Circuits in Graphs and the Hamiltonian Index - LISA · Circuits in Graphs and the Hamiltonian Index...

Circuits in Graphs and the Hamiltonian Index

Liming Xiong

2001

Ph.D. thesisUniversity of Twente

Also available in print:www.tup.utwente.nl/uk/catalogue/technical/circuitsingraphs

T w e n t e U n i v e r s i t y P r e s s

http://www.tup.utwente.nl/uk/catalogue/technical/circuitsingraphs

Circuits in Graphs and the Hamiltonian Index

Publisher: Twente University Press,

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Cover design: Jo Molenaar, [deel 4] ontwerpers, Enschede

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© L. Xiong, Enschede, 2001

No part of this work may be reproduced by print, photocopy or any other means

without the permission in writing from the publisher.

ISBN 9036516196

T w e n t e U n i v e r s i t y P r e s s

CIRCUITS IN GRAPHSAND

THE HAMILTONIAN INDEX

PROEFSCHRIFT

ter verkrijging vande graad van doctor aan de Universiteit Twente,

op gezag van de rector magnificus,prof.dr. F.A. van Vught,

volgens besluit van het College voor Promotiesin het openbaar te verdedigen

op woensdag 13 juni 2001 te 13.15 uur

door

Liming Xionggeboren op 8 april 1965

te Anyi, P.R. China

Dit proefschrift is goedgekeurd door de promotoren

prof.dr. C. Hoedeprof.dr. X. Li

en de assistent-promotor

dr.ir. H.J. Broersma

If more distant views are what you desire, you simply have to go up higher.

(Chinese proverb)

(www.reliefweb.int/w/map.nsf/wByCLatest/843776DE959CC44485256A240063F4C3?Opendocument)

To my parents

Preface

This thesis is the result of almost five years of research in the field of hamiltoniangraph theory between September 1996 and April, 2001. After an introductorychapter the reader will find eleven chapters that contain more or less independenttopics within this research field. The papers that together underly this thesis arelisted below.

Papers underlying this thesis

[1] L. Xiong, J. Wang, Z. Li and M. Li,Pancyclic line graphs, in: Combinaorics,Graph Theory, Algorithms and Applications (edited by Y. Alavi, Don R. Lick andJ. Liu, proceedings of the third China-U.S.A. conference on graph theory, Beijing1993, 399-403.[2] L. Xiong, The line graph of a graph with diameter at most two is either pan-cyclic or a cycle of length 4 or 5. J. Jiangxi Normal University, Vol.21, 1 (1997)22-25.[3] L. Xiong, Edge degree conditions for subpancyclicity in line graphs, DiscreteMathematics 188 (1998) 225-232.[4] L. Xiong, On subpancyclic line graphs, J. Graph Theory 27 (1998) 67-74.[5] L. Xiong, H.J. Broersma, C. Hoede and X. Li,Degree sums and subpancyclic-ity in line graphs, to apear in Discrete Mathematics.[6] L. Xiong, The hamiltonian index of a graph, to appear in Graphs and Combi-natorics.[7] H.J. Broersma and L. Xiong,On minimum degree conditions for supereuleriangraphs, to appear in Discrete Applied Mathematics.[8] L. Xiong and Z. Liu,Hamiltonian iterated line graphs, submited.[9] L. Xiong, H.J. Broersma, C. Hoede and X. Li,The hamiltonian index and itsbranch-bonds, submitted.[10] H.J. Broersma and L. Xiong,Connected even factors with degree restrictions,submitted .[11] L. Xiong, H.J. Broersma and C. Hoede,Subpancyclicity in line graph anddegree sums of vertices along a path, preprint.

Acknowledgement

Although many people think of mathematics as a solitary activity, this is definitelynot the case. For instance, this thesis would have been nonexistent without the co-operation and support of many people.

The roots of my interest in graph theory go back more than a decade. At that timeI was a bachelor’s degree student in the Jiangxi Normal University. I followedmy graph theory course from Zhanhong Liu. It was beyong doubt the stimulatingenthusiasm of my supervisor, Zhongxiang Li, and Jianfang Wang that activatedmy further interest in graph theory when I was a master’s degree student at theUniversity of Science and Technology Beijing. The discussions with Mingchu Liformed the actual beginning of my research in graph theory.

This thesis was begining to take shape when Kees Hoede invited me to work in hisgroup. Xueliang Li’s encouragement fastened my progress. The common interestin the same topics with Hajo Broersma made me very happy during my stay inTwente.

Furthermore, I enjoyed the company of Shenggui Zhang, Xiaodong Liu, Hao Sun,Baozhu Guo, Kiyoshi Yoshimoto, Hongjian Lai and Guisheng Yi.

Above all I must thank my family, including my parents, my wife, sisters andbrother. In particular, I am glad that my daughter Xiangyu Xiong’s progress wentparallel to my progress. I am very grateful to my wife Lili Wen for her skillfuland accurate typing.

Contents

1 Introduction 11.1 Iterated line graphs and the hamiltonian index . . . .. . . . . . . 31.2 Subpancyclic line graphs .. . . . . . . . . . . . . . . . . . . . . 7

1.2.1 Degree conditions for subpancyclicity in line graphs . . . 71.2.2 Pancyclic line graphs involving its diameter .. . . . . . . 10

1.3 Spanning circuits in graphs . . . . . . . . . . . . . . . . . . . . . 111.3.1 Connected even [2;2k− 2]-factors . . . . . . . . . . . . . 111.3.2 Supereulerian graphs . . . . . .. . . . . . . . . . . . . . 13

2 Hamiltonian iterated line graphs 172.1 Characterization of graphs with iterated line

graphs that are hamiltonian. . . . . . . . . . . . . . . . . . . . . 192.2 Methods for determining the hamiltonian index of a graph . . . . 27

2.2.1 Split blocks of a graph . . . . .. . . . . . . . . . . . . . 272.2.2 The contraction of a graph . . .. . . . . . . . . . . . . . 32

2.3 Upper bounds for the hamiltonian index of a graph .. . . . . . . 33

3 The hamiltonian index of a graph and its branch-bonds 373.1 Branch-bonds. . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.2 A reduction method for determining the hamiltonian index of a

graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.3 Sharp upper and lower bounds forh.G/ . . . . . . . . . . . . . . 443.4 Analysis of known results . . . . . . . . . . . . . . . . . . . . . . 47

4 The hamiltonian index of a graph 514.1 A sharp upper bound . . .. . . . . . . . . . . . . . . . . . . . . 524.2 The hamiltonian index of the complement of a graph. . . . . . . 56

i

ii CONTENTS

5 Edge degree conditions for subpancyclicity in line graphs 655.1 Proof of the main theorems . . . . . . . . . . . . . . . . . . . . . 66

6 On subpancyclic line graphs 756.1 Preliminary Results . . . . . .. . . . . . . . . . . . . . . . . . . 766.2 Proof of Theorem 6.3 . . . . . . . . . . . . . . . . . . . . . . . . 78

7 Degree sums and subpancyclicity in line graphs 877.1 Proof of Theorem 7.3 . . . . . . . . . . . . . . . . . . . . . . . . 89

8 Subpancyclicity in the line graph of a graph with large degree sums 1038.1 Proof of Theorem 8.3 . . . . . . . . . . . . . . . . . . . . . . . . 104

9 Pancyclic line graphs 1179.1 Some Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . 1189.2 An alternative proof of Theorem 9.1 and some lemmas. . . . . . . 1229.3 Proof of the main results . . . . . . . . . . . . . . . . . . . . . . 125

10 Connected even factors with degree restrictions 12910.1 Connected even [2, 2k]-factors. . . . . . . . . . . . . . . . . . . 13010.2 Every supereulerian claw-free graph contains a connected even

[2,4]-factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

11 A note on minimum degree conditions for supereulerian graphs 14111.1 Catlin’s reduction method . . .. . . . . . . . . . . . . . . . . . . 14311.2 Main result and its consequences . . . . . .. . . . . . . . . . . . 14411.3 Remarks on minimum degree condition . .. . . . . . . . . . . . 148

Bibliography 149

Index 155

Summary 157

Chinese Summary 159

Curriculum Vitae 161

Chapter 1

Introduction

We begin with a short informal introduction to the topics of this thesis. In 1891the Danish mathematician Julius Petersen (1839-1910) published a paper on thefactorization of regular graphs. This was the first paper in the history of mathemat-ics to explicitly contain fundamental results in graph theory (see [43] and [44]).He formulated the problem in terms of a certain structure, in such a way that hewas not only able to deal with the particular problem, but also to give a generalsolution for other problems of the same type. Such a structure is called agraph,consisting of vertices and edges. The earliest known paper related to graph theoryis undoubtedly that ofL: Euler [30], published in 1736, in which he solved theproblem of the“Seven Bridges of Königsberg”. He showed in this paper that it isimpossible to cross each of the seven bridges of Königsberg exactly once duringa walk through the city (The city of Königsberg is now named Kaliningrad). Bya circuit we mean a connected subgraph in which every vertex has even degree.A eulerian circuitis a circuit which traverses every edge of a graph exactly once.The Konigsberg Bridge Problem is then equivalent to the problem of determininga eulerian circuit of a graph corresponding to the bridges and areas of the cityof Konigsberg. A related problem is known as the“Chinese postman problem”,since it was first considered by a Chinese mathematician, Kuan (1962) [35].

A hamiltonian cyclein a graph is a circuit which traverses every vertex of thegraph exactly once. A graph which contains a hamiltonian cycle is called ahamil-tonian graph. Hamiltonian graphs are named after William Rowan Hamilton, al-though they were studied earlier by Kirkman. In 1856, Hamilton invented a math-ematical game, the“Icosian Game”, consisting of a dodecahedron. Each of itstwenty vertices was labeled with the name of a city and the problems was to find

1

2

a hamiltonian cycle in this dodecahedron graph, to make a“Voyage around theworld”.

The line graph transformation is probably the most interesting of all graph trans-formations, and it is certainly the most widely studied. Much of this activity wasstimulated by Ore’s discussion of line graphs, and problems about them, in [47].The line graph concept is quite natural, and has been introduced in several ways,see Whitney[56]. Herethe line graphL.G/ of a graphG hasE.G/ as its vertex setand two vertices are adjacent inL.G/ if and only if they are adjacent as edges inG.

In this thesis we concentrate on the existence of certain circuits in graphs andline graphs. Harary and Nash-Williams characterized the graphs for which linegraphs are hamiltonian. One can apply their characterization in two ways. One isto investigate the cycle structure of line graphs and the other is to investigate thehamiltonian index of a graph. This will be discussed in Chapter 2 up to Chapter 9.

In Chapter 2 up to Chapter 4, we concentrate on thehamiltonian indexof a graph,i.e., the minimumm such thatthe m-iterated line graphLm.G/ of a graphG ishamiltonian. HereLm.G/ is defined recursively byL0.G/ = G and Lm.G/ =L.Lm−1.G//. We first characterize the graphs for whichm-iterated line graphsare hamiltonian. Using this we give some methods to determine thehamiltonianindexof a graph and derive some upper and lower bounds. These bounds involvetwo parameters of the graph, namely itsbranch-bondand itsdiameter. We willalso discuss the relationship between the hamiltonian index of a graph and that ofits complement.

The next topic in this thesis concerns degree sum conditions. We mainly concen-trate on degree sums, along a path in graphs, that are sufficient for pancyclicity inhamiltonian line graphs. In Chapter 5 up to Chapter 9, we will obtain results onthe so-calledsubpancyclic graphs.

A concept that generalizes both hamiltonian cycles and spanning eulerian sub-graphs is the concept of aconnected even[2;2k]-factor. In Chapters 10 and 11,we concentrate on degree sum conditions guaranteeing the existence of connectedeven [2;2k]-factors and introduce a new type of condition involving the degreesum ofk+ 1 disjoint vertices in a graph that have a common neighbor.

In the remainder of this introduction we describe our results and indicate some

1.1 Iterated line graphs and the hamiltonian index 3

connections with related older results. Most of the terminology we use in thisthesis is standard and can be found in any textbook on graph theory. We useBondy and Murty [8] as our main source for terminology and notation. We givea brief outline of the main results of this thesis without introducing all relevantterminology. The reader is referred to the separate chapters for more details andproofs.

1.1 Iterated line graphs and the hamiltonian index

There is a close relationship between dominating eulerian subgraphs of a graphand hamiltonian cycles in its line graph.

Theorem 1.1 (Harary and Nash-Williams [32]) The line graph L.G/ of a con-nected graph G with at least three edges is hamiltonian if and only if G has adominating eulerian subgraph.

There are various applications of the above characterization of hamiltonian linegraphs, (see e.g.[3], [5], [26], [24]). In Chapter 2, we will characterize thosegraphs for which then-iterated graphs are hamiltonian forn ≥ 2. We will needsome terminology, abranch etc.A branchof a graph is a nontrivial path the endsof which do not have degree 2 and the internal vertices of which, if any, have de-gree two. B.G/ denotes the set of branches ofG. Define B1.G/ = {b ∈ B.G/ :one of ends ofb has degree one}: Let EUn.G/ denote the set of subgraphs satis-fying the following five conditions for every elementH in EUn.G/:

• Any vertex ofH has even degree, i.e.,dH.x/≡ 0 (mod 2) for anyx∈ V.H/;

• {v ∈ V.H/ : dH.v/ = 0} ⊆ {v ∈ V.G/ : dG.v/ ≥ 3} ⊆ V.H/;

• dG.H1; H − H1/ ≤ n− 1 for any componentH1 of H;

• |E.b/| ≤ n+ 1 for any branchb in B.G/\BH.G/;

• |E.b/| ≤ n for any branchb in B1.G/.

Theorem 1.2 Let G be a connected graph with at least three edges and let n≥ 2be an integer. Then the n-th iterated line graph L.G/ of G is hamiltonian if andonly if EUn.G/ is nonempty.

4

Theorem 1.2 is not a generalization of Theorem 1.1 since for any subgraph inEUn.G/ we have{v ∈ V.H/ : dH.v/ = 0} ⊆ {v ∈ V.G/ : dG.v/ ≥ 3} ⊆ V.H/:

The investigation of thehamiltonian indexof a graph was initiated by Chartrand[22] who introduced the definition of the hamiltonian index,h.G/. He showedthat thehamiltonian indexexists for any connected graph other than a path.

Theorem 1.3 (Chartrand [22]) Let G be a connected graph of order n other thana path. Then h.G/ ≤ n− 3:

Theorem 1.4 (Chartrand and Wall [23]) Let G be a connected graph withŽ.G/≥3. Then h.G/ ≤ 2.

A set S of branches in a graphG is called abranch-cutif the subgraph obtainedfrom G by deleting the branches ofS, i.e., by deleting all edges and internalvertices of each branch ofS, has more components thanG. A minimal branch-cutis called abranch-bond. Thelength of a branch-bondis the length of a smallestbranch in it. A branch-bond is called odd if it contains an odd number of branches.Defineh1.G/ to be the largest value of the length of a branch-bond which containsonly one branch such that one of its ends has degree one inG; defineh2.G/ to bethe largest value of the length of a branch-bond which contains only one branchsuch that its ends have degree at least three inG; defineh3.G/ to be the largestvalue of the length of an odd branch-bond which have at least three branchesin G. In Chapter 3, we will determine a lower bound and upper bound for thehamiltonian index, based onh1;h2; andh3.

Theorem 1.5 Let G be a connected graph with h.G/ ≥ 1. Then

h.G/ ≥ max{h1.G/;h2.G/+ 1;h3.G/− 1}:

Theorem 1.6 Let G be a connected graph that is neither a 2-cycle nor a path.Then

h.G/ ≤ max{h1.G/;h2.G/+ 1;h3.G/+ 1}:

We also construct a class of graphs showing thath.G/may reach all integers fromh3.G/− 1 toh3.G/+ 1. This upper bound generalizes and improves many upperbounds on the hamiltonian index of graphs in literature see [1], [20], [36], [49].Chartrand and Wall [23] showed a formula to determine the hamiltonian index ofa tree, their result is a consequence of Theorems 1.5 and 1.6.

1.1 Iterated line graphs and the hamiltonian index 5

Corollary 1.1 (Chartrand and Wall [23]) Let T be a tree that is not a path. Then

h.T/ = max{h1.T/;h2.T/+ 1}:

A generalization of Corollary 1.1 gave in Chapter 2, is the following.

Theorem 1.7 Let G be a connected graph and let SB1; SB2; : : : ; SBt denote allsplit blocks of G. Then

h.G/ = max{h.SB1/;h.SB2/; : : : ;h.SBt/; k.G/}:

Herek.G/ = 0 if G is 2-connected,k.G/ = 1 if G is not 2-connected and has nobranch-bond, andk.G/ = max{h1.G/;h2.G/+ 1} otherwise. See Chapter 2 for adetailed definition of “split blocks”.

Catlin[19] developed a reduction method for determining whether a graph has aspanning closed trail(a connected even factor). Using his reduction method andTheorem 1.1, many upper bounds for the hamiltonian index of a graph have beenestablished (see e.g. [20], [36], [49]). However, the lower iterated line graph mustbe considered using Theorem 1.1 to study it. Using Theorem 1.2, we only need toconsider the original graph. In presenting the proofs of Theorems 1.5 and 1.6, wegive a refinement of Catlin’s reduction method.

Theorem 1.8 Let G be a connected graph other than a path, and let G1; G2; · · · ;Gk be all nontrivial components of G[{v : dG.v/ ≥ 3}] − {e : e is a nontrivial cutedge of G}. If h.G/ ≥ 2, then

h.G/ = h.G=={G1;G2; : : : ;Gk}/:

Here theattach-contractionG==H is the graph obtained fromG by contracting alledges ofH and the attachment of two new edge-disjoint branches of length two.The following method in Chapter 2 is another reduction technique.

Theorem 1.9 Let G be a connected graph and let b1;b2; : : : ;bk be all branchesof length at least 2 in G. If h.G/ ≥ 4; then

h.G/ = h.G={b1;b2; : : : ;bk}/+ 1:

6

HereG={b1;b2; : : : ;bm} is obtained fromG by replacingbi by a new branch oflength|E.bi /| − 1 for eachi ∈ {1;2; : : : ;m}: From the graph given in Chapter 2,it follows that the condition, that the hamiltonian index of a graph is not less thanfour, is necessary. Since Theorem 1.2 is not a generalization of Theorem 1.1, thefollowing conjecture seems to be trustworthy.

Conjecture 1.1 The problem of determining whether the hamiltonian index of agraph exceeds two is polynomial.

Although we do not answer the conjecture, by Theorem 1.9 we only need to con-sider the complexity of determining whether the hamiltonian index is 2 or 3. UsingCatlin’s reduction method, Sara˘zin obtained a sharp upper bound.

Theorem 1.10 (Sarazin [49]) Let G be a connected simple graph of order n otherthan a path. Then

h.G/ ≤ n−1.G/:Clearlydia.G/− 1≤ n−1.G/ (see [63]). In Chapter 4, we will give the follow-ing improvement of Theorem 1.10.

Theorem 1.11 Let G be a connected graph other than a path. Then

h.G/ ≤ dia.G/− 1;

and the upper bound is sharp.

The following result due to Veldman is a special case of Theorem 1.11.

Theorem 1.12 (Veldman [55]) Let G be a connected graph with at least threeedges and such that dia.G/ ≤ 2: Then h.G/ ≤ 1:

In Chapter 4 we also consider the hamiltonian index of the complement of a graph.We obtain a Nordhaus-Gaddum inequality for the hamiltonian index. Our mainresults are the following.

Theorem 1.13 Let G be a simple graph with at least 61 vertices and letG denotethe complement of G such that neither G norG is a path. Then

(a) min{h.G/;h.G/} ≤ 1;

(b) if min{h.G/;h.G/} = 1; then max{h.G/;h.G/} ≤ .n+ 1/=2.

1.2 Subpancyclic line graphs 7

The equality holds if and only if one ofG or G is isomorphic to the graph obtainedby identifying one end-vertex of a path of lengtht− 1 with exactly one vertex ofa complete graph of ordert.

The following Nordhaus-Gaddum inequality can be obtained from Theorem 1.13.

Theorem 1.14 Let G be a simple graph with at least 61 vertices and let G denotethe complement of G such that neither G norG is a path. Then

h.G/+ h.G/ ≤ n− 3 and

h.G/h.G/ ≤ .n− 1/=2:

1.2 Subpancyclic line graphs

1.2.1 Degree conditions for subpancyclicity in line graphs

The earliest degree condition for a graph to contain a hamiltonian cycle was givenby Dirac in 1952.

Theorem 1.15 (Dirac [29]) If G is a graph on n≥ 3 vertices such that d.u/ ≥n=2 for every vertex u in G, then G contains a hamiltonian cycle, and the result isbest possible.

Ore investigated the same question involving the degree sum of two nonadjacentvertices and gave the following generalization of Theorem 1.15.

Theorem 1.16 (Ore[46]) If G is a graph on n≥ 3 vertices such that d.u/ +d.v/ ≥ n for every pair u; v of nonadjacent vertices, then G contains a hamil-tonian cycle.

Clark[28] and Van Blankenet al.[5] investigated the same question concerninghamiltonicity of line graphs. A more recent and best possible result is the follow-ing.

Theorem 1.17 (Veldman [54]) Let G be a connected simple graph of order n suchthat every cut edge of G is incident with a vertex of degree one. If n is sufficientlylarge and d.u/ + d.v/ > .2n+ 10/=5 for any edge uv ∈ E.G/, then L.G/ ishamiltonian.

8

A graphG is calledsubpancyclicif it contains a cycle of length 3 up to the cir-cumference. A graph is calledpancyclicif it is subpancyclic and hamiltonian.Bondy[6] was the first author who investigated pancyclic graphs. He proved thatthe graphs satisfying the conditions of Theorem 1.15 are pancyclic except com-plete bipartitle graphs. In [7], the following metaconjecture was stated.

Bondy’s metaconjecture[7]. Almost any nontrivial condition on a graph whichimplies that the graph is hamiltonian also implies that it is pancyclic, except pos-sibly for a simple family of exceptional graphs.

In Chapter 5 up to Chapter 8 we present degree sum conditions along a path guar-anteeing thatL.G/ is subpancyclic. Here²k.G/ is the minimum degree sum ofvertices such that they are in a path of lengthk− 1; k≥ 1: These conditions showthat degree sum conditions along paths of a graph which ensure that its line graphis subpancyclic are considerably weaker than those required to ensure that its linegraph is hamiltonian. This observation supports Bondy’s metaconjecture.

Theorem 1.18 Let G be a connected graph of order n. If G satisfies one of thefollowing conditions.

(i) ²2.G/ > .√

8n+ 1+ 1/=2 and n≥ 600;

(ii) ²3.G/ > .n+ 6/=2 and n≥ 76;

(iii) ²4.G/ > .2n+ 16/=3 and n≥ 76,

then L.G/ is subpancyclic, and the conditions are all best possible even under thecondition that L.G/ is hamiltonian.

Corollary 1.2 Let G be a connected graph of order n. If G satisfies one of (i),(ii), (iii), then L.G/ is pancyclic if and only if L.G/ is hamiltonian.

The exceptional graphs in Corollary 1.2 are defined in Chapter 9. Corollary 1.2shows that the graphs in Theorem 1.17 are pancyclic. From results in [5] it followsthat there exists a constantA such that ifŽ ≥ An1=3 andL.G/ is hamiltonian, thenL.G/ is pancyclic. Also, there exists a constantB and an infinite family of graphsG such thatŽ ≥ Bn1=3 andL.G/ is hamiltonian but not pancyclic. The followingconjecture is stated in [5].

1.2 Subpancyclic line graphs 9

Conjecture 1.2 (Van Blanken et al: [5]) Let G be a graph of order n and mini-mum degreeŽ such that L.G/ is hamiltonian and

Ž3+ 2Ž2+ 7Ž+ 6> 6n:

Then L.G/ is pancyclic unless G is isomorphic to C4, C5 or the Petersen graph.

In general, results involving degree sums are direct generalizations of results in-volving the minimum degree of the graph. Theorem 1.18 shows an exception tothis rule. Comparing Theorem 1.18 (i) and (ii), we observe that results involvingdegree sums of the vertices along a 2-path or a 3-path do not immediately implythe corresponding results involving the minimum edge degree. In Chapter 6, wewill deal with subpancyclic line graphs of graphs with girth at least 5.

Theorem 1.19 Let G be a graph of order n.n ≥ 450/, girth at least 5, and sup-pose²2.G/ ≥

√2n+ 1. Then L.G/ is subpancyclic.

One can show that even for graphs with large girth the degree condition in The-orem 1.19 is best possible. In Chapter 8 we investigate the exceptional graphs of(iii) in Theorem 1.18.

Brandt and Veldman [9] investigated the subpancyclicity of graphs with²2.G/ ≥n:

Theorem 1.20 (Brandt and Veldman [9]) Let G be a graph of order n such that²2.G/ ≥ n. Then G is subpancyclic, unless G is a complete bipartite graph.

Certainly, we cannot replace “subpancyclic” by “hamiltonian”, as the conditionapplies to every complete bipartite graph, hence also to the starK1;n−1, whichcontains no cycle at all. But for its line graph things are different.

Theorem 1.21 (Clark[28]) Let G be a graph of order n with²2.G/ ≥ n. ThenL.G/ is hamiltonian.

Theorem 1.22 Let G be a connected graph of order n.n ≥ 146/. If ²4.G/ ≥.n+ 10/=2, then L.G/ is subpancyclic unless G is isomorphic to the exceptionalgraph F shown in the following, and the result is best possible.

10

The exceptional graphF is defined as follows: Letn ≡ 1.mod3/, andC1; C2;

· · · ; C.n−1/=3 be .n− 1/=3 edge-disjoint cycles of length 4.F is obtained fromthose cycles such thatC1;C2; · · · ; C.n−1/=3 have exactly one common vertex inF and E.F/ = E.C1/ ∪ E.C2/ ∪ · · · ∪ E.C.n−1/=3/: Obviously |V.F/| = n and²4.F/ = 2n+16

3 .

Theorem 1.22 shows that when we exclude an exceptional graph, the conditioninvolving degree sums of the vertices along a 3-path which ensures that its linegraph is subpancyclic will be greatly improved (replace2n

3 by n2) and it is almost

the same as the condition involving degree sums of the vertices along a 2-path(comparing with Theorem 1.18). Recently, Trommel, Veldman and Verschut ob-tained a consequence of Theorem 1.18.

Corollary 1.3 (Trommelet al.[53]) Let G be a line graph on at least 100577

vertices. IfŽ >√

2n+ 14 − 3

2, then G is subpancyclic.

They also investigated the same question for claw-free graphs.

Theorem 1.23 (Trommel et al:[53]) Let G be a claw-free graph on at least 5vertices. IfŽ >

√3n+ 1 , then G is subpancyclic and the result is best possible.

1.2.2 Pancyclic line graphs involving its diameter

In Chapter 9, we present a pancyclicity version of Theorem 1.12

Theorem 1.24 Let G be a graph. If dia.G/≤ 2, then L.G/ is either pancyclic ora cycle of length 4 or 5.

The following result is similar to Theorem 1.24.

Theorem 1.25 Let G be a graph with at least one cycle. If d.e; f / ≤ 1 for anypair of edges e; f in G, then L.G/ is either pancyclic or a cycle of length 4 or 5.

The extremal graphs of the above theorems are also presented in Chapter 9.

1.3 Spanning circuits in graphs 11

1.3 Spanning circuits in graphs

1.3.1 Connected even[2;2k− 2]-factors

A connected even factor of a graph is a connected spanning subgraph such thatevery vertex has even degree, i.e., it is a spanning circuit. Such a graph is calledsupereulerian: A connected even [2;2k]-factor is a connected even factor withdegrees at most 2k. A [2, 2]-factor is usually called a 2-factor. Clearly a hamil-tonian graph is a connected 2-factor, and a supereulerian graph onn vertices con-tains a connected even [2;2bn=2c]-factor.

Cai, Benhocineet al.and Catlin considered the degree sum conditions for a graphto have a connected even factor.

Theorem 1.26 (Cai[14], Benhocineet al.[3], Catlin[18])Let G be a 2-edge-connected graph on n vertices. If G satisfies one of the follow-ing conditions:

(a) Ž ≥ n=5;

(b) ¦2 ≥ .2n+ 3/=3 and n≥ 3;

(c) ¦3 ≥ n+ 1 and n≥ 6;

then G has a connected even factor.

Here ¦k.G/ = min{∑ki=1 d.xi / : {x1; · · · ; xk} is an independent set ofk vertices}

if the independence number ofG is not less thank;∞ otherwise.In Chapter 10 we consider special independent sets ofk vertices having a commonneighbor.

Theorem 1.27 Let G be a connected simple graph with n≥ 3 vertices and letk be an integer. If d.x1/+ d.x2/+ : : : + d.xk/ ≥ n for any independent set ofk vertices{x1; x2; : : : ; xk} in G such that they have a common neighbor, then Gcontains a connected even[2;2k− 2]-factor if and only if G has a connected evenfactor.

Theorem 1.27 is best possible with respect to the degree sum condition. Theextremal graphG.k/ is the graph obtained by identifying exactly one vertex ofk complete subgraphs of order at least 3. Combining Theorem 1.26 and Theo-rem 1.27, we obtain the following results.

12

Theorem 1.28 Let G be a 2-edge-connected graph of order n and k∈ {2;3;4;5}.If Ž.G/ ≥ n=k; and n≥ 6; then G contains a connected even[2;2k− 2]-factor.

Theorem 1.29 Let G be a 2-edge-connected graph of order n;n≥ 6. If G satisfiesone of the following conditions:

(a) ¦2 ≥ .2n+ 3/=3;

(b) ¦3 ≥ n+ 1;

then G contains a connected even [2,4]-factor.

Note that Theorem 1.16 is the casek= 2 of Theorem 1.29.

The last consequence of Theorem 1.27 that we want to mention is the following.

Corollary 1.4 Let G be a 2-edge-connected graph of order n≥ 3 and let k be aninteger. If d.x1/+ d.x2/ ≥ 2n

k for any independent set of two vertices{x1; x2} inG such that they have a common adjactcent vertex, then G contains a connectedeven[2;2k− 2]-factor if and only if G has a connected even factor.

Based on Theorem 1.26 and Corollary 1.4, we pose the following conjecture thatwill be best possible, if it is true.

Conjecture 1.3 If the degree sum of an independent set of k vertices in G, suchthat they have a common neighbor, is greater than n+ 1, then G contains a con-nected even[2;2k− 2]-factor.

If we consider aK1;s+1-free graph, then we obtain that a supereulerian graph con-tains a connected even [2;2s− 2/]-factor for s≥ 2:

Theorem 1.30 Every supereulerian K1;s+1-free graph contains a connected even[2;2s− 2]-factor for s≥ 2:

Theorem 1.31 Every 4-edge-connected claw-free graph has a connected even[2,4]-factor.

The graphG.k/ also shows that the edge-connectivity bound cannot be relaxed.The following result is a consequence of Theorem 1.31, since every 4-connectedgraph is 4-edge-connected.

Corollary 1.5 (Broersmaet al. [13]) Every 4-connected claw-free graph has aconnected even [2,4]-factor.

1.3 Spanning circuits in graphs 13

1.3.2 Supereulerian graphs

A supereulerian graphis a graph having a spanning circuit, i.e., a graph having aconnected even factor.

In Chapter 11, we improve a result of Catlin and Li [21]. The reduction ofG is thegraph obtained fromG by successively contracting some edge-disjoint subgraphsH1; H2; · · · ; Hk. A graph is reduced if it is the reduction of some graph.

Theorem 1.32 Let G be a simple 2-edge-connected graph on n vertices. If forevery minimal edge cut of G S⊆ E.G/with |S| ≤ 3 we have that every componentof G− S has order at least.n−2/=5> 2, then exactly one of the following holds:

• G is supereulerian;

• The reduction G′ of G is isomorphic to K2;5 such that each preimage of avertex with degree two in G′ has exactly order.n− 1/=5;

• The reduction G′ of G is isomorphic to K2;3 such that each vertex of G′

corresponds to a preimage of G which has order at least.n− 2/=5.

We also obtain the following results.

Corollary 1.6 Let G be a simple 2-edge-connected graph on n vertices. IfŽ.G/≥4 and min{max{d.x/;d.y/} : xy∈ E.G/} ≥ .n− 7/=5> 1; then G satisfies theconclusion of Theorem 1.32.

The following known result is a consequence of Corollary 1.6.

Theorem 1.33 (Catlin and Li[21]) Let G be a simple graph of order with½.G/ ≥2. If Ž.G/ ≥ 4 and if

²2.G/ ≥ .2n− 10/=5;

then exactly one of the following holds:

(a) G is supereulerian.

(b) The reduction G′ of G is isomorphic to K2;3 such that each vertex of G′

corresponds to a preimage in G with order exactly n=5.

In Chapter 11 a discussion on minimum degree is given. We show that we cannotrelax the lower bound four on the minimum degree in the above results.

Notation and terminology

Throughout Chapter 2 up to Chapter 4, we use [8] for terminology and notation notdefined here and consider only loopless finite graphs. LetG be a graph. For eachintegeri > 0 defineVi.G/ = {v ∈ V.G/ : dG.v/ = i} andW.G/ = V.G/\V2.G/.As in [20], a branchin G is a nontrivial path with ends inW.G/ and internal ver-tices, if any, that have degree 2 (and thus are not inW.G/). If a branch has length1, then it has no internal vertices. We denote byB.G/ the set of branches ofGand byB1.G/ the subset ofB.G/ in which every branch has an end inV1.G/. Forany subgraphH of G, we denote byBH.G/ the set of branches ofG with edgesthat are all inH. For any two subgraphsH1 and H2 of G, define thedistancedG.H1; H2/ betweenH1 and H2 to be the minimal distancedG.v1; v2/ such thatv1 ∈ V.H1/ andv2 ∈ V.H2/. A graph is called acircuit if it is connected andevery vertex has an even degree. Note that by this definition (the trivial subgraphinduced by) a single vertex is also a circuit.

The line graph ofG, hasE.G/ as its vertex set, where two vertices are adjacent inL.G/ if and only if the corresponding edges are adjacent inG. Them-th iteratedline graphLm.G/ is defined recursively byL0.G/ = G, Lm.G/ = L.Lm−1.G//whereL1.G/ denotesL.G/.

Finally, EUk.G/ denotes the set of those subgraphsH of a graphG that satisfythe following conditions:

(I) dH.x/ ≡ 0 (mod 2) for everyx ∈ V.H/;

(II) V0.H/ ⊆1.G/⋃i=3

Vi.G/ ⊆ V.H/;

(III) dG.H1; H − H1/ ≤ k− 1 for every componentH1 of H;

(IV) |E.b/| ≤ k+ 1 for every branchb ∈ B.G/ with E.b/∩ E.H/ = ∅;

15

16

(V) |E.b/| ≤ k for every branchb ∈ B1.G/.

EUn.G/ will play an important role in our main result, which is Theorem 2.5.

Chapter 2

Hamiltonian iterated line graphs

All results in this chapter are related to the well-studied concept of the line graphoperation on graphs.

Harary and Nash-Williams characterized those graphsG for which L.G/ is hamil-tonian.

Theorem 2.1 (Harary and Nash-Williams [32]) Let G be a connected graph withat least three edges. Then L.G/ is hamiltonian if and only if G has a closed trailT such that each edge of G is incident with at least one vertex of T.

It follows from Theorem 2.1 that the line graph of a hamiltonian graph is hamil-tonian, while the converse is not true in general. The following corollary is alsoimmediate.

Corollary 2.1 Let G be a graph with at least 3 edges. If G has a spanning closedtrail, then L.G/ is hamiltonian.

Theorem 2.1 has been used by many authors to investigate the cyclic propertiesof line graphs. In fact, the paper [32] in which they presented Theorem 2.1, hasbeen cited frome 1995 to 1998 in at least 12 published papers that are covered bythe CompuMath Citation Index. If one thinks about Theorem 2.1, Corollary 2.1,and the line graph operation more carefully, it becomes natural to believe that formost graphs, after applying the line graph operation iteratively a finite number oftimes, the resulting graph will become hamiltonian. Two natural questions thencan be raised.

17

18

(1) For which graphs is this indeed the case?

(2) If this is the case for a graphG; what is the smallest number of iterationsthat will yield a hamiltonian graph?

In order to investigate this kind of questions, Chartrand [22] considered then-iterated line graphLn.G/ of G and introduced thehamiltonian index of a graph,denoted byh.G/, i.e., the minimum numbern such thatLn.G/ is hamiltonian.Here then-iterated line graphLn.G/ of a graph is defined to beL.Ln−1.G//,whereL1.G/ denotes the line graphL.G/ of G, andLn−1.G/ is assumed to havea nonempty edge set. In fact, he also gave another proof of Theorem 2.1. Heshowed that for any graphG other than a path, the hamiltonian index ofG exists.With the aid of Theorem 2.1, Chartrand and Wall [23] determined the hamiltonianindex of a tree other than a path, and showed that ifG is connected and has acycle of lengthl , thenh.G/ ≤ |V.G/| − l : They also showed thath.G/ ≤ 2 forany connected graphG with minimum degreeŽ.G/ ≥ 3: Kapoor and Stewart [34]determinedh.G/ for a graphG that is homeomorphic toK2;n; for n≥ 3:

Catlin [19] developed a reduction method to investigatesupereulerian graphs, i.e.,graphs that have a spanning closed trail. For a connected subgraphH of G; letG=H denote the graph obtained fromG by contractingH to a single vertex anddeleting any resulting loops. A graphH is calledcollapsibleif for every evensubsetS⊆ V.H/; there is a subgraphT of H such thatH − E.T/ is connectedand the set of odd degree vertices ofT is S:

Theorem 2.2 (Catlin [19]) Let H be a collapsible subgraph of G: Then G issupereulerian if and only if G=H is supereulerian.

After Catlin introduced thisreduction method, many results about hamiltonianline graphs have been derived; for surveys see [17] and [26]. Zhan [66] usedCatlin’s method and Theorem 2.1 to prove that every 7-connected line graph ishamiltonian. An interesting conjecture related to this result, that was posed byThomassen [52], is still open and reads as follows: Every 4-connected line graphis hamiltonian.

Catlin’s reduction method was also used to investigate the hamiltonian index ofa graph. Lai [36] and Catlinet al. [20] used Catlin’s method to give some upperbounds onh.G/ that are related to so-calledbranches;we will come back to thislater. Sara˘zin [49] used Catlin’s method to show that the hamiltonian index of a

2.1 Characterization of graphs with iterated line graphs that are hamiltonian19

simple graphG other than a path, is at most|V.G/|−1.G/; where1.G/ denotesthe maximum degree ofG:

Theorem 2.1 is a good tool for investigating cyclic properties of line graphs. How-ever, when one uses it to investigate the (hamiltonian) cycles in then-iterated linegraph of a graph, closed trails in its.n− 1/-iterated line graph should be con-sidered. Since it is not convenient to examine.n− 1/-iterated line graphs whenn ≥ 2; this leads to a natural question: for any integern ≥ 2; does there exist acharacterization of those graphsG for which Ln.G/ is hamiltonian? This was alsomentioned in [15]. The answer is affirmative. We will give such a characteriza-tion in Section 1. As its application, in Section 2 we will examine the hamiltonianindex of a graph and give two methods for determining it. One of them resemblesCatlin’s reduction method. We also present some new upper bounds on the hamil-tonian index in Section 3. Our results enhance various results on the hamiltonianindex known earlier.

2.1 Characterization of graphs with iterated linegraphs that are hamiltonian

Our aim in this section is to give a characterization of graphs with iterated linegraphs that are hamiltonian. Our main result, Theorem 2.5, is a direct conse-quence of Theorem 2.3 and Theorem 2.4.

We start with a close relationship betweenEUk.L.G// andEUk+1.G/; the proofof which will be postponed.

Theorem 2.3 Let G be a connected graph and k≥ 1 be an integer. ThenEUk.L.G// 6= ∅ if and only if EUk+1.G/ 6= ∅:

We will use Theorem 2.1 to characterize graphs with 2-iterated line graphs thatare hamiltonian. The proof of this will also be postponed.

Theorem 2.4 Let G be a connected graph with at least three edges. Then L2.G/is hamiltonian if and only if EU2.G/ 6= ∅.

Using Theorem 2.3 and Theorem 2.4, one easily derives the following main resultby induction.

20

Theorem 2.5 Let G be a connected graph with at least three edges and n≥ 2:Then Ln.G/ is hamiltonian if and only if EUn.G/ 6= ∅:

Comparing Theorem 2.1 with Theorem 2.5, one might think thatL.G/ is hamil-tonian if and only ifEU1.G/ is nonempty. Unfortunately, this is not true becauseevery subgraph inEU1.G/ should satisfy (II). For example, Figure 2.1 shows thatw is a vertex of degree 4 but does not belong to the unique circuitC = G0−wsuch thatE.C/= E.G0/: HenceEU1.G0/ is empty, butL.G0/ is hamiltonian, byTheorem 2.1.

'

&

$

%

•

•

•

• •

• •

• •

• •

• •

w

v

u

x y

Figure 2.1: A graphG0 with wu; wv; wx; wy ∈ E.G0/:

The following theorem is a consequence of Theorem 2.5.


Theorem 2.6 For n ≥ 2; Ln.G/ is hamiltonian if and only if there exists exactlyone component G1 of G such that EUn.G1/ 6= ∅, and any other component of Gis a path of length at most n− 1.

In order to prove Theorem 2.3 and Theorem 2.4, we first present some auxiliaryresults. We omit the proof of the following lemma since it is a slight modifica-tion of the proof of Theorem 2.1 [32]. We first introduce a notation related toLemma 2.1. For any subgraphC of L.G/, by S.G;C/ we denote the collection ofcircuits H of G, such thatL.G[E.H/]/ containsC, andC contains all elementsof E.H/. Here and throughout,G[S] denotes the subgraph ofG induced byS;whereS⊆ V.G/ or S⊆ E.G/.

Lemma 2.1

A. If C is a cycle of L.G/ with |E.C/| ≥ 3, then S.G;C/ is nonempty.

B. If G has a circuit H such thatE.H/ has at least three edges, then L.G/ hasa cycle C with V.C/ = E.H/.

The following lemma is known.

Lemma 2.2 (Beineke [2]) K1:3 is not an induced subgraph of the line graph ofany graph.

Lemma 2.3 Let b= u1u2 · · ·us.s ≥ 3/ be a path of G and ei = uiui+1. Thenb ∈ B.G/ if and only if b′ = e1e2 · · ·es−1 ∈ B.L.G//.

Proof. b= u1u2 · · ·us = G[{e1;e2; · · · ;es−1}] ∈ B.G/⇐⇒ u1;us ∈ W.G/ anddG.ui / = 2 for i ∈ {2;3; · · · ; s− 1} ⇐⇒ e1;es−1 ∈ W.L.G// anddL.G/.ei / = 2for i ∈ {2;3; · · · ; s− 2} ⇐⇒ b′ = e1e2 · · ·es−1 ∈ B.L.G//:�

Lemma 2.4 Let H be a subgraph of G in EUk.G/ with a minimum number ofcomponents. Then there exist no multiple edges inE.H1/ ∩ E.H2/ for any twocomponents H1 and H2 of H.

Proof. Otherwise there would exist two componentsH1; H2 of H and edgese1;e2

in E.H1/ ∩ E.H2/ with the same set of endvertices. One can easily check thatH ′ = H+{e1;e2} ∈ EUk.G/; which is a contradiction becauseH ′ contains fewercomponents thanH: �

A eulerian subgraphof G is a circuit which contains at least one cycle of length atleast 3.

22

Lemma 2.5 Let G be a connected graph and C be a eulerian subgraph of the linegraph L.G/. Then there exists a subgraph H of G with

(1) dH.x/ ≡ 0(mod 2) for every x∈ V.H/;

(2) dG.x/ ≥ 3 for every vertex x∈ V.G/ with dH.x/ = 0;

(3) for any two components H0; H00 of H, there exists a sequence of compo-nents H0 = H1; H2; · · · ; Hs = H00 of H such that dG.Hi; Hi+1/ ≤ 1 fori ∈ {1;2; · · · ; s− 1};

(4) L.G[E.H/]/ contains C, and C contains all elements of E.H/.

Proof. SinceC is a eulerian subgraph ofL.G/ and L.G/ is a simple graph, we

can letC1;C2; · · · ;Cm be the edge-disjoint cycles withC=m⋃

i=1Ci.

By Lemma 2.1A, we can findm subgraphsF1; F2; · · · ; Fm of G such thatFi ∈S.G;Ci/ for i ∈ {1;2; · · · ;m}. Hence, there existmi edge-disjoint cyclesDi;1; Di;2;

· · · ; Di;mi (possibly, formi = 1; Di;1 might be a single vertex) such thatFi =mi⋃j=1

Di; j. Define

H ′ =m⋃

i=1

mi⋃j=1

Di; j:

For anye∈ E.H ′/, let

r H ′.e/ = |{C′ : e∈ E.C′/ andC′ ∈m⋃

i=1

mi⋃j=1

{Di; j}}|:

We construct a subgraphH of G from H ′ as follows:

V.H/ = V.H ′/ andE.H/ = E.H ′/\{e∈ E.H ′/ : r H ′.e/ ≡ 0.mod2/}:Next we will prove thatH satisfies (1) to (4).

For anx ∈ V.Di; j/; the cycleDi; j is counted exactly twice in∑

e∈EH′ .x/r H ′.e/

which is therefore an even number. If we denoteEi.x/ = {e∈ EH ′.x/ : r H ′.e/ ≡i.mod2/}.i = 0;1/, then∑

e∈EH′ .x/r H ′.e/ =

∑e∈E0.x/

r H ′.e/+∑

e∈E1.x/

r H ′.e/;


and so∑

e∈E1.x/r H ′.e/ is even, which implies thatdH.x/ = |E1.x/| is even. Thus

(1) holds.

ObviouslydG.w/ ≥ 2 for allw ∈ V.H/. If there were aw ∈ V.H/with dG.w/=2 anddH.w/ = 0, then we would have two cyclesD′; D′′ in the set{Di; j} suchthat e1;e2 ∈ E.D′/ ∩ E.D′′/; wheree1;e2 are the two edges incident tow. Butthen there would exist two cyclesCp;Cq having the edgee1e2 in common in theline graphL.G/, contrary to the choice of the cyclesCi. This proves (2).

SinceH ′ =m⋃

i=1Fi andFi ∈ S.G;Ci/; L.G[E.H ′/]/ containsC=

m⋃i=1

Ci which con-

tains all elements ofE.H ′/. However,V.H/ = V.H ′/ implies that E.H/ =E.H ′/, henceE.H/ ⊆ V.C/ ⊆ E.H/ and (4) holds.

Suppose thatH has a componentH∗ with dG.H∗; H − H∗/ ≥ 2. ThenE.H∗/∩E.H − H∗/ would be empty andC disconnected. This contradiction shows that(3) is true forH, too, which completes the proof of Lemma 2.5.�

Now we can present the proofs of Theorem 2.3 and Theorem 2.4.

Proof of Theorem 2.3.

Supposing thatEUk+1.G/ 6= ∅, we choose anH ∈ EUk+1.G/ with a minimumnumber of components which we denote byC1; · · · ;Ct.

By Lemma 2.1B, we can find a cycleC′i of L.G/with V.C′i /= E.Ci /.i = 1; · · · ; t/.HenceC′i is a cycle ofL.G/ with length at least 3 sinceH ∈ EUk+1.G/. Let

H ′ =t⋃

i=1C′i . We will prove thatH ′ ∈ EUk.L.G//.

Since1.G/⋃i=3

Vi.G/ ⊆ V.H/ andV.H′/ =t⋃

i=1E.Ci /,

1.L.G//⋃i=3

Vi.L.G// ⊆ V.H′/:

SincedG.Ci ;Cj/≥ 1; by Lemma 2.4,E.C′i /∩ E.C′j /=∅ for {i; j} ⊆ {1;2; · · · ; t}with i 6= j; which implies thatH ′ satisfies (I).

24

Obviously H ′ contains no isolated vertex by definition ofH, henceH ′ satisfies(II).

Take an arbitraryi ∈ {1; · · · ; t}. By the choice ofH, it follows thatdG.Ci; H −Ci/ ≤ k. Let P = xu1 · · ·usy be a shortest path fromCi to H − Ci, wherex ∈V.Ci/; y∈ V.H−Ci / ands≤ k−1. Evidently,L.P/ is a path fromC′i to H ′ −C′iwith lengths, thusdL.G/.C′i ; H ′ −C′i / ≤ k− 1; which implies that (III) holds forH ′.

Since H satisfies (IV) and (V), using Lemma 2.3 one can easily check thatH ′

satisfies (IV) and (V).

Conversely, assumeEUk.L.G// 6= ∅. Let H be a subgraph ofL.G/ in EUk.L.G//with a minimum number of isolated vertices. ThenH contains no isolated ver-tices. For, supposeC1= {e0} is an isolated vertex ofH, then by (II),dL.G/.e0/ ≥ 3and by Lemma 2.2, there existe1;e2 ∈ NL.G/.e0/ such thate1e2 ∈ E.L.G//. Nowwe construct a subgraphH0 of L.G/ as follows.

H0 ={

H + {e0e1;e1e2;e2e0}; if e1e2 6∈ E.H/;H + {e0e1;e0e2} − {e1e2}; if e1e2 ∈ E.H/:

ObviouslyH0 ∈ EUk.L.G// has fewer isolated vertices thanH has, a contradic-tion.

Let H1; H2; · · · ; Hm be the components ofH. SinceH ∈ EUk.L.G// andH con-tains no isolated vertices,Hi is a eulerian subgraph ofL.G/ for i ∈ {1;2; · · · ;m}.Hence for anyHi.i ∈ {1;2; · · · ;m}/, by Lemma 2.5, there exists a subgraphCi ofG satisfying (1) to (4). Set

C= .1.G/⋃i=3

Vi.G//∪ .m⋃

i=1

Ci/:

We will prove thatC ∈ EUk+1.G/.

SinceV.Hi/∩ V.Hj/ = ∅ for {i; j} ⊆ {1;2; · · · ;m} with i 6= j, E.Ci /∩ E.Cj / =∅. It follows thatdC.x/ ≡ 0(mod 2) for everyx ∈ V.C/; which implies thatCsatisfies (I). SinceCi satisfies (2),dG.x/ ≥ 3 for everyx ∈ V.C/ with dC.x/ = 0:


Thus (II) holds.

Since1.L.G//⋃

i=3Vi.L.G//⊆ V.H/, dG.x;G[V.C/ \ {x}]/ ≤ k for every vertexx in C

with dC.x/= 0. Take an arbitraryi ∈ {1;2; · · · ;m}: By the choice ofH, it followsthat dL.G/.Hi; H − Hi/ ≤ k− 1: Let P = e1e2 · · ·es be a shortest path fromHi

to H − Hi, wheree1 ∈ V.Hi/ ⊆ E.Ci / andes ∈ V.H − Hi/ ⊆ E.C− Ci/, ands≤ k. Sinceet andet+1 are two adjacent edges inG for eacht ∈ {1;2; · · · ; s− 1},it follows thatG[{e1;e2; · · · ;es}] is connected. Hence

dG.Ci;C−Ci / ≤ |E.G[{e1;e2; · · · ;es}]/| ≤ s≤ k;

which implies thatC satisfies (III) by Lemma 2.5.

SinceH satisfies (III) to (V), using Lemma 2.3 one can easily check thatC satis-fies (IV) and (V). It follows thatC ∈ EUk+1.G/. �


Supposing thatEU2.G/ 6= ∅ , we choose anH ∈ EU2.G/ with a minimum num-ber of components that are denoted byH1; H2; · · · ; Ht.

Since H ∈ EU2.G/; |E.Hi/| ≥ 3 for i ∈ {1;2; · · · ; t}: Hence, by Lemma 2.1B,we can find a cycleCi of L.G/ with length at least 3 such thatV.Ci/ = E.Hi /,for i ∈ {1;2; · · · ; t}. Let

C=t⋃

i=1

Ci:

By Lemma 2.4,C1;C2; · · · ;Ct aret edge-disjoint cycles inL.G/. HenceC is asubgraph ofL.G/ satisfying (I). SincedG.H1; H − H1/ ≤ 1 for any subgraphH1

of H, C is a connected subgraph ofL.G/. By Lemma 2.3 and sinceH ∈ EU2.G/,any branchb ∈ B.L.G// with E.b/ ∩ E.C/ = ∅ has length at most 2 and anybranch inB1.L.G// has length at most 1. SinceH satisfies (II),

1.L.G//⋃i=3

Vi.L.G// ⊆ V.C/:

26

HenceE.C/ = E.L.G// which implies thatL2.G/ is hamiltonian by Theorem2.1.

Conversely, suppose thatL2.G/ is hamiltonian. By Theorem 2.1, there exists acircuit C of L.G/ such thatE.L.G// = E.C/: Select aC with a maximum num-ber of vertices of degree at least 3. Then

Claim 1.1.L.G//⋃

i=3Vi.L.G// ⊆ V.C/.

Proof of Claim 1. Otherwise lete0 ∈ .1.L.G//⋃

i=3Vi.L.G///\V.C/. By Lemma 2.2,

there exist two distinct verticese1;e2 ∈ NL.G/.e0/ such thate1e2 ∈ E.L.G//.SinceE.C/ = E.L.G// ande0 6∈ V.C/, {e1;e2} ⊆ V.C/. Now we construct asubgraphC0 of L.G/ as follows,

C0 ={

C+ {e0e1;e0e2} − {e1e2}; if e1e2 ∈ E.C/;C+ {e0e1;e0e2;e1e2}; if e1e2 6∈ E.C/:

ObviouslyC0 is a circuit such thatE.L.G//= E.C0/, butC0 contradicts the max-imality of C. This completes the proof of Claim 1.

HenceC is a eulerian subgraph ofL.G/ sinceL.G/ is a simple graph. By Lemma2.5,G has a subgraphH satisfying (1) to (4).

Claim 2. dG.x; H/ ≤ 1 for anyx ∈1.G/⋃i=3

Vi.G/.

Proof of Claim 2. If G is either a star or a cycle, then the conclusion holds.

If G is neither a star nor a cycle, thenEG.x/ ∩ .1.L.G//⋃

i=3Vi.L.G/// 6= ∅ for every

vertexx in1.G/⋃i=3

Vi.G/. Hence by Claim 1 and (4), there exists an edgeex such that

ex ∈ EG.x/∩ .1.L.G//⋃

i=3

Vi.L.G/// ⊆ V.C/ ⊆ E.H/:

This implies thatex has an endvertex inH. This completes the proof Claim 2.

2.2 Methods for determining the hamiltonian index of a graph 27

We will prove thatH ′ = H ∪ .1.G/⋃i=3

Vi.G// ∈ EU2.G/. Claim 2 and property (3)

of H imply that dG.H ′1; H ′ − H ′1/ ≤ 1 for every subgraphH ′1 of H ′; thus H ′

satisfies (III). It follows from Lemma 2.3 andE.C/ = E.L.G// that |E.b/| ≤ 3for b ∈ B.G/ with E.b/ ∩ E.H/ = ∅ and |E.b/| ≤ 2 for b ∈ B1.G/. HenceH ′ ∈ EU2.G/. �

2.2 Methods for determining the hamiltonian indexof a graph

In this section, we will give two methods for determining the hamiltonian indexof a graph.Define

CB.G/ = {b ∈ B.G/ : any edge of b is a cut edge of G} and

CB1.G/ = B1.G/:

One can easily see thatCB.G/\CB1.G/ is the set of bridge-paths ofG andCB1.G/ is the set of its end-paths (see [48]).

As in [22], if L0.G/ stands forG, then we define the hamiltonian indexh.G/ of agraphG to be

h.G/ = min{n : Ln.G/ is hamiltonian}:Since the hamiltonian index does not exist for paths and 2-cycles, we will excludethem in the rest of this section. Thus,G will always stand for a connected graphother than a path or a 2-cycle in this section.

2.2.1 Split blocks of a graph

Define k.G/ = 0 if G is 2-connected;k.G/ = 1 if G is not 2-connected andCB.G/= ∅; k.G/=max{max{|E.b/|+1 : b∈CB.G/ \CB1.G/};max{|E.b/| :b ∈ CB1.G/}}; otherwise.

Chartrand and Wall obtained the hamiltonian index of a tree.

Theorem 2.7 (Chartrand and Wall [23]) Let T be a tree. Then

h.T/ = k.T/:

28

A block of a graphG is a maximal connected subgraph which contains no cutvertex of itself. Ablock is a 2-connected subgraph ofG with maximum verticesnumber. A block ofG is called anacyclic blockif it is a single edge ofG and acyclic blockotherwise. Recently, Sara˘zin generalized the above result as follows:

Theorem 2.8 (Sarazin [48]) If every cyclic block of G is hamiltonian, then

h.G/ = k.G/:

In this subsection, we will characterize those graphsG for which h.G/ = k.G/.To do this, for each cyclic blockB of G, we construct asplit blockSBfrom B asfollows:

(a) split each vertexx ∈ V2.B/∩ .1.G/⋃i=3

Vi.G// into a trianglex1x2x3 in SB;

(b) replace the two edgesux andvx (say) inE.B/ by ux1 andvx2 in E.SB/:

This construction is illustrated in Figure 2.2.1.

Let G′ denote the resulting graph obtained by performing (a) and (b). DefineS.G′/ = {F′ ⊆ G′ : F′ has no vertices of odd degree, and if a triangle createdby performing (a) has a vertex inF′; then all vertices of the triangle are inF′ andhave degree two inF′}: Then there exists a one-to-one correspondence8 betweenany subgraphF′ in S.G′/ and the subgraph with even degrees,F = 8.F′/; of G;which is obtained by contracting all triangles inF′ created in step (a).

Let SB1; SB2; · · · ; SBt be all split blocks ofG: For two branchesb1 ∈ B.G/ and

b2 ∈t⋃

i=1B.SBi/; we sayb1 = b2 if the internal vertices ofb1 andb2 coincide, and

if the endvertices either coincide, or the endvertices ofb2 belong to triangles ob-tained from endvertices ofb1 via construction of split blocks.

The following lemma is immediate.

Lemma 2.6 Let SB1; SB2; · · · ; SBt be all split blocks of G. Then

B.G/ \CB.G/ =t⋃

i=1

.B.SBi/ \ B2.SBi//;

where B2.SBi / is the set of branches of SBi of length 2 that are contained intriangles resulting from the construction of split blocks.


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30

The following lemma is necessary for our proof.

Lemma 2.7 Let G be a graph such that h.G/ ≥ 2 and let H be a subgraph of Gin EUh.G/.G/. For F ⊆ H, if p is a path from F to H− F such that|E.p/| ≥ 2and the internal vertices of p are not in V.H/, then p∈ B.G/.

Proof. This follows fromH satisfying (I), (II) and|E.p/| ≥ 2. �

Lemma 2.8 Let G be a connected graph and let SB1; SB2; · · · ; SBt be all splitblocks of G. Then

h.G/ ≥ max{h.SB1/;h.SB2/; · · · ;h.SBt/; k.G/}:Proof. Clearly h.G/ ≥ k.G/. It remains to prove thath.G/ ≥ h.SBi/ for anyi ∈ {1;2; · · · ; t}. If h.G/ = 0, thenG itself is a single block and the lemmafollows. If h.G/= 1, thenk.G/≤ 1. Hence the lemma follows from Theorem 2.1.Next we assume thath.G/ ≥ 2; which implies that there exists a subgraphH inEUh.G/.G/ by Theorem 2.5. ObviouslyH is a union of subgraphs in different

blocks, i.e.,H =t⋃

i=1Hi; whereHi ∈ Bi. Let

H ′i = 8−1.Hi /:

We will prove thatH ′i ∈ EUh.G/.SBi/. ClearlyH ′i satisfies (I) and (II). By Lemma2.6, H ′i satisfies (IV) and (V). It remains to show thatH ′i satisfies (III), i.e.,dG.F′; H ′i − F′/ ≤ h.G/− 1 for each componentF′ ⊆ H ′i : If this were not true,there would exist anH ′i with a componentF′ such thatdG.F′; H ′i − F′/≥ h.G/≥2. It follows from (II) and the definition ofH ′i that any shortest path fromF′ toH ′i − F′ is in B.SBi/\B2.SBi/. By Lemma 2.6,p is in B.G/ \ CB.G/. LetF = 8.F′/. Since any path fromF′ to H ′i − F′ is also a path fromF to Hi − F,p is a shortest such path. Hence|E.p/| ≥ h.G/. On the other hand, sinceH ∈ EUh.G/.G/, |E.p/| ≤ h.G/− 1; which is a contradiction. This implies thatH ′i satisfies (III) for eachi ∈ {1;2; · · · ; t}: ThereforeH ′i ∈ EUh.G/.SBi /, and itfollows thath.SBi / ≤ h.G/ by Theorem 2.5.�

Now we can state our main results of this subsection.

Theorem 2.9 Let G be a connected graph and let SB1; SB2; · · · ; SBt be all splitblocks of G. Then

h.G/ = max{h.SB1/;h.SB2/; · · · ;h.SBt/; k.G/}:


Proof. Let

m.G/ = max{h.SB1/;h.SB2/; · · · ;h.SBt/; k.G/}:By Lemma 2.8, we only need to prove thath.G/ ≤ m.G/. If m.G/ = 0, whichimplies thatk.G/ = 0, thenG has only one split block of itself. Thus the theoremfollows. If m.G/ = 1, which implies thatk.G/ ≤ 1, then the theorem follows byTheorem 2.1 and Lemma 2.8. So we only need to consider the case thatm.G/≥ 2.

By Theorem 2.5, for anyi ∈ {1;2; · · · ; t}, there exists a subgraphH ′i such thatH ′i ∈ EUm.G/.SBi/ andH ′i contains all vertices in triangles created by performing(a). SinceH ′i satisfies (I),H ′i ∈ S.G′/. Let

H =t⋃

i=1

8.H ′i /:

We will prove thatH ∈ EUm.G/.G/: Since

E.H ′i /∩ E.H ′j / = ∅ for {i; j} ⊆ {1;2; · · · ; t} with i 6= j;

H satisfies (I). ObviouslyH satisfies (II). Using Lemma 2.6, we obtain thatHsatisfies (IV) and (V).

It remains to prove thatdG.F; H − F/ ≤ m.G/− 1 for any subgraphF ⊆ H. Ifthis were not the case, then there would exist a subgraphF of H with dG.F; H −F/ ≥ m.G/ ≥ 2. It follows from Lemma 2.7 and the definition ofk.G/ that anyshortest pathp of G from F to H − F is in B.G/ \ CB.G/. By Lemma 2.6,

p is int⋃

i=1.B.SBi/\B2.SBi//: Without loss of generality, we may assume thatp

is in B.SB1/ \ B2.SB1/. Let H1 = 8.H ′1/ and F′ = 8−1.F ∩ H1/. Since ev-ery path fromF′ to H ′1− F′ is also a path fromF to H1− F, p is a shortest suchpath. Hence|E.p/| ≥m.G/. On the other hand, byH ′1 ∈ EUm.G/.SB1/; |E.p/| ≤m.G/− 1; which is a contradiction. This implies thatH satisfies (III). SoH ∈EUm.G/.G/; implying thath.G/ ≤ m.G/ by Theorem 2.5.�

We conclude this subsection with a characterization of graphsG for whichh.G/=k.G/.

Corollary 2.2 Let G be a connected graph and let SB1; SB2; · · · ; SBt be allthe split blocks of G. Then h.G/ = k.G/ if and only if h.SBi / ≤ k.G/ fori ∈ {1;2; · · · ; t}.

32

RemarkIt is not difficult to determinek.G/ of a graphG. By Theorem 2.9, we can deter-mine the hamiltonian index of a graph by first determining the hamiltonian indicesof its split blocks. Since each split block of a connected graph is 2-connected, weonly need to consider graphs of connectivity at least two.

2.2.2 The contraction of a graph

Catlin [19] developed a reduction method for determining whether a graphG hasa spanning circuit. Using this reduction method and Theorem 2.1, several authorsobtained good bounds forh.G/ (see [20], [26], [36] and [49]). Here we give asimilar reduction method for determiningh.G/ graphsG with h.G/ ≥ 4.

For{b1;b2; · · · ;bm} ⊆ B.G/ with |E.bi /| ≥ 2 for eachi ∈ {1;2; · · · ;m}, the con-traction ofG is defined to be a graph, denoted byG=={b1;b2; · · · ;bm}, which isobtained fromG by contracting an edge ofbi, i.e., replacingbi by a new branchof length|E.bi /| − 1, for eachi ∈ {1;2; · · · ;m}.

Theorem 2.10 Let G be a connected graph and let b1;b2; · · · ;bm be all branchesof length at least 2 in G. If h.G/ ≥ 4, then

.∗/ h.G/ = h.G=={b1;b2; · · · ;bm}/+ 1:

Proof. Let G′ = G=={b1;b2; · · · ;bm}. Clearlyh.G′/ ≤ h.G/, by Theorem 2.5.If h.G′/ ≤ 1, then there exists a connected subgraphH ′ in which every vertexhas even degree such thatE.G′/ = E.H ′/, by Theorem 2.1. Letb′1;b

′2; · · · ;b′m

be the branches ofG′ corresponding to the branchesb1;b2; · · · ;bm, respectively.Let H ′′ be the subgraph ofG obtained fromH ′ by replacingb′1;b

′2; · · · ;b′m by

b1;b2; · · · ;bm, respectively. ByH we denote the subgraph with

V.H/ = V.H ′′/∪ .1.G/⋃i=3

Vi.G// and E.H/ = E.H′′/:

One can easily check thatH ∈ EU3.G/. Henceh.G/ ≤ 3 by Theorem 2.5, acontradiction implying thath.G′/ ≥ 2. It follows from Theorem 2.5 andh.G/ ≥h.G′/ ≥ 2 that EUh.G/.G/ 6= ∅ and EUh.G′/.G′/ 6= ∅. Take any subgraphH ∈EUh.G/.G/ and let H ′ be the subgraph ofG′ corresponding toH. It followsfrom Lemma 2.7 and the definition ofG′ that H ′ ∈ EUh.G/−1.G′/: Henceh.G′/ ≤

2.3 Upper bounds for the hamiltonian index of a graph 33

h.G/− 1 by Theorem 2.5. Similarly, take any subgraphH ′ ∈ EUh.G′/.G′/ and letH be the subgraph ofG corresponding toH ′: It follows from Lemma 2.7 and thedefinition of G′ that H ∈ EUh.G′/+1.G/. Henceh.G/ ≤ h.G′/+ 1 by Theorem2.5. Thus (*) is true.�

Remark. The condition in Theorem 2.10 is best possible in the following sense:there exists a family of graphs with hamiltonian index 3 for which (*) does nothold. LetC = u1u2 · · ·u3s · · ·ut be a cycle of length at leastt; t ≥ 3s+ 1≥ 13;and letw; v1; v2; v3 be four vertices not belonging toC. Let G0 be the graph with

V.G0/ = V.C/∪ {w; v1; v2; v3}

andE.G0/ = E.C/∪ {wv1; v1us; wv2; v2u2s; wv3; v3u3s}:

One can easily check thath.G0/= 3 but that its contraction has hamiltonian index1, which implies that (*) does not necessarily hold for a graph with hamiltonianindex 3.

The complexity of determining the hamiltonian index (not exceeding 1) of a graphis NP-complete [4]. So far, we do not know how difficult it is to determine thehamiltonian index (exceeding 1) of a graph. However we conjecture that this ispolynomial. By Theorem 2.10, we only need to consider the complexity of deter-mining whether the hamiltonian index is 2 or 3.

2.3 Upper bounds for the hamiltonian index of agraph

In this section, we will give some upper bounds on the hamiltonian index of agraph.

For every connected graphG with1.G/ ≥ 3; define

B0.G/ = {b ∈ B.G/ : G[V.b/] is a cycle o f G}

andk= max{|E.b/| : b ∈ B.G/ \ B0.G/}:

34

Now for eachb ∈ B0.G/; denote byC.b/ the cycle induced byV.b/: We take asubgraphH of G with

V.H/ = .⋃

b∈B0.G/

V.b//∪ .1.G/⋃i=3

Vi.G//

andE.H/ =

⋃b∈B0.G/

.E.b/ \ {e : |{b : e∈ C.b/}| ≡ 0.mod2/}/:

It is easily seen thatH ∈ EUk+1.G/. Hence we obtain the next result.

Theorem 2.11 Let G be a connected graph that is not a path. Then

h.G/ ≤ max{|E.b/| : b ∈ B.G/ \ B0.G/} + 1:

In order to show that the upper bound in Theorem 2.11 is sharp, we construct agraphG0 as follows: Letp be a path of lengthk; k ≥ 1; and letC1 , C2 be twocycles.G0 is obtained by identifying the two end-vertices ofp with two verticesof C1 andC2; respectively. By Theorem 2.5,Lk+1.G0/ is hamiltonian butLk.G0/

is not.

We will present some corollaries of Theorem 2.11. Corollary 2.3 is in fact strongerthan the result in [36].

Corollary 2.3 Let G be a simple connected graph that is not a path. Then

h.G/ ≤ max{|E.b/| : b ∈ B.G/ \ B0.G/} + 1:

Corollary 2.4 (Chartrand and Wall [23]) If G is a connected graph such thatŽ.G/ ≥ 3, then

h.G/ ≤ 2:

Next, we give a simple proof of the following known result.

Theorem 2.12 (Sarazin [49]) If G is a connected simple graph with1.G/ ≥ 3,then

h.G/ ≤ |V.G/| −1.G/:

2.3 Upper bounds for the hamiltonian index of a graph 35

Proof. Letw be a vertex ofG with dG.w/ = 1.G/.

First we defineH ′ as follows:

V.H ′/ =1.G/⋃i=3

Vi.G/

andE.H ′/ = ∅:

Now let H = H ′ ∪ H ′′; whereH ′′ is a maximal circuit ofG throughw (i.e., thereis no circuitK such thatK 6= H ′′ andK containsH ′′/:

SinceG is a connected simple graph, it follows thatH ∈ EU|V.G/|−1.G/.G/. Henceby Theorem 2.5,h.G/ ≤ |V.G/| −1.G/. �

Chapter 3

The hamiltonian index of a graphand its branch-bonds

The hamiltonian index of a graphG, denoted byh.G/, is the minimumm suchthat Lm.G/ is hamiltonian. Chartrand [22] showed that if a connected graphGis not a path, then the hamiltonian index ofG exists. In [23], a formula for thehamiltonian index of a tree other than a path was established. There exist manyupper bounds in literature (see [20], [23], [36], [49]. The following are the simplerbounds.

Theorem 3.1 (Lai [36]) Let G be a connected simple graph that is not a path, andlet l be the length of a longest branch of G which is not contained in a 3-cycle.Then h.G/ ≤ l + 1.

Theorem 3.2 (Sarazin [49]) Let G be a connected simple graph on n verticesother than a path. Then h.G/ ≤ n−1.G/.Note that the graph in Theorem 3.2 must be simple, which is not mentioned in[48]. These known bounds are based on the following characterization of hamil-tonian line graphs obtained in [32].

Theorem 3.3 (Harary and Nash-Williams [32])Let G be a graph with at least three edges. Then h.G/ ≤ 1 if and only if Ghas a connected subgraph H in which every vertex has even degree such thatdG.e; H/ = 0 for any edge e∈ E.G/.

37

38

Xiong and Liu [60] characterized the graphs withn-iterated line graphs that arehamiltonian, for integern≥ 2.

Theorem 3.4 ([60])Let G be a connected graph that is not a 2-cycle and let n be an integer at leasttwo. Then h.G/ ≤ n if and only if EUn.G/ 6= ∅. Where EUn.G/ denotes the setof those graphs H of G which satisfy the following conditions:

(i) any vertex of H has even degree in H;

(ii) V0.H/ ⊆1.G/⋃i=3

Vi.G/ ⊆ V.H/;

(iii) d G.H1; H − H1/ ≤ n− 1 for any component H1 of H;

(iv) |E.b/| ≤ n+ 1 for any branch b in B.G/\BG.H/;(v) |E.b/| ≤ n for any branch in B1.G/:

Using Theorem 3.4, Xiong improved Theorem 3.2 as follows.

Theorem 3.5 ([59]).Let G be a connected graph other than a path. Then h.G/ ≤ dia.G/− 1.

It is important to investigate whether the line graph of a graph is hamiltonian,i.e., h.G/ ≤ 1. Since the line graph of a hamiltonian graph is again hamiltonian,the study of these graphs withh.G/ ≥ 2 is equivalent to that of the graphs withh.G/ ≤ 1. Motivated by these observations, and in an attempt to improve existingresults including Theorem 3.5, we will give a reduction method to determine thehamiltonian index of a graph withh.G/ ≥ 2 in Section 3.2. Using this methodwe will give a sharp lower bound and a sharp upper bound ofh.G/ such that thedistance of the two bounds is exactly 2 in Section 3.3. Our results generalizeresults known earlier in [1], [4], [20], [49]. In the next section, we will introducea terminology called branch-bond that involves our bounds.

3.1 Branch-bonds

For any subsetS of B.G/, we denote byG− S the subgraph obtained fromG[E.G/\E.S/] by deleting all internal vertices of degree 2 in any branch ofS. A

3.2 A reduction method for determining the hamiltonian index of a graph39

subsetSof B.G/ is called a branch cut ifG− Shas more components thanG has.A minimal branch cut is called a branch-bond. IfG is connected, then a branchcut S of G is a minimal subset ofB.G/ such thatG− S is disconnected. It iseasily shown that, for a connected graphG, a subsetSof B.G/ is a branch-bondif and only if G− Shas exactly two components. We denote byBB.G/ the set ofbranch-bonds ofG. A connected graphG is eulerian if every vertex ofG has evendegree. The following characterization of eulerian graphs is well-known [42].

Theorem 3.6 (McKee[42])A connected graph is eulerian if and only if each bond contains an even numberof edges.

The following characterization of eulerian graphs involving branch-bonds followsfrom Theorem 3.6.

Theorem 3.7 A connected graph is eulerian if and only if each branch-bond con-tains an even number of branches.

3.2 A reduction method for determining the hamil-tonian index of a graph

Before presenting our main results, we first introduce some additional notation.Catlin [19] developed a reduction method for determining whether a graphG has aspanning closed trail. This method needs a tool, the so-called graph contractions.Let G be a graph and letH be a subgraph ofG. For this we give a refinementof Catlin’s reduction method. Thecontractionof H in G, denoted byG=H, isthe graph obtained fromG by contracting all edges ofH, i.e., replacingH by anew vertexvH, which is called contracted vertex inG=H, such that the numberof edges inG=H joining anyv ∈ V.G/\V.H/ to vH in G=H equals the numberof edges joiningv to H in G. Note that contractions may also result in loops andmultiple adjacencies, but that loops can be avoided if the subgraphH is inducedby a vertex subset. The following lemma follows from Theorem 3.7 and is neededfor our proof of main results.

Lemma 3.1 If G is a eulerian graph and H is a subgraph of G, then G=H is alsoa eulerian graph.

40

Catlin’s reduction method and Theorem 3.3 are useful in the study of the hamil-tonian index as seen in [20], [36], [48] and [49]. However, we must consider thelower iterated line graph when we want to do that.

In the construction, the contraction graphG=H may have lower hamiltonian in-dex thanG. For example the graphG obtained from aK2;3 by replacing eachvertex of degree 2 in theK2;3 by a triangleK3; and letH be one of these newlyaddedK3’s. ThenG=H has an eulerian subgraph containing all vertices exceptthe vertex representingH. Therefore,G=H has hamiltonian index 1 butG hashamiltonian index 2. In order to use Theorem 3.4 we must guarantee that if agraph has hamiltonian index at least two then the contraction must have that also.We do this by attaching two new edge-disjoint branches of length two at the con-tracted vertexvH. The attachment-contractionG==H is the graph obtained fromG=H by attaching two new edge-disjoint branchesb′H;b

′′H of length two atvH,

wherevH ′ is the contracted vertex inG=H so that the vertices of these branchesare not inG or G==H(exceptvH, in the case ofG==H ), i.e. identifyingvH to ex-actly one end ofb′H;b

′′H such thatb′H;b

′′H ∈ B1.G==H/. If G hask vertex disjoint

nontrivial subgraphsH1, H2; · · · ; Hk, then the{H1; H2; · · · ; Hk}-contraction ofG, denoted byG={H1; H2; · · · ; Hk} , is the graph obtained fromG by contractingH1; H2; · · · ; Hk , respectively, and the{H1; H2; · · · Hk}-attachment-contraction ofG, denoted byG=={H1; H2; · · ·Hk}, is the graph obtained fromG by attachment-contractingH1; H2; · · · ; Hk respectively. Let

{H ′1; H ′2; · · · ; H ′s} ⊆ {H1; H2; · · · ; Hk}

be a subset of subgraphs. LetvH ′1; vH ′2; · · · ; vH ′s denote the vertices in

G=={H1; H2; · · · ; Hk}

onto which the subgraphsH ′1; H ′2; · · · ; H ′s are attachment-contracted respectively.Note that a contracted vertex is viewed as a vertex inG={H1; H2; · · · ; Hk} as wellas a subgraph inG. For branchesb1 ∈ B.G/ andb2 ∈ B.G=={H1; H2; · · · ; Hk}),we sayb1 = b2 if the internal vertices ofb1 contain the internal vertices ofb2, andif the ends ofb1 belong to the contracted vertices of the ends ofb2.

Now we can state the main result of this section.

Theorem 3.8 Let G be a connected graph other than a path, and let G1; · · · ;Gk

be all nontrivial components of G[{v : dG.v/ ≥ 3}]−{e : e is a nontrivial cut edge


of G} . If h.G/ ≥ 2, then

h.G/ = h.G=={G1;G2; · · · ;Gk}/:Proof. Theorem 3.4 will be used. Note thatEUk.G/ is the same as in Theorem3.4. In order to prove that a subgraphH belongs toEUk.G/; we only need tocheck thatH satisfies all conditions in the definition ofEUk.G/. That is, theseconditions hold for the graphG and the integerk:

Let G′ = G=={G1;G2; · · · ;Gk} . The following claim is straightforward.

Claim 1. G andG′ have the same branch set of length at least 2 and the samenontrivial cut edges set, but{b′G1

;b′′G1;b′G2

;b′′G2; · · · ;b′Gk

;b′′Gk} ⊆ B1.G′/\B.G/:

First, we will prove thath.G/ ≤ h.G′/.

Take H ∈ EUh.G/.G/. By (ii), H contains all vertices ofk⋃

i=1V.Gi/: We setHi =

H[V.Gi/] for i ∈ {1;2; · · · ; k} and letH ′ = H={H1; H2; · · · ; Hk}. ObviouslyHis a subgraph ofG′ and H contains all vertices of{vG1; vG2; · · · ; vGk}: We willprove thatH ′ ∈ EUh.G/.G/; which implies thath.G′/ ≤ h.G/. By H satisfying(i), H is an union of eulerian subgraphs inG. Hence it follows from Lemma 3.1that H ′ is also union of eulerian subgraphs ofG′; which implies thatH ′ satisfies(i). It follows that H ′ satisfies (ii) fromH satisfying (ii). In order to prove thatH ′ satisfies (iii), it suffices to consider the case that for a subgraphK ′ ⊆ H ′ wehavedG′.K ′; H ′ − K ′/≥ 2. LetK = H[V′K ∪ V′′K] whereV′K = V.K ′/∩ V.G/ andV′′K = V.K ′/ ∩ {vG1; vG2; · · · ; vGk} is a set of contracted vertices. One can easilysee thatK is a subgraph ofH and any shortest pathP in G from K to H − Khas ends of degree at least 3 inG. So P′ = G′[E.P/ ∩ E.G′/] is a path fromK ′

to H ′ − K ′ in G′. Hence sinceH satisfies (iii),dG′.K ′; H ′ − K ′/ ≤ |E.P′/| ≤|E.P/| = dG.K; H − K/ ≤ h.G/− 1. SoH ′ satisfies (iii). By Claim 1,H ′ sat-isfies both (iv) and (v). HenceH ′ ∈ EUh.G/.G′/ which implies thath.G′/≤ h.G/.

It remains to prove thath.G/ ≤ h.G′/. Obviouslyh.G′/ ≥ 2. Hence by Theorem3.4, we can takeH ′ ∈ EUh.G′/.G′/. We will construct a subgraph inEUh.G′/.G/from H ′. SinceH ′ satisfies (ii), and by the definition ofG1;G2; · · · ; Gk;H ′ con-tains all vertices of{vG1; vG2; · · · ; vGk}:Set

Vbi.H′/ = {x ∈ V.Gi/ : x is an endvertex of a branch ofBH ′.G/}

42

for i ∈ {1;2; · · · ; k} and

Vb =k⋃

i=1

Vbi.H/:

We denote byR.x/ the number of branches ofBH ′.G/, one of which hasx as anend vertex. Set

V jbi = {x ∈ Vbi.H

′/ : R.x/ ≡ j.mod2/}and Vjb =

k⋃i=1

V jbi f or i ∈ {1;2}:

SinceH ′ satisfies (i),∑x∈V1

bi

R.x/+∑x∈V2

bi

R.x/ =∑x∈Vbi

R.x/ = dH ′.vGi /

is even. Since∑

x∈V2bi

R.x/ is even, it follows that∑

x∈V1bi

R.x/ is also even. Thus|V1bi|

is even.

Without loss of generality, assume

V1bi = {ui

1; vi1;u

i2; v

i2; · · · ;ui

si; vi

si}:

SinceGi is connected, we can select a shortest path, denoted byp.uij ; v

ij/, between

uij andvi

j in Gi for i ∈ {1;2; · · · ; si} . Set

P.V1b / =

k⋃i=1

Si⋃j=1

{p.uij ; v

ij/}:

Let H be the subgraph ofG with the following vertex set

V.H/ = .k⋃

i=1

V.Gi//⋃.V.H ′/\{vG1; vG2; · · · ; vGk}/

and edge set

E.H/ = E.H ′/⋃{e∈

k⋃i=1

E.Gi / : |{p ∈ P.V1b / : e∈ E.p/}| ≡ 1.mod2/}:

We will prove thatH ∈ EUh.G′/.G/; i.e., by verifying thatH satisfies the condi-tions in the definition ofEUh.G′/.G/ for a graphG and integerh.G′/. First we


prove thatH satisfies (i).

Defining Ex.G/ = {e∈ E.G/ : e is an edge that is incident withx}, we have

d.

k⋃i=1

Gi /⋂

H.x/ =

2|{p ∈ P.V1b / : x ∈ V.P/}|

−1−∑e∈Ex.G/2[1

2|{p ∈ P.V1b / : e∈ E.P/}|];

if x ∈ V1b

2|{p ∈ P.V1b / : x ∈ V.P/}|

−∑e∈Ex.G/2[1

2|{p ∈ P.V1b / : e∈ E.P/}|];if x ∈ V.G/\V.V1

b /:

Hence, for any vertexx ∈ V.H/⋂.

k⋃i=1.Gi /; we have that

dH.x/ = d k⋃i=1.Gi /

⋂H.x/+ R.x/

is even. For anyx ∈ V.H/\k⋃

i=1.Gi /; we havedH.x/ = dG.x/ = 2. SoH satisfies

(i). Since H ′ satisfies (ii),H satisfies (ii). By Claim 1,H satisfies both (iv) and(v).In order to prove thatH satisfies (iii), we only need to consider a subgraphK ofH such thatdG.K; H − K/ ≥ 2; sinceH.G/ ≤ 2. Hence, since

V.Gi/ ⊆1.G/⋃i=3

Vi.G/ ⊆ V.H/ f or i ∈ {1;2; · · · ; k};

V.K/∩ V.Gi/ is either empty orV.Gi/ for i ∈ {1;2; · · · ; k} . Let K1; K2; · · · ; Kc

be all nontrivial components of

K[{v : dk.v/ ≥ 3}] − {e : e is a cut edge of G}:We obtain thatK ′ = K={K1; K2; · · · ; Kc} is a subgraph ofH ′. Let p′ = x′u1u2:::ut y′

be a shortest path fromK ′ to H ′ − K ′ in G′. Since{vG1; vG2; · · · ; vGk} ⊆ V.H ′/;

{u1;u2; · · · ;ut} ∩ {vG1; vG2; · · · ; vGk} = ∅:Hence{u1;u2; · · · ;ut} ⊆ V.G/. By the selection ofK ′ and H, there exist twoverticesx ∈ V.K/ and y ∈ V.H − K/ such thatxu1;ut y ∈ E.G/. Hencep =xu1u2 · · ·ut y is a path fromK to H − K; which implies that

dG.K; H − K/ ≤ |E.p/| = |E.p′/| = dG′.K′; H ′ − K ′/ ≤ h.G′/− 1:

HenceH ∈ EUh.G′/.G/ which implies thath.G/ ≤ h.G′/: �

44

3.3 Sharp upper and lower bounds forh.G/

A branch-bond is called odd if it consists of an odd number of branches. Thelength of a branch-bondS∈ BB.G/, denoted byl .S/, is the length of a shortestbranch in it. DefineBB1.G/ to be the set of branch-bonds, one of which containsonly one branch such that one of its ends has degree one inG, and defineBB2.G/to be the set of branch-bonds, one of which contains only one branch such thatits ends have degree at least three inG, and defineBB3.G/ to be the set of oddbranch-bonds, one of which has at least three branches inG respectively. Define

hi.G/ = {max{l .S/ : S∈ BBi.G/} for i ∈ {1;2;3};

if BBi.G/ is not empty0; otherwise.

If F1 andF2 are two subsets ofE.G/, thenH + F1− F2 denotes the subgraph ofG obtained fromG[.E.H/

⋃F1/\F2] by adding to its vertex set any vertices of

1.G/⋃i=3

Vi.G/ which were not already in its vertex set, and so that any vertices added

are isolated vertices inH + F1− F2:

The following lower bound forh.G/ involving odd branch-bonds can now begiven.

Theorem 3.9 Let G be a connected graph with h.G/ ≥ 1. Then

h.G/ ≥ max{h1.G/;h2.G/+ 1;h3.G/− 1} (3.1)

Proof. If h.G/= 1 then, by Theorem 3.3,h1.G/ ≤ 1;h2.G/ ≤ 0 andh3.G/ ≤ 2;i.e., (3.1) is true. So we can assume thatG is a connected graph withh.G/ ≥ 2:We can take anySi ∈ BBi.G/ such thathi.G/ = l .Si / for i ∈ {1;2;3}: For anysubgraphH ∈ EUh.G/.G/; E.b/∩ E.H/ = ∅ for anyb ∈ S1∪ S2 and there existsat least a branchb ∈ S3 such thatE.b/ ∩ E.H/ = ∅: Hence by Theorem 3.4, weobtainh.G/ ≥ h1.G/ by (v), h.G/ ≥ h2.G/+ 1 by (iii) and h.G/ ≥ h3.G/− 1by (iv). Soh.G/ ≥ max{h1.G/;h2.G/+ 1;h3.G/− 1}; i.e., (3.1) holds.�

We can construct an extremal graph for the equality (3.1). For an integert ≥1, let P1; P2; · · · ; P2t+3 be 2t + 3 vertex disjoint paths and letKa; Kb be twovertex disjoint complete graphs of order at least 3. Taking two verticesu ∈

3.3 Sharp upper and lower bounds forh.G/ 45

V.Ka/ andv ∈ V.Kb/, we construct a graphG0 by identifying exactly one end ofP1; P2; · · · ; P2t+1 respectively, identifyingu and another end ofP1; P2; · · · ; P2t+1,identifying exactly one end ofP2t+2; P2t+3, respectively, identifyingv and anotherend ofP2t+2 such thatP1; P2; · · · ; P2t+3; Ka; Kb are edge-disjoint subgraphs ofG0.Set

k1.G0/ = max{h1.G0/;h2.G0/+ l ;h3.G0/− l}:Obviouslyh1.G0/ = |E.p2t+3/|;h2.G0/ = |E.P2t+2/| and

h3.G0/ = min{|E.P1/|; |E.P2/|; |E.P2t+1/|}:

One can easily see thatKa∪ Kb ∈ EUk1.G0/.G0/:By Theorem 3.4,h.G0/≤ k1.G0/.From the proof of (3.1),h.G0/ ≥ k1.G0/. So h.G0/ = max{h1.G0/;h2.G0/ +1;h3.G0/− l}:�

Now we state our upper bound forh.G/.

Theorem 3.10 Let G be a connected graph that is not a path. Then

h.G/ ≤ max{h1.G/;h2.G/+ 1;h3.G/+ 1} (3.2)

Proof. Let k.G/ = max{h1.G/;h2.G/+ 1;h3.G/+ 1}: Obviouslyk.G/ ≥ 1.If k.G/ = 1, i.e., h1.G/ ≤ 1 andh2.G/ = h3.G/ = 0, then, by Theorem 3.7,G[V.G/\V1.G/] is eulerian. Hence, using Theorem 3.3, we obtainh.G/ ≤ 1;i.e., (3.2) is true.

So we assume thath.G/≤ 2 andk.G/≤ 2. By Theorem 3.4, it suffices to considerthe graphG such thatG[{v : dG.v/ ≥ 3}] − {e : e is a nontrivial cut edge of G}has no nontrivial component. LetH be a subgraph inEUh.G/.G/ such thatH con-tains as many branches as possible, one of which has a number of edges greaterthank.G/− 1. Then we can prove the following.

Claim 1. If S is a branch-bond inBB.G/ such that it contain at least threebranches, then there exists no branchb ∈ S\BH.G/ such that|E.b/| ≥ k.G/.

Proof of Claim 1. Otherwise there exists a branchb0 ≥ B.G/\BH.G/ and abranch bondS∈ BB3.G/ such that|E.b0/| ≥ k.G/ andb0 ∈ S\BH.G/. Obvi-ouslyb0 has two endsu andv (say). Now we can select a branch-bond, denotedby S.u;b0/, such that it containsb0 and any branch ofS.u;b0/ has the endu:

46

In order to obtain a contradiction, we will take a cycle ofG that containsb0 by thefollowing algorithm.

Algorithm b0.

1. If |S.u;b0/| ≡ 0(mod2), then (by Theorem 3.7, we can) select a branchb1 ∈S.u;b0/\.BH.G/ ∪ {b0}/. Otherwise (we can) select a branchb1. 6= b0/ ∈S.u;b0/ with |E.b1/| = l .S.u;b0// ≤ h3.G/ and letu1. 6= u/ be anotherend ofb1. If u1 = v, thent := 1 and stop. Otherwisei := 1.

2. Select a branch-bondS.u;ui;b0/ in G which containsb0 but b1;b2; · · · ;bi

such that eitheru or ui is an end of a branch inS.u;ui;b0/. If |S.u;ui;b0/| ≡0.mod2/, then (by Theorem 3.7, we can) select a branch

bi+1 ∈ S.u;ui;b0/\.BH.G/∪ {b0}/:Otherwise (we can) select a branchbi+1 ∈ S.u;ui;b0/ such thatbi+1 6= b0

and|E.bi+1/| = l .S.u;ui;b0// ≥ h3.G/, and letui+1 be the end ofbi+1 thatis neitheru nor ui.

3. If ui+1= v, thent := i + 1 and stop. Otherwise replacei by i + 1 and returnto step 2.

Since|B.G/| is finite, Algorithmb0 will stop after a finite number of steps. Since

ut = v, G[u⋃

i=0E.bi /] is connected. Hence we obtain the following.

Claim 2. G[u⋃

i=0E.bi /] has a cycle ofG, denoted byC0, which containsb0:

Now we construct a subgraphH ′ ⊆ G as follows:

H ′ = H + E.C0/\E.H/− .E.H/∩ E.C0/:

By the selection of{b1;b2; · · · ;bt};|E.b/| ≤ h3.G/ ≤ k.G/− 1 f or b ∈ BH.G/∩ {b1;b2; · · · ;bt}:

Hence, by Claim 2,H ′ satisfies (iii)-(iv). ObviouslyH ′ satisfies both (i) and (ii),and this impliesH ′ ≤ EUh.G/.G/. But H ′ contains more branches thanH does,one of which has edges number greater thank.G/− 1; and this contradicts theselection ofH, which completes the proof of Claim 1.�

3.4 Analysis of known results 47

For any branchb of G, if G[E.b/] is not a cycle ofG, then there exists a branchbondS∈ BB.G/ with b ∈ S. Hence, by Claim 1 and the selection ofk.G/; H ∈EUk.G/.G/ which implies thath.G/ ≤ k.G/:�

We can construct a family of extremal graphs for Theorem 3.10. From the ex-tremal graph of Theorem 3.9, we only need to construct an extremal graphG0

with h.G0/ = h3.G0/+ 1. In fact, in the following construction we can constructa family of graphsG0 such thath.G0/ can take all integers betweenh3.G0/− 1andh3.G0/+ 1. Let k ≥ 1 be an integer and letH = K2;2k+1 be a complete bi-partite graph withV1.H/= {x; y} andV2.H/= {u1;u2; · · · ;u2k+1} . Let 1≤ l1�l2 ≤ l3� l4 be four integers. ObtainG0 by subdividingxu1; xu2; · · · ; xu2k into 2kbranches of lengthl4; yu1; yu2; · · · ; yu2k into 2k branches of lengthl1; xu2k+1 intoa branchb of length l2; yu2k+1 into a branchb′ of length l3 respectively and byreplacing each vertex ofV2.H/ by a K4. One can easily see thath3.G0/ = l3 andthatG0[E.G0/\.E.b/ ∪ E.b′//] has a subgraph inEUmax{l3−1;l2+1}.G0/ . By The-orem 3.4,h.G0/ ≤ max{l3− 1; l2+ 1}: By an argument similar to the one in theproof of (3.1),h.G0/ ≥max{l3− 1; l2+ 1} . Henceh.G0/ = max{l3− 1; l2+ 1}:Clearly

h.G0/ = max{L3− 1; l2+ 1} = l3− 1; if l2 ≤ l3− 2

l3; if l2 = l3− 1l3+ 1; if l2 = l3:

Henceh.G0/ may have all integers fromh3.G0/− 1 to h3.G0/+ 1 according todifferent integersl2 andl3.

3.4 Analysis of known results

Theorems 3.5 and 3.10 show two upper bounds for the hamiltonian index of agraph. Clearlyhi.G/ ≤ dia.G/ for i ∈ {1;2;3} and there exists a graph with largediameter and smallh3.G/: For example, the graph obtained by replacing eachedge of a path by an odd branch-bond, which contains at least three branches.Hence the upper bound in Theorem 3.5 is not better than the one in Theorem 3.10.It seems that the upper bound in Theorem 3.10 is better than the one in Theorem3.5. However this is not true. We investigate the graphFt obtained fromK2;2t+1 byreplacing each edge ofK2;2t+1 by a path of lengths. Clearlyh3.Ft/= s= dia.Ft/

but h.Ft/ = s− 1= h3.G/− 1= dia.Ft/− 1.

48

The following relation between the two bounds in Theorems 3.5 and 3.10 is ob-tained.

Theorem 3.11 Let G be a connected graph other that a path with h.G/ ∈ 1. Ifh3.G/ = dia.G/, then

h.G/ = dia.G/− 1:

Proof. Follows from Theorems 3.5 and 3.9.�

Obviously Theorem 3.1 is a consequence of Theorem 3.10. Although Theorem3.2 is not a consequence of Theorem 3.10, one easily checks thathi.G/ ≤ n−1.G/ for any i ∈ {1;2;3}: Hence, from Theorem 3.10, we have thath.G/ ≤n−1.G/+ 1. Moreover ifh3.G/ ≤ n−1.G/, then Theorem 3.10 is better thanTheorem 3.2. The following consequences of Theorem 3.10 are easily obtained.�

Corollary 3.1 (Catlin et al. [20]) Let G be a connected graph that is neither apath nor a 2-cycle. Then

h.G/ ≤ max{v1;v2}⊆W.G/

minP

X.P/+ 1

where X.P/ denotes the length|E.b/| of the longest branch b in BP.G/ and P isa subgraph induced by all branches in G whose end-vertices are u andv.

Proof. Let Sbe a branch bond inBB.G/with l .S/=max{h2.G/+1;h3.G/+1}.Then any path ofG between two verticesu andv in two components ofG− Srespectively must have a branch inS. Hence

max{h1.G/;h2.G/+ 1;h3.G/+ 1} ≤ max{v1;v2}⊆W.G/

minP

X.P/+ 1:

This relation and Theorem 3.10 give Corollary 3.1.�

Corollary 3.2 (Chartrand and Wall [23]). If T is a tree which is not a path, then

h.T/ = max{h1.T/;h2.T/+ 1}:Proof. If T is a tree, thenh3.T/ = 0: Hence by Theorems 3.9 and 3.10, we obtainCorollary 3.2 is true.�

3.4 Analysis of known results 49

Corollary 3.3 (Balakrishnan and Paulraja [1]) Let G be a connected graph withat least four edges. If the only 2-degree cut sets of G are the singleton subsetswhich are neighbors of end vertices of G, then h.G/ ≤ 2.

Proof. One can easily check thath1.G/ ≤ 2;h2.G/ ≤ 1 andh3.G/ ≤ 1. Hencethis corollary follows from Theorem 3.10.

Corollary 3.4 (Lesniak-Foster and Williamson [39]) Let G be a connected graphwith at least four edges. If every vertex of degree two is adjacent to an end vertex,then h.G/ ≤ 2.

Proof. From the condition of this corollary, we knowh1.G/ ≤ 2;h2.G/ = 0 andh3.G/ = 0. Hence this corollary follows from Theorem 3.10.�

Corollary 3.5 (G. Chartrand [22]) Let G be a connected graph other than a path.If Ž.G/ ≥ 3, then h.G/ ≤ 2.

Proof. Obvious.�

Chapter 4

The hamiltonian index of a graph

Harary and Nash-Williams characterized hamiltonian line graphs as follows.

Theorem 4.1 (Harary and Nash-Williams[32]) Let G be a connected graph. Thenthe line graph L.G/ is hamiltonian if and only if G has a circuit C such thatE.G/ = {e∈ E.G/ : dG.e;C/ = 0}:Theorem 4.1 is a tool for investigating problems concerning cycles in line graphs.As such it is a tool to invertigate the hamiltonian index of a graph. Recently, Xiongand Liu characterized those graphsG for which Ln.G/.n ≥ 2/ is hamiltonian.HereVi.H/ = {v ∈ V.H/ : dH.v/ = i}:Theorem 4.2 (Xiong and Liu[60]) Let G be a connected graph and n≥ 2: Thenthe n− th iterated graph Ln.G/ is hamiltonian if and only if EUn.G/ 6= ∅, whereEUn.G/ denotes the set of those subgraphs H of a graph G which satisfy thefollowing conditions:

(i) any vertex of H has even degree in H, i.e., dH.x/ ≡ 0.mod2/ for any x∈V.H/;

(ii) V0.H/ ⊆1.G/⋃i=3

Vi.G/ ⊆ V.H/;

(iii) d G.H1; H − H1/ ≤ n− 1 for any subgraph H1 of H;

(iv) |E.b/| ≤ n+ 1 for any branch b∈ B.G/\BG.H/;

51

52

(v) |E.b/| ≤ n for any branch b∈ B1.G/.

In Section 4.1, we will use Theorem 4.2 to show that the hamiltonian index ofa connected graph other than a path is less than its diameter which improves theresults of Catlin et al.[20] and Sara˘zin [49]. In Section 2, we will consider thehamiltonian index of the complement of a graph.

4.1 A sharp upper bound

In this section, our main purpose is to improve the following result.

Theorem 4.3 (Catlin et al.[20]) Let G be a connected graph that is neither a pathnor a 2-cycle. Then

h.G/ ≤ dia.G/:

Using the reduction method, Sara˘zin [49] recently gave a sharp upper bound forthe hamiltonian index of a connected simple graph other than a path.

Theorem 4.4 (Sarazin [49]) Let G be a connected simple graph of order n otherthan a path. Then

h.G/ ≤ n−1.G/:One can easily see thatdia.G/− 1≤ n−1.G/[63]: Using Theorem 4.2, Xiongand Liu[60] gave a simple proof of Theorem 4.4 and here we improve it. Therelated result obtained by Veldman [55].

Theorem 4.5 [55] Let G be a connected graph with at least three edges and suchthat dia.G/ ≤ 2: Then L.G/ is hamiltonian, i.e., h.G/ ≤ 1.

Theorem 4.5 is the case thatdia.G/ ≤ 2 of the following theorem.

Theorem 4.6 Let G be a connected graph other than a path. Then

h.G/ ≤ dia.G/− 1

and the upper bound is sharp.

Proof . We only need to consider the case thatdia.G/ ≥ 3 sinceG is a completegraph ifdia.G/ = 1 andh.G/ ≤ 1 if dia.G/ = 2 by Theorem 4.5. DefineH0 as

the subgraph ofG havingV.H0/ =1.G/⋃i=3

Vi.G/ andE.H0/ = ∅. And then define

4.1 A sharp upper bound 53

H as subgraph ofG obtained fromH0 by adding toH0 as many vertices ofV2.G/as possible(along with the least number of appropriate edges) so thatH satisfiesboth (i) and ( ii). Next we will prove thatH ∈ EUdia.G/−1.G/. We only need toprove the following three claims.

Claim 1. dG.H1; H − H1/ ≤ dia.G/− 2 for any subgraphH1 ⊆ H.

Proof of Claim 1. Let p= u1u2 · · ·us be a shortest path fromH1 to H− H1 withu1 ∈ V.H1/ andus ∈ V.H − H1/. Obviously{u2;u3; · · · ;us−1} ∩ V.H/ = ∅: ByH satisfying (ii),

{u2;u3; · · · ;us−1} ⊆ V2.G/: (4.1)

SupposedG.H1; H − H1/ ≥ dia.G/− 1, i.e.,".p/ ≥ dia.G/− 1≥ 2: SinceHsatisfies both (i) and (ii) ands= ".p/ + 1 ≥ 3; {u1;us} ∩ V2.G/ = ∅ and p ∈B.G/\B1.G/: Hence we have

{u1;us} ⊆1.G/⋃i=3

Vi.G/: (4.2)

By (4.1) and (4.2), we can select two edgese1 = u1x ande2 = usy of G such thatx 6= y and{x; y}∩V.p/= ∅. Let p′ be a shortest path fromx to y in G. Obviously".p/− 2≤ ".p′/ ≤ ".p/+ 2.If ".p′/ = ".p/+ 2, then".p′/ = dG.H1; H − H1/+ 2≥ dia.G/+ 1, a contra-diction.

If ".p′/ = ".p/ − 2, thenu1 p′us is a shortest path fromH1 to H − H1. By Hsatisfying (ii),V.p′/ ⊆ V2.G/. Therefore, by (4.1),H ′ = H + E.p/∪ E.u1 p′us/

is a subgraph ofG which satisfies both (i) and (ii) but it contains more vertices inV2.G/ thanH does, a contradiction.

So we can assume".p/− 1≤ ".p′/ ≤ ".p/+ 1. Hence

V.p/∩ V.p′/= ∅; |V.p′/∩ V2.G/| ≥ ".p/−1 and|V.p′/∩ V.H/| ≤ 3: (4.3)

We will obtain contradictions by considering the following cases.

Case 1. ".p/ ≥ 3:

By (4.1), (4.2) and (4.3),

54

H ′ = H + .E.p/ ∪ E.u1 p′us//− .E.u1 p′us/∩ E.H//

is a subgraph ofG which satisfies both (i) and (ii) but it contains more vertices inV2.G/ thanH does, a contradiction.

Case 2. ".p/ = 2:

By (4.1), (4.2) and (4.3),

|V.p′/∩ V2.G/∩ V.H/| ≤ 2:

We will distinguish the following two subcases.

Case 2.1.|V.p′/∩ V2.G/∩ V.H/| ≤ 1:

By (4.1), (4.2) and (4.3),

H ′ = H + .E.p/ ∪ E.u1 p′us//− .E.u1 p′us/∩ E.H//

is a subgraph ofG which satisfies both (i) and (ii) but it contains more verticesthanH does, a contradiction.

Case 2.2.|V.p′/∩ V2.G/∩ V.H/| = 2:

By (4.1), (4.2) and (4.3),

|V.p′/∩ V2.G/| = 3 and |V.p′/∩ .1.G/⋃i=3

Vi.G//| = 1: (4.4)

Without loss of generality, we letp′ = xx1x2y and{x; x1} ⊆ V2.G/∩ V.H/; x2 ∈1.G/⋃i=3

Vi.G/ ⊆ V.H/; y ∈ V2.G/\V.H/: Hence, by (4.2), there exists a vertexy′ ∈V.G/ such thatu3y′ ∈ E.G/ andy′ 6∈ {y;u2}:Hence, sincedG.x; y′/ ≤ dia.G/ ≤ s= 3; either x2y′ ∈ E.G/ or dG.u1; y′/ ≤2. If x2y′ ∈ E.G/, thenx2y′u3 is a shortest path fromH1 to H − H1: So y′ ∈V2.G/\V.H/: Hence by (4.1), (4.2), (4.3) and (4.4),

H ′ = H + E.u3yx2y′u3/

4.1 A sharp upper bound 55

is a subgraph ofG which satisfies both (i) and (ii), but it contains more vertices inV2.G/ thanH does, a contradiction.If dG.u1; y′/ ≤ 2, then letp.u1; y′/ be a shortest path fromu1 to y′ in G. Hence

H ′ = H + .E.p/ ∪ {usy′} ∪ E.p.u1; y′///− ..{usy

′} ∪ E.p.u1; y′///∩ E.H//

is a subgraph ofG which satisfies both (i) and (ii) but it contains more vertices inV2.G/ thanH does, a contradiction.This completes the proof of Claim 1.

Claim 2. |E.b/| ≤ dia.G/− 1 for anyb ∈ B1.G/:

Proof of Claim 2. Otherwise there exists a branchb ∈ B1.G/ with |E.b/| ≥dia.G/ ≥ 3: Let x be the end ofb with dG.x/ ≥ 3 andy be the end ofb withdG.y/ = 1. Then there exists a vertexz∈ V.G/\V.b/ such thatxz∈ E.G/. Ob-viously dG.y; z/ = |E.b/| + 1≥ dia.G/+ 1 which is a contradiction. This priveClaim 2.

Claim 3. |E.b/| ≤ dia.G/ for anyb ∈ B.G/\BG.H/:

Proof of Claim 3. Otherwise there exists a branchb∈ B.G/with E.b/∩ E.H/=∅ such that|E.b/| ≥ dia.G/+ 1. By Claim 2,b 6∈ B1.G/. Without loss of gen-erality, we assume thatu andv are two endvertices ofb, let p be a shortest pathbetweenu andv in G. Then|E.p/| ≤ dia.G/ ≤ |E.b/| − 1. Hence

H ′ = H + .E.b/∪ E.p//− .E.p/ ∩ E.H//

is a subgraph ofG which satisfies both (i) and (ii) but it contains more vertices inV2.G/ thanH does, a contradiction. This completes the proof of Claim 3.

By Claim 1, Claim 2 and Claim 3,H ∈ EUdia.G/−1.G/. Hence by Theorem 4.2,h.G/ ≤ dia.G/− 1:

For any integerd ≥ 3 we can construct a graphGd to show that the upper boundis sharp. LetKs.s≥ 3/ and Kt.t ≥ 3/ be two complete graphs, and letpd be apath of lengthd− 2 such thatKs, Kt and pd are three edge -disjoint graphs. NowGd is obtained by identifying two endvertices ofpd with one vertex ofKs andone ofKt respectively. By Theorem 4.2,h.Gd/ ≤ dia.Gd/− 1= d− 1 since theedgeless graphH with V.H/ = V.Ks/ ∪ V.Kt/ such thatH ∈ EUd−1.Gd/: One

56

can easily see that for eachi ∈ {0;1; · · · ;d− 2}; Li.Gd/ is not hamiltonian andthush.Gd/ ≥ d− 1= dia.Gd/− 1. Henceh.Gd/ = dia.Gd/− 1: �

Note. The graphGd constructed in this way shows that the upper bounddia.Gd/−1= d− 1 for the hamiltonian index of a graphGd is superior to the upper boundn−1.G/= s+ t+ d−3−max{s; t} =min{s; t}+ d−3: In [20], it was claimedthat Theorem 4.3 is best possible. Our result shows the falsity of this claim.

4.2 The hamiltonian index of the complement of agraph

In this section, we consider the hamiltonian index of thecomplement ofa graph.Before presenting our results, we need to state some useful results.

Theorem 4.7 (Lai[36]) Let G be a simple graph with at least 61 vertices, and letG denote the complement of G. One of the following holds.

• Either G orG is supereulerian.

• Both G andG have a vertex of degree 1.

• Either G or G is contractible to a K2;t for some odd integer t≥ 3, and theother one has either one or two vertices of degree 1.

• Either G orG is contractible to a K1;p for some integer p≥ 1, and the otherone has exactly one isolated vertex.

Theorem 4.8 [29] If G is a simple graph of order n≥ 3 andŽ.G/ ≥ n=2, then Gis hamiltonian, i.e., h.G/ = 0:

Theorem 4.9 [60] Let G be a connected simple graph other than a path. Then

h.G/ ≤ max{|E.b/| : b ∈ B.G/\{b ∈ B.G/ : G[V.b/] is a cycle of G}} + 1.

Our main results constitute are the following.

Theorem 4.10 Let G andG be two connected simple graphs of order n≥ 61other than paths. Then

4.2 The hamiltonian index of the complement of a graph 57

(a)min{h.G/;h.G/} ≤ 1 (4.5)

and the bound is sharp.

(b) If min{h.G/;h.G/} = 1; then

max{h.G/;h.G/} ≤ .n− 1/=2 (4.6)

and the equality holds if and only if one of G andG is isomorphic to thegraph Ft obtained by identifying one endvertex of a path Pt of length t− 1with exactly one vertex of a complete graph Kt of order t, where|V.Ft/| =n= 2t− 1≥ 61:

Proof(a) By Theorem 4.7, one ofG andG either is supereulerian or has one vertex ofdegree 1. If one ofG andG is supereulerian, then by Theorem 4.1 its hamiltonianindex is at most one. Hence (4.5) holds.

It remains to consider the case that one ofG andG has one vertex of degree 1.Without loss of generality, we assume thatG has one vertexv with dG.v/= 1 anduv ∈ E.G/. SinceG is connected,dG.u/ ≤ n− 2:

Hence we distinguish the following two cases.

Case 1.dG.u/ ≤ n− 3:

We can take two distinct verticesx; y in V.G/\{u; v} such thatxu; yu 6∈ E.G/, i.e.,xu; yu∈ E.G/. Let C be a circuit ofG with {vx; xu;uy; yv} ⊆ E.C/ such thatCcontains as many edges inE.G/ as possible. ThendG.e;C/= 0 for anye∈ E.G/.OtherwiseG has an edgee0 = u1u2 such thatdG.e0;C/ = 1. Since{vu1; vu2} ⊆E.G/, C′ = C+ {vu1; vu2;u1u2} is also a circuit such that{vx; xu;uy; yv} ⊆E.C′/ but C′ contains more edges thanC does. ThusdG.e;C/ = 0 for anye∈ E.G/. Hence, by Theorem 4.1,h.G/ ≤ 1; which implies that (4.5) holds.

Case 2.dG.u/ = n− 2:

Letw be the vertex withuw 6∈ E.G/.

58

We consider the following three subcases.

Subcase 2.1.dG.w/ ≤ n− 5:

We can take two verticesx; y in V.G/\{u; v;w} such thatwx; wy 6∈ E.G/;i.e.,wx;wy ∈ E.G/. Let C be a circuit ofG with {wx; xv; vy; yw} ⊆ E.C/ such thatC contains as many edges inE.G/ as possible. Then by an argument similar tothe one in the proof of Case 1,dG.e;C/ = 0 for anye∈ E.G/. Hence, by Theo-rem 4.1,h.G/ ≤ 1; which implies that (4.5) holds.

Subcase 2.2.dG.w/ = n− 4.

Let x be the vertex inV.G/\{u; v;w} with xw 6∈ E.G/. Hencetu; tw ∈ E.G/for any t ∈ V.G/\{u; v;w; x}. Let C be a circuit ofG such thatC contains asmany edges inE.G/ as possible. One can easily see thatdG.e;C/ = 0 for anye∈ E.G/. Hence by Theorem 4.1,h.G/ ≤ 1 which implies that (4.5) holds.

Subcase 2.3.dG.w/ = n− 3.

Hencexu; xw ∈ E.G/ for anyx ∈ V.G/\{u; v;w}. By an argument similar to theone in the proof of subcase 2.2,h.G/ ≤ 1. This implies that (4.5) holds.The sharpness of (4.5) will be shown in the proof of (b).

(b) If min{h.G/;h.G/} = 1, say,h.G/ = min{h.G/;h.G/} = 1, then

1.G/ > n=2− 1: (4.7)

OtherwiseŽ.G/= n−1−1.G/≥ n=2, and, orem 4.8,h.G/= 0, a contradiction.Setk.G/ = max{|E.b/| : b ∈ B.G/\{b ∈ B.G/ : G[V.b/]is a cycle ofG}}Let b ∈ B.G/\{b ∈ B.G/ : G[V.b/] is a cycle ofG} such that|E.b/| = k.G/.

Now we distinguish the following two cases.

Case 1.k.G/ ≤ .n− 4/=2.

By Theorem 4.9,h.G/ ≤ k.G/+ 1≤ .n− 2/=2. Hence (4.6) holds.


Case 2.k.G/ ≥ .n− 3/=2:

Before proceeding, we prove the following two claims.

Claim 1. If b ∈ B1.G/, thenk.G/ ≤ .n− 1/=2.

Proof of Claim 1. Otherwisek.G/ ≥ n=2. It is easily seen that1.G/ ≤ n−k.G/ ≤ n=2. By (4.7), k.G/ ≤ n−1.G/ < n− .n=2− 1/ ≤ .n+ 2/=2 and1.G/ ≥ .n− 1/=2. Thusk.G/ ≤ .n+ 1/=2 and1.G/ + k.G/ = n. Withoutloss of generality, we assume thatb= xuk.G/uk.G/−1 · · ·u2u1 with dG.u1/ = 1 andV.G/\V.b/ = {v1; v2; · · · ; v1.G/−1}, then

C={

u1xu2v1u3v2 · · ·uk.G/v1.G/−1u1; if 1.G/ = n=2u1xu2v1u3v2 · · ·v1.G/−1uk.G/u1; if 1.G/ = .n− 1/=2:

is a hamiltonian cycle ofG, i.e., h.G/ = 0, a contradiction. This completes theproof of Claim 1.

Claim 2. If b ∈ B.G/\B1.G/; thenG[E.G/\E.b/] has only one nontrivial com-ponent.

Proof of Claim 2. OtherwiseG[E.G/\E.b/] has two nontrivial components. Ifk.G/≥ .n−2/=2;then1.G/≤ n− .k.G/+2/≤ n− ..n−2/=2+2/= n=2−1;which contradicts (4.7). If.n−3/=2≤ k.G/ < .n−2/=2, i.e.,k.G/= .n−3/=2,then letG1 andG2 be the two nontrivial components ofG[E.G/\E.b/]. Withoutloss of generality, we may assume that

s= |V.G1/| ≤ |V.G2/| = .n+ 5/=2− s;V.G1/ = {u1;u2; · · · ;us};b= usus+1 · · ·us+.n−3/=2 andV.G2/ = {us+.n−3/=2;us+.n−1/=2; · · · ;un}.

Hence 3≤ s≤ .n+ 5/=4, and thusu.n+1/=2 ∈ V.b/\.V.G1/∪ V.G2//. So

{unu.n+1/=2;u.n+1/=2u1} ⊆ E.G/:

HenceC= u1u.n+3/=2u2u.n+5/=2 · · ·u.n−1/=2unu.n+1/=2u1

is a hamiltonian cycle ofG, i.e., h.G/ = 0, a contradiction that completes theproof of Claim 2.

60

Now we will distinguish two subcases.

Subcase 2.1.b ∈ B1.G/.

By Claim 1,k.G/ ≤ .n− 1/=2. By (4.7) and.n− 3/=2≤ k.G/ ≤ .n− 1/=2, the

edgeless graphH ′ with V.H ′/ =1.G/⋃i=3

Vi.G/ is a subgraph inEUk.G/.G/. So by

Theorem 4.2,h.G/ ≤ k.G/ ≤ .n− 1/=2, i.e., (4.6) holds.

Subcase 2.2.b 6∈ B1.G/; i:e:;b ∈ B.G/\B1.G/.

By Claim 2,G[E.G/\E.b/] has exactly one nontrivial component. Let

b= u1u2 · · ·uk.G/+1 ∈ B.G/\B1.G/;

then there exists a longest pathp= p.u1;uk.G/+1/ from u1 to uk.G/+1 in the sub-graphG[E.G/\E.b/] of G: Let H′ be a subgraph ofG with V.H ′/ = V.b/ ∪V.p/∪ .

1.G/⋃i=3

Vi.G// andE.H ′/= E.b/∪ E.p/. By (4.7) andk.G/ ≥ .n− 3/=2,

|V.G/\.⋃

v∈V.H ′ /N.v//| ≤ n− ..k.G/− 2/+1.G/+ 1/ ≤ 3:

HenceH ′ is a subgraph inEU7.G/. Therefore, by Theorem 4.2,h.G/ ≤ 7 whichimplies that (4.6) holds forn≥ 61.

Clearly, Ft is a graph of ordern = 2t − 1. The subgraphH of Ft; for whichV.H/ = V.Kt/ and E.H/ = ∅, is an element ofEUt−1.Ft/: By Theorem 4.2,ham.Ft/≤ t−1= .n−1/=2. One can easily see that for eachi ∈ {0;1;2; · · · ; t−2}; Li.Ft/ is not hamiltonian and thush.Ft/ ≥ t− 1= .n− 1/=2. Henceh.Ft/ =.n− 1/=2: Since the number of components ofFt − .V.Pt/\.V.Pt/ ∩ V.Kt///

is greater than|V.Pt/| − 1; Ft is not hamiltonian. Henceh.Ft/ ≥ 1. By (4.5),h.Ft/= 1. This also shows the sharpness of (4.5).

On the other hand, if min{h.G/;h.G/} = 1 andmax{h.G/;h.G/} = .n− 1/=2;say,h.G/ = .n− 1/=2 andh.G/ = 1 then, by the proof of (4.6),k.G/ = .n−1/=2 and G has a branchb ∈ B1.G/ of length k.G/. Let G1 be the uniquenontrivial component ofG[E.G/\E.b/]. Then G1 is isomorphic to a completegraph of order.n+ 1/=2. OtherwiseG1 has two verticesu1 and u2 such that


u1u2 6∈ E.G1/, i.e., u1u2 ∈ E.G/. Without loss of generality, we letV.G1/ ={u1;u2; · · · ;u.n−1/=2;u.n+1/=2} andb= uiu.n+3/=2u.n+5/=2 · · ·un whereui ∈ V.G1/.

If ui ∈ {u1;u2}; say,ui = u1, then

C= unu1u2un−1u3un−2u4 · · ·u.n−1/=2u.n+3/=2u.n+1/=2un

is a hamiltonian cycle ofG; which contradictsh.G/ = 1:

If ui ∈ {u3;u4; · · · ;u.n+1/=2}, say,ui = u3; then

C= unu3un−1u1u2un−2u4un−3u5un−4 · · ·u.n−1/=2u.n+3/=2u.n+1/=2un

is a hamiltonian cycle ofG; which contradictsh.G/ = 1. This shows thatG1 isisomorphic to a complete graph of order.n+ 1/=2. HenceG is isomorphic toFt

wheret = .n+ 1/=2.

By Theorem 4.10, we can obtain the following Nordhaus-Gaddum inequalities.

Theorem 4.11 If G andG are two connected simple graphs of order n≥ 61otherthan paths and cycles, then

(c) If 1.G/ ≤ n=2− 1, then

0≤ h.G/+ h.G/ ≤ n−1.G/ ≤ n− 3

h.G/ h.G/ = 0;

(d) If 1.G/ > n=2− 1, then

0≤ h.G/+ h.G/ ≤ n− 3

0≤ h.G/ h.G/ ≤ .n− 1/=2;

and all bounds are sharp.

Proof.(c) If 1.G/ ≤ n=2− 1, thenŽ.G/ = n− 1−1.G/ ≥ n=2: By Theorem 4.8,h.G/ = 0: (c) follows from Theorem 4.4. Clearly, (c) is sharp by Theorem 4.4.(d) This follows from Theorem 4.4 and the proof of Theorem 4.10. One can easilysee that the lower bounds of (d) are sharp. The sharpness of the upper bounds of(d) follows from Theorem 4.4 and Theorem 4.10.�

Notation and Terminology

Throughout Chapter 5 up to Chapter 9, We follow the notation of Bondy andMurty [8], unless otherwise stated. The graphs considered in these chapters arefinite simple graphs.

Let G be a finite simple graph, andH be a subgraph ofG. ThenV.H/ andE.H/denote the set of vertices and edges ofH; and E.H/ denotes the set of edges ofG that are incident with vertices ofH: We denote|V.H/|; |E.H/| and|E.H/|by v.H/; ".H/; ".H/; respectively. IfS is a subgraph ofH or a subset ofV.H/;then thedegreeof S in H; denoted bydH.S/; is defined to be the degree sum ofvertices inS; i.e., dH.S/ =

∑u∈V.S/ dH.u/; or justd.S/ if G= H: Thedistance

between two verticesx andy, denoted byd.x; y/; is the length of a shortest pathbetweenx and y: The distance between two subgraphs G1;G2 of H; denotedby dH.G1;G2/; is defined to bemin{dH.x; y/ : x ∈ V.G1/ andy ∈ V.G2/}: Thediameterof H is the maximum distance between two vertices ofH: Thegirth ofa graph is the minimum length of a shortest cycle of a graph.

We useH − E.S/ to denote the edge induced subgraphH[E.H/\E.S/] of H.We denote the nontrivial component ofH by H − S if H − E.S/ has at mostone nontrivial component ofH: By Ck we denote the cycle of lengthk: Let½.G/ = {k : G has aCk}: We usecr.G/ to denote thecircumf erenceof G; i.e.,the length of a longest cycle ofG: G is saidsubpancyclicif ½.G/ = [3; cr.G/] ={3;4; · · · ; cr.G/}: G is calledpancyclicif it is subpancyclic and hamiltonian.Cis called acircuit of a graphG if C is a eulerian subgraph ofG, i.e., a connectedsubgraph in which every vertex has even degree. Note that by this definition (thetrivial subgraph induced by) a single vertex is also a circuit.

Define

²i.G/ = min{d.P/ : P is a path of lengthi − 1 in G}:

63

64

ObviouslyŽ.G/ = ²1.G/: As introduced in [5], letfi.n/ be the smallest integersuch that for any graphG of ordern with ²i.G/ > fi.n/; the line graphL.G/ ofG is pancyclic wheneverL.G/ is hamiltonian.

Chapter 5

Edge degree conditions forsubpancyclicity in line graphs

A natural question is the following: how large should the minimum degreeŽ.G/of a graphG be in order that the Hamiltonian line graphL.G/ is guaranteed to bepancyclic?

From results in [5] it follows that there exists a constantA such that ifL.G/ ishamiltonian andŽ.G/ > An1=3, thenL.G/ is pancyclic. Also, there exists a con-stantB and an infinite family of graphsG such thatŽ.G/ > Bn1=3 and L.G/ ishamiltonian but not pancyclic.

Here we consider a similar question concerning edge degree.Our main results are the following.

Theorem 5.1 Let G be a graph of order n.n ≥ 72/; girth at least 5, and with²2.G/ >

√2n+ 1 (or ²2

2.G/−1> 2n). Then L.G/ is subpancyclic and the resultis best possible.

Theorem 5.2 Let G be a graph of order n.n≥ 72/; girth at least 4, and²22.G/−

²2 > 2n: Then L.G/ is subpancyclic and the result is best possible.

In general, results involving degree sums are direct derivatives from results involv-ing the minimum degree of the graph. Theorem 5.1 and 5.2 show an exception tothis rule (compare the results in [5] mentioned above).

65

66

The main theorems in the chapter deal with graphs with girth at least 5 (Theorem5.1 ) and girth at least 4 (Theorem 5.2). Since the necessary degree condition forsubpancyclicity in Theorem 5.1 is a little weaker than the degree condition in The-orem 5.2, an obvious question is whether one can expect an even weaker conditionfor graphs with girth at leastk for somek ≥ 6: But, in fact, the extremal graphsgiven at the end of the proof of Theorem 5.1 can be chosen to have arbitrarilylarge girth. This shows that even for graphs with large girth the degree conditionin Theorem 5.1 is best posssible.

5.1 Proof of the main theorems

Before proving Theorems 5.1 and 5.2 we introduce some additional terminologyand notation, and state a number of preliminary results.

By acircuit of a graphG we will mean a Eulerian subgraph ofG, i.e., a connectedsubgraph in which every vertex has even degree. Note that by this definition (thetrivial subgraph induced by ) a single vertex is also a circuit. IfC is a circuit ofG, thenE.C/ denotes the set of edges ofG incident with at least one vertex ofC,we write".C/ for |E.C/| and".C/ for |E.C/|: Let Cm denote a cycle of lengthm.

Harary and Nash-Williams [32] characterized Hamiltonian line graphs.

Theorem 5.3 (Harary and Nash-Williams [32] ). The line graph L.G/ of a graphG is Hamiltonian if and only if G contains a circuit C such that".C/= |E.G/| ≥3:

From Theorem 5.3 one can easily prove a more general results (see e.g., [12]).

Theorem 5.4 The line graph L.G/ of a graph G contains a cycle of length k≥ 3if and only if G contains a circuit C such that".C/ ≤ k≤ ".C/:A key lemma for our proof of Theorems 5.1 and 5.3 is the following.

Lemma 5.1 Let G be a graph of order n; girth at least 4, and with minimum edgedegree²2 ≥ 8 such that L.G/ contains Cm+1 but not Cm. Then

m≤ 4n− ²2

²2:

5.1 Proof of the main theorems 67

Proof. Let G satisfy the hypothesis of the lemma. By Theorem 5.4,G containsa circuit C with ".C/ ≤ m+ 1≤ ".C/: In fact, ".C/ = m+ 1; otherwiseL.G/containsCm. By Theorem 5.4,".C/ ≥ 1.G/+ 2≥ .²2.G/+ 4/=2≥ 6: SinceCis a circuit, there exist edge- disjoint cyclesD1; D2; · · · ; Dr such that

C=r⋃

i=1

Di:

We distinguish the following cases.

Case 1.r = 1:

ThenC is a cycle of lengthm+ 1:

Case 1.1.C has a chord.

Let C′ be a longest cycle among all cycles that contain exactly one chord ofCwhile the remaining edges belong toC:

In∑

e∈E.C′/d.e/; every edge inE.C′/ is counted at most four times. Hence,

".C′/ ≥ ".C′/²2=4> ".C/²2=8= .m+ 1/²2=8≥ m+ 1:

On the other hand,".C′/ ≤ m: ThusL.G/ containsCm, a contradiction.

Case 1.2.C has no chord.

Since²2≥ 8;C cannot be a Hamilton cycle ofG: Let u be a vertex inV.G/\V.C/:If u is adjacent to at least three vertices ofC, thenG contains a cycleC′ with".C/=2< ".C′/ ≤ m and we obtain a contradiction as in Case 1.1. Hence,u isadjacent to at most two vertices ofC: Define p as the number of edges incidentwith exactly one vertex ofC: We then have

p≤ 2|V.G/\V.C/| = 2.n−m− 1/:

On the other hand, sinceG has no chords,

p=∑

e∈E.C/

.d.e/− 4//=2≥ .m+ 1/.²2− 4/=2:

68

It follows that.m+ 1/.²2− 4/=2≤ 2.n−m− 1/ or, equivalently,

m≤ 4n− ²2

²2:

This settles Case 1.

Case 2.r ≥ 2:

Let H be the graph withV.H/= {D1; D2; · · · ; Dr} andDi D j ∈ E.H/ iff V.Di/∩V.Dj/ 6= ∅.i 6= j/: SinceH is connected, at least two vertices ofH are not cut ver-tices ofH. Equivalently, there are at least two values ofj for which

⋃1≤i≤r;i 6= j Di

is a connected subgraph ofG, and hence a circuit ofG.

Assume, without loss of generality, thatC′ =⋃ri=2 Di andC′′ = D1∪

⋃ri=3 Di are

circuits ofG: We have

".C′/ ≥ |E.C′/| + |E.D1/∩ E.C′/| + |E.D2− V.C′′//|.²2− 4/=4

= |E.C/| − |E.D1− V.C′//| + |E.D2− V.C′′//|.²2− 4/=4:

On the other hand, sinceL.G/ does not containCm;

".C/ ≤ m− 1= ".C/− 2:

It follows that |E.D1 − V.C′//| ≥ |E.D2 − V.C′′//|.Ž′ − 4/=4+ 2 and hence,since²2 ≥ 8;

|E.D1− V.C′//| > |E.D2− V.C′′//|:But then by symmetry we also have

|E.D2− V.C′′//| > |E.D1− V.C′//|:This settles Case 2.�

In a way similar to the proof of Lemma 5.1, one easily proves the following.

Lemma 5.2 Let G be a graph of order n, girth at least 5, and with minimum edgedegreeŽ ≥ 8 such that L.G/ contains Cm+1 but not Cm: Then

m≤ 2n− ²2+ 2²2− 2

:


The proof of our main results also need the following lemmas.

Lemma 5.3 Let G be a graph of order n with²2 >√

2n+ 1.n ≥ 72/: Theng.G/ ≤ 1.G/+ 1:

Proof. Sincen≥ 72, we have²2 ≥ 13: Let G satisfy the hypothesis of the lemmaand letC be a shortest cycle ofG.

We distinguish the following two cases.

Case 1.Ž.G/ = 1:Obviously,1.G/ ≥ ²2− 1. SetE1.C/ = {e= uv ∈ E.C/ : eitheru or v is ad-jacent to an isolated vertex inG− V.C/}; E2.C/ = E.C/\E1.C/ andV2.C/ =V.G[E2.C/]/:

We have the following claim.

d.e/ ≥ .²2− 1/+ 2= ²2+ 1 f or any edge e∈ E1.C/: (5.1)

Assuming".C/ ≥ 1.C/+ 2, we thus have".C/ ≥ ²2+ 1≥ 14, implying that

|{uv ∈ E.G/ : v ∈ V.C/}| ≤ 1 for anyu ∈ V.G/\V.C/: (5.2)

By the choice ofC we obtain that, for any vertexx∈ V2.C/; there exists a set{pi}of d.x/− 2 paths of length 2 such thatV.pi /∩ V.C/= {x} andV.pi /∩ V.pj /={x}.i 6= j/:

Defining P1 as the set of such paths, we thus obtain that any pair of paths inP1

have at most one common vertex inG and

|P1| ≥∑

e∈E2.C/

.d.e/− 4//=2≥ |E2.C/|=2: (5.3)

Using (5.1)-(5.3) and²2 >√

2n+ 1; we obtain

n ≥ .∑

e∈E.C/

.d.e/− 4//=2+ ".C/+ |P1|

≥ .∑

e∈E1.C/

d.e/+∑

e∈E2.C/

d.e//=2+ |P1| − ".C/

≥ ..²2+ 1/|E1.C/| + ²2|E2.C/|/=2+ |E2.C/|=2− ".C/= .Ž+ 1/.|E1.C/| + |E2.C/|/=2− ".C/= ".C/.²2− 1/=2≥ .²2+ 1/.²2− 1/=2> n;

70

a contradiction.

Case 2.Ž.G/ > 1:

Assuming".C/ ≥ 1.G/ + 2; we have".C/ ≥ .²2 + 4/=2 ≥ 172 ; implying that

".C/≥ 9: By the choice ofC andŽ.G/ > 1 we have that, for any vertexx∈ V.C/;there exists a set{pi} of d.x/− 2 paths of length 2 such thatV.pi /∩ V.C/= {x}andV.pi /∩ V.pj / = {x}.i 6= j/:

Defing P2 as the set of such paths, we obtain that any pair of paths inP2 have atmost one common vertex inG and|P2| ≥

∑e∈V.C/.d.u/− 2/; implying that

n ≥ 2∑

u∈V.C/

.d.u/− 2/+ ".C/

= 2.∑

e∈E.C/

.d.e/− 4//=2+ ".C/

≥ ..²2− 4/+ 1/".C/

≥ .²2− 3/.²2+ 4/=2> n;

a contradiction.�

Lemma 5.4 Let G be a graph of order n; girth at least 5, and with g.G/.²2 −2/2+ 2.²2− 2/ ≥ 4n: Then

½.L.G// = [g.G/; cr.L.G//]:

Proof. Let G satisfy the hypothesis of the lemma and letC be a shortest cycle ofG, i.e.,".C/ = g.G/: We have

".C/ = .∑

e∈E.C/

.d.e/− 4///=2+ ".C/

= .∑

e∈E.C/

d.e//=2− ".C/

≥ .²2− 2/".C/=2

= g.G/.Ž− 2/=2:

(5.4)

By g.G/.²2− 2/2+ 2.²2− 2/ ≥ 4n;

g.G/.²2− 2/=2≥ 2n=.²2− 2/− 1: (5.5)


Using (5.4) and (5.5), Lemma 5.2′ and Theorem 5.4, we obtain

½.L.G// = [g.G/; cr.L.G//];

which completes the proof of Lemma 5.4.�

Now we turn to the proofs of our main results.


Let G satisfy the hypothesis of Theorem 5.1 andC be a shortest cycle ofG, i.e.,".C/ = g.G/: From Lemmas 5.3, 5.4 and Theorem 5.4, we have thatL.G/ issubpancyclic.

Next, we construct a family of graphs of ordern with ²2 =√

2n+ 1 such thattheir line graps are hamiltonian but not pancyclic.

For any integerd = 2k+ 1 ≥ 3; define the graphGd as follows. Now we letC= u1u2 · · ·ud+1u1 be a cycle of lengthd+ 1 and letH1; H2; · · · ; Hk+1 bek+ 1copies of the empty graph of orderd− 3 such thatC; H1; H2; · · · ; Hk+1 are pair-wise disjoint.

Now, Gd is obtained fromC∪ .∪k+1i=1 Hi/ by joining each vertex ofHi to u2i; for

i = 1;2; · · · ; k+ 1: We have

²2.Gd/ = d; (5.6)

and

|V.Gd/| = .d+ 1/+ .k+ 1/.d− 3/ = .d2− 1/=2: (5.7)

Clearly,C is a cycle ofG with ".C/ = |E.G/|; hence

L.Gd/ is hamiltonian: (5.8)

Obviously,L.Gd/ does not containCd and hence

L.Gd/ is not pancyclic: (5.9)

Using (5.6) to (5.9), we obtain that Theorem 5.1 is best possible in the sense thatthe condition²′ >

√2n+ 1 cannot be relaxed, even under the condition thatL.G/

72

is hamiltonian.�

Proof of Theorem 5.2

Let G satisfy the hypothesis of the theorem. Assume thatL.G/ is not subpan-cyclic. Setm= max{i ≤ cr.L.G// : L.G/ does not containCi}: Then m ≤cr.L.G// − 1 and L.G/ containsCm+1: By Theorem 5.4,G contains a circuitC with ".C/ ≤ m+ 1< ".C/: In fact,".C/ = m+ 1; otherwiseL.G/ contains aCm: Using arguments, similar to those in the proof of Lemma 5.2, we obtain

Claim 1. C is a cycle without any chord andC cannot be a hamiltonian cycle ofG.

Letw be a vertex inV.G/\V.C/: It is easy to see that

Claim 2. w is adjacent to at most two vertices ofC and the distance (inC) be-tween any pair ofN.w/∩ V.C/ is 2.

OtherwiseG contains a cycleC′ with ".C/=2< ".C′/ ≤m; and we obtain a con-tradiction as in Case 1.1 of Lemma 5.1.

By Theorem 5.1, we only need to consider the case thatg.G/ = 4: Let C4 be a4-cycle ofG. By arguments similar to that in that inthe proof of Lemma 5.4, wehave".C4/ ≥ 2.²2− 2/: By Theorem 5.4,½.L.G// ⊆ [3;2²2− 4]: Clearly,

".C/ ≥ .2²2− 4/+ 2= 2²2− 2: (5.10)

Using (5.10), Claim 1, Claim 2 and²22− ²2 > 2n; we obtain

n ≥ .∑

e∈E.C/

.d.e/− 4//=4+ ".C/

= .∑

e∈E.C/

.d.e//=4≥ ".C/²2=4

≥ 2.²2− 1/²2=4> n;

a contradiction, which implies thatL.G/ is subpancyclic.

Now we construct a family of graphs of ordern with ²22.G/− ²2.G/ = 2n such

that their line graphs are hamiltonian but not pancyclic.


For any integerd= 2k+ 1≥ 3; define the graphGd as follows.Let C= u1u2 · · ·u4ku1 be a cycle of length 4k and letH1; H2; · · · ; Hk bek copiesof 2k−3 isolated vertices such thatC; H1; H2; · · · ; Hk are pairwise disjoint. Now,

Gd is obtained fromC∪ .k⋃

i=1Hi/ by joining each vertex ofHi to u4i−3 andu4i−1,

for i = 1;2; · · · ; k: We have

²2.G/ = d (5.11)

and

|V.Gd/| = 4k+ k.d− 4/ = .d2− d/=2: (5.12)

Clearly,C is a cycle ofG with ".C/ = |E.G/|; hence

L.Gd/ is hamiltonian. (5.13)

Obviously,L.Gd/ does not containC4k−1 = C2d−3 and hence

L.Gd/ is not pancyclic: (5.14)

Using (5.11)-(5.14), we obtain that Theorem 5.2 is best possible in the sense thatthe condition²2

2− ²2 > 2n cannot be relaxed, even under the condition thatL.G/is hmiltonian.�

From the proofs of Theorems 5.1 and 5.2, two important corollaries can easily beobtained.

Corollary 5.1 Let G be a graph of order n≥ 72; girth at least 5, and with²2 >

√2n+ 1: Then the line grpaph L.G/ of G is pancyclic ,whenever L.G/

is hmiltonian, and the result is best possible.

Corollary 5.2 Let G be a graph of order n≥ 72; girth at least 4, and with²2

2 − ²2 > 2n: Then the line graph L.G/ of G is pancyclic whenever L.G/ ishmiltonian, and the result is best possible.

Chapter 6

On subpancyclic line graphs

Van Blankenet al.[5] prove that ifG is a graph onn vertices such that its line graphL.G/ is hamiltonian and such that the minimum vertex degreeŽ.G/ ≥ An1=3 for acertain constantA, thenL.G/ is pancyclic. Moreover, they showed that the resultis essentially best possible, up to the multiplicative constantA:

The author gave two best possible conditions for a graph with girth at least 4 or 5so that its line graphL.G/ is subpancyclic as follows:

Theorem 6.1 [57] Let G be a graph of order n.n≥ 72/, with girth at least 5, andwith ²2.G/ >

√2n+ 1: Then L.G/ is subpancyclic .

Theorem 6.2 [57] Let G be a graph of order n.n≥ 72/, with girth at least 4, andwith ²2

2.G/− ²2.G/ > 2n: Then L.G/ is subpancyclic.

In this chapter we give a generalization, which is also best possible, of Theorem6.2.

Theorem 6.3 Let G be a graph of order n.n≥ 600/ with ²22.G/− ²2.G/ > 2n:

Then L.G/ is subpancyclic.

From Theorem 6.3, the following corollary can be obtained.

Corollary 6.1 Let G be a graph of order n.n≥ 600/ with ²22.G/− ²2.G/ > 2n:

Then L.G/ is pancyclic whenever L.G/ is hamiltonian, and the result is bestpossible.

75

76

In order to show that the condition in Corollary 6.1 can not be relaxed, we con-struct a graphGd as follows. For any integerd= 2k+1≥ 3; let C= u1u2 · · ·u4ku1

be a cycle of length 4k and letH1; H2; · · · ; Hk be pairwise disjoint copies of theempty graph of order 2k− 3; also disjoint fromC: Now Gd is obtained fromC∪ .∪k

i=1Hi/ by joining each vertex ofHi to u4i−3 andu4i−1, for i = 1;2; · · · ; k:Obviously²2.Gd/ = 2k+ 1= d; |V.Gd/| = 2k2+ k = .d2− d/=2 and".C/ =|E.Gd/|; henceL.Gd/ is hamiltonian but it is not pancyclic since 4k− 1= 2d−3 6∈ ½.L.Gd//:

Corollary 6.1 improves the following results of [3].

Theorem 6.4 If G is a hamiltonian graph of order n.n> 12/ with ²2.G/ > n=2;then L.G/ is pancyclic.

Corollary 6.1 also proves that the hamiltonian line graphs of these graphs involv-ing edge degree conditions in [28] and [54] are pancyclic. Here we give oneexample as follows.

Theorem 6.5 Let G be a connected simple graph of order n such that every cutedge of G is incident with a vertex of degree one. If n is sufficiently large and²2.G/ > .2n− 10/=5; then L.G/ is pancylic.

6.1 Preliminary Results

Before proving Theorem 6.3 we introduce some additional terminology and nota-tion, and state a number of preliminary results.

By acircuit of a graphG we will mean an eulerian subgraph ofG, i.e., a connectedsubgraph in which every vertex has even degree. Note that by this definition (thetrivial subgraph induced by ) a single vertex is also a circuit.IfC is a circuit ofG, thenE.C/ denotes the set of edges ofG incident with at least one vertex ofC: We write".C/ for |E.C/| and".C/ for |E.C/|: For any subgraphH of G, letN.H/ = ∪u∈V.H/N.u/:

Harary and Nash-Williams [32] characterized hamiltonian line graphs. One caneasily prove a more general result(see, e.g., [12]).

Theorem 6.6 The line graph L.G/ of a graph G contains a cycle of length k≥ 3if and only if G contains a circuit C such that".C/ ≤ k≤ ".C/:

6.1 Preliminary Results 77

In a way similar to that in the proof of Lemma 6.4 in [5], we can prove the follow-ing:

Lemma 6.1 Let G be a graph of order n and minimum edge degree²2 ≥ 8 suchthat L.G/ contains Cm+1 but not Cm: Then

m≤ .6n− ²2− 2/=.²2+ 2/:

The following lemma is necessary for our proof.

Lemma 6.2 Suppose G contains a eulerian subgraph H on k vertices such thatG[V.H/] contains a spanning subgraph F with at least one edge and degree atmost 2. Then we have

".H/ ≥ l²2=2− ".H/ ≥ .l²2− k.k− 1//=2:

Proof. A straightforward calculation shows that

".H/ =∑

x∈V.H/

.dG.x//− dH.x//+ ".H/

=∑

x∈V.H/

dG.x/−∑

x∈V.H/

dH.x/+ ".H/

≥∑

xy∈E.F/

.dG.x/+ dG.y//=2− ".H/ ≥ l²2=2− ".H/:

The result follows since".H/ = |E.H/| ≤ k.k− 1/=2: �

Lemma 6.3 Let G be a graph of order n.n≥ 600/ with g.G/ = 3 and²22.G/−

²2.G/ > 2n: If G contains a cycle H of order k, for some k∈ {7;8;9;10}; suchthat G[V.H/] contains a spanning subgraph F with k edges and degree at most2, then L.G/ is subpancyclic.

Proof. Let G andH satisfy the hypothesis of the lemma. This means that²2≥ 36:By g.G/ = 3; we can take a cycleC of G with ".C/ = 3: Using Lemma 6.2, weobtain".C/ ≥ .3²2− 6/=2: Hence, by Theorem 6.6 andn≥ 600;

½.L.G// ⊇ [3; .3²2− 6/=2] ⊇ [3;51]: (6.1)

Using Lemma 6.2, we have

".H/ ≥ .k²2− k.k− 1//=2: (6.2)

78

By (6.1), (6.2), the fact that".H/ ≤ k.k− 1/=2≤ 45 and Theorem 6.6,

½.L.G// ⊇ [3; .k²2− k.k− 1//=2]: (6.3)

By Lemma 6.1 and Theorem 6.6,

½.L.G// ⊇ [6n=.²2+ 2/; cr.L.G//]: (6.4)

By ²22.G/− ²2.G/ > 2n andn≥ 600,

6n=.²2+ 2/ ≤ k.²2− .k− 1//=2 f or k ∈ {7;8;9;10}: (6.5)

Using (6.2), (6.4), and (6.5), we obtain½.L.G// = [3; cr.L.G//], i.e., L.G/ issubpancyclic.�

6.2 Proof of Theorem 6.3

Let G satisfy the hypotheses of Theorem 6.3. By Theorems 6.1 and 6.2, we onlyneed to consider the case thatg.G/ = 3:

Assume thatL.G/ is not subpancyclic. Setm= max{i ≤ cr.L.G// : L.G/ doesnot containCi}: Thenm≤ cr.L.G//− 1 andL.G/ containsCm+1: By Theorem6.6, G contains a circuitC with ".C/ ≤ m+ 1 ≤ ".C/: In fact ".C/ = m+ 1,otherwiseL.G/ containsCm:

Take a cycleC′ of G with ".C′/= g.G/= 3: Using Theorem 6.6 and Lemma 6.2,we obtain

".C/ ≥ ".C′/+ 2≥ 3²2=2− 3+ 2= .3²2− 2/=2: (6.6)

Using ideas from the proof of Lemma 6.1, we obtain the following claims:

Claim 1. C is a cycle without any chord andC can not be a hamiltonian cycle ofG:

Claim 2. There does not exist a cycleC′ with ".C/=2< ".C′/ ≤ m in G. Hence,

|N.w/ ∩ V.C/| ≤ 3 f or anyw ∈ V.G/\V.C/: (6.7)

and

6.2 Proof of Theorem 6.3 79

if w ∈ V.G/\V.C/ and{x; y} ⊆ N.w/ ∩ V.C/;

thenx andy have distance at most 2 on the cycleC:(6.8)

SetVi = {w ∈ V.G/\V.C/ : |N.w/ ∩ V.C/| = i}:

By (6.7),

N.C/\V.C/ =3⋃

i=1

Vi : (6.9)

Let C = u1u2 · · ·um+1u1; andu−i andu+i be the vertices preceding and followingui onC. Set

S3;i = {w ∈ V3 : N.w/ ∩ V.C/ = {u−i ;ui;u+i }} f or i ∈ {1;2; · · · ;m+ 1}:

By Lemma 6.3, we obtain

Claim 3. G does not contain a circuitH of orderk, for anyk ∈ {7;8;9;10}; suchthatG[V.H/] contains a spanning subgraphF with k edges and degree at most 2.

Hence

Claim 4. If |S3;i| ≥ 2; then|N.x/ ∩ {u−i ;ui;u+i }| ∈ {0;3} for anyx ∈ V3:

SetEs;t.C/ = {e= uv ∈ E.C/ : |{w ∈ Vs : {u; v} ⊆ N.w/}| ≥ t}

and

E0s;t.C/ = {e= uv ∈ E.C/ : |{w ∈ Vs : {u; v} ∩ N.w/ 6= ∅}| ≥ t}:

We obtain

Claim 5. If E3;2.C/ = ∅, thenG does not contain a circuitH of order 5 such thatG[V.H/] containsC5 or C3+ e0; wheree0 ∈ E.G/:

80

Proof of Claim 5. If G contains a circuitH of order 5 such thatG[V.H/] containsa spanning subgraphC3+ e0 wheree0 ∈ E.G/; then in the same way as in theproof of Lemma 6.2, we obtain

".H/ =∑

x∈V.H/

dG.x/− ".H/

=∑

x∈V.C3/

dG.x/+ dG.e0/− ".H/

≥ 3²2=2+ ²2− .5× 4/=2= .5²2− 20/=2:

So by Theorem 6.6,

".C/ ≥ ".H/+ 2≥ .5²2− 16/=2: (6.10)

SinceE3;2.G/ = ∅; using Claim 1, (6.10) and²22− ²2 > 2n.n≥ 600/ we obtain

n ≥∑

e∈E.C/

.d.e//− 6/=4+ ".C/

≥ .²2− 6/".C/=4+ ".C/ = .²2− 2/".C/=4

≥ .²2− 2/.5²2− 16/=8> n;

a contradiction. Similarly we obtain a contradiction ifG contains a cycle of length5 as an induced subgraph. This completes the proof of Claim 5.�

SetE′s;t.C/ = {e= uv ∈ Es;t.C/ : {w ∈ Vs : d.w/ = s and{u; v} ⊆ N.w/} 6= ∅} andE′′s;t.C/ = Es;t.C/\E′s;t.C/:

Obviously,

d.e/ ≥ 2.²2− s/ f or s∈ {2;3} and e∈ E′s;1.C/: (6.11)

Let

Fk.G/ = {e= uv ∈ E.G/ : u ∈ Vk ∩ N.Ek;3/ andv ∈ V.G/\N.C/}:

We distinguish the following five cases:


Case 1.E3;3 6= ∅:

HenceG contains a eulerian subgraphH1 such thatC6 is a spanning subgraph ofH1. So, by Theorem 6.6 and Lemma 6.2,

".C/ ≥ ".H1/+ 2≥ 3²2− 13: (6.12)

By Claims 2 and 3, we have

|N.e/\N.C/| ≥ ²2− 6 f or e∈ F3.G/: (6.13)

and

.N.e1/\N.C// ∩ .N.e2/\N.C// = ∅ f or {e1;e2} ⊆ F3.G/

such thate1 and e2 are not adjacent inG:(6.14)

Hence, by (6.13), (6.14) and Claim 4,

|V.G/\N.C/| ≥ 3.²2− 6/|E′′3;3.C/|=2: (6.15)

By Claim 4, every vertex inV3 is counted at most four times in∑

e∈E3;3.C/d.e/ and

at most two times in∑

e∈E.C/\E3;3.C/d.e/: Every vertex inN.C/\.V3 ∪ V.C// is

counted at most four times in∑

e∈E.C/d.e/: Hence, using Claim 1, Claim 3, (6.9),

(6.11), (6.12), (6.15) and²22− ²2 > 2n.n≥ 600/; we obtain

n = |N.C/\V.C/| + |V.C/| + |V.G/\V.C/|≥ .

∑e∈E′3;3.C/

.d.e/− 4/=4+∑

e∈E′′3;3.C/

.d.e/− 4/=4

+∑

e∈E.C/\E3;3.C/

.d.e/− 8/=2/=2+ ".C/+ 3.²2− 6/|E′′3;3.C/|=2

≥ ..2.²2− 3/− 4/|E′3;3.C/|=4+ .²2− 4/E′′3;3.C/|=4+ .²2− 8/.".C/− |E3;3.C/|/=2/=2+ ".C/+ 3.²2− 6/|E′′3;3.C/|=2= .²2− 8+ 4/".C/=4− .²2− 8/|E3;3.C/|=4+ .²2− 5/|E′3;3.C/|=4+ .11²2− 76/|E′′3;3.C/|=8≥ .²2− 4/".C/=4− .²2− 8/|E3;3.C/|=4+ .²2− 5/.|E′3;3.C/| + |E′3;3′.C/|/=4≥ .²2− 4/".C/=4≥ .²2− 4/.3²2− 13/=4> n;

82

a contradiction.

Case 2.E3;3.C/ = ∅ andE3;2.C/ 6= ∅:

HenceG contains a eulerian subgraphH2 such thatC5 is a spanning subgraph ofH2: So, by Theorem 6.6 and Lemma 6.2,

".C/ ≥ ".H2/+ 2≥ .5²2− 16/=2: (6.16)

Using Claim 1, (6.9), (6.16) and²22− ²2 > 2n.n≥ 600/; we obtain

n≥∑

e∈E.C/

.d.e/− 8/=4+ ".C/ ≥ .²2− 4/".C/=4≥ .5²22− 36²2+ 64/=8> n;

a contradiction.

Case 3.E3;2.C/ = ∅ andE3;1.C/ 6= ∅:

HenceG contains a eulerian subgraphH2 such thatC4 is a spanning subgraph ofH3: So, by Theorem 6.6 and Lemma 6.2,

".C/ ≥ ".H3/+ 2> 2²2− 4: (6.17)

By Claim 3 and Claim 5, we obtain

Claim 6. If |S3;i| = 1; thenN.V2/∩ {u−i ;ui;u+i } = ∅:

SinceE3;1.C/ 6= ∅;|E0

3;1.C/| ≥ 4: (6.18)

By Claim 6, every vertex inV2 is counted at most once in∑

e∈E03;1.C/

d.e/ and at most

two times in∑

e∈E.C/\E03;1.C/

d.e/: Hence using Claim 1, Claim 6, (6.9), (6.17), (6.18)

and²22− ²2 > 2n.n≥ 600/; we obtain

n ≥ .∑

e∈E03;1.C/

.d.e/− 6/+∑

e∈E.C/\E03;1.C/

.d.e/− 4/=2/=2+ ".C/

≥ ..²2− 6/|E03;1.C/| + .²2− 4/.".C/− |E3;1.C/|/=2/=2+ ".C/

= ²2".C/=4+ .²2− 8/|E03;1.C/|=4

≥ ²2.2²2− 4/=4+ .²2− 8/ = .²22− 16/=2> n;


a contradiction.

Case 4.E3;1.C/ = ∅ andG contains a copyH4 of C4 as an induced subgraph.

By Theorem 7 and an argument similar to the one in the proof of Lemma 6.2,

".C/ ≥ ".H4/+ 2≥ 2²2− 4+ 2= 2²2− 2: (6.19)

Using Claim 1, (6.9), (6.19) and²22− ²2 > 2n; we obtain

n ≥ .∑

e∈E.C/

.d.e/− 4/=4+ ".C/ ≥ .²2− 4/".C/=4+ ".C/

≥ .²2− 4/.2²2− 2/=4+ 2²2− 2= .²22− ²2/=2> n;

a contradiction.

Case 5.E3;1.C/ = ∅ andG does not contain a copyH4 of C4 as an induced sub-graph.

Hence, by (6.8), we obtain that ifw ∈ V.G/\V.C/ and{x; y} ⊆ N.w/ ∩ V.C/;thenx andy have distance one on the cycle. By Claim 5,

|.N.x/ ∪ N.y// ∩ V.C/| 6= 1 f or {x; y} ⊆ V2: (6.20)

If E2;3.C/ 6= ∅ then, using Claim 5, we obtain thatG contains a eulerian subgraphH5 such thatC4 is a spanning subgraph ofH5 andG[V.H5/] contains exactly 5edges. Hence, by Theorem 6.6 and Lemma 6.2,

".C/ ≥ ".H5/+ 2≥ 4²2=2− 5+ 2= 2²2− 3: (6.21)

From Claim 2, Claim 3 and Claim 5, we have

|N.e/\N.C/| ≥ ²2− 3 f or e∈ F2.G/ (6.22)

and

.N.e1/\N.C//∩ .N.e2/\N.C// = ∅for {e1;e2} ⊆ F2.G/ such thate1 ande2 are not adjacent inG:

(6.23)

84

Hence, by (6.20), (6.22) and (6.23),

|V.G/\N.C/| ≥ .²2− 3/|E′′2;3.C/|: (6.24)

We distinguish the following three subcases.

Subcase 5.1.|E2;3.C/| ≥ 2:

Using Claim 1, (6.9), (6.11), (6.20), (6.21), (6.24) and²22− ²2 > 2n.n ≥ 600/;

we obtain

n ≥ .∑

e∈E′2;3.C/

.d.e/− 4/+∑

e∈E′′2;3.C/

.d.e/− 4/+∑

e∈E.C/\E2;3.C/

.d.e/− 4//=4

+ ".C/+ .²2− 3/|E′′2;3.C/|≥ .2.²2− 2/− 4/|E′2;3.C/|=4+ .²2− 4/|E′′2;3.C/|=4+ .²2− 4/.".C/

− |E2;3.C/|/=4+ ".C/+ .²2− 3/|E′′2;3.C/|≥ ²2".C/=4+ .²2− 4/|E′2;3.C/|=2+ .²2− 4/|E′′2;3.C/|=2− .²2− 4/|E2;3.C/|=4≥ ²2".C/=4+ .²2− 4/|E2;3.C/|=4≥ ²2.2²2− 3/=4+ .²− 4/=2> n;

a contradiction.

Subcase 5.2.|E2;3.C/| = 1:

Hence|E0

2;3| = 3: (6.25)

Using Claim 1, (6.9), (6.20), (6.21), (6.25) and²22− ²2 > 2n.n≥ 600/; we obtain

n ≥ .∑

e∈E02;3.C/

.d.e/− 4/=2+∑

e∈E.C/\E02;3.C/

.d.e/− 8//=2+ ".C/

≥ .²2− 4/|E02;3.C/|=4+ .²2− 8/.".C/− |E0

2;3.C/|/=2+ ".C/= .²2− 6/".C/=2− .3²2− 36/=4

≥ .²2− 6/.2²2− 3/=2− .3²2− 36/=4> n;


a contradiction.

Subcase 5.3.E2;3.C/ = ∅:

Using Claim 1, (6.6), (6.9), (6.20) and²22− ²2 > 2n.n≥ 600/; we obtain

n ≥∑

e∈E.C/

.d.e/− 8/=2+ ".C/

≥ .²2− 8/".C/+ ".C/≥ .²2− 6/.3²2− 2/=4> n;

a contradiction.�

Chapter 7

Degree sums and subpancyclicity inline graphs

Harary and Nash-Williams characterized those graphs whose line graphs are hamil-tonian.

Theorem 7.1 (Harary and Nash-Williams [32])The line graph L.G/ of a graph G is hamiltonian if and only if G contains a circuitC such that".C/ = ".G/ ≥ 3:

A more general result is the following.

Theorem 7.2 [12]The line graph L.G/ of a graph G contains a cycle of length k≥ 3 if and only ifG contains a circuit C such that".C/ ≤ k≤ ".C/.

Van Blankenet al. [5] proved that f1.n/ has an order of magnitude:O.n1=3/. Itwas shown thatf2.n/ = [.

√8n+ 1+ 1/=2] (if n ≥ 600) [58] and thatf4.n/ ≤

n− 1 (if n ≥ 40) [62]. In this chapter, we obtain that ifn ≥ 76, then f3.n/ =[.n+ 6/=2] and f4.n/ = [.2n+ 16/=3]. Those results show thatfi.n/ has a inter-esting order of magnitude then1=.4−i/ for 0≤ i ≤ 2.

Moreover, we give the following more general results.

87

88

Theorem 7.3 Let G be a graph of order n.n ≥ 76/. If G satisfies one of thefollowing conditions:

(i) ²3.G/ > .n+ 6/=2;(ii) ²4.G/ > .2n+ 16/=3;

then L.G/ is subpancyclic and the results are all best possible.

The condition.i / and .ii / of Theorem 3 can not be improved in the followingsense.

Let s= .m− 2/=2.m≡ 0.mod2// and t = .n− 1/=3.n ≡ 1.mod3//. Definetwo graphsG1 of order m and G2 of order n as follows, respectively : ver-

tices setsV.G1/ = .s⋃

i=1{ui; vi}/

⋃{x; y} andV.G2/ = .t⋃

i=1{xi; yi; zi}/

⋃{w}; edge

setsE.G1/ =s⋃

i=1{xui;uivi; vi y} and E.G2/ =

t⋃i=1{wxi; xi yi ; yizi; ziw}. It is obvi-

ous thatG1 andG2 are two graphs such that²3.G1/ = s+ 4= .m+ 6/=2 and²4.G2/ = 2t + 6 = .2n+ 16/=3, respectively. But, by Theorem 7.2, 3s− 1 ∈[3; ".G1/] \½.L.G1// and 4t−1∈ [3; ".G2/] \½.L.G2//which imply thatL.G1/

andL.G2/ are not subpancyclic.

It follows from Theorem 7.1 thatL.G1/ andL.G2/ are all hamiltonian. This alsoimplies the following corollary.

Corollary 7.1 If n ≥ 76; then f3.n/ = [.n+ 6/=2] and f4.n/ = [.2n+ 16/=3].

Corollary 7.1 improves the results of [51], [62] and shows that the graphs in [40],[54] and [64] are pancyclic. We only give one example as follows.

Theorem 7.4 [40]. Let G be a simple graph with".G/ ≥ 3 and let G not be a

3-path. If4∑

i=1d.ui / ≥ 2n− 2 for any four vertices such that u1u2;u3u4 ∈ E.G/;

then L.G/ is hamiltonian and the result is best possible.

Combining Theorems 7.3 and 7.4 we obtain the following.


Corollary 7.2 If G is a simple graph of order n≥ 76such that4∑

i=1d.ui / ≥ 2n− 2

for any four vertices with u1u2;u3u4 ∈ E.G/, then L.G/ is pancyclic.

Corollary 7.2 supports the famous metaconjecture of Bondy (see e.g.[6]) that al-most every nontrivial condition which implies that a graph is hamiltonian alsoimplies that the graph is pancyclic. The results shown in the above as well as in[11], [19], [40] and [54] show that degree sums conditions along paths required ona graph which ensure that its line graph is subpancyclic are considerably weakerthan those required to ensure that its line graph is hamiltonian. This sheds somelight on the metaconjecture.

In general, results involving degree sums are directly derived from results involv-ing the minimum degree of the graph. The results in [58] shows an exception tothis rule. Our results show that results involving degree sums of the vertices alonga 2-path or a 3-path do not imply immediately the corresponding results involvingthe minimum edge degree.


We will give the proof by contradiction.

AssumingG is a graph of ordern which satisfies the conditions of Theorem 7.3but its line graphL.G/ is not subpancyclic, we let

k= max{i : i ∈ [3; cr.L.G//] \ ½.L.G//}:

Then it follows from Theorem 7.2 that

Claim 1. G does not contain a circuitC0 with ".C0/ ≤ k≤ ".C0/.

It is obvious thatL.G/ contains a cycleCk+1 of length k+ 1: It follows fromTheorem 2 thatG contains a circuitC with ".C/ ≤ k+ 1≤ ".C/. By Claim 1,".C/= k+1: SinceC is a circuit, there exist edge-disjoint cyclesD1; D2; · · · ; Dr

such thatC=r⋃

i=1Di andr is maximized.

90

Hence,

if r ≥ 2; then|V.Di/∩ V.Dj/| ≤ 2 for {i; j} ⊆ {1;2; · · · ; r}: (7.1)

Let U Pi.C/ = {P : P is a path of lengthi in C}.

Proof of (i) in Theorem 7.3.

Since²3.G/ > .n+ 6/=2≥ 41;

".C/ = k+ 1≥ 1.G/+ 2≥ ²3.G/=3+ 2> .n+ 18/=6≥ 14: (7.2)

We will consider the following two cases:

Case 1. r = 1; i.e.,C is a cycle of lengthk+ 1.

First, we show a needed claim.

Claim 2. G does not contain a cycleC′ with ".C/=2< ".C′/ ≤ k:

Proof of Claim 2. Otherwise, in∑

P∈U P2.C/d.P/; every edge inE.C′/ is counted at

most 6 times. Hence, by (7.2) and.i /,

".C′/ ≥∑

P∈U P2.C/

.d.P/− 6/=6+ ".C′/

≥ .²3− 6/".C′/=6+ ".C′/= ²3".C

′/=6

> ²3".C/=12≥ k+ 1:

On the other hand,".C′/ ≤ k: Theorem 7.2 implies thatL.G/ contains aCk, acontradiction, which provies Claim 2.

So,C has no chord. By²3 ≥ 42;C cannot be a hamiltonian cycle ofG. Let u bea vertex inV.G/ \ V.C/. By Claim 2,u is adjacent to at most three vertices ofC.Hence, by (7.2),

".C/ ≤ 3|V.G/ \ V.C/| + ".C/ = 3.n− ".C//+ ".C/ < .8n− 18/=3: (7.3)


On the other hand, sinceC has no chord,

".C/ ≥∑

P∈U P2.C/

.d.P/− 6/=3+ ".C/

≥ .²3− 6/".C/=3+ ".C/= .²3− 3/".C/=3> n.n+ 18/=36;

which contradicts (7.3), sincen≥ 76; and proves Case 1.

Case 2.r ≥ 2.

Let H be the graph withV.H/ = {D1; D2; · · · ; Dr} and Di D j ∈ E.H/ if andonly if V.Di/∩ V.Dj/ 6= ∅. SinceC is a circuit,H is connected. Without loss ofgenerality, we assume thatD1 andDr are two vertices ofH such that

dH.D1; Dr / = dia.H/: (7.4)

Obviously each of{D1; Dr} is not cut-vertex ofH, henceC1 =r⋃

i=2Di andCr =

r−1⋃i=1

Di are two circuits ofG.

Let

E1.Di / = E.Di /∩ E.Ci/ andE2.Di / = E.Di/ \ E1.Di /

V1.Di / = V.Di/∩ V.Ci/ andV2.Di/ = {u; v : uv ∈ E2.Di /};wherei ∈ {1; r}:

For any pathPof C, letd2.P/= d.P/−dC.P/. Since".Ci /≥ ".C/−|E2.Di/| =k+ 1− |E2.Di /|;

|V2.Di /| − 1≥ |E2.Di /| ≥ 2 for i ∈ {1; r}: (7.5)

OtherwiseCi is a circuit with".Ci/ ≤ k≤ ".Ci / which contradicts Claim 1.

Since".Ct/ ≥ ".C/− |E2.Dt/| + |E.Ds/ \ E.C/|,

|E.Ds/ \ E.C/| ≤ |E2.Dt/| − 2 (7.6)

92

where{s; t} = {1; r}: OtherwiseCt is a circuit with".Ct/ ≤ k ≤ ".Ct/; whichagain contradicts Claim 1.

We now present two more claims:

Claim 3. Taking any pathP= uvw of C with uv ∈ E2.Ds/; we obtain

dC.w/ > n=2− |E2.Dt/| + 1 (7.7)

and|E2.Dt/| = |V2.Dt/|=2 anddC.w/ > n=2− |V2.Dt/|=2+ 1 (7.8)

where{s; t} = {1; r}:

Proof of Claim 3. Let P= uvw be a path ofC with uv ∈ E2.Ds/. Then

|E.Ds/ \ E.C/| ≥ d.u/+ d.v/− 4+ d2.w/:

Hence, by (7.6),

d.u/+ d.v/+ d2.w/ ≤ |E2.Dt/| − 2+ 4= |E2.Dt/| + 2:

Sinced.u/+ d.v/+ d.w/ > .n+ 6/=2;

dC.w/ > .n+ 6/=2− .d.u/+ d.v/+ d2.w// ≥ .n+ 6/=2− .|E2.Dt/| + 2/;

i.e., (7.7) is true.

In order to obtain (7.8), we only need prove the following claim.

Each component ofC[E2.D1/∪ E2.Dr /] is a path of length one. (7.9)

Otherwise, there would exist ans ∈ {1; r} and a pathP0 = u0v0w0x of Ds suchthat{u0; v0; w0} ⊆ V2.Ds/ andx ∈ V1.Ds/. By (7.5) and (7.7),

dC.w0/ > n=2− |V2.Dt/| + 2 where{s; t} = {1; r}: (7.10)

By (7.10) anddC.w0/ = 2; |V2.Dt/| > n=2≥ 38: Hence by (7.1) and|V2.Dt/| >38; there exists a pathP′0 = u′0v

′0w′0 in Dt such thatu′0v

′0 ∈ E2.Dt/ andw′0 6∈

V1.Ds/:

Obviously,


|NC.w0/∩ NC.w′0/| ≤ 1: (7.11)

From (7.5) and (7.7), we obtain

dC.w′0/ > n=2− |V2.Ds/| + 2: (7.12)

Hence, by (7.10) and (7.12),

dC.w0/+ dC.w′0/ > n− |V2.D1/| − |V2.Dr /| + 4: (7.13)

On the other hand,

|.V2.D1/∪ V2.Dr // \ .NC.w0/∪ NC.w′0//| ≥ |V2.D1/| + |V2.Dr /| − 5: (7.14)

Using (7.11) and (7.14), we obtain

dC.w0/+ dC.w′0/ = |NC.w0/∪ NC.w

′0/| + |NC.w0/∩ NC.w

′0/|

≤ n− 2− .|V2.D1/| + |V2.Dr /| − 5/+ 1

≤ n− .|V2.D1/| + |V2.Dr /|/+ 4:

(7.15)

(7.13) and (7.15) are contradictory. This implies that (7.8) and (7.9) are true,which completes the proof of Claim 3.

Claim 4. There exist two verticesw ∈ V1.D1/∩ NC.V2.D1// andw′ ∈ V1.Dr /∩NC.V2.Dr // such thatdC.w;w

′/ ≥ 2:

Proof of Claim 4. Let w1uvw2 and x1u′v′x2 be two paths inD1 and Dr re-spectively, such thatuv ∈ E2.D1/ andu′v′ ∈ E2.Dr /: It follows from (7.9) that{w1; w2; x1; x2} ⊆ V1.D1/∪ V1.Dr /: It is easy to see that there exist two verticesw ∈ {w1; w2} andw′ ∈ {x1; x2} with ww′ 6∈ E.C/: Otherwise

C′ ={

C− E.x1u′v′x2x1/; if w1 ∈ {x1; x2};C− E.x1u′v′x2w1x1/; otherwise,

is a circuit with".C′/≤ k≤ ".C′/; a contradiction which implies thatdC.w;w′/≥

2 and proves Claim 4.

Due to Claim 4, we only need consider the following two subcases:

94

Case 2.1.There exist two verticesw ∈ V1.D1/∩ NC.V2.D1// andw′ ∈ V1.Dr /∩NC.V2.Dr // such thatdC.w;w

′/ ≥ 3:

ThenNC.w/∩ NC.w′/ = ∅: Obviously,

|.V2.D1/∩ V2.Dr // \ .NC.w/∪ NC.w′//| ≥ |V2.D1/| + |V2.Dr /| − 8:

Therefore,

dC.w/+ dC.w′/ = |NC.w/∪ NC.w

′/| + |NC.w/ ∩ NC.w′/|

≤ n− 2− .|V2.D1/| + |V2.Dr /| − 8/

= n+ 6− |V2.D1/| − |V2.Dr /|:(7.16)

By (7.8),

dC.w/+ dC.w′/ > n− .|V2.D1/| + |V2.Dr /|/=2+ 2: (7.17)


|V2.D1/| + |V2.Dr /| < 8;

which is a contradiction, since|V2.Di/| = 2|E2.Di/| ≥ 4 for i ∈ {1; r} by (7.5)and (7.8).

Case 2.2. dC.w;w′/ ≤ 2 for any pair of vertices{w;w′} with w ∈ V1.D1/ ∩

NC.V2.D1// andw′ ∈ V1.Dr /∩ NC.V2.Dr //:

We will prove thatV.D1/ ∩ V.Dr / = ∅. Otherwise, we can take a vertexw0 ∈V.D1/ ∩ V.Dr /; let w0w1w2 · · ·whuvw andw0u1u2 · · ·uf u′v′w′ be two paths inD1 andDr respectively, such that{uv;u′v′} ⊆ E2.D1/∪ E2.Dr / and

{w0; w1; w2; · · · ; wh} ∩ V2.D1/ = {w0;u1;u2; · · · ;uf } ∩ V2.Dr / = ∅:In a way similar to that of the proof of Claim 4, we obtainwhw

′ 6∈ E.C/ andufw 6∈ E.C/: By (7.9), {wh; w;uf ; w

′} ⊆ .V1.D1/ ∪ V1.Dr // ∩ .NC.V2.D1/ ∪NC.V2.Dr ///: Let PC.x; y/ denote a shortest path betweenx andy in C. Hence|V.PC.x; y//| ≤ 2 for x ∈ {wh; w} andy ∈ {uf ; w

′}: Since bothdia.H/ = 1 anddH.D1; Dr / = 1;

C′ = C− .E.PC.wh; w′//∪ E.PC.w;w

′//∪ E.whuvw//


is a circuit with".C′/ ≤ k≤ ".C′/; which contradicts Claim 1.

So V.D1/ ∩ V.Dr / = ∅: By Claim 4, we can take two distinct verticesw;w′

with dC.w;w′/ = 2 such thatw ∈ V1.D1/ ∩ NC.V2.D1// andw′ ∈ V1.Dr / ∩

NC.V2.Dr //: Obviously

|.V2.D1/∪ V2.Dr // \ .NC.w/∪ NC.w′//| ≥ |V2.D1/| + |V2.Dr /| − 4:

Therefore, ifþ = |NC.w/ ∩ NC.w′/| < 4; then


′/| + |NC.w/∩ NC.w′/|

≤ n− 2− .|V2.D1/| + |V2.Dr /| − 4/+ þ≤ n− |V2.D1/| − |V2.Dr /| + 6:

(7.18)


|V2.D1/| + |V2.Dr /| < 8;

which is a contradiction since|V2.Di/| = 2|E2.Di /| ≥ 4 for i ∈ {1; r} by (7.5) and(7.8).

If þ = |NC.w/ ∩ NC.w′/| ≥ 4; then {wx; xw′; wy; yw′} is an nontrivial cutset

of C for any pair of vertices{x; y} ⊆ NC.w/ ∩ NC.w′/. OtherwiseC′ = C−

E.xw′ywx/ is a circuit such that".C′/ ≤ k ≤ ".C′/, a contradiction. Hence wecan take an nontrivial component ofC− E.xw′ywx/, denoted byQ.x; y/; whichdoes not containw andw′. Hence,

|V.Q.x; y//| ≥ 4: (7.19)

OtherwiseQ′.x; y/ = C− Q.x; y/ is a circuit such that

".Q′.x; y// ≤ k≤ ".Q′.x; y//;

a contradiction.Obviously,

|V.Q.x; y// \ .NC.w/∪ NC.w′//| ≥ 2

96

and

V.Q.a;b//∩ V.Q.c;d// = ∅for any four vertices{a;b; c;d} ⊆ NC.w/∩ NC.w

′/:This implies, from (7.19), that

|V.G/ \ .NC.w/∪ NC.w′//| ≥ .|V2.D1/| + |V2.Dr /| − 4/+ 2+ [þ=2]× 2

≥ |V2.D1/| + |V2.Dr /| + þ− 3:

Therefore


′/| + |NC.w/ ∩ NC.w′/|

≤ n− .|V2.D1/| + |V2.Dr /| + þ− 3/+ þ≤ n− .|V2.D1/| + |V2.Dr /|/+ 3:

(7.20)


|V2.D1/| + |V2.Dr /| < 2;

which contradicts (7.5).

This completes the proof of (i) in Theorem 7.3.

The proof of (ii) in Theorem 7.3.

Since²4.G/ > .2n+ 16/=3≥ 56;

".C/ = k+ 1≥ 1.G/+ 2≥ ²4.G/=4+ 2> .n+ 20/=6≥ 16: (7.21)

We will consider the following two cases:

Case 1.r = 1;i.e.,C is a cycle of lengthk+ 1.

First, we show that Claim 2 is also true here.

Otherwise, in∑

P∈U P3.C/d.P/, every edge inE.C′/ is counted at most 8. Hence by

(7.21) and (ii),


".C′/ ≥∑

P∈U P3.C/

.d.P/− 8/=8+ ".C′/

≥ .²4− 8/".C′/=8+ ".C′/= ²4".C

′/=8≥ ²4".C/=16≥ k+ 1:

On the other hand,".C′/ ≤ k. Theorem 7.2 implies thatL.G/ contains aCk; acontradiction. This shows that Claim 2 is true.

SoC has no chord. By²4 ≥ 57;C cannot be a hamiltonian cycle ofG. Let u be avertex inV.G/ \ V.C/. By Claim 2,u is adjacent to at most three vertices ofC.Hence, by (7.21),

".C/ ≤ 3|V.G/ \ V.C/| + ".C/ = 3.n− ".C//+ ".C/ < .8n− 20/=3: (7.22)

On the other hand, sinceC has no chord, using (7.21) we obtain

".C/ ≥∑

P∈U P3.C/

.d.P/− 8/=4+ ".C/

≥ .²4− 8/".C/=4+ ".C/ = .²4− 4/".C/=4

> .n+ 20/.n+ 2/=36;

which contradicts (7.22) sincen≥ 76:

Case 2.r ≥ 2:

As in the proof of.i / in Theorem 7.3,H is defined to be the graph withV.H/ ={D1; D2; · · · ; Dr} and Di D j ∈ E.H/ if and only if V.Di/ ∩ V.Dj/ 6= ∅; andD1; Dr are two vertices ofH with (7.4). Hence (7.5) and (7.6) are also true here.

By (7.5),|V.Di/| ≥ 4 for i ∈ {1; r}. Hence we can prove the following claim.

Claim 5. Let P be a path of length 3 inDs. We obtain

dC.P/ > .2n+ 22/=3− |E2.Dt/| (7.23)

anddC.P/ > .2n+ 25/=3− |V2.Dt/|; (7.24)

98

where{s; t} = {1; r}.

Proof of Claim 5. Let P be a path of length 3 inDs. Then

|E.Ds/ \ E.C/| ≥ d.P/− dC.P/:

Hence, by (7.6) and (ii),dC.P/ > .2n+16/=3− .|E2.Dt/|−2/= .2n+22/=3−|E2.Dt/|; i.e., (7.23) is true. (7.24) is easily obtained from (7.5) and (7.23). Thissettles Claim 5.

We consider the following two subcases to obtain contradictions.

Subcase 2.1.There exist two pathsP= uvxyandP′ = u′v′x′y′ of length 3 inD1

andDr respectively, such that{uv;u′v′} ⊆ E2.D1/∪ E2.Dr / andV.P/∩ V.P′/=∅.

LetS= {x; y; x′; y′};

Ni.C/ = {w ∈ V.C/ : |NC.w/∩ S| = i} for i ∈ {1;2;3;4};

N2;1 = {w ∈ N2.C/ : |NC.w/ ∩ {x; y}| = |NC.w/∩ {x′; y′}| = 1};

N2;2 = {w ∈ N2.C/ : |NC.w/ ∩ {x; y}| = 2 and|NC.w/∩ {x′; y′}| = 0};

N2;3 = {w ∈ N2.C/ : |NC.w/ ∩ {x′; y′}| = 2 and|NC.w/ ∩ {x; y}| = 0};

M1 = NC.x/ ∩ NC.x′/∩ N2;

M2 = NC.x/ ∩ NC.y′/∩ N2;

M3 = NC.y/ ∩ NC.x′/∩ N2;

M4 = NC.y/ ∩ NC.y′/∩ N2;


nj = |N2; j| for j ∈ {1;2;3} andmi = |Mi| for i ∈ {1;2;3;4}:It is easy to see that

N2.C/ = N2;1∪ N2;2∪ N2;3 (7.25)

and

|N2| =3∑

i=1

ni andn1 =4∑

i=1

mi: (7.26)

We now prove the following three claims.

Claim 6. |N3∪ N4| ≤ 1:

Proof of Claim 6. Otherwise, let{w;w′} ⊆ N3∪ N4. Obviously

{w;w′} ⊆ .NC.x/∩ NC.y// ∪ .NC.x′/∩ NC.y

′//:

Without loss of generality, we assume thatwx; wy ∈ E.C/. Hence

C′ = C− {wx; wy; xy}is a circuit with".C/− 3= ".C′/ ≤ k≤ ".C′/; a contradiction, proving Claim 6.�

Claim 7. For i ∈ {1;2;3;4}, if mi ≥ 3, then there exist at leastmi −1 cut verticesof C in Mi.

Proof of Claim 7. For any pair of vertices{w;w′} ⊆ Mi;C[S∪ {w;w′}] has a4-cycle, denoted byC.w;w′/, which contains the vertices in{w;w′} but not theedgeww′. It is easy to see thatC′ = C− E.C.w;w′// has at least two nontrivialcomponents inC. Otherwise".C/− 4= ".C′/ ≤ k≤ ".C′/ = ".C/; which con-tradicts Claim 1.

Sincemi ≥ 3, for any pair of vertices{w;w′} ⊆ Mi; w;w′ is not in the same com-

ponent ofC− E.C.w;w′//; which does not contain any element ofS. Hencethere exist at leastmi − 1 cut-vertices ofC in Mi.�

Let Wi denote the cut-vertex set ofC in Mi. Taking any elementxi of Wi , we cantake a nontrivial componentC− {xi} , denoted byQi;x, which does not contain

100

any element ofS for i ∈ {1;2;3;4}. It is easy to see that

|V.Qi;x/| ≥ 3 and|{Qi;x : x ∈Wi}| = |Wi | for i ∈ {1;2;3;4}: (7.27)

Otherwise,C′i;x = C− Qi;x is a circuit with".C/− 3= ".C′i;x/ ≤ k ≤ ".C′i;x/; acontradiction, proving Claim 7.

Claim 8. For j ∈ {2;3}, if nj ≥ 2; then each vertex ofN2; j is a cut-vertex ofC.

Proof of Claim 8. Otherwise, there would exist aj ∈ {2;3}, say, j = 2; such thatN2;2 has a vertexw0 which is not a cut-vertex ofC: Hence

C′ = C− {w0x; w0y; xy}

has exactly one nontrivial componentC′′, and such thatC′′ is a circuit with".C′′/≤k≤ ".C′′/, a contradiction, proving Claim 8.

Let W′ denote the cut-vertices set ofC in N2;2 ∪ N2;3 such that for any elementy ∈W′;C− y has a nontrivial component which does not contain any element ofS. It is easy to see that

|W′| ≥ n2+ n3− 2: (7.28)

Hence, taking any elementy ∈W′, we can take a nontrivial component ofC− y,denoted byQ′y, which does not contain any element ofS. It is easy to see that

|V.Q′y/| ≥ 3 for y ∈W′ and|{Q′y : y ∈W′}| = |W′|: (7.29)

OtherwiseC′y = C− Q′y is a circuit such that".C′y/ ≤ k≤ ".C′y/; a contradiction.Obviously,

A∩ B= ∅ for any pair of

{A; B} ⊆ {V2.D1/;V2.Dr /} ∪ {Q′x : x ∈W′}⋃.

4⋃i=1

{Qi;x : x ∈Wi}/:(7.30)

Let I1 = {i : 1≤ i ≤ 4 and mi ≥ 3} and I2 = {i : 1≤ i ≤ 4 andmi ≤ 2}.


Using Claim 6, Claim 7, Claim 8 and (7.25)-(7.30), we obtain

dC.S/ = |4⋃

i=1Ni| + |N2| + 2|N3| + 3|N4|

≤ |4⋃

i=1Ni| +

3∑i=1

ni + 3

≤

n− .|V2.D1/| + |V2.Dr /| − 12+∑i∈ I1

3.mi − 1/+ 3.n2+ n3− 2//+∑3

i=1 ni + 3; if n2+ n3 ≥ 2

n− .|V2.D1/| + |V2.Dr /| − 12+∑i∈ I1

3.mi − 1//+3∑

i=1ni + 3;otherwise

≤

n− |V2.D1/| − |V2.Dr /| − 3

∑i∈ I1

mi − 2.n2+ n3/+∑4

i=1 mi

+3|I1| + 21; if n2+ n3 ≥ 2n− |V2.D1/| − |V2.Dr /| − 3

∑i∈ I1

mi +∑4

i=1 mi + .n2+ n3/+ 15;otherwise

=

n− |V2.D1/| − |V2.Dr /| − 2

∑i∈ I1

mi − 2.n2+ n3/

+∑i∈ I2mi + 3|I1| + 21; if n2+ n3 ≥ 2

n− |V2.D1/| − |V2.Dr /| − 2∑

i∈ I1mi +

∑i∈ I2

mi + .n2+ n3/+ 15;otherwise.

Hence we obtain

dC.S/ ≤ n− |V2.D1/| − |V2.Dr /| + 24: (7.31)

On the other hand, by (7.24),

dC.S/ > .4n+ 50/=3− |V2.D1/| − |V2.Dr /|: (7.32)

By n≥ 76; (7.31) and (7.32) are contradictory. This settles Subcase 2.1.

Subcase 2.2.If P= uvxy and P′ = u′v′x′y′ are two paths of length 3 inD1 andDr; respectively, such that{uv;u′v′} ⊂ E2.D1/∪ E2.Dr /; thenV.P/∩ V.P′/ 6=∅:

It then follows, from (7.5), that|V.D1/| = |V.Dr /| = 4 and|V.D1/∩V.Dr /| = 1:By (7.4),dH.Di; D j/= dia.H/= 1 for {i; j} = {1;2; · · · ; r}: This implies thatHis a complete graph. If there exists aj with |V.Dj/| ≥ 5; thenD1 andD j play thesame roles asD1 andDr. We derive a contradiction in a way similar to that in the

102

proof of Subcase 2.1. So|V.Dj /| ≤ 4 for j ∈ {2;3; · · · ; r − 1}: It is easy to seethat

|V.Di/| = 4 and|V.Di /∩ V.Dj/| = 1 for {i; j} ⊆ {1;2; · · · ; r}: (7.33)

Otherwise,C′i = C− Di is a circuit such that".C′i / ≤ ".C/− 3≤ k ≤ ".C′i / =".C/− 1 for i ∈ {1;2; · · · ; r}, which contradicts Claim 1.

Let P = uvxy be a path ofD1 with {uv; vx} = E2.D1/: By (7.5) and (7.33),|V2.Dr /| = 3: Hence by (7.24),

dC.y/ > .2n+ 16/=3− 6= .2n− 2/=3: (7.34)

On the other hand, by (7.33),dC.y/ ≤ n− 1− dC.y/=2; i.e.,dC.y/ ≤ .2n− 2/=3which contradicts (7.34).

This completes the proofs of (ii) of Theorem 7.3.�

Chapter 8

Subpancyclicity in the line graph ofa graph with large degree sums

Van Blankenet al.[5] prove that f1.n/ has the order of magnitudeO.n1=3). Thefollowing results are obtained.

Theorem 8.1 Let G be a connected graph of order n: If G satisfies one of thefollowing conditions:

(i) [58] ²2.G/ > .√

8n+ 1+ 1/=2 and n≥ 600;

(ii) [[61] ²4.G/ > .n+ 6/=2 and n≥ 76;(iii) [61] ²4.G/ > .2n+ 16/=3 and n≥ 76;

then L.G/ is subpancyclic and the results are all best possible.

Trommel,et al.showed the consequence of Theorem 8.1 (i) for large line graphs.

Corollary 8.1 (Trommel et al. [53]). Let G be a line graph on at least 100577vertices. If

Ž > .√

8n+ 1− 3/=2;

then G is subpancyclic.

Theorem 8.1 shows that the graphs in [11], [28] are pancyclic. Results related toTheorem 8.1 have appeared in [53].

103

104

Theorem 8.2 (Trommel et al. [53]). Let G be a claw-free graph on at least 5vertices. IfŽ >

√3n+ 1− 2, then G is subpancyclic.

In this chapter, we will characterize the graphG; with ²4.G/ = f4.G/; such thatL.G/ is not subpancyclic.

Theorem 8.3 Let G be a connected graph of order n.n≥ 146/. If ²4.G/ > .n+10/=2, then its line graph L.G/ is subpancyclic unless G is isomorphic to anexceptional graph F shown in the following and the result is best possible.

The exceptional graphF is defined as follows: Letn≡ 1.mod3/, and letC1; · · · ;C.n−1/=3 be .n− 1/=3 edge-disjoint cycles of length 4.F is obtained from thosecycles such thatC1;C2; · · · ;C.n−1/=3 have exactly one common vertex inF and letE.F/ = E.C1/∪ E.C2/∪ · · · ∪ E.C.n−1/=3/: Obviously|V.F/| = n and²4.F/ =f4.n/.

In general, conditions involving results on cycles can be slightly improved whenwe exclude an exceptional graph. But Theorem 8.3 shows that when we excludean exceptional graph the condition involving degree sums of the vertices along a3-path which ensure that its line graph is subpancyclic will be greatly improved(replace2n

3 with n2) and the condition is are almost the same as the condition in-

volving degree sums of the vertices along a 2-path (mentioned in Theorem 8.1).


Before we present the proof of our main result, we introduce some additional ter-minology and notation, and state a number of preliminary results.

By acircuit of a graphG we will mean a eulerian subgraph ofG, i.e., a connectedsubgraph in which every vertex has even degree. Note that by this definition (thetrivial subgraph induced by ) a single vertex is also a circuit. IfC is a circuit ofG, thenE.C/ denotes the set of edges ofG incident with at least one vertex ofC.The distance dH.G1;G2/ between two subgraphsG1 andG2 of H is defined tobe min{dH.v1; v2/ : v1 ∈ V.G1/ andv2 ∈ V.G2/}. Thediameterof a connectedsubgraphH, denoted bydia.H/, is defined to be max{dH.u; v/ : u; v ∈ V.H/}.We write ".C/ for |E.C/| and ".C/ for E.C/: For any subgraphH of G, letN.H/ = ∪u∈V.H/N.u/.


Harary and Nash-Williams [32] characterized the graph whose line graphs arehamiltonian. One can easily prove a more general result (see, e.g., [12]).

Theorem 8.4 (Broersma [12]) The line graph L.G/ of a graph G contains a cycleof length k≥ 3 if and only if G contains a circuit C such that".C/ ≤ k≤ ".C/:


We will complete the proof by contradiction.AssumingG is a graph of ordern which satisfies the conditions of Theorem 8.3but its line graphL.G/ is not subpancyclic, we can define

k= max{i : i ∈ [3; cr.L.G//]\½.L.G//}:

Hence, by Theorem 8.4, we obtain

Claim 1. G does not contain a circuitC0 with

".C0/ ≤ k≤ ".C0/:

Obviously,L.G/ contains a cycleCk+1 of lengthk+ 1. Hence, by Theorem 8.4,we obtain thatG contains a circuitC with ".C/ ≤ k+ 1 ≤ ".C/. By Claim 1,".C/ = k+ 1. SinceC is a circuit, there exist edge-disjoint cyclesD1; D2; :::; Dr

such thatr⋃

i=1Di andr is maximized.

Hence,

If r ≥ 2; then|V.Di/∩ V.Dj/| ≤ 2 for {i; j} ⊆ {1;2; :::; r}: (8.1)

LetU Pi.C/= {P : P is a path of lengthi−1 in C} . Since²4.G/ > .n+10/=2≥78,

".C/ = k+ 1≥ 1.G/+ 2≥ ²4.G/=4+ 2> .n+ 26/=8≥ 21: (8.2)

If r = 1, i.e.,C is a cycle of lengthk+ 1, then we obtain

Claim 2. G does not contain a cycleC′ with ".C/=2< ".C′/ ≤ k.

106

Proof. Otherwise, in∑

P∈U P4.C/d.P/, every edge inE.C′/ is counted at most 8

times. Hence, by (8.2) and²4.G/ ≥ .n+ 10/=2≥ 78;

".C′/ ≥∑

P∈U P4.C′/.d.P/− 8/=8+ ".C′/

≥ .²4− 8/".C′/=8+ ".C′/ = ²4".C′/=8

≥ ²4".C/=16

≥ k+ 1:

On the other hand,".C′/ ≤ k. Thus L.G/ contains aCk; a contradiction, thatproves Claim 2.

SoC has no chord. Since²4 ≥ 78;C cannot be a hamiltonian cycle ofG. Let ube a vertex inV.G/\V.C/. By Claim 2,u is adjacent to at most three vertices ofC. Hence, by (8.2),

".C/ ≤ 3|V.G/\V.C/| + ".C/= 3.n− ".C//+ ".C/< .11n− 26/=4:

(8.3)

On the other hand, sinceC has no chord,

".C/ ≥∑

P∈U P4.C/

.d.P/− 8/=4+ ".C/

≥ .²4− 8/".C/=4+ ".C/= .²4− 4/".C/=4

≥ .n2+ 28n+ 52/=64;

which contradicts (8.3) andn≥ 146. This implies thatr 6= 1:

So it suffices to consider the caser ≥ 2.

Let H be the graph withV.H/ = {D1; D2; :::; Dr} andDi D j ∈ E.H/ if and onlyif V.Di/ ∩ V.Dj/ 6= ∅. SinceC is a circuit, H is connected. Without loss ofgenerality, we assume thatD1 andDr are two vertices ofH such that

dH.D1; Dr / = dia.H/: (8.4)

Hence, any element of{Di; Dr} is not cut vertex ofH, so


C1 =r⋃

i=2Di andCr =

r−1⋃i=1

Di are two circuits ofG. Let

E1.Di/ = E.Di/∩ E.Ci / andE2.Di/ = E.Di/\E1.Di /

andV1.Di / = V.Di/∩ V.Ci/ and V2.Di / = {u; v : u ∈ E2.Di /}

wherei ∈ {1; r}:For any pathP of C, let d2.P/ = d.P/− dC.P/:Since".Ci / ≥ ".C/− |E2.Di /| = k+ 1− |E2.Di /|,

|V2.Di /| − 1≥ |E2.Di /| ≥ 2 (8.5)

wherei ∈ {1; r}. Otherwise".Ci / ≤ k≤ ".Ci / which contradicts Claim 1.

Since".Ct/ ≥ ".C/− |E2.Dt/| + |E.Ds/\E.C/|; we have

|E.Ds/\E.C/| ≤ |E2.Dt/| − 2; (8.6)

where{s; t} = {1; r}. Otherwise".Ct/ ≤ k≤ ".Ct/ which contradicts Claim 1.

We now prove the following claim.

Claim 3. Let P be a path of length 3 inDs. We obtain

dC.P/ > .n+ 14/=2− |E2.Dt/| (8.7)

and

|E2.Dt/| ≤ 2|V2.Dt/|=3 anddC.P/ > .n+ 14/=2− 2|V2.Dt/|=3; (8.8)

where{s; t} = {1; r}:

Proof. Let P be a path of length 3 inDs. Then

|E.Ds/\E.C/| ≥ d.P/− dC.P/:

Hence, by (8.6) and²3.G/ > .n+ 10/=2;

dC.P/ > .n+ 10/=2− .|E2.Dt/| − 2/;

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i.e., (8.7) is true.

In order to obtain (8.8), it suffices to prove the following claim.

Each component ofC[E2.D1/∪ E2.Dr /] is a path of length at most two. (8.9)

Otherwise, there must exist ans∈ {1; r} and a pathP0 = u0v0x0y0 of Ds such that{u0; v0; x0; y0} ⊆ V2.Ds/: By (8.5) and (8.7),

dC.P0/ > .n+ 16/=2− |V2.Dt/|; (8.10)

where{s; t} = {1; r}:

SincedC.P0/ = 8; |V2.Dt/| > n=2≥ 78. Hence there exists a pathP′0 = u′0v′0x′0y′0

in Dt such thatu′0v′0 ∈ E2.Dt/ and{x′0; y′0} ∩ V1.Ds/ = ∅:

For anyx ∈ NC.x′0/∩ NC.y′0/, C− x has at least a nontrivial component, denotedby Qx, which does not contain any vertex ofDt. Otherwise".C′/ ≤ k ≤ ".C′/;whereC′ = C− {xx′0; xy′0; x

′0y′0}; a contradiction.

It is easy to see that|V.Qx/| ≥ 3: (8.11)

Otherwise".C′/ ≤ k≤ ".C′/; whereC′ = C− Qx.

Let B denote the cut-vertex set ofNC.x′0/∩ NC.y′0/ such that for anyx∈ B;C− xhas a nontrivial component, denoted byQx, which does not contain any vertex ofV.D1/∪ V.Dr /.Set

þ = |NC.x′0/∩ NC.y

′0/|:

Obviously|{Qx : x ∈ B}| = |B| ≥ þ− 1: (8.12)


dC.x′0/+ dC.y′0/ = |NC.x′0/∪ NC.y′0/| + |NC.x′0/∩ NC.y′0/|≤{

n− .|V2.D1/| + |V2.Dr /| − 4/+ 1; if þ ≤ 1;n− .|V2.D1/| + |V2.Dr /| − 2+ 3.þ− 1//+ þ; if þ ≥ 2:

≤ n− .|V2.D1/| + |V2.Dr /|/+ 5≤ n− n=2− |V2.Dt/| + 5= .n+ 10/=2− |V2.Ds/|;


which contradicts (8.10) anddC.u′0/ = dC.v′0/ = 2. This implies that (8.8) and

(8.9) are true and settles Claim 3.

Next we will consider the following three cases to obtain contradictions.

Case 1.dia.H/ ≥ 2:

This implies thatV.D1/∩ V.Dr / = ∅:

We can take two pathsP = uvxy and P′ = u′v′x′y′ of length 3, inD1 and Dr

respectively, with{uv;u′v′} ⊆ E2.D1/ ∪ E2.Dr / and{x; x′} ⊆ V1.D1/∪ V1.Dr /;

such thatV.P/∩ V.P′/ = ∅:Let

S= {x; y; x′; y′};Ni = {u ∈ V.C/ : |NC.u/∩ S| = i};M1 = ..NC.x/∩ NC.y// ∪ .NC.x′/∩ NC.y′///∩ N2;

M2 = N2\M1;

ni = |Ni | and mi = |Mi|:We now prove three claims with respect to Case 1.

Claim 4. |N3∪ N4| ≤ 1:

Proof. Otherwise, letw;w′ ∈ N3∪ N4: Obviously,

w;w′ ∈ .NC.x/∩ NC.y// ∪ .NC.x′/∩ NC.y

′//:

Without loss of genenality, we assume thatwx; wy ∈ E.C/: HenceC′ = C−{wx; wy; xy} is a circuit with".C′/ = ".C/− 3≤ k≤ ".C′/, a contradiction, set-tles Claim 4.

Claim 5. Each element ofM1 is cut vertex ofC.

Proof. Otherwise, there exists a vertexw ∈ M1, say,w ∈ NC.x/ ∩ NC.y/ ∩ N2,which is not cut vertex ofC. HenceC′ = C− {wx; wy; xy} is a circuit with".C/− 3= ".C′/ ≤ k≤ ".C′/; a contradiction, settling Claim 5.

Let W1 denote the cut vertex set ofC in M1 such that for anyz∈ W1;C− z has anontrivial component which does not contain any element ofS. By Claim 2, we

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obtain|W1| ≥ m1− 2; (8.13)

andIf either N3∪ N4 6= ∅ or m2 6= 0; then|W1| = m1: (8.14)

Claim 6. If m2 ≥ 3, then for any pair of vertices{w;w′} of M2, a cycle ofC[{w;w′} ∪ ..NC.w/ ∪ NC.w

′// ∩ S/], which does not containww′, is a non-trivial cutset ofC.

Proof. Otherwise,C[{w;w′} ∪ ..NC.w/ ∪ NC.w′// ∩ S/] has a cycleC′ which

does not containww′, such that".C′′/ ≤ k ≤ ".C′′/ whereC′′ = C−C′, a con-tradiction, settling Claim 6.

Let W2 denote the cut vertex set ofC in M2 such that for anyy ∈W2;C− y has anontrivial component which does not contain any element ofS. By Claim 6, weobtain

|W2| ≥ m2− 2; (8.15)

if n4 = 1; then|W2| = m2 (8.16)

andif n3 = 1; then|W2| = m2− 1: (8.17)

For y∈W1∪W2, let Qy denote the nontrivial component ofC− y which does notcontain any element ofS. Then it is easy to see that

|V.Qy/| ≥ 3: (8.18)

Otherwise,".C′/ ≤ k≤ ".C′/; whereC′y = C− Qy. We also obtain

|{Qy : y ∈W1∪W2}| = |W1 ∪W2|: (8.19)

If N3∪ N4 6= ∅, then using Claims 3, 4, 5, 6 and (8.14)-(8.19), we obtain

dC.S/ = |4⋃

i=1Ni| + n2+ 2n3+ 3n4

≤ n− .|V2.D1/| + |V2.Dr /| − 4+ 3n2/+ n2+ 3; if n4 = 1;

n− .|V2.D1/| + |V2.Dr /| − 5+ 3.n2− 1//+ n2+ 2; if n2 ≥ 2;n3 = 1;n− .|V2.D1/| + |V2.Dr /| − 5/+ 3; if n2 ≤ 1;n3 = 1;

≤ n− .|V2.D1/| + |V2.Dr /|/+ 8:


If N3∪ N4 = ∅; then using Claim 5, Claim 6 and (8.13), (8.15), (8.18), (8.19), weobtain

dC.S/ = |4⋃

i=1Ni| + n2+ 2n3+ 3n4

≤

n− .|V2.D1/| + |V2.Dr /| − 5+ 3.m1− 2/+ 3.m2− 2//+m1+m2; if m1 ≥ 2;m2 ≤ 3;

n− .|V2.D1/| + |V2.Dr /| − 5+ 3.m1− 2//+m1+ 2;if m1 ≥ 2;m2 ≥ 2;

n− .|V2.D1/| + |V2.Dr /| − 5+ 3.m2− 2//+ 1+m2;

if m1 ≤ 1;m2 ≥ 3;n− .|V2.D1/| + |V2.Dr /| − 5/+ 3;

if m1 = 1;m2 ≤ 2;n− .|V2.D1/| + |V2.Dr /| − 6/+ 2;

if m1 = 0;m2 ≤ 2;≤ n− .|V2.D1/| + |V2.Dr /|/+ 8:

HencedC.S/ ≤ n− |V2.D1/| − |V2.Dr /| + 8: (8.20)

On the other hand, by (8.8),

dC.S/ > n+ 6− 2.|V2.D1/| + |V2.Dr /|/=3: (8.21)


|V2.D1/| + |V2.Dr /| < 6

which contradicts (8.5) and proves Claim 1.

Case 2.dia.H/ = 1 and|V.D1/∩ V.Dr /| = 1:

Now H is a complete graph.

Let V.D1/∩ V.Dr / = {y}: We will consider the following two subcases.

Subcase 2.1.|V.Di/| ≥ 5 for i ∈ {1; r}:

112

Now, we can take two pathsP= u′v′x′y′ andP′ = u′′v′′x′′y′′ of length 3, inD1 andDr respectively, such that{uv;u′v′} ⊆ E2.D1/∪ E2.Dr / andV.P/∩ V.P′/ = ∅.By (8.9),|V.P/∩ V1.D1/| ≥ 1 and|V.P′/∩ V1.Dr /| ≥ 1. In a way similar to thatused in the proof of Case 1, we derive contradictions.

Subcase 2.2.There exists aDi.i ∈ {1; r}/, say,D1, such that|V.D1/| = 4.

Now we can take a pathP= uvxy in D1 such that{uv; vx} = E2.D1/.

Claim 7. |V.Dr /| = 4:

Proof. Otherwise, by (8.9), we can take a pathP′ = u′v′x′y′ in Dr such thatu′v′ ∈ E2.Dr /; y 6∈ {x′; y′} and|{x′; y′} ∩ V2.Dr /| ≤ 1. This proves Claim 7.

In a way similar to that used for Claim 4 and Claim 5, we obtain

Claim 8. n3 = n4 = 0 andn2 ≤ 1:

If |E2.Dr /| = 2 or 3 then, by (8.7),

dC.S/ > n+ 6− |E2.D1/| − |E2.Dr /| = n+ 1: (8.22)

On the other hand, by Claim 8,

dC.S/ = |4⋃

i=1

Ni| + n2+ 2n3+ 3n4 ≤ n+ 1;


If |E2.Dr /| ≥ 4 then, using (8.8), we obtain|V2.Dr /| ≥ 3|E2.Dr /|=2 ≥ 6. Thisimplies that, by (8.5),

|V2.D1/| + |V2.Dr /| ≥ 9: (8.23)

On the other hand, using Claim 8, we obtain

dC.S/ = |4⋃

i=1

Ni| + n2+ 2n3+ 3n4;

≤ n− .|V2.D1/| + |V2.Dr /| − 8/+ 1

= n− |V2.D1/| − |V2.Dr /| + 9;


implying that, by (8.21),

|V2.D1/| + |V2.Dr /| < 3.9− 6/ = 9

which contradicts (8.23). This proves Claim 7.

If there exists aDi.i ∈ {2;3; : : : ; r − 1}/, say,D2, such that|V.D2/| ≥ 5, thenD2

plays the same role asDr in subcase above. In a way similar to that using in theproof of the above subcase, we detrive contradictions.

So|V.Di /| = 4 for i ∈ {1;2; : : : ; r}.

Next we prove thatG∼= F.

If there exists a vertexx ∈ V.C/ such thatdC.x/ < d.x/, thenCi = C− Di.x 6∈V.Di/ unlessx= y/ is a circuit with".Ci / ≤ k≤ ".Ci /, a contradiction.

SodC.u/ = dG.u/ for anyu ∈ V.C/. SinceG is connected,G∼= F. This provesCase 2.

Case 3.dia.H/ = 1 and|V.D1/∩ V.Dr /| = 2:

So H is a complete graph.

Let V.D1/∩ V.Dr / = {u; v}: Then there exist four pathsP1; P2; P3; P4 such thatD1 = P1∪ P2, Dr = P3∪ P4, andV.Pi /∩ V.Pj / = {u; v}.i 6= j/:If there exists a pair of{Ps; Pt} and a Di where i ∈ {2;3; : : : ; r − 1}; say D2,such that|.V.Ps/∪ V.Pt//∩ V.D2/| ≤ 1 where{s; t} ⊆ {1;2;3;4} then letD′1=Ps∪ Pt; D′r = .D1∪ Dr /− .E.Ps/∪ E.Pt// andD′i = D j for j ∈ {2;3; : : : ; r−1}:Let H ′ be a graph with vertex setV.H ′/ = {D′1; D′2; · · · ; D′r}, D′i D

′j ∈ E.H ′/ if

and only if V.D′i / ∩ V.D′j/ 6= ∅: Obviously H ′ is a complete graph. Note thatD′1 and D′2 in H ′ play the same role asD1 and Dr in H respectively. Since|V.D′1/ ∩ V.D′2/| ≤ 1, we derive contradictions in a way similar to that used inthe proof of Case 1 or Case 2.

Hence, using (8.1), we obtain

Claim 9. |.V.Ps/ ∪ V.Pt//∩ V.Di/| = 2 for any pair of{s; t} ⊆ {1;2;3;4} and

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i ∈ {2;3; · · · ; r − 1} .

Furthermore, we can prove the following claim.

Claim 10. {u; v} ⊆ V.Di/ for i ∈ {2;3; · · · ; r − 1}:

Proof of Claim 10. Otherwise, there exists aDi, say D2, such that|{u; v} ∩V.D2/| ≤ 1:

If |{u; v} ∩ V.D2/| = 1, sayu ∈ V.D2/, then, by (8.1), there exist two pathsPs

and Pt such thatV.Ps/ ∩ V.D2/ = V.Pt/ ∩ V.D2/ = {u} wheres ∈ {1;2} andt ∈ {3;4}; which contradicts Claim 9.

So{u; v} ∩ V.D2/ = ∅.

If |V.Pj /∩ V.D2/| = 1 for j ∈ {1;2;3;4}, then there exist four edge-disjoint cy-clesC1;C2;C3;C4 in D1∪ D2∪ Dr such thatD1∪ D2∪ Dr = C1∪C2∪C3∪C4

which contradicts the maximum ofr .

So there exists aPi, sayP1; such that|V.P1/∩ V.D2/| = 0 or 2.By (8.1) and{u; v} ∩ V.D2/ = ∅, there exists ans ∈ {3;4} such that|V.Ps/ ∩V.D2/| ≤ 1.

If |V.P1/ ∩ V.D2/| = 2, then by (8.1),|.V.Ps/ ∪ V.P2// ∩ V.D2/| ≤ 1; whichcontradicts Claim 9.

If |V.P1/∩ V.D2/| = 0, then|.V.Ps/∪ V.P1//∩ V.D2/| ≤ 1; which contradictsClaim 9.

This completes the proof of Claim 10.

It follows from Claim 10 that there existdC.u/ = dC.v/ = 2r edge-disjoint paths

P1; P2; · · · ; P2r such thatC=2r⋃

i=1Pi andV.Pi /∩ V.Pj / = {u; v}.i 6= j/:

Hence, by (8.9), we obtain

|V.Pi /| ≤ 5 for i ∈ {1;2; · · · ;2r}: (8.24)


From this, it is easy to see that

|E2.D1/| + |E2.Dr /| ≤ 4+min{4; |{i : V.Pi/| = 5}|}: (8.25)

By (8.5), there exist two pathsP′1 and P′2 of length 3, in D1 and Dr respec-tively, such that each path contains exactly one vertex inV1.D1/ ∪ V1.Dr /, andu ∈ V.P′1/ andv ∈ V.P′2/. So, by (8.7),

dC.u/+ dC.v/ > n+ 2− |E2.D1/| − |E2.Dr /|: (8.26)

If there exists a pair of{i; j} such that|V.Pi /| ≤ 3 and|V.Pj /| ≤ 4, thenC′ =C− .Pi ∪ Pj/ is a circuit with".C′/≤ k≤ ".C′/, This contradicts Claim 1. Hencewe obtain

{.i; j/ : |V.Pi /| ≤ 3 and|V.Pj /| ≤ 4} = ∅: (8.27)

Without loss of generality, we can assume that|V.P1/|; |V.P2/|; · · · ; |V.P2r /| isan increasing sequence andDi = P2i−1∪ P2i for i ∈ {1;2; · · · ; r}:

Claim 11. If |V.P1/| = 2, i.e.,uv ∈ E.C/, thenr ≥ 4.

Proof of Claim 11. Otherwiser = 2 or r = 3. By (8.24) and (8.27),|V.Pi /| = 5for i ∈ {2;3; · · · ;2r}. Sincen ≥ 146, there exists a vertexx of C with dC.x/ ≤d.x/ − 1. Hence there exists a circuitC′ such that".C′/ ≤ k ≤ ".C′/, whereC′ = C− .Pi ∪ P1/.Pi 6= P1, andx 6∈ V.Pi / unlessx ∈ {u; v}/: This contradictsClaim 1 and proves Claim 11.

Next, we obtain some inequalities which contradict (8.25).

If 3 ≤ |V.P1/| ≤ 5 then, using (8.27), we obtain

dC.u/+ dC.v/ ≤ n− 2− |{i : |V.Pi /| = 5}|: (8.28)

Using (8.26) and (8.28), we obtain an inequality that contradicts (8.25).

If |V.P1/| = 2, i.e., uv ∈ E.C/ then, using (8.24) and (8.27), we obtain that|V.Pi /| = 5 for i ≥ 2, and

dC.u/+ dC.v/ ≤ n− |{i : |V.Pi /| = 5}|: (8.29)

Using (8.26), (8.27), (8.29) and Claim 11, we obtain

|E2.D1/| + |E2.Dr /| > 2+ |{i : |V.Pi /| = 5}| = 2+ .2r − 1/ ≥ 9;

116


The result in Theorem 8.3 is best possible in the sense that the condition²4.G/ >.n+ 10/=2 can not be relaxed, even under the condition thatL.G/ is hamiltonian.To see this, we will construct a graphG0 as follows:

Let s= .n− 2/=2.n≡ 2.mod4//. Define a graphG0 of ordern with vertex set

V.G0/ = {u1; v1;u2; v2; · · · ;us; vs; x; y} and edge setE.G/ =s⋃

i=1{xui;uivi; vi y}.

Clearly G0 is a graph such that²4.G0/ = s+ 6= .m+ 10/=2: Theorem 8.4 im-plies thatL.G/ is hamiltonian and 3s− 1 ∈ [3; ".G0/]\½.L.G0// and which im-plies thatL.G0/ is not (sub)pancyclic.

Chapter 9

Pancyclic line graphs

Veldman considered the condition on the diameter of a graph for its line graph tobe hamiltonian.

Theorem 9.1 (Veldman [54]) Let G be a connected graph with at least threeedges and such that dia.G/ ≤ 2. Then L.G/ is hamiltonian.

The main result on hamiltonian line graph is due to to Harary and Nash-Williams.

Theorem 9.2 (Harary and Nash-Williams [32]) The line graph L.G/ of a con-nected graph G with at least three edges is hamiltonian if and only if G has adominating eulerian subgraph H withE.H/ = E.G/:

The pancyclic version of the above result was obtained by Broersma.

Theorem 9.3 (Broersma[12]) The line graph L.G/ of a graph G contains a cycleof length k≥ 3 if and only if G contains a circuit C such that".C/ ≤ k≤ ".C/:

In this chapter, we improve Theorem 9.1.

Theorem 9.4 Let G be a graph. If dia.G/ ≤ 2; then L.G/ is either pancyclic ora cycle of length 4 or 5.

In a sense, Theorem 9.4 is best possible. Lett ≥ 2 be an integer. DefineGt to bethe graph with vertex set

V.Gt/ = {u1;u2; · · · ;ut} ∪ {v1; v2; · · · ; vt} ∪ {x; y}

117

118

and edge set

E.Gt/ =t⋃

i=1

{xui;uivi; vi y}:

It is easy to check thatdia.G/ = 3:

By Theorem 9.2, we obtain that ift ≡ 1.mod2/, thenL.Gt/ is hamiltonian butGt does not contain a circuitC such that|E.C/| ≤ 3t − 1≤ |E.C/|. Hence byTheorem 10.3,L.Gt/ does not contain a cycle of length 3t − 1. HenceL.Gt/ isneither pancyclic nor a cycle of length 4 or 5.

We also obtain a similar result on the distance between two edges.

Theorem 9.5 Let G be a graph of order n with at least one cycle. If d.e1;e2/ ≤ 1for any pair of edges e1, e2 in G, then L.G/ is either pancyclic or a cycle of length4 or 5.

The graphGt also shows that Theorem 9.5 is best possible since max{ d.e; f / :e; f are two edges ofG} =3.

The conditions in Theorem 9.4 and 9.5 do not include each other. For example, letCi;1 = ui;1ui;2ui;3 for i ∈ {1;2; · · · ; t}: the graphG1 obtained by identifying theseverticesu1;1;u2;1; · · · ;ut;1 satisfies the condition of Theorem 9.4 but Theorem 9.5.Similarly, the graphG2 obtained by attaching every vertex on a cycle of length 3some edges satisfies the condition of Theorem 9.5 but not that of Theorem 9.4.

We will give another proof of Theorem 9.1. The same idea is then used to proveTheorem 9.5.

9.1 Some Lemmas

We begin with the following lemmas.The first lemma is obvious.

Lemma 9.1 If G contains a circuit C, then½.L.G// ⊇ [".C/; ".C/]:

The following Lemma is necessary for our proof of Theorem 9.4. HereV.H/ ={v ∈ V.G/ : dG.v; H/ ≤ 1} for any subgraphH ⊆ G:

9.1 Some Lemmas 119

Lemma 9.2 Let G be a graph with at least one cycle. If dia.G/ ≤ 2; then Gcontains a circuit H withV.H/ = V.G/; i:e:;dG.v; H/ ≤ 1; for anyv ∈ V.G/:

Proof. Let C be the shortest cycle ofG. Sincedia.G/ ≤ 2; d.v;C/ ≤ 2 for anyvertex ofG: Let k be a positive integer and let

Nk.S/ = {u ∈ V.G/ : d.u; S/ = k} and

N′1.C/ = N1.C/∩ N1.N2.C//:

We obtain3≤ ".C/ ≤ 4: (9.1)

Otherwise, if".C/= 5; sinceG 6∼= C5, thenN′k.C/ has a certexw with d.w;C/=3, a contradiction. If".C/ ≥ 6; thenC has two vertices such that their distance isgreater than 2, a contradiction. By (9.1), we select a subset ofN2.C/; denoted byN′2.C/, such thatN2.C/⊆ N1.N1.C/∩ N1.N′2.C//: and|N1.x/∩ N′2.C/| = 1 foranyx ∈ N1.C/∩ N1.N′2.C//:

We have the following claims.

Claim 1. Let C be a cycle with".C/ = 3: Let x ∈ N′1.C/. If |N1.x/∩ V.C/| = 1,then for anyw ∈ N1.x/ ∩ N2.C/ andu ∈ V.C/\N1.x/; there exists at least onevertexy; y 6= x, such thaty ∈ N1.w/ ∩ N1.u/:

Proof of Claim 1. Sincew ∈ N2.C/ and u ∈ V.C/;d.w;u/ ≥ 2: Obviously,d.w;u/ ≤ dia.G/ ≤ 2: Henced.w;u/ = 2: Since |N1.x/ ∩ V.C/| = 1: Theremust exist a vertexy; y 6= x such thaty ∈ N1.w/∩ N1.u/: This proves Claim 1.

Claim 2. Let C be a cycle with".C/ = 4: .N1.w/ ∩ N1.u//\N1.N1.x// 6= ∅ foranyx ∈ N′1.C/;w ∈ N1.x/∩ N2.C/ andu ∈ V.C/\N1.x/:

Proof of Claim 2. Sincew ∈ N1.x/∩ N2.C/ andu ∈ V.C/\N1.x/; d.w;u/ ≥ 2:Obviously,d.w;u/ ≤ dia.G/ ≤ 2: Henced.w;u/ = 2: So there must exist a ver-tex y ∈ N1.w/ ∩ N1.u/: SinceG do not contain any triangle,y 6∈ N1.N1.x//:Hence.N1.w/ ∩ N1.u/\N1.N1.x/// 6= ∅: This proves Claim 2.

For the proof of Lemma 9.2, we distinguish two cases.

120

Case 1.".C/ = 3:

Let C= u1u2u3u1: Set

V1;1 = {v ∈ N′1.C/ : vu2; vu3 ∈ E.G/}:V2;1 = {v ∈ N′1.C/ : vu1; vu2 ∈ E.G/}\V1;1:

V3;1 = {v ∈ N′1.C/ : vu1; vu2 ∈ E.G/}\.V1;1∪ V2;1/:

and K=3⋃

i=1

Vi;1:

For i ∈ {1;2;3}; set

Si = {{x; y; z} : for anyz∈ N′2.C/\N1.K/; there exists verticesx andy;x ∈ N1.uj /∩ N1.z/ andy ∈ N1.uk/∩ N1.z/; such that{i; j; k} = {1;2;3}}:

SetVi;2 =

⋃s∈Si

s and Vi = Vi;1∪ Vi;2:

ObviouslyVi ∩ Vj = ∅ f or {i; j} ⊂ {1;2;3}:

We define a subgraphH of G withvertex set

V.H/ = V.C/∪ V1∪ V2∪ V3

and edge setE.H/ = E1.H/∪ E2.H/∪ E3.H/;

whereE1.H/ = {vui; vuj : v ∈ Vk and {i; j; k} = {1;2;3}}:

E2.H/ = {zx; zy; xuj; yuk : {x; y; z} ∈ Si and {i; j; k} = {1;2;3}}and

E3.H/ = E.C/; if |V1| ≡ |V2| ≡ |V3| . mod2 /

E.C/\{uiuj;uiuk}; if |Vi| ≡ 0; |Vj| ≡ |Vk| ≡ 1 . mod2/E.C/\{ujuk}; if |Vj| ≡ |Vk| ≡ 0; |Vi| . mod2/;

where{i; j; k} = {1;2;3}:

9.1 Some Lemmas 121

It is easy to see thatH is a eulerian subgraph ofG. HenceV.H/ = V.G/ byClaim 1.

Case 2.".C/ = 4:

Let C= u1u2u3u4u1: Set

V′i;i+2= {v ∈ Ni.C/ : vui; vui+2 ∈ E.G/}; f or i = 1;2:

ObviouslyV′1;3∩ V′2;4 = ∅: Set

K =2⋃

i=1

V′i;i+2; for i = 1;2:

Set

Si;i+2= {{x; y; z} : for any z∈ N′2.C/\N1.K/ there alway exist two verticesx andy such thatx ∈ N1.ui /∪ N1.z/ andy ∈ N1.ui+2/∩ N1.z/}:

andV′′i;i+2=

⋃s∈Si;i+2

s and Vi;i+2= V′i;i+2

⋃V′′i;i+2:

By Claim 2, if N2.C/ 6= ∅; thenVi;i+2 6= ∅.i = 1;2/: We construct a subgraphH ′

of G as follows.vertex setV.H ′/= V.C/∪ V1;3∪ V2;4 and edge setE.H ′/ = E1.H ′/∪ E2.H ′/∪E3.H ′/;where

E1.H′/ = {vui; vui+2 : v ∈ V′i;i+2; i ∈ {1;2}};

E2.H′/ = {zx; zy; xui; yui+2 : {x; y; z} ∈ Si;i+1; i ∈ {1;2}}

and

E3.H′/=

E.C/; if |V1;3| ≡ |V2;4| ≡ 0 . mod 2/;{u1u2;u3u4}; if |V1;3| ≡ |V2;4| ≡ 1 .mod 2/;{uiuj;ujui+2}; if |Vi;i+2| ≡ 1; |Vj; j+2| ≡ 0 .{mod 2/{i; j} = {1;2}:

We construct subgraphH of G as follows.If N2.C/ = ∅ thenH = C. OtherwiseH = H ′: By Claim 2, H is a eulerian sub-graph ofG such thatV.H/ = V.G/:

122

'

&

$

%"!# �

�� @

@@@• x •w1

•y′

•u •v

• y

•

C

Figure 9.1: Replace some vertex

9.2 An alternative proof of Theorem 9.1 and somelemmas.

We give another proof of Theorem 9.1. By Lemma 9.2, we can suppose thatCis a circuit withV.C/ = V.G/ such thatC contains as many vertices as possible.We will prove thatE.C/ = E.G/. Otherwise, lete= uv ∈ E.G/\E.G/; i.e.,u; v 6∈ V.G/: By V.C/ = V.G/;N.w/ ∩ V.C/ 6= ∅ for anyw ∈ {u; v}: So, wecan select two verticesx∈ V.C/∩ N.u/ andy∈ V.C/∩ N.v/ such thatdC.x; y/is a minimum.By dia.G/ ≤ 2, dG.x; y/ ≤ 2: Let p= xw1y be the shortest path betweenx andy: Set

C′ = C+ {xu;uv; vy} + .E.G/\E.C// ∩ {xw1; w1y} − .E.C/∩ {xw1; wy}/:If xw1; w1y ∈ E.C/ andw1 is a cut vertex ofC, then either there exists a ver-tex y′ ∈ NC.w1/\{x; y}; such thatvy′ ∈ E.G/, or dG.v; y′/ = 2 for any y′ ∈NC.w/\{x; y}, i.e., there exists a pathy′u′v with u′ ∈ V.G/\V.C/: Now, replacey by y′ or replaceu by u′. We obtainC′, see Figure 9.2.

It is easy to seeC′ is a circuit withV.C′/ = V.G/; But |V.C′/ > |V.C/|; a con-tradiction. SoL.G/ is hamiltonian.

9.2 An alternative proof of Theorem 9.1 and some lemmas. 123

The following lemmas are necessary for our proof of Theorem 9.5.

Lemma 9.3 Let G be a connected graph of order n, that contains at least onecycle. If d.e1;e2/ ≤ 1; for any pair of edges e1;e2 ∈ G and G6∼= C4 or C5, then½.L.G// ⊇ [3;n].

Proof. SinceG contains a cycle, letC be a shortest cycle with lengtht. Sinced.e1;e2/ ≤ 1 for any pair of edgese1;e2 in G, d.v;C/ ≤ 2 for everyv ∈ V.G/andG[V.G/\V.C/] is an independent set. Let

Nk.S/ = {u : d.u; S/ = k} for the positive integerk

and letN′1.C/ = N1.C/∩ N1.N2.C//:

Sinced.e1;e2/ ≤ 1; for anyv ∈ N′1.C/, there exist two vertices inC adjacent tov and t ≤ 4. Otherwise,t = 5 which implies thatG contains a 5-cycle. SinceG 6≡ C5, there exists an edgee on C and an edgee′ incident withC such thatd.e;e′/ = 2, a contradiction. Ift ≥ 6; there exist two edgese ande′ on C suchthatd.e;e′/ = 2, a contradiction. Therefor we consider the following two cases:

Case 1.t = 3.

Let C= u1u2u3u1. Let

V1 = {v ∈ N′1.C/ : vu2; vu3 ∈ E.G/};V2 = {v ∈ N′1.C/ : vu1; vu3 ∈ E.G/}\V1;

V3 = {v ∈ N′1.C/ : vu1; vu2 ∈ E.G/}\.V1 ∪ V2/;

thenN′1.C/ = V1∪ V2∪ V3;and Vi ∩ Vj = ∅.i 6= j/;

We construct a subgraphH of G with

V.H/ = N′1.C/∪ V.C/ and E.H/ = E1.H/∪ E2.H/;

whereE1.H/ = {vus; vut : v ∈ Vi and {s; t; i} = {1;2;3}} and

E2.H/ = E.C/; if |V1| ≡ |V2| ≡ |V3|.mod2/

E.C/\{uiuj : uiuk}; if |Vi| = 0; |Vj| ≡ |Vk| ≡ 1.mod2/E.C/\{ujuk}; if |Vj| ≡ |Vk| ≡ 0 and |Vi| ≡ 0.mod2/;

124

where{i; j; k} = {1;2;3}. It is easy to see thatH is a eulerian subgraph ofG.

Case 2.t = 4.

Let C= u1u2u3u4u1. Let

Vi;i+2= {v : v ∈ N′1.C/ andvui; vui+2 ∈ E.G/} f or i = 1;2:

Obviously,V1;3∩ V2;4 = ∅. Sinced.e1;e2/ ≤ 1 for any pair of edgese1;e2 in GandG does not contain a 3-cycle, we know that if|Vj; j+2| ≥ 1 and|Vi;i+2| = 0,thend.ui / or d.ui+2/ is 2, where{i; j} = {1;2}. We construct a subgraphH ′ of Gwith

V.H ′/ = V.C/∪ V1;3∪ V2;4 and E.H ′/ = E1.H′/∪ E2.H

′/;

whereE1.H

′/ = {vui; vui+2 : v ∈ Vi;i+2; i = 1;2}and

E2.H′/ =

E.C/; if |V1;3| ≡ |V2:4| ≡ 0.mod2/;{u1u2;u3u4}; if |V1;3| ≡ |V2:4| ≡ 1.mod2/;{uiuj;ujui+2}; if |Vi;i+2| ≡ 1; |Vj; j+2| ≡ 0.mod2/;

and |Vj; j+2| ≥ 2{i; j} = {1;2};{us+2ui;us+2ui+2}; if |Vi;i+2| ≡ 1; .mod2/; |Vj; j+2| = 0; .2:1/

and d.us/ = 2 where {i; j} = {1;2}; s∈ { j; j + 2}:Obviously,H ′ is a subgraph ofG and every vertex ofH ′ has even degree inH ′.

Now, we construct a subgraphH of G from H ′ as follows:

H ={

H ′ − us; if .2:1/ is true;H ′ otherwise.

It is easy to see thatH is a eulerian subgraph ofG.

SinceH is a eulerian subgraph ofG, let C′ be a eulerian circuit ofH. Obviously,C′ is a vertex-D-circuit ofG and E.C′/ ⊆ E.C/: Thus".C′/ ≥ n and".C′/ ≤".C/. By Theorem 9.1, we have that½.L.G// ⊇ [".C′/; ".C′/] ∪ [".C/; ".C/] ⊇[".C/;n].

By the assumption of Lemma 9.3, ift = 4, then1.G/ ≥ 3. That is , 3∈ ½.L.G//,so that½.L.G// ⊇ [3;n]:

9.3 Proof of the main results 125

Lemma 9.4 Let G be a connected graph of order n with at least one cycle. Ifd.e1;e2/ ≤ 1 for any pair of edges e1;e2 ∈ G and G 6� C4 or C5, then L.G/ ishamiltonian.

Proof. From Lemma 9.3, we can suppose thatC is a vertex-D-circuit ofG suchthatC contains as many vertices as possible. We will prove thatC is a D-circuitof G. Otherwise, lete= uv ∈ E.G/, with u; v 6∈ V.C/. Then

V.C/∩ N.w/ 6= ∅ f or anyvertexw ∈ {u; v}:So, we can select two verticesx ∈ V.C/∩ N.u/ andy ∈ V.C/∩ N.v/ such thatdC.x; y/ is minimal.

Let P= xw1w2 · · ·wsy be a shortest path inG. It is easy to see thats≤ 1. Other-wise,s≥ 2: Sinced.wsy; xu/ ≤ 1, uws ∈ E.G/, a contradiction. Let

C′ = C+ {xu;uv; vy} + .E.G/\E.C// ∩ {xw1; w1y} − E.C/∩ {xw1; w1y}:It is easy to see thatC′ is a vertex-D-circuit ofG but that|V.C′/| > |V.C/|; (Ifxw1; w1y ∈ E.C/ andw1 is a cutvertex ofC, then eitherux′ or vy′, sayvy′, is inE.G/, and we can replacey by y′, see Figure 9.2). This is a contradiction. SoL.G/ is hamiltonian.

9.3 Proof of the main results

In this section we will prove Theorem 9.4 and Theorem 9.5.


Assume thatG satisfies the condition butL.G/ is neither pancyclic nor a cycle oflength 4 or 5.

If G does not contain any cycle, then bydia.G/≤ 2;G∼= K1;v.G/−1. ClearlyL.G/is pancyclic. HenceG contains at least a cycle. Set

k= max{i : i ∈ [3; ".G/]\½.L.G//}:Obviously L.G/ contain a cycle of lengthk+ 1:By Lemma 9.3,k < ".G/: ByTheorem 9.3,G contains a circuitC with ".C/ ≤ k+ 1≤ ".C/: Clearly".C/ =

126

k+ 1:So we can considerC=r⋃

i=1Di; whereDi is a cycle ofG andr is maximal.

Hence

E.Di /∩ E.Dj / = ∅ and |V.Di /∩ V.Dj/| ≤ 2;

for {i; j} ∈ {1;2; · · · ; r}:

We distinguish two cases.

Case 1.r = 1:

Clearly ".C/ ≥ 4: SinceG is not a cycle of length 4 or 5,".C/ ≥ 6: Let C =u1u2 · · ·u".C/u1 and let p.u1;u4/ be a shortest path betweenu1 and u4: Sinced.u;u4/ ≤ dia.G/ ≤ 2; ".p.u1;u4// ≤ 2: Hence

C′ = p.u1;u4/u4 · · ·u".C/u1

is a circuit with".C′/ ≤ k≤ ".C′/; a contracdiction.

Case 2.r ≥ 2:

Let H be the graph withV.H/ = {D1; D2; · · · ; Dr}; Di D j ∈ E.H/ if and only ifV.Di/ ∩ V.Dj/ 6= ∅. SinceC is a circuit ofG, H is connected. Therefore, wecan assume, without loss of generality, thatD1; Dr are leaves of a spanning tree

of H. So C1 =r⋃

i=2Di andCr =

r−1⋃i=1

Di are two circuits ofG. For i ∈ {1; r}, de-

fine E′.Di/ = E.Di /\.E.Di /∩ E.Ci //. Obviously,V′.Di /∩ V′.Dj / = ∅, where{i; j} = {1; r}. Since".Ci / ≥ ".C/− |E′.Di/| = k+ 1− |E′.Di /|; |E′.Di /| ≥ 2,wherei = 1; r . Otherwise,".Ci / ≤ k≤ ".Ci /, a contradiction. Thus|V′.Di/| ≥ 3for i ∈ {1; r}. Takeuv ∈ E′.Di /. It is easy to see thatD1 ∪ Dr has a path oflength at least 3 betweenu andu3; denoted byP0 = uu′u3(if d.u;u3/ = 1;thenp0 = uu3). Let C′ be the nontrivial component of the induced subgraph

G[.E.C/∪ .E.P0/∩ E.G/\E.C///\.{uv; vu2;u2u3} ∪ .E.P0/∩ E.C///]:

ThenC′ is a circuit ofG such that

".C/− 3≤ ".C′/ ≤ k≤ ".C/− ≤ ".C′/;

9.3 Proof of the main results 127

a contradiction.


Assume thatG is a graph satisfying the conditions of Theorem 9.5, butL.G/ isnot pancyclic. Let

k= max{i : i ∈ [3; ".G/]\½.G/}:By Lemmas 9.3 and 9.4, we obtain thatn< k< ".G/ andL.G/ contains a cycleCk+1 of lengthk+ 1. From Theorem 9.3, it follows thatG contains a circuitCsuch that".C/ ≤ k+ 1≤ ".C/. Obviously,".C/ = k+ 1. We can assume thatC=⋃r

i=1 Di, whereDi is a cycle ofG andr is maximized. SoE.Di /∩ E.Dj/=∅for {i; j} ⊆ {1;2; · · · ; r}: From Lemma 9.4 we know thatr ≥ 2.

Let H be the graph withV.H/ = {D1; D2; · · · ; Dr}; Di D j ∈ E.H/ if and onlyif V.Di/ ∩ V.Dj/ 6= ∅. Obviously,H is connected. Therefore, we can assume,without loss of generality, thatD1; Dr are leaves of a spanning tree ofH. SoC1 =

⋃ri=2 Di and Cr =

⋃r−1i=1 Di are two circuits ofG. For i ∈ {1; r}, define

E′.Di/ = E.Di/\.E.Di / ∩ E.Ci //. Obviously, V′.Di / ∩ V′.Dj/ = ∅, where{i; j} = {1; r}. Since".Ci / ≥ ".C/− |E′.Di /| = k+ 1− |E′.Di/|; |E′.Di/| ≥ 2,wherei = 1; r . Otherwise,".Ci / ≤ k≤ ".Ci /, a contradiction.Thus|V′.Di /| ≥ 3;for i ∈ {1; r}. Let uv ∈ E′.Di /. It is easy to see thatC has a pathP= uvu1u2 · · ·us

such thats≥ 4 andC[C+ {uu3} − {vu1;u1u2;u2u3}] has at most one nontriv-ial component. Sinced.uv;u2u3/ = 1; {uui; vui : i ∈ {2;3}} ∩ E.G/ 6= ∅, sayuu3 ∈ E.G/. Obviously,uu3 6∈ E.C/. Let C′ = C+ {uu3} − {uv; vu1;u1u2;u2u3}and letC′′ be the nontrivial component ofC′. Then C′ is a circuit such that".C/− 4+ 1= ".C′′/− 2+ 1≤ k≤ ".C′′/, a contradiction.

Chapter 10

Connected even factors with degreerestrictions

In this Chapter 10 we use the following special notation and terminology. For agraphG; a setSof vertices inG is independentif they are pairwise nonadjacentin G: An independent set with cardinalityt is called at-independent set.By Þ.G/we denote the number of vertices in maximal independent set ofG. We define¦k.G/ = min{∑k

i=1 d.ui / : {u1;u2; · · · ;uk}is a k-independent set ofG} if k ≤ Þ;otherwise we set¦k.G/ =∞:

A trail is a finite sequenceT = u0e1u1e2 · · ·erur; whose terms are alternately ver-tices and edges, withei = ui−1ui.1≤ i ≤ r /; and where the edges are distinct. AgraphG is supereulerianif G has a spanning closed trail.

A spanning subgraphH of a graphG is called afactorof G; and is calledk-factorif all vertices of H have degreek in H, and is called aconnected even factorifall vertices have even degree andH is connected. Hence a hamiltonian cycle isa connected 2-factor. ClearlyG contains a connected even factor if and only ifG is supereulerian. Aconnected even[2; k]-factorof a graph is a connected evenfactor with maximum degree at mostk:

The earliest degree condition for a graph to contain a connected 2-factor was givenby Dirac in 1952.

Theorem 10.1 (Dirac[29]) If G is a graph on n≥ 3 vertices such thatŽ.G/ ≥n=2, then G contains a hamiltonian cycle, i.e., a connected 2-factor.

129

130

Ore gave the following generalization of Theorem 10.1.

Theorem 10.2 (Ore[46]). If G is a graph on n≥ 3 vertices such that¦2.G/ ≥ n,then G contains a hamiltonian cycle, i.e., a connected 2-factor.

Many results on supereulerian graphs, i.e., those graphs having a connected evenfactor, have appeared in the literature [1] [4]-[7] We list some of them here.

Theorem 10.3 (Cai[14]) If G is a 2-edge-connected graph withŽ.G/≥ n=5, thenG is supereulerian, i.e., G contains a connected even factor.

Theorem 10.4 (Benhocine et al. [12]) Let G be a 2-edge-connected graph onn≥ 3. If ¦2.G/ ≥ .2n+ 3/=3, then G has a connected even factor.

Theorem 10.5 (Catlin [16]) Let G be a 2-edge-connected graph of order n≥ 6with¦3.G/≥ n+1, then G is supereulerian, i.e., contains a connected even factor.

Recently Broersmaet al. [13] invesigated connected even [2;4]-factors of 4-connected claw-free graphs, i.e., graphs not containing aK1;3 as an induced sub-graph. In Section 10.1, we obtain some stronger results than Theorem 10.3, The-orem 10.4, Theorem 10.5 on connected even factors with degree restrictions interm ofŽ.G/ and¦k.G/. In Section 10.2, we will show that every supereulerianclaw-free graph contains a connected even [2;4]-factor.

10.1 Connected even [2, 2k]-factors

The main result in this section is the following.

Theorem 10.6 Let G be a connected graph of order n≥ 3 and let k be a pos-itive integer. If d.x1/ + d.x2/ + · · · + d.xk/ ≥ n for every k-independent set{x1; x2; · · · ; xk} of G such that x1; x2; · · · ; xk have a common neighbour, then Gcontains a connected even[2;2k−2]-factor if and only if G has a connected evenfactor.

One can easily derive the following corollaries.

Corollary 10.1 Let G be a connected graph of order n≥ 3 and let k≥ 2 be aninteger. If¦k.G/ ≥ n, then G contains a connected even[2;2k− 2]-factor if andonly if G has a connected even factor.

10.1 Connected even [2, 2k]-factors 131

Corollary 10.2 Let G be a connected graph of order n≥ 3 and let k be an integer.If d.x1/+d.x2/≥ 2n=k for any 2-independent set{x1; x2} in G such that they havea common adjacent vertex, then G contains a connected even[2;2k− 2]-factor ifand only if G has a connected even factor.

Proof. For anyk-independent set{x1; x2; · · · ; xk} in G such that they have a com-mon adjacent vertex, by the assumption,

d.xi /+ d.xj / ≥ 2n=k f or any{xi; xj} ⊆ {x1; x2; · · · ; xk}:

Hence 2.d.x1/+ d.x2/+ · · · + d.xk// ≥ 2n implies thatG satisfies the conditionof Theorem 10.6. So Corollary 10.2 follows from Theorem 10.6.

Theorem 10.6 is best possible with respect to the degree sum conditions. LetG.k/be the graph obtained by identifying exactly one vertex ofk complete subgraphsof order at least 3. One can easily see thatG.k/ is a connected graph such thatd.u1/+ d.u2/+ · · · + d.uk/ = n− 1 for everyk-independent set{u1;u2; · · · ;uk}such that they have a common neighbor, and thatG.k/ contains a connected evenfactor but no connected even [2;2k− 2]-factor.

Combining Theorem 10.4, Theorem 10.5 and Theorem 10.6, one can easily obtainthe following results.

Theorem 10.7 Let G be a 2-edge-connected graph on n≥ 3. If ¦2.G/ ≥ .2n+3/=3, then G has a connected even[2;4]-factor.

Theorem 10.8 Let G be a 2-edge-connected simple graph of order n≥ 6 with¦3.G/ ≥ n+ 1, then G contains a connected even [2, 4]-factor.

Combining Theorems 10.3 and 10.6, we obtain a Dirac-type condition for theexistence of a connected even [2;2k]-factor.

Theorem 10.9 Let G be a 2-edge-connected graph of order n and k∈ {3;4;5}.If Ž.G/ ≥ n=k, then G contains a connected even[2;2k− 2]-factor.

Note that Theorem 10.1 is the casek= 2 of Theorem 10.9. The graphG.k/ showsthat Theorem 10.9 is also best possible. We make the following conjecture.

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Conjecture 10.1 Let G be a 2-edge-connected graph of order n≥ 6 and letk ∈ {2;3;4;5}. If d.x1/ + d.x2/ + · · · + d.xk/ ≥ n for any k-independent set{x1; x2; · · · ; xk} in G such that x1; x2; · · · ; xk have a common neighbor, then Gcontains a connected even[2;2k− 2]-factor.

Before proving Theorem 10.6 we introduce some additional terminology and no-tation.For anyk-independent setT = {t1; t2; · · · ; tk} such that they have a common ad-jacent vertexswe denoteW.T : s/= ∪{i; j}⊆{1;2;···k}.NG.ti /∩ NG.t j //\{s}. For anyconnected even factor F ofG, we use the notationn.G; F;1/ to denote the num-ber of vertices inF with maximum degree1.F/. For anyS⊆ V.F/; we denoteM.S/ = {t ∈ V.G/ : dF.t/ = 2 andNF.t/ ⊆ S}.

If S⊆ V.G/, thenNG.S/ denotes the neighborhood ofS, i.e., the set of all verticesin V.G/\Swhich are adjacent to at least one vertex inS. Let

Ni.S/ = {t ∈ V.G/\V.S/ : |{u∈ .S/ : tu∈ E.G/}| = i} f or a positive integer i:

Then for anyk vertices setT = {t1; t2; · · · ; tk} of G, we have

N.T/ =k⋃

i=1

Ni.T/ and N0.T/ = V.G/\.T ∪ N.T//:


We only need to prove the sufficiency of Theorem 10.6. SupposeG has a con-nected even factor, and letF be a connected even factor ofG, with 1.F/ andn.G; F;1/ minimum. We will prove that1.F/ ≤ 2k− 2, i.e., F is a connectedeven [2;2k− 2]-factor. Otherwise, letw be a vertex withdF.w/ =1.F/ ≥ 6 andNF.w/ ⊇ {u1; v1;u2; v2; · · · ;uk; vk} such thatui; vi are in the same cycle ofF foreachi, respectively. Set

U = {u1; v1;u2; v2; · · · ;uk; vk}:Now we can choosek verticesx1; x2; · · · ; xk in U as a vertex setX such that

xi ∈ {ui; vi} and|M.X/| ≤ |M.U\X/|: (10.1)


By minimality of n.G; F;1/,

E.G[ X]/ = ∅: (10.2)

Otherwise, without loss of generality, we may assume thatx1x2 ∈ E.G/, so that

F′ = F + {x1x2} ∩ .E.G/\E.F//− {x1x2; x1w;wy2} ∩ E.F/

is a connected even factor, but thenn.G; F′;1/ = n.G; F;1/− 1, a contradic-tion. Hence{x1; x2; · · · ; xk} is a k-independent set ofG such that they have acommon adjacent vertexw. By the hypothesis of Theorem 10.6,

d.x1/+ d.x2/+ · · · + d.xk/ ≥ n: (10.3)

The following claims are necessary for our proof.

Claim 1. For u ∈W.X : w/, exactly one of the following holds:

(i) {ux1;ux2; · · · ;uxk} ∩ E.F/ = ∅ and|NF.u/| ≥ 2k− 2;

(ii) |{ux1;ux2; · · · ;uxk}∩ E.G/| = |{ux1;ux2; · · · ;uxk}∩ E.F/| = 2 and F−.{x1u; x2u; · · · ; xku} ∩ E.F// has two components.

Proof of Claim 1. If |{ux1;ux2; · · · ;uxk} ∩ E.F/| ≥ 3; say{ux1;ux2;ux3} ⊆E.F/ then, by (10.1),

F′ = F − {ux1;ux2; x1w; x2w}is a connected even factor withn.G; F′;1/ = n.G; F;1/− 1; a contradiction.

If there exist two edges in{ux1;ux2; · · · ;uxk}, sayux1;ux2, such thatux1 ∈ E.F/and ux2 ∈ E.G/\E.F/, then F′ = F + {ux2} − {ux1; x1w; x2w} is a connectedeven factor withn.G; F′;1/ = n.G; F;1/− 1, a contradiction.

Thus either{ux1;ux2; · · · ;uxk} ∩ E.F/ = ∅ or |{ux1;ux2; · · · ;uxk} ∩ E.G/| =|{ux1;ux2; · · · ;uxk} ∩ E.F/| = 2.

If {ux1;ux2; · · · ;uxk} ∩ E.F/ = ∅, say{ux1;ux2} ⊆ E.G/\E.F/, then (i) holds.Otherwise, sinceF is an even factor ofG, |NF.u/| ≤ 2k− 4: Hence,

F′ = F+ {ux1;ux2} − {x1w; x2w}

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is a connected even factor withn.G; F′;1/ = n.G; F;1/− 1; a contradiction.

If |{ux1;ux2; · · · ; xuk} ∩ E.G/| = 2 and|{ux1;ux2; · · · ;uxk} ∩ E.F/| = 2, say{ux1;ux2} ⊆ E.F/, then (ii) holds. Otherwise

F′ = F − {ux1;ux2; x1w; x2w}is a connected even factor withn.G; F′;1/ = n.G; F;1/− 1, a contradiction.This proves Claim 1.

SetV1 = {u ∈W.X : w/ : .i / holds}

andV2 = {u ∈W.X : w/ : .ii / holds and dF.u/ ≥ 4}:

Claim 2. For anyu ∈ V1, if there exist two verticesx ∈ {x1; x2; · · · ; xk} andy ∈ NF.u/ such thatxy ∈ E.G/, then xy ∈ E.F/ and {uy; yx} is an edge-cut-set ofF .

Proof of Claim 2. If there exist two verticesx ∈ {x1; x2; · · · ; xk} andy ∈ NF.u/;respectively, such thatxy∈ E.F/, then sinceu ∈ V1, without loss of generality,we may assume thatux1;ux2 ∈ E.G/\E.F/ andx 6= x1. Hence

F′ = F + {ux1; xy} − {uy; x1w; xw}

is a connected even factor withn.G; F′;1/ ≤ n.G; F;1/− 1, a contradiction.Hence,xy∈ E.F/.

If {uy; xy} is not an edge-cut-set ofF, then

F′ = F+ {ux1} − {uy; xy; xw; x1w}

is a connected even factor withn.G; F′;1/ ≤ n.G; F;1/− 1, a contradictionthat completes the proof of Claim 2.

Claim 3.|{y ∈ NF.u/ : dF.y/ = 2 andy ∈ NF.W.X : w/ ∪ X/\{u}}| ≤ k− 1 forany vertexu ∈ V1.


Proof of Claim 3. For any vertexu ∈ V1, we set

Y.u/ = {y ∈ NF.u/ : dF.y/ = 2 and y∈ NF..W.X : w/∪ X/\{u}}/ 6= ∅}:

By u∈ V1, without loss of generality, we may assume thatux1;ux2 ∈ E.G/\E.F/.

We will prove that|Y.u/| ≤ k− 1, i.e., Claim 3 is true. Otherwise,|Y.u/| ≥ k,say

{y1; y2; · · · ; yk} ⊆ Y.u/:

SetY= {y1; y2; · · · ; yk}.

If there exists an edge inG[Y], sayy1y2 ∈ E.G/\E.F/, then

F′ = F+ {ux1;ux2; y1y2} − {uy1;uy2; x1w; x2w}

is a connected even factor withn.G; F′;1/ ≤ n.G; F;1/− 1, a contradiction.

Hence{y1; y2; · · · ; yk} is ak-independent set ofG such that they have a commonadjacent vertexu. By the hypothesis of Theorem 10.6,

d.y1/+ d.y2/+ · · · + d.yk/ ≥ n: (10.4)

In a way similar to that used inthe proof of Claim 1, bydF.y/ = 2; for any y ∈ Y;we obtain the following.

|NF.v/| ≥ 2k− 2 for anyv ∈W.Y : u/: (10.5)

We can also obtain the following.

For anyv ∈W.Y : u/; there do not exist two vertices

z∈ NF.v/ andy ∈ {y1; y2; · · · ; yk} such thatyz∈ E.G/:(10.6)

Otherwise, without loss of generality, we may assume thaty1z ∈ E.G/; withz∈ NF.v/\Y. SincedF.y1/ = 2; y1z 6∈ E.F/.

By the definition ofW.Y : u/, we can choose a vertexy′ ∈ Y with vy′ ∈ E.G/\E.F/.Hence,

F′ = F+ {y1z; vy′;ux1;ux2} − {vz;uy1;uy′; x1w; x2w}

136

is a connected even factor withn.G; F′;1/ ≤ n.G; F;1/− 1, a contradiction.This implies that (10.6) is true.

SincedF.y/ = 2; for y ∈ {y1; y2; · · · ; yk},.U ∪ {w}\X/∩ NF.Y/ = ∅:

We can also prove the following.

.U ∪ {w}/\NF.y// ∩ NG.y/ = ∅ for y ∈ Y: (10.7)

If w ∈k⋃

i=1.NG.y1/\NF.yi // , sayw ∈ NG.yi /\NF.y1/, then

F′ = F+ {wy1;ux1;ux2} − {uy1; wy1; wy2}is a connected even factor withn.G; F′;1/ ≤ n.G; F;1/− 1, a contradiction.

If there exists a vertexy in Y such that.U\NF.y// ∩ NG.y/ 6= ∅, say u1y1 ∈E.G/\E.F/ andu1 6= x1, then

F′ = F+ {u1y1;ux2} − {u1w; x2w;uy1}is a connected even factor withn.G; F′;1/ ≤ n.G; F;1/− 1, a contradictionthat implies (10.7) is true.

If there exist three verticesW.Y : u/∪ V1; such that they have a common adjacentvertex inF, then in a way similar to that used in the proof of (10.7), we can find aconnected even factorF′ with n.G; F′;1/ ≤ n.G; F;1/− 1, a contradiction thatimplies the following.

There exist at most two vertices inW.Y : u/∪ V1 such that

they have a common adjacent vertex inF:(10.8)

Hence, by (10.5) to (10.8),

| ∪v∈W.Y:u/ .NF.v/\Y/| ≥ .2k− 2/|W.Y : u/|: (10.9)

Sinceu 6∈W.Y : u/∪ N1.Y/∪ N0.Y/; by (10.7)-(10.9) andY⊆ N0.Y/,

|N0.Y/| ≥ .k− 1/|W.Y : u/| + k+ 1: (10.10)


Sinceu 6∈W.Y : u/∪ N1.Y/∪ N0.Y/,

|W.Y : u/| + |N1.Y/| + |N0.Y/| = n− 1: (10.11)

By (10.10) and (10.11),

|N1.Y/| ≤ n− k|W.Y : u/| − k− 1: (10.12)

Hence by (10.10) and (10.12),

d.y1/+ d.y2/+ · · · + d.yk/ ≤ k|W.Y : u/| + |N1.Y/| + k≤ n− 2;

which contradicts (10.4). This completes the proof of Claim 3.

In a way similar to that used in the proof of Claim 3, we obtain

Claim 4. M.U\X/ ⊆ N0.X/.

By Claim 4 and (10.1), we can prove

Claim 5. |N0.X/\X| ≥ .k− 1/|V1| + |V2| + |M.X/|:

Proof of Claim 5. By the definition ofV2, one can easily see that

NF.V2/∩ N.X/ = ∅: (10.13)

By the definition ofV2,

|NF.v/\X| ≥ 2; for v ∈ V2: (10.14)

Obviously,.NF.v/∩ NF.v

′//\X = ∅; for {v; v′} ⊆ V2: (10.15)

In a way similar to that used in the proof of (ii) in Claim 1, we obtain

|{u ∈ V1 : yu∈ E.F/}| ≤ 2 for anyy ∈ NF.V1/\X; (10.16)

and if the equality in (10.16) holds, thenF− {yu∈ E.F/ : u ∈ V1} has two com-ponents.

138

Set

V3 = {y ∈ NF.V1/ : F − {yu∈ E.F/ : u ∈ V1∪ X}has two nontrivial components ofF}:

Obviously for anyy ∈ V3, we havedF.y/ ≥ 4, and by (ii) in Claim 2 and (10.16),

|{yu∈ E.F/ : F − {yu∈ E.F/ : u ∈ V1∪ X}has two nontrivial components ofF}| = 2:

Hence,|NF.v/\.N.X/ ∪ X/| ≥ 2for any vertexv ∈ V3: (10.17)

In a way similar that in the proof of (10.7), we obtain that

any vertex betweenNF.V1/\.N.X/ ∪ X/ and.NF.V2∪ V3//\.N.X/∪ X/

must have degree 2 inF:(10.18)

Using Claims 1-3 and (10.13)-(10.18), we obtain

|NF.V1∪ V2∪ V3/\.N.X/∪ X/| ≥ .k− 1/|V1| + |V2|: (10.19)

By Claim 1 and Claim 2, we have

NF.V1∪ V2∪ V3/\.N.X/∪ X/ ⊆ N0.X/: (10.20)

In a way similar to that used inthe proof of (ii), one can easily see that

M.U\X/∩ NF.V1∪ V2∪ V3/ = ∅: (10.21)

Hence, using Claim 4 and (10.1), (10.19)-(10.20), we obtain the proof of Claim 5.

SinceX ⊆ N0.X/, by Claims 4-5 , (10.1) and (10.19),

|N0.X/| ≥ .k− 1/|V1| + |V2| + |M.X/| + k: (10.22)

Sincew 6∈W.X : w/∪ N1.X/∪ N0.X/;

|W.X : w/| + |N1.X/| + |N0.X/| = n− 1: (10.23)

10.2 Every supereulerian claw-free graph contains a connected even [2,4]-factor139

Obviously|W.X : w/| = |V1| + |V2| + |M.X/|. Hence, by (10.22) and (10.23),

|N1.X/| ≤ n− 1− k|V1| − 2|V2| − 2|M.X/| − k: (10.24)

Obviously, |{x ∈ X : t ∈ M.X/ and xt ∈ E.G/}| = 2 for any t ∈ M.X/: Hnceusing Claim 1, we obtain

d.x1/+ d.x2/+ · · · + d.xk/ ≤ k|V1| + 2|V2| + 2|M.X/| + |N1.X/| + k:

Hence, by (10.22) and (10.24),

d.x1/+ d.x2/+ · · · + d.xk/ ≤ n− 1;

which contradicts (10.3) .

10.2 Every supereulerian claw-free graph containsa connected even [2,4]-factor

In a way similar to that used in the proof of of Theorem 10.6, we show the follow-ing theorem is true.

Theorem 10.10Every supereulerian claw-free graph G contains a connectedeven[2;4]-factor.

Proof. Let G be a supereulerian claw-free graph. We can letF be a connectedeven factor ofG with1.F/ andn.G; F;1/minimum. We will prove that1.F/≤4, i.e., F is a connected even [2;4]-factor. Otherwise, letw be the vertex withdF.w/ =1.F/ ≥ 6 andNF.w/ ⊇ {u1; v1;u2; v2;u3; v3} such thatui; vi are in thesame cycle ofF for eachi; respectively. By the claw-freeness ofG, there existsan edgeuiv j ∈ E.G/ with i 6= j , say,u1v2 ∈ E.G/. Hence,

F′ = F + .{u1v2} ∩ .E.G/\E.F///− .{u1v2;u1w;wv2} ∩ E.F//

is a connected even factor, butn.G; F′;1/ = n.G; F;1/− 1, a contradiction.

In a way similar to that used in the proof of Theorem 10.10, we obtain the follow-ing result.

140

Theorem 10.11Every supereulerian K1;s-free graph contains a connected even[2;2s− 2/]-factor for s≥ 2.

From the construction ofG.k/; it follows that there exists a family of graphs thathave any large edge-connectivity but no connected even [2;2k]-factor. The situ-ation is quite different if we consider claw-free graphs. Although from the con-struction ofG.k/ it also follows that there exists a family of graphs such that theyhave any large edge-connectivity but no connected even [2;4]-factor, we showthat there exists a small edge-connectivity guarantee that the claw-free graph hasa connected even [2;4]-factor. It is known that every 4-edge-connected graph issupereulerian [33]. Hence the following theorem follows from Theorem 10.10.

Theorem 10.12Every 4-edge-connected claw-free graph has a connected even[2;4]-factor.

We give a similar example as in [27] to show that Theorem 10.12 is best possiblein term of the edge-connectivity. LetP10 denote the Petersen graph and letSP10

denote the graph obtained fromP10 by replacing each vertex ofP10 by a clique oforder at least 4 and replacing each edge ofP10 by a path of length 2, then the linegraphL.SP10/ of SP10 is 3-edge-connected and claw-free butL.SP10/ containsno connected even [2, 4]-factor.

It follows that ak-edge-connected claw-free graph can be one connected graphfrom the graph which is obtained by identifying exactly two vertices one vertexof two cliques of order at leastk+ 1; respectively, for any integerk ≥ 2. So thefollowing known result [12] is a very weaker consequence of Theorem 10.12 sinceany 4-connected graph is 4-edge-connected.

Corollary 10.3 (Broersma et al. [12])Every 4-connected claw-free graph has a connected even[2;4]-factor.

Chapter 11

A note on minimum degreeconditions for supereulerian graphs

We use [8] for terminology and notation not defined here and consider finite loop-less graphs only. As special notation and terminology. LetG be a graph. We use½.G/, 1.G/ andŽ.G/ to denote the edge-connectivity, the maximum degree andthe minimum degree ofG, respectively. IfE.G/ 6= ∅, thenthe edge degreeof G,denoted by²2.G/, is defined as min{d.x/+ d.y/ | xy∈ E.G/}. Let O.G/ denotethe set of all vertices ofG with odd degrees. A eulerian graph is a connected graphG with O.G/ = ∅ (henceK1 is a eulerian graph). A graph is calledsupereulerianif it has a spanning eulerian subgraph. A subgraphH of a graphG is dominatingif G− V.H/ is edgeless, i.e. if every edge ofG is incident with at least one vertexof H.

Theline graph of a graphG, denoted byL.G/, hasE.G/ as its vertex set, wheretwo vertices inL.G/ are adjacent if and only if the corresponding edges inG areadjacent. There is a close relationship between a dominating eulerian subgraph inG and hamiltonian cycles inL.G/.

Theorem 11.1 (Harary and Nash-William [32]) Let G be a graph with|E.G/| ≥3. Then L.G/ is hamiltonian if and only if G has a dominating eulerian subgraph.

Various sufficient conditions for the existence of supereulerian graphs and domi-nating eulerian subgraphs in terms of²2.G/ have been derived (See, e.g. [3],[16]–[19], [24]–[27]).

141

142

From Theorem 11.1 one easily sees that a supereulerian graph has a hamiltonianline graph. Simple examples show that not every graph with a hamiltonian linegraph is supereulerian [17]. Veldman proved the following, which was conjec-tured in [3]. HereD1.G/ denotes the set of vertices ofG with degree one.

Theorem 11.2 (Veldman[54]) If G is a simple graph of order n with½.G−D1.G// ≥ 2 and if

²2.G/ >25

n− 2; (11.1)

then, for n sufficiently large, L.G/ is hamiltonian.

If (11.1) holds, then we have

min{max{d.x/;d.y/} | xy∈ E.G/} > 15

n− 1: (11.2)

Therefore, it is natural to investigate whether (11.1) can be replaced by (11.2). Laiasked this problem. He obtained the following result with a slightly better lowerbound.

Theorem 11.3 (Lai[38]) If G is a simple graph of order n with½.G− D1.G// ≥2 and ifmin{max{d.x/;d.y/} | xy∈ E.G/} ≥ n

5−1, then, for n sufficiently large,L.G/ is hamiltonian, unless G is in a class of well-characterized graphs.

Since in this chapter the above result plays a minor role, we refrain from explic-itly describing the exceptional graphs. We only mention here that the exceptionalgraphs can be contracted to one of seven graphs, includingK2;3 andK2;5, in a waysimilar to that described in the next two sections.

It is in its turn natural to investigate whether the minimum degree condition in theabove theorem (combined with a necessary condition for the existence of a span-ning eulerian subgraph, e.g.½.G/ ≥ 2) guarantees a spanning eulerian subgraphin G instead of a dominating eulerian subgraph. This is indeed the case. We showthat in fact a slightly weaker condition is sufficient for a 2-edge-connected graph,with minimum degree at least 4, to be supereulerian, with again some exceptionalclasses.

11.1 Catlin’s reduction method 143

Theorem 11.4 Let G be a simple graph with½.G/ ≥ 2 and with n> 12vertices.If Ž.G/ ≥ 4 and if

min{max{d.x/;d.y/} | xy∈ E.G/} ≥ n− 25− 1; (11.3)


(a) G is supereulerian;

(b) The reduction G′ of G is isomorphic to K2;5 such that each pre-image of avertex with degree 2 in G′ has exactly order.n− 2/=5 in G;

(c) The reduction G′ of G is isomorphic to K2;3 such that each vertex of G′

corresponds to a pre-image in G of order at leastn−25 .

Before we present a proof of this result as well as related results, we have to definewhat we mean by the reduction of a graphG. For this purpose we give a shortdescription of Catlin’s reduction method in Section 11.1. We present our resultsand proofs in Section 3. Our main result (Theorem 11.7 in Section 11.2) impliesseveral known and new results on dominating eulerian subgraphs and supereule-rian graphs of minimum degree at least 4. The proofs are similar to the proofsof Catlin and Li in [21]. In Section 11.3 we show that we cannot relax our lowerbound four on the minimum degree in the above result.

11.1 Catlin’s reduction method

Let G be a graph and letH be a connected subgraph ofG. G=H denotes thegraph obtained fromG by contractingH, i.e. by replacingH by a vertexvH suchthat the number of edges inG=H joining anyv ∈ V.G/− V.H/ to vH in G=Hequals the number of edges joiningv in G to H. A graphG is contractible to agraphG′ if G contains pairwise vertex-disjoint connected subgraphsH1; : : : ; Hk

withk⋃

i=1V.Hi/ = V.G/ such thatG′ is obtained fromG by successively con-

tracting H1; H2; : : : ; Hk. The subgraphHi of G is called the pre-image of thevertexvHi of G′; the vertexvHi is called trivial if Hi contains precisely one vertex.i = 1;2; : : : ; k/. A graph iscollapsibleif, for every even subsetX of V.G/; thereexists a spanning connected subgraphGX of G such thatX = O.GX/. In particu-lar, K1 is collapsible. Note that any collapsible graphG is supereulerian since∅ is

144

an even subset ofV.G/. Catlin [19] showed that every graphG has a unique col-lection of pairwise vertex-disjoint maximal collapsible subgraphsH1; H2; : : : ; Hk

such thatk⋃

i=1V.Hi/ = V.G/. The reduction ofG is the graph obtained fromG by

successively contractingH1; H2; : : : ; Hk. A graph is reduced if it is the reductionof some graph.

The following results from [19] are necessary for the proofs of our results.

Theorem 11.5 (Catlin[19]) Let G be a connected graph and let G′ be the reduc-tion of G. Then G is supereulerian if and only if G′ is supereulerian.

Theorem 11.6 (Catlin[19]) Let G be a nontrivial graph and let V3= {v ∈ V.G/ |d.v/ ≤ 3}. If G is a reduced graph, then each of the following holds:

(a) G is a simple graph.

(b) G has no cycle of length less than four.

(c) If ½.G/ ≥ 2, then either|V3| = 4 and G is eulerian or|V3| ≥ 5.

11.2 Main result and its consequences

In our proof of the following theorem, we use Catlin’s reduction method. Theorem6 is the key to our proof.

Theorem 11.7 Let G be a simple graph of order n with½.G/ ≥ 2. If for everyminimal edge cut S⊆ E.G/; with |S| ≤ 3;we have that every component of G− Shas order at least.n− 2/=5> 2, then exactly one of the following holds:


(b) The reduction G′ of G is isomorphic to K2;5 such that each pre-image of avertex with degree 2 in G′ has exactly order.n− 2/=5 in G.

(c) The reduction G′ of G is isomorphic to K2;3 such that each vertex of G′

corresponds to a pre-image in G with order at least.n− 2/=5.

11.2 Main result and its consequences 145

Proof. Let G′ be the reduction ofG. If G′ = K1, then G is supereulerian byTheorem 5. Next supposeG′ 6= K1. ThenG′ is 2-edge-connected and nontrivial.By (c) of Theorem 11.6, it is sufficient to consider the case that|V3| ≥ 5. Letv1; v2; : : : ; v5 be vertices ofV.G′/ in V3, i.e., d.vi/ ≤ 3 for eachi. The corre-sponding pre-images areH1; H2; : : : ; H5. EachHi is joined to the rest ofG bya minimal edge cut consisting ofd.vi/; d.vi/ ≤ 3; edges. By the hypothesis ofTheorem 11.7,|V.Hi/| ≥ .n− 2/=5 and so

n= |V.G/| ≥5∑

i=1

|V.Hi/| ≥ n− 2:

Hence,5≤ |V.G′/| ≤ 7:

If |V3| ≥ 6, we would similarly obtainn≥ 65.n− 2/, hencen≤ 12, contradicting

65.n− 2/ > 12. So|V3| = 5.

We distinguish three cases to complete the proof.

Case 1.|V.G′/| = 5.

By (b) of Theorem 11.6 and since½.G′/ ≥ 2,1.G′/ ≤ 3 and there exist at mosttwo vertices ofG′ with degree 3.G′ cannot have exactly one vertex of degreethree. HenceG′ has exactly two vertices of degree 3, and, by (b) of Theorem 11.6and since½.G′/ ≥ 2, G′ = K2;3. In this case,G′ satisfies (c) of Theorem 11.7.

Case 2.|V.G′/| = 6.

Let u ∈ V.G′/\{v1; v2; : : : ; v5}.By (b) of Theorem 11.6 and sinceG′ is 2-edge-connected,dG′.u/ = 4. LetNG′.u/ = {v1; v2; v3; v4}. SinceŽ ≥ ½ ≥ 2, we obtain thatv1v5; v2v5; v3v5; v4v5 ∈E.G′/. ThusG′ is eulerian. SoG is supereulerian by Theorem 11.5.

Case 3.|V.G′/| = 7.

Let {u; v} = V.G′/\{v1; v2; : : : ; v5}. ClearlydG′.u/ ≥ 4 anddG′.v/ ≥ 4.

By (b) of Theorem 11.6,uv 6∈ E.G′/. Hence

|NG′.u/∩ NG′.u/| ≥ 3:

146

Since½.G′/ ≥ 2 andG′ contains no 3-cycle,

|NG′.u/∩ NG′.v/| 6= 4:

We distinguish the following two subcases.

Subcase 3.1.|NG′.u/∩ NG′.v/| = 3.

Without loss of generality, we assume that

NG′.u/∩ NG′.v/ = {v1; v2; v3}; v4 ∈ NG′.u/ and v5 ∈ NG′.v/:

Since G′ is 2-edge-connected and nontrivial, by (b) of Theorem 11.6,v4v5 ∈E.G′/. HenceG′ is eulerian, implying thatG is supereulerian by Theorem 11.5.

Subcase 3.2.|NG′.u/∩ NG′.v/| = 5.

By (b) of Theorem 11.6,G′ = K2;5. Now G′ satisfies (b) of Theorem 11.7.We now present some consequences of the main Theorem 11.7.

Corollary 11.1 Let G be a simple graph of order n with½.G/ ≥ 2. If for everyminumal edge cut S⊆ E.G/ with |S| ≤ 3 we have that every component of G− Shas order greater thann5, then G is supereulerian.

Proof. Let G satisfy the hypothesis of Corollary 11.1. ThenG satisfies the hy-pothesis of Theorem 11.7 and satisfies neither (b) nor (c) of Theorem 11.7. SoGis supereulerian.

Next we can present a proof of Theorem 11.4.

Proof of Theorem 11.4.Let G be a graph satisfying the hypothesis of Theorem11.4. It is sufficient to show thatG satisfies the hypothesis of Theorem 11.7.Let S be a minimal edge cut ofG with |S| ≤ 3, and letG1 and G2 be the twocomponents ofG− S with |V.G1/| ≤ |V.G2/|. It is sufficient to prove that|V.G1/| ≥ .n− 2/=5. SinceŽ.G/ ≥ 4, G1 has at least an edge, sayuv, suchthat both ofu; v are not incident with any edge ofS. By (11.3),

|V.G1/| ≥max{d.u/;d.v/} + 1≥ n− 25− 1+ 1= n− 2

5:

ThusG satisfies the hypothesis of Theorem 11.7.

Obviously, Theorem 11.4 improves the following result (for graphs on more than12 vertices).

11.2 Main result and its consequences 147

Theorem 11.8 (Catlin et al.) [21]) Let G be a simple graph of order n≥ 12with½.G/ ≥ 2. If Ž.G/ ≥ 4 and if

²2.G/ ≥ 2n5− 2;



(b) The reduction G′ of G is isomorphic to K2;3 such that each vertex of G′

corresponds to a pre-image in G with order exactly n=5.

We present some other consequences of Theorem 11.4 and Theorem 11.7.

Corollary 11.2 Let G be a simple graph of order n> 12with Ž.G/ ≥ 4. If L.G/is 4-connected, then G is supereulerian.

Proof. One easily checks thatG satisfies the hypothesis of Theorem 11.7 and itsatisfies neither (b) nor (c) of Theorem 11.7. SoG is supereulerian.

Corollary 11.2 supports the conjecture due to Thomassen that every 4-connectedline graph is hamiltonian. A related result by Jaeger implies that Corollary 11.2is true even without the assumption on the order, since if the line graphL.G/ of agraphG with Ž.G/ ≥ 4 is 4-connected, thenG itself must be 4-edge-connected.

Theorem 11.9 (Jaeger) [33] Every 4-edge-connected graph is supereulerian.

The final consequence of our main results we want to mention is the following.

Corollary 11.3 (Cai) [14] Let G be a 2-edge-connected simple graph of ordern> 20. If Ž.G/ > n

5 − 1, then either G is supereulerian or the reduction of G isisomorphic to K2;3, where every pre-image of the vertices of K2;3 is either Kn

5or

Kn5− e.

Proof. Let G satisfy the hypothesis of this corollary. ThenG satisfies the hy-pothesis of Theorem 11.5 and does not satisfy (b) of Theorem 11.4. SinceŽ.G/ > n

5 − 1; every pre-image of the vertices ofK2;3 is eitherKn5

or Kn5− e.

Corollary 11.3 follows.

148

11.3 Remarks on minimum degree condition

From our main results and its consequences, one may wonder whether the mini-mum degree conditionŽ.G/ ≥ 4 is crucial or not for our conclusions. One mightexpect that the same conclusions hold without this restriction on the minimumdegree. However, there exist graphs with a large minimum degree that are notsupereulerian.K2;3 with the vertices replaced by large complete subgraphs is suchan example that appears in Corollary 11.3. As remarked in the previous section,Corollary 11.2 supports the conjecture due to Thomassen that every 4-connectedline graph is hamiltonian. More recently, Broersma, Kriesell and Ryj´acek [13]have shown that this conjecture is equivalent to seemingly weaker conjectures inwhich the conclusion is replaced by the conclusion that there exists a spanningsubgraph consisting of a bounded number of paths. From Corollary 11.2, onemight expect that a stronger conjecture holds, namely that if the line graph of agraphG is 4-connected, thenG is supereulerian. But this is not true:K2;n−2 (withn ≥ 7 odd) is an exception. Similarly, we have examples showing that the min-imum degree restrictionŽ.G/ ≥ 4 is necessary for results of the above type forsupereulerian graphs (and hamiltonian line graphs).

Minimum degree 3

The following result on graphs with minimum degree at least 3 has been obtainedby Veldman.

Theorem 11.10 (Veldman [54]) Let G be a 2-edge-connected simple graph oforder n such thatŽ.G/ ≥ 3 and

²2.G/ ≥ 2.17

n− 1/:

If n is sufficiently large, then either G is supereulerian or G is contractible to K2;3.

Comparing the above result with Theorem 11.2, the condition of Theorem 11.10is considerably weaker, but one has to exclude all graphs that are contractible toK2;3. One might expect that the condition in (11.2) can be used instead, with thesame exceptional graphs related toK2;3. This is not the case. We can constructmany other exceptional graphs. See Figure 11.1 for a class of examples. Here theblack vertices represent large complete subgraphs, e.g. all isomorphic toKn−2

4.

Minimum degree 2

Within the class of graphs with minimum degree at least four, Theorem 11.4 im-proves the following best possible result of Catlin [16].

11.3 Remarks on minimum degree condition 149

Figure 11.1: Not supereulerian graph, with minimum degree 3.

Theorem 11.11 (Catlin [16]) Let G be a 2-edge-connected simple graph of ordern such that

²2.G/ ≥ 23.n+ 1/:

Then either G is supereulerian or G is isormorphic to K2;n−2 and n is odd.

Without a restriction on the minimum degree, we can construct many graphsGwith a large lower bound on min{max{d.x/;d.y/} | xy∈ E.G/}, but such thatGis not supereulerian. In Figure 11.2 we give a class of examples. Here we choosen and devide the vertices in such a way that the four vertices with degree largerthan 2 get degreen−4

2 ≡ 1.mod2/:More generally, one can start with aP3, P4, C3 or C4 and replace all edges by anumber of internally-disjoint paths of length 2 in such a way that

(i) the resulting graph is 2-edge-connected,

(ii) at least one (hence two) of the vertices of the original graph gets an odddegree and

(iii) all vertices of the original graph get a degree at leastn2 −3; wheren denotes

the number of vertices of the resulting graph after the replacement of edgesby paths.

Note that in case one starts with aC3, the degree of the vertices of the originalC3

could be made close to2n3 :

150

odd degree (n−4)/2

Figure 11.2: Not supereulerian graph, with minimum degree two.

Bibliography

[1] R. Balakrishnan and P. Paulraja,Line graphs subdivision graphs, Journal ofCombinatorics, Information and System Sciences, Vol.10, Nos.1-2 (1985)33-35.

[2] L.W. Beineke,Characterization of derived graphs, J. Comb. Theory, 9(1970) 129-135.

[3] A. Benhocine, L. Clark, N. K¨ohler, H.J. Veldman,On circuits and pancyclicline graphs, J. Graph Theory, 10 (1986) 411-425.

[4] A.A. Bertossi,The edge hamiltonian path problem is NP-complete, Inform.Process. Lett., 13 (1981) 157-159.

[5] E. Van Blanken, J. van den Heuvel and H.J. Veldman,Pancyclicity of hamil-tonian line graphs, Discrete Math., 138 (1995) 379-385.

[6] J.A. Bondy,Pancyclic graphs I, J. Combin. Theory B, 11 (1971) 80-84.[7] J.A. Bondy,Pancyclic graphs, Proc. Second Louisiana Conference on Com-

binatorics, Graph Theory and Computing (R.C. Mullin et al., eds.), Con-gressus Numerantium 3(1971) 181-187.

[8] J.A. Bondy and U.S.R. Murty,Graph Theory with Applications, AmericanElsevier, New York, (1996).

[9] S. Brandt and H.J. Veldman,Degree sums for edges and cycle lengths ingraphs, J. Graph Theory, 25 (1997) 253-256.

[10] S. Brandt, R. Faudree and W. Goddard,Weakly pancyclic graphs, J. GraphTheory, 27 (1998) 141-173.

[11] R.A. Brualdi and R.F. Shong,Hamiltonian line graphs, J. Graph Theory, 5(1981) 304-307.

[12] H.J. Broersma,Subgraph conditions for dominating circuits in graphs andpancyclicity of line graphs, Ars Combinatoria, 23 (1987) 5-12.

[13] H.J. Broersma, M. Kriesell and Z. Ryj´acek,On factors of 4-connected claw-free graphs, to appear in J. Graph Theory.

[14] X.T. Cai, A sufficient condition of minimum degree for a graph to have an

151

152 Bibliography

S-circuit, Chin. Ann. Math., 4A (1989) 117-122.[15] M. Capobianco and J.C. Molluzzo,Examples and counterexamples in graph

theory,Elsevier North-Holland, 1978.[16] P.A. Catlin,Spanning eulerian subgraphs and matchings, Discrete Math.,

76 (1989) 95-116.[17] P.A. Catlin,Supereulerian graphs, a survey, J. Graph Theory, 16 (1992)

177-196.[18] P.A. Catlin,Spanning trails, J. Graph Theory 11, (1987) 161-167.[19] P.A. Catlin,A reduction method to find spanning eulerian subgraphs, J.

Graph Theory, 12 (1988) 29-45.[20] P.A. Catlin, Iqbalunnisa, T.N. Janakiraman and N. Srinivasan,Hamilton

cycles and closed trails in iterated line graphs, J. Graph Theory, 14 (1990)347-364.

[21] C.A. Catlin and X.W. Li,Supereulerian graphs of minimum degree at least4, Advances in Math., 28 (1999) 65-68.

[22] G. Chartrand,On hamiltonian line graphs, Trans. Amer. Math. Soc., 134(1968) 559-566.

[23] G. Chartrand and C.E. Wall,On the hamiltonian index of a graph, StudiaSci. Math. Hung., 8 (1973) 43-48.

[24] Z.-H. Chen,A degree condition for spanning eulerian subgraphs, J. GraphTheory, 17 (1993) 5-21.

[25] Z.-H. Chen,Supereulerian graphs and the Petersen Graph, J. of Comb.Math. and Comb. Computing, 9 (1991) 70-89.

[26] Z.-H. Chen and L.-J Lai,Reduction techniques for supereulerian graphsand related topics- a survey. In: Combinatorics and Graph Theory’95, Pro-ceedings of the Summer School and International Conference on Combina-torics. Hefei.53-69.

[27] Z.-H. Chen, H.J. Lai and G.Q. Weng, Jackson’s conjecture on eulerian sub-graphs, in: Combinatorics, Graph Theory, Algorithms and Applications,Proceedings of the third China-USA international Conference, Beijing, June1-5, (1993), 53-58.

[28] L. Clark,On hamiltonian line graphs, J.Graph Theory, 8 (1984) 303-307.[29] G.A. Dirac,Some theorems on abstract graphs, Proc. London Math. Soc.,

(3) 2 (1952) 69-81.[30] L. Euler, Solutio problematis ad geometriam situs pertinentis. Comment.

Academiae Sci. I. Petropolitanae, 8 (1736) 128-140.[31] W.R. Hamilton,Letter to John T. Graves on the Icosian, 19 Oct.1856.

In: (H. Halberstam and R.E. Ingram, eds.), The mathematica papers of

Bibliography 153

Sir William Rowan Hamilton, Vol. (Algebra). Cambridge University Press,(1931) 612-625.

[32] F. Harary and C.St. J.A. Nash-Williams,On eulerian and hamiltoniangraphs and line graphs, Canad. Math. Bull., 8 (1965) 701-709.

[33] F. Jaeger,A note on subeulerian graphs, J. Graph Theory, 3 (1979) 91-93.[34] S.F. Kapoor and M.J. Stewart,A note on the hamiltonian index of a graph,

Studia Sci. Math. Hungar. 8 (1973) 307-308.[35] M.-K. Kuan, Graphic programming using odd or even points, Chinese

Math., 1 (1962) 273-277.[36] H.J. Lai,On the hamiltonian index, Discrete Math., 69(1988) 43-53.[37] H.J. Lai,Complementary supereulerian graphs, J. Graph Theory, 17 (1993)

263-273.[38] H.-J. Lai, Eulerian subgraphs containing given vertices and hamiltonian

line graphs, Discete Math., 178 (1998) 93-107.[39] L. Lesniak-Foster and J.E. Williamson,On spanning and dominating cir-

cuits in graphs. Can. Math. Bull., 20 (1997) 215-220.[40] C.F. Liu, L.C. Zhou and H. Liu,A result on hamiltonian line graphs, Acta

Math. Appl. Sincia, Vol.9, No.1 (1986) 81-84.[41] M.M. Mattews and D.P. Summer,Hamiltonian results in K1;3-free graphs,

J. Graph Theory, 8 (1984) 139-146.[42] T.A. Mckee,Recharacterizing eulerian: intimations of new duality, Dis-

crete Math., 51 (1984) 237-242.[43] H.M. Mulder, Julius Petersen’s theory of regular graphs,Discrete Math.,

100 (1992) 157-175.[44] H.M. Mulder,To see the history for the trees,Congressus Numerantium 64

(1988) 26-43.[45] L. Nebesk´y, On eulerian subgraphs of complementary graphs, Czechoslo-

vak Mathematics Journal, 29 (1979) 298-302.[46] O. Ore,Note on hamiltonian circuits, Amer. Math. Monthly, 67(1960) 55.[47] O. Ore,Graphs and subgraphs, Trans. Am. Math. Sol., 84(1957) 109-136.[48] M.L. Sarazin, On the hamiltonian index of a graph, Discrete Math., 122

(1993) 373-376.[49] L. Sarazin, A simple upper bound for the hamiltonian index of a graph,

Discrete Math., 134 (1994) 85-91.[50] E.A. Nordhaus and J.W. Gaddum,On complementary graphs, Amer.Math.

Monthly, 63 (1956)175-177.[51] W.J. Song, X.W. Li and T.X. Yao,A sufficient condition for pancyclicity of

line graph, J. Nanjing University, 27 (1991) 35-37.

154 Bibliography

[52] C. Thomassen,Reflections on graph theory, J. Graph Theory, 10 (1986)309-324.

[53] H. Trommel, H.J. Veldman and A. Verschut,Pancyclicity of claw-free hamil-tonian graphs, Discrete Math., 197/198 (1999) 781-789.

[54] H.J. Veldman,On dominating and spanning circuit in graphs, DiscreteMath., 124 (1994) 229-239.

[55] H.J. Veldman,A result on hamiltonian line graphs involving restrictions oninduced subgraphs, J. Graph Theory, 12 (1988) 413-420.

[56] Whitney,On the abstract properties of linear dependence, Amer. J. Math.,57 (1935) 509-533.

[57] L. Xiong, Edge degree conditions for subpancyclicity in line graphs, Dis-crete Math., 188 (1998) 225-232.

[58] L. Xiong, On subpancyclic line graphs, J. Graph Theory, 27 (1998) 67-74.[59] L. Xiong, The hamiltonian index of a graph,Graphs and Combinatorics,

To appear.[60] L. Xiong and Z. Liu,Hamiltonian iterated line graphs, preprint.[61] L. Xiong, H.J. Broersma, C. Hoede and Xueliang Li, Degree sum and sub-

pancyclicity in line graphs, Discrete Math. to appear.[62] L.Xiong, J. Wang, X. Dai,Pancyclicity of hamiltonian line graphs, J. Jiangxi

Normal University, 19 (1995) 140-148.[63] S. Xu,Relations between parameters of a graph, Discrete Math., 89 (1991)

65-88.[64] L.C. Zhao and C.F. Liu,A sufficient condition for hamiltonian line graphs,

Math. Applicata, Vol.3., No.1 (1990) 22-26.[65] S.M. Zhan, hamiltonian connectedness of line graphs, Ars Combinatoria

22 (1986) 89-95.[66] S.M. Zhan,On hamiltonian line graphs and connectivity, Discrete Math.,

89 (1991) 89-95.

Index

contraction 39attach-contraction 5block 28

acyclic 28cyclic 28split block 28

branch 3-cut 4-bond 4length of a branch-bond 4

Chinese postman problem 1circuit 1

eulerian 1circumference 63edge degree 141diameter 63distance 15cycle

hamiltonian 1factor 129

k- 129connected even 129connected even [2, k] 129

collapsible 18girth 63graph 1

complement of 56hamiltonian 1eulerian 1line 2

subpancyclic 8supereulerian 13pancyclic 8iterated line 2dominating subgraph 141

hamiltonian index 2reduction method 18independent 129

set 129trail 129

closed 5Voyage around the world 2

155

156 INDEX

Summary

This thesis contains some results on circuits and the hamiltonian index of a graph.In the first part of the thesis, Chapter 2 through 4, we concentrate on the topic ofthe hamiltonian index of a graph. In the second part, Chapters 5 through 9, wefocus on the topic of the degree sum along paths for subpancyclic line graphs.The results in these chapters are all related to line graphs and are best possible.In Chapter 10, we obtain some results on so-called connected even factors withdegree restriction. In the last chapter, we obtain some degree conditions for su-pereulerian graphs. These results are also best possible.

In Chapter 1 we present an introduction to the topics of this thesis and give anoverview of the main results obtained, together with some connections with olderresults. Specific terminology and notation can be found just before each topic.

In Chapter 2, we give a characterization of graphs with hamiltoniann-iterated linegraphs forn ≥ 2. Using it we present two methods for determining the hamil-tonian index of a graph. We also obtain some upper bounds for the hamiltonianindex.

In Chapter 3, we show another reduction method for determining the hamiltonianindex of a graph. Using thus we present an upper bound and a lower bound for it.The distance of the two bounds is at most two.

Sarazin showed that the hamiltonian index of a graphG; other than a path, is atmost the number of vertices ofG minus the maximum degree ofG: In Chapter 4,we improve the upper bound to be the diameter ofG minus one. We also presenta relation between the hamiltonian index of a graph and its complement.

In Chapter 5, we give some conditions of edge degree sums for subpancyclic line

157

158 Summary

graphs. This chapter deals with girth at least 4. The results for girth 5 differ con-siderally from that for girth 4. Our results show that even for graphs with largegirth the degree condition is also best possible. We generalize the results in Chap-ter 4 to more general graphs in Chapter 6.

In Chapter 7, we give conditions of degree sums of vertices along paths of length3 and 4. In Chapter 8 we deal with a condition of degree sums along a path oflength 4. The results in Chapter 8 show that the condition involving degree sumsof the vertices along a 4-path, which ensures that its line graph is subpancyclic,will be much stronger, when we exclude an exceptional graph. In fact the condo-tion is then almost as strong as the condition involving degree sums of the verticesalong a 3-path.

In Chapter 9 we show that every line graph of a graph with diameter at most 2is either pancyclic or a cycle of length 4 or 5. This improves the results of Veld-man. We also show that every line graph of a graphG such that any pair of edgeswith distance between them at most 1 is either pancyclic or a cycle of length 4 or 5.

In Chapter 10 we show that if the degree sum of anyk vertices in a independentset, such that they have a common neighbor, is at least the number of vertices ofthe graphG, thenG contains a connected even [2;2k− 2]-factor if and only ifGhas a connected even factor. This result improves and generalizes some old results.

In Chapter 11 we present a min-max degree sum condition for supereuleriangraphs.

Curriculum Vitae

The author of this thesis was born on April 8, 1965, at Anyi in Jiangxi Provincein the south of China. From 1977 to 1983 he attended grammar school at Jing’anin Jiangxi Province. From September 1983 to July 1987, he studied mathematicsat the Jiangxi Normal University. He worked as a teacher at the Department ofMathematics in the Yinchun Normal College. In September 1990, he started tostudy graph theory and its applications under the supervision of Zhangxiang Li atthe University of Science and Technology Beijing. His master’s thesis, entitledThe Vertex Arboricity of a Graph,was awarded as“excellent master’s degreepaper” by the same university. After obtaining the master’s degree, he startedto work at the Department of Mathematics in the Jiangxi Normal University in1993. In 1998, he was appointed associate professor. He has finished a project ofthe Natural Science Fund of Jiangxi Province.

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Circuits in Graphs and the Hamiltonian Index - LISA · Circuits in Graphs and the Hamiltonian Index...

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Transcript of Circuits in Graphs and the Hamiltonian Index - LISA · Circuits in Graphs and the Hamiltonian Index...