Edge disjoint Hamilton cycles in graphs

Edge Disjoint HamiltonCycles in Graphs

Guojun LiDEPARTMENT OF MATHEMATICS

SHANDONG UNIVERSITY

JINAN 250100,

P. R. CHINA

Received November 4, 1998

Abstract: Let G be a graph of order n and k� 0 an integer. It is con-jectured in [8] that if for any two vertices u and v of a 2(k� 1)-connectedgraph G,dG(u,v�� 2 implies that max{d(u;G),d(v;G)}� (n/2)� 2k, then G hask� 1 edge disjoint Hamilton cycles. This conjecture is true for k� 0,1 (seecf. [3] and [8]). It will be proved in this paper that the conjecture is true forevery integer k� 0. ß 2000 John Wiley & Sons, Inc. J Graph Theory 35: 8±20, 2000

Keywords: Graph; Fan 2k-type graph; edge disjoint Hamilton cycles

1. INTRODUCTION

All graphs under consideration are undirected, ®nite, and simple. Let G be agraph. We use the symbols V�G� and E�G�, respectively, to denote the vertex setand edge set of G, and d�x; G� to denote the degree of x in G, and ��G� to denotethe minimum degree of G. The complement of G, denoted �G, is a graph withvertex set V�G� of which two vertices are adjacent iff they are not adjacent in G.

ÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐ

Contract grant sponsor: AROContract grant number: DAAH04-96-1-0233Contract grant sponsors: National Science Foundation of China; DoctoralDiscipline Foundation of China.

ß 2000 John Wiley & Sons, Inc.

For U � V�G�, we use G�U� to denote the subgraph of G induced by U. Fora 2 V�G�;A � V�G�;B � V�G� ÿ A, we set

NA�a; G� � fx 2 A : ax 2 E�G�g; dA�a; G� � jNA�a; G�j;NA�B; G� �

[y2B

NA�y; G�; dA�B; G� � jNA�B; G�j;

EG�A;B� � fxy 2 E�G� : x 2 A and y 2 Bg; eG�A;B� � jEG�A;B�j:

Other terminology and notation unde®ned here can be found in [2].The following condition is suf®cient for a graph to have a Hamilton cycle is

due to G. Fan [3].

Theorem 1 [3]. Let G be a 2-connected graph of order n. If for any two verticesu and v of G, dG�u; v� � 2 implies that maxfd�u; G�; d�v; G�g � �n=2�, then G

has a Hamilton cycle.

In 1994, S. Zhou got a suf®cient condition for a graph to have two edgedisjoint Hamilton cycles.

Theorem 2 [8]. Let G be a 4-connected graph of order n. If for any two verticesu and v of G, dG�u; v� � 2 implies that maxfd�u; G�; d�v; G�g � �n=2� � 2, then

G has two edge disjoint Hamilton cycles.

In general, for nonnegative integer k, S. Zhou conjectured in his paper [8] thatif for any two vertices u and v of a 2�k � 1�-connected graph G; dG�u; v� � 2implies that maxfd�u; G�; d�v; G�g � �n=2� � 2k, then G has k � 1 edge disjointHamilton cycles. Simultaneously, the graphs satisfying the above degreecondition were refered to as Fan 2k-type graphs. So the above conjecture canbe written as follows.

Conjecture 1 [8]. Every 2�k � 1�-connected Fan 2k-type graph has k � 1 edgedisjoint Hamilton cycles.

Theorem 1 and Theorem 2 imply that the conjecture is true for k � 0; 1. Beforejustifying Conjecture 1 for general k, we would like to recall two other interestingresults (cf [4] and [5]) concerning edge disjoint Hamilton cycels. In particular, thetechnique used in [5] is useful in solving Conjecture 1.

Theorem 3 [4]. Let G be a graph of order n and k be a positive integer.

If d�u; G� � d�v; G� � n� 2k ÿ 2 holds for each pair of nonadjacent verticesu and v, then G has k edge disjoint Hamilton cycles whenever n is suf®ciently

large.

Theorem 4 [5]. Let G be a graph of order n and k be a positive integer. If2k � 1 � ��G� � 2k � 2; n � 8k2 ÿ 5 and d�u; G� � d�v; G� � n for each pair of

nonadjacent vertices u and v, then G has k edge disjoint Hamilton cycles.

The following theorem will be used frequently in the proof of main results.

HAMILTON CYCLES IN GRAPHS 9

Theorem 5 [1]. Kn�n � 2k � 1� has k edge disjoint Hamilton cycles;Kn�n � 2k� has k edge disjoint Hamilton paths having any given k pairs of

vertices, which are mutually disjoint, as their endvertices respectively.

2. PRELIMINARIES

In this section, we suppose that G is a 2�k � 1�-connected graph, andV�G� � X [ Y is a partition of V�G� with jXj < 2�k � 1� and jY j � 2�k � 1�.Further, we suppose that both G�x� and G�Y � are complete. Let us nowdemonstrate a number of properties of such a graph which are useful for us tojustify Conjecture 1.

Proposition 1. Let G be a 2�k � 1�-connected graph, and let V�G� � X [ Y be

a partition of V�G� with jXj < 2�k � 1�; jY j � 2�k � 1�, and both of G�X� andG�Y � be complete. If, for any vertex x 2 X; d�x; G� � 2k � 3, then there is a vertex

y 2 NY�x; G� such that Gÿ fxyg is still 2�k � 1�-connected.

Proof. Suppose that there exists x 2 X with d�x; G� � 2k � 3 such thatGÿ fxyg is not 2�k � 1�-connected for any y 2 NY�x; G�. Since jXj < 2�k � 1�and d�x; G� � 2k � 3, we can take y; z 2 NY�x; G� with y 6� z. Let S � V�G� be acutset of Gÿ fxyg with jSj � 2k � 1. Then we can write V�G� ÿ S � A [ B sothat A 6� ;;B 6� ;;A \ B � ;, and EGÿfxyg�A;B� � ;. Since G is 2�k � 1�-connected, we may assume x 2 A and y 2 B.

Similarly, let T be a cutset of Gÿ fxzg with jTj � 2k � 1, and writeV�G� ÿ T � C [ D so that C 6� ;;D 6� ;; x 2 C; z 2 D;C \ D � ;, andEGÿfxzg�C;D� � ;. Since G�X� and G�Y � are complete, B;D � Y and A;C � X,and hence B \ C � A \ D � ;. Since x 2 C; xy 2 E�Gÿ fxzg�, and y 2 B, weget y 2 B \ T from B \ C � ;, and we similarly get z 2 S \ D. LetP � �S \ T� [ �S \ C� [ �A \ T� and P � �S \ T� [ �S \ D� [ �B \ T�. Then inGÿ fxy; xzg;P separates A \ C and B [ D. If A \ C � fxg, then this impliesjPj � d�x; G� ÿ 2 � 2k � 1; If A \ C 6� fxg, then in G;P [ fxg separates

10 JOURNAL OF GRAPH THEORY

A \ C ÿ fxg and B [ D, and hence jPj � 2�k � 1� ÿ 1 by the assumptionthat G is 2�k � 1�-connected. Thus in either case, jPj � 2k � 1. SincejPj � jQj � jSj � jT j � 4k � 2, this implies jQj � 2k � 1. On the other hand, Qseparates B \ D and A [ C in Gÿ fxy; xzg. Since y; z 2 Q, this means that Q

separates B \ D and A [ C in G. Since jQj � 2k � 1, this forces B \ D � ;. SinceY � �S [ B� \ �T [ D� by the assumption that G�Y � is complete, we now obtainjYj � j�S [ B� \ �T [ D�j � jQj � 2k � 1, which contradicts the assumption thatjYj � 2�k � 1�. Proposition 1 is proved. &

Proposition 2. Let G be a graph, and let V�G� � X [W [ Z be a partition ofV�G� with jX [W j � 2h and jZj � 2. Suppose that G�W [ Z� is complete, and

dZ�x; G� � 1 �1�

for all x 2 X. Suppose further that

dX�y; G� � h �2�

for all y 2 Z. Then X [W can be partitioned into h pairs fu1; v1g; . . . ; fuh; vhgsuch that dZ�fui; vig; G� � 2 for all 1 � i � h:

Proof. We proceed by induction of h. If h � 0, there is nothing to be proved.Thus let h � 1, and assume the proposition holds for hÿ 1. LetZ0 � fy 2 ZjdX�y; G� > 0g. If Z0 � ;, then X � ; by (1), and hence theproposition immediately follows from the assumption that G�W [ Z� is complete.Thus we may assume Z0 6� ;. Assume ®rst that jZ0j � 1. Write Z0 � fy0g. Thenby (1) and (2), jXj � dX�y0; G� � h, and hence W 6� ;. Take u1 2 X and v1 2 W .Then dZ�fu1; v1g; G� � jZj � 2. Further, applying the induction hypothesis toGÿ fu1; v1g, we see that �X [W� ÿ fu1; v1g can be partitioned into hÿ 1 pairshaving the desired property. Assume now that jZ0j � 2. Choose y0; z0 2 Z0 withy0 6� z0 so that dX�y0; G� � dX�z0; G� and so that dX�y; G� � dX�z0; G� for ally 2 Z ÿ fy0; z0g. Then by (1),

dX�y; G� � jXj3

� �� 2h

3

� �� hÿ 1 �3�

for all y 2 Z ÿ fy0; z0g. Take u1 2 NX�y0; G� and v1 2 NX�z0; G�. ThendZ�fu1; v1g; G� � 2. Further by (3), we can apply the induction hypothesis toGÿ fu1; v1g, to see that �X [W� ÿ fu1; v1g can be partitioned into hÿ 1 pairshaving the desired property. Proposition 2 is proved. &

Proposition 3. Let G be a 2�k � 1�-connected graph, and let V�G� � Y [ X be

a partition of V�G� with X 6� ; and jY j � 2�k � 1�. If G�X� and G�Y� are bothcomplete, then G has k � 1 Hamilton cycles C1;C2; . . . ;Ck�1 such that

E�Ci� \ E�Cj� � E�G�Y �� for any i, j with i 6� j.


Proof. The proof is by contradiction. Let G be a counterexample withminimal number of edges. Then jXj < 2k � 2 since otherwise the 2�k � 1�-connectedness of G guarrantees that there exists a matching with at least 2k � 2edges between X and Y, and then Theorem 5 implies that G has k � 1 edgedisjoint Hamilton cycles. Similarly, if jXj � 1, then it follows from Theorem 5that G has k � 1 edge disjoint Hamilton cycles, which contradicts the assumptionthat G is a counterexample. Thus we also have jXj � 2.

Now that jXj < 2k � 2 and jYj � 2k � 2, we can choose a proper subset W of Ysuch that jX [Wj � 2k � 2, and that eG�X;W� is as large as possible. For thesake of convenience, we write Z � Y ÿW . From jY j � 2k � 2 � jX [W j, we getjZj � jXj � 2.

Since G is a counterexample with minimal number of edges, by Proposition 1,we see that d�x; G� � 2k � 2 for all x 2 X.

Now we label the vertices in Z and W, respectively, as follows.Z : y1; y2; . . . ; ys

W : w1;w2; . . . ;wt

such that

dX�y1; G� � dX�y2; G� � � � � � dX�ys; G�:

Subsequently, we construct a sequence of graphs as follows:

X11 � fx 2 X : xy1 2 E�G� and xw1 62 E�G�g;G11 � �Gÿ EG�X11; y1�� [ E�G�X11;w1�;X12 � fx 2 X : xy1 2 E�G11� and xw2 62 E�G11�g;G12 � �G11 ÿ EG11

�X12; y1�� [ E�G11�X12;w2�;

..

.

X1t � fx 2 X : xy1 2 E�G1;tÿ1� and xwt 62 E�G1;tÿ1�g;G1t � �G1;tÿ1 ÿ EG1;tÿ1

�X1t; y1�� [ E�G1;tÿ1�X1t;wt�;

X21 � fx 2 X : xy2 2 E�G1t� and xw1 62 E�G1t�g;G21 � �G1t ÿ EG1t

�X21; y2�� [ E�G1t�X21;w1�;

X22 � fx 2 X : xy2 2 E�G21� and xw2 62 E�G21�g;G22 � �G21 ÿ EG21

�X22; y2�� [ E�G21�X22;w2�;

..

.

X2t � fx 2 X : xy2 2 E�G2;tÿ1� and xwt 62 E�G2;tÿ1�g;G2t � �G2;tÿ1 ÿ EG2;tÿ1

�X2t; y2�� [ E�G1;tÿ1�X2t;wt�;

..

. ... ..

.

Xs1 � fx 2 X : xys 2 E�Gsÿ1;t� and xw1 62 E�Gsÿ1;t�g;Gs1 � �Gsÿ1;t ÿ EGsÿ1;t

�Xs1; ys�� [ E�Gsÿ1;t�Xs1;w1�;

Xs2 � fx 2 X : xys 2 E�Gs1� and xw2 62 E�Gs1�g;Gs2 � �Gs1 ÿ EGs1

�Xs2; ys�� [ E�Gs1�Xs2;w2�;

..

.

Xst � fx 2 X : xys 2 E�Gs;tÿ1� and xwt 62 E�Gs;tÿ1�g;Gst � �Gs;tÿ1 ÿ EGs;tÿ1

�Xst; ys�� [ E�Gs;tÿ1�Xst;wt�:


It follows from the construction of Gst that Gst�X [W � is complete, and thatdZ�x; Gst� � 1 for all x 2 X since d�x; Gst� � 2k � 2 and jX [W j � 2k � 2, andthat Gst�Y � is still complete.

The following fact follows instantly from the construction of Gst.

Claim 1. Let x and yi be two vertices respectively in X and Z, and xyi be an edge

of Gst. If yj 2 Z; j 6� i, is adjacent to x in G then j < i.

The following claim will play an important role in the rest of the proof ofProposition 3.

Claim 2. Gst has k � 1 Hamilton cycles C1;C2; . . . ;Ck�1 such that

E�Ci� \ E�Cj� � E�Gst�Z��for any i,j with i 6� j:

Proof. The proof is by contradiction. Suppose that the claim is not true.Noting that Gst satis®es all conditions of Proposition 2 apart from it's condition(2), we see that there is a vertex yi0 2 Z such that dX�yi0 ;G� � k � 2. Sinceotherwise, Proposition 2 guarantees that X [W can be partitioned into k � 1 pairsfu1; v1g; . . . ; fuk�1; vk�1g such that dZ�fui; vig; G� � 2 for all 1 � i � k � 1, andTheorem 5 ensures that G contains k � 1 Hamilton cycles satisfying (4), acontradiction. Let

M � NX�yi0 ; Gst�; m � jMj;L � X ÿM; l � jLj:

By the arrangement of vertices of Z, we get dX�yi;G� � dX�yi0 ; G� �dX�yi0 ; Gst� � m for all i < i0. Set

U � fyi 2 Z ÿ fyi0g : NM�yi;G� 6� ;g:By Claim 1, U � fy1; y2; . . . ; yi0ÿ1

g, and hence dX�y; G� � m for all y 2 U.If jUj � 2k ÿ t ÿ l, since jU [W [ L [ fyi0gj � 2k ÿ t ÿ l� t � l� 1 �

2k � 1 and jY j � 2k � 2, we know that U [W [ L [ fyi0g is a cut set of G,and so that ��G� � jU [W [ L [ fyi0gj � 2k � 1, contradicting the fact that G is2�k � 1�-connected. So we may assume that jUj � 2k ÿ t ÿ l� 1.

Keeping t � l� m � 2k � 2 and m � k � 2 in mind, we know that t � l � k.By the fact that we have chosen W so that eG�W ;X� is as large as possible, wehave that dX�w; G� � dX�yi0 ; G� � dX�yi0 ; Gst� � m for all w 2 W . It follows that

tjXj � eGst�X;W� � eG�W ;X� � jUjm� dX�yi0 ; G� ÿ jXj

� tm� �2k ÿ t ÿ l� 1�m� mÿ jXj� tm� �k � 1�m� mÿ �2k � 2ÿ t�� m� 1�t � km� 2mÿ 2k ÿ 2 � �m� 1�t � km� 2�k � 2� ÿ 2k ÿ 2

> 2mt � t�2k � 4� > tjXj:


This ®nal contradiction means that Gst has k � 1 Hamilton cycles satisfying (4).Claim 2 is proved. &

By Claim 2, we can choose p and q such that(a) Gpq has k � 1 Hamilton cycles C1;C2; . . . ;Ck�1 satisfying

E�Ci� \ E�Cj� � E�Gpq�Z��

for any i, j with i 6� j.(b) subject to (a), p is as small as possible;(c) subject to (a) and (b), q is as small as possible.

Since G contains no k � 1 Hamilton cycles satisfying (1), we know that G 6� Gpq.Set

G0 � Gp;qÿ1; q > 1;Gpÿ1;t; q � 1:

�Then it follows from the generation of Gpq that Gpq � �G0 ÿ EG0 �Xpq; yp��[E�G0 �Xpq;wq�. By the choice of p and q, we know that Xpq 6� ;. For the sake ofconvenience, we use X1; y1 and w1, respectively, to denote Xpq; yp and wq. LetX1 � fx1; x2; . . . ; xdg. Then xiy1 2 E�G0� and xiw1 62 E�G0� for 1 � i � d. WriteX2 � fz 2 Xjzy1; zw1 2 E�G0�g. Now for the cycle

Ci � a1a2 � � � ana1;

where al � w1; as � y1; of Gpq, we use Qi and Qi�k�1 respectively to denote thepaths alal�1 � � � as and asas�1 � � � al. Then Q1;Q2; . . . ;Qk�1;Qk�2; . . . ;Q2�k�1� are2�k � 1� paths in Gpq connecting y1 and w1, and they satisfy

E�Qi� \ E�Qj� � E�Gpq�Z��:

Note that for 1 � i � 2�k � 1�, the vertices adjacent to y1 and w1 in Qi

respectively are determined uniquely. Set

G0 � G0 ÿ[2�k�1�

i�1

�E�Qi� \ �E�G0� ÿ E�G0�Z��:

We are going to construct k � 1 Hamilton cycles, O1;O2; . . . ;Ok�1, in G0 suchthat E�Oi� \ E�Oj� � E�G0�Z�� by replacing in C1;C2; . . . ;Ck�1 the edges in[2�k�1�

i�1 E�Qi� ÿ E�G0�, which will contradict the choice of p and q.By the generation of Gpq;[2�k�1�

i�1 E�Qi� ÿ E�G0� � EGpq�w1;X1�. Suppose there

is some xj 2 X1 such that xjw1 2 Qj1 . We let zj;1 � xj, and put

Hj1 � �Qj1 ÿ fzj;1w1; zj;2y1g� [ fzj;2w1; zj;1y1g;


where zj;2 is the vertex joined to y1 on Qj1 . Since zj;2y1 2 E�Gpq�, we havezj;2 2 X2 [ Y and hence zj;2w1 2 E�G0� \ E�Gpq�. Thus Hj1 is a path in G0

connecting y1 and w1. Therefore, Hj1 [ Qk�1�j1 is a Hamilton cycle of�Gpq ÿ fzj;1w1g� [ fzj;1y1g.

Now suppose zj;2w1 is an edge of some Qj2 and zj;2 2 X. Note that zj;2 2 X2. Weput

Hj2 � �Qj2 ÿ fzj;2w1; zj;3y1g� [ fzj;3w1; zj;2y1g;

where zj;3 is the vertex adjacent to y1 on Qj2 ; and so on.In general, we have found zj;1; zj;2; . . . ; zj;s� j� and Qj1 ;Qj2 ; . . . ;Qjs� j�ÿ1

; s� j� � 2,such that

(i) zj;1 � xj 2 X1; zj;l 2 X2, for 2 � l � s� j� ÿ 1;(ii) zj;lw1 and zj;l�1y1 are on Qjl ; 1 � l;� s� j� ÿ 1;

(iii) zj;l 6� zj;d and Qjl 6� Qjd , for any 1 � l 6� d � s� j� ÿ 1; and(iv) subject to (i), (ii) and (iii), s� j� is as large as possible.

Now we claim that

zj;s� j� 62 fzj;l : 1 � l � s� j� ÿ 1g: �5�

In fact, if zj;s� j� � zj;l for some 1 � l � s� j� ÿ 1, then from zj;s� j�y1 2 E�Gpq�and zj;1y1 62 E�Gpq�, we get l 6� 1, and hence it is clear from the property (ii) thatQjs� j�ÿ1

� Qjlÿ1and hence zj;s� j�ÿ1 � zj;lÿ1, contradicting the property (iii).

Moreover, if zj;s�j� 2 X2 and there exists some Qjs� j� such that zj;s� j�w1 2 Qjs� j� ,then there exists some zj;s� j��1 on Qjs�j� such that zj;s� j��1y1 2 Qjs� j� . IfQjs� j� 6� Qjl ; 1 � l � s� j� ÿ 1, we have a contradiction to the property (iv). Andif Qjs� j� � Qjl for some l; 1 � l � s� j� ÿ 1, then by the uniqueness of the verticesjoined to w1 and y1, respectively, on each Qi, we have zj;s� j� � zj;l, contradictingthe formula (5).

The above argument and the fact that zj;s� j�y1 2 E�Qjs� j�ÿ1� show that either

zj;s� j�2Y , or zj;s� j� 2 X2 and zj;s� j�w1 2 E�G0�. Thus we de®ne in general

Hjl � �Qjl ÿ fzj;lw1; zj;l�1y1g� [ fzj;l�1w1; zj;ly1g; for 1 � l � s� j� ÿ 1: �6�

We conclude from this de®nition that these Hjl ; 1 � l � s� j� ÿ 1, are pathsconnecting y1 and w1 and pairwise having no common edge in E�G0� ÿ E�G�Z��.Moreover, since fzj;ly1; zj;l�1w1 : 1 � l � s� j� ÿ 1g � [s� j�ÿ1

l�1 E�Qjl� [ E�G0�,each of Hjl ; 1 � l � s� j� ÿ 1, has in E�G0� ÿ E�G�Z�� no common edge withany of the remaining Qi; i 2 f1; 2; . . . ; 2�k � 1�g ÿ f j1; j2; . . . ; js� j�ÿ1g.

If xi and xj are two different elements of X1 such that xiw1 is on some Qi1

and xjw1 is on some Qj1 , then we may get xi � zi;1; zi;2; . . . ; zi;s�i�;Qi;1;Qi;2; . . . ;Qi;s�i�ÿ1 and xj � zj;1; zj;2; . . . ; zj;s� j�;Qj1 ;Qj2 ; . . . ;Qjs� j�ÿ1

respectively, bythe above procedure.


We claim that zi;l 6� zj;d for any 1 � l � s�i� and any 1 � d � s� j� (except thatit is possible that zi;s�i� � zj;s� j� if zi;s�i�; zj;s� j� 2 Y�, and Qil 6� Qjd for any1 � l � s�i� ÿ 1 and any 1 � d � s� j� ÿ 1. Otherwise, suppose that l is theminimal number such that there exist some d; 1 � d � s� j�, with zi;l � zj;d. Ifd � 1, then since zi;ry1 2 E�Gpq� for each 2 � r � s�i� and zj;1y1 62 E�Gpq�, weget l � 1, which contradicts the assumption that xi 6� xj. Thus d 6� 1. Similarlyl 6� 1. Now, since the vertices adjacent to w1 and y1, respectively, on each Qi arerespectively unique, it is easy to see from the property (ii) that Qilÿ1

� Qjdÿ1

(unless l � s�i�; d � s� j� and zi;s�i�; zj;s� j� 2 Y�, and hence zi;lÿ1 � zj;dÿ1, contra-dicting the choice of l.

Based on the above discussion, for any xj in X1 such that xjw1 is on some Qj1,we de®ne Hjl ; 1 � l � s� j� ÿ 1, as in (6). Put Hj � Qj for the remaining Qj thatdo not appear in the above procedure. Then we obtain H1;H2; . . . ;H2�k�1� in G0,which are pairwise edge disjoint in E�G0� ÿ E�G�Z��. By this de®nition and theabove argument, we put

Oi � Hi [ Hi�k�1; for 1 � i � k � 1;

which are k � 1 Hamilton cycles of G0 such that E�Oi� \ E�Oj� � E�G0�Z��,contradicting the choice of p and q. Proposition 3 is proved. &

3. PROOF OF CONJECTURE 1

In this section we turn to justify the following theorem which was everconjectured by Zhou [8].

Theorem 6. Every 2�k � 1�-connected Fan 2k-type graph has k � 1 edge

disjoint Hamilton cycles.

Proof. The proof is by contradiction. Suppose that G is a 2�k � 1�-connectedFan 2k-type graph, but G contains no k � 1 edge disjoint Hamilton cycles.Furthermore, let G be a counterexample with the maximum number of edges, i.e.,the graph G itself contains no k � 1 edge disjoint Hamilton cycles, but for eachpair of nonadjacent vertices u and v, either G� fuvg contains k � 1 edge disjointHamilton cycles or G� fuvg is no longer a Fan 2k-type graph. Put

Y � v 2 V�G� : d�v; G� � n

2� 2k

n o:

Since G is 2�k � 1�-connected, n � 2k � 3. By Theorem 5 and the fact that G is acounterexample, we know that G is not complete, and that Y 6� ;.

The proof of Theorem 6 can be divided into a series of the following lemmas.

Lemma 1. G[Y] is complete.


Proof. Suppose, to the contrary, that there are two vertices x; y 2 Y such thatxy 62 E�G�. It is easy to verify that G� fxyg is still a Fan 2k-type graph. By theassumption that G is a maximum counterexample, G� fxyg must have k � 1edge disjoint Hamilton cycles C1;C2; . . . ;Ck�1, one of them, say Ck�1, passesthrough the edge fxyg in G� fxyg. Let

G0 � Gÿ[ki�1

E�Ci�:

Then d�x; G0� � d�y; G0� � d�x; G� � d�y; G� ÿ 4k � n. By the proof of Ore'stheorem [6], there are two consecutive vertices u and v along the pathCk�1 ÿ fxyg from the starting vertex x to the terminating vertex y such thatboth xv and yu are edges of G0, and therefore Ck�1 ÿ fxy; uvg � fxv; yug is aHamilton cycle in G0 and so in G which is certainly disjoint from the cyclesC1;C2; . . . ;Ck, contradicting the assumption that G is a counterexample. Nowwe write &

X � V�G� ÿ Y :

By Lemma 1, Theorem 5, and the fact that G is a counterexample, we knowthat X 6� ;. We use Gi � �Xi;Ei�; 1 � i � !, to denote the components ofG�X�, and let Yi � NY�Xi; G�; 1 � i � !. By the hypothesis, there is not apair of vertices in X at distance two. Consequently, the following lemma istrivial.

Lemma 2. (1) Gi�1 � i � !� is complete; (2) Yi \ Yj � ;; i 6� j:

To justify Theorem 6, the following lemma is useful.

Lemma 3. jYij � 2�k � 1�; i � 1; 2; . . . ; !:

Proof. Suppose, to the contrary, that jYij < 2�k � 1�. Then since G is2�k � 1�-connected and Xi 6� ;, we know that ! � 1, i.e., Xi � X and Yi � Y . G

being incomplete guarrantees that there is at least one vertex y 2 Y which is notadjacent to some vertex in X. Therefore, we get the inequality

jY j � d�y; G� � 1ÿ �jXj ÿ 1� � n

2� 2k ÿ jXj � 2:

It follows from jXj � jY j � n that n � 4k � 4, and from jY j � 2k � 1 thatjXj � nÿ jY j � 4k � 4ÿ �2k � 1� � 2k � 3 � jY j � 2. So we obtain theinequality

jY jjXj ÿ 1 � ÿ 2

jXj.Using the equality

Px2X dY�x; G� �Py2Y dX�y; G�, we can derive the

following inequality jXj maxx2X dY�x; G� � jY j miny2YdX�y; G�:


If follows from the above inequalityjY jjXj ÿ 1 � ÿ 2

jXj < 0 that

maxx2X

dY�x; G� � miny2Y

dX�y; G� � jYjjXj ÿ 1

� �miny2Y

dX�y; G�

� miny2Y

dX�y; G� � jYjjXj ÿ 1

� �jXj

� miny2Y

dX�y; G� � jYj ÿ jXj:

Then a contradiction will be derived from the series of following inequalities.

n

2� 2k > max

x2Xd�x; G� � jXj ÿ 1�max

x2XdY�x; G�

� jXj ÿ 1�miny2Y

dX�y; G� � jY j ÿ jXj� jYj ÿ 1�min

y2YdX�y; G�

� miny2Y

d�y; G� � n

2� 2k: &

The following lemma, which will be the key for justifying Theorem 6, followsinstantly from Proposition 3.

Lemma 4. For i � 1; 2; . . . ; !;G�Yi [ Xi� has k � 1 Hamilton cycles

Ci1 ;Ci2 ; . . . ;Cik�1such that

E�Cis� \ E�Cit� � E�G�Yi��

for any s,t with s 6� t.

Lemma 5.

E�Ci; j� \ E�G�Y �� 6� ;; i � 1; 2; . . . ; !; j � 1; 2; . . . ; k � 1:

Proof. If jXij � 2k � 2, by the proof of Proposition 3, each Hamilton cycleCi;j; j � 1; 2; . . . ; k � 1 in G�Xi [ Yi� is constituted by two independent edgesbetween Xi and Yi, and a Hamilton path in G�Xi�, and a Hamilton path in G�Yi�. SoE�Ci;j� \ E�G�Y �� 6� ;; j � 1; 2; . . . ; k � 1:

If jXij < 2k � 2, since jYij � 2k � 2 there must be two consecutive vertices oneach cycle Ci; j; j � 1; 2; . . . ; k � 1 being in Yi. Hence E�Ci;j� \ E�G�Y �� 6� ;;j � 1; 2; . . . ; k � 1. &

Lemma 6. If M is a matching of Kn, then Kn has a Hamilton cycle passing

through all edges of M.


Lemma 7. G contains k � 1 Hamilton cycles C1;C2; . . . ;Ck�1 such that

E�Ci� \ E�Cj� � E�G�Y �� 7�

Proof. In Lemma 5, let j � 1. Then Lemma 5 guarantees that there existsy0i;1y00i;1 2 E�G�Yi�� \ E�Ci;1� for 1 � i � !. Write Y0 � Y ÿ [!i�1Yi.

Then by Lemma 6, G�Y0 [ �S!

i�1fy0i;1; y00i;1g�� has a Hamilton cycle passingthrough the edges fy0i;1y00i;1 : 1 � i � !g. Now for each i; 1 � i � !, we substitutethe path Ci;1 ÿ fy0i;1y00i;1g for the edge y0i;1y00i;1 to get a Hamilton cycle C1 of G. Forj � 2; . . . ; k � 1, using the same method, we can then get the cycles C2; . . . ;Ck�1.By Lemma 3, we can see that the k � 1 cycles are just our desired. &

Lemma 8. G has k � 1 edge disjoint Hamilton cycles if and only if G has k � 1Hamilton cycles satisfying (7).

Proof. It suf®ces to show the suf®ciency. Let C1;C2; . . . ;Ck�1 be k � 1Hamilton cycles satisfying (7) such that p �Pi6�j jE�Ci� \ E�Cj� \ E�G�Y ��j isas small as possible. We assume, without loss of generality, that p 6� 0 andxy 2 E�C1� \ E�C2� \ E�G�Y��. Set

G0 � Gÿ[k�1

i�2

Ci;

Then dG0 �x� � dG0 �y� � n. By the proof of Ore's theorem [6], we know that thereare two consecutive vertices u and v on C1 such that both xu and yv are the edgesof G0 and C01 � �C1 ÿ fxy; uvg� � fxu; yvg is a Hamilton cycle. LetC0i � Ci; 2 � i � k � 1. Then C01;C

02; . . . ;C0k�1 are k � 1 Hamilton cycles of G

satisfying (7) andXi6�j

jE�C0i� \ E�C0j� \ E�G�Y ��j � pÿ 1 < p;

contradicting the choice of C1;C2; . . . ;Ck�1. &

Lemma 7, together with Lemma 8, ensures that G has k � 1 edge disjointHamilton cycles, violating our assumption at the beginning. Theorem 6 is proved.

&

We close this paper by pointing out that the 2�k � 1�-connectedness inTheorem 6 is essential. For example, consider the graph G consisting of twovertex disjoint complete graphs K2k�1 and Kp�p � 6k � 3� and a matchingwith 2k � 1 edges between the vertex sets of K2k�1 and Kp. Evidently G is a2k � 1-connected Fan 2k-type graph, but contains no k � 1 edge disjointHamilton cycles.


ACKNOWLEDGMENT

I am sincerely grateful to an anonymous referee who presented novel short proofsfor Proposition 1 and Proposition 2, and many useful suggestions for the polish ofthe manuscript.

REFERENCES

[1] L. W. Beineke and R. L. Wilson, Selected Topics in Graph Theory, AcademicPress, New York, 1978, 161 p.

[2] J. A. Bondy, U. S. R. Murty, Graph Theory with Applications, MacMillan,London, 1976.

[3] G. H. Fan, New suf®cient condition for cycles in graphs, J Combin TheoryB37 (1984), 221±227.

[4] R. J. Faudree, C. C. Rousseau, and R. H. Schelp, Edge disjoint Hamiltoniancycles. Graph Theory with Applications to Algorithms and Computer Science(Kalamazoo, Mich; 1984), Wiley-Interscience New York (1985) 231±249,MR 87c:05104.

[5] H. Li, Edge disjoint cycles in graphs, J Graph Theory 3 (1989), 313±322.

[6] O. Ore, Note on Hamilton circuits. Am Math Monthly 67 (1960), 55.

[7] F. Tian, A short proof of a theorem about the circumference of a graph,J Combin Theory B45 (1988), 373±375.

[8] S. Zhou, Disjoint Hamiltonian cycles in Fan 2k-type graphs, J Graph Theory 3(1989), 313±322.


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Edge disjoint Hamilton cycles in graphs