Two sufficient conditions for dominating cycles

Two Sufficient Conditionsfor Dominating Cycles

Mei Lu,1 Huiqing Liu,2 and Feng Tian2

1DEPARTMENT OF MATHEMATICAL SCIENCES

TSINGHUA UNIVERSITY

BEIJING 100084, CHINA

E-mail: [email protected]

2INSTITUTE OF SYSTEMS SCIENCE

ACADEMY OF MATHEMATICS AND SYSTEMS SCIENCE

CHINESE ACADEMY OF SCIENCES

BEIJING 100080, CHINA

E-mail: [email protected] and [email protected]

Received April 8, 2002; Revised August 31, 2004

Published online in Wiley InterScience(www.interscience.wiley.com).

DOI 10.1002/jgt.20070

Abstract: A cycle C of a graph G is dominating if each component of GnCis edgeless. In the paper, we will give two sufficient conditions for eachlongest cycle of a 3-connected graph to be a dominating cycle.� 2005 Wiley Periodicals, Inc. J Graph Theory 49: 135–150, 2005

Keywords: dominating cycle; connectivity; insertible vertex

1. INTRODUCTION

We use Bondy and Murty [4] for terminology and notation not defined here and

consider simple undirected graphs only. Let G be a graph. If S � VðGÞ, then

NðSÞ denotes the neighbors of S. For a subgraph H of G and S � VðGÞ, let

——————————————————

Contract grant sponsor: NNSFC (to M.L.); Contract grant number: 60172005;Contract grant sponsor: NNSFC.

� 2005 Wiley Periodicals, Inc.

135

NHðSÞ ¼ NðSÞ \ VðHÞ. When H ¼ G, we write NðSÞ instead of NGðSÞ. If

S ¼ fs1; s2; . . . ; slg, then NHðSÞ is written as NHðs1; s2; . . . ; slÞ. For a graph G,

we denote by �ðGÞ and �ðGÞ the independence number and the connectivity of G,

respectively. We define �kðGÞ by the minimum value of the degree sum of any k

independent vertices of G if k � �ðGÞ; if k > �ðGÞ, we set �kðGÞ ¼ þ1.

A cycle C of G is said to be a D�-cycle, if jHj < � for any component H of

GnC. Obviously, a D1-cycle is a hamiltonian cycle and a D2-cycle is a dominating

cycle. For a graph G, we denote by pðGÞ and cðGÞ the order of a longest path and

the order of a longest cycle in G, respectively. Let diffðGÞ ¼ pðGÞ � cðGÞ. Then a

connected graph G is hamiltonian if and only if diffðGÞ ¼ 0 and if diffðGÞ � 1,

then each longest cycle of G is a dominating cycle.

Bondy [3] gives the following sufficient condition for each longest cycle of a

2-connected graph G to be a dominating cycle.

Theorem 1 [3]. Let G be a 2-connected graph of order n � 3 with �3ðGÞ �nþ 2. Then each longest cycle of G is a dominating cycle.

A much stronger result is given in [6].

Theorem 2 [6]. A 2-connected graph G of order n with �3ðGÞ � nþ 2 satisfies

diffðGÞ � 1.

When involved in connectivity, we have the following theorem.

Theorem 3 [2]. Let G be a 2-connected graph on n vertices with �3ðGÞ �nþ �. Then G is hamiltonian.

A short proof of Theorem 3 was given in [10]. In [8], a problem proposed by

Tian in [9] is solved and a related conjecture is proposed.

Theorem 4 [8]. Let G be a 3-connected graph of order n � 3 with

�4ðGÞ � nþ 2�. Then G contains a longest cycle which is a dominating cycle.

Conjecture A. Let G be a k-connected graph of order n with �kþ1ðGÞ �nþ ðk � 1Þ�. Then each longest cycle of G is a Dk�1-cycle.

In the paper, we show that under the condition of Theorem 4, each longest

cycle is a dominating cycle.

Theorem 5. Let G be a 3-connected graph of order n � 3 with �4ðGÞ � nþ 2�.Then each longest cycle of G is a dominating cycle.

Theorems 3 and 5 show that Conjecture A holds when k ¼ 2 and 3.

Li et al. proposed a conjecture about the difference diffðGÞ as follows.

Conjecture B [7]. Let G be a 3-connected graph of order n. If �4ðGÞ � 43nþ 5

3,

then diffðGÞ � 1.

And we will propose a problem also about the difference diffðGÞ and is

supported by Theorem 5.

136 JOURNAL OF GRAPH THEORY

Problem C. Let G be a 3-connected graph of order n � 3 with

�4ðGÞ � nþ 2�. Is it true that diffðGÞ � 1?

The following class of graphs shows the sharpness of Conjecture B. Let

G ¼ kK1 þ ðk þ 1ÞK2, (k � 3). Then jGj ¼ n ¼ 3k þ 2 and �4ðGÞ ¼ 4ðk þ 1Þ ¼43nþ 4

3, but diffðGÞ ¼ 2.

In the paper, we obtain the following result which supports Conjecture B.

Theorem 6. Let G be a 3-connected graph of order n � 13. If �4ðGÞ � 43nþ 5

3,

then each longest cycle of G is a dominating cycle.

The class of graphs described above also shows that Theorem 6 is best

possible.

Next, we will use two examples to show that the conditions of Theorems 5 and

6 are independent.

Example 1. Let G ¼ kK1 þ ðkK2 [ K1Þ with k � 3. Then n ¼ 3k þ 1, �ðGÞ ¼ k

and �4ðGÞ ¼ 43nþ 5

3. Hence by Theorem 6 each longest cycle of G is a

dominating cycle. But nþ 2� ¼ 5n�23

, that is, the conditions of Theorem 5 are not

satisfied.

Example 2. Let F ¼ ðG1 þ G2Þ þ Kc2r with G1 ¼ G2 ¼ Kr; where r � 3. The

graph G is defined as: VðGÞ ¼ VðFÞ [ VðK2rÞ and EðGÞ ¼ EðFÞ [ EðK2rÞ [fuv : u 2 VðK2rÞ; v 2 VðG1Þg. It is easy to check that n ¼ 6r, �ðGÞ ¼ r and

�4ðGÞ ¼ 8r. Obviously, �4ðGÞ ¼ nþ 2�. Hence by Theorem 5 each longest cycle

of G is a dominating cycle. But �4ðGÞ ¼ 4n3< 4nþ5

3. That is, the conditions of

Theorem 6 are not satisfied.

Let C ¼ v1v2 � � � vtv1 be a cycle of G with a given orientation. For vi 2 VðCÞ,we use vþi ; v

�i to denote the successor and predecessor of vi on C, respectively.

The i-th successor and i-th predecessor of a vertex v on C are denoted by vþi and

v�i, respectively. If A � VðCÞ, then A� and Aþ are the sets fv� : v 2 Ag and

fvþ : v 2 Ag, respectively. If u; v 2 VðCÞ, we denote by uCv the subpath

uuþ � � � v�v of C. The same subpath, in reverse order, is denoted by vCu. We will

consider uCv and vCu both as paths and as vertex sets. We also use analogous

notations for a path P in G.

2. LEMMAS

In this section, we give some lemmas which will be used in Sections 3 and 4.

Lemma 1 [11]. Let G be a 3-connected graph of order n � 3 with �4 � nþ 6

and C be a longest cycle of G. Then C is a D3-cycle.

Next, we assume that G is a 3-connected nonhamiltonian graph. Let C be a

longest cycle of G with a given orientation and H a component of GnC. Assume

that VðHÞ ¼ fu; vg. Suppose that NCðHÞ ¼ fx1; x2; . . . ; xtg, and the occurrence

CONDITIONS FOR DOMINATING CYCLES 137

of the vertices on C agrees with the given orientation of C. A vertex u 2 xþi Cx�iþ1

is insertible if there exist vertices v; vþ 2 xiþ1Cxi such that uv; uvþ 2 EðGÞ;then vvþ 2 EðGÞ is an insertion edge of u. We denote the set of insertion

edges of u by IðuÞ, and denote the first noninsertible vertex occurring on xþi Cx�iþ1

by ui. Let Ui ¼ xþi Cui, T ¼ fu1; u2; . . . ; utg, Tu ¼ fui : xi 2 NCðuÞg and Tv ¼fuj : xj 2 NCðvÞg. Then we have the following lemmas.

Lemma 2 [1][5]. (1) ui exists.

(2) ui 62 NðuÞ [ NðvÞ for any ui 2 T.

(3) There is no path whose internal vertices (if any) in GnC joining a vertex

of Ui and a vertex of Uj, and IðxÞ \ IðyÞ ¼ ; for all x 2 xþi Cu�i and

y 2 xþj Cu�j . In particular, GnT is connected.

(4) For any ui; uj 2 T ; ujw 62 EðGÞ whenever uiw� 2 EðGÞ and w 2 uþj Cxi.

By using the proof of Lemma 2 ([1]), we can show the following lemma

similarly.

Lemma 3. Let ui 2 Tu and uj 2 Tv. Then ð1Þ uiuþj 62 EðGÞ and uþi uj 62 EðGÞ;

ð2Þ ujw 62 EðGÞ for w 2 uþj Cxi and uiw�2 2 EðGÞ; ð3Þ uiw 62 EðGÞ for w 2 uþi Cxj

and ujw�2 2 EðGÞ.

Proof. We only give the proof of (2), and parts (1) and (3) can be proved in a

similar way.

(2) Assume ujw 2 EðGÞ for some w 2 uþj Cxi and uiw�2 2 EðGÞ. By Lemma 2,

ww�;w�w�2 62 IðuÞ for each u 2 ðxþi Cu�i Þ [ ðxþj Cu�j Þ. Then we can insert the

vertices of ðxþi Cu�i Þ [ ðxþj Cu�j Þ in the cycle xiCwujCw�2uiCxjvuxi to derive a

contradiction with the choice of C. &

Lemma 4. If ui 2 Tu ðui 2 Tv , resp.Þ and uj; uk 2 Tv ðuj; uk 2 Tu, resp.Þ, then

dðuiÞ þ dðujÞ þ dðukÞ � n� jW j þ 1;

where W � ðNðuÞ \ NðvÞÞnNðui; uj; ukÞ.

Proof. Set C1 ¼ uiCu�j , C2 ¼ ujCu

�k , C3 ¼ ukCu

�i , and Wh ¼ W \ Ch,

1 � h � 3. Since W�ðNðuÞ\ NðvÞÞnNðui; uj; ukÞ, fw;wþ;wþ2g\ Nðui; uj; ukÞ ¼; for each w 2 W by Lemmas 2 and 3.

(1) N�2C1ðuiÞ [ NC1

ðujÞ [ N�C1ðukÞ � ðC1nW1Þ [ fu�i g, and N�2

C1ðuiÞ, NC1

ðujÞ and

N�C1ðukÞ are pairwise disjoint by Lemmas 2 and 3. Hence

dC1ðuiÞ þ dC1

ðujÞ þ dC1ðukÞ � jC1j � jW1j þ 1: ð2:1Þ

(2) NC3ðuiÞ [ N�

C3ðujÞ [ N�2

C3ðukÞ � ðC3nW3Þ [ fu�k g, and NC3

ðuiÞ, N�C3ðujÞ and

N�2C3ðukÞ are pairwise disjoint by Lemmas 2 and 3. Hence

dC3ðuiÞ þ dC3



(3) Now we want to show that

dC2ðuiÞ þ dC2


First we note that if uk 2 Tv \ Tu, then N�C2ðuiÞ [ N�2

C2ðujÞ [ NC2

ðukÞ �ðC2nW2Þ [ fu�j g, and N�2

C2ðujÞ, N�

C2ðuiÞ, and NC2

ðukÞ are pairwise disjoint by

Lemmas 2 and 3, and hence ð2:3Þ holds. So in the following proof, we assume

that uk 2 TvnTu.

Claim 2.1. xk =2 W2:

Proof. Since uk 2 TvnTu, we have xk 2 NðvÞnNðuÞ. Thus xk =2 W2: &

Set P ¼ uþj Cxk ¼ blbl�1 � � � b1. For each r, 1 � r � l, define AðrÞ ¼fb1; b2; . . . ; brg and W2ðrÞ ¼ W2 \ AðrÞ. Let U0

k ¼ ðUknfukgÞ [ fujg. Then

W2 ¼ W2ðlÞ and C2 ¼ AðlÞ [ U0k. By Lemma 2, NðuiÞ \ U0

k ¼ ; and

NðujÞ \ U0k ¼ ;. For each r, 1 � r � l, if

dAðrÞðuiÞ þ dAðrÞðujÞ þ dAðrÞðukÞ � jAðrÞj � jW2ðrÞj þ 2; ð2:4Þthen

dC2ðuiÞ þ dC2

ðujÞ þ dC2ðukÞ ¼ dAðlÞðuiÞ þ dAðlÞðujÞ þ dAðlÞðukÞ þ dU0

kðukÞ

� jAðlÞj � jW2ðlÞj þ 2 þ jU0kj � 1 ¼ jC2j � jW2j þ 1:

Hence, we just need to show that ð2:4Þ holds for each r, 1 � r � l.

Before giving the proof of (2.4), we point out the following facts.

Since b1 ¼ xk 2 NðvÞ, we get b2; b3 =2 NðuÞ, and hence we get

Fact 1. b2; b3 =2 W2, and hence if bt 2 W2, then t � 4.

By the definition of W and C being the longest cycle, we have

Fact 2. If bt 2 W2, then bt 62 Nðui; uj; ukÞ; btþ1; btþ2 =2 W2; and bt�2; bt�1 =2W2 [ Nðui; uj; ukÞ:

It is easy to check that (2.4) holds for r � 3; (in this case, jW2ðrÞj ¼ 0 by

Fact 1). Next we assume that ð2:4Þ is true for AðsÞ, where 3 � s � r < l. We will

prove that ð2:4Þ is true for Aðr þ 1Þ. Let J ¼ fui; uj; ukg.

Claim 2.2. fbrþ1; brg \W2 ¼ ;:Proof. (i) If brþ1 2 W2, (in this case, r � 3 by Fact 1), then jW2ðrÞj ¼

jW2ðr þ 1Þj � 1 and dJðbrþ1Þ ¼ 0. By induction,

dAðrþ1ÞðuiÞ þ dAðrþ1ÞðujÞ þ dAðrþ1ÞðukÞ¼ dAðrÞðuiÞ þ dAðrÞðujÞ þ dAðrÞðukÞ þ dJðbrþ1Þ� jAðrÞj � jW2ðrÞj þ 2 ¼ jAðr þ 1Þj � jW2ðr þ 1Þj þ 2;

and hence (2.4) holds for Aðr þ 1Þ.


(ii) If br 2 W2, (in this case, r � 4), then brþ1 =2 W2 and br�2; br�1 =2W2 [ Nðui; uj; ukÞ by Fact 2. Thus jW2ðr � 3Þj ¼ jW2ðr þ 1Þj � 1 andPrþ1

t¼r�2 dJðbtÞ ¼ dJðbrþ1Þ � 3: By induction,

dAðrþ1ÞðuiÞ þ dAðrþ1ÞðujÞ þ dAðrþ1ÞðukÞ¼ dAðr�3ÞðuiÞ þ dAðr�3ÞðujÞ þ dAðr�3ÞðukÞ þ dJðbrþ1Þ� jAðr � 3Þj � jW2ðr � 3Þj þ 5 ¼ jAðr þ 1Þj � jW2ðr þ 1Þj þ 2;

and hence (2.4) holds for Aðr þ 1Þ. &

Claim 2.3. dJðbrþ1Þ � 2.

Proof. Suppose dJðbrþ1Þ � 1. Then by Claim 2.2, jW2ðrÞj ¼ jW2ðr þ 1Þj. By

induction,

dAðrþ1ÞðuiÞ þ dAðrþ1ÞðujÞ þ dAðrþ1ÞðukÞ¼ dAðrÞðuiÞ þ dAðrÞðujÞ þ dAðrÞðukÞ þ dJðbrþ1Þ� jAðrÞj � jW2ðrÞj þ 3 ¼ jAðr þ 1Þj � jW2ðr þ 1Þj þ 2;

and hence (2.4) holds for Aðr þ 1Þ. &

Claim 2.4. br�1 =2 W2.

Proof. Suppose that br�1 2 W2, (in this case, r � 5). Noting that

jW2ðr � 3Þj ¼ jW2ðr þ 1Þj � 1 by Claim 2.2 andPrþ1

t¼r�2 dJðbtÞ ¼ dJðbrÞþdJðbrþ1Þ � 3 by the definition of uk; ui and C being the longest cycle, we get

that (2.4) holds for Aðr þ 1Þ by a similar argument as Claim 2.2. &

Now by Claim 2.3, we finish the proof by considering the following two cases.

Case 1. dJðbrþ1Þ ¼ 2.

In this case, we first have that brþ1uk =2 EðGÞ. Otherwise, by Lemma 2 and the

definition of uk, br 62 Nðui; uj; ukÞ. Thus jW2ðr � 1Þj ¼ jW2ðr þ 1Þj by Claim 2.2,

and hence ð2:4Þ holds by induction. Hence, we assume that brþ1 2 NðuiÞ \ NðujÞand then br 62 Nðui; ujÞ. Note that if br 2 NðukÞ, then br�1 =2 Nðui; uj; ukÞ. Thus

jW2ðr � tÞj ¼ jW2ðr þ 1Þj by Claims 2.2 and 2.4, where t ¼ 1 if br =2 NðukÞ, and

t ¼ 2 if br 2 NðukÞ, and ð2:4Þ holds for Aðr þ 1Þ by induction.

Case 2. dJðbrþ1Þ ¼ 3.

In this case, br =2 Nðui; uj; ukÞ and br�1 =2 Nðui; ujÞ by Lemma 3. If

br�1 =2 NðukÞ, then jW2ðr � 2Þj ¼ jW2ðr þ 1Þj by Claims 2.2 and 2.4, and hence

ð2:4Þ holds. On the other hand, if br�1 2 NðukÞ, then br�2 =2 Nðui; uj; ukÞ, and

hencePrþ1

t¼r�2 dJðbtÞ ¼ 4. We may assume that r � 4. Note that jW2ðr � tÞj ¼jW2ðr þ 1Þj � ðt � 3Þ, where t ¼ 3 if br�2 =2 W2 and t ¼ 4 if br�2 2 W2, (in this

case, r � 6). Thus ð2:4Þ holds for Aðr þ 1Þ by induction.


By (2.1), (2.2), (2.3) and noting dRðuiÞ þ dRðujÞ þ dRðukÞ � n� jCj � 2, we

have dðuiÞ þ dðujÞ þ dðukÞ � n� jW j þ 1: &

The proof of the following lemma is very trivial, so we will omit it here.

Lemma 5. Let P be a path and vi; vj 62 VðPÞ. If NPðviÞ \ N�P ðvjÞ ¼ ;, then

dPðviÞ þ dPðvjÞ � jPj þ 1.

3. PROOF OF THEOREM 6

Suppose Theorem 6 fails to hold, then there exists a longest cycle C in G which is

not a dominating cycle. Since n � 13, �4ðGÞ � 43nþ 5

3� nþ 6. By Lemma 1,

each component of GnC has at most two vertices. Let R ¼ GnC. Then we can

choose a component H of R with H ¼ fu; vg. Give the cycle C an orientation.

Suppose that NCðu; vÞ ¼ fx1; x2; . . . ; xtg, and the occurrence of the vertices on C

agrees with the given orientation of C. Denote the first noninsertible vertex

occurring on xþi Cx�iþ1 by ui. The other notations such as Ui; T ;Tu, and Tv are the

same as those in Section 2. By Lemma 2, we know that T [ fug and T [ fvg are

both independent sets of G. Since G is 3-connected, jTuj � 2, jTvj � 2 and

jTu [ Tvj � 3. Choose ui; uj; uk 2 T with ui 2 Tu and uj; uk 2 Tv. By Lemma 4

and letting W ¼ ;, we have the following claim.

Claim 3.1. dðuiÞ þ dðujÞ þ dðukÞ � nþ 1.

Now we have another claim as follows.

Claim 3.2. dðujÞ þ dðukÞ þ dðuÞ � nþ 1.

Proof. Let P1 ¼ uþj Cxk and P2 ¼ uþk Cxj. By Lemmas 2 and 3, we have

N�P1ðujÞ \ NP1

ðukÞ ¼ ; and uh =2 N�P1ðujÞ [ NP1

ðukÞ for uh 2 Tu \ P1. Similarly,

N�P2ðukÞ \ NP2

ðujÞ ¼ ; and uh =2 N�P2ðukÞ [ NP2

ðujÞ for uh 2 Tu \ P2. Hence, by

Lemmas 2 and 5,

dCðujÞ þ dCðukÞ ¼X2

l¼1

ðdPlðujÞ þ dPl

ðukÞÞ þ dUjðujÞ þ dUk

ðukÞ � dCðuÞ þ 2

� ðjP1j þ 1Þ þ ðjP2j þ 1Þ þ ðjUjj � 1Þ þ ðjUkj � 1Þ � dCðuÞ þ 2

¼ jCj � dCðuÞ þ 2:

Since dRðujÞ þ dRðukÞ þ dRðuÞ � n� jCj � 1, Claim 3.2 holds. &

Claim 3.3. dðuiÞ þ dðujÞ þ dðuÞ � nþ 1.

Proof. Let P1 ¼ uþi Cxj and P2 ¼ uþj Cxi. By Lemma 2, we have

N�P1ðuiÞ \ NP1

ðujÞ ¼ ;. If uþh 2 NðuiÞ for some uh 2 Tu \ P1, then uþ2h =2 NðuiÞ

by the definition of ui. By Lemma 3, uþh =2 NðujÞ. Hence, we have that


uh 62 N�P1ðuiÞ [ NP1

ðujÞ when uþh =2 NðuiÞ and uþh 62 N�P1ðuiÞ [ NP1

ðujÞ when

uþh 2 NðuiÞ. On the other hand, by Lemmas 2 and 3, N�P2ðujÞ \ NP2

ðuiÞ ¼ ; and

uh 62 N�P2ðujÞ [ NP2

ðuiÞ for uh 2 Tu \ P2. Hence, by Lemmas 2 and 5, we have that

dCðuiÞ þ dCðujÞ ¼X2

l¼1

ðdPlðuiÞ þ dPl

ðujÞÞ þ dUiðuiÞ þ dUj

ðujÞ � dCðuÞ þ 2

� ðjP1j þ 1Þ þ ðjP2j þ 1Þ þ ðjUij � 1Þ þ ðjUjj � 1Þ � dCðuÞ þ 2

¼ jCj � dCðuÞ þ 2:

Since dRðuiÞ þ dRðujÞ þ dRðuÞ � n� jCj � 1, Claim 3.3 holds. &

By a similar argument as Claim 3.3, we obtain

Claim 3.4. dðuiÞ þ dðukÞ þ dðuÞ � nþ 1.

Since fu; ui; uj; ukg is an independent set and by Claims 3.1–3.4, we have

dðuiÞ þ dðujÞ þ dðukÞ þ dðuÞ � 4

3nþ 4

3;

a contradiction. &

4. PROOF OF THEOREM 5

The main ideas of the proof of Theorem 5 come from [10] and [8].

Let G be a graph of order n with connectivity � � 3 satisfying �4ðGÞ �nþ 2�, and G has a longest cycle C which is not a dominating cycle.

We denote GnC by R. Since � � 3 and by Lemma 1, each component of GnChas at most two vertices. Let H be chosen as in Section 2. Suppose that

NCðu; vÞ ¼ fx1; x2; . . . ; xtg, and the occurrence of the vertices on C agrees with

the given orientation of C. Denote the first noninsertible vertex occurring on

xþi Cx�iþ1 by ui. The other notations such as T , Tu, and Tv are the same as those in

Section 2. By Lemma 2, T [ fug and T [ fvg are both independent sets of G. Let

S be any vertex cut set with jSj ¼ �, and B1;B2; . . . ;Bp; p � 2 the components of

GnS, where p is the number of components of GnS. By Lemma 2(3), GnT is

connected.

Obviously, we have the following claim, which will be used in the next often.

Claim 4.0. Let vs 2 Bi and vt =2 Bi for some i; 1 � i � p. If there exists a path

Pðvs; vtÞ in G joining vs and vt, then S \ Pðvs; vtÞ 6¼ ;.

Claim 4.1. minfdðuÞ; dðvÞg � �þ 2.

Proof. Suppose that dðuÞ � �þ 1. Since G is 3-connected, jTj � 3 and

minfdðuÞ; dðvÞg � 3. So such three vertices as those of Lemma 4 exist. Assume,

without loss of generality, that ui 2 Tu and uj; uk 2 Tv, then fu; ui; uj; ukg is an


independent set. By Lemma 4 and letting W ¼ ;, we have dðuÞ þ dðuiÞ þ dðujÞþdðukÞ � nþ 1 þ dðuÞ � nþ �þ 2; a contradiction. Therefore, dðuÞ � �þ 2.

Similarly, dðvÞ � �þ 2. &

The proofs of Claims 6 and 5 in [8] can be used to prove the following

Claims 4.2 and 4.3, respectively. For the sake of completeness, we still provide

the proofs of them below.

Claim 4.2. If Bi \ T ¼ ;, then Bi \ fu; vg ¼ ;, where 1 � i � p.

Proof. Otherwise, without loss of generality, let u 2 Bi. By Claim 4.1, we

may assume that dðuÞ ¼ d þ 1, where d � �þ 1. Suppose that NCðuÞ ¼fx1; x2; . . . ; xdg. Thus we have d paths: uxjCuj; 1 � j � d, which are from u to

vertices of T , and pairwise disjoint except for u. Since Bi \ T ¼ ;, each of these d

paths contains at least one vertex of S by Claim 4.0. So jSj � d � �þ 1, a

contradiction. &

Claim 4.3. jfBi : Bi \ T 6¼ ;; 1 � i � pgj � 2:

Proof. Otherwise, without loss of generality, suppose that Bi \ T 6¼ ;; i ¼ 1;2; 3: Clearly,

dðyiÞ � jBij þ jSj � jðBi [ SÞ \ ðT [ fu; vgÞj;

where yi 2 Bi \ T; i ¼ 1; 2; 3. So we have

dðy1Þ þ dðy2Þ þ dðy3Þ �X3

i¼1

jBij þ 3jSj �X3

i¼1

jðBi [ SÞ \ ðT [ fu; vgÞj

� nþ 2jSj �Xp

i¼4

jBij � j[3

i¼1

ðBi [ SÞ \ ðT [ fu; vgÞj

� nþ 2jSj � ðjTj þ 2Þ � nþ 2�� dðuÞ � 1:

That is, dðy1Þ þ dðy2Þ þ dðy3Þ þ dðuÞ � nþ 2�� 1; a contradiction. &

By Claim 4.3 we may assume, without loss of generality, that jBi \ T j ¼ 0 for

i � 3. Then by Claim 4.2, we get T [ fu; vg � B1 [ B2 [ S. Since GnT is

connected, we have S 6� T , and hence jS \ Tj � �� 1.

Claim 4.4. If jS \ T j ¼ �� 1 and u; v 2 Br, then jBs \ Tj ¼ 0, where

fr; sg ¼ f1; 2g.

Proof. Assume, without loss of generality, that u; v 2 B1. Suppose that

jB2 \ T j � 1 and let ui 2 B2 \ T . Since u; v 2 B1 and ui 2 B2, S \ ðxiCu�i Þ 6¼ ;and S \ ðuþi Cxiþ1Þ 6¼ ;. Thus jSnTj � 2. Since jS \ T j ¼ �� 1, we have that

jSj � �þ 1, a contradiction. &


Claim 4.5. If jBr \ Tj � 2 for some r 2 f1; 2g, then there exists some s,

1 � s � 2, such that jBs \ Tuj � 2 or jBs \ Tvj � 2.

Proof. Assume, without loss of generality, that jB1 \ T j � 2. Suppose, to the

contrary, that jBi \ Tuj � 1 and jBi \ Tvj � 1 for i ¼ 1; 2. Then jB1 \ T j ¼ 2, and

we can assume that jB1 \ ðTunTvÞj ¼ jB1 \ ðTvnTuÞj ¼ 1. Obviously, jB2 \ T j �jB2 \ Tuj þ jB2 \ Tvj � 2. Moreover, when jB2 \ Tj ¼ 2, we can assume that

jB2 \ ðTunTvÞj ¼ jB2 \ ðTvnTuÞj ¼ 1.

(a) jS \ T j ¼ �� 1, and hence jSnT j ¼ 1.

Note that T � S [ B1 [ B2. Since jT j � dCðuÞ þ jB1 \ ðTvnTuÞj þ jB2 \ðTvnTuÞj � �þ 2 þ jB2 \ ðTvnTuÞj; we get that jS \ Tj ¼ jT j � jB1 \ T j �jB2 \ T j � �þ jB2 \ ðTvnTuÞj � jB2 \ Tj � �� 1. Therefore, jS \ T j ¼ �� 1.

(b) u; v 2 B1.

By (a), we may assume, without loss of generality, that u =2 S. If u 2 B2, then

v =2 B1. Let ui 2 B1 \ ðTunTvÞ and uj 2 B1 \ ðTvnTuÞ. Thus S \ ðuxiCu�i Þ 6¼ ;and S \ ðvxjCu�j Þ 6¼ ;, which contradict jSnTj ¼ 1. Thus u 2 B1 and hence

v 2 B1 [ S. If v 2 S, then SnT ¼ fvg. Thus we get jB2 \ Tuj ¼ 0, and hence

jTuj ¼ jS \ Tuj þ jB1 \ Tuj þ jB2 \ Tuj � ð�� 1Þ þ 1 ¼ �; which contradicts

Claim 4.1. Thus v 2 B1.

By (a), (b), and Claim 4.4, we have jB2 \ T j ¼ 0: But in this case,

jTj � dCðuÞ þ jfujgj � �þ 2 and then jS \ T j � ð�þ 2Þ � 2 ¼ �. This contra-

dicts that jS \ T j � �� 1: &

Claim 4.6. If ui; uj 2 B1 \ Tu, (ui; uj 2 B1 \ Tv , resp.), then

dCðuiÞ þ dCðujÞ � jCj � jB2 \ Cj þ jB2 \ Tj � dðvÞ þ �þ 1;

ðdCðuiÞ þ dCðujÞ � jCj � jB2 \ Cj þ jB2 \ Tj � dðuÞ þ �þ 1; resp:Þ:

Symmetrical statements of Claim 4.6 hold when exchanging B1 and B2.

Proof. Set P1 ¼ uiCu�j , P2 ¼ ujCu

�i , Th

v ¼ Tv \ Ph and Bh2 ¼ B2 \ Ph,

ðh ¼ 1; 2Þ.By Lemma 2, we have N�

P1ðuiÞ \ NP1

ðujÞ ¼ ;. By Lemma 3, we have

N�P1ðuiÞ [ NP1

ðujÞ � ðP1nðB12 [ ðT1

v nfuigÞÞÞ [ P�1ðuiÞ; ð4:1Þ

where P�1ðuiÞ ¼ fw 2 B1

2 : wþ 2 NP1ðuiÞg.

Similarly, we have

N�P2ðujÞ [ NP2

ðuiÞ � ðP2nðB22 [ ðT2

v nfujgÞÞÞ [ P�2ðujÞ; ð4:2Þ


where P�2ðujÞ ¼ fw 2 B2

2 : wþ 2 NP2ðujÞg.

Set y1 ¼ ui and y2 ¼ uj. Thus by (4.1) and (4.2), we have

dCðuiÞ þ dCðujÞ

�X2

h¼1

ðjPhj � jBh2j � jTh

v nfyhgj þ jBh2 \ ðTh

v nfyhgÞj þ jP�hðyhÞjÞ

� jCj � jB2 \ Cj þ jB2 \ Tvj � jTvnfui; ujgj þ jP�1ðuiÞj þ jP�

2ðujÞj: ð4:3Þ

For convenience, we set P�1 ¼ P�

1ðuiÞ and P�2 ¼ P�

2ðujÞ. Now, we will prove the

following facts.

Fact 1. jP�1j þ jP�

2j � jS \ Cj � 1. Moreover, jP�1j þ jP�

2j � jS \ Cj � 2 if P�h 6¼

; for h ¼ 1; 2.

Proof of Fact 1. (a) By the definition of P�1 and ui 2 B1, we get that ðP�

1Þþ �

S \ P1, and hence jP�1j ¼ jðP�

1Þþj � jS \ P1j. Similarly, we have jP�

2j � jS \ P2j.(b) Suppose that P�

1 6¼ ;. Thus B2 \ P1 6¼ ;. We denote by w the first vertex of

P1 occurring in B2. From Claim 4.0 we obtain that w� 2 SnðP�1Þ

þ, and hence

jðS \ P1ÞnðP�1Þ

þj � 1: Thus jP�1j � jS \ P1j � 1. Similarly, if P�

2 6¼ ; then jP�2j �

jS \ P2j � 1.

Note that since �ðGÞ � 3, we have jS \ Cj � 1 by Lemma 1. Thus, Fact 1

follows from (a) and (b) immediately. &

Note that jTvj ¼ dðvÞ � 1, jB2 \ Tvj � jB2 \ Tj. Thus by Fact 1 and (4.3) we

may assume, without loss of generality, that

(a) jS \ Cj ¼ �; and hence S � C:(b) jP�

1j ¼ jS \ Cj � 1¼�� 1 and jP�2j ¼ 0:

(c) ui; uj 2 Tv; (otherwise, assume, without loss of generality, that ui =2 Tv,

then Tvnfui; ujg ¼ Tvnfujg. Thus the claim holds by (4.3) and Fact 1.)

Since jTvj ¼ dðvÞ � 1, by (4.3) and (a)–(c), we have

dCðuiÞ þ dCðujÞ � jCj � jB2 \ Cj þ jB2 \ Tvj � dðvÞ þ �þ 2: ð4:4Þ

We set P�1 ¼ fw1;w2; . . . ;w��1g and their occurrence on C agrees with the

given orientation of C. Since ðP�1Þ

þ � S, we get that jSnðP�1Þ

þj ¼ 1 by (b). Since

ui 2 B1 and w1 2 B2, there exists one and only one vertex, denoted by w0, in

uþi Cw�1 such that w0 2 SnðP�

1Þþ

. Thus we get that

(d) S ¼ fw0;wþ1 ;w

þ2 ; . . . ;w

þ��1g � uþi Cxj:

Noting that ui; uj 2 B1 and w1;w2; . . . ;w��1 2 B2, by (d) and Claim 4.0, we

have the following fact.


Fact 2. ðwþ0 Cw1Þ [ ð[��2

t¼1 ðwþ2t Cwtþ1ÞÞ � B2 and wþ2

��1Cw�0 � B1, and hence

C ¼ ðB1 \ CÞ [ ðB2 \ CÞ [ S.

Fact 3. B1 \ Tu � B1 \ Tv .

Proof of Fact 3. Otherwise, suppose that there exists a vertex uk 2 B1 \ðTunTvÞ. Then uk 6¼ ui; uj by (c). By Lemmas 2 and 3, uk 62 N�

P1ðuiÞ [ NP1

ðujÞ (or

uk 62 N�P2ðujÞ [ NP2

ðuiÞ, resp.). Hence, we can replace (4.1) (or (4.2), resp.) by

N�P1ðuiÞ [ NP1


v nfuigÞ [ fukgÞÞ [ P�1ðuiÞ; ð4:10Þ

(or

N�P2ðujÞ [ NP2

ðuiÞ � ðP2nðB22 [ ðT2

v nfujgÞ [ fukgÞÞ [ P�2ðujÞ; ð4:20Þ

resp.).

Thus applying a similar argument to (4.10) and (4.2), (or (4.1) and (4.20), resp.),

we have

dCðuiÞ þ dCðujÞ � jCj � jB2 \ Cj þ jB2 \ Tvj � dðvÞ þ �þ 2 � jfukgj;

and hence the claim holds. &

Fact 4. jB2 \ Tvj � 1.

Proof of Fact 4. Otherwise, suppose that jB2 \ Tvj � 2. Let ue; uf 2 B2 \ Tv.

Now as the beginning of the proof, we define Ph; Thu ;B

h1; ðh ¼ 1; 2Þ and

P�1ðueÞ;P�

2ðuf Þ for ue; uf . By a similar argument, we have the following inequality

similar to (4.3):

dCðueÞ þ dCðuf Þ � jCj � jB1 \ Cj þ jB1 \ Tuj � jTunfue; ufgjþ jP�

1ðueÞj þ jP�2ðuf Þj:

Similar to Fact 1, we also have jP�1ðueÞj þ jP�

2ðuf Þj � jS \ Cj � 1. Thus, we get

the following inequality similar to (4.4):

dCðueÞ þ dCðuf Þ � jCj � jB1 \ Cj þ jB1 \ Tuj � dðuÞ þ �þ 2:

Therefore,

dCðuiÞ þ dCðujÞ þ dCðueÞ þ dCðuf Þ � 2jCj � jB1 \ Cj � jB2 \ Cj þ jB1 \ Tujþ jB2 \ Tvj � dðvÞ � dðuÞ þ 2�þ 4:


On the other hand, we have

dRðuiÞ þ dRðujÞ þ dRðueÞ þ dRðuf Þ � n� jCj � 2:

By Fact 2 and Fact 3, we have

dðuiÞ þ dðujÞ þ dðueÞ þ dðuf Þ � nþ jCj � jB1 \ Cj � jB2 \ Cj þ jB1 \ Tujþ jB2 \ Tvj � dðvÞ � dðuÞ þ 2�þ 2

� nþ ðjCj � jB1 \ Cj � jB2 \ CjÞ þ ðjB1 \ Tvjþ jB2 \ TvjÞ � dðvÞ � dðuÞ þ 2�þ 2

� nþ jS \ Cj þ jTvj � dðvÞ � dðuÞ þ 2�þ 2

¼ nþ 3�� dðuÞ þ 1:

Thus, by Claim 4.1, we have that dðuiÞ þ dðujÞ þ dðueÞ þ dðuf Þ � nþ 2�� 1; a

contradiction. &

Fact 5. ujwþ1 =2 EðGÞ.

Proof of Fact 5. Otherwise, suppose ujwþ1 2 EðGÞ: In this case, we first show

that w1w2 =2 EðGÞ. By Fact 2, we have w1;w2 2 B2 and ðxþi Cu�i Þ [ ðxþj Cu�j Þ �B1. Thus IðaÞ 6¼ fwi;w

þi g for all a 2 ðxþi Cu�i Þ [ ðxþj Cu�j Þ, where i ¼ 1; 2.

If w1w2 2 EðGÞ, then, by inserting the vertices ðxþi Cu�i Þ [ ðxþj Cu�j Þ into the

cycle xiCujwþ1 Cw2w1Cuiw

þ2 Cxjuvxi, we can have a cycle longer than C, a con-

tradiction. Thus w1w2 =2 EðGÞ, and hence we have

dCðw1Þ � jB2 \ Cj þ jSj � 2: ð4:5Þ

By (d) and Fact 2, xi 2 B1 and hence v 2 B1. Since w1 2 B2 \ C, the set fui; uj;v;w1g is independent. By Fact 4, (4.4) and (4.5), we have dCðuiÞ þ dCðujÞþdCðw1Þ þ dðvÞ � jCj þ 2�þ 1: Noting that S � C, we get dRðuiÞ þ dRðujÞþdRðw1Þ � n� jCj � 2: Therefore, we obtain that dðuiÞ þ dðujÞ þ dðw1Þ þ dðvÞ �nþ 2�� 1; a contradiction. &

Now, we complete the proof of Claim 4.6. By Fact 2, we have that wþ21 2 B2,

and hence uiwþ21 =2 EðGÞ; and by Fact 5, we get wþ

1 =2 NP1ðujÞ. Thus wþ

1 =2N�P1ðuiÞ [ NP1

ðujÞ. On the other hand, since wþ1 2 S and uiw

þ1 2 EðGÞ,

wþ1 =2 B1

2 [ Tv. Thus, we can replace (4.1) by

N�P1ðuiÞ [ NP1


vnfuigÞ [ fwþ1 gÞÞ [ P�

1ðuiÞ: ð4:100Þ

Applying a similar argument to (4.100) and (4.2), we have

dCðuiÞ þ dCðujÞ � jCj � jB2 \ Cj þ jB2 \ Tvj þ ðjP�1j þ jP�

2jÞ � jTvj þ 2 � jfwþ1 gj

� jCj � jB2 \ Cj þ jB2 \ T j þ ð�� 1Þ � dðvÞ þ 2

¼ jCj � jB2 \ Cj þ jB2 \ T j � dðvÞ þ �þ 1;

which is just that we want to prove. &


Claim 4.7. jTnSj � 2. Moreover, if jTnSj ¼ 2, then jS \ Tj ¼ �� 1 and

jBr \ Tj ¼ 0 for some r 2 f1; 2g.

Proof. If NCðuÞ 6¼ NCðvÞ, then jT j � jTvj þ jTunTvj � �þ 2 by Claim 4.1.

Note that jS \ T j � �� 1 and hence jTnSj � 3.

If NCðuÞ ¼ NCðvÞ, then jT jð¼ jTuj ¼ jTvjÞ � �þ 1, and hence jTnSj � 2.

When jTnSj ¼ 2, we have jTj ¼ �þ 1 and jS \ Tj ¼ �� 1. Assume, without

loss of generality, that jB1 \ T j 6¼ 0. Let ui 2 B1 \ T . We will show that u; v 2 B1.

Since NCðuÞ ¼ NCðvÞ and jS \ T j ¼ �� 1, u; v 62 S. If u 2 B2, then

S \ ðxiCu�i Þ 6¼ ; and S \ ðuþi Cxiþ1Þ 6¼ ; which contradicts to jSj ¼ �. Hence

u 2 B1 and similarly, v 2 B1. By Claim 4.4, we have jB2 \ T j ¼ 0. &

Claim 4.8. jBr \ Tj ¼ 0 for some r 2 f1; 2g.

Proof. If jTnSj ¼ 2, then jBr \ Tj ¼ 2 for some r 2 f1; 2g by Claim 4.7 and

thus Claim 4.8 holds. So we suppose that jTnSj � 3. Thus for some r 2 f1; 2g;jBr \ Tj � 2. By Claim 4.5, we have that, for some s 2 f1; 2g; jBs \ Tuj � 2

or jBs \ Tvj � 2. Assume, without loss of generality, that jB1 \ Tuj � 2. Let

ui; uj 2 B1 \ Tu. We will show that jB2 \ T j ¼ 0: Otherwise, let uk 2 B2 \ T .

Clearly dCðukÞ � jB2 \ Cj þ jS \ Cj � jB2 \ Tj; and by Claim 4.6 and jS \ Cj ��, dCðuiÞ þ dCðujÞ þ dCðukÞ � jCj þ 2�� dðvÞ þ 1: Note that dRðuiÞ þ dRðujÞþdRðukÞ � n� 2 � jCj: Therefore, dðuiÞþ dðujÞ þ dðukÞ� n� 2 þ 2�� dðvÞ þ 1.

Hence, dðuiÞ þ dðujÞ þ dðukÞ þ dðvÞ � nþ 2�� 1; a contradiction. Thus jB2 \T j ¼ 0. &

Noting that for each k � 3, jBk \ Tj ¼ 0, by Claims 4.8 and 4.2, we have

fu; vg [ T � B1 [ S or fu; vg [ T � B2 [ S.

Claim 4.9. If jBr \ Tj ¼ 0, then BrnðNðuÞ \ NðvÞÞ 6¼ ;, where r 2 f1; 2g.

Proof. Assume, without loss of generality, that jB2 \ Tj ¼ 0. Thus, we have

fu; vg [ T � B1 [ S. We want to show that B2nðNðuÞ \ NðvÞÞ 6¼ ;. Otherwise,

assume that B2 � NðuÞ \ NðvÞ. Then u; v 2 S and thus jS \ Tj � �� 2. Since

jTuj � �þ 1, jTvj � �þ 1 and T � S [ B1, jB1 \ Tuj � 3 and jB1 \ Tvj � 3.

Hence, we can assume that ui 2 B1 \ Tu and uj; uk 2 B1 \ Tv . Noting that

B2 � ðNðuÞ \ NðvÞÞnNðui; uj; ukÞ, we get dðuiÞ þ dðujÞ þ dðukÞ � nþ 1 � jB2jby letting W ¼ B2 in Lemma 4. Let x 2 B2, then fx; ui; uj; ukg is an independent

set and dðxÞ � jB2j � 1 þ �; hence dðuiÞ þ dðujÞ þ dðukÞ þ dðxÞ � nþ �; a

contradiction. &

We assume, without loss of generality, that B2 \ T ¼ ; by Claim 4.8, and

hence u; v 2 S [ B1 by Claim 4.2, and B2nðNðuÞ \ NðvÞÞ 6¼ ; by Claim 4.9.

Now we complete the proof of Theorem 5 by considering two cases.

Case 1. B2nC 6¼ ;.

Let x 2 B2nC. Since B2 \ fu; vg ¼ ;, x =2 fu; vg. Since dRðxÞ � 1 by Lemma 1,

dðxÞ � jB2 \ Cj þ jS \ Cj þ 1. By Claim 4.7, jT \ B1j � 2 and then we have that


jB1 \ Tuj � 2 or jB1 \ Tuj � 2 by Claim 4.5. Note that for any ui; uj 2 B1 \ Tu or

ui; uj 2 B1 \ Tv, dRðuiÞ þ dRðujÞ � n� 3 � jCj; since x 2 B2nC.

If jB1 \ Tuj � 2, then we have dCðuiÞ þ dCðujÞ � jCj � jB2 \ Cj þ �� dðvÞþ1 by Claim 4.6. Thus dðuiÞ þ dðujÞ � n� jB2 \ Cj þ �� dðvÞ � 2: Therefore,

we have dðuiÞ þ dðujÞ þ dðxÞ � nþ 2�� dðvÞ � 1: That is, dðuiÞ þ dðujÞþdðxÞ þ dðvÞ � nþ 2�� 1, a contradiction to the fact that fui; uj; x; vg is an

independent set.

If jB1 \ Tvj � 2, then we have dCðuiÞ þ dCðujÞ � jCj � jB2 \ Cj þ �� dðuÞþ1 by Claim 4.6. Thus dðuiÞ þ dðujÞ � n� jB2 \ Cj þ �� dðuÞ � 2: Therefore,

we have dðuiÞ þ dðujÞ þ dðxÞ � nþ 2�� dðuÞ � 1: That is, dðuiÞ þ dðujÞþdðxÞ þ dðuÞ � nþ 2�� 1, a contradiction to the fact that fui; uj; x; ug is an

independent set.

Case 2. B2nC ¼ ;.

Let x 2 B2nðNðuÞ \ NðvÞÞ. Assume, without loss of generality, that x =2 NðuÞ.Clearly dðxÞ � jB2j þ jSj � 1 ¼ jB2 \ Cj þ jSj � 1:

Since jS \ Tj � �� 1, jTvj � �þ 1 and Tv � S [ B1, we have jB1 \ Tvj � 2.

Assume ui; uj 2 B1 \ Tv, then fui; uj; x; ug is an independent set. Note that for

ui; uj 2 B1 \ Tv, dRðuiÞ þ dRðujÞ � n� 2 � jCj: By Claim 4.6, we have dCðuiÞþdCðujÞ � jCj � jB2 \ Cj þ �� dðuÞ þ 1: Thus dðuiÞ þ dðujÞ � n� jB2 \ Cj þ�� dðuÞ � 1: Hence dðuiÞ þ dðujÞ þ dðxÞ � nþ 2�� dðuÞ � 2. That is,

dðuiÞ þ dðujÞ þ dðxÞ þ dðuÞ � nþ 2�� 2, our final contradiction. &

ACKNOWLEDGMENTS

Many thanks to the anonymous referees for their many helpful comments and

suggestions, which have considerably improved the presentation of the paper.

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Two sufficient conditions for dominating cycles

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