}. Thus d(u) < µ and then d(vs+1) ≥µ. By Claim 3, N(P [vs+1, vk−1]) ⊆ P [vs, vk+1]. By the same argument as (3), wecan show that N(P [vs+1, vk−1]) ⊆ P [vs, vk]. Thus G − {vs, vk} is not connected. Soi3 = k + 1. �

Claim 5. If vs+1vk−1 ∈ E(G) and d(vk−1) < µ, then 〈P [vs+1, vk−1]〉G is a completesubgraph and |P [vs+1, vk−1]| < µ − 2.

Proof. Since vs+1vk−1 ∈ E(G), N(vs+2) ⊆ V (P ). Since vsy, xvk ∈ E(G), wehave N(vs+1) ∪ N(vk−1) ⊆ V (P ). By Claim 2(2), N [vs+1] ∪ N [vk−1] ∪ N [vs+2] ⊆P [vs, vk+1]. If vs+2vk−1 /∈ E(G), then d(vs+2) ≥ µ by d(vk−1) < µ andd(vk−1, vs+2) = 2. But by Claim 1, d(vs+2) ≤ |P [vs, vk+1]|−2 ≤ µ−1, a contradic-tion. Thus vs+2vk−1 ∈ E(G). Similarly, we can show that P [vs+1, vk−2] ⊆ N(vk−1)

and N(P [vs+1, vk−1]) ⊆ P [vs, vk+1]. Thus |P [vs+1, vk−1]| < µ − 2 by d(vk−1) < µ.Since |P [vs, vk+1]| ≤ µ + 1, we have 〈P [vs+1, vk−1]〉G is a complete subgraph byLemma 1. �

Claim 6. If d(vk−1) < µ and d(vs+1) < µ, then vk−1vs+1 ∈ E(G).

Proof. Note that N [vk−1] ∪ N [vs+1] ⊆ P [vs, vk+1]. Assume vk−1vs+1 /∈ E(G).Denote i∗ = min{i > s : vivk−1 ∈ E(G)}, j∗ = max{j < k : vjvs+1 ∈ E(G)}.Then i∗ �= s + 1 and j∗ �= k − 1. By an argument similar to that proof ofClaim 5, P [vi∗ , vk−2] ⊆ N(vk−1), P [vs+2, vj∗ ] ⊆ N(vs+1) and N(P (vi∗ , vk−2]) ∪N(P [vs+2, vj∗)) ⊆ P [vs, vk+1]. If i∗ ≤ j∗, then vj∗vk−1 ∈ E(G). This impliesd(vs+1, vk−1) = 2, which contradicts with d(vk−1) < µ and d(vs+1) < µ. Hencei∗ > j∗. By Claim 1 and i∗ > j∗, d(u) < µ for each u ∈ P [vs+1, vj∗−1] ∪P [vi∗+1, vk−1]. Thus E(P [vs+1, vj∗), P (vi∗ , vk−1]) = ∅.

Now, we will claim that N(P [vs+1, vj∗−1]) ∩ P [vj∗+1, vi∗ ] ⊆ {vj∗+1}. Assumethere exist u ∈ P [vs+1, vj∗−1] and v ∈ P [vj∗+2, vi∗ ] such that uv ∈ E(G). Then u �=vs+1 by the choice of j∗. The path v−Pu+vs+1PuvPyvsPx shows N(v−) ⊆ V (P ).By Claim 2(2), N(v−) ⊆ P [vs, vk+1]. By Claim 1 and v−vk−1 /∈ E(G), d(v−) < µ

and then v−u ∈ E(G). But in the case, d(vs+1, v−) = 2, d(vs+1) < µ and d(v−) < µ,

a contradiction. Hence N(P [vs+1, vj∗−1]) ∩ P [vj∗+1, vi∗ ] ⊆ {vj∗+1} and similarly,N(P [vi∗+1, vk−1]) ∩ P [vj∗ , vi∗−1] ⊆ {vi∗−1}.

Next, we will show that if there exists u ∈ P [vs+1, vj∗−1] such that uvj∗+1 ∈E(G), then N(vj∗) ⊆ P [vs, vj∗+1] ∪ {vk, vk+1}. By the path vj∗Pu+vs+1Puvj∗+1

PyvsPx and Claim 2(2), N(vj∗) ⊆ P [vs, vk+1] \ {vk−1}. Then d(vj∗) < µ byClaim 1. If there exists v ∈ P [vj∗+2, vk−1] such that vj∗v ∈ E(G), then v /∈P(vi∗ , vk−1] (otherwise, d(vs+1, v) = 2, but d(vs+1) < µ, d(v) < µ) and the path

Fan-Type Theorem for Long Cycles Containing a Specified Edge 499

v−Pvj∗+1uPvs+1u+Pvj∗vPyvsPx and Claim 2(2) show N(v−) ⊆ P [vs, vk+1]. By

v−vs+1 /∈ E(G), d(v−) < µ. Since d(vj∗) < µ, v−vj∗ ∈ E(G). But in the case,d(vs+1, v

−) = 2, d(vs+1) < µ and d(v−) < µ, a contradiction. Similarly, if thereexists v ∈ P [vi∗+1, vk−1] such that vvi∗−1 ∈ E(G), then N(vi∗) ⊆ P [vi∗−1, vk+1] ∪{vs}.

Finally, we will complete the proof of Claim 6. If there exists u ∈ N(vs) ∩P [vi∗+1, vk−1], then d(vs+1, u) = 2, but d(vs+1) < µ and d(u) < µ, a contra-diction. Hence N(vs) ∩ P [vi∗+1, vk−1] = ∅. Similarly, N(vk) ∩ P [vs+1, vj∗−1] =∅. If N(vk+1) ∩ P(vi∗ , vk−1] �= ∅, then N(vk+1) ∩ P [vs+1, vj∗) = ∅. Thus whenN(u) ∩ P [vj∗+1, vi∗ ] = ∅ for any u ∈ P [vs+1, vj∗−1], G − {vs, vj∗} has at leasttwo components, a contradiction. If N(u) ∩ P [vj∗+1, vi∗ ] = {vj∗+1} for some u ∈P [vs+1, vj∗−1], then d(vj∗) < µ by the proof above. Thus vkvj∗ , vk+1vj∗ /∈ E(G).Hence G−{vs, vj∗+1} is non-connected, a contradiction. If N(vk+1)∩P(vi∗ , vk−1] =∅, then we can similarly show that G−{vi∗ , vk} or G−{vi∗−1, vk} is non-connected,a contradiction. Hence Claim 6 holds. �

By Claims 2(3), 3, 4, 5 and 6, we have 〈N [x] \ {vs, vk, vk+1}〉G, 〈N [y] \ {vs, vk,

vk+1}〉G and 〈P [vs+1, vk−1]〉G are complete subgraphs. Thus N(u) ⊆ V (P ) for anyu ∈ V (P ) \ {vs, vk, vk+1} by the choice of P . Let H be a component of G \ V (P ).Then NP (V (H)) = {vs, vk, vk+1}.

Claim 7. H is a complete subgraph and |V (H)| ≤ µ − 2

Proof. Suppose that H is not a complete subgraph. Set H ∗ = 〈V (H)∪ {vs, vk}〉G +vsvk. Then H ∗ is 2-connected, but it is not a complete subgraph. By Lemma 4, thereexists a path Q connecting vs and vk of length at least µ − 1 in H ∗. Choose Q suchthat the length of Q is as long as possible. Let Q = vsQ

∗vk, where V (Q∗) ⊆ V (H)

and denote Q∗ = u1u2 · · · uh. Then h ≥ µ − 2.If d(vk−1) < µ and d(vs+1) < µ, then 〈P [vs+1, vk−1]〉G is a complete subgraph

and |P [vs+1, vk−1]| < µ − 2 by Claims 6 and 5. Replacing P [vs+1, vk−1] by Q∗, wecan have a longer path containing e than P , a contradiction.

Hence we can assume that d(vk−1) ≥ µ or d(vs+1) ≥ µ.By Claim 3, 〈P [vs+1, vk−1]〉G is a complete subgraph and |P [vs+1, vk−1]| =

µ − 2. Set P ′ = v0PvsQ∗vkPvt . By the choice of P , |V (P )| = |V (P ′)| and then

|V (Q∗)| = µ − 2. By the same argument as Claims 3, 5 and 6, we can show that〈V (Q∗)〉G is a complete subgraph. Since H is not a complete subgraph, there existsu ∈ V (H) \ V (Q∗) such that NQ∗(u) �= ∅.

By G being 3-connected and NP (V (H)) = {vs, vk, vk+1}, u can be joined tovs and vk by two internally-disjoint paths P1 = P1[u, vs ] and P2 = P2[u, vk] with(V (P1) ∪ V (P2)) \ {vs, vk} ⊆ V (H). Assume that V (Pj ) ∩ V (Q∗) �= ∅ for j = 1, 2.Denote by xj the first vertex in V (Pj )∩V (Q∗) along Pj . Assume, without loss of gen-erality, thatx1 = up, x2 = uq(p < q). Then the pathvsu1Q

∗upP1uP2uqQ∗up+1uq+1

Q∗uhvk would be longer than Q, a contradiction. Thus we have V (P1)∩V (Q∗) = ∅or V (P2)∩V (Q∗) = ∅. Assume, without loss of generality, that V (P1)∩V (Q∗) = ∅.Let ui ∈ NQ∗(u). Then the path vsP1uuiQ

∗u1ui+1Q

∗uhvk would be longer than Q

in H ∗, a contradiction.

500 M. Lu et al.

Thus H is a complete subgraph and then |V (H)| ≤ |P [vs+1, vk−1]| = µ − 2. �

Let H1, . . . , Hl (l ≥ 1) be the components of G \ V (P ). Then Hi is a completesubgraph with |V (Hi)| ≤ µ − 2 by Claim 7. Assume |V (H1)| ≤ · · · ≤ |V (Hl)|. Ifthere exist Hi and Hj (i �= j ) such that |V (Hi)| < µ − 2 and |V (Hj )| < µ − 2, thend(u) < µ for all u ∈ V (Hi) ∪ V (Hj ). Let ui ∈ NHi

(vs) and uj ∈ NHj(vs). Then

d(ui, uj ) = 2 but d(ui) < µ and d(uj ) < µ, a contradiction. Hence |V (H1)| ≤µ − 2 and |V (Hi)| = µ − 2 for 2 ≤ i ≤ l if l ≥ 2. Thus G is a spanning sub-graph of K3 + (lKµ−2 ∪ Ks), where n = l(µ − 2) + s + 3, 1 ≤ s ≤ µ − 2 andV (K3) = {vs, vk, vk+1}. �

The proof of the following lemma contains in the proof of Theorem 16 in [5].

Lemma 7 [5]. Let P = P [x, y] be a path that satisfies the condition of Lemma 3 andP(G; P [x, ym], e) �= ∅.

(1) If |P [y+m, y]| ≥ 2, then |NP [x] ∪ NP [y]| ≤ ce(G).

(2) If |NP [x] ∪ NP [y]| ≤ ce(G), then ce(G) ≥ min{|V (G)|, d(x) + d(y)}.

3. Proof of Theorem 6

Let G be a 3-connected noncomplete graph, e ∈ E(G) and P = P [x, y] be a pathin G that satisfies the condition of Lemma 3. Set P = v0v1 · · · vt , where v0 = x,vkvk+1 = e and vt = y. It follows from Lemma 5 that P(G; P [v0, vc], e) �= ∅ forsome vertex vc of V (P ). Choose vc so as to maximize c. By Lemma 6, we canassume that c < t . Assume that ce(G) < min{|V (G)|, 2µ}. Then by the choice ofP , N(x) ∪ N(y) ⊆ V (P ).

Let P = {Pi [xi, yi ] : 1 ≤ i ≤ m} be a subvine on P [v0, vc] in G with the endvertex ym = vc. If |P [y+

m, y]| ≥ 2, then ce(G) ≥ d(x) + d(y) by Lemma 7. Thusce(G) ≥ 2µ by Lemma 3, a contradiction. Hence we have vc+1 = vt . Consider thecycle C = C(P, P) and assume that P is chosen so that

(1) e ∈ E(C) ∩ E(P [v0, vc]);(2) subject to (1), E(C) ∩ E(P [v0, vc]) is maximal.

Set y0 = x1 = v0 = x and xm+1 = ym = vc. Then E(C) ∩ E(P [v0, vc]) =∪m

i=1E(P [yi−1, xi+1]). Since e = vkvk+1 ∈ E(C) ∩ E(P [v0, vc]), there exists aninteger r, 1 ≤ r ≤ m, such that e ∈ E(P [yr−1, xr+1]). Set vi0 = yr−1 if N(v0) ∩P [yr−1, vk] = ∅; otherwise, set vi0 = vp, where p = max{i ≤ k : vi ∈ N(v0)}.Denote W = P [vi0 , vk]. Then we have

Fact 1 [5]. N [v0] = N [v0] ∩ P [v0, vc] ⊆ V (C).

Fact 2 [5]. V (C) ∩ P [vc+1, vt ] = ∅ and N [v0] ∩ Int (W) = ∅.

Fact 3 [5]. If R := R[v, vt ] is a path in G−vc connecting (V (C)\{vc})∪P [v0, vc−1]and vc+1 = vt , then v ∈ V (W).

By the choice of P and Fact 3, N(vt ) ⊆ V (W) ∪ {vc}. Since V (W) ∪ {vc} ⊆V (C), we have N [v0] ∪ N(vt ) ⊆ V (C), i.e., N [x] ∪ N(y) ⊆ V (C). Since G is

Fan-Type Theorem for Long Cycles Containing a Specified Edge 501

3-connected, |N(vt ) ∩ V (W)| ≥ 2. If there exist two vertices vj , vj+1 ∈ W such thatvjvt , vj+1vt ∈ E(G), then we have a cycle C′ = (C − {vjvj+1}) + {vjvt , vj+1vt }.Obviously, N [v0] ∪ N [vt ] ⊆ V (C′). Hence ce(G) ≥ |V (C′)| ≥ |N [v0] ∪ N [vt ]| =|N [x]∪N [y]|, a contradiction with Lemmas 7(2) and 3. Thus we have |Int (W)| = 1but vi0+1vt /∈ E(G) or |Int (W)| ≥ 2 and for any v ∈ Int (W) ∩ N(vt ), v+ /∈ N(vt )

or v− /∈ N(vt ). Then there exists a vertex v ∈ Int (W) such that v /∈ N(v0) ∪ N(vt ).Thus ce(G) ≥ |V (C)| ≥ |N [v0] ∪ N(vt )| + 1 = |N [x] ∪ N [y]|. By Lemmas 7(2) and3, ce(G) ≥ 2µ, a contradiction. This completes the proof of our Theorem.

Acknowledgements. Many thanks to the anonymous referees for their many helpful com-ments and suggestions, which have considerably improved the presentation of the paper.

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Received: October 10, 2002Final version received: April 4, 2005