Relative length of longest paths and cycles in 3-connected graphs

Relative Length of LongestPaths and Cycles in3-Connected Graphs

Rao Li,1 Akira Saito,2� and R.H. Schelp3

1DEPARTMENT OF COMPUTER AND INFORMATION SCIENCE

GEORGIA SOUTHWESTERN STATE UNIVERSITY

AMERICUS, GA 31709

E-mail: [email protected]

2DEPARTMENT OF MATHEMATICS

NIHON UNIVERSITY

SAKURAJOSUI 3-25-40

SETAGAYA-KU, TOKYO 156-8550, JAPAN

E-mail: [email protected] OF MATHEMATICAL SCIENCES

THE UNIVERSITY OF MEMPHIS

MEMPHIS, TN 38152

E-mail: [email protected]

Received September 28, 1998

Abstract: For a graph G, let p(G) denote the order of a longest path in Gand c(G) the order of a longest cycle in G, respectively. We show that if G isa 3-connected graph of order n such that

P4i�1 degG xi � 3

2 n �1 for everyindependent set {x1, x2, x3, x4}, then G satis®es c(G)� p(G)ÿ 1. Using thisresult, we give several lower bounds to the circumference of a 3-connectedgraph. ß 2001 John Wiley & Sons, Inc. J Graph Theory 37: 137±156, 2001

Keywords: relative length; longest path; cycle; 3-connected graph

ÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐ

�Correspondence to: Akira Saito, Department of Mathematics, Nihon University,Sakurajosui 3-25-40, Setagaya-ku, Tokyo 156-8550, Japan.

ß 2001 John Wiley & Sons, Inc.

1. INTRODUCTION

For a graph-theoretic notation, we refer the reader to [3]. In particular, we denoteby ��G� and ��G� the independence number and the minimum degree of a graphG, respectively.

Let G be a graph and let k be a positive integer. Then, we de®ne �k�G� by

�k�G� �min fPx2S degG x : S is an independent set if k � ��G�;

of order k in Gg�1 if k > ��G�:

8<:For a graph G, we denote by p�G� the order of a longest path in G. If G has acycle, then we denote by c�G� the order of a longest cycle in G, which is oftencalled the circumference of G.

The main interest of this paper is the difference diff�G� � p�G� ÿ c�G�. If G isa connected graph which is not a tree, then diff�G� � 0. On the other hand, forintegers m� 2 and k � 2, G � Kk � �k � 1�Km is a k-connected graph withdiff�G� � m. Thus, diff�G� can be arbitrarily large. However, it is easy to see thata connected graph G is Hamiltonian if and only if diff�G� � 0.

Bondy [2] gave the following suf®cient condition for a k-connected graph G tobe Hamiltonian, or for diff�G� to be zero.

Theorem A (Bondy [2]). Let G be a k-connected graph of order n. If

�k�1�G� > 12�k � 1��nÿ 1�, then diff�G� � 0. &

If G � Kk;k�1, then n � 2k � 1 and �k�1�G� � k�k � 1� � 12�k � 1��nÿ 1�.

However, G is not Hamiltonian. Thus, Theorem A is sharp in this sense. If we putk � 2, a 2-connected graph G of order n with �3�G� � 1

2�3nÿ 2� is Hamiltonian,

and the bound 12�3nÿ 2� is sharp. However, Bauer et al. [1] and Bondy [2]

observed that a 2-connected graph G of order n with �3�G� � n� 2 still satis®esa number of `̀ Hamiltonian-like'' properties. Enomoto et al. [4] showed that theseproperties come from the inequality diff�G� � 1, and that this inequality isguaranteed whenever �3�G� � n� 2.

Theorem B (Enomoto et al. [4]). A 2-connected graph G of order n with

�3�G� � n� 2 satis®es diff�G� � 1.

In this paper, we investigate the relation between �4�G� and diff�G� in3-connected graphs G. If we put k� 3 in Theorem A, �4�G� � 2nÿ 1 guaranteesdiff�G� � 0. In this paper, we give a smaller bound to �4�G� to assure diff�G�� 1.

Theorem 1. Let G be a 3-connected graph of order n. If �4�G� � 32

n� 1, thendiff�G� � 1.

Since the inequality diff�G� � 1 implies a number of cycle-related properties,Theorem 1 can be a useful tool when we study a 3-connected graph G of order n

138 JOURNAL OF GRAPH THEORY

with �4�G� � 32

n� 1. We ®rst demonstrate its usefulness by showing severalapplications in Section 2. Then we give a proof of Theorem 1 in Section 3. InSection 4, we give concluding remarks.

For a graph G and x 2 V�G�, we denote by NG�x� the neighborhood of x in G.Thus deg G x � jNG�x�j. For S � V�G�, let NG�S� � [v2SNG�v�: We, sometimes,identify a graph and its vertex set if there is no fear of confusion. For example, ifH is a subgraph of G, we, sometimes, write x 2 H instead of x 2 V�H�. When weconsider a path or a cycle, we always assign orientation. Let P � x0x1 � � � xm be apath. Then we call x0 and xm the starting vertex and the terminal vertex of P,respectively. If u 2 V�P�, then u��P� denotes the successor of u on P and uÿ�P� itspredecessor. If we do not have to specify the path P, we sometimes omit `̀ �P�''and write u� and uÿ, respectively. For S � V�P�, let S� � fv� : v 2 Sg. For xi,xj 2 V�P� with i � j, we denote that subpath xixi�1 � � � xj by xi P

!xj. The same path

traversed in the opposite direction is denoted by xj P

xi. We also use the samenotation for cycles. For A, B � V�G� with A \ B � ;, we denote by eG�A, B� thenumber of the edges ab 2 E�G� with a 2 A and b 2 B. If jAj � 1, say A � fag,we write eG�a;B� instead of eG�fag;B�.

2. APPLICATIONS OF THEOREM 1

A longest cycle in a 2-connected graph G with p�G� ÿ c�G� � 1 has thefollowing properties, from which we can derive various lower bounds to thecircumference of G.

Lemma 2. Let G be a 2-connected graph with p�G� ÿ c�G� � 1. Then

(1) every longest cycle is dominating, and(2) let C be a longest cycle of G and let A � V�G� ÿ V�C�. Then NG�A�� [ A

is independent.

Proof. Since p�G� ÿ c�G� 6� 0, G is not Hamiltonian. Let C be a longestcycle of G and let A � V�G� ÿ V�C�.

(1) If C is not dominating, A has a pair of adjacent vertices u and v. Since G isconnected, we can choose such uv 2 E�G� so that NG�v� \ V�C� 6� ;, sayw 2 NG�v� \ V�C�. Then P � uvw C

!wÿ is a path with jPj � jCj � 2 �

c�G� � 2. This contradicts the assumption p�G� � c�G� � 1.(2) By (1), A is independent and NG�A�� is well de®ned. Assume uv 2 E�G�

for some u 2 A, and v 2 NG�A��, say v 2 NG�a�� with a 2 A. Then vÿ 2NG�a�. Let T � uv C

!vÿa. If u 6� a, then T is a path of order c�G� � 2. If

u � a, then T is a cycle of order c�G� � 1. Thus, we have a contradiction ineither case. Assume uv 2 E�G� for some u, v 2 NG�A��. Let u 2 NG�u0��and v 2 NG�v0�� with u0, v0 2 A. Let T � u0uÿC

vu C!

vÿv0. Then T is apath of order c�G� � 2 if u0 6� v0 and a cycle of order c�G� � 1 if u0 � v0,

LENGTH OF LONGEST PATHS AND CYCLES 139

respectively. Therefore, we have a contradiction in either case, and we seethat A [ NG�A�� is independent. &

Bauer et al. [1] gave the following bound to the circumference of a2-connected graph with large �3�G�.Theorem C (Bauer et al. [1]). Let G be a 2-connected graph of order n with

�3�G� � n� 2. Then c�G� � minfn; nÿ ��G� � ��G�g.Now we see that the same conclusion holds for a 3-connected graph with large

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

2n� 1.

Then c�G� � minfn; nÿ ��G� � ��G�g. In particular, if ��G� � ��G�, then G isHamiltonian.

Proof. Assume G is not Hamiltonian. Let C be a longest cycle of G and letA � V�G� ÿ V�C�. Then A 6� ;. Let x 2 A. Then NG�x�� NG�A��. Since�4�G� � 3

2n� 1 and G is not Hamiltonian, p�G� ÿ c�G� � 1 by Theorem 1 and

hence A [ NG�x�� is independent by Lemma 2. Thus, we have

��G� � jA [ NG�x��j � jAj � jNG�x�j � jAj � degG x � nÿ jCj � ��G�:

Therefore, we have jCj � nÿ ��G� � ��G�: &

Lemma 4. Let c be an integer with c � 2. If G is a non-Hamiltonian3-connected graph of order n with �4�G� � 1

2�3n� c� and ��G� � 7ÿ c, then G

has a longest cycle C and a vertex x 2 V�G� ÿ V�C� with j V�G�ÿ�V�C�� [ �NG�x�� \ NG�x�ÿ�j � 4.

Proof. Assume, to the contrary,

j V �G� ÿ V �C�� [ �NG�x�� \NG�x�ÿ�j � 3

for every choice of a longest cycle C and a vertex x 2 V�G� ÿ V�C�. Since G isnot Hamiltonian and �4�G� � 1

2�3n� c� � 3

2n� 1, we have p�G� ÿ c�G� � 1 by

Theorem 1.

Claim 1. For every longest cycle C and for each x 2 V�G� ÿ V�C�,

degG x � 1

3�n� 3ÿ 2s�;

where s � jV�G� ÿ V�C�j.Proof. Let S � V�G� ÿ V�C�, and let NG�x� � fx1; . . . ; xdg �d � degG x�.

By Lemma 2, S is independent, and hence NG�x� � V�C�. We mayassume x1; . . . ; xd appear in the consecutive order along C. Let Ii � xi C

!xi�1


�1 � i � d ÿ 1� and Id � xd C!

x1. Since C is a longest cycle, jIij � 3 for each i,1 � i � d. Let I � fIi : jIij � 3g and m � jIj. If Ii 2 I , say Ii � xivixi�1, thenvi 2 NG�x�ÿ \ NG�x��. Hence by the assumption m � 3ÿ jSj. Since

jCj �Xd

i�1

�jIij ÿ 1� � 2m� 3�d ÿ m� � 3d ÿ m � 3 degG xÿ m

� 3 degG xÿ 3� jSj;

and jCj � nÿ jSj, we have nÿ jSj � 3 degG xÿ 3� jSj, or degG x � 13�n� 3ÿ 2jSj�. &

Let C be a longest cycle, and let S � V�G� ÿ V�C�. Let x 2 S andNG�x� � fx1; . . . ; xdg. We may assume x1; . . . ; xd appear in the consecutiveorder along C. Take two different neighbors xk, xl in G and consider x�k and x�l .By Lemma 2, NG�x�k � [ NG�x�l � � V�C�. Let A1 � NG�x�k �ÿ \ x�k C

!xl;A2 �

NG�x�k � \ x�l C!

xk, B1 � NG�x�l � \ x�k C!

xl, B2 � NG�x�l �ÿ \ x�l C!

xk, A � A1 [ A2

and B � B1 [ B2. Since C is a longest cycle, it is easy to see that jAj � degG x�k ,jBj � degG x�l and A \ B � ;. In particular,

degG x�k � degG x�l � jA [ Bj � jCj � nÿ jSj:

Let J � fj : 1 � j � d; fxj; x�j g � A [ Bg.

Claim 2. For each j 2 J ÿ fk; lg; degG x�j � 13�n� 3ÿ 2jSj�.

Proof. By symmetry, we may assume xj 2 x�k C!

xÿl . By the de®nition of

J; fxj; x�j g � A [ B. By Lemma 2, x�j 62 NG�x�k � [ NG�x�l �. This implies

x��j 2 NG�x�k � and xj 2 NG�x�l �. Let C0 � xxk C

x�l xj C

x�k x��j C!

xlx. Then C0 isa longest cycle with V�G�ÿ V�C0� � �Sÿ fxg� [ fx�j g. Therefore, by Claim 1,we have degG x�j � 1

3�n� 3ÿ 2jSj�. &

Let T � fx�j : j 2 J ÿ fk; lgg.Claim 3. jS [ T j � 2.

Proof. Assume jS [ T j � 3. Take three distinct vertices s1, s2 and s3 fromS [ T . Then by Claims 1 and 2,

P3i�1 degG si � n� 3ÿ 2jSj. By Lemma 2,

fx�k ; s1; s2; s3g is an independent set of order 4, and henceP3

i�1 degG si�degG x�k � 3

2n� 1. Thus, degG x�k � 1

2nÿ 2� 2jSj. Similarly, degG x�l � 1

2nÿ

2� 2jSj. Therefore,

nÿ jSj � degG x�k � degG x�l � nÿ 4� 4jSj;

or 5jSj � 4. This implies S � ;, a contradiction. &


Let J0 � f1; . . . ; dg ÿ J. Then for each j 2 J0, there exists a vertexzj 2 fxj; x

�j g ÿ �A [ B�. Note zj1 6� zj2 if j1 6� j2. Let Z � fzj : j 2 J0g. Then

A [ B � V�C� ÿ Z. Therefore, degG x�k � degG x�l � jCj ÿ jZj � nÿ jSj ÿ jZj.Since jZj � d ÿ jJj � degG xÿ �jTj � 2�, we have

degG x�k � degG x�l � nÿ jSj ÿ degG x� jT j � 2;

or degG x�k � degG x�l � degG x � nÿ jSj � jTj � 2. Since d � 3, by taking�k; l� � �1; 2�, �2; 3� and �3; 1�,

degG x�1 � degG x�2 � degG x � nÿ jSj � jT j � 2;

degG x�2 � degG x�3 � degG x � nÿ jSj � jT j � 2;

degG x�3 � degG x�1 � degG x � nÿ jSj � jT j � 2:

By taking the sum of these inequalities,

2X3

i�1

degG x�i � degG x

!� degG x � 3nÿ 3jSj � 3jT j � 6:

Since fx; x�1 ; x�2 ; x�3 g is independent,P3

i�1 degG x�i � degG x � 12�3n� c�.

Therefore, degG x� 3n� c � 3nÿ 3jSj � 3jTj � 6, or

degG x � 3jT j ÿ 3jSj � 6ÿ c � 3�2ÿ jSj� ÿ 3jSj � 6ÿ c

� 12ÿ 6jSj ÿ c � 6ÿ c

since jSj � 1. This contradicts the assumption. Therefore, the lemma follows.&

Since the minimum degree of a 3-connected graph is at least three, thecondition ��G� � 7ÿ c in Lemma 4 is trivial if c � 4. Thus, we have thefollowing corollary.

Corollary 5. Let G be a non-Hamiltonian 3-connected graph of order n with�4�G� � 3

2n� 2, then G has a longest cycle C and a vertex x 2 V�G� ÿ V�C�

with j V�G� ÿ V�C�� [ NG�x�� \ NG�x�ÿÿ �j � 4.

Lemma 6. Let G be a graph with p�G� ÿ c�G� � 1 and let C be a longest cycleof G and x be a vertex in V�G� ÿ V�C�. Let S � V�G� ÿ V�C�� [�NG�x�� \ NG�x�ÿ� and T � NG�S�. Then

(1) T � V�C�,(2) T \ T� � ;, and(3) jCj � 2jTj.


Proof. First, (1) immediately follows from Lemma 2. Assume T \ T� 6� ;,say u 2 T \ T�. Then fu; uÿg � T and u 2 NG�a� and uÿ 2 NG�b� for somea; b 2 S. By Lemma 2, fa; bg 6� V�G� ÿ V�C�.

Assume a 2 V�G� ÿ V�C� and b 2 NG�x�� \ NG�x�ÿ. If a 6� x, thenau C!

bÿxb�C!

uÿb is a path of order c�G� � 2. If a � x, then uÿ and b are pairof adjacent vertices in NG�x�ÿ. This contradicts Lemma 2. Similarly, we have acontradiction when a 2 NG�x�� \ NG�x�ÿ and b 2 V�G� ÿ V�C�.

Assume a; b 2 NG�x�� \ NG�x�ÿ and a 6� b. Since ab 62 E�G�, u 2 a��

C!

bÿ [ b��C!

aÿ. If u 2 a��C!

bÿ, then a C!

uÿb C!

aÿxbÿC

ua is a cycle which

is longer than C. If u 2 b��C!

aÿ, then a C!

buÿC

b�xaÿC

ua is a cycle which islonger than C. Therefore, we have a contradiction in either case.

Finally, assume a � b 2 NG�x�� \ NG�x�ÿ. Then xa�C!

uÿau C!

aÿx is a cyclewhich is longer than C. This is contradiction and (2) follows. The inequality (3)immediately follows from (2). &

There are a number of ways to evaluate jTj in Lemma 6, each of which gives alower bound to the circumference. We give two examples here.

Theorem 7. Let c 2 f2; 3; 4g and let G be a 3-connected graph of order n with�4�G� � 1

2�3n� c� and ��G� � 7ÿ c. Then c�G� � minfn; 1

2�4�G�g.

Proof. Assume G is not Hamiltonian. Then p�G� ÿ c�G� � 1 by Theorem 1.By Lemma 4, there exists a longest cycle C and a vertex x 2 V�G� ÿ V�C�with j V�G� ÿ V�C�� [ �NG�x�� \ NG�x�ÿ�j � 4. Let S � V�G� ÿ V�C�� [�NG�x�� \ NG�x�ÿ� and T � NG�S�. By Lemma 6, T � V�C� and jCj � 2jTj.Choose four distinct vertices v1, v2, v3, v4 from S and let S0 � fv1; v2; v3; v4g.Then S0 is an independent set and NG�S0� � T , which implies �4�G� �

P4i�1

degG vi � eG�S0;T�. On the other hand, since jS0j � 4; eG�S0; T� �P

t2T

eG�t; S0� � 4jT j. Therefore, we have jT j � 14�4�G� and hence c�G� � 1

2�4�G�.

&

Note that, while the above theorem implies Theorem A for k � 3, a strongerresult was obtained by Fournier and Fraisse [6].

Theorem 8. Let c 2 f2; 3; 4g and l > 0, and let G be a 3-connected graph oforder n with minimum degree at least 7ÿ c. If

X4

i�1

degG xi � max3n� c

2; l�

\4i�1

NG�Xi��

��( )

for every independent set fx1; x2; x3; x4g of order 4, then c�G� � 23

l.

Proof. Assume G is not Hamiltonian. By Theorem 1, p�G� ÿ c�G� � 1. Thenby Lemma 6, there exists a longest cycle C and a vertex x 2 V�G� ÿ V�C�such that jCj � 2jT j, where S � V�G� ÿ V�C�� [ �NG�x�� \ NG�x�ÿ� and


T � NG�S�. By Lemma 4, S is an independent set of order at least 4. Choose fourdistinct vertices x1, x2, x3 and x4 form S and let S0 � fx1; x2; x3; x4g. Since

NG�S0� � T ,P4

i�1 degG xi � eG�S0; T�. Let Ti � fy 2 T : eG�y; S0� � ig. Then

eG�S0;T� �P4

i�1 ijTij. Since T4 �T4

i�1 NG�xi�,

X4

i�1

degG xi ÿ\4i�1

NG�xi��

�� X3

i�1

ijTij � 3jT4j � 3jT j:

By the assumption

3jT j �X4

i�1

degG xi ÿ\4i�1

NG�xi��

�� l;

and we have jCj � 2jT j � 23

l. &

As an immediate corollary, we have the following.

Corollary 9. Let G be a 3-connected graph of order n. If

X4

i�1

degG xi � 3

2n�

\4i�1

NG�xi��

�� 2

for every independent set fx1; x2; x3; x4g of order 4, then G is Hamiltonian.

Let G � Kk � �k � 1�K1 with k � 3. Then jGj � n � 2k � 1, and everyset of four independent vertices fx1; x2; x3; x4g satis®es

P4i�1 degG x iÿ

jT4i�1 NG�xi�j � 3k � 3

2nÿ 3

2. However, G is not Hamiltonian. Thus, Corollary 9

is almost best possible in this sense.

3. PROOF OF THEOREM 1

In this section, we give a proof of Theorem 1. Though the following lemma is aneasy observation, it is often used in the proof of Theorem 1.

Lemma 10. Let P be a longest path of a connected non-Hamiltonian graph G,

and let x and y be the starting vertex and the terminal vertex of P, respectively.Then

(1) NG�x� [ NG�y� � V�P� �and NG�x�ÿ is well de®ned ).(2) NG�x�ÿ \ NG�y� � ;. In particular, xy 6� E�G�.


(3) NG�x�ÿ [ NG�y� � V�P� ÿ fyg and hence degG x� degG y � jPj ÿ 1.

Proof of Theorem 1. Assume diff�G� � 2. Let P be the set of all the longestpaths in G. For P 2 P, let xP and yP be the starting vertex and the terminal vertexof P, respectively.

Case 1. y0 2 xP P!

xÿ0 for some P 2 P, x0 2 NG�xP� and y0 2 NG�yP�:We choose such P 2 P so that x0 P

!yP is as short as possible. Let x � xP and

y � yP. By Lemma 10, NG�x� [ NG�y� � V�P� and xy 62 E�G�.Let x1 � x�0 and y1 � y�0 . If jx1 P

!yj � 1, then C � x P

!x0x is a cycle with

jCj � jPj ÿ 1 � p�G� ÿ 1. This contradicts the assumption. Hence jx1 P!

yj � 2.We ®rst claim the following.

Claim 1. NG�x� [ NG�y1� � x P!

x0.

Proof. Since x P!

y0y P

y1 2 P, NG�y1� � V�P� by Lemma 10. Assume

NG�x� [ NG�y1� 6� x P!

x0. Then NG�x� [ NG�y1�� \ x1 P!

y 6� ;, say u 2 NG�x� [�NG�y1�� \ x1 P

!y. If u 2 NG�x�, then u P

!y is shorter than x0 P

!y. This contradicts

the choice of �P; x0; y0� If u 2 NG�y1�, then let P0 � y1 P!

x0x P!

y0y P

x1. Then

P0 2 P, u 2 NG�y1�; x0 2 NG�x1� \ y1 P!

0uÿ�P0� and u P!

0x1 is shorter than x0 P!

y.

This contradicts the choice of �P; x0; y0�. &

Since G is 3-connected, there exists a path Q connecting x1 P!

y andx P!

yÿ0 [ y1 P!

xÿ0 in Gÿ fx0; y0g. Let a and b the starting vertex and the terminalvertex of Q, respectively Choose Q so that a is as close to y as possible alongP (possibly a� y). We can apply Claim 1 to xÿ0 P

xx0 P!

y to seeNG�xÿ0 � [ NG�yÿ0 � � x P

!x0. Thus, b 2 x� P

!yÿÿ0 [ y�1 P

!xÿÿ0 .

Claim 2. The set fx; y; y1; b�g is independent.

Proof. Since b 6� x0; y0, we have b� 6� x1; y1. By Claim 1, y 62 NG�x� [NG�y1�. Since x P

!y0y P

y1 2 P, xy1 62 E�G� by Lemma 10(2). If xb� 2 E�G�,let P0 � aÿ�P� P

b�x P!

bQ

a P!

y. Then jP0j � jPj, a 2 NG�aÿ�P��, y0 2 NG�y� \aÿ�P� P

!0aÿ�P0� and a P

!0y � a P

!y is shorter than x0 P

!y. If yb� 2 E�G�,

let P0 � x P!

bQ

a P!

yb� P!

aÿ�P�. Then jP0j � jPj, x0 2 NG�x�, a 2 NG�aÿ�P�� \x P!

0xÿ�P0�0 and x0 P

!0aÿ�P� is shorter than x0 P

!y. Hence in either case, we have a

contradiction. Thus, xb�, yb� 62 E�G�.Assume y1b� 2 E�G�. If b 2 y�1 P

!xÿÿ0 , let P0 � a��P� P

!yy0 P

xx0 P

b�y1 P!

bQ

a P

x1 (if a � y, we assume a��P� � y0�. Then jP0j � jPj, a 2 NG�a��P��,x0 2 NG�x1� \ a��P� P


!0x1 is shorter than x0 P

!y. If b 2 x� P

!yÿÿ0 ,

let P0 � a��P� P!

yy0 P

b�y1 P!

x0x P!

bQ

a P!

x1 (again, if a � y, then we assume

a��P� � y0�. Then jP0j � jPj, a 2 NG�a��P��, x0 2 NG�x1� \ a��P� P!

aÿ�P0� and


a P!

0x1 is shorter than x0 P!

y. Hence, we have a contradiction in either case, andwe have y1b� 62 E�G�.

Therefore, we have a contradiction in each case, and fx; y; y1; b�g is an

independent set. &

Claim 3. Let R be a path starting from x0 with V�R� � V�x0 P!

y�. Let z be

the terminal vertex of R. Suppose NG�z� \ x P!

xÿ0 6� ;, say z0 2 NG�z� \ x P!

xÿ0 .Then

degG x� degG z� degG z��P�0 � n� 1:

Proof. Let z1 � z��P�0 . Since x P

!x0 R!

z 2 P, NG�z� � V�P�. By applyingthe same arguments as in the proof of Claim 1 to x P

!x0 R!

z, we have NG�x� [NG�z1� � x P

!x0. Furthermore, we apply the same arguments as those in the proof

of Claim 2 to see that fx; z; z1g is an independent set. Let C � x P!

x0x. Thenx��C�0 � x. Let

A1 � NG�x�ÿ \ x P!

z0; A2 � NG�x� \ z1 P!

x0;

B1 � NG�z1� \ x P!

z0; B2 � NG�z1�ÿ \ z1 P!

x0;

and let A � A1 [ A2 and B � B1 [ B2. Since A��C�1 � NG�x� \ x� P

!z1 and

fx; z1g \ NG�x� � ;, NG�x� � A��C�1 [ A2 (disjoint). Therefore,

jAj � jA1 [ A2j � jA1j � jA2j � A��C�1

�� jA2j

� A��C�1 [ A2

�� jNG�x�j � degG x:

Similarly, we have jBj � degG z1.Assume A \ B 6� ;, say u 2 A \ B. Suppose u 2 x P

!z0. Then u 2 A1 \ B1.

Since z1 62 NG�x�, u 2 x P!

zÿ0 and hence u� 2 NG�x� \ x� P!

z0. Let C0 �x P!

uz1 P!

x0 R!

zz0 P

u�x. Then C0 is a cycle with jC0j � jPj � p�G�. This is acontradiction. A similar argument leads us to a contradiction if u 2 z1 P

!x0. Hence

we have A \ B � ;.Assume fu; u��C�g � NG�z� for some u 2 x P

!x0. Since x 62 NG�z�, u 6� x0. Let

P0 � x P!

uzu� P!

x0 R!

zÿ�R�. Then P 2 P, x0 2 NG�x�, z 2 NG�zÿ�R�� \ x P!

0xÿ�P0�0

and x0 R!

zÿ�R� is shorter than x0 P!

y. This contradicts the choice of P. Therefore,

�� fu; u�g 6� NG�z� for each u 2 x P!

x0:

We claim fu; u�g 6� A [ B for each u 2 NG�z� ÿ fx0; z0g. Assume fu; u�g �A [ B for some u 2 NG�z� ÿ fx0; z0g. Suppose u 2 x P

!zÿ0 . By (�), u 6� zÿ0 and

hence u 2 x P!

zÿÿ0 . If u� 2 B, then u� 2 B1. Let P0 � x P!

uzz0 P

u�z1 P!

x0 R!

zÿ�R�.


Then P0 2 P, x0 2 NG�x�, z 2 NG�zÿ�R�� \ x P!

0xÿ�P0�0 and x0 R

!zÿ�R� is shorter

than x0 P!

y. This is a contradiction. Hence u� 62 B. If u 2 A, then u 2 A1. LetC0 � x P

!uz R

x0 P

u�x. Then C0 is a cycle with jC0j � jPj � p�G�. This is acontradiction. Hence we have u 62 A. Therefore, we have u 2 B and u� 2 A,which imply u 2 NG�z1� and u�� 2 NG�x�.

Now, let C0 � x P!

uz1 P!

x0 R!

zz0 P

u��x. Then C0 is a cycle with jC0j �jPj ÿ 1 � p�G� ÿ 1. This is again a contradiction. By a similar argument, wehave a contradiction if u 2 z1 P

!xÿ0 . Thus, we have fu; u�g 6� A [ B for each

u 2 NG�z� ÿ fx0; z0g.For each u 2 �NG�z� \ x P

!x0� ÿ fx0; z0g, let f �u� be a vertex in fu; u��C�gÿ

�A [ B�. By (�), f �u1� 6� f �u2� for each pair of distinct vertices u1, u2 in�NG�z� \ x P

!x0� ÿ fx0; z0g. Let

D1 � ff �u� : u 2 �NG�z� \ x P!

x0� ÿ fx0; z0gg;D2 � NG�z� \ x

��R�0 R!

z;

and D � D1 [ D2. Then jDj � degG zÿ 2 and �A [ B� \ D � ;. Since z 62 A[B [ D,

nÿ 1 � jA [ B [ Dj � jAj � jBj � jDj � degG x� degG z1 � degG zÿ 2;

or degG x� degG z1 � degG z � n� 1. &

By Claim 3,

degG x� degG y1 � degG y � n� 1:

On the other hand, since fx; y1; y; b�g is independent,

degG x� degG y1 � degG y� degG b� � 3

2n� 1:

Therefore, we have degG b� � 12

n.

Claim 4. degG y1 <12

n and degG x < 12

n.

Proof. First, we prove degG y1 <12

n. Suppose b 2 y�1 P!

xÿÿ0 . By Claim 1,NG�y1� � x P

!x0. Let

A1 � NG�y1�ÿ \ y1 P!

b; A2 � NG�y1� \ �b� P!

x0 [ x P!

y0�;B1 � NG�b�� \ y1 P

!b; B2 � NG�b��ÿ \ �b� P

!x0 [ x P

!y0�;

B3 � NG�b�� ÿ x P!

x0;


and let A � A1 [ A2 and B � B1 [ B2 [ B3. Since A�1 � NG�y1� \ y�1 P!

b� and

fy1; b�g \ NG�y1� � ;, NG�y1� � A�1 [ A2 (disjoint), and hence jAj � degG y1.

Similarly, jBj � degG b�. Assume A \ B 6� ;, say u 2 A \ B. Then u 2 x P!

x0 by

Claim 1. If u 2 y1 P!

b, then u 6� b and u� 2 y�1 P!

b. Let P0 � x P!

y0y

P

aQ!

b P

u�y1 P!

ub� P!

aÿ�P�. Then jP0j � jPj, x0 2 NG�x�, a 2 NG�aÿ�P��\x P!



!y. If u 2 b� P

!xÿ0 , then

u� 2 b�� P!

x0. Let P0 � x P!

y0y P

aQ!

b P

y1u P

b�u� P!

aÿ�P�. Then jP0j �jPj; x0 2 NG�x�, a 2 NG�aÿ�P�� \ x P

!0xÿ�P0�0 and x0 P

!0aÿ�P� is shorter than

x0 P!

y. If u � x0, then u� � x1 2 NG�b��. If a 6� x1, let P0 � aÿ�P� P

x1b�

P!

x0x P!

bQ

a P!

y. Then jP0j � jPj, a 2 NG�aÿ�P��, y0 2 NG�y�\ aÿ�P� P!

0aÿ�P0�

and a P!

0y � a P!

y is shorter than x0 P!

y. If a � x1, let P0 � b� P!

x0x P!

bQ

a P!

y.

Then jP0j � jPj, x1 2 NG�b��, y0 2 NG�y�\ b� P!


!0y � x1 P

!y

is shorter than x0 P!

y. If u 2 x P!

y0, then u 6� y0 and hence u� 2 x� P!

y0.

Let P0 � x P!

uy1 P!

bQ

a P!

yy0 P

u�b� P!

aÿ�P�. Then jP0j � jPj, x0 2 NG�x�,a 2 NG �aÿ�P�� \ x P



!y. Therefore, we

have a contradiction in each case, and we have A \ B � ;. Since A [ B �V�G� ÿ fyg,

nÿ 1 � jA [ Bj � jAj � jBj � degG y1 � degG b� � degG y1 � 1

2n;

and we have degG y1 <12

n.

Next, suppose b 2 x� P!

yÿÿ0 . Let

A1 � NG�y1�ÿ \ �y1 P!

x0 [ x P!

b�; A2 � NG�y1� \ b� P!

y0;

B1 � NG�b�� \ �y1 P!

x0 [ x P!

b�; B2 � NG�b��ÿ \ b� P!

y0;

B3 � NG�b��ÿ ÿ x P!

x0;

and Let A � A1 [ A2 and B � B1 [ B2 [ B3. Then jAj � degG y1 and jBj �degG b�. Assume A \ B 6� ;, say u 2 A \ B. If u 2 y1 P

!x0, let P0 � x P

!bQ

a P!

yy0 P

b�u P

y1u� P!




!y. If u 2 x P

!b, then let P0 �

x P!

ub� P!

y0y P

aQ!

b P

u�y1 P!




!y. If u 2 b� P

!y0, then let

P0 � x P!

bQ

a P!

yy0 P

u�b� P!

uy1 P!

aÿ�P�. Then jP0j � jPj, x0 2 NG�x�, a 2NG�aÿ�P�� \ x P



!y. Therefore, we have

a contradiction in each case, and hence A \ B � ;. Since A [ B � V�G� ÿ fyg,


nÿ 1 � jA [ Bj � jAj � jBj � degG y1 � degG b� � degG y1 � 1

2n;

and hence we have degG y1 <12

n.By considering y1 P

!x0x P!

y0y P

x1 instead of P, we have degG x < 12

n by thesame argument. &

Claim 5. a 6� y and aÿy 62 E�G�.Proof. First, assume a � y. Then since P 2 P, NG�y� � V�P� and hence

Q � ab. By Claim 3, we have degG x� degG y� degG b� � n� 1. On theother hand, since fx; y; y1; b

�g is independent by Claim 2, we havedegG x� degG y� degG y1 � degG b� � 3

2n� 1. Thus, we have degG y1 � 1

2n.

This contradicts Claim 4, and hence we have a 6� y.Next assume aÿy 2 E�G�. Let R � x0 P

!aÿy P

a. Since x P!

x0 R!

a 2 P,NG�a� � V�P� and hence Q � ab. Then R and b satisfy the assumption ofClaim 3, and degG x� degG a� degG b� � n� 1. Then we can applyClaim 2 to x P

!x0 R!

a, we see that fx; a; b�; y1g is an independent set andhence

degG x� degG a� degG b� � degG y1 � 3

2n� 1:

Therefore, we have degG y1 � 12

n. This again contradicts Claim 4. &

Claim 6. NG�x1� \ �y1 P!

xÿ0 [ x P!

yÿ0 � � ;. In particular, a 6� x1.

Proof. Assume NG�x1� \ �y1 P!

xÿ0 [ x P!

yÿ0 � 6� ;, says u 2 NG�x1� \ �y1 P!

xÿ0[ x P!

yÿ0 �. Let P0 � u� P!

x0x P!

ux1 P!

y. Then P0 2 P. Note u� 6� y1 since u 6� y0.By applying Claims 2 and 3, we see that fu�; y; y1; xg is independent anddegG u� � degG y� degG y1 � n� 1. Therefore, we have degG x � 1

2n. This

contradicts Claim 4.If a � x1, then since y1 P

!x0x P!

y0y P

x1 2 P, Q � x1b and hence b 2NG�x1� \ �y1 P

!xÿ0 [ x P

!yÿ0 �. This is a contradiction. &

Claim 7. x1y 62 E�G�.Proof. Assume x1y 2 E�G�. Since a 2 x�1 P

!yÿ and aÿy 62 E�G� by Claim 5,

there exists a vertex w 2 x�1 P!

aÿ such that x1 P!

wÿ � NG�y� and w 62 NG�y�.Then fx; y; y1;wg is an independent set by Claims 1 and 2. SincedegG x� degG y� degG y1 � n� 1 by Claim 3, degG w � 1

2n. Since x P

!wÿy

P

w 2 P, NG�w� � V�P�. Let


A1 � NG�w� \ x P!

wÿ; A2 � NG�w�ÿ \ w P!

y;

B1 � NP�b��ÿ \ x P!

wÿ; B2 � NP�b�� \ w P!

y;

B3 � NG�b�� ÿ V�P�;

and let A � A1 [ A2, and B � B1 [ B2 [ B3. Since A�2 � NP�w� \ w� P!

y and

w 62 NP�w�, NP�w� � A1 [ A�2 (disjoint) and jAj � degG w. If wb� 2 E�G�,let P0 � x P

!bQ

a P

wb� P!

wÿy P

a��P�. Then jP0j � jPj, x0 2 NG�x�, a 2NG�a��P�� \ x P


!0a��P� � x0 P

!wÿy P

a��P� is shorter than x0 P!

y,

a contradiction. Hence wb� 62 E�G�. Since B�1 � NP�b�� \ x� P!

w and

fw; xg\ NP�b�� ;, NG�b�� B�1 [ B2 [ B3 (disjoint) and hence jBj �degG B�.

We claim A \ B � ;. Assume A \ B 6� ;, say u 2 A \ B. Suppose u 2 x P!

xÿ0 .Then u 2 NG�w� and u� 2 NG�b��. However, we can apply Claim 2 toP0 � x P

!wÿy P

w to see that fx;w; u�; b�g is an independent set since b 6� x; u.This is a contradiction. Suppose u � x0. Then x1 2 NG�b��. Since b� 2�y1 P!

xÿ0 [ x P!

yÿ0 �, this contradicts Claim 6.Suppose u 2 x1 P

!wÿ: Then u� 2 x�1 P

!w \ NG�b��. Since u 2 x1 P

!wÿ, u 2

NP�y�: Let

P0 � x P!

bQ

a P

u�b� P!

uy P

a�:

Then jP0j � jPj, x0 2 NG�x�, a 2 NG�a�� \ x P!


!0a� � x0 P

!uy

P

a� is shorter than x0 P!

y. This is a contradiction.Suppose u 2 w P

!aÿ. Then u 2 NG�b�� and u� 2 NG�w�. Since w 62 NG�b��,

u 2 w� P!

aÿ. Let

P0 � x P!

bQ

a P

u�w P!

ub� P!

wÿy P

a�:

Then jP0j � jPj, x0 2 NG�x�, a 2 NG�a�� \ x P!


!0a� � x0 P

!wÿy

P

a� is shorter than x0 P!

y. This is a contradiction.Suppose u 2 a P

!y. Then u 2 NG�b�� and u� 2 NG�w� Since y 62 NG�b��,

u 6� y. Let

P0 � x P!

bQ

a P!

ub� P!

wÿy P

u�w P!

aÿ:

Then jP0j � jPj, x0 2 NG�x�, a 2 NG�aÿ� \ x P!


!0aÿ � x0 P

!wÿy

P

u�w P!

aÿ is shorter than x0 P!

y, a contradiction.Therefore, A \ B � ;. Since y 62 A [ B,


nÿ 1 � jA [ Bj � jAj � jBj � degG w� degG b� � 1

2n� 1

2n � n:

This is a contradiction, and the claim follows. &

Since x1 62 NG�x� [ NG�y1�, fx; y; x1; y1g is an independent set. SincedegG x� degG y� degG y1 � n� 1, degG x1 � 1

2n. Since y1 P

!x0x P!

y0y P

x1 2 P,NG�x1� � V�P�. Let

A1 � NG�x1� \ x P!

x0; A2 � NG�x1�ÿ \ x1 P!

y;

B1 � NP�b��ÿ \ x P!

x0; B2 � NP�b�� \ x1 P!

y;

B3 � NG�b�� ÿ V�P�;

and let A � A1 [ A2 and B � B1 [ B2 [ B3. Since y 62 NP�x1� by Claim 7,

NG�x1� � A1 [ A�2 (disjoint), and jAj � degG x1. Similarly, we have NG�b�� B�1 [ B2 [ B3 (disjoint) and jBj � degG b�. Note A1 � fx0; y0g by Claim 6.

Assume A \ B 6� ;, say u 2 A \ B. Then u 2 V�P�. If u 2 x P!

xÿ0 , then u � y0

and y1 2 NG�b��. This contradicts Claim 2. If u � x0, then x1b� 2 E�G�. Since

b� 2 y��1 P!

xÿ0 [ x�� P!

yÿ0 , this contradicts Claim 6.

Suppose u 2 x1 P!

aÿ. Then u� 2 NG�x1� \ x�1 P!

a. Let

P0 � aÿ P

u�x1 P!

ub� P!

x0x P!

bQ

a P!

y if u 2 x1 P!

aÿÿ;

x1 P!

aÿb� P!

x0x P!

bQ

a P!

y if u � aÿ:

(

If u 2 x1 P!

aÿÿ, then jP0j � jPj, a 2 NG�aÿ�, y0 2 NG�y� \ aÿ P!

0aÿ�P0� anda P!

0y � a P!

y is shorter than x0 P!

y, a contradiction. If u � aÿ, then jP0j � jPj,a 2 NG�x1�, y0 2 NG�y� \ x1 P


!0y is shorter than x0 P

!y, again a

contradiction.

Suppose u � a. Then a 2 NG�b�� and a� 2 NG�x1�. If b 2 y�1 P!

xÿÿ0 ,

let P0 � y1 P!

bQ

ab� P!

x0x P!

y0y P

a�x1 P!

aÿ. If b 2 x� P!

yÿÿ0 , let P0 � y1 P!

x0x P!

bQ

ab� P!

y0y P

a�x1 P!

aÿ. Then in either case, jP0j � jPj, y0 2 NG�y1�,a 2 NG�aÿ� \ y1 P

!0yÿ�P0�0 and y0 P

!0aÿ � y0yPa�x1 P

!aÿ is shorter than x0 P

!y.

This is a contradiction.

Finally, suppose u 2 a� P!

y. In this case, ub� is a path of length one whichconnects x1 P

!y and y1 P

!xÿ0 [ x P

!yÿ0 . Since we have chosen Q so that a is as close

to y as possible, this is a contradiction, and hence we have A \ B � ;.Since y 62 A [ B by Claim 2, we have

nÿ 1 � jA [ Bj � jAj � jBj � degG x1 � degG b� � 1

2n� 1

2n � n:

This is a contradiction, and the theorem is proved in this case.


Case 2. For every choice of P 2 P, x0 2 NG�xP� and y0 2 NG�yP�, x0 2 x P!

y0.

Choose P 2 P; x0 2 NG�xP�; y0 2 NG�yP� so that jxP P!

x0j � jy0 P!

yPj is as largeas possible. Let x � xP and y � yP. By the assumption of the case,jx P!

x0j � jy0 P!

yj � jPj � 1 � p�G� � 1.

Case 2.1. jx P!

x0j � jy0 P!

yj � jPj ÿ 1

In this case x�0 P!

yÿ0 6� ;, say z 2 x�0 P!

yÿ0 . Then fx; y; zg is independent. LetA � NG�x�ÿ, B � NG�y�� and D � NG�z�. By the choice of P, A � x P

!xÿ0 and

B � y�0 P!

y and hence A \ B � ;. Assume A \ D 6� ;, say u 2 A \ D. Thenu 2 x P

!xÿ0 . Let P0 � u P

xu� P!

y. Then P 2 P, z 2 NG�u�, y0 2 NG�y� and

ju P!

0zj � jy0 P!

0yj � jx P!

zj � jy0 P!

yj > jx P!

x0j � jy0 P!

yj:

This contradicts the choice of P. Therefore, A \ D � ;. By symmetry, B \ D � ;.Since jAj � degG x, jBj � degG y and jDj � degG z and z 62 A [ B [ D,

nÿ 1 � jA [ B [ Dj � jAj � jBj � jDj � degG x� degG y� degG z:

Assume x P!

xÿ0 � NG�x�ÿ. Since G is 3-connected, there exists a path Q whichconnects x P

!xÿ0 and x�0 P

!y in Gÿ x0. Let a and b be the starting and the

terminal vertices of Q, respectively, with a 2 x P!

xÿ0 , b 2 x�0 P!

y andV�Q� \ V�P� � fa; bg. Then a� 2 NG�x� and P0 � a P

xa� P!

y 2 P, whichimplies NG�a� � V�P�. Thus Q � ab and b 2 x�0 P

!y. Then

ja P!

0bj � jy0 P!

0yj � jx P!

bj � jy0 P!

yj > jx P!

x0j � jy0 P!

yj;

a contradiction. Therefore, x P!

xÿ0 6� NG�x�ÿ, or x� P!

x0 6� NG�x�.Choose w 2 x� P

!xÿ0 ÿ NG�x� so that w P

!x0 is as short as possible. Then

w� 2 NG�x� and w 2 A. By the choice of w; xw 62 E�G�. Therefore, fx;w; z; yg isan independent set. Since �4�G� � 3

2n� 1,

degG x� degG w� degG z� degG y � 3

2n� 1:

Hence, we have degG w � 12

n� 2. By symmetry, we have y�0 P!

yÿ 6� NG�y�.Choose v 2 y�0 P

!yÿ ÿ NG�y� so that y0 P

!v is as short as possible. Then fx; z; v; yg

is independent, and we have degG v � 12

n� 2. Now let

Q � w P

xw� P!

vÿy P

v:

Then Q 2 P with degG w� degG v � n� 4 > jPj. This contradicts Lemma10(3), and the theorem is proved in this case.


Case 2.2. jx P!

x0j � jy0 P!

yj � jPjIn this case y0 � x�0 . Since G is 3-connected, there exists a path Q which

connects x P!

xÿ0 and y�0 P!

y in Gÿ fx0; y0g. Let a and b be the starting and theterminal vertices of Q. We may assume a 2 x P

!xÿ0 and b 2 y�0 P

!y (possibly

a � xÿ0 or b � y�0 ). Then

P0 � a� P!

x0x P!

aQ!

b P!

yy0 P!

bÿ

is a path containing V�P�. Therefore, Q � ab and NG�a�� [ NG�bÿ� � V�P�.Thus, we have shown that there is an edge ab with a 2 x P

!xÿ0 and b 2 y�0 P

!y.

Choose such ab so that a P!

b is as short as possible.If a� 2 NG�x�, let P1 � a P

xa� P!

y. Then P1 2 P, b 2 NG�a� andy0 2 a P

!1bÿ�P1�. This contradicts the assumption of Case 2. Thus a�x 62 E�G�.

Similarly, bÿy 62 E�G�. In particular, a 6� xÿ0 and b 6� y�0 .Therefore, from the choice of ab, a�bÿ 62 E�G�. Thus, fx; a�; bÿ; yg is an

independent set, and

degG x� degG a� � degG bÿ � degG y � 3

2n� 1: �1�

Let A1 � NG�a�� \ x P!

a, A2 � NG�a��ÿ \ a� P!

y and A � A1 [ A2. Notefx; yg \ NP�a�� ;. Since Aÿ1 � NP�a�� \ x P

!aÿ, A�2 � NP�a�� \ a�� P

!y and

a� 62 NP�a��, NG�a�� Aÿ1 [ A�2 [ fag (disjoint). Hence

jAj � jA1 [ A2j � jA1j � jA2j � jAÿ1 j � jA�2 j � jAÿ1 [ A�2 j� jNG�a�� ÿ fagj � degG a� ÿ 1:

By the assumption of the case, NG�x� \ NG�y� � NP�x� \ NP�y� � ;. Assume

NG�x� \ A 6� ;, say u 2 NG�x� \ A. If u 2 x P!

a, then u 2 NP�a��, or

uÿ 2 NP�a��. Let P1 � a P

ux P!

uÿa� P!

y. Then P1 2 P, b 2 NG�a� and

y0 2 a P!

1bÿ�P1�. This contradicts the assumption of Case 2. If u 2 a� P!

y, then

since u 2 NG�x�, u 2 a� P!

x0 and u� 2 NP�a�� \ a�� P!

x�0 . Let P1 �a P

xu P

a�u� P!

y. Then P1 2 P, b 2 NG�a� and y0 2 a P!

1bÿ�P1�. This again

contradicts the assumption of Case 2. Therefore, A \ NG�x� � ;. Next assume

A \ NG�y� 6� ;, say u 2 A \ NG�y�. Then u 2 y0 P!

y and u 2 A2. Let P1 �x P!

uy P

u�. Then P1 2 P and a� 2 NG�u�� \ x P!

1xÿ�P1�0 . This contradicts the

assumption of Case 2. Therefore, A \ NG�y� � ;. Since fx; yg \ �A [ NG�x�[NG�y�� ;,

nÿ 2 � jA [ NG�x� [ NG�y�j � jAj � jNG�x�j � jNG�y�j� degG a� � degG x� degG yÿ 1:


Hence, we have degG a� � degG x� degG y � nÿ 1. Then by (1), wehave degG bÿ � 1

2n� 2. Similarly, we have degG a� � 1

2n� 2. Then

a� P!

x0x P!

ab P!

yy0 P!

bÿ 2 P and degG a� � degG bÿ � n� 4. This contradictsLemma 10(3).

Case 2.3. jx P!

x0j � jy0 P!

yj � jPj � 1

In this case x0 � y0. Since G is 3-connected, there exists a path Q whichconnects x P

!xÿ0 and x�0 P

!y in Gÿ x0. Let a and b be the starting vertex and

the terminal vertex of Q, respectively, with a 2 x P!

xÿ0 , b 2 x�0 P!

y andV�Q� \ V�P� � fa; bg. We choose Q so that a P

!b is as short as possible. If

bÿy 2 E�G�, let P0 � x P!

bÿy P

b. Then P0 2 P and hence Q � ab. This impliesa 2 NG�b� \x P

!0xÿ�P0�0 , contradicting the assumption of Case 2. Hence

bÿy 62 E�G�. In particular, b 6� x�0 . Similarly, a�x 62 E�G� and a 6� xÿ0 . Sincebÿ 2 x�0 P

!y, bÿx 62 E�G�. Thus, fx; bÿ; yg is an independent set. Let

B1 � NP�bÿ�� \ x P!

bÿ; B2 � NP�bÿ�ÿ \ b P!

y; B3 � NG�bÿ� ÿ V�P�;

and B � B1 [ B2 [ B3. Then Bÿ1 � NP�bÿ� \ x P!

bÿÿ and B�2 � NP�bÿ� \ b� P!

y.Since bÿ 62 NP�bÿ� and b 2 NP�bÿ�, NG�bÿ� � Bÿ1 [ B�2 [ B3 [ fbg (disjoint)and

jBj � jB1 [ B2 [ B3j � jB1j � jB2j � jB3j� jBÿ1 j � jB�2 j � jB3j � jNG�bÿ� ÿ fbgj � degG bÿ ÿ 1:

If NG�x� \ B 6� ;, say u 2 NG�x� \ B, then u 2 B1 since u 2 x P!

x0. LetP0 � uÿ P

xu P!

y. Then P0 2 P and bÿ 2 NP�uÿ� \ x�0 P!

0y, contradicting theassumption of Case 2. Therefore, NG�x� \ B � ;. Next, assume NG�y� \ B 6� ;,say u 2 NG�y� \ B. Then u 2 x0 P

!y. If u 2 x0 P

!bÿ, then u 2 B1 and

uÿ 2 NG�bÿ� \ xÿ0 P!

bÿÿ. Let P0 � x P!

uÿbÿ P

uy P

b. Then, since P0 2 P,NG�b� � V�P0� � V�P�. Since V�Q� \ V�P� � fa; bg;Q � ab. Furthermore,a 2 NG�b� \ x P

!0xÿ0 . This contradicts the assumption of Case 2. If u 2 b P

!y,

then u 2 B2 and u� 2 NG�bÿ� \ b� P!

y. Let P0 � x P!

bÿu� P!

yu P

b. ThenP0 2 P, Q � ab and a 2 NG�b� \ x P

!0xÿ0 . This again contradicts the assumption

of Case 2. Hence, we have B \ NG�y� � ;.Since fx; yg \ �NG�x� [ NG�y� [ B� � ; and NG�x� \ NG�y� � fx0g;

nÿ 2 � jNG�x� [ NG�y� [ Bj � jNG�x� [ NG�y�j � jBj� jNG�x�j � jNG�y�j ÿ 1� jBj� degG x� degG y� degG bÿ ÿ 2:

Thus, degG x� degG y� degG bÿ � n. Since a�x 62 E�G�, there exists a vertexx1 2 x� P

!xÿ0 ÿ NG�x� with x�1 P

!x0 � NG�x�. Since x0 6� x1, x1y 62 E�G�. If


x1bÿ 2 E�G�, then x�1 2 NG�x� \ B1, a contradiction. Therefore, x1bÿ 62 E�G�.Hence fx; y; x1; b

ÿg is an independent set, and

degG x� degG y� degG bÿ � degG x1 � 3

2n� 1:

This implies degG x1 � 12

n� 1. Similarly, there exists a vertex y1 2 x�0 P!

yÿ ÿNG�y� with x0 P

!yÿ1 � NG�y�. Then, by symmetry, we have degG y1 � 1

2n� 1. Let

P1 � x1 P

xx�1 P!

yÿ1 y P

y1. Then P1 2 P and degG x1 � degG y1 � n� 2. Thiscontradicts Lemma 10. Therefore, the theorem follows. &

4. CONCLUDING REMARKS

We do not know the sharpness of Theorem 1. Actually, we even do notknow whether the coef®cient 3

2of n in Theorem 1 is best possible. Let

G � kK1 � �k � 1�K2 �k � 3�. Then diff�G� � 2, jGj � n � 3k � 2, and�4�G� � 4�K � 1� � 4

3n� 4

3. Therefore, the sharp coef®cient of n may be 4

3.

Liu et al. [7] (see also [5]) introduced the notion of a k-LTW sequence. Since�2

3; 4

3; 2; 2� is a 4-LTW sequence, their result implies Corollary 9. Actually, we

suspect that there exists a suf®cient condition in terms of 4-LTW sequences for3-connected graphs G to satisfy diff�G� � 1, which is stronger than Theorem 1.However, we have made no progress in this direction so far.

Finally, as a related problem, we would like to raise the following question: Isthere a ®xed connectivity k � 3 such that every (suf®ciently large) k-connectedgraph contains a longest path containing all the vertices of some longest cycle?Again, we have made no progress regarding this question.

ACKNOWLEDGMENT

The authors thank the referees for their helpful comments. This research wascarried out while the second author visited the Department of MathematicalSciences, The University of Memphis. He is grateful for the hospitality extendedduring his stay.

REFERENCES

[1] D. Bauer, H. J. Veldman, A. Morgana and E. F. Schmeichel, Long cycles ingraphs with large degree sums, Discrete Math 79 (1989/1990), 59±70.

[2] J. A. Bondy, Longest paths and cycles in graphs of high degree, ResearchReport CORR 80-16, University of Waterloo, Waterloo, Ontario, 1980.

[3] G. Chartrand and L. Lesniak, Graphs and Digraphs, 2nd Edition, Wadsworth& Brooks/Cole, Monterey, CA, 1986.


[4] H. Enomoto, J. van den Heuvel, A. Kaneko and A. Saito, Relative length oflong paths and cycles in graphs with large degree sums, J Graph Theory 20(1995), 213±225.

[5] O. Favaron, E. Flandrin, H. Li, Y. Liu, F. Tian and Z. Wu, Sequences, clawsand cyclability of graphs, J Graph Theory 21 (1996), 357±369.

[6] I. Fournier and P. Fraisse, On a conjecture of Bondy, J Combin Theory Ser B39 (1985), 17±25.

[7] Y. Liu, F. Tian and Z. Wu, A suf®cient condition for k-H-nice and �k � 1�-HC-nice sequences, J Nanjing Normal Univ. 17 (1) (1994).


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Relative length of longest paths and cycles in 3-connected graphs