Eulerian Graphs and Related Topics: Part 1, Volume 1 (Annals of Discrete Mathematics, Volume 45)

EULERIAN GRAPHS AND RELATED TOPICS Part 1, Volume 1 ANNALS OF DISCRETE MATHEMATICS 45 General Editor: Peter L. HAMM...

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EULERIAN GRAPHS AND RELATED TOPICS Part 1, Volume 1

ANNALS OF DISCRETE MATHEMATICS

45

General Editor: Peter L. HAMMER Rutgers University, New Brunswick, NJ, U.S.A.

Advisory Editors: C. BERGE, Universite de Paris, France M.A. HARRISON, University of California, Berkeley, CA, U.S.A. K KLEE, University of Washington, Seattle, WA, U.S.A. J.H. VAN LINT California Institute of Technology,Pasadena, CA, U.S.A. G.C. ROTA, Massachusetts Institute of Technology,Cambridge, M.A., U.S.A.

EULERIAN GRAPHS AND RELATED TOPICS Part 1, Volume 1

Herbert FLEISCHNER Institute for Information Processing Austrian Academy of Sciences Vienna, Austria

1990

NORTH-HOLLAND -AMSTERDAM

NEW YORK

OXFORD

TOKYO

ELSEVIER SCIENCE PUBLISHERS B.V. Sara Burgerhartstraat 25 P.O. Box 21 1, 1000 AE Amsterdam, The Netherlands Distributors for the U.S.A. and Canada:

ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 655 Avenue of the Americas New York, N.Y. 10010, U.S.A.

ISBN: 0 444 88395 9

0 ELSEVIER SCIENCE PUBLISHERS B.V., 1990 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the publisher, Elsevier Science Publishers 6.V. / Physical Sciences and Engineering Division, PO. Box 703, 1000 AC Amsterdam, The Netherlands. Special regulations for readers in the U.S.A. - This publication has been registered with the Copyright Clearance Center Inc. (CCC), Salem, Massachusetts. Information can be obtained from the CCC about conditions under which photocopies of parts of this publication may be made in the U.S.A. All other copyright questions, including photocopying outside of the U.S.A., should be referred to the publisher. No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter o f products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein.

PRINTED IN THE NETHERLANDS

T o those who cannot be refrained

from asking questions. Therefore, also to my son.

Portrait of Leonard Euler (1707-1783) by the Alsatian painter F’ranz Bernhard F’rey (1716-1806). The portrait must have been painted during Euler’s years in Berlin (1741-1766); for he had lost one eye in 1735 (during his first stay in St. Petersbug), whereas he lost the second eye in 1766.

vii

PREFACE

Since writing my Ph.D. thesis, hamiltonian and eulerian graph theory have been the main topics of my research. Until 1975 I put more emphasis on hamiltonian graph theory; since then, however, problems in eulerian graph theory and related questions have been central to my work. This shift in research emphasis from hamiltonian to eulerian graphs was initiated by a problem posed to me by Gert Sabidussi (Universitk de Montrdal) in the summer of 1975 when I was in Montrkal undertaking joint research with him. At that time neither Sabidussi nor I could foresee the relatively wide range of applications that a solution to his problem (henceforth called Sabidussi’s conjecture) would have. During the academic year 1976/77 I was able to prove a generalization of Sabidussi’s conjecture for planar eulerian graphs (henceforth called the compatibility result); from 1979 onwards, a series of various applications in the theory of planar graphs were discovered (partly just by myself, partly in co-operation with Bill Jackson, the ‘Chinese Postman’ Guan Meigu, and - recently - And& Frank). In the course of these discoveries, it transpired that in various proofs, the application of the Four-Color Theorem (or the assumption of the validity of the Four-Color Conjecture, if you wish) could be avoided by using the compatibility result. ‘1 Unfortunately, there is no simple analogy to the compatibility result in the case of non-planar graphs, however, in recent joint work with Bill Jackson we have put forward a conjecture which contains as a special ‘1 At this point a remark seems appropriate: I do not have any opinion as to whether the Appel-Haken-proof of the Four-Color Conjecture is correct (thus establishing the Four-Color Theorem) or not. I would prefer t o see a proof of that conjecture/theorem which is not computer-aided (but which graph theorist wouldn’t prefer that ?), or at least it might be desirable for various independent teams t o try and produce computer-aided proofs along the lines adopted by K. Appel and W. Haken (but who has the time and money for it ?).

...

Vlll

Preface

case an equivalent formulation of the cycle-double-cover conjecture. This and the preceding statements indicate why Sabidussi’s conjecture and generalizations thereof, but also its solution in the planar case (we call this whole complex of conjectures and results the compatibility problem), as well as its applications and implications, assume a dominant position in the second part of the book. The academic year 1977/78 was proclaimed graph theory year at the Department of Mathematics at Memphis State University (MSU), Tennessee. Following an invitation by R. Faudree and R. Schelp I spent the fall semester there, during which I gave a three-hour course on eulerian graph theory. At that time I was familiar with A. Kotzig’s work in that field of research, and, apart from the compatibility problem I had done research on a special type of eulerian trails which had been extended by my first Ph.D. student S. Regner. In presenting the material at MSU, it occurred to me for the first time that as a topic eulerian graphs constituted more than a mere accumulation of results. Tendencies towards the development of theories on eulerian graphs became visible for me. Consequently, when I learned from L.W. Beineke and R.J. Wilson at the graph theory meeting in Oberwolfach in 1979, that they were preparing Selected Topics in Graph Theory 2 ([BEIN83a]), I asked them if they were interested in a survey article on eulerian graphs. They doubted whether there was sufficient material to justify such an article, but agreed that I should try to write a contribution to [BEIN83a]. In preparing this survey article I was astonished to learn how much material had in fact accumulated on the subject Eulerian Graphs and Related Topics. By 1980, a f i s t version of the survey article Eulerian Graphs was ready; after a first criticism by the editors of [BEIN83a], we agreed that I should write a second extended version from which the editors would produce a condensed version (which finally appeared in [BEIN83a]). This extended version was 90 typewritten pages long (ten of which were taken up by a bibliography using 124 references). After the 90-page version of Eulerian Graphs (originally named Towards Theories on Eulerian Graphs) had been delivered to the editors of [BEIN83a], it occurred to me that there was enough material to justify writing a book on Eulerian Graphs and Related Topics that would be four to five times as long. This book is the result of checking further into the literature which has become available during the first half of the eighties. Its structure follows partly that of the survey article Eulerian Graphs as presented in [BEIN83a] (with some chapters added and greater

Preface

ix

emphasis added to others). In the preparation of the final material for the book, but also in the course of typing it, Mrs. M. Wenger has been of great help (through her, I saved literally hundreds of hours that would have gone otherwise into copying, typing, and producing a catalogue of the material treated in this book). Mr. Erich Wenger produced most of the figures, and wrote an earlier qualifying thesis (as part of his qualifying to become a high school teacher) in which he treated some of the material treated in this book (viz. the Chinese Postman Problem and on the Theorem of Amitsur and Levitzki). His brother Emanuel Wenger provided invaluable services akin to those of a technical organizer; being a student of mine and familiar with a large part of my research, he made valuable comments. Other (former) students of mine, Mrs. M. Music and Mr. J. Galambfalvy de Geges had written theses (similar to Erich Wenger’s) on arbitrarily traceable graphs, and enumeration in eulerian graph theory, respectively. My colleagues Mr. L. Dimitrov and Mr. R. Thaller helped me mXnically in producing the index and in correcting parts of the manuscript. Many of my research colleagues, especially those with whom I undertook joint research, contributed - directly or indirectly - to this book: Be it through valuable suggestions and/or comments; be it through their own experience in writing books and/or articles; be it through explanations of material to which they introduced me; be it through moral support; be it through papers they sent me - they all helped me, and be it only to a minor degree, in completing the book (which, nevertheless, I had to write). I wish to name in alphabetical order those who were of crucial importance to this work: Lowell Beineke, Adrian J. Bondy, AndrAs Frank, Bill Jackson, Franqois Jaeger, Charles H.C. Little, Crispin St.J.A. NashWilliams, Mike D. Plummer, Gert Sabidussi (who also called my attention to F.B. Frey’s portrait of Euler), and Robin Wilson. My warmest thanks to them and all the others. I hope the book shows that I made proper use of their help. Finally I wish to express my gratitude to Mr. Peter Lillie (a British citizen living in Vienna) who read almost the entire book from the linguistic point of view. Any shortcomings in this respect are entirely my fault.

This Page Intentionally Left Blank

xi

CONTENTS

I.

. . . . . . . . . . . . . . . . . . . . . . . . vii INTRODUCTION . . . . . . . . . . . . . . . . . . 1.1

11.

THREE PILLARS OF EULERIAN GRAPH THEORY

PREFACE

. 11.1

Solution of a Problem Concerning the Geometry of Position . . . . . . . . . . . . . . . . . . . . . 11.12 On the Possibility of Traversing a Line Complex Without Repetition or Interruption . . . . . . . . . . . . . 11.23

. . . . . . . . . . BASIC CONCEPTS AND PRELIMINARY RESULTS . Mixed Graphs and Their Basic Parts . . . . . . . . . From 0 . Veblen’s “Analysis situs”

111. 111.1.

11.26

111.1 111.2

111.2.

Some Relations Between Graphs and (Mixed) (Di)graphs. Subgraphs . . . . . . . . . . . . . . . . . . . . 111.9

111.3.

. . . . . . . . . Walks, Trails, Paths, Cycles, Trees; Connectivity . . .

111.4. 111.5. 111.6.

Graphs Derived from a Given Graph

111.13

111.18

Compatibility, Cyclic Order of and Corresponding Eulerian Trails . . . . . . . . . . . . . . . . . . 111.460 Matchings, 1-Factors, 2-Factors, 1-Factorizations, 2-Factorizations, Bipartite Graphs . . . . . . . . . . . . 111.43

. . . . . 111.50 111.8. Coloring Plane Graphs . . . . . . . . . . . . . . . 111.660 111.9. Hamiltonian Cycles . . . . . . . . . . . . . . . . 111.62 111.10. Incidence and Adjacency Matrices, Flows and Tensions . 111.63 111.11. Algorithms and Their Complexity . . . . . . . . . . 111.72 111.12. Final Remarks . . . . . . . . . . . . . . . . . . 111.75 111.7.

Surface Embeddings of Graphs; Isomorphisms

xii

Contents

IV.

CHARACTERIZATION THEOREMS AND COROLLARIES . . . . . . . . . . . . . . . . . . . . .

IV.l. IV.2. IV.3. IV.4.

Graphs .

.

. . . . Digraphs . . . . . . . . Mixed Graphs . . . . . . Exercises . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . . . . . . . . . . . . . . , . . .

IV.l

.. IV.l . IV.8

. IV.13 . IV. 20

V.

EULER REVISITED AND AN OUTLOOK ON SOME GENERALIZATIONS . . . . . . . . . . . . . . . . V.l

V.l.

Trail Decomposition, Path/Cycle Decomposition

V.2.

Parity Results . . . . . . . . . . . . . . . . . . . . V.3

V.3.

Double Tracings . . . . . .

V.4.

Crossing the Border:

V.5.

Exercises . . . . . . . . . . . . . . . . . . . . . . V.7

.

. . . V.l

. . . . . . . . . . . . . V.4 Detachments of Graphs . . . . . . V.5

. . VI.l VI.l. Eulerian Trails Avoiding Certain Transitions . . . . . VI.l VI.l.l. P(D)-Compatible Eulerian Trails in Digraphs . . . . . VI. 7 VI.

VARIOUS TYPES OF EULERIAN TRAILS

. .

.

VI. 1.2. Aneulerian Trails in Bieulerian Digraphs and Bieulerian Orientations of Graphs . . . . . . . . . . . . . . VI.17 VI. 1.3. Do-Favoring Eulerian Trails in Digraphs VI.2.

. . . . . . .

Pairwise Compatible Eulerian Trails . . . . . . . . . VI.34

VI.2.1. Pairwise Compatible Eulerian Trails in Digraphs VI.3.

VI.25

A-Trails in Plane Graphs . . . . . . . .

. . . VI.59

. . . . . .

VI.69

VI.3.1. The Duality between A-Trails in Plane Eulerian Graphs and Hamiltonian Cycles in Plane Cubic Graphs . . . VI.I 1 1 VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs . VI.145 VI.3.3. How to Find A-Trails: Some Complexity Considerations and Proposals for Some Algorithms . . . . . . . . VI.156 An A-Trail Algorithm for Arbitrary Plane Eulerian Graphs . . . . . . . . . . . . . . . . . . . .

VI.165

Contents

...

Xlll

VI.3.4. Final Remarks on Non-Intersecting Eulerian Trails and A.Trails, and another Problem . . . . . . . . . . VI.1g5

. . . . . . . . . . . . . . . . . . . .

VI. 170

VI.4.

Exercises

VII .

TRANSFORMATIONS OF EULERIAN TRAILS . . . VII.l

VII.l.

Transforming Arbitrary Eulerian Trails in Graphs . . . VII.2

VII.2.

Transforming Eulerian Trails of a Special Type . . . . v11.9

VII.2.1. Applications to Special Types of Eulerian Trails and q-Transformations . . . . . . . . . . . . . . . . VII.23 VII.3.

Transformation of E.ulerian Trails in Digraphs . . . .

VII.45

. . . . . . VII.51 VII.5. Exercises . . . . . . . . . . . . . . . . . . . . VII.53 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . A.1 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . B.l VII.4.

Final Remarks and Some Open Problems


I.1

Chapter I

INTRODUCTION

When Ddnes Konig’s book Theorie der endlichen und unendlichen Graphen (The Theory of Finite and Infinite Graphs) [KONI36a]. was first published in 1936, it took only 248 pages (excluding the preface, contents and bibliography) to present the larger part of graph theoretical material which had evolved in the course of 200 years since the publication of Leonhard Euler’s paper on the solution of the Konigsberg Bridges Pro b Zem. Actually, Euler had presented his paper to the Academy of Sciences in St. Petersburg (now Leningrad) on August 26, 1735, but it was only in 1741 that the Commentarii Academiae Scientiarum Imperialis Petropolitanae containing Euler’s article under the title Solutio problematis ad geometriam situs pertinentis were published. On the other hand, since the Commentarii were dated 1736, that is generally considered the date of birth of the branch of mathematics known today as Graph Theory. However, as D. Konig pointed out in the preface, the treatment of graph theory in his book was restricted in essence to what he called absolute graphs (today one would rather say abstract graphs), while - with a few exceptions - topological graph theory (he calls it r e h t i v e graph theory) and enumerative graph theory were not considered. Since the appearance of D. Konig’s book, but mainly since the end of World War 11, graph theory has developed at an ever-increasing pace. This has led to the founding of more and more journals which specialize in publishing articles on combinatorid problems, about half of which (if not more) are graph theoretical papers. The Journal of Graph Theory, for example came into being in 1977. This upsurge in graph theoretical findings is reflected not only in the growing number of books on graph theory, but also in the fact that many of these books concentrate on special aspects of graph theory, such as topological graph theory, algebraic

1.2

I. Introduction

graph theory and - as a more recent tendency - algorithmic graph theory (thus satisfying a need of computer scientists). Graph theory can thus be seen to have followed the general course of development of any science. Originally breaking away from a more general field of research (the subtitle of D. Konig’s book is Kombinatorische Topologie der Strekkenkomplexe - Combinatorial Topology of One-Dimensional Compleses) , it breaks into various parts according to the development of results and methods into certain directions. In the case of graph theory, this development has been illustrated - in a concise form - by the publication of Selected Topics in Graph Theory and Selected Topics in Graph Theory 2 ([BEIN78a, BEIN83a]), containing 22 survey articles by various authors and edited by L.W. Beineke and R.J. Wilson. A third volume has just been published (The usefulness of graph theory in various sciences is illustrated in Applications of Graph Theory (same editors as above) containing 1 2 survey articles). Euler’s article on the bridges of Konigsberg (now Kaliningrad) did not descend on the scientific community as a deus ez machina. This is made quite clear by Euler himself when, in his article, he refers to Leibniz:

In addition t o that branch of geometry which is concerned with magnitudes and which has always received the greatest attention, there is another branch, previously almost unknown, which Leibniz first mentioned, calling it the geometry of position [geometria situs]. This branch is concerned only with the determination of position and its properties; it does not involve measurements, nor calculations made with them... Euler went on to remark that it was not yet quite clear which problems were relevant to the geometry of position, or what methods should be used in solving them, but he certainly regarded the Konigsberg bridges as such a problem, especially since its solution “involved only position, and no calculation was of any use”, [WILSSSa]. In fact, as early as 1679, Leibniz made the following statement in a letter to Huygens (we quote from [WILS85a]): I a m not content with algebra, in that it yields neither the shortest proofs nor the most beautiful constructions of geometry. Consequently, in view of this I consider that we need yet another kind of analysis, geometric or linear, which deals directly with position, just as algebra deals with magnitude ... By introducing the term analysis situs (analysis of position) he did not lay the foundations of a new mathematical field of research, but rather

I. Introduction

1.3

pointed into a general direction along which some of the research should progress (The reader interested in a historical account of the term analysis situs is referred to R. Wilson’s article [WILS85a]). We shall make some more remarks on the history of graph theory in the next chapters, but it should be mentioned at this juncture that D. Konig’s book is probably the richest single source for the early history of graph theory (with the word early we mean the development of graph theory until the appearance of D. Konig’s book in 1936). But why a book on eulerian graphs ? Is it because graph theory (and especially Euler’s article) has recently celebrated its 250th anniversary ? The proximity of publication to that anniversary is purely co-incidental (I had originally planned to finish this book by the end of March, 1983). However, as already pointed out in the preface, there has not only been a rapid growth in the number of papers on eulerian graphs, but also tendencies towards theories unifying various results on this topic. These two facts can be considered a necessary condition for writing a book on Eulerian graphs and related topics, and together with the interest shown by many colleagues justify its being written. Moreover, as I have outlined above, this book is in line with the general tendency in graph theory which, slowly but surely, over the past twenty years has been breaking into various parts. In the next chapter, the original versions of three papers are presented which form - in the opinion of most graph theorists - the main roots of eulerian graph theory. A large part of the book is devoted to those results which are more or less directly related to the concepts developed in these papers. However, I believe that to restrict oneself to this part of eulerian graph theory would be too narrow an approach in comparison with current graph theory developments. This point of view also prevailed already in my survey article Eulerian Graphs as presented in [BEIN83a]. On the other hand, this point of view poses the problem of determining the material to be treated in a book like this. Because this book is the first to focus on the treatment of eulerian graphs, I decided to cover as broad a range of topics as possible. Some I have treated in greater detail than others and on occasion I have merely touched on various aspects as though in a survey. This has the disadvantage, of course, that in some instances the reader who wants to know more about the corresponding topic, has to resort to other books or even the original papers quoted in this book. Moreover, it sometimes overlaps

1.4

I. Introduction

with other branches of graph theory covered in [BEIN78a, BEIN83al. This becomes apparent in the chapters on the Chinese Postman Problem and in those places where 1-factorizations of graphs play a certain role, as well as in the chapters on enumeration, coloring, and some other places. But such overlappings are - generally speaking - more or less inevitable precisely because of present developments in graph theory. In lining up the material for the book I checked hundreds of articles as to their suitability. From the above, it is clear that many of the references are not treated extensively in this book. However, one aim of the book is to indicate various directions of current research. Thus, while some readers might get the impression that the references are too numerous when compared with the various topics treated, the large number of references has the advantage of helping interested readers to reach out in various directions beyond the limits of the book. Despite my having checked so many articles I hope that I did not overlook important contributions t o eulerian graph theory. On the other hand, my survey article contained some shortcomings in this respect. Last, but not least, this book is self-contained because it is intended to reach as broad a public as possible. Therefore, Chapter I11 deals with the basics of graph theory to the extent required for the subsequent chapters. I admit that I find it annoying that many monographs require a more or less broad knowledge of a particular branch of mathematics before one can understand the matter presented. By the same token, statements such as “as can be easily seen”, “it now follows easily”, “it is trivial to see”, and the like are mostly avoided. Many mathematicians (including myself) have more than once encountered situations in proofs where it took paper, pencil, plus half an hour or longer to see something easily. Hence I have not hesitated to include many figures in this book to illustrate situations and arguments, without using figures instead of logical arguments. Nevertheless, in several instances whole proofs of results are left to the reader as not too difficult exercises. Consequently, this book contains enough material for both an undergraduate or graduate graph theory course which emphasizes eulerian graphs. Hence it can be read by mathematicians not yet familiar with graph theory. But it is also of interest to researchers in graph theory because it contains many recent results, some of which are only partial solutions on yet unsolved more general problems; and a fair number of conjectures have been included as well.

I. Introduction

1.5

A few words on algorithms and complexity studies. Various problems (such as finding eulerim trails, cycle decompositions, postman tours and walks through labyrinths) are also addressed algorithmically. However, it is not the aim of this book to produce an algorithm following theoretical considerations wherever possible. The question of complexity is treated similarly. From a theoretical point of view, it appears more important to know whether certain problems can be solved in polynomial time or not; in this book, it is of secondary relevance whether the complexity of an algorithm is O(n) or O(n2). I know that many colleagues (particularly computer scientists or graph theorists leaning in this direction) will seriously criticize this point.

I would be glad to receive critical response (positive, negative or mixed) from anyone, because it could contribute to improving my further work. I promise to respond t o all my critics.


11.1

Chapter I1

THREE PILLARS OF EULERIAN GRAPH THEORY There exist various translations of Euler’s article [EULER36a] on the Konigsberg Bridges Problem and of Hierholzer’s article [HIER73a] on constructing an eulerian trail in a connected eulerian graph. However, in what follows I present my own translation of these two artic1es.l) I decided to do so because the translations I found had one disadvantage: they were ‘up to date’ translations, so t o say, i.e., translations which disregard to a larger or lesser extent the style in which papers where written at the time of their appearance. Thus, these translations are inaccurate from a historical point of view and therefore - unintentionally - produce a distorted picture of the road to knowledge and how it was formulated ‘back in the old days’. Hence it is clear that my translation of Euler’s article is not a belated home exercise of a former high school student who was taught latin for six years; nor did the translations arise out of fear to get into a conflict with copy rights. For a historical account of Euler’s article we refer the interested reader to [WILS85a, WILS86al and [SACH86a, SACH86bl.

I wish t o express my gratitude to Mr. H. Reitterer of the Austrian Academy of Sciences, Vienna, who checked my translation of Euler’s article.

11.2

11. Three Pillars of Eulerian Graph Theory

SOLUTIO PROBLEMATIS AD GEOMETRIAM SITUS PERTINENTIS Commenhtio 53 indicis ENESTUOEMUNI Comment.nrii academiae scieuliaruin Yetropolitanae 8 (1736), 1741, p. 138-140

1. Praeter illam geometriae partem, quae circa quantitates versatur et omni tempore summo studio est exculta, alterius partis etianinum admoduni ignotae primus mentionem fecit LEIBNKTZIUS'), quam Geometriani situs vocavit. Ista pars ab ips0 in solo situ determinando situsque proprietatibus eruendis occupata esse statuitur; in quo negotio neque ad quantitates respiciendum neque calculo quantitatum utendum sit. Cuiusmodi autem problemata ad h a w situs geometriam pertineant et quali methodo in iis resolvendis uti oporteat, non satis est definitum. Quamobrem, cum nuper problematis cuiusdam mentio esset fncta, quod quidem ad geometriam pertinere videbatux, at ita erat comparatum, u t neque determinationem quantitatum requireret neque solutionem calculi quantitatum ope admitteret, id ad geometriam situs referre haud dubitavi, praesertim quod in eius solutione solus situs in consideiationem veniat, calculus vero nullius prorsus sit usus. Methodum ergo meam, quam ad huius generis problemata solvenda inveni, tanquam specimen Geometriae situs hic exponere constitui. 1) Vide epistolam R G. LELLINIZ (1646-1716) ad CER HUYGENS (1629-1695) scriptam d. 8. Sept. 1679, LEZBNIZENS dfathematische Schiiften, herausg. von C. I. GEUHAUDT,Erste Abt , Bd. 2, Epistoia, qua HUYOENS respondit d. Berlin 1850, p. 17-20, imprimis p. 19, Beilage p. 20-25. 22. Nov. 1679, invenitur ibidem p. 27. Vide porro G. LEIBNIZ, Dc analysi situs, LEIBXIZENS MailteL. G. D. mathhc Schlriften, Zweite Abt., Bd. 1, Haile 1858, p. 178-183. LMONUAUUI EULRMI Opera omnin

I7

Commentationes algebrnicae

1


2

SOLUTIO PBOBLEMdTIS

11.3

rl29

2. Problema autem hoc, quod mihi satis notum esse perhibebatur, erat sequens: Regiomonti in Borussia esse insulam A, der Kneiphof dictam, fluviumque eam cingentem in duos dividi ramos, quemadmodum ex figura (Fig. 1) videre licet; ramos vero huius fluvii septem instructos esse pontibus a, b, c, d, e, f e t 9. Circa hos pontes iam ista proponebatur quaestio, num quis cursum ita instituere queat, ut per singulos pontes semel et non plus quam semel transeat. Hocque fieri posse, mihi dictum est, alios negare alios dubitare; neminem vero affirmare. Ego ex hoc mihi sequens maxime generale formavi problema: quaecunque sit fluvii figura et distributio in ramos atque quicunque fuerit numerus pontium, invenire, utrum per singulos pontes semel tantum transiri queat an vero secus. A

Fig. 1.

3. Quod quidem ad problema Regiomontanum de septem pontibus attinet, id resolvi posset facienda perfecta enumeratione omninm cursuum, qui institui possunt; ex his enim innotesceret, num qnis cursus satisfaceret an vero nullus. Hic vero aolvendi modus propter tantum combinationum numerum et nimis esset difficilis atque operosus et in aliis quaestionibus de multo pluribus pontibus ne quidem adhiberi posset. Hoc porro mod0 si operatio ad finem perducatur, multa inveniuntur, quae non erant in quaestione; in quo procul dubio tantae difficultatis causa consistit. Quamobrem missa hac me-


11.4

129-1311

AD GEOMETRIBM SITUS PERTINEXTIS

3

thodo in aliam inquisivi, quae plus non largiatur, quam ostendat, utrum talis cursus institui queat an secus; talem enim methodum multo simpliciorem fore sum suspicatus. 4. Innititur autem tota mea methodus idoneo mod0 singulos pontiuln

transitus designandi, in quo utor litteris maiusculis A, B, C, D singulis regionibus adscriptis, quae flumine sunt separatae. Ita, si quis ex regione A in regionem B transmigrat per pontem a sive b , hunc transitum denoto litteris A B , quarum prior praebet regionem, ex qua exierat viator, posterior vero dat regionem, in quam pontem transgressus pervenit. Si deinceps viator ex regione B abeat in regionem D per pontem f , hic transitus repmesentabitur litteris B D ; duos autem hos transitus successive institutos A B et B D denoto tantum tribus litteris A B D , quia media B designat tam regionem, in quam primo transitu yervenit, quam regionem, ex qua alter0 transitu exit. 5. Simili mod0 si viator ex regione D progrediatur in regionem C per pontem 9, hos tres transitus successive factos quatuor litteris A B D C denotabo. Ex his enim quatuor litteris ABDC intelligetur viatorem primo in regione A existentem transiisse in regionem B , hinc esse progressum in regionem D ex hacque ultra esse profectum in C; cum vero hae regiones fluviis sint a se invicem separatae, necesse est, ut viator tres pontes transierit. Sic transitus per quatuor pontes successive instituti quinque litteris denotabuntur; et si viator trans quotcunque pontes eat, eius migratio per litterarum numerum, qui Unitate est maior quam numerus pontium, denotabitur. Quare transitus per septem pontes ad designandum octo requirit litteras. 6. In hoc designandi mod0 non respicio, per quos pontes transitus sit factus, sed si idem transitus ex una regione in aliam per plures pontes fieri potest, perinde est, per quemnam transeat, mod0 in designatam regionem perveniat. Ex quo intelligitur, si cursus per septem figurae pontes ita institui posset, ut per singulos semel ideoque per nullum bis transeatur, hunc curs u m octo litteris repraesentari posse easque litteras ita esse debere dispositas, ut immediata litterarum A et B successio bis occurrat, quia sunt duo pontes a et b has regiones A et B iungentes; simili mod0 successio litterarum A et C quoque debet bis occurrere in illa octo litterarum eerie; deinde successio litterarum A et D semel occurret eimiliterque successio litterarum B et D itemque C et D semel occurrat necesse est.


4

SOLUTIO PROBLEMATIS

11.5

[131--133

7. Quaestio ergo huc reducitur, u t ex quatuor litteris A , B , C et 1) series octo litterarum formetur, in qua omnes illae successiones toties occurrant, quoties est praeceptum. Antequam autem ad talem dispositionem opera adhibeatur, ostendi convenit, utrum tali mod0 hae litterae disponi queant an non. Si enim demonstrari poterit talem dispositionem omnino fieri non possel inutilis erit omnis labor, qui ad hoc efficiendum locaretur. Qunmobrem regulam investigavi, cuius ope tam pro hac quaestione quam pro omnibus similibus facile discerni queat num Ldis litterarum dispositio locum habere queat.

8. Consider0 ad huiusmodi regulam inveniendam unicam regionem A, in quam quotcunque pontes a, b, c, d etc. conducant (Fig. 2). Horum pontium contemplor primo unicum a , qui ad regionem A ducat; si nunc viator per hunc poutem transeat, vel ante transitum ewe debuit in regione A vel post transiturn in A perveniet; quare in supra stabilito transitus designandi modo oportet, ut littera A semel occurrat. Si tres pontes, puta a, b, c, in regionem A conducant et viator per omnes tres transeat, tum in designatione eius migrationis littera A bis occurret, sive ex A initio cursum instituerit sive minus. Simili mod0 si quinque pontes in 9 conducant, in designatione transitus per eos omnes littera A ter occurrere debet. Atque si numerus pontium fuerit quicunque numerus impar, turn, si is unitate augeatur, eius dimidium dabit, quot vicibus littera A occurrere debeat.

A

Fig. 2.

9. In casu igitur pontium tsanseundorum Regiomontano (Fig. l), quia in insulam A quinque pontes deducunt a, 13, c, d, e, necesse est, ut in designatione transitus per hos pontes littera A ter occurrat. Deinde littera B, quia in regionem B tres pontes conducunt, bis debet occurrere similique mod0 littera D bis debet occurrese atque etiam littera C bis. In sene ergo octo litterarum, quibus transitus per septem pontes deberet designari, littera A ter adesse deberet, litterarum vero B, C et D unaquaeque bis; id quod in serie octo litterarum omnino fieri nequit. Ex quo perspicuum est per septem pontes Regiomontanos tadem transituni institui non posse.


11.6

133-134

I

AD GEOMETRJAhl SITUS PERTIKENTIS

5

10. Simili mod0 de omni alio casu pontiiim, si quidem numerus pontium, qui in qua.mque regionem conducit, fuerit impar, iudicari potest, an per singulos pontes transitus semel fieri queat. Si enim evenit, ut summa omnium vicium, quibus singulae litterae occurrere debent, aequalis sit nunieio omnium pontium unitate aucto, tum talis transitus fieri potest; sin autem, ut in nostro exemplo accidit, summa omnium vicium maior fuerit numero pontium unitate aucto: t~urntalis transitus nequaquam institiii potest. Regula autem, qurtm dedi pro numero vicium A ex numero pontium in regionem A deducentium inveniendo, aeque valet, sive omnes pontes ex una regione B, ut in figura pig. 2) repraesentatur, ducant sive ex diversis; tanturn enim regionem A consider0 et inquiro, quot vicibus littera A occurrere debeat.

11. Si autem numerus pontium, qui in regionem A conducunt, fuerit par, tum circa transiturn per singulos notandurn est, utrum initio viator cursum suum ex regione A instituerit an non. Si enim duo pontes in A conducant et viator ex A cursum inceperit, turn littera A bis occurrere debet; semel enim adesse debet ad designandurn exitum ex A per alterum pontem et semel quoque ad designandurn reditum in A per alterum pontem. Sin autem viator ex alia regione cursum inceperit, tum semel tantum littera A occurret; semel enirn posita tam adventurn in A quam exitum inde denotabit, ut huiusmodi cursus designare statui.

12. Conducant iam quatuor pontes in regionem A et viator ex A cursum incipiat; turn in designatione totius cursus littera A ter adesse debebit, si quidem per singulos semel transierit. At si ex alia regione ambulare inceperit, turn bis tanturn littera A occurret. Si sex pontes ad regionem A conducant, tum littera A, si ex A initium eundi est sumpturn, quater OCCUIret, at si non ex A initio exierit viator, turn ter tantum occurrere debebit. Quare generaliter: si numerus pontium fuerit pm, turn eius dimidium dat numerum vicium, quibus littera A occurrere debet, si initium non est in regione A sumpturn; dimidium vero unitate auctum dabit numerum vicium, quoties littera A occurrere debet, initio cursus in ipsa regione A surnpto.

13. Quia autem in tali cursu nonnisi ex una regione initiurn fieri potest, ideo ex numero pontium, qui in quamvis regionem deducunt, ita numerum vicium, quoties littera quamque regionem denotass occurrere debet, dethio, ut


6

SOLUTIO PROBLEMATIS

11.7

[134--136

sumam numeri pontium unitate aucti dimidium, si nnmerus pontium fuerit impar; ipsius vero numen pontium medietatam, si fuerit par. Deinde si numerus omnium vicinm adaequet numerum pontium unitate auctum, tum transitus desideratus succedit, at initium ex regione, in quam impar pontium numerus ducit, capi debet. Sin autem numerus omnium vicium fuerit unitate minor quam pontium numerus unitate auctus, tum transitus succedet incipiendo ex regione, in quam pax pontium numerus ducit, qnia hoc mod0 vicium numerus unitate est augendus.

14. Proposita ergo quacunque aquae pontiumque fignra ad investigandum, num quis per singulos semel transire queat, sequenti modo operationem instituo. Primo singulas regones aqua a se invicem diremptas litteris A, B, C etc. designo. Secundo sumo omnium pontium numerum eumque unitate augeo atque sequenti operationi praefigo. Tertio singulis litteris A, B, C etc. sibi subscriptis cuilibet adscribo numerum pontium ad eam regionem deducentium. Quaito ens litteras, quae pares adscriptos habent numeros, sign0 asterisco. Quint0 singulorum horum numerorum parium dimidia adiicio, imparium vero unitate auctorum dimidia ipsis adscribo. Sexto hos numeros ultimo scriptos in unam summam coniicio; quae summa si vel unitate minor fuerit vel aequalis numero supra praefixo, qui est numerus pontium unitate auctus, tum concludo transitum desideratum perfici posse. Roc vero est tenendum, si summa inventa fuerit unitate minor quam numerus supra positus, tum initium ambulationis ex regione asterisco notata fieri debere; contra vero ex regione non signata, si summa fuerit aequalis numero praescripto. Ita ergo pro cam Regiomontano operationem instituo, u t sequitur:

4

5

3

Bl

3

2

c,

3

2

D,

3

2

Quia ergo plus prodit quam 8, huinsmodi transitus potest.

nequaquam fieri

11.8


1361

AD GEOMETRIAM SITUS PERTINENTIS

7

15. Sint duae insulae A et B aqua circumdatae, qua cum q u a cornmunicent quatuor fluvii , quemadmodum figura (lhg. 3) repraesentat. Traiecto porro sint super q u a m insulas circumdantem et fluvios quindecim pontes a, b, c, d etc. et quaeritur, num quis cursum ita instituere queat, ut per

Fig. 3.

omnes pontes transeat, per nullum autem plus quam semel. Design0 ergo primum omnes regiones, quae aqua a ee invicem eunt separatae, litteris A, B, C,D, E, F, cuiusmodi ergo sunt sex regiones. Dein numerum pontium 15 unitate augeo et summam 16 sequenti operationi praefigo. 16

A*, 8 B*, 4

c*,

(4 2

4

2

D, 3 El 5

2

3


8

SOLUTIO PROBLEMATIS

11.9

[137-138

Tertio litteras A, B, C etc. sibi invicem subscribo e t ad quamque numerum pontium, qui in eam regionem ducunt, pono, u t ad A octo ducunt pontes, ad B quatuor etc. Quarto litteras, quae pares adiunctos habent numeros, asterisco noto. Quinto in tertiam columnam scribo parium numerorum dimidia, impares vero unitate mgeo et semisses sppono. Sexto tertiae columnae numeros invicem addo et obtineo siimmam 16; quae cum aequalis sit numero supra posito 16, sequitur transitum desiderato mod0 fieri posse, si modo cursus vel ex regione D vel E incipiatur, quippe quae non sunt asterisco notatae. Cursus autem ita fieri poterit

EaFb BcFd AeFf Cg Ah Ci DkAm En A p Bo El D , ubi inter litteras maiusculas pontes simul collocavi, per quos fit transitus. 16. Hac igitur ratione facile erit in casu quam maxime composito iudicare, utrum transitus per omneb pontes semel tantum fieri queat an non. Hoc tamen adhuc multo faciliorem tradam modum idem dignoscendi, qui ex hoc ips0 mod0 non difficulter eruotur, postquam sequentes observationes in medium protulero. Priino autem observo omnes numeros pontium singulis litteris A, B, C etc. sdscriptos simul sumptos duplo msiores esse toto pontium numero. Huius rei ratio est, quod in hoc computo, quo pontes omnes in, datam regionem ducentes numerantur, quilibet pons bis numeretur; refertur enim quisque pons ad utramque regionem, quas iungit. 17. Sequitur ergo ex hac observatione sunimam omniuin pontium, qui in singulas regiones conducunt, esse numerum parem, quia eius dimidium pontium numero aequatur. Fieri ergo non potest., u t inter numeros pontium in quamlibet regionem ducentium unicus sit impar; neque etiam, ut tres sint impares, neque quinque etc. Quare si qui pontium numeri litteris A , B, C etc. adscripti sunt impares, necesse est, u t eorum numerus sit p a ; ita in exemplo Regiomontano quatuor erant pontinm numeri impares litteris regionum A, B, C, D adscripti, uti ex § 14 videre licet; atque in exemplo praecedente, Q 15, duo tantum sunt numeri impares, litteris D et E adscripti. 18. Cum summa omnium numerorum litteris A, B, C etc. adiunctorum aequet duplum pontium numerum, manifestum est illam summam binario auctam et per 2 divisam dare numerum operationi praefixum. Si igitur


11.10

138-1393

9

AD GEOJIETRIAY SITUS PERTINENTIS

Omnes nunieri litteris A, B, C, D etc. adscripti fuerint pares et eorwn singulorum medietates capiantur ad numeros tertiae columnae obtiiiendos , erit horum numerorum summa unitate minor quam numerus praefisus. Quamobrem his casibus semper transitus per onines pontes fieri potest. In quacunque enim regione cursus incipiatur, ea habebit pontes numero pares ad se conducentes, uti requiritur. Sic in exemplo Regiomontano fieri potest, u t quis per omnes pontes bis transgrediatur; quilibet enim pons quasi in duos erit divisus numerusque pontium in quamvis regionem ducentium erit par. 19. Praeterea, si duo tantum numeri litteris A , B , C etc. adscripti fuerint impares, reliqui vero omnes pares, tum semper desideratus transitus succedet, si mod0 cursus ex regione, ad quam pontium impar numerus tendit, incipiatur. Si enim pares numeri bisecentur atque etiam impares unitate aucti, uti praeceptum est, summa harum medietaturn unitate erit maior quam numerus pontium ideoque aequalis ipsi numero praefixo. Ex hocque porro perspicitur, si quatuor vel sex vel octo etc. fuerint numeri impares in secunda columna, tum summam numerorum tertiae columnae maiorem fore numero praefiro eumque excedere vel unitate vel binario vel ternario etc. et idcirco transitus fieri nequit.

20. Casu ergo quocunque proposito statim facillime poterit cognosci, utrum transitus per omnes pontes semel institui qiieat an non, ope huius regulae: Si fuerint plures duabus regiones, ad p a s ducentiunz pontiunt gtunrerus est inzpar, tun1 certo affirinari potest talem trunsitum mm dari. Si autem ad dtuls tantuni regwnes ducentium pontium nuinems cst inpar, tu9x transitus fieri poterit, s i mod0 cursus in altera ? u a m reyionutn incapiatur. Si denipue nulla omnino fuerit re*, ad p a m pontes nutnero impres conducant, tm transitus desiderato d o ittstitui poterit, in puacunque regime ambulandi inaium ponatur. Hac igitur data regula problemati proposito plenissime satisfit. LEOXEAXDI EULE.I 0-

omnia

I T Commentationei dgebraicae

2

10


11.11

SOLUTIO PROBLEMATIS AD GEOMETRIAM SITUS PERTINENTIS

[140

21. Quando autem inventum fuerit talem tran~ituminstitui posse, quaestio superest, quomodo cursus sit dirigendus. Pro hoc sequenti utor regula: tollantur cogitatione, quoties fien potest, bini pontes, qui ex una regione in aliam ducunt, quo pacto pontium numerus vehementer ylerumque diminuetur; turn quaeratur, quod facile fiet, cursus desideratus per pontes reliquos; quo invent0 pontes cogitatione sublati hunc ipsum cursum non multum turbabunt, id quod paululum attendenti statim patebit; neque opus esse iudico plura ad cursus reipsa formandos praecipere.

11.12

11. Three Pillars of Eulerian Graph Theory From Leonard Euler’s Complete Works, 17, Algebraic Studies

Solution of a Problem Concerning the Geometry of Position Study 53 of Enestroemianus’ index Notices of the Petersburg Academy of Sciences 8 (1736), 1741, pp.128-140.

1. In addition to that part of geometry which is concerned with quantities and which has always aroused particular interest, there is another - still virtually unknown - part which was mentioned first by Leibniz and b>him called Geometry of position. This part of geometry is concerned precisely with that which can be determined by position alone, and with the investigation of properties of position; in this matter one should neither be concerned with quantities nor their calculation. However, the kinds of problems belonging t o this geometry of position and the methods used for their solution, have not been sufficiently defined. Therefore, when a problem recently arose which seemed to be basically geometrical. but was of such a nature that it neither required the determination of qwantities nor admitted a solution by means of calculating quantities, I was convinced it belonged to the Geometry of Position, especially since only position could be used to solve it, while computation was of no use whatsoever. Therefore, I have decided to esplain here the method which I derived to solve problems of this kind, as an example of the Geometry of Position.

2. The problem, I am told is quite well known and relates to the following. In Iionigsberg in Prussia is an island, called der Kneiphof, and the river surrounding it is divided into two arms, as one can see from the figure (Fig. 1). The arms of this river, however, are crossed by seven bridges a , b, c , d, e , f and g . Concerning these bridges, the question has been posed as to whether one can take a walk in such a way that one walks across all the single bridges once, but only once. I was told that some deny that this can be done and others doubt it but nobody confirms it. From this I formulated for myself the following general problem: whatever the shape of the river and its distribution of arms may be, and whatever the number of bridges crossing them may be, to discover whether one can cross the individual bridges once and only once. 3. Of course, it is possible to solve Konigsberg’s problem of the seven bridges by making a complete list of all walks which can be taken, whence

Solution of a Problem Concerning the Geometry of Position

11.13

it would transpire whether some walk is satisfactory or none is. However. because of the large number of combinations, this way of solving the problem would be extremely difficult and laborious. Moreover, it could not be applied to other questions concerning many more bridges. Furthermore. if, in this way, the work could be brought to a conclusion, many irrelevant factors would arise; therein without doubt lies the reason for the difficulty. Therefore, having dismissed this method, I looked for another which would merely decide n-hether a walk as required is possible; for I suspected that such a method would be much simpler. 4. My whole method is based on a proper way of denoting the single

crossings of the bridges. I use the capital letters A , B , C, D to denote the single regions which are separated by the river. Accordingly, if one moves from region A to region B yia the bridge a or b, I denote this crossing by the letters AB. The first letter denotes the region whence the traveler came, whereas the second letter indicates the region at which he arrives after crossing the bridge. If, thereafter, the traveler goes from region B t o region D via bridge f, this transition will be described by the letters B D ; however, I denote these two successive transitions AB and B D only by the three letters ABD because the middle letter B denotes the region reached by the first crossing, as well as the region left by the second. 5 . Similarly, if the traveler advances from region D to region C via the bridge 9, then I shall denote these three successive transitions by four letters ABDC. From these four letters ABDC, one will see that the traveler who first appears in region A, has crossed over to region B , whence he has progressed to region D , and thence advanced further to C'. However, since these regions are separated by rivers, the traveler must cross three bridges. Similarly, crossing four bridges successively will be denoted by five letters; and whatever number of bridges the traveler crosses, this path will be denoted by a number of letters which is one unit larger than the number of bridges. Therefore, the traversal of seven bridges will require eight letters for its description.

6. In this kind of notation I do not consider which bridge is crossed. However, if one can cross from one region to another via several bridges, it does not matter which bridge is taken, as long as one arrives in the designated region. It thus becomes evident that if one could wall- across the seven bridges so as to cross each bridge once and only once, the wall; could be represented by eight letters. The order of these letters would have to be such that the immediate succession of the letters A and B

11.14


would occur twice, because two bridges a and b join the regions A and B . Similarly, the succession of the letters A and C would occur twice in this sequence of eight letters, the succession of the letters A and D would occur once and, similarly, the succession of the letters B and D and also C and D would necessarily occur once. 7. Thus the question is reduced to whether by the four letters A , B , C and D a sequence of eight letters can be formed in which all these successions occur as often as required. But before trying to find such an arrangement, one should show whether these letters can be so arranged. For: if it can be shown that such an arrangement is impossible, all the work invested in its construction will be useless. Therefore I have investigated a rule, by virtue of which it can be easily decided - both for this question and for all similar ones - whether such an arrangement of the letters can be realized.

8. To discover a rule of this kind I consider a particular region A to which an arbitrary number of bridges a , b, c, d , etc. may lead (Fig. 2). Of these bridges I consider first a particular a which leads to region A. If the traveler now crosses this bridge, then either he had to be in region A prior to crossing or he will arrive in A after crossing. Therefore, in order to denote the crossing as described above, it is necessary that the letter A occurs once. If three bridges - a, b and c, for example - lead to region A , and the traveler crosses all three of them, the letter A will occur twice in the description of his crossings, regardless of whether or not he initially started his walk from A. Similarly, if five bridges lead to A , the letter A must occur three times in the description of their traversal. And if the number of bridges were an arbitrary odd number, then, if it were increased by one, half of it would indicate how often the letter A must occur. 9. Consequently - in the Konigsberg case of the bridges to be crossed (Fig. 1) - because five bridges a, b, c, d, e lead to the island A , it is neces-

sary that the letter A occur three times in the description of the traversal of these bridges. Furthermore, the letter B must occur twice because three bridges lead to region B , and similarly, the letter D and also the letter C must occur twice. Therefore, in the sequence of eight letters by which the walk across the seven bridges should be described, the letter A should appear three times, but each of the letters B , C, D twice. This, however, cannot be done at all in a sequence of eight letters. Hence it is evident that such a walk cannot be undertaken across the seven bridges

Soltition of a Problem Concerning the Geometry of Posittion

11.15

of Konigsberg. 10. For every other case of bridges - assuming of course, that the number of bridges leading to each region is odd, it can be decided in a similar manner whether a walk crossing all the bridges only once, can be made. For, if it is the case that the total number of occurrences of letters is equal to the number of bridges increased by one, then such a wall; is possible. If, however, as happened in our example, the total number of times [letters occur] is greater than the number of bridges increased by one, then the walk is impossible. Nevertheless the rule which I offered for deducing the number of appearances of the letter A from the number of bridges leading to region A , is valid equally whether all bridges lead across from one region B as shown in the figure (Fig. 2), or they lead from various regions, for I consider only region A and inquire how often the letter A must occur.

11. If, however, the number of bridges which lead to region A , is even, one has to consider whether the traveler started in A. If two bridges lead to A and the traveler began his walk from A , the letter A must occur twice. The first time it denotes the departure from A via one bridge and the second time the return to A via the other bridge. If, however, the traveler started the walk from another region, the letter A would occur only once; for, counted once, it will denote an arrival in A as well as a departure from A, according to my description of such walks. 12. If we assume that four bridges lead to region A , and the traveler begins his wall; from A, then the letter A will have to appear three times in the description of the whole walk, if he crosses each bridge only once. But if he should start his walk in another region, the letter A would occur only twice. If sis bridges lead to region A , the letter A will occur four times if the walk starts in A; but if the traveler does not leave from A initially, A will occur only three times. Therefore, in general, if the number of bridges is even, half that number gives the number of times the letter A must occur, provided that the walk is not begun in region A. However, half the number of bridges plus one gives the number of times the letter A must occur, if the walk is begun in this same region A. 13. In such a walk the initial step can be taken in only one region. From the number of bridges which lead to an arbitrary region, I therefore determine the number of times the letter denoting a particular region must occur, as half the number of bridges increased by one if the number of bridges should be odd, but half this number of bridges if it should be

11.16


even. Consequently, if the number of all appearances [of letters] equals the number of bridges increased by one, the desired walk is possible but must begin in that region to which an odd number of bridges lead. If, however, the number of all appearances [of letters] should be one less than the number of bridges increased by one, the walk is possible if started from a region to which an even number of bridges lead, because in this way the number of appearances [of letters] is increased by one.

14. So, having discussed an arbitrary configuration of water and bridges, I introduce an operation in the following way in order to investigate whether someone can cross each bridge just once. First, I denote the individual regions which are separated from each other by water, by the letters A , B , C, etc. Second, I take the number of all bridges and increase it by one, and write this number down for the following operation. Third, after the single letters A , B , C , etc. have been listed, I write nest to each of them the number of bridges leading to the region. Fourth, I marl; by an asterisk those letters which have even numbers written nest t o them. Fifth, I attach half of each of these even numbers, while I write one half of the odd numbers increased by one next to those. Sixth, I sum up these numbers written last; if this sum should be equal to or one less than the number written at the top (which is the number of bridges increased by one) I conclude that the wall; is possible. It should be noted, however, that if the calculated sum is one less than the number written at the top, the walk must start from a region marked with an asterisk, but if the sum is equal to that number, then the walk must start from an unmarked region. Therefore, in the Konigsberg case I undertake the operation as follows: Number of bridges 7, therefore 8 is valid.

A, B,

c,

D,

Bridges 5 3 3 3

3 2 2 2

Because this yields more than 8, the walk is not possible.

15. Let A and B be two islands surrounded by water, and four rivers connected with this water as shown in the figure (Fig. 3). Furthermore, let fifteen bridges a , b , c , d , etc. lead across the water surrounding the


11.17

islands and across the rivers, and the question arises as to whether one can cross all the bridges just once. I first denote all the regions which are separated by water from each other, by the letters A , B , C , D,E , F since there are six regions. The number of the bridges, which is 15, I then increase by one and write the sum 16 at the top of the following table.

16

A*,

8

B*,

4 4

D, E, F*,

3 5

c*,

6

4 2 2 2 3 3 16

Third, I list the letters A, B , C, etc., and write next to each the number of bridges which lead to the region (thus eight bridges lead to A , four to B , etc). Fourth, I mark with an asterisk the letters which are associated with even numbers. Fifth, I write in the third column one half of each even number, while I increase the odd ones by one and add one half [of each of these numbers to the column]. Sixth, I add the single numbers of the third column and obtain 16 as the sum; since this sum equals the number 16 written at the top (of the column), it follows that a walk can be taken in the desired manner only if the walk starts from either region D or E which are not marked by an asterisk. Such a walk can be made, however, as follows

EaFbBcFdAeF f CgAhCiDkAmEnApBoElD

,

where at the same time I have put between the capital letters the bridges across which one should walk. 16. By such considerations, it will be easy to decide in the most complicated cases whether a walk over all the bridges crossing each only once, can be made. However, I should like to convey a much easier way of doing this. It can be easily deduced from the above approach after I have acquainted the reader with the following observations. But, first I observe that all numbers of bridges written next to the letters A , B , C, etc., taken together, are twice the total number of bridges. The reason for this is that in this computation every bridge is counted twice in order

11.18


that all bridges leading to a given region are counted; for, every bridge leads to each of the two regions which it joins.

17. From this observation, it therefore follows that the sum of all bridges leading to each region, is an even number because half of it equals the number of bridges. Therefore it cannot happen that among the numbers of bridges leading into any region, there is just one odd number; nor can it happen that three numbers are odd, nor five, etc. Therefore, if the numbers of bridges written next to the letters A , B , C, etc., are odd, it follows that their sum is even. Thus, in the example of Konigsberg, there were four odd numbers of bridges written next to the letters of regions A , B , C, D , as can be seen from paragraph 14; and in the preceding example, paragraph 15, there were only two odd numbers, written nest t o the letters D and E . 18. Since the sum of all numbers written next to the letters A, B , C, etc., equals twice the number of bridges, it is clear that this sum, after being increased by two and then divided by two, yields the number written down at the top of the calculation. If, therefore, all numbers written next to the letters A , B , C , D , etc., are even and the halves of each of these numbers are understood as the numbers to be obtained for the third column, then the sum of these numbers will be one less than the number written at the top. Therefore, in these cases a walk across all the bridges can be made. For, in whatever region the walk should begin, there will be an even number of bridges leading to it, as required. Thus, in the Konigsberg example, it can be done by someone walking across all bridges twice; for, every bridge could be divided into two, so to say, and the number of bridges leading into every region would then be even. 19. Furthermore, if only two numbers, written next to the letters A , B , C, etc., are odd while all the remaining ones are even, the desired wallc is always possible, but must start from a region to which an odd number of bridges lead. For, if the even numbers, as well as the odd ones increased by one, are bisected as prescribed, then the sum of these halves will be one more than the number of bridges and therefore equal to the number written at the top.

From this it can also be seen that if four, or six or eight, etc. odd numbers should appear in the second column, the sum of the numbers in the third column will be larger than the number at the top, exceeding it by one, two or three, etc. Therefore a walk is not possible.


11.19

20. Therefore, in every possible case, one can immediately and very easily decide, with the help of the following rule, whether or not a walk

without repetition can be taken across all the bridges:

If there are more than two regions with a n odd number of bridges leading t o them, it can be declared with certainty that such a walk is impossible. IL however, there are only two regions with a n odd number of bridges leading t o them, a walk is possible provided the walk starts in one of these two regions.

If, finally, there is n o region at all with a n odd number of bridges leading t o it, a walk in the desired manner is possible and can begin in any regi0.n. Consequently by this given rule the posed problem can be completely solved. 21. However, once it has been discovered that such a walk is possible, the question still remains how a walk should be arranged. For this I use the following rule: delete mentally as often as possible a pair of bridges which lead from one region to another. This will drastically reduce the number of bridges in most cases. Then a desired walk across the remaining bridges

can easily be found. The pairs of bridges, deleted mentally above, will not disturb this walk very much. That such is the case becomes immediately clear with a little thought; therefore I deem it unnecessary to put forth still more on the arrangements of the walks.


11.20

30

CRR. W I B N E U .

Man verfolgt dabei zweckmiisjig die Axe des Weges shtt seiner Randlinie. Jene Miiglichkeit beruht auf der Wahrheit, dms so lange man den Ausgang noch nicht erreicht hat, ein bereits durchhufeues Stiick der Wegeaxe notbwendig von noch niclit beschriebenen Theilen derselben getroffen werden muss, weil sonst jeiies Stack in sich a b g c schlossen w b e und mit dem’Eingangswege uicht zusammenhinge. Nail markire 6kh daher den Weg, den man zuracklegt nebst dern Sinnca, in welchem es geschieht. Sobnld man auf einen schon markirten Weg stiisst, kehre man urn und durebschreite den s&on beschriebenen Weg in umgekehrtem Shne. -Dsman, wenn mon nidt ablenkte, denselben hierbei in seiner ganzen Ausdehnung nochmals zuriicklegen wiirde, so muss man nothwendig hierbei auf eineii nocb nicht markirten einmihdenden Weg treffen, den man d a m verfolge, bis man wieder auf einen markirten t a . Hier kehre man wider um und verfahre wie vorher. Es werden dadurch stets neue Wegtlieile zu den kschriebencn zugefiigt,. so dilss man nach einer endlicheu Zeit das game Labyrinth durchwandern wiirde und so jedeqfirlls den Busgang fiiide, weiin er nicht schon vorher erreicht worden wulre. C a r l s r u h e , December 1871.

Ueber die Blbglichkeit , eineii Liuienzug ohne Wiederliolung und ohne Unterbrechung zu umfahren. Yon CARL HIEREOLZER. M i t g e t h e i l t v o n CHR.W I m x m *).

In einem beliebig verschluugeuen Linienzuge luijgen Zwcigc eiiies Punktes diejenigen verschiedenen Theile des Buges heiqen, auf welchen man den fraglichen Punkt verlassen ksnu. Ein Punkt mit rnelirercn Zweigen heisse ein K f i o h p d t , der so vielfach genanut \verde, nls *) Die folgeude Uuteraucbung trug der leider so frGh dew Dieuste der Wissenschnft durch den Tod entrissene Priratdocent Dr. Hierholeer dahier (seat. 13.3ept. 1871) einem Kreise befreundeter Mathemotiker vor. Urn J e vor Vergcascnheit zu bcwahren. m u d e ne bei dem Mange1 jeder echriftlichen Aufzeichnuug aus dem Gedkhtniss wieder bergestellt werden, wm icli unter Bcihilfc meines verebdn Collegeu Lliroth durch dau Folgende mbglichst gctren auwiifiihren such&.

11. Three Pillars of Eulerian Graph Theory Deber eine Aufgabe der Geomotria situs.

11.21

31

die A n d l der Zweige angiebt, und je nnch dieser Anzalil als gerad oder ungerad genannt seiu soll. Ein gem6hdicher Doppelpunkt wHre hiernach ein vierfacher Knotenpunkt, ein gew8hnlicher P u n k t kann als ein zweifacher und eine freie Endigung als ein einfacher Knotenpunkt bezeichnet werden. Wmn ein finitmug in cinent Zugc umfdwcn w c r h X-ann, ohnc dass i r g d ein Linkstuck wachrfml~dw-dhufm wird, so ?what a. entwede- keincn ode- ~ w cungcradc i Kmtenymn;l;te. Wenn mrui beini Durchlaufen irgend einen P u n k t iiberschreitct so sind dadurch zwei Zweige eiues Knotenpunktes beschrieben, und da keine Strecke zweimnl durchlaufen werden SOU, muss ein Punkt, den man i m Ganzen nmal iiberschreitet, eiu Snfacher Knoteupunkt sein. Ein Punkt kann dalier nur daun ein ungerader Knotenpuukt sein, wenn er beim Durchlaufen einmal nicht Rberschritten wird, d. h. wenn er Anfnngs- oder Endpunkt ist. Wenn man daher beim Durchlaufen zu dem Ausgangsyuuktc bei dem Schlussc zuriickkehrt, so k6nneu nur gerade Knotenpunkte vorhanden sein; wenn niclit, so siiid Ausgangs- und Endpuukt uiigerade Knotenpunkte. Umgekehrt: Wenn ein msantntnzlaangender Linieneug kcinen odcr .ma' ugagerade Knohpnkte enthalt, so kann er in &an Zup untfahren wcrdm. Denn a) hat man irgend einen Theil des Linienzuges umfahren, so ist jeder Knotenpunkt im noch nicht durchfahrenen Theile gerad oder ungerad, wie er es im ganzen Zuge war; nur der Anfangs- nnd Endpunkt des durchlaufenen Stticks kehren ihren Charakter um, ausser wenn sie zusammenfallen. Denn durch das Durchlaufen eines Punktes werden zwei Zweige, durch den Anfang und das Ende des Durchlaufens j e ein Zweig ausgeschlossen. b) Hat man einen Linienzug in einem ungeraden Knotenpunkt zu umfahren angefangen, so kann man nur in einem andem ungeraden Knotenpunkte stecken bleiben. Denn in einem geraden Knotenpunkte hat man bei jedesmaligem Uurchlaufen zwei Zweige ausgeschlossen, so dass bei erneutem Aukommen in demselben wenigstens noch ein Zweig zum Verlassen fibng bleibt. Der Anfangspunkt ist nber durch den Beginu zu einem geraden Knotenpunkte verwmdelt worden, so dass auch in ihm ein Gteckenbieiben nicht miiglich ist. > man dngegen einen Linienzug in einem geraden Knotenpunkte zu umfahren nu, so kann man auch in diesem stecken bleiben, indem er durch dcii Beginn zu einem nngeraden verwmdelt murde. c) Hat nun ein Linienzug zwei ungerade Knotwpunkte, so beginne mnn dns Umfahren in einem derselben; man wird d a m nothwendig in dem andern stecken bleiben. Der zuriickgelcgte Linienzug ist in diesem Falle ein offener. Hat der gegebene Linienzug dagegeu keiiie

,


11.22

32

CnR.

'WIMER.

ungeraden Knotenpunhcte, so beginne man das Umfahren in eiiieni beliebigen Pmkte, der also eiu gerader Knotenpunkt ist; man wird dann nothwendig im Ausgangspunkte stecken bleiben. Der zuriickgelegte Linienzug ist in diesem F d l e ein gescldossener. d) Bleibt dabei ein Theil b undurchlaufen, 80 kann derselbe nur gerade Knotenpnnkte enthalteji, weil die swei etwa vorhandeneu uugeraden Knotenpunkte durch das erste Umfahren ausgeschlossen wurdeii, und die tibrigen Knotenpunkte ihren Charakter behielten. %ugleich muss b n i t dem schon beschriebenen Zuge a durch wenigstens einen gemeinsamen h n k t zusammenhiingen, weil sonst der Zug in mehrerc nicht zusamrnenhiingende zerfallen wtirde. Geht man beim Umfahren des a in einem solchen Punkte P des Zusammenhangs auf b fiber, BO muss man nothwendig auf b in P stecken bleiben, uod kann von da das Umfahren des a so fortsetzen, wie es friiher geschehen war. Auf dieselbe Weise hfngt man jedes noch nicht umfahrene Stiick nn die schon umfahrenen an und beschreibt so die game Linie in einem Zuge. Es ergiebt sich aoch folgeuder Satz: Ein Liiiicnzug kunn nuleine gerade Amah2 ungwadcr KnohptnX-le besitem. Denii schaltet man durch Umfahren ein Linienstiick BUS, i d e m man in einem ungeraden Knotenpunkte beginnt nnd so lange weiter geht, bis man stecken bleibt, was wieder in einem ungeraden Knotenpunkte geschehen muss, und entfernt dadurch zwei ungerde Knotenpuiikte; so kann man durch Wiederholung die Auzahl der ungeraden Knotenpunkte auf weniger als zwei vermindern. Dieser Rest kann aber nicht Eins, sondern muss Null sein; denn wenn ein Zug uur einen ungeraden Knotenpunh+ beslsse und man wiirde in ihm anfangen den Zug zu unifahren, so kiinute man nie zu Ende kommeii, da dies nur in einem aiidern ungeraden Knotenpunkte niijglich ist. Die Zahl der ungeraden Knotenpunkte des urspriinglichen Linienzuges ist daher eiue gerade. C a r l s r u h e , im December 1871. Aum. der Hed. Der wesentlichc Inhalt des Voretekenden, nur in kiirzcrer D:wtellung, eum Tbeil obnc nahere AusTuhrung der Beweisc, findct rich iu der leidcr wenig b e b n t e n Abhandlung von Listing, Vorsfudieic sur Y'opoloyie, welche in den GIttinger Studien (Enter &I. Wittingen , 1847) erechienen ist. Viellcicht kanii der vomtehende Aufsatz dw~u dieneu, die Aufnlerksamkeit der Geoineter auf diese rrucli in vicleii mderu Bedelilingen inhdtreiclie ArLeit wieder hinziilenken.

Traversing a Line Complex Without Repetition or Interruption

11.23

On the Possibility of Traversing a Line Complex Without Repetition or Interruption by Carl Hierholzer. Communicated by Chr. Wiener*) In an arbitrarily interwoven line complex1), the different parts of the complex may be called branches of a point. Along them one can leave the point in question. A point with several branches should be termed a node. It is called as manyfold as the number of branches, and depending on this number it is either even or odd. An arbitrary double-point is thus a four-fold node, an ordinary point can be described as a two-fold node, and a free end as a one-fold node.

If a line complex can be traversed in one stroke without any part of a line being repeatedly traversed, then it has either two odd nodes or none. If one passes through any point during the traversal then two branches of a node are consumed in that way, and because no line should be traversed twice, a point which one crosses altogether n times must be a 2n-fold node. A point can therefore be an odd node only if it is not crossed once in the traversal, i.e., only if it is the initial or terminal point. Therefore, if in the traversal one finally returns to the initial point, only even nodes can be present; if not, both the initial and terminal points are odd nodes. Conversely: if a connected h e complex contains t w o odd nodes or none, t h e n it can be traversed in one stroke. *) Privatdozent Dr. Hierholzer, whose service t o science, unfortunately, was ended by his early death ( 13 September 1871), lectured on the following investigation t o a circle of fellow mathematicians. In order t o be saved from oblivion, the work had to be restored from memory because of the lack of any written notes. With the help of my honored colleague Liiroth I have tried to do this as accurately as possible in the following article.

I use the translation line complex for Linienzug because the term line system as used in [BIGG7Ga] is an inaccurate translation from a linguistic point of view (although from a graph theoretical point of view it is probably the best possible). The term ‘one-dimensional complex’ as I found it in a mathematical dictionary is an exact translation for ‘Linienzug’, but it does not seem t o comply with the standard notation used in Hierholzer’s times. In fact, Listing uses already the term ‘Lineaxkomplexion’ (linear complex) more than twenty years before the appearance of Hierholzer’s paper; [LIST48a].

11.24


For: a) If one has traversed any part of the line complex, then every node in the part not yet traversed is even or odd, as it was in the whole complex; only the initial and the terminal point of the traversed part change their parity, unless they coincide. For, two branches are consumed by the traversal of a point, and one branch is consumed at each of the beginning and the end of the traversal.

b) If one has started to traverse a line complex at an odd node then one can be stopped only at another odd node. For, one has consumed two branches at an even node by each traversal of it, so that after repeated arrivals at the same [node], there is at least one branch left for leaving it. The initial point has been transformed, however, into an even node by the beginning [of the traversal], so that there also it is impossible to be stopped. If one starts, however, to traverse a line complex at an even node, then one can be stopped there, for it has been transformed into an odd one by the beginning [of the traversal]. c ) Now, if a line complex has two odd nodes, start the traversal at one of them; necessarily, one will stop at the other. The covered line complex is an open one in this case. On the other hand, if the given line complex has no odd nodes, then begin the traversal at an arbitrary point which is thus an even node; necessarily, one will then finish at the initial point. The covered line complex is a closed one in this case.

d) If thereby a part b remains untraversed, then that same [part] must contain only even nodes because the two possibly existing odd nodes have been consumed by the first traversal and the remaining nodes have retained their parity. At the same time, b must be connected to the already traversed trail a by at least one common point, for otherwise the complex would break into several parts not connected together. If, in the traversal of a, one passes into b at a point of connection P , then necessarily one has to finish in b at P, and can then continue the traversal of a as before. In this way one attaches every part not yet traversed to the ones already traversed, and thus the whole complex is traversed in one stroke. Still another theorem follows: a line complex can have only an even number of odd nodes. Indeed, if one traverses part of the line complex by starting at an odd node and continuing as long as possible, one finishes at an odd node, and thus uses two odd nodes. By repetition one can diminish the number of odd nodes to less than two. This remainder, however, cannot be one, but must be zero. For if a complex were to

Traversing a Line Complex Without Repetition or Interruption

11.25

have only one odd node, and one were to begin to traverse the complex there, then one would never come to an end, for this is possible only at another odd node. The number of odd nodes in the original line complex is therefore even. Carlsruhe, December 1871. Editor’s note. The essence of the foregoing, but in a shorter form partly without details of proofs, is found in the regrettably little known treatise Vorstudien tur Topologie [Preliminary Studies on Topology] by Listing, which appeared in Gottinger Studien (First Vol., Gottingen 1847).2) Perhaps the foregoing article can serve t o direct the attention of geometers towards this work again. I t is rich in content in many other respects as well.

We finish this chapter with a quotation from Oswald Veblen’s article [VEBL3la]. Although it is generally accepted as one of the basic papers in the development of graph theory, it contains a minor flaw. For it is not true in general that “Each of these 1-circuits [of Ci] is coincident with/[corresponds to] a 1-circuit of C,, . . .” (see below). Thus our third pillar has a minor crack. However, Veblen showed already in his paper [VEBL12a] how this ‘crack’ can be removed: “. . .For let us start with an arbitrary edge . . . and describe a path among the edges . . .Whenever there is an edge by which this path approaches a vertes, since the number of . . .edges at this vertes is even, there is a[n] . . .edge by which the path can go away. Hence the path may be continued till it intersects itself. A portion of the path then forms a closed circuit. If this be removed there are still an even number of . . . edges at each vertex. Another circuit may be removed and so on till all edges are accounted for.”

2,

See also $17 of Euler’s article (H.F.).

11.26


Oswald Veblen, ANALYSIS SITUS

15-16 pp.

* * * * * One-dimensional Circuits. 22. A connected linear graph each vertex of which is an end of two and only two l-cells is called a one-dimensional circuit or a l-Circuit. By the theorems of 9 5 any closed curve is decomposed by any finite set of points on it into a 1-circuit. Conversely, it is easy to see that the set of all points on a l-circuit is a simple closed curve. It is obvious, further, that any linear graph such t h a t each vertex is a n end of two and only two l-cells is either a l-circuit or a set of l-circuits no two of which have a point in common. Consider a linear graph C1 such that each vertex is an end of a n even number of edges. Let us denote by 2ni the number of edges incident with each vertex a?. The edges incident with each vertex a? may be grouped arbitrarily in ni pairs no two of which have a n edge in common; let these pairs of edges be called the pairs associated with the vertex aP. Let C1' be a graph coincident with C1 in such a way that (1) there is one and only one point of CI'on each point of C1 which is not a vertex and ( 2 ) there are ni vertices of C1' on each vertex aio of C1 each of these vertices of C1' being incident only with the two edges of C1' which coincide with a pair associated with a?. The linear graph 01' has just two edges incident with each of its vertices and therefore consists of a number of l-circuits. Each of these l-circuits is coincident with a l-circuit of C1, and no two of the l-circuits of C1 thus determined have a l-cell in common. Hence C1 consists of a number of l-circuits which have only a finite number of O-cells in common. It is obvious t h a t a linear graph composed of a number of closed curves having only a finite number of points in common has a n even number of l-cells incident with each vertex. Hence a necessary and s u . n t condition that C1 consist of a number of l-circuits having only O-cells in common is that each O-cell of C1 be incident with an men number of l-cells.

* * * * *

111.1

Chapter I11

BASIC CONCEPTS AND PRELIMINARY RESULTS Most of this chapter can be found in any introductory text book on graph theory. However, since I want to reach as broad a readership as possible I have decided “to start from scratch’). Moreover, for various reasons which will become apparent in the course of studying the following chapters) it seems feasible and even necessary to treat certain graph theoretical concepts in a more differentiated way than is usually done (e.g. o p e n and closed edges, half-edges, a.s.0. - see below). This approach necessitates defining, as a first step, certain basic concepts in a relatively formalistic manner. Later on (in this and following chapters), formalizations are avoided wherever possible, giving way to a more heuristic approach to the material presented in this book, the reason being that all too often formalisms exhaust the reader far too soon. As far as this chapter is concerned, most of the elementary results are stated without proof; they can either be easily proved by a more esyerienced student, or their proofs can be easily found in any introduction to graph theory. The main reason for this procedure is, however, the desire to keep this chapter as short as possible. By the same token, many less elementary results will also be stated without proof; these proofs can be found in the references quoted. I recommend [HARA69a, BONDi6a) BEHZ79a, BEINi8a, BEIK83aI as introductions to graph theory. As for mathematical knowledge other than graph theory, it is assumed that the reader knows some set theory,’) basics in algebra and some topology of surfaces (particularly of the plane). As for algorithms, the complexity of algorithms and the complexity of decision problems, we shall in many instances proceed in a somewhat sketchy manner throughout the book and refer the interested reader to the appropriate literature ‘1 This and the second volume deal with finite graphs only; will be dealt with - to a lesser extent - in a third volume.

infinite graphs

111. Basic Concepts and Preliminary Results

111.2

where the details can be studied. Therefore, the basic concepts with respect to algorithms and their complexity will not be defined explicitly. Let us now leap into the water.

111.1. Mixed Graphs and Their Basic Parts Definition 111.1. Let there be given four mutually disjoint sets V,Ii,I 1.

Remark 111.16. For a walk FV in H consider the subsequences TIT, and W, consisting esclusively of the edges/arcs of W and of its vertices, respectively,

VV, = b,, b,, . . . ,b,

, W, = 210, vl,. . .,?J, .

It follows from Definition 111.9.1) that W can be reconstructed if one only knows VVl , and the same holds true for W, if H is simple. However, even if H is not simple but a digraph, it may sometimes be sufficient to just consider an arbitrary arc a of the form ( ~ l ~ , ~ l rather ~ + ~ ) than a fised arc joining zli to zli+l . In any case we also shall refer to a sequence Wl or W, defined above as a walk, and the same applies to special types of wallcs such as trails, paths, cycles, a.s.0. For example, this enables us to view a cycle decomposition as a partition of K ( H ) , provided one views the elements of VV, and K ( H ) simultaneously as closed (open) edges/arcs. Similarly, we shall not distinguish between a cycle C or a path P as defined in Definition 111.9.3) and the mixed graph ( K ( C ) ) H( ,K ( P ) ) H respective1y.l) While this convention may cause a decrease in the rigidity of proofs, it definitely does not diminish their clarity. By the same token, if the above VV, corresponds to a path we shall sometimes write (W,) to denote this path (going even further by writing W, as an unordered set, provided no confusion arises). In the case of a cycle we use (W, - {un}) instead of (W,).

Definition 111.17. For a mixed graph H = V U K let V, KO K be given.

c

V, and

1)We call V, ( K O a ) separating vertex (edge/arc) set if c(H-V,)

>

) H , and c ( H ) ( c ( H - K , ) > c ( H ) ) .We also say that Vo ( K O separates '1 Most authors do not distinguish between a cycle (path) and a graph induced by a cycle (path).

111.4. Walks, Trails, Paths, Cycles, Trees; Connectivity

111.23

that the empty set separates H if H is disconnected. We may also use the shorter term vertex cut for V, , and edge cut or cut set for K O . A minimal cut set I c ( H ) and V, # 8 # K O ,we call V, U KO a mixed cut (the word 'mixed' refers to V, U KO containing both vertices and edges/arcs). An n-cut is a cut set of size n. Note that a non-empty coboundary is a separating edge/arc set. 2) In the special case where I KOI= 1 and I S ( H ) may very well hold true. On the other hand, in the case of 3-regular graphs (which are also called cubic graphs) a logical upper bound for XJH) is the girth g ( G H )of G, , i.e. the length of a smallest cycle (of G,). For the same class of graphs we even have the following.

Lemma 111.38. Let G be a 3-regular graph such that K(G)< 3. Then K ( G )= X(G) = X,(G). Proof. If G is disconnected, then every component of G is 3-regular and has thus a cycle. This implies 0 = K(G)= X(G) = X,(G). If K(G) = 1, then each cut vertex v, of G is incident with a bridge of G. This is clear if one of the blocks containing v, is a loop. In the case where G - v, has two or more components, it follows that for at least one of them, C, say, precisely one e E EVcjoins vc to C, in G. That is, e separates C, from G - V(C,).X(G) = 1 follows. Moreover, no component of G - e is a tree: otherwise, such a tree would be nontrivial, thus having two end-vertices which would be joined to G - V ( C , ) by four edges, an obvious contradiction. Thus, every component of G - e has a cycle, and we have 1 = K ( G )= X(G) = X,(G). Finally suppose K ( G )= 2. It follows that G is loopless and has a vertex cut (2,y}. A component C of G - {x,y} is joined in G to both x and y since K ( G )> 1. Thus C 'absorbs' at least two edges of E,UE,. Let C' be a component of G - {x,y} - C. If C or C' is joined by two edges to x and y, then these two edges constitute a 2-cut E,. Assuming that both C and C' are joined by at least and therefore, precisely three edges to x and y, we may conclude that w.1.o.g. C is joined to x by precisely one edge e and C' is joined to y by precisely one edge f . Now { e , f } is a 2-cut E, such that x and y belong to different components of G-E,. 2 = K ( G )= X(G) follows. However, no component C, of G-E, is a tree, where E, is an arbitrary 2cut; otherwise, we may conclude as before that C,, is joined to G - V(C,) by at least four edges: again a contradiction. Thus X(G) 5 X,(G) 5 2 from which 2 = K(G)= X(G) = X,(G) follows. This proves the lemma.


111.3s

G Figure 111.12. A 4-regular graph G with K ( G )= 3, X(G) = 4, X,(G) = 6.

We note in passing that Lemma 111.38 no longer holds if we replace 3 with 4. This is exemplified by Figure 111.12. Interestingly, Theorem 111.35 has its correspondence concerning edgeconnectivity (see [HARA69a, Theorem 5.111 for its independent discovery by various authors).

Theorem 111.39. Let H be a mixed graph. The following statements are true. 1) For z, y E V ( H ) , X(z,y) is the maximum number of pairwise edge-disjoint paths in G, joining x and y, and X(z,y) 2 X ( H ) . 2)

H is k-edge-connected if and only if every pair of vertices can be joined by at least k edge-disjoint paths in G, .

We finish this section with an inequality relating to edge-connectivity and the maximum number of pairwise edge-disjoin t spanning trees. Based on work by W.T. Tutte and C.St.J.A. Nash-Williams, [TUTTGla,

111.39

111.4. Walks, Trails, Paths, Cycles, Trees; Connectivity

NASHGla] (see also [NASH64a] for an extension to the maximum number of pairwise edge-disjoint forests), the following has been proved in [KUND74a, Theorem 21. Lemma 111.40 (Kundu's Lemma). Let k E N and let G be a k-edgeconnected loopless graph. Then the number of pairwise edge-disjoint spanning trees of G is at least denotes the least integer not less than x).

1 9 1 ([XI

Proof. The basis for the proof is the following result of W.T. Tutte and C.St.J.A. Nash-Williams (see [TUTTGla, NASHGla]): For a loopless graph G let P = {V',V", . . . ,V ( ? ) }be a partition of V ( G ) . Denote qp = qGP where Gp = (. . .(Gvr)vrr . . .)vc,., is considered loopless. Then G has k edge-disjoint spanning trees if and only if

(4

> k(r - 1)

QP -

for every partition P of V ( G ) ( r =I P I may vary as P varies). Now let nT(G) denote the maximum number of pairwise edge-disjoint spanning trees of G. Re-interpreting inequality (*) we can say that there must be a partition Po of V ( G ) ,r, :=I PoI such that

Considering GPOand applying Lemma 111.4 we obtain

Consequently, GPOhas a vertex X, whose valency is at most A, = 2 . By definition of G P O ,Ex,C E(Gpo) can be interpreted as a coboundary E ( V ( i ) )C E ( G ) where V ( i )E Po corresponds to x, , and

k IIE(V(")I=IEzo15 A,

(2)

since G is k-edge-connected. Using (1) and (2) we obtain U

k 5 2 = 2(n,(G) TO

That is, n,(G)

+ 1)

*

r,-1 2 -< 2(n,(G) TO

r0

> $ - 1. The lemma now follows.

+ 1)

.

111.40


111.5. Compatibility, Cyclic Order of KG and Corresponding Eulerian Trails The concept of compatibility, one of the central themes of this monograph has occupied the largest part of my research since 1975. Several other colleagues (e.g., B. Jackson, F. Jaeger, A. Bouchet) have also been attracted by this topic. Its origin lies in questions raised and dealt with first by A. Kotzig (eulerian trails) and then by G. Sabidussi (cycle decompositions). The term compatible, explained next, is due to G. Sabidussi.') Definition 111.41. 1) Consider Ii': in a mixed graph H = V U I p.

111.7. Surface Embeddings of Graphs; Isomorphisms

111.53

course, a graph embedded in some surface can be looked at ‘from outside’ or ‘from inside’ that surface which implies that O+(G) becomes 0 - ( G ) if we switch our position from the outside to the inside of the surface. This observation also explains why the definition of unique embeddability involves O+- and 0--isomorphisms. However, if we have an embedding G, of a planar 3-connected graph G on the torus, the corresponding O+(G,) may satisfy O+(G) # O+(G,) # O-(G), where O+(G) and O-(G) refer to any plane embedding of G. In fact, G is not uniquely embeddable on the torus. Whence the question arises which cyclic orderings of E,* can be obtained from embeddings of G on various surfaces. A general answer is given by the next theorem (see [BEHZ79a, Theorem 5.111).

Theorem 111.53. Let a non-trivial connected graph G be given together with 0+(G). There exists a surface F and a 2-cell embedding G, of G on F such that G =$G1. We will frequently use the following result regarding plane graphs.

Lemma 111.54. Let G be a plane graph. The following is true. 1) If G is 2-connected then every face boundary of G is a cycle of G. 2) If K ( G )= 1 then bd(F) is connected for any face F of G and there is a face F, such that every covering walk W, of bd(F,) passes a vertex at least twice. Such vertex is a cut vertex of G. If W, passes e E E(bd(F,)) at least twice then e is a bridge of G. Conversely, if TJ (e) is a cut vertex (bridge) of G then there is some face F, , such that TJ E V(bd(F,)) (e E E(bd(F,))) and every covering walk of bd(F,) passes TJ (e) at least twice.

3) K ( G )= 0 if and only if a face F, exists such that bd(F,) is disconnected. We note in passing that Lemma 111.54 cannot be extended to include toroidal graphs (i.e. graphs embeddable or embedded in the torus). We say a graph G is of orientable genus r (non-orientable genus r ) if it is embeddable on a sphere with r handles ( r crosscaps) whereas it is not embeddable on a sphere with s handles (s crosscaps) for s < r . We denote the orientable genus of a graph G by r(G), the non-orientable genus of G by T(G). In fact, every graph has an orientable as well as a nonorientable genus. Planar graphs, i.e. graphs for which r(G) = T(G) = 0, are characterized as follows.


111.54

Theorem 111.55. (Kuratowski’s Theorem). A graph G is planar if and only if it has no subgraph homeomorphic to K3,3or K5. We refer to [THOM8lb] which contains short proofs of this theorem but also many references to various other proofs as well as various other characterizations of planar graphs; see also [ROSS76b, HOLT8la, CHEN8lal. For a refinement of Kuratowski’s theorem, see [THOM84a]. In general, given a connected graph G 2-cell embedded in some surface F having precisely T handles but no crosscaps (we write y ( F ) = T ) , or having precisely T crosscaps but no handles (we write T ( F ) = T ) , the numbers pG,qG, and fG (= the number of faces of G) are related by Euler’s Polyhedron Formula.

Theorem 111.56. Let G be a connected graph 2-cell embedded in a surface F satisfying y ( 3 ) = r or y ( F )= T . Then pG

2 - 2r

-

-I-fG =

{2 -

T

if if

F is orientable F is non-orientable .

The number 2 - 2r (2 - T ) is called the Euler characteristic x ( 3 ) of the orientable (non-orientable) surface F.We define x,(G) := x ( F ) for a 2-cell embedding of G on 3 and note that x ( F ) 2 0 for just four surfaces: the sphere and the torus which are orientable, whereas the projective plane (one crosscap) and the Klein bottle (two crosscaps) are non-orient able. In any case, Euler’s Polyhedron Formula (Theorem 111.56) and 2-cell embeddings G of a connected graph Go on a surface F in general permit us to draw important conclusions. Namely, looking at the set of the shortest covering walks FV(bd(F))of the boundaries of the faces F of G we observe that every edge e is used twice: either because it belongs to bd(F,) and bd(F2), F, # F 2 , and is thus used precisely once in both W ( b d ( F , ) ) and W(bd(F,)), or F , say, lies on ‘both sides’ of e , in which case e is used twice in W ( b d ( F ) ) .Consequently, if we denote by F ( G ) the set of faces of G and define Fi:= { F E F ( G ) / I ( W ( b d ( F ) ) = ) i}, fi :=IFi 1, then we have 00 03 i= 1

i= 1

(note that in the above sums fi # 0 only for finitely many i since G is finite). F E Fiis called an i-gonal face. We may sometimes use G~


instead of fi . Now, if 6(G)2 3 and if : = q G , f := fc P:=PG,

fj

= 0, 1 5 i

111.55

I 6, then we haye for

m

m

Applying now Theorem 111.56 we obtain

2 1 ~ 3 ( G=) 21p - 21q

+ 21f

5 14q - 21q + 6q = -q

which is impossible if x7(G) 2 0 and implies q 5 21 I x,(G) I if x7(G) < 0. A similar argument s h o w that if 6(G) 2 3, x7(G) 2 0 and f i = 0, 1 5 i I 5, then x,(G) = 0, f = f 6 , and 3p = 29, i.e. G is 3-regular. Summarizing parts of these considerations we obtain the following.

Corollary 111.57. Let 3 be a surface, x ( 3 ) its Euler characteristic. There are only finitely many connected graphs Go with b(Go)2 3 (possibly none) such that a 2-cell embedding G of Go on 3 has no i-gonal face, 1 5 i 5 6. In particular, if x ( 3 ) 2 0 then G must have an i-gonal face for some i E {1,2,3,4,5, 6}, and, more specifically, if x ( 3 ) > 0 then G must have an i-gonal face for some i E {1,2,3,4,5}. However, some of the preceding considerations applied to connected 3regular graphs G lead to

That is, even if x7(G) and all f i , i # 6, are known (with all but one f i , i # 6, and x7(G)determining the remaining fi,i # 6), this equation gives us no clue concerning the number fs. In fact, the following is true.

Lemma 111.58. Let F be an orientable or non-orientable surface of genus T , and let G3(.F) be the set of all connected bridgeless 3-regular graphs 2-cell embedded in F. Then an infinite subset G:(.F) C G3(.F) exists such that for any G,,G2 E G:, G, # G,,we have G , G,, whereas the number of i-gonal faces is the same in both G, and G, for every i # 6.

!+

A sketchy proof. Replace every edge e = sy with a hexagon C" whose z, x,, yl,y, yz , where xi,yi # vertices are labeled counterclockwise by zl,


111.56

V ( G ) ,i = 1,2, and G , := G E G3(F)has been arbitrarily chosen. View the edges of C, joining xi with yj+,, i = 1 , 2 (putting 3 = l), as 'long' edges, the others as 'short' edges. For e, f E E, identify 5 E C" with 2 E Cf,and identify precisely one x i E C" with xi+, E Cf,i = 1 , 2 (putting 3 = 1). The corresponding edges of C" and Cf are also identified. It follows that the resulting graph G, is 3-regular, and that there is a 1-1correspondence between i-gonal faces of G, and i-gonal faces of G, for every i # 6. In fact, a 6-gonal face of G , (if such a face exists) corresponds in G, to a 6-gonal face other than any of these C", e E E(G,). We also have G , $ G, since pG2 = 4pGl . By repeating the above procedure a set G:(.F) as described in the statement of the lemma, can readily be constructed (we note, however, that there may be other constructions as well). Looliing at Euler's Polyhedron Formula as well as at the numbers di := IV,(G)I and f i , 1 5 i < 00, a duality is suggested by this formula which leads to the next definition.

Definition 111.59. Let G be a graph embedded in a surface F. Define the geometrical dual (or simply dual) graph D ( G ) embedded in F as well in the following way. There are bijections Q : V ( D ( G ) )t F ( G ) and P : E ( D ( G ) )+ E ( G ) such that a)

VF E

F whenever Q ( v ~=) F ;

n ( Z - h ( e ) ) I= 1 whenever P ( e F F r ) = e , while P ( e F F r ) # e implies e F p ne = 0 , where /3 is such that e F F , E E,, nEVFr if and only if / 3 ( e F F r ) E E ( b d ( F )n bd(F')). This includes the case where b)

I e F F # n B I=I

eFFr

F = F' lies on both sides of e in which case e F F is a loop. It follows from this definition that G ~3 D ( D ( G ) ) ;thus we simply write G = D ( D ( G ) ) . In particular, Q is such that i-valent vertices of D ( G ) correspond to i-gonal faces of G. Thus j-gonal faces of D(G) correspond to j-valent vertices of G. This does not preclude D ( G ) having loops and/or multiple edges although G may be a simple graph. However, if G is a plane 3-connected graph, then D(G) is simple. If G # K4 is a connected, plane 3-regular graph and pG > 2, then K ( D ( G ) )= X,(G). We observe that vF E V ( D ( G ) )is quite often called the capital of F . We also note that G N D ( G ) for G = K4 ; thus we call G self-dual if

G N D(G). We shall need another (embedded) graph derived from an embedded graph.


111.57

Definition 111.60. Let G be a graph embedded in some surface 3 such that every face boundary is a cycle. Define the radial graph, denoted by R(G) and also embedded in 3, in the following way. V ( G )c V ( R ( G ) ) , and there are bijections a : V ( R ( G ) )+ V ( G )U F(G) and PF : E,, -+ V (b d (F ))for every F E F(G) such that a) a(.) = 'L' for v E V ( G )and

E F whenever a ( v F )= F ;

'L'~

b) for ' L ' ~satisfying a ( v F ) = F and e E E,,, we have hR(e) = { v F , P F ( e ) }( h R denotes the incidence function for R(G)). That is, vF is joined by precisely one edge to each 'L' E V(bd(F)). Note that R(G) is bipartite in any case. We introduce some additional graphs. A graph G embedded in a surface 3 is called a triangulation of F if F3(G) = F(G). Note that even a triangulation of the plane need not be a simple graph. Moreover, triangulations are the duals of cubic graphs, and an eulerian triangulation of the plane is the dual of a cubic, bipartite, plane graph. We identify any plane embedding of the 1-skeletons of any of the five platonic solids (tetrahedron, cube, octahedron, dodecahedron, icosahedron) or any other 3-dimensional convex polyhedron with the corresponding convex polyhedron. This is admissible because of a theorem of Steinitz and Rademacher (see, for example, [HARA69a, Theorem 11.61 or [BOND76a, p.1351). In particular, we observe that K4 is the graph of the tetrahedron. We denote by 0, the graph of the octahedron (see Figure 111.3), by the graph of the n-sided pyramid and observe that W,, is also called the nwheel (see Figure 111.16). Similarly, we shall speak of the graph of the n-sided double pyramid and note that 0, is the 4-sided double pyramid. Finally, a cycle C of length n embedded in the plane is called n-gon, and we consider a plane graph G to be a triangulation of an n-gon C if C = bd(F,) for some Fo E F(G) and F3(G) = F ( G ) - { F , } . Thus a triangulation of the plane can be viewed as a triangulation of a triangle, and note that n-wheels are special types of triangulations of n-gons.

w,,

Many considerations concerning plane graphs (and some of the preceding results) implicitly use the nest (topological) result which we shall invoke explicitly on various occasions. Theorem III.61 ( J o r d a n Curve T h e o r e m ) . Every simple closed curve C in the (euclidean) plane E2 decomposes E2 into precisely three sets E' = C u intC u extC


111.58

"6

Figure 111.16. The 5-wheel and the 6-wheel. Note that

w,=

such that intC (eztC)is a bounded (unbounded) region, and C, nC for every (continuous) curve C, joining 2 E intC with y E eztC.

#0

Consequently, if G is a two-connected plane graph, all but one face of G are bounded regions; the remaining face is called the unbounded face and denoted by F,. However, every plane graph has this property. In the case of outerplane graphs we shall assume in most of the instances that V ( G )= V(bd(F,)). Corollary III.6l.a. Let C be a cycle of the plane graph G , 21 E V ( G ) , and suppose the open edges e , f E E, lie on different sides of C. Then e and f belong to different face boundaries of G. Definition 111.62. Call a cycle C of a graph G separating cycle if either V ( C )is a vertex cut or if G is embedded in the plane and both intC and eztC contain an open edge of G (note that C can be viewed as a simple closed curve). If C is a simple closed curve in the plane such that C f l G C V ( G ) and if both intC and extC contain an open edge of G, then we shall also refer to C as a separating cycle.

Note that for a plane 2-connected graph G , every cycle C C G satisfying C @ { b d ( F ) / F E F ( G ) } is by this definition a separating cycle.


111.59

We finish this section with an observation concerning plane triangulations.

Proposition 111.63. Let G be a triangulation of the plane, and let {V,, V,} be a partition of V(G). The following is true. 1) Both (V,) and (V,) are acyclic if and only if they are trees. 2 ) If (V,) is disconnected, then (V,) contains a separating cycle.

Proof. Suppose (V,) is disconnected. Consider a component G, of (V,). Viewing G, as a plane graph in its own right, we may conclude that the boundary C of its unbounded face is well defined. Now consider V‘ c V(G), the set of vertices adjacent to elements of V(G,). Since G, is a component of (V,), V’ 5 V, follows. Since G is finite, G, can be chosen such that (V,) - G, C eztC,, where C, is a Jordan curve satisfying G, c intC, and C, n V(G) = 8. C,can be chosen to be a cycle for of D ( G ) such that e E E(C,) implies e E E(bd(F,,)) n E(bd(FW2)) some 21, E V(G,) , 21, E V’. By this choice of C,, it follows that if A = bd(F,), FA E F ( G ) , satisfies C,n A # 0, then C, n FA is an open curve C, whose end-points lie on elements of E ( A )and 1 1. Assume a E N U (0) exists such that k > 2 a and b(G) 2 k - a. Then G is hamilt onian.

Proof. We proceed indirectly. Since the corollary holds if G II I 0,

1) implies 3). We proceed by induction on a(G) as in the above proof until we reach G,,, (in the case G = K,, S = 0 is the unique cycle

IV. 2

IV. Characterization Theorems and Corollaries

decomposition of G). By induction, G,,, has a cycle decomposition S,,, . Every C E S,,, not containing v , , ~is a cycle in G as well. Let C, E S,,, denote the cycle with w,,, E V(Cl,2). If w $ V(Cl,2), then Cl,2 corresponds in G to a cycle Ci,, with w E V(Ci,2).Otherwise, VjC,,,) 2 {v,,,, v} and C, corresponds to a subgraph H of G in which w is 4-valent and the other virtices are 2-valent. Hence H can be decomposed into two cycles C,,C, with ei E E(C,),i = 1 , 2 (just start a run in w along el and stop after reaching v again; this gives C,, and C, = H - C,).

,

Consequently, we obtain a cycle decomposition of G by defining in the case 21 $ V(C,,,)

and in the case v E V(C,,,)

2) implies 1).Consider a run through a fixed eulerian trail T starting at w E V(G). For T = w, we have G = I 0. For every x E V(G), x # w (if such x exists), every time we reach x in that run along some edge we leave x along some other edge. Since T uses every edge of G exactly once we have a pairing of the edges incident with x where every pair consists of an incoming and an outgoing edge. Consequently, d(v) is even. But also for w we have a pairing of the edges incident with 21: one pair consists of the first and the last edge traversed by T , and the other pairs are defined the same way as above for x # v.

3) implies 1). Let S = {Cl,. . .,C,} be a fixed cycle decomposition of G (possibly S = 0). For every 21 E V(G) and every i = 1,.. .,t , we have dCj(v) = 0 or 2 depending on whether $ V(C,) or w E V ( C i ) . Consequently, since 1

E ( G )=

u

W

i

)

i=l

we have

t

d(v) =

Ed&)

f O(mod2)

i= 1

This finishes the proof of Theorem IV.l.

.

IV.l. Graphs

IV.3

We note in passing that in the proofs of the preceding two implications, a loop zrzr or xx was counted twice which is in accordance with the convention that a loop contributes two units to the degree of a vertex We also note that if G is disconnected, then it is eulerian if and only if it has a cycle decomposition (see Exercise IV.1). Thus the concept of a cycle decomposition is, in fact, a more general concept when compared with the concept of an eulerian trail. On the other hand, if G is connected, then the equivalence of these two concepts manifests itself in each block of G (see Exercise 1V.l.b) and c)). The next two characterizations of eulerian graphs can be deduced by arguments similar to those used in the proof of Theorem IV.l; their proofs are therefore left to the reader as exercises. Corollary IV.2.(see also [KOTZ68a, Lemma 11) A graph G is eulerian if and only if G has a decomposition into closed trails. Corollary IV.3.A graph G is eulerian if and only if it has an eulerian orientation. Another characterization result can easily be proved by using the fact that the number of odd vertices in a graph is even (see Exercise IV.4). Corollary IV.4.A graph is of even size.

G is eulerian if and only if every cut set of G

The following characterization of eulerian graphs is the outcome of more recent independent research by Toida [TOID73a] and McKee [McKE84a]. The proof presented here is independent of their work. For another proof of this theorem see [WOOD88a].

Theorem IV.5. A graph G is eulerian if and only if every edge of G belongs to an odd number of cycles of G. Proof. Suficiency. Suppose for a graph G that each edge e belongs t o an odd number of cycles; call this number ye.

We assume G to be loopless because a loop belongs to exactly one cycle, and the deletion of loops does not alter either the parity of any degree or the parity of the number of cycles an edge uv,u # zr, belongs to. Let S = {Cl,. . .,C,} be the set of cycles of G. Now replace in G each edge e by -ye edges and replace in each of the cycles containing e the edge e with one of the -ye new edges. Thus we arrive at a new graph Go and a

IV.4


set So = {C:,. . . ,Ck} of cycles which is, in fact, a cycle decomposition of Go by the above replacement procedure. By Theorem IV.1, Go is eulerian, i.e., dGo(v) = O(mod2) for every II

E V ( G o )= V ( G ) .

However, for E, = {el,. ye,

. .,e d } C E ( G ) we have

1( m o d 2 ) for i = 1,.. . ,d, by hypothesis,

implying d

0

d G o ( v )=

Ere; d ( m o d 2 )

.

i= 1

Consequently, d = d ( v ) is even for every

II

E V ( G ) ;i.e., G is eulerian.

Necessity. Suppose G is eulerian. We proceed by induction on p =I V (G ) I. If p = 1, then every edge is a loop and hence belongs to exactly one cycle, i.e., to an odd number of cycles. - In what follows, we assume w.1.o.g. that G is connected. Suppose p > 1. Since G is connected there exists an edge e = xy with x # y. X(e) 2 1 denotes the multiplicity of e. Form the eulerian graph G, by contracting the edge e and denote the vertex of G, replacing the vertices x , y E V ( G ) , with z. The edges incident with a in G, can be partitioned into three classes E,(z),E,(y), E,(zy) where E,(x)consists of the edges of G incident with 2 but not with y, E,(y) being defined analogously, and E,(xy) is the set of loops corresponding in G to the X - 1 edges incident with x and y but different from e (see Figure IV.1). We now observe that any cycle C ( e )of G containing e corresponds in G, either to a loop in E,(xy) or to a cycle containing some edge of E,(z) and some edge of E,(y). We denote a cycle of the latter type by C(ei,E,(y)) where ei E E,(x). Furthermore, let k = lEz(x)l, let C ( e i , e j ) denote a cycle of G, containing the two different edges ei, e j E E,(x),and let yG(e) ( ~ ~ , ( e denote ;)) the number of cycles of G (G,) containing the edge e (ei E E,(z)). By hypothesis we have

k + X = dG(x) and by induction, yG,(ei)

0 (m0d2)

= 1 (rnod2)

.

,

IV.l. Graphs

IV. 5

/

Figure IV.l.The graph G, Therefore we obtain (by observing that IE,(zy)l = X - 1)

= x - 1+ k

3l(mod2)

EL,

(note that in rG,(ei) every cycle C(ei, ej) is counted once in yGe( e i ) and once in -yG,(ej)). Hence the number of cycles of G containing e E E(G), is odd. This finishes the proof of the theorem. Theorem IV.5 enables us to prove another characterization of eulerian graphs in the form of a parity result on the number of cycle decompositions of an eulerian graph (see also [BOND86a, Corollary 3.43).

Corollary IV.6. A graph is eulerian if and only if it has an odd number of cycle decompositions. Proof. If a graph G has an odd number of cycle decompositions, then it has at least one and is therefore, by Theorem IV.1 applied to each component of G, eulerian. Hence assume G to be eulerian; let e E E ( G ) be arbitrarily chosen, and let C,, . . .,C, be the cycles containing e. By Theorem IV.5, T 1(mod2). We proceed by induction on q = IE(G)I.

=

IV.6


If q = 0, then S = 0 is the unique cycle decomposition of G;hence assume > 0.

q

If Gi = G - Ci = 0 for some i, then T = 1 and G is a cycle with S = {C,} as its unique cycle decomposition. Assume, therefore, that Gi # 0, 1 5 i 5 T . By induction, G ihas an odd number of cycle decompositions (note that Gineed not be connected even if G is). This yields an odd number of cycle decompositions of G containing Ci.Denote this number by s(Ci),and denote by s(G) the number of cycle decompositions of G. Consequently, T

s(G) =

s(Ci)

T

1 (mod2)

.

i= 1

Corollary IV.6 now follows.

I would like to point out that it was through the article [LESN86a] that I became familiar with the papers [TOID73a, McKE84aI. The same is true for the following characterization of eulerian graphs which is due to Shank, [SHAN79a]. The proof presented here differs from Shank’s proof. Theorem IV.7. A connected graph G is eulerian if and only if the number of subsets of E ( G ) (including the empty set) each of which is contained in a spanning tree of G, is odd.

Proof. Let G be an arbitrary connected graph. Let rG denote the number defined in the statement of the theorem, and let T G ( e )denote the number of the corresponding subsets of E(G) containing e. We proceed by induction on q =I E ( G )I. Since a loop is a cycle it cannot be contained in any spanning tree of G. Hence we assume w.1.o.g. that G is loopless. Consequently, if p =I V ( G )I= 1, then G = K,, q = 0, and E(G) = 0 is the only subset contained in the unique spanning tree G = Kl. Hence we assume p > 1,q > 0 (Note that K , is eulerian). Let e = zy be an arbitrary edge. Form G, by contracting e with V(G,) - V ( G )replacing x and y. G, has fewer edges than G, and so does G-e. G , is connected since G is connected, and G-e is disconnected if and only if e is a bridge of G. z E

The subsets of E ( G ) each of which is contained in some spanning tree of G , are divided into two classes: in the first class are the sets containing e , in the second those not containing e.

IV.l. Graphs

IV .7

A set in the first class, M say, corresponds to M’ := &I - { e } in G, with M’ contained in some spanning tree of G,; and &I’ c E(G,) belonging to a spanning tree of G, corresponds to M = M‘U { e } with M belonging to the first class. Also, note the 1-1-correspondence between spanning trees of G containing e and spanning trees of G,. Consequently, we have by the above and by induction

with

r

1(mod 2)

if and only if G, is eulerian

.

(2)

Furthermore, if N c E ( G ) belongs to the second class, then N belongs to a spanning tree of G - e if and only if G - e is connected, i.e., if and only if e is not a bridge of G. Thus we distinguish between two cases. 1) e is not a bridge of G. Then we conclude from the above

2) e is a bridge of G. Then e is contained in every spanning tree of G (see Theorem 111.18.2) and Theorem III.21), and any N C E ( G )belonging to the second class is an N’ c E(G,) contained in a spanning tree of G, and vice versa. Thus, we have in this case

Consequently, in case 2) we have rGE 0 (mod2), in accordance with the statement of the theorem since G cannot be eulerian (an eulerian graph has no bridges by Corollary IV.4), regardless of whether G, is eulerian or not. To finish the proof of the theorem we consider case 1) and assume first that G is eulerian. Then Ge is eulerian as well, but G - e is not eulerian. By induction and (2), (3) we then have

in accordance with the statement of the theorem.

IV.8


Finally assume G not to be eulerian. If none of G, and G - e is eulerian, then induction plus (2) and (3) yield

If however, one of G, and G - e is eulerian, then both are eulerian; for in this case z and y are the only odd vertices of G. Consequently,

i.e.,

rG

0 ( m o d 2 ) if G is not eulerian.

This finishes the proof of the theorem. We remark that Theorem IV.5 and Corollary IV.6 have been derived in [BONDSSa] in a more general setting where certain paths and cycles are being counted. On the other hand, Theorem IV.5 can be deduced from a result on matroids; this has been done in [McKE84a 1. Toida, having proved in [TOID73a] only the necessity of the condition of Theorem IV.5, deduces his result from a result on paths in connected graphs with precisely two odd vertices. In the next chapter, we shall obtain this result as a consequence of Theorem IV.5 - Welsh, [WELS69a], considering the characterization of eulerian graphs by cycle decompositions, extends this concept to binary matroids and shows that a binary matroid is eulerian if and only if its dual matroid is bipartite.

IV.2. Digraphs We now turn to the characterization of eulerian digraphs. Our next theorem is the analogue to Theorem IV.l (see [KONI36a, p.29 3. Theorem IV.8. For a weakly connected digraph D , any two of the following statements are equivalent:

1) i d ( v ) = od(v) for every v E V ( D ) (i.e., D is eulerian). 2) D has an eulerian trail T . 3) D has a cycle decomposition S.

IV.2. Digraphs

IV.9

Proof. 1) implies 2). Proceeding by induction on

o ( D )=

c

(od(v)- 1) = g - p

,

uEV(D)

we first note that if a ( D ) 5 0, then D = K,, or D is a cycle C since D is weakly connected; starting at any v E V ( D ) ,either T = v, or the run through C is uniquely determined by the orientation of C ,and this run defines an eulerian trail T of D. Hence we assume o ( D ) > 0; consider any v E V ( D ) with i d ( v ) = od(v) > 1, and let a, = ( v , , ~ ) be any arc incident to v, and let a, = ( u , v 2 ) , a 3= (v,v3) be any two arcs incident from v. Both D,,, and Dl,3 are eulerian, and by the Splitting Lemma at least one of them is weakly connected; say, Dl,2is weakly connected. By induction, since U(D,,~.)< a ( D ) , D,,, has an eulerian trail T1,2which induces an eulerian trail of D the same way as in the undirected case. 1) implies 3). Again we proceed by induction on a ( D ) . Since the case a ( D ) = 0 can be argued the same way as in the proof of the preceding implication (with S = 0 if A ( D ) = 0), therefore, we assume o ( D ) > 0. As in the proof of the preceding implication we apply the Splitting Lemma and assume w.1.o.g. that D,,, is weakly connected. Let S,,, be a cycle decomposition of D,,,. Arguing as in the proof of Theorem IV.l we conclude that the cycle C{,2 E S,,? corresponds in D either to a cycle C,,,or to a subdigraph of D which is the arc-disjoint union of two cycles C,,C2. Hence, also for digraphs we obtain a cycle decomposition S by defining or

depending on whether Ci,, corresponds to a cycle in D or not. To prove that 2) implies 1) and 3) implies 1) one can proceed along the same line of arguments as in the proof of Theorem IV.l. We leave it therefore to the reader to finish the proof of Theorem IV.8 (Exercise IV.5.). We note in passing that Corollary IV.2 can be restated for digraphs as well, while, in general, this is not true for Corollary IV.3; for, any 4-edge-connected eulerian graph G can be oriented in such a way that the corresponding digraph D is not eulerian although D is strongly connected (just take an eulerian orientation of G and reverse the orientation

IV. 10


of precisely one arc; see also Esercise IV.6). Next we prove a result which is the analogue to Corollary IV.4.

Corollary IV.9. A digraph D is eulerian if and only if every cut set A, C A ( D ) with bipartition V,,V, contains as many arcs incident from V, as it has arcs incident from V, (in other words: u + ( q ) = u-(V;.) for i = l,2).

Proof. Suppose the condition stated in the corollary holds for arbitrary A,. Then it must also hold for A, = A , - A,; i.e., where V, = {v} or V, = {v} for arbitrary v E V(G).Say, V, = {v}. By hypothesis, there are as many arcs of A, - A, incident from v as there are arcs of A, incident from V ( D )- {v}, i.e., incident to v. That is, od(v) = i d ( v ) for arbitrary ZJ E V ( D ) ,i.e., D is eulerian (note that a loop in v contributes a unit to both id(v) and od(v)). Conversely, if D is eulerian, then consider an arbitrary cut set A,. Since the condition is fulfilled vacuously if A, = 0, assume A, # 0. Form D(A,), the tie-up of D with respect to A, (see Figure IV.2.).

u Figure IV.2. The eulerian digraph D and its tie-up D(A,).

For i = 1,2, we conclude from the fact that odDi(v)= odD(v>= id,(v) = idDi(v)for every v E y,that o d D , ( z j )= i d D i ( z i ) holds as well (see Corollary III.5a). Since an arc of A, is incident from V;., to V;. respectively,

IV.2. Digraphs

IV.11

if and only if the corresponding arc in A,; has this property, we conclude that A, contains as many arcs incident from Vias it contains arcs incident to , (I namely i d D i ( z i ) = odD,(zi),i= 1,2 (note that this equation is independent of i because idD,(z,) = OdD,(Z,). This finishes the proof of Corollary IV.9.

However, Theorem IV.8 also implies the following Corollary whose proof is left as an exercise. Corollary N . l O . Every weakly connected eulerian digraph is strongly connected. However, Theorem IV.5 has no analogue to digraphs. This can be seen from any eulerian orientation of 1 u - ( X ) , then every e E E ( X ) corresponds in D to an arc incident to X ; otherwise e corresponds to an arc incident from X . Consequently, if H ; and H; are the two disjoint (mixed) (di)graphs resulting from the tie-up of H* (see also Figure IV.2) with respect to a critical cut set Cx of H * , and if Hi* has an eulerian orientation D;,i= 1,2, then an eulerian orientation of H* is induced in a natural way by 0; and 0; (see also Exercise IV.9). Note that for {z?} = V(H,*)- V ( H ) ,f H I ( z z f ) = f H I ( X )(even if Cx is not critical). .)If H* satisfies the cut condition, then G*, the graph underlying H * , is eulerian; consequently Gf, the graph underlying Hi*,i = 1 , 2 , is eulerian; hence fHr(Y) is even for any Y C V ( H ; * ) . i

Now we prove

H has n o non-trivial critical cut sets.

(1)

Suppose it does. Then there exists X C V ( H ) with IX I> 1, I V ( H )I> 1, and f H ( X )= 0. Let Hi with { z i } = V ( H i )- V ( H ) be the (mixed) (di)graphs of the tie-up with respect to Cx,i = 1 , 2 .

X

Suppose f H . ( Y )< 0 for some Y C V ( H , ) with zi 6 Y (see a), c ) ) , i E {1,2}. +hen Y c V ( H ) and C, is also a cut set of H by the defini) f H i ( Y2) 0. Th'is contradiction tion of Cx and a tie-up; and f H ( Y = shows that H , and H , satisfy the cut condition. Since Cx is non-trivial, I V ( H J I 2; otherwise let H,,, = H with x,,, = x. Let H , = (H,,, - x,,,) U {uw}. It follows from the choice of H that it suffices to show that H , satisfies the cut condition: an eulerian orientation of H, corresponds in a natural way to an eulerian orientation of H,,, and hence to an eulerian orientation of H ; and V(H,)5 V ( H ) ,IE(H,) I=IE(H) I -1. f o r every

IV. 16


f H ( x )>_ 2 implies fHo(z)2 0 if x E V ( H o ) ;but suppose there exists X c V ( H o )with z @ X such that fHo(X) < 0 (see a), and note that f H o ( X )is even anyway). Let C, denote the corresponding cut set in

Ho ) 0, If uw &I C, then C, = C, is also a cut set of H,,, with f H , , 2 ( X ‘ < where X‘ = X or X’ = XU(cc1,,} depending on whether I {u,w}nX I= 0 *

or = 2.

If uw E C,, then w.1.o.g. u E X , w $ X ,and C i = (Cx-{uw})U{ux1,2} is a cut set of H,,, with fH1,*(X) < 0 (in this case x,,, # X’= X). Consequently, in both cases we obtain X’C V ( H )defining a cut set with f H ( X ’ ) < 0 unless uw @ C, and xl,, E X‘. We now conclude that u,w E X,X‘ - {xl,,} defines in H a cut set C = C, U {ux,wx},and {xl,,}) 5 0. Consequently f H ( X ’ - {xl,,}) = 0. Therefore, thus fH(X’C is critical. Since u,w E X,x E V ( H )- X ,it follows from (1) that {x} = V ( H )- X. This and a) yield fH(x) = 0, a contradiction which finishes the proof of (2). Therefore, the critical cut sets of H are the sets C{vl where 21 E V ( H ) is arbitrary. Since E ( H ) # 8, there must exist x E V ( H ) with E, # 8. Since C{zlis critical, in the build-up of an eulerian orientation of H the orientation of every e E E, is defined by o d H ( x )- i d H ( x )(see b)). Correspondingly, replace e = zy with a E {(z, y), (y, z)} for a fixed e E E,, thus obtaining H‘ with V ( H ’ ) = V ( H ) and I E(H’) 1 4 and T is an eulerian trail of D, then a T-compatible eulerian trail T‘ of D exists.

Proof. In view of Corollary VI.15 it suffices to construct from D a connected eulerian digraph D‘ without 4- and 6-valent vertices with X(D’) being induced by X, such that no transition of T at a 6-valent vertex of D defines A: where x is 2-valent in D’. In fact we can do better by constructing D’ in such a way that X(D’) defines an eulerian trail of D‘. For this purpose we assume D to have a 6-valent vertex v (otherwise, by Corollary VI.15, nothing has to be proved). Write T in the form

+ ,- - - , e- 3,v,e+3 , . . . , x T = x , . . . , e l , v,e t , - - . , e-2,v,e2 where X(v) = { { e i ( v ) - ,ei(v)+}li= 1,2,3} ei(v)+ E (At)+, i = 1,2,3. Then

X, with e j ( v ) - E ( A : ) - ,

is an eulerian trail of D which follows T everywhere except in where it behaves like a T-compatible eulerian trail of D. Doing this step by step for every 6-valent vertex of D we finally obtain an eulerian trail T’ which coincides with T in all but the 6-valent vertices; and in these vertices it behaves like a T-compatible eulerian trail of D. Denoting by D’ the digraph obtained from D by replacing every 6-valent vertex with three 2-valent vertices such that T‘ corresponds to an eulerian trail of D‘, we have arrived at an eulerian digraph without 4- and 6-valent vertices with a system of transitions X ( 0 ’ ) induced by T and defining an eulerian trail of D’. By the construction of D‘, X(0’) respectively, any eulerian trail of D’ resulting from the application of Corollary VI.15 to D’, corresponds to a T-compatible eulerian trail of D. Corollary VI.16 now follows. It is clear that Corollary VI.16 cannot be extended to include eulerian digraphs with 4-valent vertices (see the discussion preceding Definition

VI.1.1. P(D)-Compatible Eulerian Trails in Digraphs

VI.15

VI.12). It is not clear which 2-regular digraphs have an eulerian trail compatible with a given eulerian trail; a related problem will be considered in subsection VI.1.2. We continue our search for results on digraphs analogous to the results previously established for graphs. As for Corollary VI.2, it is clear that if we proceed as in the proof of that Corollary, then we can write D = D,U D, (with D, and Db being arc-disjoint), while the degree condition is being transformed into

1’) idDr(.) = odDb(v), odD,(v) = idDb(.) for every

‘u

E

V(D)

(any ‘red’ arc incident to ‘u is matched by a ‘blue’ arc incident from ’u, and vice versa). However, statement 2 ) of Corollary VI.2 is no longer valid for eulerian digraphs: to see this, consider any connected eulerian digraph D with an even number of arcs and a 4-valent cut vertex x (such digraphs exist ! - see Exercise VI.9). Then we can write

D = D,U D,,D,nD, = x, and Di is connected for i = 1,2. Thus idDl(.) = idD,(x) = o d D l ( x ) = odD,(x) = 1. Consider any bluered coloring of the arcs in accordance with an eulerian trail of D,and interchange the color classes on D,. Now assume D has been chosen s O(mod2), i = 1,2, and w.1.o.g. the arc in such a way that I A( D i) 1 incident to x in D, has been colored red. Then 1’)(see above) is still fulfilled for this new blue-red coloring, but we have

that is, every eulerian trail of D arriving at x via the red arc in D, must continue from x along the red arc in D,. Generally speaking, we are faced with the (somewhat more general) problem of having to define P ( v ) for any v E V ( D ) ,where P ( v ) consists of two classes: the blue half-arcs incident to and from v forming the first, the corresponding red half-arcs the second; moreover, P(v) satisfies 1’). Let P ( D ) = UP(,). Question: when does D have a P(D)-compatible eulerian trail ? The answer to this question is given in the following theorem for which we need some more terminology.

VI. 16

VI. Various Types of Eulerian Trails

Let D be an eulerian digraph, and for every v E V ( D ) ,let P ( v ) be a partition of A: into two classes,

Moreover, for i = 1,2, let Pi(.) be written as

Pi(,)= Pi,l(v) U Pi,2(v) with Pi,l(v) n Pi,2(2)) =0 where Pi,l(v) contains precisely the half-arcs of Pi(.) which are incident to v (hence Pi,,contains only half-arcs incident from TI). L,et P ( D ) = U V E V ( D )P ( v ) , and let a new digraph D , be derived from D in the following way: every 21 E V ( D )is replaced by two vertices v1,2 and t ~ , ,and ~ , the incidences in D, are defined by

Theorem VI.17. Let D be a connected eulerian digraph and let P ( D ) be defined as above. Then the following statements are equivalent. 1) D has a P(D)-compatible eulerian trail. 2) D , has an eulerian trail.

Proof. If D has a P(D)-compatible eulerian trail T , then, by definition of P ( D ) ,if T reaches an arbitrary v E V ( D )via an half-arc belonging to Pi(u), then this half-arc belongs to Pj,l(~), and T must leave 21 via an halfarc belonging to Pj+l,2(u) where we put i 1 = 1 if i = 2. Consequently, because of the definition of the incidences in D,, T defines a unique eulerian trail of D,.

+

Conversely, if D , has an eulerian trail T p , then T p defines a unique eulerian trail T of D by definition of D,. Moreover, by this definition it follows that the half-arcs defining a transition of T p in 211,2 or 212!1 belong to different classes of P ( v ) for any v E V ( D ) . That is, T is P(D)-compatible. However, Corollary VI.3 has an analogue for digraphs. Corollary VI.18. Let D be a k-regular digraph having an even number of arcs. Then D can be decomposed into two arc-disjoint subdigraphs D, and D, such that d D i ( v ) = k for every v E V ( D ) and i = 1 , 2 . In particular, if k = 2, then D is the arc-disjoint union of two 1-factors.

VI.1.2. Aneulerian Trails, Bieulerian Digraphs and Orientations VI. l i

Proof. The first part of the corollary is trivial - just use the bluered coloring of the arcs of D as described above. As for the second part, split every vertex v into two 2-valent vertices v- and v+ such that odDo(zl-) = idDo(.+) = 0 for the digraph Do thus obtained. By this definition, Dois bipartite since every vertex of Do is either a source or a sink. Consequently, each component of Do has an even number of arcs. Hence the arcs can be 2-colored with blue and red such that no vertex of Do is incident with two arcs of the same color. Considering this arccoloring in D we have at every vertex z1 E V ( D ) idGb(v)= 0dGb(w)= idG,(v) = 0dG,(u) = 1. That is, each of G, and G, is a 1-factor of D. We note in passing that a generalization of the idea just used will be employed in the next subsection. Corollaries VI.6 - VI.8 can be reformulated for digraphs as well provided the orientation of the arcs alternates in @ ( A ; ) . We state the most general result only, leaving its proof as an exercise (Exercise VI.10).

Corollary VI.19. Let D be a connected eulerian digraph with a prescribed order @(At) of the half-arcs incident with w , for every w E V ( D ) , such that two consecutive elements in O+(A:) are not both incident to (from, respectively) w. Then D has a non-intersecting eulerian trail. Note. It may be an interesting research problem to reformulate some of the preceding results for mixed graphs. Also, on the basis of the results obtained so far, it should not be too difficult to reformulate for digraphs what has been said concerning open trails, trail decompositions and path/cycle decompositions in graphs. We did not go into details here because we want to concentrate on graphs.

VI.1.2. Aneulerian Trails in Bieulerian Digraphs and Bieulerian Orientations of Graphs Definition VI.20. Let D be a connected digraph whose valencies are all even, and let TG be an eulerian trail of the graph G underlying D. The sequence To in D corresponding to TG is called an antidirected eulerian trail, in short an aneulerian trail, if for every section a i , vi+l, ai+l of T D , a; and are both either incident to zli+l or incident from v ; + ~ . D is called aneulerian if it has an aneulerian trail. If D is eulerian and has an aneulerian trail, then we call D a bieulerian digraph.


VI. 18

The following is an immediate consequence of Definition VI.20.

Corollary VI.21. If D is a bieulerian digraph, then for every o E V ( D ) , id,(v) = odD(o) O(mod2) and, consequently, I A ( D ) 10 (mod2).

=

Note that if D is just aneulerian, then the first part of the conclusion of Corollary VI.21 is false in general, while the second part, I A ( D ) 10 (mod 2), remains valid.

As a generalization of the construction of Do in the proof of Corollary VI.18, we define for an arbitrary digraph D the digraph 07 as obtained from D by splitting every o E V ( D ) into o-,o+ E V ( D 7 ) and by letting o - , vs respectively, be incident with the elements of (At)-, (A:)+ respectively.

Theorem VI.22. Let D be an eulerian digraph. The following statements are equivalent. 1) d,(v)

3

O(mod4) for every o E V ( D ) ,and DT is connected.

2) D is bieulerian. Proof. If d,(v) 0 (mod4), then id,(v) = odD(v) E 0 ( m o d 2 ) , hence GT , the graph underlying D+ , is eulerian by definition of DT . Since D+ is connected by hypothesis, so is GT. An eulerian trail of GT corresponds to an aneulerian trail of D+ and to an aneulerian trail of D by definition of D+. Hence D is a connected eulerian digraph; hence it has an eulerian trail. That is, it has an aneulerian trail as well as an eulerian trail; i.e., D is bieulerian.

Conversely, if D is bieulerian, then d,(v) 5 O(mod4) for every o E V ( D )by Corollary VI.21. By definition of DT, an aneulerian trail of D corresponds to an aneulerian trail of DT; hence 07 is connected. This finishes the proof of the theorem. The two-regular digraph on two vertices shows that the condition for D+ to be connected, cannot be deleted in the statement of Theorem VI.22. In fact, Theorem VI.22 is only a special case of the next result.

Theorem VI.23. Let D be an eulerian digraph with d,(v) 0 (mod4) for every o E V ( D ) . Then D has a unique decomposition S,, into maximal aneulerian subdigraphs of D , and I s,, /=c(D7). Proof. Consider 0 7 , and let its components be denoted by C,,. . . ,C,, k 2 1. For any i E { 1,.. .,k}, let Gi be the graph underlying Ci.

VI.1.2.

Aneulerian Trails, Bieulerian Digraphs and Orientations VI. 19

Since D is eulerian and d D ( v ) G 0 (mod4) for every v E V ( D ) ,we have d(v+) = d(v-) O(rnod2) in 07 for every v+,v- E V ( D 7 ) . Consequently, Gi is eulerian, and an eulerian trail of Gi corresponds to an aneulerian trail of Ciby definition of DT. Consequently, if D, . . . D, are the subdigraphs of D corresponding to C, , . . . ,C, respectively, then S ,, = {D,, . . . ,D,} is a decomposition of D into aneulerian subdigraphs (note that there is a 1 - 1 correspondence between aneulerian trails of Ciand aneulerian trails of Di, 1 _< i 5 k). Moreover, precisely because of the definition of 0 7 , and because C,,. . . ,C, are its components, S,, is uniquely determined and Diis a maximal aneulerian subdigraph of D for i = 1,.. .,k. Consequently) I S ,, I= c ( D 7 ) . This finishes the proof of the theorem. )

)

We observe that if D is 2-regular, then the aneulerian subdigraphs of D are necessarily the maximal aneulerian subdigraphs of D simply because the vertices of 07 are 2-valent. Which 4-regular graphs have a bieulerian orientation)i.e. an eulerian orientation admitting an aneulerian trail ? The answer we offer is not a good one from an algorithmic point of view; but it relates 4-regular graphs and 3-regular graphs in a way which will play an important role in the context of the Compatibility Problem (We note that the above question was originally stated in [BERM78b]).

Theorem VI.24. A 4-regular graph G, has a bieulerian orientation if and only if there is a hamiltonian, bipartite, 3-regular graph G, with hamiltonian cycle H such that G, N G,/L where L = E(G,) - E ( H ) . Proof. If G, has a bieulerian orientation) then fix such an orientation) call it 0,) and construct the bipartite digraph DT as before. The aneulerian trail of 0, guarantees that D+ is connected. Therefore) for H being the cycle of the underlying graph Go,G, = Go U {v+v-/v E V ( D , ) } is a 3-regular graph with a hamiltonian cycle H ; by the construction of D+ and G,, the latter is bipartite as well, and G, N G J L where L = {v+v-/v E V(D,)} = E(G3) - E ( H ) . Conversely) if G, N G,/L, and if H = G, - E(L) is a hamiltonian cycle of the 3-regular graph G,, then orient the edges of H such that H contains no path of length 2; i.e., the digraph Do thus obtained from H has only sources and sinks. The bipartition of V ( D o )defined by the sources, sinks respectively, coincides with the bipartition of V(G,) since Do is connected. Thus, every e E L joins a source with a sink since G,

VI.20


is bipartite. Hence, D, obtained from Do by identifying the same pairs of vertices as in forming G,/L from G,, is 2-regular, and a run along the aneulerian trail of Do corresponds to an aneulerian trail of D,. Consequently, since G, is underlying 0, by hypothesis and by construction, G, has a bieulerian orientation. The theorem now follows. Thus, G , being bipartite is an essential property in order that G, has a bieulerian orientation. However, G, must not be bipartite if it has such an orientation. This fact is substantiated by the next result (see also [BERM79d]).

Proposition VI.25. If G is a bipartite graph with dG(u) for every TJ E V ( G ) ,then it has no bieulerian orientation.

0 (rnod4)

Proof. Suppose an orientation D of G has an aneulerian trail To starting w.1.o.g. at v E V, along the arc (v,w)where w E V,, and {V,, V,} is the bipartition of V ( G ) = V ( D ) . Then the next arc in To is of the form (z,w), with z E V,. Then the arc following (5, w)in To is of the form (z,y) with y E V,, a.s.0. Consequently, the first component of an arc of D always belongs to V,, while the second component lies in V,. That is, D is not an eulerian orientation of G , and therefore, G cannot have a bieulerian orientation. On the other hand, we have for the class of connected eulerian graphs G where each cycle is a block of G, the following strong result. Although this is a rather narrow class of graphs, it will play an important role in the next two sections of this chapter.

Theorem VI.26. If G is a connected graph with d G ( v ) 0 (mod4) for every v E V ( G ) ,and whose cycles are the blocks of G , then every eulerian orientation of G is bieulerian.

Proof. We proceed by induction on bn(G),the number of blocks of G. In any case, bn(G) 2 2. If bn(G) = 2, then, by hypothesis, G has just one vertex incident with just two loops. Orient each of them arbitrarily in one of the two possible ways. This gives an eulerian orientation D of G; passing through one loop according to its orientation and through the other against its orientation, defines an aneulerian trail of D. Hence the theorem holds for bn(G) = 2. For bn(G) > 2 consider an end-block B, of G and its only vertex v (G cannot have 2-valent vertices by hypothesis, so every end-block of G is a loop). Depending on the degree of v we distinguish between two cases.

VI. 1.2. Aneulerian Trails, Bieulerian Digraphs and Orientations VI.21

1) dG(v) = 4. Then we form G, obtained from G - B, by suppressing in G - B, the 2-valent vertes v (see Figure VI.5).

Y

X

X

Y

G Figure VI.5. G , obtained from G by deleting the loop B , and suppressing the 2-valent vertex v of G - B,; possibly 2

= y.

G I satisfies the hypothesis of the theorem and has fewer blocks than G. Let D be an arbitrary eulerian orientation of G , and let D, be the eulerian orientation of G , induced by D. By induction, D,is bieulerian. Assuming w.1.o.g. ( 2 , ~ E ) A ( D ) , we have (v,y) E A ( D ) and (x,y) E A (D,). Considering now any aneulerian trail T, of D,, we see that T, can be extended to an aneulerian trail T of D by replacing the arc (z,y) with (2, v), v, A ( B , ) - ,v,(v, y), where A ( B , ) - means that we pass through B, against its orientation (w.1.o.g. T, passes (2,y) from x to y. Observe that for an aneulerian trail, the inverse sequence is also an aneulerian trail). 2 ) d,(v)

> 4.

Because of 1) handled already, we assume w.1.o.g. that no 4-valent vertex of G is incident with a loop of G. Consequently, if we look a t the block-cut vertex graph bc(G), then no end-vertes of bc(G)is adjacent to a 2-valent vertex of bc(G). Consequently, since bc(G) is a tree, there is a cut vertex z of G such that all blocks containing z are loops, except possibly one of them. W.1.o.g. z = v. Because 8 5 d,(v) 0 (rnodd), v is incident with two loops, B, and B, say. Now form G , = G - ( B , U B2) .

VI.22


G, satisfies the hypothesis of the theorem because 4 5 dG(v) - 4 = dG,(v). Let D be an eulerian orientation of G as in case 1)with ( 2 , ~E) A ( D ) . The orientation D, of G, induced by D is eulerian; by induction, D, is even bieulerian. Replace in any aneulerian trail T, of D, the arc (x,TJ)with (2,TJ), v,A ( B , ) - , TJ, A(B,), thus obtaining an aneulerian trail of D (A(&)- means, as above, that we pass through B, against its orientation, while A(B,) stands for passing through B, according to its orientation; also here we assume w.1.o.g that T, passes ( 2 , ~ )from ) 2 to u ) . This finishes the proof of the theorem.

Proposition VI.25 and Theorem VI.26 name classes of graphs representing the extremal cases concerning the theme of bieulerian orientations of 4regular graphs (no bieulerian orientations at all, every eulerian orientation is bieulerian). Proposition VI.25 does not account for all 4-regular graphs without bieulerian orientation. For example, if G is obtained from two 4-regular graphs Gjby subdividing an edge ei with one vertex ZI;, i = 1 , 2 , and then identifying the two 2-valent vertices TJ,and v,, then G is a 4regular graph with cut vertex TJ arising from identifying v, and v,. It is straightforward to see that G has a bieulerian orientation if and only if G, and G, have such orientations. In fact, one can prove the validity of the following statement. Lemma VI.27. A connected 4-regular graph G has a bieulerian orientation if and only if for each block B c G which is homeomorphic to a 4-regular graph B*, B* has a bieulerian orientation.

One proceeds similarly to show the validity of the next lemma. Lemma VI.28. For a connected 4-regular graph G, every eulerian orientation is bieulerian if and only if for every block B of G which is homeomorphic to a 4-regular graph B*, B* has the property that every eulerian orientation is bieulerian.

The proofs of Lemmas VI.27 and VI.28 are left as exercises (Exercise VI. 11). Note. Observe for the statements of the preceding two lemmas that G may contain cycles which are blocks of G; for those there is no homeomorphic 4-regular graph.

Because of the Lemmas VI.27 and VI.28, it suffices to deal with 2connected graphs if one wants to determine all 4-regular graphs which have bieulerian orientations, and/or the property that every eulerian

VI. 1.2. Aneulerian Trails, Bieulerian Digraphs a n d Orientations VI.23

orientation is bieulerian. The example above which has no bieulerian orientation at all, has, however, a cut vertex; but in [BERM79d], a 2connected, 4-regular graph without bieulerian orientation is given. To see that also Theorem VI.26 does not cover all graphs having the property that an eulerian orientation is necessarily bieulerian, we loolc at an arbitrary eulerian orientation D, of K5.It is a routine matter to check that D5has a hamiltonian cycle (in fact, as we shall see later, every eulerian orientation of a complete graph on an odd number of vertices, is hamiltonian). Thus if we draw such a hamiltonian cycle as the outer pentagon, then there are, formally, only two choices for orienting the inner pentagram (see Figure VI.6). These two eulerian orientations are isomorphic, though, and the vertex-sequence 1,3,2,4,3,5,4,1,5,2,1determines the aneulerian trail corresponding to these orientations. Thus the K5 has an eulerian and thus a bieulerian orientation which is unique up to isomorphisms. However, if we label the vertices of the K5beforehand, then the I(, has several eulerian orientations which are, however, bieulerian as well. In fact, the number of bieulerian orientations of any 4-regular graph containing a prescribed arc (2,y) is even, [BERM79d, Theorem 6.21.

Figure VI.6. For a given cyclic orientation of the outer pentagon of the K5,the two possible choices of orienting the inner pentagram yield isomorphic eulerian orientations of the K5,with the isomorphism indicated by the vertex labeling .

VI.24


As we have seen in Corollary VI.18, every 2-regular digraph can be decomposed into two l-factors. In general, there will be more than one such decomposition. Hence the question: what is a necessary and sufficient condition for a 2-regular digraph to have a unique l-factorization ? This question is answered as a consequence of the following proposition. Proposition VI.29. Let D be a 2-regular digraph. Then there is a 1-1correspondence between the l-factors of D and the maximal independent arc sets of 0;.

Proof. If F is a l-factor of D , then we have odF(v) = i d F ( v ) = 1 for every v E V ( D ) . That is, one of the arcs incident with v in F is incident from v+, the other to v- (if a loop is a component of F , then these two arcs are identical). Hence F corresponds in 0 3 to an independent arc set covering all vertices of D+, i.e. to a maximal independent arc set, of DT. This arc set is uniquely determined by the construction of DT , Conversely, let M be a maximal independent arc set of DT. We have IA(C)I = O(mod2) for every cycle C of G because DT is bipartite D;

by construction. Moreover, since D is 2-regular, the vertices of 0; are 2-valent. It follows that M covers all vertices of 0;. Consequently, for every v E V ( D ) there is precisely one arc of M incident from v+ and precisely one arc of M incident to v-. That is, M corresponds to a uniquely determined subgraph F of D satisfying i d F ( v ) = o d F ( v )= 1; i.e., F is a l-factor of D.Proposition VI.29 now follows.

For a 2-regular digraph D with l-factor F,D - F is a l-factor F’ of D as well; and M‘ = A(DT) - M is a maximal independent arc set of 0; if M is such an arc set; and if M corresponds to F , then M’ corresponds to F’. This follows from the definition of D 7 . Moreover, if DT has at least two components, one of which is C,say, then we have for the above M and M’ that

M“ = ( M n A @ ) ) u (M’ n (DT - C)) is a maximal independent arc set of DT different from M and M’. These considerations together with Proposition VI.29 yield the following relation between the concept of a l-factorization and that of being bieulerian.

Corollary VI.30. A 2-regular digraph has a unique l-factorization if and only if it is bieulerian.

VI. 1.3. Do-Favoring Eulerian Trails in Digraphs

VI.25

We note in passing that Proposition VI.29 is only a different way of stating and proving [BERM78b, Theorem 2.21; Corollary VI.30 appears as a statement in the introduction of [BERM79d], but it also follows from [BERM78b, Theorems 2.3 and 2.41. These are summarized as follows.

Corollary VI.31. Let D be a 2-regular digraph. Then the number of 1-factors of D equals 2c(D;) and, consequently, the number of 1factorizations of D equals 2c(DT)-1. Corollary VI.31 follows easily from the proof of Proposition VI.29 and the subsequent considerations; its proof is therefore left as an exercise. The paper [BERM78b] finishes with the question: is it true that a 4valent graph with a n odd circuit can be directed so as t o be aneulerian ? The question has a trivial positive answer. For, if G is any connected eulerian graph on an even number of edges, then let T be any eulerian trail of G and orient the edges of G alternatingly following the orientation induced by T or in the opposite direction. This yields an aneulerian orientation of G. Apparently, in the above question the word aneulerian should be replaced with bieulerian. But then the answer is negative as the following example shows: take three copies of any 4-regular graph G which has no bieulerian orientation (e.g., by Proposition VI.25 G can be chosen to be bipartite). Subdivide in each of the three copies an edge. Thus we have three graphs G,, G,, G, with just one 2-valent vertex a, b, c respectively, and all the other vertices are 4-valent. Consequently, G+ = G, U G, U G, U {ab, bc, cn} is a 4-valent graph having a triangle (i.e. an odd cycle); but G+ cannot have a bieulerian orientation because G, has no bieulerian orientation (see Lemma VI.27).

VI.1.3. D,-Favoring Eulerian Trails in Digraphs We finish this section with another type of restriction on eulerian trails in digraphs (see [BERK78a]).

Definition VI.32. Let D be a connected eulerian digraph, and let Do be a subdigraph of D. An eulerian trail T of D is called Do-favoringif and only if for every ZI E V ( D ) ,T traverses every arc of Do incident from 21 before it traverses any arc of D, := D - Do incident from 21.

Of course, every connected eulerian digraph D can be written as D = DoUD, with A ( D , ) n A ( D , )= 0; just take D, = D , V ( D o )= V ( D )

VI.26


and A(Do)= 8. In this case any eulerian trail of D is Do-favoring.') Also, Definition VI.32 indicates that the existence of a D,-favoring eulerian trail T may depend on the choice of an initial vertex for T . We prove two theorems on the existence of Do-favoring eulerian trails depending on the structure of D, := D - Do (see Proposition 111.24).

Theorem VI.33. Let D be a connected eulerian digraph, and for given v E V ( D ) let Do C D be chosen such that D, = D - Do is a spanning in-tree of D with root v. Then a D,-favoring eulerian trail starting (and ending) at v exists. Conversely, if T is an eulerian trail of D starting (and ending) at 21, and if we mark at every w E V ( D ) ,w # v, the last arc of T incident from w,then D,, the subdigraph of D induced by the marked arcs, is a spanning in-tree with root v (and hence T is a ( D - D,)-favoring eulerian trail of 0). Proof. Let Do C D be chosen such that D, = D - Do is a spanning in-tree of D with root v. Construct T by starting at vertex v with any arc (v,x), choose any arc of Do incident from x,if such arc exists; choose the arc of D , incident from x, otherwise. Continue this way until this procedure terminates at some y E V ( D ) . Then y = v; otherwise, T contains more arcs incident to y than it contains arcs incident from y, contradicting D being eulerian. Suppose T does not contain all arcs of D. Then let z be a vertex incident with arcs not contained in T . Since D is eulerian and T is a closed trail, idD-*(z) = o ~ ~ - ~#( 0.z Moreover, ) z # v by the very construction of T. By definition of D,, there is a path P ( z , v ) C_ D,joining z to v. Write

p ( z > = *, ( z 7

u l ) ,u l ,

-

* * 7

uk:,(uk, v),

;

possibly z = uk and u1 = v (i.e., P ( z , v ) may contain just one arc). By the construction of T it follows that ( z , u l ) is not contained in T ; therefore, also (u1,u2)is not contained in T (note that (u1,u2)can be contained in T only if all arcs incident to u1 are contained in T), a.s.0. In particular, (uk,v) is not contained in T , contradicting the fact that i d T ( v ) = odT(v>= i d D ( v ) = odD(v). Thus, T contains all arcs of D.

1' This observation is contained in [BERK78a] where the author attributes Theorem IV.8 to I.J. Good, an error apparently inherited from [I 2 , then there are at least m a x ( 2 , k - 1) pairwise compatible eulerian trails, [JACK87b]. In the same paper, B. Jackson put forward the following conjecture which is the starting point for the results of this section and their discussion. Conjecture VI.36. If G is a connected eulerian graph with S(G) = 2k > 2 , then G contains at least 21; - 2 pairwise compatible eulerian trails.

This conjecture has been verified in [FLEISSa] for the class of eulerian graphs G where each block of G is a cycle. Such graphs demonstrate that the bound 21; - 2 as quoted in the above conjecture, is best possible in general: for, if the edge e is incident with a cut vertex v and d ( v ) = 2k, and i f f is the only other edge incident with v and belonging to the same block as e, then {e(v), f(v)} is a separating transition and can thus not be my visit t o China in the summer of 1987. This question was asked by Prof. Li Qiao, Hefei (Anhui Province), during the same visit. Recently, G. Sabidussi asked me the same question.

VI.2. Pairwise Compatible Eulerian Trails

VI.35

contained in X , where T is an arbitrary eulerian trail of G. Consequently, in this case there are a t most 2 1 - 2 possible choices for e(u) to form a transition with other half-edges incident with u. But of course there are instances where a graph contains 21; - 1 pairwise compatible eulerian trails. This is exhibited in Figure VI.8 for k = 2.

Figure VI.8. G with 6(G) = 4 and three pairwise compatible eulerian trails TI = u,1, w, 2, u,3 , w , 4, u;T2 = U , 1,W, 3, U , 4, W, 2, U ; T3 = U , 1,W, 4, U , 2, W, 3, U .

In fact, one might conjecture that if one strengthens the hypothesis of Conjecture VI.36 by additionally assuming G to be 2-connected, then one can even find 21; - 1 pairwise compatible eulerian trails. Figure VI.8 points into this direction. In the case where G = 1 2. Suppose every cycle of G is a block of G. Then G contains b(G) - 2 pairwise compatible eulerian trails, and among any S(G) - 1 eulerian trails of G there are at least two which are not compatible with each other.

Proof. The second part of the conclusion of the theorem follows from the discussion performed immediately after the statement of Conjecture VI.36; hence it suffices to show that under the given hypothesis, G contains b(G) - 2 pairwise compatible eulerian trails. We proceed by induction on p =I V ( G )I. Suppose p = 1. Since S(G) > 2, the only vertex v of G is incident with m 2 2 loops. Thus we have the situation discussed after Figure VI.8 and preceding the statement of Theorem VI.37. There it was shown that the case p = 1 is equivalent to Theorem VI.37. Hence Theorem VI.38 is valid for p = 1. For p > 1 we proceed algorithmically by repeatedly applying Theorem VI.37. We start with an eulerian trail T of the eulerian graph G with V ( G ) p > 1 and b(G) > 2, and consider the cycle C, with E(C,) = E(G) whose traversal in one of the two possible directions corresponds to To = T . Denote V(G)= {q, .. . , u p } .

I

I=

Let G, be obtained from C, by identifying the vertices v,,,,. . . ,' u , , ~ ~of C, where 2k, = d(v,) 2 S(G),and where these k, vertices correspond to the replacement of u1 E V(G) by k, 2-valent vertices following To. Hence G, has precisely one vertex of vdency exceeding 2, namely u,. We note that the very structure of G (every cycle of G is a block of G) implies that for i = 1 (with G,,, = G,) two half-edges incident with ui belong to the same block of G,,, if and only i f these half-edges belong t o the same block of G. (*> We rephrase statement (*) in a more descriptive way by saying that G,,, and G have the same block distribution in vi.


VI.47

By the discussion preceding Theorem VI.37 and by this theorem itself, G, has eulerian trails Tl,l,T1,2,. . .,T1,2k1 -2 which necessarily define the same transitions in V ( G )- {v,} (since these vertices correspond to sets of 2-valent vertices in G I ) ,whereas in v1 they behave compatibly (i.e., a pair of half-edges adjacent in v1 defines a transition for at most one Tl,j, 1 5 j 5 2k, - 2). For n = b(G), let T,,,,. . .,Tl,n-2be the first n - 2 eulerian trails constructed above (if 2k1 = n, then take all of them), and for j = 1 , . . . ,n - 2, let C,,j be the cycle corresponding to For each j E (1,. . .,n - 2) we construct G2,j from Cl,j by identifying the vertices v ~ ,. .~.,,v 2 , k 2 where 2k2 = d ( v 2 ) 2 6(G) = n (see the construction of G , from Co above). We conclude that statement (*) holds for i = 2 as well; i.e., G2,jand G have the same block distribution in v2, and therefore, all these n - 2 graphs G2,jhave the same block distribution in v2 (observe that the cycles of G containing vl correspond to one then the pairs of halfcycle in each G 2 , j ) . Hence, if we look at edges incident with v2 and defining the blocks of G2,j are the same for every j E { 1, .. .,n - 2 ) and therefore define a unique 1-factor L of li;kz. Consequently, the application of Theorem VI.37 plus the discussion preceding this theorem yield n - 2 5 2k2 - 2 = d(v2) - 2 eulerian trails T2,P - * - 9 T 2 p - 2 of G which define the same transitions in V ( G )- {v, ,v2} (namely, the transitions of To),but behave compatibly in {vl, v2}.') Next we consider the cycles C2,j corresponding to T2,jand construct correspondingly the graphs G3,j from C 2 , jj, = 1,.. .,n - 2. Now we conclude that statement (*) holds for i = 3, i.e., the graphs G3,j. all have the same block distribution in v3. Consequently, we obtain eulerian trails T3,1, - * 7 T3,n-2 which behave like To in V ( G )- {v,, v2, v3} but behave compatibly in {v1 ,v2,v3}. If p = 2 or p = 3, then the theorem follows from the above. For p > 3, we obtain, by repeated application of the above argument, eulerian trails T p - l , , , - * * 7 T p - L n - 2 of G which behave compatibly precisely at V ( G ){ u p } while in v p they still behave like To. Constructing G,,j from the cycle CP-lJ . corresponding to Tp-l,j,j = 1 , . . .,n - 2 , and observing that these To be more precise one could first view Tz,jas an eulerian trail of G 2 , j , with Tz,i chosen from the 2k2 2 painvise compatible eulerian trails of G2,j, which in turn have been obtained by the application of Theorem VI.37 with respect to G2,j, j = 1,. . .,n - 2. However, one has to start with a fixed 1-factorization of KzkZsatisfying Theorem VI.37.

-

VI.48


n - 2 graphs have the same block distribution in up, we apply Theorem VI.37 to the corresponding where 2kp = d(vp) 2 S(G) = n to obtain eulerian trails TP,,,.. . ,Tp,n-2of G which now behave compatibly on all of V ( G ) .That is, T P , , ,... ,Tp,n-2are 6(G)- 2 eulerian trails of G which are pairwise compatible. This finishes the proof of the theorem. It should be noted, however, that a long-standing conjecture of Kotzig, [KOTZ64a,KOTZ64b], and others says that K2m has a 1-factorization {L,, . . . ,L 2 m - l } such that L , U L j is a hamiltonian cycle for i # j, 1 5 i, j 5 2m - 1. It follows immediately that the validity of this conjecit is ture implies Theorem VI.37; for, because of the symmetry of no real restriction concerning the 1-factorization to contain a prescribed 1-factor as expressed in the statement of Theorem VI.37. The above conjecture, however, has only been proved so far for some special cases: for those cases where 2m - 1 is a prime, m is a prime, [KOTZ64a], or 2m = 16,28,36,50,244 or 344 (for detailed references on this matter, see [MEND85a, SEAH87a, STIN87al; for 2m = 12 all non-isomorphic solutions of this problem can be found in [PETR80a])l). On the other hand, the proof of Theorem VI.38 as presented here shows that Theorem VI.37 and Theorem VI.38 are equivalent (see also the statement preceding Theorem VI.37). We now turn to the question whether it is possible to use the approach employed in the proof of Theorem VI.38, in order to prove the whole of Conjecture VI.36. Take an eulerian trail To of the connected eulerian graph G without 2-valent vertices and consider C,,the cycle corresponding to To;produce G, as in the proof of Theorem VI.38, apply Theorem VI.38 to G,, a.s.0. The problem we are faced with, however, is that for some i 2 2 the n - 2 graphs G i j may no longer have the same block distribution in v i (see the proof of Theorem VI.38). This problem can be illustrated by the graph of Figure VI.8 and its eulerian trails Tl and T2: starting with the eulerian trail Tl which assumes the role played by To in the proof of Theorem VI.38, and with Co being the cycle corresponding to TI, one obtains from C, the graph G, with v, = w as its sole vertex of valency exceeding 2. GI has two eulerian trails which behave compatibly in w, namely = '1,4, 1,W,2,212,3,3,W,4,211,4

'1 A. Rosa informed me recently that various authors solved Kotzig's conjecture for 2m = 126,170,730,1332,1370,1850,2198,3126,6860.


VI.49

and

T1,, = '1,4, 1,w,3, 212,3, 2, w ,4, v , , ~ ( v , , ~and v2,3 denote the 2-valent vertices corresponding to the splitting of 21 following TI). From T,,' and TI,, we obtain the graphs G,,, and G,,? whose sole vertex with valency exceeding 2 is 21. But in G,,, the blocks are defined by the edge sets {1,2} and {3,4}, while in G,,, the blocks are defined by the edge sets {1,3} and {2,4}. Hence G,,, and GI,, have different block distributions in v. However, the problem just discussed leads us to the following conjecture (first put forward in [FLEI86a]) whose validity would prove Conjecture VI.36 and which can be viewed as a generalization of Theorem VI.37.')

Conjecture VI.39. Let n = 2m > 2 be a positive even integer, and let Lo,L', . . . ,L("-,) be 1-factors of K , such that Lo U L(i) is a hamiltonian cycle of I(, for i = 1,.. . ,n - 2. Then there is a 1-factorization { L l , . . . , L n - l } of Kn such that Li U L ( i ) is a hamiltonian cycle for i = 1,..., n - 2 . We observe that in the statement of Conjecture VI.39, the 1-factors L(;),1 5 i 5 n-2, need not all be different, nor does L(') # L ( j ) , 1 _< i < j 5 n - 2, imply L(i)n L ( j )= 0. This takes account of the fact that in a proof of Conjecture VI.36 analogous to that of Theorem VI.38, for some i >, 2, two graphs Gi,jl and Gt,j2may or may not have the same block distribution; and if they have hfferent block distributions, they still may contain some identical block-defining pair of half-edges incident with vi.

If we have L' = . . , = L(n-2),then Conjecture VI.39 reduces to Theorem VI.37 with L = L' = L,-l. Observe that in this case the 1-factor Lo becomes irrelevant: for it does not play any role in the conclusion of Conjecture VI.39; thus, for Lo to form a hamiltonian cycle together with L = L ' = Ln-l is tantamount to saying "let L be any 1-factor of I 2 be arbitrarily chosen (if such v exists). Then G has an A-trail inducing a 1-splitting in v. Proof. If G has only 2-valent vertices it is a cycle; a run through this cycle represents an A-trail, whence we can assume that G has at least one vertex whose valency exceeds 2. We observe that G being a 2-connected simple graph implies that in this case G has at least three vertices with a valency exceeding 2. Consequently, the smallest graph Go other than a cycle satisfying the hypothesis of the theorem, has precisely three 4valent vertices and three 2-valent vertices; it can be interpreted as being obtained from a plane embedding of the octahedron graph by deleting the edges of the outer face boundary (see Figure VI.22). A 1-splitting in u renders the graph GO,,,having precisely two 4-valent vertices x,y which are necessarily cut vertices of GO,,,. In fact, Gt,l satisfies the hypothesis of Lemma VI.58, whence we can conclude that Go has an A-trail To inducing a 1-splitting in v and a 2-splitting in each of x and y. Using the same argument, we can conclude that if G is an outerplane graph homeomorphic to GO, G has an A-trail T inducing a 1-splitting in the chosen vertex v and a 2-splitting in the other two 4-valent vertices. Now we can proceed by induction. Consider a graph G satisfying the hypothesis of the theorem and having n > 3 vertices of valency exceeding 2, and assume the validity of the theorem for all corresponding graphs H having at most n - 1 such vertices. Consider in G the chosen vertex v and O+(v) = (ei,e;, .. .,e&) where d = d(v) > 2, and w.1.o.g. assume that e l , e d E E(bd(Fo))where Fo is the outer face of G. Now form H = Gv,l and denote Po = bd(Fo)- v. In any case, H is connected, eulerian and outerplane. If we denote the cut vertices of H by xl,.. .,x, according to the order in which they appear in Po,it follows from the argument used in proving property 3) of Lemma VI.61, that xi and xj belong to the same block of H , if and only if I i - j I= 1. Hence we can denote the blocks of H by


VI.86

V

Figure VI.22. The 2-connected outerplane eulerian graph Go having an A-trail Towhich induces a 1-splitting in v. B,, . . . ,B,+l such that z1E V ( B , ) ,x, E V(B,+,) and xi-l,xiE V ( B i ) for 2 5 i 5 r. As for the distribution of the j = 1,.. . ,k, d(v) = 2k, in these blocks B,, i = 1,.. .,T 1, we have

+

I {J1), t A k ) } n V ( B J I=]{&I, Jk)}n v(B,+,) I= 1 since G is 2-connected and because of the definition of dj), j = 1,.. . ,k. W.l.o.g., dl)E V ( B l ) ,dk)E V(B,+,) (observe that the notation for Bi,i = 1, . . . ,r 1, can be chosen either “from the left to the right” or “from the right to the left”). For j = 2 , . . . ,k - 1, let ij denote the index such that v(j) E V ( B i j ) .Of course, there may be blocks of H not containing any &I,j = 1,.. .,k; but for j # rn we have ij # ,i because among the four vertices of Po incident with the elements of E,,cj)U E,(,, , there are two cut vertices of H , each of which separates the remaining two vertices from each other (see the proof of Lemma VI.61, property 3), and observe that G is a simple graph and that a vertex of Po incident with some e E E, - {el, ed}, is a cut vertex of H by Lemma VI.61, property 1)). It follows from the outerplane embedding of G and the definition In addition to the of d j ) , j = 1,.. .,k, that if j < rn, then ij < .,i xi,i = 1,.. . ,r , we introduce zo and x,+~defined in G by vzo = el and ~ x , . +=~e d . Thus, B, n {xj/j = 0 , . . .,r 1) = {xi-l,xi}. Figure VI.23

+

+

VI.3. A-Trails in Plane Graphs

VI.8’7

illustrates the structure of H which follows from Lemma VI.61 and the above argument. V

V

G

H

Figure VI.23. The graphs G and H = GV,,.

The outer face F, of the block-chain H is a 1-face; and Po C bd(F,). Moreover, bd(F,) n E(Bi) # 0 for i = 1,.. . ,T 1 because E ( B i )n E(P,) # 0. This and the chosen notation allow the application of Lemma VI.60. Therefore, an A-trail of G inducing a 1-splitting in v necessarily induces a 2-splitting in each of x, ,. ..,2,. On the other hand, if we can show that Bihas an A-trail inducing a 2-splitting in and zi,i = 1,.. .,T 1, then H has an A-trail by Lemma VI.60, which is equivalent to saying that G has an A-trail inducing a 1-splitting in v. Hence, t o finish the proof of the theorem, it suffices to show that ( B i ) ~ z i - l , z i l , , has an A-trail for i = 1,.. . ,T 1. For this purpose, fix an arbitrary i E { 1,.. .,T 1) and consider B, . Depending on the position of B, in H and its structure, we can consider the following cases.

+

+

+

+

(I) v(B,)n { d J ) / j= 1,. . . ,k} = 8 . Denote for short B = Bi, x = xi-,, and y = xi.Let A,

= (z, v, t,), A, = (y, 21, t,) be two triangles with t ,, t , $ V ( G ) ,and construct a plane em-

bedding of the planar graph

13’ = B U A,

U A,


VI.88

such that t , and t , are 2-valent in B+ and lie in the boundary of the unbounded face of B+ (see Figure VI.24). Observe that for the unbounded face F k of B it follows that

V ( b d ( F k ) )= V ( P ( x y)) , , where P ( z ,y) = b d ( F k ) n Po ; otherwise, Po U {el,ed} = bd(Fo) is a cycle in G containing a vertex of the cycle bd(Fk) in its bounded (but not in its unbounded) region, thus contradicting the fact that G is outerplane. Thus, the embedding of B+ as defined above and illustrated by Figure VI.24 is outerplane indeed. Moreover, since B is 2-connected, we can conclude by the same reasoning that P’(x,y) := b d ( F L ) - P(x,y) contains no edge other than x y (see Figure VI.24).

X

E3+

Figure VI.24. Forming the outerplane graph B+ from the outerplane graph B by adding the triangles A, and A*.

Now suppose that B has at most n - 2 vertices of valency exceeding 2; then B+ has at most n - 1 such vertices. So, applying the theorem to B+ we obtain an A-trail T+ of B+ inducing a 1-splitting in v (the 2face-coloring of B+ is induced by the 2-face-coloring of G via the induced 2-face-coloring of B). Observing that T + ,read as an edge sequence, can also be regarded as an A-trail of B:,, we can conclude that by Lemma VI.60, B has an A-trail T such that the vertex splittings defined by X,(z) and X,(y), are 2-splittings.


VI.89

However, if B has precisely n- 1vertices of valency exceeding 2, induction cannot be applied to B+ as constructed above. So in this case we have to proceed differently. Consider the 2-face F2 of B (which, in fact, is also a 2-face of G) with z y E bd(F2),and form B- = B - {zy}. By Lemma VI.62 we can write the block-chain B- as the union of its blocks,

... U B , , where t =I V ( b d ( F 2 )I) -1, and B; n B3T # 8 if and only if I i - j I= 1. Furthermore, we can denote for j = 1,.. ., t - 1, B-=B,u

{W 3.} = B3

n B3~+l ;

wjE V(bd(F,)) - {z,y} by Lemma VI.62. In addition, denote w, = z and wt= y (w.1.o.g. z E B,,y E I?;). Note that since d,-

( z ) = 1( m o d 2 ) , if and only if

z E {z, y}

,

w e h a v e f o r j e J : = { l ,..., t } d,: 3

(wj-l)= d,:

(Wj)3

1(mod 2),

3

d B T ( u ) = O(mod2) if

u # wj-l,wj.

3

B being outerplane and V ( b d ( F 2 ) )= {wj/j= 0,. . . ,t } yields E(bd(F2))- {zy} = {wj-lwj/jE J}

.

Hence wj-lwjE E(B3T)for j E J, and by the above congruences, Bi := B j - {wj-lwj}is eulerian. Of course, E ( B j ) = 0 may hold true for some (but not all) j E J . Consider J, := { j E J / E ( B i )# 0) and observe that since B3T is a block of B - , it follows that BIT is 2-connected for j E J,; hence Bi is a connected eulerian outerplane graph. By the same reasoning used in proving Lemma VI.62, we can conclude that B; is a non-trivial block-chain; thus it has a cut vertex, and wj-l, wjbelong to different end-blocks of Bi and are not cut vertices of Bi. Summarizing the above applications of Lemma VI.62, we can express B as follows:

B = bd(F2)u

u B;

j€Jo


VI.90

=x. = w 1-1

=x.=w

0

1

t

B = Bi Figure VI.25. A block B = Bi of H , the non-eulerian blocks B; of B - {zy}, i E J, and the eulerian block-chains B; of BY - {wj-,wj},j E J,.

where Bi is a non-trivial eulerian block-chain for j E J,; any two of these block-chains have at most one vertex in common, and that vertex belongs to V (bd( F,)). Figure VI.25 illustrates this structure of B. Our aim now is to find for every j E J, two sub-trails covering B; and combine them with the cycle bd(F,) so that it results in an A-trail of B as required. For this purpose let cj be an arbitrary cut vertex of B; . Just as we defined B+ from B (see Figure VI.24), we can now form for every j E J, the 2-connected outerplane eulerian graph B r by attaching

AY)

to B; two triangles A?' = (wj-l,p,tl), = (wj,p,t,) at wj-,,wj respectively, where p , t,, t , g' V ( G )and t,, t, are 2-valent in BF. We now have wj-lin place of x, w j in place of y, p in place of z1, and Bi in place of B , if we compare this construction with the one illustrated by Figure VI.24. The fact that B is 2-connected while B(i is a block-chain (and therefore, cannot contain does not matter.

"3'

lias n: < n vertices of valency exceeding 2; this derives In any case, from the fact that B- is a non-trivial block-chain. Now, by induction we can conclude that BT has an A-trail 'T inducing in c j a 1-splitting. By


VI.91

Lemma VI.60, TT induces a 2-splitting in wj-17 w j and p (observe that (BT)cj,l is a block-chain with wjWl7wj,pas three of its cut vertices). Hence we can write T ’ as an edge sequence as follows:

+ =Pwj-17 T1( j ) w3-1 .

T‘ 3

7

t t 1 7

1p7pt,7t,wj,T2

(j)

7wjP

Ti’)and Ti’) are edge-disjoint closed trails covering Bi. T,(j)induces a 2-splitting in wj-17as does Ti’)with respect to wj. Since Ti’) and Ti’)are A-trails of the respective components of ( B i ) c,1, j it follows where

that two edges incident with u E V(BS)and which are not consecutive in O+(u), can never form a transition defined by any of Tij),T.j). By analogy to the proof of Lemma VI.60, we can now choose for every j E J, edges ej,f j , g j , h j with ej,fj E E w3-1 . n E ( B i ) , g j hj , E EwjnE(Bi) so that the trails 5”’’) and Ti’) (viewed as edge sequences) can be written in the form

Tij) - ej7Tl,j,fj7 and Ti’)= h j , T 2 , j , g j

.

We can look at the vertices wj,j E J U (0). If j , j + 1 J,, dB(wj)= 2; if j E J,, j 1 @ J,, TiJ) contains edges of Ewj while T,(j+’) does not

+

exist; if j

# J,,

j

+ 1 E J,,

Ti’’’)

contains edges of Ewj while Ti’) does

+

not exist; and if j , j 1 E J,, both Ti’) and T,(j+l)exist and contain edges of E w j . Moreover, if we look at O+(wj) in B , either

O+(wj) = (wj+lwj,wjwj-l)

>

VI.92


and taking into account the above equations concerning Ti’)and O+(wj), we conclude that

is an A-trail of B inducing in wj, j E J U {0} and, therefore, especially in wo = and wt = xi a 2-splitting. In other words, (Bi)~zi-l,zsl, has an A-trail. This settles the case (I).

(11) v(B;) n { d j ) / j = 1,.. . ,k}

# 8.

Then there exists precisely one j E (1,. . . ,k} so that v(j) E V ( B i )(see the discussion preceding Figure VI.23). Again denote for short B = B i , z = zi-l,y = xi,set v = d j ) , and let P ( x , y ) = b d ( F k ) n Po be defined as in case (I). In any case, B has fewer vertices of valency exceeding 2 than G. So, if z y $! E ( B ) , we suppress TJ to obtain B* which is of the type treated in (I);and any A-trail of B* corresponds to an A-trail of B , and vice versa. Thus, for x y # E ( B ) induction applied to B’ yields an A-trail of B as required. Whence we assume zy E E(B): B contains a triangular 2-face A = ( 2 1 7 2 , Y).

If A = B , a run through A is an A-trail of B as required because of our additional definition of a &splitting in 2-valent vertices. Assume, therefore, A # B. In this case we have precisely the situation illustrated by Figure VI.25 with t = 2, Jo = {2}, and TJ in place of w o , x in place of wl, and y having the same meaning as in Figure VI.25. Note that B - v is a 2-connected outerplane simple graph; hence B- := ( B - TJ)- {xy} = B - A is a non-trivial block-chain by Lemma VI.62. As in case (I) we find c E P ( z ,y) - {z,y} which is a cut vertex of B - , and by induction an A-trail TB of B which induces a l-splitting in c and, therefore, a 2-splitting in x and y. Thus, in all possible instances we have reduced case (11) to case (I). Theorem VI.63 now follows. It is tempting to suspect that in the hypothesis of Theorem VI.63, one can drop outerplanarity to obtain a true statement for an even larger class of eulerian graphs. Or, in more cautious terms, one could ask whether it is true that every simple 2-connected eulerian plane graph has an A-trail. In fact, when I first considered the problem of finding A-trails in plane


VI.93

eulerian graphs, I thought that 2-connectedness would be a sufficient condition for a plane eulerian graph G to admit a partition {V., Vi} of V(G) - {v} for some v E V(G), such that (Gvl,,)v;,2 does not contain a subgraph homeomorphic to the graph G, of Figure VI.17. For, if one considers the problem of finding an A-trail in a connected plane eulerian graph G from an algorithmic point of view, it follows from Theorem VI.59 and the discussion of Figure VI.17 that G has an A-trail, if and only if for every v E V(G) there is a partition {V., Vi} of V(G) - {v} with the property just described. Therefore, the graph G, of Figure VI.17 characterizes the forbidden stage that one has to avoid in trying to produce an A-trail by a sequence of S-splittings, 6 E {1,2} (here, homeomorphic includes the meaning it has in the topology of the plane. In this sense, the two graphs of Figure VI.17 have to be viewed as different). Unfortunately, there exist 2-connected plane eulerian simple graphs having only 4- and 6-valent vertices and not having any A-trail. In order to see that the graph Go of Figure VI.26 is such an example, we assume first w.1.o.g. that Go is 2-face-colored with the outer face being a 1-face. Suppose now that Go has an A-trail T. Observing that (Go)(YI,u61,2 is a disconnected graph and applying Corollary VI.57, it follows from the symmetry of Go that this graph also has an A-trail inducing a l-splitting in vl. W.1.o.g. suppose T has been chosen with this property. This implies, however, with necessity that the 6-splitting induced by T in u;, 2 5 i 5 6, satisfies the congruence i G 6 (mod 2). That is, we arrive at the graph H = ((G0){UlrV3,U5},l {VZ,V4,V6}r2 as illustrated by Figure VI.27, where T , viewed as an edge sequence, is an A-trail of H as well. Now, if one applies the corresponding &splitting to every 4valent vertex of H , one obtains a graph homeomorphic to the graph G, of Figure VI.17.') No matter which &splitting is applied to every vertex of V(Go)- {v}, 6 E {1,2}, the result is either a disconnected graph or, at best, what has been termed the forbidden stage whence we can conclude that Go cannot have an A-trail.

1

Interestingly enough, a slight change in the embedding of Go yields the plane graph GT which in fact does have an A-trail T (see Figure VI.28); the checking of the validity of this assertion is left as an exercise. '1 In fact, for every splitting of the 4-valent vertices of H which does not yield a disconnected graph, the result is - homeomorphically - the same, namely G, of Figure VI.17.

VI.94


Figure VI.26. A 2-connected plane eulerian simple graph Go with 4- and 6-valent vertices only and having no A-trail.

Hence, the existence of an A-trail in a 2-connected plane eulerian graph in general depends on the actual embedding of the underlying (abstract) planar graph, a fact we have already encountered in graphs of connectivity 1 (cf. Figures VI.17, VI.26 and VI.28). So, one might ask whether for a given 2-connected planar eulerian graph G, there exists an embedding H of G on the plane such that H has an A-trail. However, S. Regner (with the help of the author) constructed 3-connected planar eulerian graphs which had no A-trails. Since for any two embeddings H,, H 2 of such a graph we have either Ozl(G) = O z 2 ( G )or Ozl(G) = OH2(G) (see Theorem 111.52), we can conclude that in the case of 3-connected


VI.95

H Figure VI.27. H obtained from Go by applying a 6splitting to U 2 j - 2 + 6 , j = 1,2,3, 6 = 1,2.

planar eulerian graphs, the existence or non-existence of an A-trail in such a graph is independent of the actual embedding of that graph. On the other hand, in view of Lemma VI.60, it follows that if one has a planar 3-connected eulerian graph G with no A-trails, the planar eulerian graph H I = G' U G", where G' and G" are copies of G and G' n G" = { u } E V ( H , ) (hence K ( H , ) = l), does not admit an A-trail in any plane embedding of H , . Furthermore, if H2 = G' U G", where G', G" are as above and G' n G" = { u , zu} E V ( H 2 )(hence & ( H 2 )= 2), no embedding of H2 in the plane has an A-trail either. This conclusion follows directly from the next lemma (see also [REGN76a, Bemerkung 3.21). There, G t will denote the graph constructed from G, in the same way that B+ was

VI.96


Figure VI.28. GF obtained from Go of Figure VI.26 by changing the embedding of the edge vlvs only: v1v6 belongs in Go to the boundary of the outer face, in GOT to the boundary of a 2-face containing v. GF has an A-trail, while Go does not.

constructed from B in the proof of Theorem VI.63 (see Figure VI.24).

Lemma VI.64. Suppose for the plane 2-connected eulerian graph G that it can be written in the form


such that Gi is eulerian, K ( G ~2) 2 and G in G,+, =

G , n G j = Q if i + l < j l k ,

i = l , ..., k

VI.97 E V(G),

,

+

(putting k 1 = 1; if k = 2, then I G, n G, [=I {v,,,, v,,,} I= 2). Suppose further that G is 2-face-colored with the outer face F, being a l-face, and that E(bd(F,)) n E(Gi) # 8 for i = 1,.. . ,Ic. The following statements are equivalent,

1) G has an A-trail. 2) The notation can be chosen in such a way that

either G,, G, have A-trails inducing a l-splitting in v,,, and a 2splitting in v1,2,v k - l , krespectively, while for i = 2,. . . ,k - 1, G ihas an A-trail inducing 2-splittings in both v ~ - , , ~vi,;+,; , or Gi has an A-trail inducing 2-splittings in both ~ ~ - , , ~ , v ~for ,~+, i = 1,.. . , k - 1 (putting 2 1 ~ = , ~vk,,) and G i has an A-trail inducing a 2-splitting in each of u ~ - , , ~z ), ~ , v, ~ , where {v} = V4(Gi) - V4(G,).

Proof. We first observe that there are precisely two faces containing all vertices v ~ , ~ + i, ,= 1,.. . ,k , in their respective boundary, namely F, and another l-face, call it F,. This follows directly from the hypothesis of the lemma. Consequently, for i # j , 1 5 i, j 5 k , H = G~vi,..+l,vj,j+,},, is a disconnected graph: in order to see this, take simple open curves 1' Flu{vi,i+l, "j,j+l} and ccc F ~ u { v i , i + 1 7 "j,j+l } each joining } hence C = C, UC, is a simple and u ~ , ~ +C,, . nC, = { v , , ~ + v~ ~, , ~ + ,follows; closed curve containing open edges of G in its bounded and unbounded region. Therefore and because each Gj is eulerian, i = 1,.. ., k , it follows that H can be constructed (topologically) in such a way that C also contains open edges of H in its bounded and unbounded region and C n H = 8. We can thus conclude that H is disconnected (compare this with the proof of property 1) of Lemma VI.61). From these observations it now follows that if G has an A-trail T or if such T should be constructed from corresponding A-trails of Gi, i = 1,.. . ,k - 1, GZ respectively, T or the construction of T induces a 2-splitting in all but one vertex vi,;+,, i = 1,... ,k , at the most. Now suppose G has an A-trail T inducing a l-splitting in some v , , ~ + , . W.1.o.g. we may assume the notation chosen in such a way that i = k (if necessary, one applies a cyclic permutation to the subscripts in order to obtain i = k ) . Construct the plane graph H Z y from G by replacing


VI.98

with two vertices x,y $! V ( G )such that E, = EUk,,n E(Gl), E, = E U k ,n ] E(G,). Hence, H,, is a non-trivial eulerian block-chain whose blocks Hiare defined by E(Gi),i = 1,.. .,k, and whose cut vertices are the vertices v++,, i = 1,.. . ,k - 1. Another way of obtaining Hxywould ,1 which are incident with edges be to identify the 2-valent vertices in GUk,] of G, (Gk), thus yielding the new vertex x (y). It follows, therefore, that ( H Z J {,], = Guk,l,l. Carrying the 2-face-coloring of G over to H,,, we can conclude from Lemma VI.60 that any A-trail TH of Hxy induces an A-trail T j in Hi= ( E ( G i ) )i, = 1,. . .,k, such that Ti induces a 2-splitting in the vertices of Hiwhich are cut vertices of Hxy.Viewing T as an edge sequence, T is also an A-trail of H,, and (Hz,)Ix,,l,l. Consequently, in this case the existence of T in G implies that G ihas an A-trail inducing 2-splittings in T J ~ - , , vj,++, ~, for i = 2,. . .,k - 1, and both G, and Gk have an A-trail inducing a l-splitting in v k , l and a 2-splitting in u , , ~ uk-1 , ,k respectively. z]k,l

Conversely, if Gi has an A-trail as described above, Hxy,as constructed above, has an A-trail TH by Lemma VI.60, and TH induces l-splittings '= ' U k , ] , l , it follows that Guk,l,, also has in x and y. Since (Hz,){z,yl,l an A-trail, i.e., G has an A-trail inducing a l-splitting in vk,,. Now suppose that G has an A-trail T inducing a 2-splitting in TI^,^+,, i = 1,.. . ,k. Similar to the notation used in the proof of Lemma VI.60, we can choose edges e i , fi,gi,hi E E(Gi),i = 1,. . .,k, such that

"

e; E J%;+

n E(bW,)),

fi E Ev,-,,, W4Fl)),

gi E

n E(b'('m))7

hi E Eu;,;+ln E("(F1));

Eu;,i+l

and w.1.o.g. we may assume that the notation for defining G has been = (e:+,,g:, . . .,hi, f;+l,.. .). chosen in such a way that O+(V;,~+,)

Let us now look at the behavior of T in a fixed Gi, 1 5 i 5 k. Since T is an A-trail, the only transitions of T containing precisely one half-edge of G iare { e : ,gi-l}, { hi-l}, {gi, e:+l}, {hi,fi+l}. Therefore and since G is plane, T cannot be of the form

fl,

T = . . . ,e ; ,T,!,hi, . . . ,g i , T;",fi,. . .

(1)

where E(T;)U E(T,'I)= E(G;)- { e i , fi,gi,hi}. Consequently, T must be either of the form

T = . . . , e i , T ; , g i , . . . , hi,T;-, fi,.. .

(2)


(where { g i , hi} fi! X , since T does not induce a l-splitting in of the form T = . ..,ei, T i ,fj, .. . ,hi, T:*,gj,. .. ,

VI.99

or

(3) where E ( T F ) U E(TF-) = E(T,*)U E(T,**)= E(Gi) - {e,, fi,g j , hi} (we assume w.1.o.g. that ei is the first edge of Gi in T). Thus, if T is of the form (2), Ti := ei, TF,g,, h j ,TF-, fj

is an A-trail of G j inducing a 2-splitting in each of t ~ , - ~ , ~ , v ~If ,T~ is +~. of the form (3), Ti+ := ux,e,,T:, fi7xtl,tlu,vt2,t2y,hi,T~*,gi, yv ,

(4)

where z := u , - ~ , ~y , := and v , t l , t 2 @ V(G),is an A-trail of GT inducing a 2-splitting in each of v, t ~ ~ (see , ~ Figure + ~ VI.24 with G, in place of B).

To see that T must be of the form (2) for all but one G,, and that it must be of the form (3) for the remaining G;, i E (1,. . .,k}, we observe that T cannot be of the form (2) for all i = 1,. . . ,k; otherwise, we would have

i.e., T does not cover all of E(G) since {gi.ei} E X , (note that {g:,e:+,) E X,, i = 1,...,k). We can thus conclude that T must ‘turn around’ somewhere, but because T induces a 2-splitting in every Z J ~ , + +i~ ,= 1,. . . ,k, this ‘turn’ must take place within some Gj, j E (1,. . .,k}. W.1.o.g. we can assume that the first ‘turn’ takes place for j = k if T starts with ei, i E (1,. . .,k - 1). Thus T is of the form

But then

is an A-trail of G- := G - G k . Since by definition G- is a block-chain of G if k 2 3, while by hypothesis it is 2-connected for k = 2, it follows

VI. 100


from Lemma VI.60 for k 2 3, and from (6) for k = 2 respectively, that T induces an A-trail Tiin G, such that Tiinduces a 2-splitting in the cut vertices of G- and, by (6), in vk-1,k and 2 1 ~as , ~well. Moreover, Tk+, defined as Ti+ in (4), is an A-trail of G: as stated in the lemma.

To finish the proof, it suffices to show that if Ti is an A-trail of Gi for i = 1,. . ., k - 1, and if T t is an A-trail of G i as stated in the conclusion of the lemma, G has an A-trail. Define G- as above and apply Lemma VI.60 to G-; this results in an A-trail T - of G- which induces a 2-splitting in both vk-l,k and vk,l. Choosing the same notation for the definition of the A-trails Ti of Gi as above, we can write T - as defined in (6) and TZ as defined in (4) (with the subscript k replacing the subscript i). T as expressed by ( 5 ) can now be obtained by inserting in (6) the corresponding subtrails of (4);and T is an A-trail of G. Lemma VI.64 now follows. In fact, Lemma VI.64 can be generalized, however in one direction only. This generalization is the natural extension of Exercise IV.l .c), but again only in one direction. Lemma VI.65. (see [REGN76a, Lemma 1.41) If the connected plane eulerian graph G has an A-trail, then every block of G has an A-trail.

The proof of Lemma VI.65 and the construction of an example showing that the converse of Lemma VI.65 does not hold, is left as an exercise. Before we turn to the construction of planar %connected eulerian graphs with no A-trails, we have to study A-partitions. For this purpose, we may introduce some concepts. Definition VI.66. Let Fl, . . . ,F,, m > 1, be m distinct faces of the 2face-colored plane eulerian graph such that, for a fixed 6 E { 1,2}, Fiis a 6-face for i = 1, . . . ,m. Let us assume distinct vertices vl,,, v , , ~ ,. . . , exist with TI;,++^ E bd(F,) n bd(F;+,), i = 1,. ..,rn (with the subscripts read mod m). Then we call R = {Fl,.. . ,F,} a unicolored face-ring of G and L, := {vl,*, . . .,v,+} a complete set of links of R (if m = 2, then %,2 # 7-J2,1>* The following result relates A-partitions to face-rings and complete sets of links (see [FLEI74a, Theorem 1 and Corollary 11)Theorem VI.67. Let G be a 2-face-colored 2-connected plane eulerian graph without 2-valent vertices, and let {Vl,V,} be a partition of V ( G ) . The following statements are equivalent.


(1) G has an A-trail

VI.101

T whose A-partition is { V, ,V,}.

(2) For every unicolored face-ring R of G and every complete set of links LR of R, if the elements of R are 6-faces then L , V,, 6 E (17 2).

Moreover, if we replace in (1) “A-partition” with “perfect A-partition” and in ( 2 ) “LR V,” with “V, 2 LR sf V,”, then these stronger statements are equivalent as well. Proof. (1) implies (2). Suppose for some unicolored face-ring R of G, whose elements are &faces, and for some complete set of links LR that LR V,, 6 E {1,2}. We may assume w.1.o.g. that 6 = 1. We claim that H := GLRIlis disconnected. Since R is a unicolored face-ring of G and 6 = 1, distinct 1-faces F,, . . . ,F, of G exist such that

R = {Fl, . . ,F,},

8 # bd(Fi) n bd(Fi+,) 5 V(G), i = 1,. . . ,m ,

and by definition,

Now let C, C FiU bd(Fi) be a simple open curve with Ci n bd(F’) = {u~-,,~ t ~ , ; , ~ + Then ~ } . C = ,U : C, is a simple closed curve. Precisely because Fiis a 1-face, i = 1,.. .,m, C contains in its interior (the bounded region of C), as well as in its exterior (the unbounded region of C) a positive even number of open edges of Eui,i+l, i = 1,.. .,m (otherwise, Ci and Cj+, would lie in faces of different colors, or else Ci = C,+, which implies Fi = F,+,). Moreover, C n G = L , 5 V, by assumption. W.1.o.g. suppose the notation chosen in such a way that for 0+(vi,;+,) = (ei,e;, .. . ,e & , ) , d, = d ( ~ , , ~ + ,> ) 2, precisely the open edges el,e,, . . .,en; lie in the interior of C. Then 2 5 ni = 0 (mod2) and di - ni 2 2 follows from the preceding paragraph. Then we can perform the 1-splitting in vi,++,, i = 1,. . . ,m, in such a way that ~ ( ~ .1. .,,d m i ) lie entirely in the ,.. . ,v(lCi) lie entirely in the exterior of C, interior of C, while dmi+’) where m i = $ n i and ki = i d , (see Figure VI.18). Consequently, since C n G = LR C V,, we can conclude that H is disconnected. Since H is disconnected and LR C V,, it follows that G,,, is disconnected, and further that (Gv1,1)v2,2is disconnected. On the other hand,

VI.102


{V,, V,} being the A-partition of an A-trail T of G with v 6 containing precisely those vertices at which T induces a &splitting, 6 = 1,2, implies that (Gvl,!)v2,2 is a cycle (compare this with Corollary VI.57). This contradiction proves the implication. ( 2 ) implies ( 1 ) . Consider H = (Gvl ,,)& , which is a 2-regular graph. If

H is connected, then by Corollary VI.57, {V,,V,} is an A-partition obtained from some A-trail T inducing 6-splittings precisely in the elements of v6,6 = 1,2. In this case, the implication follows. Hence suppose H to be disconnected. Choose V, V ( G ) as small as possible such that Ho := (Gvi,l)vdf,2is disconnected, where Vd = V, n V,, V', = V, n V,. It follows that H , has a face F, such that bd(F,) is disconnected; w.1.o.g. F, is the unbounded face of H,. Furthermore, since we did not make any assumptions concerning the 2-face-coloring of G, we may now assume that F, is a 1-face. Now, if V', # 0, consider w E V:. We claim that already H , := (Gv,, ,1) Vi' - { w } ,2 has a disconnected face boundary (whence H , is disand by definition of a &splitting, connected): for, H , = (H1)jw1,27 5 E {1,2}, the application of the 2-splitting to a vertex of H , leaves the 1-faces of H , (homotopically) invariant (see the footnote at the end of Definition VI.54); i.e., bd(F,) is disconnected already in H,. This contradicts the choice of V,; hence V: = 0 and, therefore V, & V,. Again by the choice of V,, and since G is connected, we must have V, c VFo for VFo:= V ( ( E ( M ( F o ) ) ) Gsince ) , V, V, and H , being disconnected has a disconnected outer face. implies that G,,,Fo,l Consider C,,a component of bd(F,), define C2:= bd(F,) - C,, and let C be a simple closed curve lying in F, and such that int C 3 H,, ext C 3 H , - H I , where H , is the component of H , with H , 3 C,. The choice of V, implies that for every v E V,, some of the di) lie in int C (i.e., in H,) and some lie in ext C (i.e., in Ho - H,), i = 1,.. .,4 d c ( v ) ; otherwise, one obtains C n Gvo-{ul,l = 0 and intC n Gvo-lul,l # 0 # extC n G v o - ~ v ~ , l for some 21 E V,. W.1.o.g. we may assume that C is drawn in such a way that the transition from H, to G, viewed as a topological procedure, leaves C invariant (i-e., w.1.o.g. C is a simple closed curve in G as well compare this with Figure VI.18) and CnG = V, while (C - V,) nF(,) = 0 for every 2-face F ( 2 )of G.

C n G = V, implies that we can write V, as vO

= {'1,2

'2,3,

* * 7

vm,l)

(ill


VI. 103

where vi-l,i and vi,++,divide C into two simple open curves C: and Cy such that C: n V, = {u;-,,~,v ~ , ~ + ~C:} . { v ~ - , , ~vi,++,} , lies in a uniquely determined 1-face, call it Fi, i = 1,.. . ,m. It follows from the choice of V, that

Fi#Fj

for i # j , l < i , j < r n

,

(i2

1

and that

bd(Fi) n b d ( F j ) # 0 if and only if

Ii - j

IE

(1, rn - 1)

,

(i3)

where

Otherwise, one could draw a simple closed curve C- in the plane such that C- n F(,) = 8 for every 2-face of G,C-nG= V; c V,, and (C- V,-) n Fi = 0 for at least one but not all i E (1,.. .,m}; and csing an argument similar to the proof of the first implication, these properties of C- imply that Gv- is disconnected. Consequently, the validity of (i,) 0 ’ - (i4) classifies R = {F, ,. . .,F,} as a unicolored face-ring and V, as a complete set of links of R. Since R is a set of 1-faces and V, C V,, R and V, violate the validity of (2); hence H must be connected in any case.

,

To finish the proof of the theorem suppose now that the A-partition in (1) is a perfect A-partition, and suppose in (2) that L , fulfills the stronger relation V, 2 LR V,; call the corresponding statements (1’)and (2’).

If {V,, V,} is a perfect A-partition, then, by definition, (GVl,1)V2,2is a cycle (which corresponds to T),and so is (Gvz,l)v,,2. Denote by T* the A-trail of G corresponding to the second cycle and define V;C := V,, V; := V,. Since (1) implies (2) a twofold application of this implication yields VgC 2 L , sf V, for every unicolored face-ring R and every complete set of links LR; i.e., LR V,, 6 = 1,2. Whence we conclude that (1’) implies (2’). Observe that {V6,V:} = {q, V,} independent of the choice of 6 E {1,2}; hence one does not have to specify the color of the elements in the unicolored face-ring R. Suppose (2’) holds. Define Vgt as above. Since (2) implies (1) we obtain from the proof of this implication that both (Gvl,l)v2,2 and (Gv;,l)vg,2 = (Gv2,1)vl,2are cycles; i.e. {V,,V,} is a perfect A-partition. Theorem VI.67 now follows.

VI.104


The theorem just proved indicates that if we look at G, = (V,) C G, 6 = 1,2, where (V,,V,} is an A-partition and G satisfies the hypothesis of the theorem , then G, must have a very special structure. In order t o determine this structure, let us have a look at an arbitrary cycle C of G which is not a face boundary. It follows that both int C and ext C contain (open) edges of G (where we view C as a simple closed curve in the plane). Furthermore, both int C and eztC contain 1-faces and 2-faces as well. In particular, every e E E ( C ) belongs to some bd(F,) where F, is a &face, 6 = 1,2. Hence, considering

R, = {F,/E(bd(F,)) n E(C)# 0, F, is a 6-face},

S = 1,2 ,

we conclude that R, is a unicolored face-ring having a complete set of links LR V ( C ) ,6 = 1,2. By Theorem VI.67, V ( C )If V,, and therefore, G,, 6 = 1,2. Hence,

c

if G, contains a cycle K , K is the boundary of a &face.

(4

If we suppose that G is just a 2-face-colored plane eulerian graph with K(G)= 1 and possibly 2-valent vertices, the conclusion (*) is still valid: for an A-trail of G induces an A-trail in every block of G (see Lemma VI.65), and any cycle of G, is necessarily contained in some block of G. Moreover, if a cycle C c G contains a 2-valent vertex of G, and if C is not the face boundary of a &face, R, as above can be constructed as well, with L , 2 {v E V(C)/d,(v) > 2) as a complete set of links of R,, 5 E {1,2}. In this case, L , If V,, S E {1,2}, by Theorem VI.67; hence ( L R ) C G,, and moreover, C If G,. Summarizing these considerations we therefore arrive at the following result [REGN76a7Satz 2.11. Corollary VI.68. If T is an A-trail of the connected 2-face-colored plane eulerian graph, G, and if {V,,V,} is a corresponding A-partition of G, G, := (V,) contains a cycle C, only if C is the boundary of a 6-face, 6 E {1,2}.

We point out, however, that neither G, nor G, need to be connected. This fact is exemplified by Figure V1.29. Moreover, the converse of Corollary VI.68 does not hold even if G, is acyclic. In other words, G may have a vertex partition {V17 V,} such that G, = (V,) is acyclic, although G does not have an A-trail. This can be seen by studying the graph Go of Figure VI.26; we leave this as an exercise.


VI

V4

V4

VI.105

V

1

Figure VI.29. (a) The four-sided antiprism H , (b) an A-trail T of H , and (c) the subgraphs induced by the Apartition of T.Observe that H is 4-regular and has, except for two 4-gonal faces, triangular faces only.

The next result, however, shows that the converse of Corollary VI.68 is true provided G, is connected. For this result, however, we need some considerations on unicolored face-rings. Let us consider a unicolored face-ring R = {Fl,F 2 , .. .,Fm}, rn 2 2, of the 2-connected, 2-face-colored plane eulerian graph G. Suppose

6 d ( F i ) n b d ( F j ) = 0 for

li--j1>1, l s i , j s m ,

except if

{ i , j } = {l,rn}.

(*I

Since G is 2-connected bd(Fi) is a cycle for i = 1,.. .,rn. Choose the notation as in the proof of Theorem VI.67. Let LR be a complete set of links. With LR we associate a simple closed curve C = C(LR) as in the first part of the proof of Theorem VI.67, where C = U ~ l C i Ci , C Fi, and C n G = L,. Of course, C not only depends on the choice of Ci C Fi, but also to a greater degree on the choice of L , which is uniquely determined if and only if I bd(Fj)n bd(Fi+l)I= 1, i = 1,.. . ,m (e.g. if G is 3-connected). In any case, among all possible choices for L , and the associated curve C(L,), choose L(H0)and C, = C(L',o') in such a way that for every complete set of links L h

intCo nL', = 0

.

(**I

VI. 106


Observe that if bd(Fi) n bd(Fj+l) 2 { v ~ , ~ v:,~+,}, + ~ , one obtains for every complete set of links Lh with vi,++, E Lk another complete set of links by defining Lk = ( L k - { V ~ , ~ + , } U ) {v:,~+~}. Whence we may conclude that the choice of La“’and Co satisfying (**) is possible. We note that La“’ is uniquely determined. Similarly, let La‘’ be the uniquely determined complete set of links such that C, = C(L$’) and

eztC, n Lk = 0

(* * *)

for every complete set of links Lk. By (*) and since G is 2-connected, it follows that there are cycles C,,C, c G defined by

Now we are in a position to present another result of S. Regner’s Ph.D. thesis, (see [REGN76a, Satz 2.21). Her original proof contains a minor flaw; the proof presented here differs considerably from hers insofar the details are concerned.

Corollary VI.69. Consider a vertex partition {V17 V,} for the plane, 2-connected7 2-face-colored eulerian graph G. If G, := (V,), 6 = 1,2, is connected, and if every cycle of G, is the face boundary of a &face, {Vl,V,} is an A-partition. Proof. In view of Theorem VI.67 it suffices to show that for every unicolored face-ring R and every complete set of links L,, if the elements of R are 6-faces, LR V,, 6 E {1,2}. We proceed indirectly by choosing a unicolored face-ring R with minimal I R I whose elements are &faces, and such that for some complete set of links LR,LR C V,, 6 E {1,2}; w.1.o.g. S = 1. Denote explicitly R = {F17.. . ,Fm},rn2 2, and L, = {ul,,, . . .,v,,~}. We have bd(F,) n bd(F’) = 0 for l i - j l > 1, 1 5 i , j 5 rn, unless { i , j } = {l,rn}; this follows from the choice of R. Now choose the curves Ci71 5 i 5 rn, and define the simple closed curve C with CnG = LR C V, as in the first part of the proof of Theorem VI.67. It is precisely because of C nG = LR V, and since G, is connected that either intCnV,=0

or eztCnV,=0

;


VI.107

for, as an application of the Jordan Curve Theorem, every path P ( z ,y) in G joining x E int C with y E e z t C satisfies P ( z ,y) n L , # 0. W.l.o.g.,

intCnV,

=0 .

(1)

Using the considerations preceding the formulation of Corollary VI.69 and (1) we may even assume that LR and C have been chosen in such a way that L R = La"' and C = C,, thus satisfying (**). Let C, be the cycle defined in (* * * *). By (1) and the definition of C,,and because of the assumption L , C V,, we conclude C, C (Vl). Now, since every edge of C, is a boundary edge precisely of some element of R and of some 2-face7 and since R contains at least two elements, it follows that C, cannot be the face boundary of a 1-face. This contradiction to the hypothesis finishes the proof of the corollary. Consequently, as an application of Corollaries VI.68 and VI.69, a partition {V,, V,} of V ( G )where G is 2-connected7 plane and eulerian, is a perfect A-partition only if (V,) and (V,) are acyclic; and if (V,) and (V,) are trees, this partition is a perfect A-partition. We observe that the A-partition of the graph in Figure VI.29 is perfect, but the corresponding graphs (V,),(V,) are both disconnected forests, while the graph of Figure VI.26, although not having any A-trails, admits such vertex decomposition (see Exercise VI.20). Note that the graph of Figure VI.29 has two 4-gonal faces only, while its other faces are triangular. However, if G is an eulerian triangulation of the plane, a vertex partition {V,, V,} induces acyclic subgraphs (V,), (V,), if and only if these subgraphs are trees (see Proposition 111.63). So we arrive at the following result.

Theorem VI.70. A vertex partition { V,, V,} of an eulerian triangulation of the plane is a perfect A-partition if and only if (V,) and (V,) are trees. We can restate Theorem VI.70 without using the concept of an Apartition or a perfect A-partition, respectively. In order to see this we reconsider Corollaries VI.68 and VI.69 in the case of eulerian triangulations of the plane. In this case, these two corollaries are the converse of each other; for, if (VJ is disconnected, (Vj)has a cycle which is not a face boundary (see Proposition 111.63, and [FLEI74a, Corollary 2]), { i , d = {1,2). Now let D denote an eulerian triangulation of the plane which is given together with its 2-facecoloring; suppose it has an A-trail T with { V,, V,}

VI. 108


as the vertex partition corresponding to T . Suppose (V',) contains a cycle A , S E { 1,2}; A is the (triangular) boundary of a &face by Corollary VI.68. Denote E(A) = {el, e,, e , } . By definition, T induces a &splitting in all vertices of V(A); i.e., {{ei, ei}, { e ; , e $ } , { e y , e y } } n X, = 0. In other words, any two edges of A are not consecutive in T . We say that T separates the edges of A. Similarly, we call the A-trail T nonseparating if and only if T does not separate any face boundary of D. On the other hand, if T separates the edges of some face boundary of D , the corresponding (V',) contains a cycle (which is a triangle). Hence, we can restate Theorem VI.70 as follows.

Theorem VI.70.a. For a vertex partition {Vl, V,} of the eulerian triangulation of the plane, D , the following statements are equivalent. 1)

(V,) is a tree for i = 1,2.

2) D has a non-separating A-trail.

Observe that for defining a non-separating A-trail, one need not start with a 2-face-coloring; one can also start from the vertex partition {V+, V-} related to O+(D), O - ( D ) respectively (see Lemma VI.53). Theorems VI.70 and VI.70.a are familiar. For we know that a 2-connected plane cubic graph G, has a hamiltonian cycle if and only if its dual D = D(G,) has a vertex partition {Vl,V,} such that both (V,)and (V,) are trees (Lemma 111.74). And if G, is bipartite as well, then D = D(G,) is an eulerian triangulation of the plane in which case D has multiple edges if and only if 4 G 3 ) = 2. Hence we arrive at the following equivalence concerning hamiltonian cycles and A-trails, [FLEI74a, Theorem 21. Theorem VI.71. A 2-connected, plane, cubic, bipartite graph has a hamiltonian cycle if and only if its dual has a non-separating A-trail. There is a much more direct proof of Theorem VI.71. So far unpublished to the best of my knowledge, I attribute it to W.T. Tutte.l) Because of ') I a m not sure that W.T. Tutte knows that I know that this proof is his, but his proof is the essence of a criticism of the approach to A-trails as presented in [FLEI74a]; when the original manuscript was returned t o me together with the referee's report, Tutte's name was improperly erased. Nevertheless I decided t o adhere t o the approach as presented here and, earlier, in [FLEI74a, REGN76al; for it proved fruitful to some extent, and the theory on A-trails developed so far might be of some use for finding a proof of the conjectures on A-trails stated below.


VI. 109

its elegance and relative shortness we present it in detail (the essence of the proof is Tutte’s, the phrasing the author’s).

Tutte’s proof of Theorem VI.71. Let H be a hamiltonian cycle of the 2-connected, plane, cubic, bipartite graph G,. From a topological point of view, H constitutes a simple closed curve in the plane such that some of the open elements of L := E(G,) - E ( H ) lie in ezt H and some in int H . Write H as an edge sequence, H = e1,e2,. . .,e2k, denote by s(ej) a point of the open edge ej, j = 1,.. ., 2 k , and construct now a simple closed curve C as follows: for i = 1,.. .,k, let C2i-1 be a simple open curve joining s(e2i-l) and s(e2J such that

(and lying sufficiently close to the section of H containing the vertex incident with e2i-1, e2i). Similarly, for i = 1,.. . ,Ic let C2; be a simple open curve joining s(ezi) and s(e2i+l) such that c2i

- {~(ezi), s(e2;+1)} c e x t H

putting e2k+l = el. W.1.o.g

I Cjn e I
s(f,) ~ ( f , , ) $! E(C).Now define H by E ( H ) = {e,,, f,/v E V ( G ) } . By definition, H is a 2-regular spanning subgraph of G,, and because C is a simple closed curve with C n e = { s ( e ) } for every e E E(G,), therefore H is connected. That is, H is a hamiltonian cycle of G,. This finishes the proof.

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI. 11 1

VI.3.1. The Duality between A-Trails in Plane

Eulerian Graphs and Harniltonian Cycles in Plane Cubic Graphs With Theorems VI.70, VI.70.a and VI.71 at hand it becomes clear that there is a close relationship between A-trails in connected plane eulerian graphs (respectively in eulerian triangulations of the plane, to be more precise) and hamiltonian cycles in connected planar, cubic, bipartite graphs.

Of course, not every planar, 2-connected, cubic, bipartite graph has a hamiltonian cycle.’) The graph of Figure VI.30 (a) is such an example; we leave it as an exercise to prove that this is really the smallest such example and that it is uniquely determined. If one replaces every digon of this graph with a copy of the 3-dimensional cube Q3minus one edge, one obtains a simple 2-connected plane, cubic, bipartite graph which is not hamiltonian (see Figure VI.20.(b)). It has been shown independently and almost simultaneously, [PET08la, ASAN82al that this simple graph is in fact the smallest example of this type, and that it is uniquely determined. However, the interest in hamiltonian cycles in plane cubic bipartite graphs does not originate from the equivalence with A-trails in eulerian triangulations of the plane as expressed by Theorem VI.71. Let us recall that Tait, after proving that the Four Color Conjecture is equivalent to the conjecture that every planar, 2-connected, cubic graph has a 1-factorization, conjectured that every planar, 3-connected, cubic graph is hamiltonian (see [KONISGa, p.27-281). This conjecture, stated in 1880 and aimed at an easy proof of the Four Color Conjecture, was disproved in 1946 by Tutte who conjectured subsequently that if in Tait’s conjecture one replaces ‘planar’ by ‘bipartite’, the graphs are hamiltonian, [TUTT7la]. J.D. Horton disproved Tutte’s conjecture; his first counterexample has 96 (!) vertices (later he found one with 92 vertices, [HORT82a]). More recently, M.N. Ellingham constructed a counterexample with 78 vertices, ‘1 In contrast to the case of A-trails, the existence of a hamiltonian cycle is independent of any actual embedding. On the other hand, the planar cubic bipartite graph G, has a hamiltonian cycle, if and only if for every plane embedding H3 of G,, D(H,) has a non-separating A-trail (Theorem VI.71). This is an interesting fact, for in the case of K(G,) = 2, D(H$) $ D(H{) may hold for two embeddings H i , H!/ of G,.

VI.112


0,- ( a1.b.1 1

Figure VI.30.(a) The smallest 2-connected, plane, cubic, bipartite non-hamiltonian graph. (b) Substituting for i = 1 , 2 , 3 the digon (ui, bi) in (a) with Q3- { u i , bi}, one obtains the smallest such graph which is simple.

and in joint work Ellingham and Horton constructed a counterexample to Tutte’s conjecture on 54 vertices (for the original Horton graph, see, e.g. [BOND76a]; for the other counterexamples and more details, see [ELLI82a, ELLI83al). A short time ago, J.P. Georges and A.K. Kel’mans developed counterexamples on 50 vertices which are cyclically 4-edgeconnected, [GEOR89a, KELM88aI .I) However, the following conjecture is generally attributed to D. Barnette. According to Tutte (private communication, 1972) he has thought about this conjecture as well. Consequently, we shall call it the Barnette-Tutte Conjecture (BTC). Conjecture VI.72 (BTC). Every planar, 3-connected, cubic, bipartite graph is hamilt onian.

In view of Theorem VI.71 the BTC is equivalent to the following con’) Mr. H. Gropp drew my attention to these two articles. As he points out in [GROP89a], bipartite cubic graphs of girth 2 6 had been studied as early as 1887, but in terms of what has become known as ‘symmetric configurations’.


VI.113

jecture (observe that planar, 3-connected, cubic graphs have simple triangulations of the plane as their duals, and vice versa).

Conjecture VI.73. Every simple eulerian triangulation of the plane has a non-separating A-trail. However, the next conjecture, although appearing weaker than Conjecture VI.73, is indeed equivalent to the latter, [FLEI74a].

Conjecture VI.74. Every simple eulerian triangulation of the plane has an A-trail. The equivalence between the last two conjectures (and therefore between conjecture VI.74 and the BTC) follows from the next result. There, we consider two eulerian triangulations of the plane, Do and D,, which are related to each other in the following way. For every face boundary A = (ab, bc, ca) of Do take a copy of the plane octahedron minus the outer face boundary A, = (a, b, c, ), embed it in the interior of A and identify the corresponding vertices a, b, c (see Figure VI.31).

A C

U

‘,C,’

a1

a

b

Figure VI.31. Obtaining the eulerian triangulation of the plane D,from the eulerian triangulation of the plane Do by embedding in every face of Doa copy of the graph H , (the Figure shows this operation for one face only).

Lemma VI.75. Let Do be an eulerian triangulation of the plane, and let D , be constructed from Do as described above. Then the following statements are equivalent.

VI.114


1) Do has a non-separating A-trail. 2)

D, has an A-trail.

Proof. We first assume that Do has a non-separating A-trail and apply Theorem VI.70.a by which we can write V ( D o ) = Vf U V: such that (yo)is a tree, i = 1,2. Next consider an arbitrary face F of Do with V ( b d ( F ) )= {a,b,c}, say. Since (yo) is a tree, i = 1,2, we have V: n V ( b d ( F ) )# 8 # V: n V ( b d ( F ) ) . Hence, w.1.o.g. we can write a, b E I$ c 'E,V i where {j,k} = {1,2}. We adjoin c1 to Qo and al,bl to V j . By this extension of yo and V j , performed for every face F of Do, we obtain a vertex partition { V i , V i } of D,. It follows from the very construction of this vertex partition that ( V t )and (V:) are acyclic, hence they are trees. By Theorem VI.70.a, D, has a (non-separating) A-trail. Conversely, if D, has an A-trail, by Corollary VI.68, the corresponding A-partition {&l,V;} has the property that (5') and ( V i ) contain a cycle only if it is a face boundary of D,. Thus, we conclude for V; := V ( D o )n Vt,V: := V ( D o )n V; that ( V t ) and (V:) are acyclic. By Proposition 111.63.1) and Theorem VI.70.a, Do has a non-separating Atrail. Thus, the lemma is true. The equivalence between Conjecture VI.73 and Conjecture VI.74 now follows from Lemma VI.75 by observing that they are formulated for all simple eulerian triangulations of the plane, and that D, is simple if and only if Do is simple (note Do C 0,). The equivalence between these two conjectures is, however, matched by the corresponding equivalence in bipartite cubic graphs. For, the BTC is equivalent t o saying that every planar, 3-connected, cubic, bipartite graph has a dominating cycle. The proof of this equivalence follows along the lines of the proof of Lemma VI.75 translated into the theory of cubic graphs; it is therefore left as an exercise.

In [FLEI74a] it had been conjectured that every planar, 3-connected, eulerian graph has a n A-trail (note that in this case one need not speak of plane graphs because a change in embeddings either leaves O+(v) invariant or else replaces O+(v) with O-(v) for every vertex v). If true, this conjecture would immediately have settled Conjecture VI.74 since simple triangulations of the plane other than K3 are 3-connected. In order to construct a planar, 3-connected, eulerian graph without an


VI. 115

A-trail as S. Regner did, one proceeds in a way similar to the above construction of D, from Do. Start with a non-hamiltonian, plane, 3-connected, cubic graph G,; such graphs exist (see p. 111.64). Consider its dual D := D(G,), and denote the odd vertices of D by q ,v2, . . .,2r2k-1, 2r2k, k 2 1. Consider the set of paths P := {P1,2,. . .,P2k-l,2k}such that P2j-1,2j joins 2.’2j-1 and vzj, 1 5 j 5 k, (see Corollary V.3). For an arbitrary but fixed j , 1 5 j 5 k, denote

and for i = 1,.. . ,t , let yi denote one of the two vertices each of which defines a face boundary containing zi-,zi, and let A; be the (triangular) face boundary containing xi-, , xi,yi. Now embed a copy of the graph Hi of Figure VI.32 in the interior of A; in the same way that Ho has been embedded in the interior of A to obtain D, from Do (see Figure VI.31 above). Doing this for i = 1,.. . ,t , we obtain a plane graph D‘ having 3and 4-gond face boundaries only; and D C D’. D’ has 2k - 2 odd vertices only since

Observe that no 4-gond face boundary of D‘ contains an edge of D. Therefore, if we choose P” E P, P” # P’, we can produce the graph D” from D’ the same way we produced D’from D. Of course, a vertex corresponding to one of the above y i , may not lie in V(D);but this does not matter precisely because E(P“)n E(Q) = 0 for every 4-gond face boundary Q of D’(note: P” C 0).Now D” has 2k-4 odd vertices only. Hence, after k steps we arrive at the plane eulerian graph D ( k )having 3and 4-gonal face boundaries only; and D c D(”.Since I V(G,) I> 2, D # K3 follows; i.e. D is 3-connected, and consequently, D ( k )is 3-connected and simple (this follows from the very construction of D(’)). However, D ( k)may still contain face boundaries A which are face boundaries in D as well. For such A proceed as in the construction of D, from


VI.116

Figure VI.32. The graph Hi.

Do by embedding H,, as illustrated in Figure VI.31. The grap.h G finally obtained has the following properties: 1) G is a plane, 3-connected, eulerian graph; 2) D 3)

c G;

No face boundary of D is a face boundary of G.

Because of these properties, if G had an A-trail, the corresponding Apartition { V,, V,} of V ( G )induces a vertex partition { V:, V,l}in D which of necessity induces acyclic graphs (V:),(V,l)(see property 3) and Corollary VI.68). Consequently, since D is a simple triangulation of the plane, they are trees. That is (see Lemma 111.74), G, with D = D(G,) has a hamiltonian cycle. This contradiction to the choice of G, shows that G has no A-trail. However, the positive outcome of this construction of G from D = D(G,) is the following general observation whose complete proof is contained in the proof of Theorem VI.91 below.

Let G, be a 3-connected, planar, cubic graph. Then a plane (even a planar 3-connected) eulerian graph G with D = D(G,) C G exists such that n o face boundary of D is a face boundary of G , and G, is hamiltonian i f and only i f G has a n A-trail. (V1.A) Moreover, we may make the following observation. In the above construction of a plane, 3-connected, eulerian graph G without A-trails having


VI. 117

properties 1)) 2), 3) it is possible to proceed in such a way that every face of D contains at the most one copy of the graph Hiof Figure VI.32. This follows from [FLEI74c, Theorem 10 and its proof] (see also the discussion on regulating sets centering around Figure VI.35 below). Another way of obtaining G with that property rests on the application of the Chinese Postman Problem. This will be done in the subsection concentrating on complexity considerations. Comparing the above construction of a plane, 3-connected, eulerian graph

G from the plane, 3-connected, cubic graph G, via D = D(G,), with Conjecture VI.74, and noting that G has 3- and 4-gonal face boundaries only, one wonders whether it is possible to disprove this conjecture (and therefore, the BTC as well). All that one would need is a triangulation D, of the plane with all but two vertices having even valency, and the remaining two vertices lying on the same face boundary and having odd valency. W.1.o.g. one could assume that these two vertices, call them and xi,lie on the outer face boundary bd(F,). Then the outer face boundary of H;' := D, - bd(F,) would be 6-gonal as in the case of Hi in Figure VI.32, with zi--l and zibeing the only odd vertices of H;', and Hi having 3-gonal face boundaries only except for the outer 6-gon. The above construction starting from the non-hamiltonian G,, would then yield a simple eulerian triangulation of the plane G without A-trails. Unfortunately, such D, does not exist, [MOON65a, FLEI74bl. Lemma VI.76. There is no triangulation G of the plane having precisely two odd vertices z,y and such that zy E E(G).

Proof. We present two proofs; both are indirect. The first proof is due to Moon, the second proof was found by the author years ago, but has never been published before. 1) Suppose such G exists. Then H := G - {zy} is a plane eulerian graph with one 4-gonal face Q, and all other faces are 3-gonal. Consider a 2-face-coloring of H such that Q is a 1-face. Denote 3;= { F / F is an i-face}, i = 1 , 2 . Since every edge of H belongs to the boundary of precisely one 1-face and precisely one 2-face) we have

IE(H)I=

F€32

On the other hand, IE(bd(F))I= 3 for F

IE(bd(F))I= F€Fi

IE(bd(F))I

IE(bW)I= FE31

# Q yields

IE(bd(F))1-

1(rnod3) and FE32

0 (rnod3)

,

VI.118


an obvious contradiction to the above equation. 2 ) Assuming G exists, let H = G - bd(F,) where we assume w.1.o.g that x , y E V(bd(F,)). Consider a plane embedding of the Ir‘, and denote V(I(,) = {z1,z2,y1,y2} such that {z1y1,z2y2} is a 1-factor of K4. Let Fibe a face with zjyi E E(bd(F,)),i = 1,2; Fl # F2 follows. Embed now a copy of H in each of F,, F2 respectively and identify zi with z and yi with y in the respective copy of H , i = 1,2. Identify the vertex z # x,y

of bd(F,) n H in the corresponding copy of H with the corresponding vertex zi in bd(Fj) C K4, i = 1,2 (note: z1 E {z2,y2},z2 E {zl,yl}). The new graph GI is an eulerian triangulation of the plane because for u E V(I(,) we have

x(G1) = 3 and K4 contradiction.

c

G,.

Since x(K4) = 4, we obtain the desired

Translated into the language of cubic graphs, Lemma VI.76 says that there is no plane (2-connected) cubic graph which has precisely two odd face boundaries and such that they have an edge in common. However, there is a partial solution to the BTC, [GOOD75a]. To formulate Goodey’s result, let G, be a planar, 3-connected, cubic, bipartite graph with the following properties: 1)

I E ( b d ( F ) )IE {4,6,8} for every face F of G,;

2)

IE(bd(F))I= 8 for at most one face F of G,;

3) if there exists F8 with IE(bd(F8))I= 8, then there exists an F4 with IE(bd(F4))I= 4 and bd(F8) n bd(F4) # 8.

Let B, denote the set of all G, having the above properties. It follows from Euler’s Polyhedron Formula that G, E Bo has precisely six 4-gonal face boundaries if it has no 8-gonal face boundary, and it has precisely seven 4-gond face boundaries otherwise (see Exercise VI.23). Goodey’s result can then be stated in the following way.

Theorem VI.77. Let G, E Bo be chosen arbitrarily, and let F,, be any face with I E(bd(F,,,)) I= muz such that there is a 4-gonal face F4 with bd(F,,,) n bd(F4) # 8. Then there exists a hamiltonian cycle H of G,


VI.119

with E(bd(F,,,) n bd(F,)) E E ( H ) (note that for any two faces F’, F” in a 3-connected, planar, cubic graph, IE(bd(F’))n E(bd(F”)) {0,1}).

IE

Our aim is to translate Goodey’s proof of Theorem VI.77 into the theory of A-trails. For this purpose we want to explore first what it means in terms of 6-splittings of vertices in D(G,), if e E E(G,) belongs to a hamiltonian cycle H of G,. Disregard, for the moment, the fact that G, is bipartite. By Lemma 111.74, G, has a hamiltonian cycle H if and only if we can write V ( D )= V, U V, and ( y ) ,is a tree, i = 1,2, where D = D(G,). Moreover, by the same lemma, H contains precisely those edges which correspond in D to the elements of {v1v2/v1 E V,, 21, E V,}. That is, if we look at the face boundaries bd(Fj) of G, corresponding to vj E Q for fixed i E {1,2}, E ( H ) = UvjEll,E ( b d ( F j ) )- { g j , k E E(bd(Fj)r l bd(Fk))/vj,vk E q} (see the proof of Lemma 111.74). Now we use the fact that G, is bipartite, and hence D is a simple eulerian triangulation. Therefore, considering the perfect A-partition { V,,V,} of D corresponding to H c G, (see Theorems VI.70, VI.70.a and VI.71) we conclude for e* = xy, the edge corresponding to e E E ( H ) , that I {x,y } n V, ]=I (2,y} n V, I= 1. Moreover, since {V,, V,} is a perfect Apartition, i.e., both (Dvl,1)v2,2 and (Dv2,1)yl,2 are cycles, we may assume w.1.o.g. that x E V,,y E V,. This means in terms of 6-splittings that if D is 2-face-colored, D has a non-separating A-trail inducing a l-splitting in x and a 2-splitting in y. Now let I,denote the set of all 2-face-colored simple eulerian triangulations D of the plane (with the outer face being a l-face), having the following properties:

1’) d,(v) E {4,6,8} for every 2’)

21

E V(D);

d,(v) = 8 for at most one v E V ( D ) ;

3’) if there exists vs E V ( D )with d,(v,) 4-valent vertex in D.

= 8, then v8 is adjacent to a

As above, it follows from Euler’s Polyhedron Formula that D E 7, has precisely six 4-valent vertices, if it has no 8-valent vertex; otherwise it has precisely seven 4-valent vertices (see Exercise VI.23). We note that both B, and 7, are infinite sets; this we conclude from [GRUN67a, p.254, Theorem 21 or Lemma 111.58 and the fact that the elements of B, and 7, are in l-l-correspondence by dualization. Hence, we arrive at the next result on A-trails.

VI. 120


Figure VI.33. The eulerian triangulation of the plane, D, , obtained from the eulerian triangulation of the plane, Do, by a W,-extension (possibly a = c # d or b = d # c ) . T h e o r e m VI.77.a. Let D E 7, be chosen arbitrarily, let vmuz E V ( D ) be of maximal valency and chosen in such a way that it is adjacent to a 4-valent vertex w. Then D has a non-separating A-trail inducing a l-splitting in umaz and a 2-splitting in w. Before we prove Theorem VI.77.a we must establish some other results. For this purpose, we define a special class of eulerian triangulations of the plane. Let D be a triangulation of the plane, and let A = (a, b, c ) be a face boundary in D. Embed the graph H , as depicted in Figure VI.31 in the corresponding face of D (i.e, A is replaced with an octahedron 0,);call the resulting graph D,.We say D, results f r o m D b y a n octahedral extension (shortly 0,-extension) at A. To define a second operation assume the existence of a 4-valent vertex z, E V ( D ) and consider the 4-wheel W, defined by N*(z,); denote N(z,) = { a , b, c, d } following O+(z,) = (az,, bz,, czo,dt,) (W, can be degenerate, i.e., a = c or b = d ) . Define the triangulation of the plane D, by

We say D, results f r o m D by a W,-extension with the 4-valent vertex t o as its center (see Figure VI.33).


VI.121

Now let n be a positive integer and define Do = D , D, = D’, and D i results from Di-l by an 0 6 - or W4-extension, i = 1,.. . ,n. Then we say D’ results from D by a sequence of 0 6 - and/or W4extensions. The validity of the following statements is a consequence of the definition of the above extensions: 1) D’ is eulerian if and only if D is eulerian; 2 ) If I V ( D )I> 4,then D is simple if and only if D’ is simple.

However, the condition I V ( D )I> 4 cannot be omitted; the octahedron can be viewed as obtained from the eulerian triangulation of the plane consisting of two 4- and two 2-valent vertices, by a W4-extension. If IV(D)I > 4 and if a = c or b = d , D has two pairs of multiple edges: two edges ba and, correspondingly, two edges zoa or two edges zob. However, only one of these two pairs of multiple edges can disappear by a FIT4extension. This becomes clear from the following observations. Firstly, if Di results from D,-lby an 06-extension, E(Di_,) C E(D,), i E (1,. . . ,n}. Therefore, if I V(D,-l) I> 3, then independent of the valencies of a , b, c in Di-l (where A = ( a , b, c); see Figure VI.31), D j - { a , b, c} is disconnected for every j with i 5 j 5 n (if Di-l has a 2-valent x E { a , b, c } , then already D j - ( { a , b, c } - {x}) is disconnected, and if none of a , b, c is 2-valent in Di-l, then d D , ( y ) 2 6 for y E { a , b, c} which implies that none of a , b, c can be the center of a W4-extension in D,, i 5 k 5 n). Secondly, if D, results from D j - l by a W,-extension and if d DI.- 1 ( a ) > 4,d D‘-1 . ( c ) > 4, then Q = ( a , b , c , d ) is a separating 4-gon in every Dj, i - 1 5 j 5 n. The valency condition concerning a and c implies the existence of a vertex x 6 { a , b , c , d , z o } and, moreover, E ( Q ) c E ( D j ) ,i - 1 5 j 5 n. These considerations lead us to the following structural result. In formulating it, we follow the notation of Figures VI.31 and VI.33.

Lemma VI.78. Let D be a simple eulerian triangulation of the plane resulting from the octahedron 0 6 by a (non-empty) sequence of 0,- and/or W4-extensions. Then at least one of the following statements is true. 1) D contains a separating triangle A = ( u , b , c ) and O6 3 A as an induced subgraph such that DL := D - H , can be obtained from 0 6 by a (possibly empty) sequence of 0,- and/or W,-extensions ( H , := 0, - A);

VI. 122


2 ) D contains (at least) two 4-valent vertices zb, tg such that the corre-

sponding 4-wheels W’ ,W4/‘have the following properties: d g ( t i ) = d D ( t i ) = d D ( t y ) = dD(z!J = 4

;

E(W;)n E ( w ~ = )0 ; each of

DI, := ( D - {ti,t k } )

U { a l z ~c’th> , and

0 6 (defined analogously)

can be obtained from 0, by a (possibly empty) sequence of W,-extensions.

06-

and/or

Proof. By hypothesis, there exists a sequence of eulerian triangulations of the plane, Do,D,, . . .,D,, such that Do = 0 6 , D, = D , and D iresults from D,-,by an 0 6 - or W4-extension, i = 1,.. . ,n. Suppose n = 1. If D results from 0 6 by an O,-extension, both sides of the (uniquely determined) separating triangle A = ( a , b, c) are graphs isomorphic to Ho := O6 - A. Hence statement 1) of the lemma prevails. If, however, D results from 0 6 by a w4-extension, D is isomorphic to the 6-sided double pyramid. Hence we can partition V,(D) (which defines the 6-gonal base of this double pyramid), {Vl,V t } ,such that PI := (Vi) and P” := ( V t ) are paths of length two. Now, W‘ := (Vi U V 6 (D)) and W4/‘ := ( V t U V , ( D ) ) are 4-wheels with the properties expressed in statement 2 ) of the lemma ( t b , 2: are the 2-valent vertices of PI, PI‘ respectively, and ti,z4 t;) are the end vertices of P‘ (P”)).Now we deduce the validity of the lemma if n = 1.

(tr,

Hence consider the case n > 1, and suppose first that Diresults from D,-, by an 0,-extension for some i E (1,.. .,n}. W.1.o.g. suppose for the where this 06-extension is performed, that triangle A = (a, b, c) c Di-l A # bd(F,) where F, is the outer face of Di-,. Consequently, the graph Ho used to perform the 0,-extension at A, lies in the bounded region of A. Moreover, A separates Diand d D i ( z )2 6 for any z E { a , b, c } . Thus, every 0 6 - or W,-extension performed after the 06-extension at A, takes place entirely either ‘outside’ or ‘inside’ A. Now, if not all of these latter extensions take place on the same side of A, then consider the last one which takes place outside and the last one which takes place inside of A. Call the correspondingly resulting graphs D j and D,, i < j , k 5 n.


VI.123

If each of D j and D , results from W,-extensions of Dj-,, Dk-l respectively, statement 2) of the lemma holds for the corresponding 4-wheels Wii) and W i k )with Wi := Wi’), W t := W i k ) :this follows from the choice of j and k and the discussion preceding the formulation of Lemma VI.78, and by defining the sequences D,*,. . .,D:-l for DA, respectively D,**,. . ., for 0 6 by

D:=D,,

r = O ,...,j - 1 ,

and D,**=D,,

s=O

,..., k - 1 ,

DT = (D,+, - { z y ) , z p ) } ) u { u ( j ) z p ) , c ( j ) z p ) }r, = j , . . . ,n - 1, D,**= (D,+, - { z ,

9

22(k)})

U

{dk))zi”,c(’)zik)}, s = k,. . .,n - 1,

where zf),zi’), I!), d‘), c(‘) correspond to z,, zl, z2, u, c in Figure VI.33 for I = j,k. Observe that d D , ( z p ) ) = dDr(zp)) = d D.(’1 ( I c ) dDs()z$”) = 4 for every r E { j , . . .,n } , respectively for every s E {k,. . . ,n}. Hence suppose w.1.o.g. that D j results from Dj-, by an 0,-extension. Denote by H f ) N H, the graph used for this 0,-extension performed at the triangle A ( j ) . Since the octahedron 0;’ = A ( j )U Hi’) belongs to every D, because of the choice of j, r = j,. . .,n, we conclude that DA := D - ~ i j results ) from 0, by a sequence of 0,- and/or ~,-extensions, namely, Do,*..>Dj-1, Dj+l - H o ( j l , . . . , D n - &) Since we can use the same argument if all extensions following the 0,extension at A take place outside A (with A and Ho in place of A ( j )and Hi’)), and because of the cases solved already, we have to assume that all these later extensions take place inside A.

In particular, we may assume that D, does not result from D,-, by an 0,-extension (otherwise, for D := D,, DA := D, - H i n ) , Hi”) 11 H,, statement 1) of the lemma holds). Denote by W2 the 4-wheel associated with the W4-extension yielding D, from DnVl. Now take the minimal i E { 1,.. .,n } such that Di results from Di-l by an 0,-extension. If i = 1, H,* := D , - (H, U A ) N H , and so H,* C D,,t = 2 , ..., n; and since H , U A I I0,, we can define

D , * : = H , U A and Df-,:=D,-H,*,

t = 2 ,...,n

VI.124


to obtain D;-, = D - H,* =: DL resulting from 0 6 by a sequence of 0,and/or W4-extensions.

If i = 2, D, is isomorphic to the 6-sided double pyramid. Denote V4(D,) = {u1,v2 ,...,21,) according to O+(w) = ( w z ~ ~ , w v. .,w'u6) ~,. where V ( ( D 1)= {w, z}. W.1.o.g. this notation has been chosen in such a way that A = (215,216,w)(= ( u , b , c ) ) . But then Wi = ( { ~ ~ , 2 ) ~ , 2 1 ~ , w , 2 } ) is a 4-wheel lying entirely outside A, and Wi c D j , j = 1,.. . ,n. Hence the W4-extension yielding D , from D,-, (see above) implies that PV: lies entirely inside A; E(W i )n E(W l )= 0 follows. With 2r2, zll ,'u3, v6,u4 assuming the role played in Figure VI.33 by z o , zl, z 2 ,a, c respectively, we define

This yields DO+ 2! 0 6 , D:-, = D; (as defined in statement 2 ) of the lemma ), and Dfresults from DJ-, by the same extension yielding Dj+l from D j , j = 1,.. .,n - 1.

To finish the proof of the lemma we have to consider the case where either i > 2, or D results from 0 6 by a sequence of W,-extensions. Assume first that D j is a (4 2j)-sided double pyramid, where j = i - 1 if i > 2 exists, and j = n if D := D, results from 0, by n W4-extensions. We either find Wi,W[ similar to the case i = 2, or else we can take any two edge-disjoint 4-wheels Wi,W t of the (4 2n)-sided double pyramid. In both cases, Wi and W: satisfy statement 2) of the lemma.

+

+

Thus we finally consider the case when there is some k < i, k 5 n respectively, such that D j is a (4 2j)-sided double pyramid for 1 5 j 5 k - 1, while D, is not a double pyramid. Note that D, is a 6sided double pyramid in any case. It follows that D, contains a 4-gon Q = ( w ,ur, ~, v r + 2 ) , where {w, z} = V(Dk-1) - v4(D,-1)7 ~ D I ; - (21,) ~ = d,,(v,) - 2 = 4 for s E { T , T 2}, and where we assume the notation chosen in such a way that vtvt+, E E(D,-,) for t = 1,. . . ,2k+2 (putting v2,+3= 21,). W.1.o.g. T = 1. Thus Q = (w, x,2ry, v3) is a separating 4-gon containing no 4-valent vertex in D j , j 2 k 2 2. Thus, every 0,- or W4extension yielding D j + , from D j , j = k, .. .,n - 1, takes place entirely either inside or outside Q. Arguing along the same lines as before (with Q in place of A) we may conclude that all these extensions take place on the same side of Q. On the other hand, v2 and v5 are in D, the centers of 4, = ( N * ( v 5 )respectively, ) lying on different wheels W,"' = ( N * ( v 2 ) )WA5)

+

+


VI.125

sides of Q. Consequently we have Wi c D j and 1 V4(Wi)nV4(Dj) I= 3 for some Wi E {W,‘”’,W i 5 ) }and j = k,. . .,n. Denote by W[ the 4-wheel corresponding to the W4-extension yielding D, from D,-,.D:, defined correspondingly, results from 0 6 by the sequence, Do,. . .,Dn-,= 0 6 . As for D;, we denote Wi = ( { z ; , zi, z;, b’, d’}) and a’, c‘ in analogy to the notation of Figure VI.33, where {b’d’} = {v,, v3} and {u’, c’} = {w,x} if Wi = W i 2 ) ,while {b‘,d‘} = { w , ~and } {a’,c’} = {v3,v7} if Wi = W J 5 ) (putting v7 = q if k = 2). Now define D;:= Dj for O < j < k - l - 6 , for k - 6 6 I < n - 1, Or:= ( I l l + , - { z ~ , t ~U} {a’z~,c’z~} ) where S = 0 if Wi = W i 2 )and 5 = 1if Wi = W l 5 ) .We obtain D; = D i - , resulting from 0 6 by a sequence of 0 6 - and/or W4-extensions since D; results from D;-, by the same extension yielding Dj+l from D j , j = k - 5,... ,n - 1. Note that for k = 2 and 6 = 1 0; results from D,* by a W4-extension, and so does D, with respect to Do but the centers of these W4-extensions are different. Lemma VI.78 now follows. We now use this lemma to prove a result on A-trails which generalizes a special case of Goodey’s result.

Theorem VI.79. Let D be a 2-face-colored eulerian triangulation of the plane, D being either the octahedron itself or arising from the octahedron by a sequence of 0,- and/or W,-extensions. Let A = (x,y, z ) be an arbitrarily chosen face boundary with x,y, z E V ( D ) .D has a non-separating A-trail whose (perfect) A-partition {V,, V2}satisfies x E V,,y, .z E V,. Proof. W.1.o.g. A is the boundary of the outer face F, of D , and F, is a 1-face (note that perfect A-partitions are independent of the 2-facecolorings). If D is the octahedron, then it has a non-separating A-trail 2’; e.g., taking the graph of Figure VI.20.b) and defining V, = V’,V, = V”, we have I V,n V ( A )I= 1, I V, n V ( A )I= 2. Because of the symmetries of the octahedron we may therefore assume w.1.o.g x E V,, y, z E V,. Thus the theorem holds in the case of the octahedron. Now let D satisfy the hypothesis of the theorem and suppose D is not the octahedron. By the very construction of D and because of Lemma VI.78 we distinguish between the following two main cases. (1) D satisfies statement 1) of L e m m a VI.78. Retaining the notation of Figure VI.31 with D in place of D, we define

D;:= D - Ho

VI.126


in which A, := (a, b, c) is a face boundary. Because of the symmetry of the possible situation of A in A, U H,, we assume w.1.o.g. that if A C A, U H,, then either r = a l , y = b,, z = cl, orr=al,y=a,~=c,,orx=c,y=a,z=cl orr=bl,y=b,z=c. Now, applying induction to Dh we obtain w.1.o.g. (by symmetry if A $ AouH,, or because of the freedom to choose if A c A, UH,) a E vjo and b, c E V l , { j , k} = {1,2}, for the perfect A-partition { V f ,V:} of V(Dh) corresponding to some non-separating A-trail of DI,. Define

v , = ~ ~ U { b , , c , } , V , = V ~ U ( a 1 } ,{ j , k } = { l , 2 }

.

Consequently, { vj, V,} is a perfect A-partition as required if A $ AoUHo, or if j = 2 , k = 1 and {x,y,z} # { b , , b , c } . On the other hand, if A c A, U H , and j = 1,k = 2, {V,,V,} is a perfect A-partition as required provided {x,y, z } = {bl, b, c } ; in the other cases we use the fact that ( V f )and (Va),6 = 1,2, are trees in D; therefore, for V; := V, and V; := V,, {V;, V;} is a perfect A-partition as required. Finally, in the case j = 2, k = 1 and {r,y, z } = { b , , b, c}, we define V;, V; as before to obtain a desired perfect A-partition {V;, V;}. This finishes case (1). (2) D satisfies statement 2) of Lemma VI.78. Then D contains a 4wheel W, as depicted in Figure VI.33 (ii) with A W,. By retaining the notation of this figure with D in place of D, and DI, in place of Do, we define the reduction of D to 0; and apply induction to DI, to obtain {qo, V:} as above. Depending on the distribution of a, to,c in Vf and V: we consider two cases.

If for { j , k} = {1,2}, either { a , zo, c} C 50,or { a , z,} C Yo and c E V: (the case {c, z o } C and a E ‘V needs no extra consideration because of symmetry), then { b , d } c V’ follows of necessity. In both cases we define vj := vjo U {zl, z,}, Vk := V i , { j ,k} = {1,2}

5’

{%, V,} = {V,, V,} of D as required. Hence we have to consider the case { a , c} C vj”, z, E Vf . Then { b , d } p

to obtain a perfect A-partition

vjo because {vjo,Vf}is an A-partition. By symmetry we may assume

w.1.o.g. that b E %O, d E V’. To obtain a perfect A-partition (6,V,} = {V,, V,} of D as required by the theorem, we define in this case

3 := vj” u { z o } , v, := ( V i - (2,))

u {zl,z,}, { j ,k} = {1,2}

.


VI.127

Observing that these considerations cover d cases regardless of A n W4 = the application of Lemma VI.78, A p W4 in any case), we now conclude the validity of the theorem.

8 or A n W, # 0 (because of

Translated into the theory of cubic graphs, Theorem VI.79 is - by dualization - equivalent indeed to the following result on hamiltonian cycles (see also the considerations following the statement of Theorem V1.77).

Corollary VI.80. Let G, be a (3-connected) planar, cubic, bipartite graph which can be derived from the cube by a sequence of one or both of the following operations: 1) Replace a vertex with three mutually adjacent 4-gons; 2) Replace a 4-gon Q with three 4-gons two of which are not adja-

cent. Then, given any e E E(G,),there exists in G, a hamiltonian cycle not containing e . The proof of the next corollary follows immediately from parts of the proof of Theorem VI.79; therefore we leave it as an exercise to the reader.

Corollary VI.81. Let Do and D, be eulerian triangulations of the plane such that D, results from Do by a sequence of 0,- and/or W4extensions. Then Dohas a non-separating A-trail if and only if D, has a non-separating A-trail. We omit the translation of Corollary VI.81 into the theory of cubic graphs; it can be done the same way by which Corollary VI.80 has been obtained from Theorem VI.79. There is a procedure similar to the W,-extension to create a new eulerian triangulation of the plane D, from a given eulerian triangulation of the be given as in Figure plane Do. Namely: let the vertices a, b, c, d E V(Do) VI.33, and suppose ac E E(Do). Subdivide ac with two new vertices, z1 and z2 say, and join both t l and z2 by an edge to each of b and d. The resulting graph D, is an eulerian triangulation of the plane indeed. We call this construction of D,from Do the weak W,-extension (see Figure VI.34 below). Translated into the theory of plane cubic graphs, the weak W4-extension corresponds to the replacement of an edge e E E(G,)with Q2in such a way that the parity of I E(bd(F))I two adjacent 4-gons Ql, remains unchanged for every face F of G,. That is, if G, is bipartite, then so is the new graph.


VI. 128

However, if we rephrase Lemma VI.78 in order to cover the graphs obtained from 0, by a sequence of 0,- and/or weak W,-extensions, the resulting statement is false. That is, starting from 0, one can produce an eulerian triangulation of the plane D by a sequence of weak W4extensions such that N ( v ) n V4(D) = 0 for every v E V,(D) except for one pair of adjacent elements of V,(D) (We leave it as an exercise to construct such an example). This fact is no surprise, however, if one observes that in Corollary VI.81, the relation between A-trails in Do and A-trails in D, is bijective (see also the proof of Theorem VI.79). This observation does not hold any longer if one replaces in Corollary VI.81 the term ‘W4-extension’ with ‘weak W,-extension’. For, while it is true that a non-separating A-trail of Do can be extended to a non-separating A-trail of D , (obtained from Do by a sequence of 0,- and/or weak W4extensions), the converse may not be true for some non-separating A-trail of D,. Figure VI.34 illustrates this consideration.

DO

Figure VI.34. D, obtained from Do by a weak W4extension. If D, has a perfect A-partition {V,,V,} as indicated by the vertex labels 1 and 2, then for the corresponding partition (V’,V,} of V(D,),the graph (Vf)has a cycle containing ac, while (V:) is disconnected.

w e note that, starting from 0 6 , one reaches only certain eulerian triangulations of the plane by 06-,W4- and weak W,-extensions. For, in


VI.129

general, an eulerian triangulation of the plane need not contain adjacent 4-valent vertices. In fact, if G, is a plane, 3-connected, cubic graph of girth 5, one can find an independent set of edges, E,, such that the cubic graph G$resulting from G, by replacing every e E E, with a 4-gon Q,, is plane and, in addition, bipartite (see Figure VI.35). Such E, is called a regulating set (of edges), and every l-factor L of G, is a regulating set; for it yields in Gg a 2-factor Q* consisting of even boundary cycles, namely Q* = { Q J e E L C E(G,)} (by Theorems 111.66 and 111.67 a plane cubic graph is bipartite if and only if it has a 2-factor consisting of even boundary cycles). On the other hand, a regulating set of G, need not be a subset of a l-factor L c E(G,).’) In any case, translated into the language of eulerian graphs, the transition from G, to G$ (with the help of a regulating set E,) means transforming Do := D(G,) into an eulerian triangulation of the plane. This transformation is achieved by subdividing every e* = ac E E(D,) which corresponds to e E E, c E(G,), with a vertex zo and joining zo to both b and d where a , c, b and a , c, d are the vertices of the boundary triangles containing e* (see Figure VI.34). Moreover, since E, is an independent edge set, no pair of edges e * , f* € E ( D o ) corresponding to e, f E E,, belongs to a face boundary in Do. That is, the above transformation of Do can be performed simultaneously for all corresponding e*. This is also why in the above construction of a plane 3-connected eulerian graph without A-trails, one can proceed in such a way that every face of D contains at most one copy of the graph Hiof Figure VI.32. We point out that the concept of a regulating set has been introduced and studied in [FLEI74c]. We now present the translation of Goodey’s proof of Theorem VI.77 into the theory of A-trails. We note however, that our proof generalizes two special cases of Goodey’s proof which have been treated in [GOOD75a] in an “as can easily be seen” manner. This generalization has been expressed in Theorem VI.79. Proof of Theorem VI.77.a. We first observe a general structural property of D which follows directly from the fact that D is a simple triangulation of the plane.

a) If P4 is a path with I(P4)= 3 and V(P4)C V4(D), P4 is an induced path, and V(P4)C N ( v 8 )where d(v,) = 8, or else D is the 6-sided double ‘1 We note that Gg may be 3-connected although K(G,)= 2. This is the case, however, if and only if all cut sets Es of size 2 in G, belong to some 2 implies E, f l Es 1 in any regulating set Eo of G, (note that Es case; one way of proving this fact is, e.g., by dualization of Lemma VI.76).

I

I=

I

I#


VI. 130

G;

63

Figure VI.35. Transforming the plane cubic graph G, into a plane cubic bipartite graph G: by replacing the edges e of a certain independent edge set E, with 4-gons Q,.

pyramid, D

II

0, respectively.

Noting that the theorem is true for 0, E 7, by Theorem VI.79 we proceed by induction and distinguish between several cases depending on the distribution of the 4-valent vertices of the chosen D E I,. Case 1. D results from an O,-extension of the eulerian triangulation of the plane Do. Then Do contains a boundary triangle A = (a, b, c) at which this 0,-extension has been performed. Suppose first that A ( D ) < 8. Then d,(a) = d,(b) = d,(c) = 6 and therefore, in Do = D - H , (for H,, see Figure VI.31) we have d D o ( u )= d D o ( b ) = dDo(c) = 4. That is, Do results from an 0,-extension of some eulerian triangulation D' of the plane; a.s.0. Therefore, D results in this case from 0, by a sequence of 0,-extensions. We conclude that in this case Theorem VI.79 implies the validity of Theorem VI.77.a. On the other hand, if A ( D ) = 8, the esistence of a non-separating A-trail in D is secured by induction and Corollary VI.81; and the proof of Theorem VI.79 shows how a perfect A-partition {V', V:} of V(D,) can be extended to such a partition {V,,V,} of V ( D ) . In particular, if ug with d D ( u 8 ) = 8 lies in A, w.1.o.g. ug = u,,, = a and w = al (see the statement of Thcorem VI.77.a); whence we conclude by induction, and since the relations d D o ( u )= 6 = A ( D , ) , d D o ( b )= 4,a , !$ V(D,) hold, that a E V', b E V:.


VI.131

Therefore, we may define V,, V, in such a way that a, E V,, b, E V,, and c1 E V, if and only if c E where { j , k} = { 1,2}. Consequently, { V,,V,} is as required in any case.

6

Case 2. D results from a W4-extension of the eulerian triangulation of the plane Do. By induction and Corollary VI.81, D has a non-separating

A-trail since Do has one; and we know from the proof of Theorem VI.79 how to extend a perfect A-partition {V:, V . } of V ( D o )to such a partition {Vl, V,} of V ( D ) if TI,,,^ E E(Do - { z o } ) . If, however, {v,,,, w} n V(W4) # 8, we proceed as follows. If A ( D ) < 8, we conclude in a way similar to Case 1that D results from 0, by a sequence of W,-extensions, and therefore, Theorem VI.77.a follows from Theorem VI.79 in this case as well. If however, A ( D ) = 8, we perform different reductions depending on the position of v,, = and w in W, U { a , c} (we use the notation of Figure VI.33). Since d(v8) = 8, it follows that v8 # z i , i = 0,1,2; hence, by symmetry, we assume w.1.o.g. that TI^ E { a , d } since we necessarily have us E { a , b,c,d}. Because of property a), if min{d(a),d(c)} = 4 it follows that v8 = d since D 3 P4 3 { z o , I,, 2,). Whence we also may assume w.1.o.g. that d(a) = 4 if min{d(a),d(c)}= 4.

(i) d(a) = 4. Then TI,,,= us = d, and we necessarily have the situation as described in Figure VI.36, where

(note that d(b) = 6 of necessity), and therefore

E E ( D ) , N ( u - ) n N(c)= { d , b, b-}, N ( d ) n N ( a - ) = { a , C, dl}, ~ ( dn)N(c)= {z,, a-, d 2 } . CU-

Noting that we necessarily have d(a-) = d(c) = 6, we conclude that either D is the graph depicted in Figure VI.36, or else A = (d, ,d,, b-) is a separating triangle of D. Also, for the chosen 4-valent vertex w adjacent to TI,,,we have w E ( a , zo, z,, z 2 } . In both cases we reduce the side of A containing zo to the octahedron, thus obtaining a smaller graph D, with A ( D , ) = 6 (see Figure VI.36). Consequently, either D, N 0,, or else it results from 0, by a sequence of 0,-extensions (see Case 1). In any case, we may assume by Theorem VI.79 that b- E Vt,d,, d, E V; holds for a corresponding perfect A-partition {V:, of V ( D , - V ( i n t A ) ) .

sl}


VI.132

D

D,

Figure VI.36. 4, 218 = d.

Reducing D to D , in the case d ( a ) =

Whence we may extend {V:, V;} to a perfect A-partition V ( D )as required by the theorem by defining

{V,, V,}

of

Note that d = vma, E V, while w E V,. This solves case (i).

(ii) d ( a ) > 4. We reduce D to Do, the graph from which D results by a W,-extension. We first consider the cases where u, = u8 = a (whence w = zl),or u, = 218 = d and w = zl, and choose in Do in both cases zo = w. While a = v,, necessarily holds in the first case (here we have d D o ( a )= S), we have the freedom to choose in the second case (here we have A(Do) = 6); whence let d = vmar in the second case. By induction, V ( D o )admits a perfect A-partition {Vp,V:} with u, Vp, and zo E V;. In both of the above cases, we define

vl:=vp,

v2:=v;u{zl,z2}

if b , d E

Vl:= Vp u { z o } ,

V2:=(V; - { z 0 } ) u {z1,z2}

if b E V:, d E V:

and if b,d E V; (thus v,,

E

VP,

= a ) , we define

v,:= v .u { z o ,z , } , v, := {V,” - { z o } }u (2,)

.

,


VI.133

The case v,, = a and b E &O, d E V: is symmetric to the case b E V;, d E yoand thus needs no extra consideration. = Hence, to finish (ii) we are left with the consideration of the case v,,, 218 = d and w = zo. Let b- be the uniquely defined vertex satisfying b- E N ( b ) - { a , c, zo, zl,z 2 } . We have d D o ( b )= 4 and d D o ( b - ) E {4,6}. Observing that A ( D o )= 6 we obtain a perfect A-partition {Vf, V;} with b- E Vp, b E V: either by induction or by Theorem VI.79 (depending on whether dDo(b-) = 6 or d D o ( b - ) = 4). Note that since the four vertex sets { a , b, c, d } , { a , b-, c, d } , { a , b, c, z o } and { a , b-, c, z o } define a 4-gon each, we cannot have { a , c , d } c Ko or { a , z o , c } c yo for any i E {1,2}. Consequently, if { a , c } C yo, { z O , d } C yowhere { i , j } = {1,2}. On the other hand, if a E yo,c E vj” for i # j, we may assume by symmetry that a E V., c E V;. This implies zo E V f , d E V;; in this case, we know from the proof of Theorem VI.79 how to extend {Vf,V;} to a perfect A-partition {q,q} of V ( D ) , and by defining V, := V . , V, := V; we obtain an appropriate A-partition {V,, V,} of V ( D ) .So we are left with the case { a , c } C W.1.o.g. i = 2 (otherwise we permute Vp and V:; thus b E V: may hold). Hence d , z o E Vf. If b E V., then V, := ( V . - { z o } ) U {zl}, V, := V: U { z o , z 2 } yields an A-partition as required by the theorem. For the case b E V; we define V, := (Vp - { z o } ) U { z1,z 2 } , V, := V i U { z o } t o obtain the desired result. This finishes the considerations of (ii) and thus settles Case 2.l)

KO.

Case 3, I N ( w )nv@) I= 1. Hence D results from a weak W,-extension of the eulerian triangulation of the plane Do. Using the notation of Figure VI.34 and because of Cases 1, 2 handled already, we conclude d ( z ) 2 6 for z E ( a , b, c, d } ; by symmetry we assume w.1.o.g. that if d ( z ) = 8 , then z E {a,b}. Hence v,, E { a , b } , and by symmetry we assume w = z1 in any case.

‘1 This last case v, = v8 = d , w = zo (translated into its dual form) requires in Goodey’s proof a different type of reduction because he does not have the dual equivalent of Theorem VI.79 at hand. Note that in this case we may have dDo(b) = dDo(b-) = 4 which does not permit to conclude inductively b- E V., b E V:. On the other hand, one has to avoid the situation a , c , d E &’, zo,b E V:; for this would necessarily yield d,zo E V,, z,, z2 E V,. Of course, one could also apply Theorem VI.79 in this case such that d E V:, a, zo E to obtain the desired result; but since dDo(d) = d D o ( u )= 6 one cannot invoke induction concerning Theorem VI.77.a to obtain d E Vp, a E V;.

v;


VI. 134

We define the eulerian triangulation of the plane D, by identifying b and d in D - {z,, z,, ab, bc}, calling this new vertex z . Thus dDl ( c ) = 4 in any case, while d,,(z) E {6,8}, d D l ( a ) E {4,6} depending on the existence and position of an 8-valent vertex in D. Consequently, in D, we denote u,,, = z (either because of necessity or because we have the freedom to choose), and choose w = c. Assuming first that D, is simple we invoke induction to obtain a perfect A-partition {V., V - }of D, with z E V . , c E V-. This induces V; := (V: - { z } ) U { b , d } , V', := V i - a partition of V ( D )- {z1,z2} such that (V,) is a forest with at most two components, while (V,l)is a tree. Now we define

V,:= Vi U {z,},

v,:=v;,

V2:=V,l U {z,} V2:=V', u {z,,

V,:= V', U {z,},

4:" Vi U (2,)

2,)

if b = v,, if a E V:, if a = u,,,

and a E V,l, and a E V,l.

In all cases, {V,, V,} is an A-partition of V ( D )as required by the theorem. To finish Case 3 it remains to be shown that D,must be a simple graph. This is tantamount to showing that bd 6E ( D ) and that N ( b ) n N ( d ) = { a >71'

'27

'}'

Suppose first that bd E E(D). Then each of the sides of the triangle A* = (b, z,, d ) defines a triangulation of the plane D*for which A* is a face boundary. However, dD. (I,) E 1 (mod 2), and since all vertices of D*- V(A*) are even vertices of D*, precisely one of b and d is an odd vertex in D*. Thus D* is a triangulation of the plane with precisely two odd vertices; and they are adjacent. This contradiction to Lemma VI.76 implies bd 6 E ( D ) . Suppose there exists z E N ( b )n N ( d ) - { a , z,, z 2 ,c}. Since d is 6-valent either A = ( c , d , z ) or A = ( a , d , z ) is a face-boundary of D. Consider A, = ( t , b , z ) with E ( A ) n E(A,) = { t z } where t E { a , c } depending on which case prevails, and denote by D* the triangulation of the plane for which A, is a boundary triangle and which does not contain z,, z 2 .D* $ K3 i because of min{dD(a),d,(c)} > 4. Since we have in any case d,, ( t ) 0 (mod 2), we conclude d,, ( b ) d,, (z) E 0 (mod 2) by Lemma VI.76. Denoting by y the vertex satisfying { t , y } = { a , c } we conclude from d, y, z,, z2 6 V ( D * )and the above congruences that {d,.(b),d,.(2),d,,(t)} E {2,4}. Since d,,(z) = 2 implies that b and t separate D* and thus also D, we must have d,.(z) = 4;


VI.135

and since d f!, V ( D * ) we conclude d , ( x ) = 6. By the same to= d,.(t) = 4 follows. Moreover, since umaz E { a , b } ken, d,,(b) and y,z1,z2 $! V ( D * )we conclude u, = b and d,(b) = 8. Since d D ( a ) > 4, d,(a) = 6 follows. Hence suppose w.1.o.g. that t = c. Define {b,} = N ( b ) - (('7 ' 1 7 '2) zl, t2, c7 2)ND* ( b ) ) , {d,} = N ( d )Because of the preceding considerations on the degrees of b and z and since d is 6-valent, these equations are well-defined; moreover, zd, ,xb, E E ( D ) - E ( D * )and zd E E ( D ) imply d, = b,. That is, A, = (a,b,d,) is a triangle; and it must be a face boundary because N(b) - { a , d,} lies on the same side of A,. Consequently, d,(a) = 4; otherwise, a and d, separate D. However, d,(a) > 4. This contradiction finishes Case 3.l)

N ( w ) n V,(D) = 0. Using the notation of Figure VI.33(i) with D in place of Do, we have w = zo. Assume w.1.o.g. that v, = a; hence d D ( a ) E {6,8} and d,(x) = 6 for x E { b , c , d } . Define D, by identifying b and d in D - {zo,ab,bc}, and as in Case 3, call the new vertex z . Again, D, is an eulerian triangulation of the plane with = 47 dDl (.) = 8. d,l ( a ) E j47 6), dDl Assuming first that D , is simple we apply induction to D, to obtain an appropriate partition {Vi,V2].} of V ( D , ) with z E V:, c E V;. Similar to Case 3, this yields a partition {V:, V,l}of V ( D )- { z o } with b, d E V:, c E V,l. Consequently, if we define Case 4.

then {V,, V,} is a desired A-partition of V ( D ) . Thus, to finish Case 4 and hence the proof of Theorem VI.77.a it remains to be shown that D, is simple. We proceed similar to the corresponding considerations in Case 3 by assuming that there exists a vertex x E '1 One could have proceeded differently by considering D** := (D-D*)U A, and D* separately, applying induction t o D** and Theorem VI.79 to D* (observe that A ( D * )5 6 and that d,.(s) = 4 for s E V(A,)). However, in this instance I preferred the approach presented because it did not rely on Theorem VI.79. Unfortunately, Goodey omits the corresponding part in his proof (which is admittedly rather sketchy) and restricts connectivity considerations t o the dual of Case 4 below where he uses the dual form of Lemma VI.76. Moreover, even his restricted considerations are incomplete and contain a flaw (see the footnote at the end of the proof).

VI.136

VJ. Various Types of Eulerian Trails

N ( b )n N ( d ) - ( a , zo, c } (one shows as before that bd in place of I,).

# E ( D ) by using zo

We first show that xt $Z E ( D ) for at least one t E {a,c}. Suppose to the contrary that xa,xc E E ( D ) . Then at least one of the triangles A, = (x,d,a),and A, = (x,b,a) is not a face boundary of D , and the same is true with respect to the triangles A3 = ( x , d , c ) and A4 = (2, b, c); otherwise, d,(a) = 4, d,(c) = 4 respectively, must hold. Since b,d E & ( D ) we assume w.1.o.g that A3 is not a face boundary of D. But then we find the triangulation of the plane D* $ K3 as in the connectivity consideration in Case 3 with A3 being a face boundary of D*; and zo # V ( D * ) . By Lemma VI.76, by K ( D ) 2 3, and since b,CE V6(D),d,(Z) E {4,6},weconcluded,.(d) = d,,(C) = d,,(z) = 4; thus d D ( x )= 6. Consequently, A, is a face boundary of D and, therefore, A, separates D. Especially, the side of A, not containing t o ,contains two or four elements of E,, two elements of Ed, but no element of E, since we have d,(x) = 6 = d,. (x)+ I {xb,xa} I. That is, a and d separate D , a contradiction. Suppose now xu,xc # E ( D ) . Then the side of the 4-gon Q1:= (a, b, x,d ) not containing z o , contains precisely one element eb E Eb and precisely one element ed E Ed, and the same is true with respect to the 4-gon Q2 := (c,b, x,d). This follows from b, d E V,(D). Denote eb = bub, ed = dud. It follows from the distribution of the elements of Eb and Ed with respect to the sides of Q,, that xub,xud E E ( D ) . To see that ub = u d , we consider the triangulation D- of the 4-gon Q- := (b, x,d , zo) defined by the side of Q- not containing a. By the above considerations we have d D - ( b ) = d,-(d) = 4, d D - ( z o ) = 3. Hence d,-(z) f 1(rnod2). Let vb,v d be defined the same way we defined u b , ud above, but with respect to Q,. Observe that

Since and therefore, (b,c,vb) and (d,c,vd) are face boundaries. 2 3, we conclude v b # ud, hence o + ( c) = d,(c) = 6 and K ( D ) ( ~ v ~ , ~ d , ~ ~ ~ , ~for b ,some c v ~s, E c sv(D-). ) This in turn implies together with d,(x) 6 and d,-(x) = 1(mod2) that d D - ( x ) = 5. Since ub and u d on the one hand and vb,vd on the other hand lie on different sides of Q-,U b = ud follows of necessity. But then we necessarily have

6, and i f I E ( b d ( F ) )I< 5 for every face F and I E(bd(F’)) I= 4 f o r at most one face F’, does then G have a n A-trail ? The graph G’ of Figure VI.37.a) shows that one cannot allow for more than one such face F‘. Also, this G’ shows that specifications concerning the 2-factor Q are not promising in general (for there, Q = Q6 is a set of hexagons). However, it is true that if a (not necessarily planar) connected cubic graph G, has a 2-factor Q4each of whose cycles is a 4-gon, then G, has a hamiltonian cycle H 2 E(G,) - Q4 [KOTZ68c, Theorem 61. This follows as an application of Corollary VI.6. It is therefore left to the reader to prove this result. The above question (F), respectively the equivalence expressed in Lemma VI.84 can just as well be treated in terms of A-trails in eulerian triangulations of the plane.

Theorem VI.85. Let G, be a 2-connected, plane, bipartite, cubic graph with its natural 1-factorization L,and let D = D(G,) be given together with the 3-vertex-coloring {Cl,C,,C,} corresponding to the (natural) 3-face-coloring of G, (see Theorem 111.67). The following statements are equivalent.

1) G, has a hamiltonian cycle H containing L , E C.

2) V(D)has a perfect A-partition {V,, V,} satisfying the equations v, = c;UC,, v, = cl’uc,, ucl’=

c;

c,.

Proof. We consider H as a simple closed curve in the euclidean plane. For every e E L,, since e E E(H)nE(bd(F,))nE(bd(F,)) for some 2-face F2 and 3-face F3 of the natural 3-face-coloring of G,, we conclude that F, E int H if and only if F3 E e s t H .


VI.142

a)

G'

b)

G;

Figure VI.37. a) A 6-regular multigraph G' without Atrail having no n-gonal face boundaries for n > 4, and having precisely two 4-gond face boundaries. b) The cubic bipartite graph G$ satisfying G' = G $ / Q , (&, is defined by the boundaries of the faces marked Q,). Gi is hamiltonian; but G$ has no hamiltonian cycle H with E(GL) - Qs c E(H).

Suppose for el, fl E L , that the corresponding i-faces F', F" with el E E(bd(F')),fi E E(bd(F")), lie on different sides of H,iE (2,3}. By choosing el and fl as lying on H as close to each other as possible, we may assume that el and fi are joined by an edge g in H . W.1.o.g. g E L, E L. Denote by e 3 , f 3 E L, the edges adjacent to g and el , f l respectively. The open edges ei,fi cannot lie on the same side of H ; otherwise, E(bd(F'))nE(bd(F"))# 0, contradicting the fact that F' and F" are both i-faces. On the other hand, e:, fi lying on different sides of H implies that {e,, g, fi} c E(bd(F*))and {el,9,f,} c E(bd(F**))for


VI.143

some faces F* and F**. However, since in the natural 3-edge-coloring C every face boundary contains edges of only two classes of C, while each of bd(F*) and bd(F**)contains edges of all three classes of C,we obtain a contradiction which implies that either a l l 2-faces lie in ezt H , or they all lie in int H . Suppose w.1.o.g that the 2-faces lie in int H . It follows from the argument at the beginning of the proof that the 3-faces lie in ext H . Denoting by Ci (Cy)the necessarily non-empty set of 1-faces lying in int H ( e z t H ) , and by further denoting V, := V(T,),V, := V ( T 2 )where T’,and T, are the trees in D(G,) defined by the faces of G, lying in i n t ( H ) , e z t ( H ) respectively, we obtain V, = Ci U C,,V, = Cr U C, where Ci U C; = C,,Ci n Cy = 0 by definition. Moreover, since G, is bipartite, D := D(G,) is an eulerian triangulation of the plane and so {V,, V,} is a perfect A-partition of V ( D )by Theorem VI.70. Conversely, suppose the perfect A-partition {V,, V,} of V ( D ) is given as described in Z), and let Ti = ( y ) i, = 1,2, be the corresponding trees. For each i E { 1,2}, the edges of G, belonging to exactly one face boundary among the faces represented in y, define a hamiltonian cycle H in G,, with H resulting from V, as well as from V,. Since

V, = Ci u C,,i.e., C, n V, = 0 , and since

L, = { e E E(bd(F,)) n E(bd(F,))/Fj is a j-face, j = 2,s) we conclude that every edge of L , belongs to precisely one boundary of a face represented in V,. That is, L, c E ( H ) . This finishes the proof of the theorem. Consequently, questions on the existence of A-trails in plane eulerian graphs c m , in principle, be treated as questions on the existence of special types of A-trails (perfect A-partitions, respectively) in eulerian triangulations of the plane, or by the same token, can be treated as questions on the existence of special types of Hamiltonian cycles in plane, bipartite, cubic graphs. This leads us to the question why one should deal with A-trails in plane eulerian graphs instead of considering right away hamiltonian problems in plane, bipartite, cubic graphs (particularly in view of the equivalence between the BTC and Conjecture VI.73, as well as between Goodey’s result (Theorem VI.77) and Theorem VI.77.a). History s e e m to justify the ‘hamiltonian approach’, and in section VI.2 we proved a

VI. 144


result on hamiltonian decompositions to solve an eulerian problem. However, as we shall see in the context of compatible cycle decompositions, dealing with certain problems in eulerian graphs and translating corresponding results into the theory of cubic graphs has proved to be a very fruitful approach indeed. In the particular case of A-trails it is my opinion that the ‘eulerian approach’ may eventually be just as fruitful. This opinion results not only from many years of experience in dealing with both eulerian and hamiltonian problems, but also from a methodological difference concerning A-trails and hamiltonian cycles in the respective classes of graphs. Namely: finding eulerian trai2s of whatever kind always involves aZZ edges of the graph, while finding a hamiltonian cycle requires the proper arrangement of certain edges and rejecting, so to say, the remaining edges. The latter problem becomes, therefore, more complicated if the graph under consideration has only vertices of low degree (in particular if it is 3-regular). As a support for this point of view, compare Ore’s Theorem (Theorem 111.75) with the BTC (Conjecture VI.72)). We finish this subsection by exhibiting a relation between arbitrary plane 2-connected graphs and certain eulerian triangulations of the plane having a non-separating A-trail ([REGN76a, Satz 5.1]), respectively plane bipartite cubic graphs having a hamiltonian cycle. Namely: let G be plane and 2-connected, and consider GA := S(G) U R ( S ( G ) )

.

It follows from this definition that GA is an eulerian triangulation of the plane (note that S(G) is bipartite); and it is a simple graph precisely because of K ( G )2 2. Again by definition, the 3-vertex-coloring of G, (which is uniquely determined up to permutation of the color classes) is given by the partition {V,, V,, V,} of V(G,) where V1 := V ( G ) ,

V, := V ( S ( G ) )- V(G),

V3 := V ( R ( S ( G ) )) V(S(G))

(note that V, is an independent set already in S(G), V, is an independent set by definition of S(G), and V, is an independent set because every face of S(G) contains precisely one vertex of R(S(G))). Observing that V, is a set of 4-valent vertices in GA, we may conclude from Theorem VI.67 that H := ((GA)v,,1)v,,2 is connected and has only 2- and 4-valent vertices; hence it has an A-trad TH which can be interpreted as an A-trail T of GA. It follows from this definition of T that T is non-separating. Hence G, := D(GA) is a hamiltonian, plane, bipartite, cubic graph with 4 G 3 ) = 3.

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs VI.145

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs Returning to the discussion preceding Theorem VI.85 we are faced with the question into which direction one should go to determine classes of plane eulerian graphs admitting A-trails. As we have seen above, and posed equivalently, specifications as to the degrees of the vertices of a color class in the 3-vertes-coloring of an eulerian triangulation of the plane do not seem to be promising, unless every vertex of the color class is 4-valent. But there is a parameter which plays an essential role in many graph theoretical problems, and especially in the theory of planar graphs connectivity. As we have seen before, 3-connected, planar, eulerian graphs exist which do not admit A-trails. Although, in our construction, the triangulation of the plane we started from can be 5-connected (as there are non-hamiltonian, planar, cyclically 5-edge-connected, cubic graphs), the outcome is always a 3- but not 4-connected, planar, eulerian graph. I have tried to modify the construction in order to obtain a 4-connected, planar, eulerian graph without A-trails which satisfies the equivalence in statement (V1.A) (see the discussion following Lemma VI.75)) but without success. Hence I pose the following conjecture. Conjecture VI.86. Every planar, 4-connected, eulerian graph has an A-trail. The above lack of success is not the only reason for posing this conjecture. For it follows from Tutte’s ‘Bridge Theorem’ (Theorem 111.70) that every planar 4-connected graph is hamiltonian (Corollary 111.71). And this is a very strong property indeed. Moreover, it has been shown that the problem of finding a hamiltonian cycle in a 4-connected planar graph, is polynomial. More precisely, an algorithm based on the proof of Tutte’s ‘Bridge Theorem’ (Theorem 111.70) has been developed whose “time and space-complexity . . . is overvalued in O ( n3) ”,where n is the number of edges, [GOUY82a, Theorem 21. This is in sharp contrast to the fact that deciding the existence of hamiltonian cycles in planar, 3-connected, cubic graphs is an NP-complete problem, [GARE76a]; and this fact will play a certain role in our considerations on the complexity of determining whether a planar, eulerian graph has an A-trail. Moreover, finding hamiltonian cycles in bipartite cubic planar graphs is an NP-complete problem as well, [PLES83a]; however, this result does include all such 2-connected

VI.146


graphs (which are not so interesting in view of the BTC and, via dualization, of little significance to the consideration of A-trails). On the other hand, it was shown in [NISH83a] that the problem of determining a hamiltonian walk (shortest closed walk covering all vertices) in triangulations of the plane, can be solved in O ( p 2 )time, and that the length of a hamiltonian walk is at most ( p - 3) provided p 2 9. Thus, in the case of 4-connected triangulations of the plane, one can find a hamiltonian cycle even in O ( p 2 )time. More recently, a linear algorithm for finding a hamiltonian cycle in such graphs has been developed in [ASANS4a], and in the same year N. Chiba and T. Nishizeki published such algorithm for arbitrary 4-connected planar graphs (see, e.g., [NISHSSa, p.182- 1841. The existence of hamiltonian cycles in 4-connected triangulations of the plane follows from Tutte’s result, but was proved already by H. Whitney in the thirties (see Corollary 111.72). Note that a triangulation of the plane need not be Hamiltonian - just apply an 0,-extension for every face of such a triangulation). These considerations can be viewed as positively supporting Conjecture VI. 86.‘1

3

A hamiltonian cycle H in a simple plane graph G defines two simple 2connected outerplane graphs G , ,G, satisfying G,UG, = G and G , nG, = H . 2 ) For, viewing H as a simple closed curve in the plane and a plane graph as a certain point set, one can define G , and G, by

V ( G , ) = V(G,) = V ( G ) E(G,)= { e E E(G)/e n int H = 0 ) , E(G,)= {fE E ( G ) / f n ext H = 0) . Now, if both G, and G, are eulerian, they have by Theorem VI.63, Atrails T, and T,, respectively. Hence one is inclined t o believe that this suffices to construct an A-trail T of G starting from T, and T,. A. Rosa once told me during a stroll that in his opinion (and I hope I quote him correctly), there are two categories of conjectures: firstly those where “everyone knows that they must be correct, but one cannot yet prove them”, and secondly those “which create surprise if they are correct”. I believe that this is a very good description of how (how many ?) mathematicians relate to conjectures (in my own research I was faced with both situations). As far as Conjecture VI.86 is concerned, I would place it in-between these two categories. 2, Of course, for precisely one of G , and G,, the ‘outer face’ is the bounded region defined by H ; but it is no problem t o re-embed the corresponding Gi, iE { 1,2}, without altering O+(Gi)- simply start with an embedding of G on the sphere.


Unfortunately, for a 4-connected, planar, eulerian graph G to have a hamiltonian cycle H does not guarantee an easy way of finding an A-trail in G . This general statement is illustrated by the following discussion in which we say that the hamiltonian cycle H of G has the A-property if any pair of edges { e , f } adjacent in H belongs to a face boundary of G (or, equivalently, e’ and f’ are neighbours in O+(v) where e‘,f’ E Ec; cf. [REGN76a, Definition 4.2.1.1). As we shall see below if G admits such an H having the A-property, it is possible indeed to combine Atrails T, and T, in G , and G,, respectively, to obtain an A-trail in G. In fact, in this case one can start from arbitrary A-trails. However, we first study A-partitions in outerplane eulerian graphs. The following result, [REGN76a, Satz 4.1.2.1, gives a simple and precise characterization of such A-partitions (this result will be more relevant, however, in establishing a formula for the number of A-trails in outerplane graphs. - See Volume 2, Chapter IX).

Theorem VI.87. Let G be a simple 2-face-colored, 2-connected, outerplane, eulerian graph whose outer face F, is a 1-face. Then the following statements are equivalent. 1) G has an A-trail.

2) V ( G )admits a partition {V,, V,} such that I V(bd(F,))nV1 I= 1 for every 1-face F, # F,. Proof. If G is a cycle, then the equivalence between 1) and 2) is vacuously true; for in this case, F, # F, does not exist. Hence we assume in the following discussion that A ( G ) > 2 holds. That is, at least one 1-face Fl # F, exists. a) Let T be an A-trail of G , and let F, # F, be arbitrarily chosen. Moreover, let { V,, V,} denote any A-partition defined by 2’. Since Fl is a 1-face we must have I V(bd(F,))n V, I> 0; this follows exclusively from the fact A ( G ) > 2; for in this case, V(bd(F,))n V, = 8 implies that G,,, is disconnected with (E(bd(F,)))as a component. Suppose now V(bd(F,))nVlZI {v,w}. Then R = {F,, F,} is a unicolored face-ring with L , = {v,w} as a complete set of links. The elements of R are 1-faces, and L R C Vl follows from the assumption. By Theorem VI.67, however, this contradicts the fact that {Vl, V,} is an A-partition. b) Suppose the partition {Vl, V,} of V ( G )satisfies I V(bd(F,))n V, I= 1 for every 1-face F, # F,. Consider an arbitrary unicolored face-ring R

VI.148


whose elements are &faces for fixed 6 E { 1,2}, and let LR be a complete set of links of R. By Theorem VI.67, it suffices to show LR v6.Suppose to the contrary that for some such R and LR we have LR C v6. Suppose first 6 = 1. Since I R I> 1 it follows that Fl E R for some Fl # F,, and by definition I V(bd(F,)) n LR I= 2. Consequently, I V ( b d ( F l ) )n V, I> 1 which violates 2). Now suppose 6 = 2. Among the possible choices for R and LR consider one with minimal I LR 1; i.e., L C LR implies that there is no unicolored face-ring R' having L as a complete set of links. We want to show that LR = V ( b d ( F * ) )for some 1-face F* # F,.

IE

It follows from the minimality of LR that I bd(F') n bd(F") n LR {0,1} for different elements Ft7F" E R. Hence we can espress R and LR in the following form:

bd(Fi)n bd(Fj)= 0 for j - i > 1, 1 5 i

< j 5 m, { i , j } # (1, m } .

As in the first part of the proof of Theorem VI.67, we construct a simple closed curve C with m

GnC=L,

.

and C - L R C U F , i= 1

The unbounded region of C, ext C, contains F,, and since G is outerplane it follows that int C n V ( G ) = 0. Consequently, the path joining ? J ~ - ~ and , ~ ?J;,;+~ in bd(Fi) whose open edges lie in int C must satisfy pi,i+l

and therefore,

int

{'i-I,i?

'i,i+I}

7


Hence,

(LR) C int C U LR and ( L R )is a cycle. This, A(e) = l ( e E E ( G ) ) ,and the minimality of LR imply that

(intC - E( ( L R ) ) )n G = 0

.

That is, adjacent edges in ( L R ) are consecutive in O+(v), where v is their common end-vertex. Thus ( L R )is a face boundary bd(F*). Since E ( ( L R )n ) E(bd(Fj))# 8 for i = 1,.. . ,m, it follows of necessity that F* is a l-face; and F* # F, because ( L R )n ezt C = 0. Thus, we have found a l-face F* with

V(bd(F*))= LR

V,,

i.e.,

IV(bd(F*))n V, I= 0 .

This contradiction to statement 2) finishes the case 6 = 2; whence we conclude that LR V,, 6 E { 1,2}, for every unicolored face-ring R whose elements are 6-faces, with LR being a complete set of links of R. By Theorem VI.67, G has an A-trail whose A-partition is {Vl,V,}. Now we prove the result indicated at the end of the discussion preceding Theorem VI.87.

Theorem VI.88. ([REGN76a, Satz 4.2.1.1). Let G be a simple plane, eulerian graph with a hamiltonian cycle H having the A-property. Then G has an A-trail. Proof. We can write G as the union of two 2-connected, outerplane, eulerian graphs G , and G, with G,nG2 = H (see the discussion preceding Theorem VI.87). Let T, and T2 be A-trails of G,,G, respectively; by Theorem VI.63, T, and T2 exist (we start with a 2-face-coloring of G and apply Theorem VI.63 to Gi whose 2-face-coloring is induced by that of G, i = 1,2. Thus, the face of G j whose boundary contains all vertices of G, has either color 1 or 2 depending on the value of i E {1,2}). Because of the A-property of H we have for arbitrary v E V ( G )

Consider now {V:, V i } ,an A-partition corresponding to Ti in Gi, i = 1,2, and define

Wj’ := T/j; n ( V ( G i )- V . ( G J ) for arbitrary i , j E {1,2}

.


VI. 150

This definition together with V j n Vj = 8 and (*) yield

~ j n’W: = 8 (observe that Wj

( j , i ) # (k,z), { i , j , k, Z} C_ {I,2)

for

2 V,(G,)

(**)

for { i , k } = {1,2}).

Consequently, {PVj/i,j = 1,2} is a partition of V ( G )- V,(G); therefore, W j := Wj’ U Wj”, j = 1,2, defines a partition of V ( G )- V,(G). Now we observe that a 2-valent vertex 21, cannot belong to any complete set of links LR of any unicolored face-ring R; for 21, belongs to precisely one &face, 6 = 1,2. Hence, if we can show for such R and LR that LR g W,, provided the elements of R are &faces, 6 E {1,2}, then for any V: V,(G) an A-partition {V,, V,} of V ( G )is obtained by defining V, := W , U V:, V, := PV, U (V,(G) - V:) (see Theorem VI.67).

s

So, let R and LR be as above. If all elements of R are &faces of Gi, i E {1,2}, then the existence of Tiimplies LR g Wi by Theorem VI.67, 6 E {1,2}. Since W i 2 V,(Gj) for { i , j } = {1,2}, LR g W, follows. Hence we have to assume that some elements of R are faces of G, and some are faces of G,; and all of them are &faces, 6 E { 1,2}. We further specify the situation by assuming w.1.o.g. that the “outer” face F 1 ( F 2 ) of G, (G,) has color 6 (6 1) where we put 6 + 1 = 1 if 6 = 2 (see footnote 2, in the discussion preceding Theorem VI.87). Moreover, denote R = {F,,. . . ,F,; m > 1) and LR = ( V ~ , ~ , V .~ ., ~ , (see the proof of Theorem VI.87, part b)). By relabeling the elements of R and LR if necessary we may assume w.1.o.g. that bd(F,) 2 G,, bd(F,) 2 G,. Define

+

i, = min{i/l < i 5 m A bd(F,)

s G,}

and

i, = i, - 1 .

Then we obtain in G, a unicolored face-ring R‘ whose elements are 6faces, and a corresponding complete set of links L‘ = LR, by defining

R’ := {F,,. . .,Fil ,F1}, L‘ = {2r1,,,

. . ., v ~ , , ~ ~ , v .~ , ~ }

We have L‘ E LR and L’ V(G,). If we had L , C W,, this and W i C V,(G,) would imply L’ E W i C V J . But Theorem VI.67, applied to G,, says that the existence of TIand its A-partition {V:, V i } implies L‘ V i . Thus, we must have L , g W,. Again by Theorem VI.67 and the above definition of V, and V,, this means that {V,, V,} is an A-partition of G; i.e., G has an A-trail. Theorem VI.88 now follows.

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs VI. 1.51

However, not every A-trail T of G is obtainable from A-trails T, and T, of G , and G, respectively, as described in the proof of Theorem VI.8g That is, some A-partition {V,, V,} of V ( G )may not be an ,4-partition of V ( G , ) = V ( G )for i = 1,2. This can be seen already in the case of 0, which we can view as obtained from the cycle c 6 = (u,, . . . , 216) by adding A, = (u,,u3, v5) and A, = ( u 2 ,u4,216). Embed A, in the bounded region of c6; thus A, lies in the unbounded region of c6. We assume the outer face of the 2-face-colored 0 6 to be a 1-face. Defining H := c6, G , := H U A,, G, := H U A2, and V, := {u,, v3,2 ] 6 } , V, := {u,,u4, us} we conclude that although {V,, V,} is a (perfect) A-partition of V ( G ) ,{V,, V,} is not an A-partition of V ( G , ) = V ( G ) ,nor of V(G,) = V ( G )(note that H has is disconnected, and so is ( G 2 ) { v z , V 6 } , l . the A-ProPertY). For, (G1){v12vs},2 However, if an A-partition {V,, V,} of V ( G )is an A-partition of V ( G i )as well, i E {1,2}, we may say that the latter is an induced A-partition of the former. Correspondingly, we can say that the A-trail Tiof Gi is induced by the A-trail T of G. The graph of Figure VI.38 shows an eulerian triangulation D of the plane together with a hamiltonian cycle H such that the corresponding graphs G,, G, satisfying G , U G, = D and G , n G, = H (see the discussion preceding Theorem VI.87), are eulerian. We discuss this graph D: H does not have the A-property since ( V ( G , ) - V,(G,)) n (V(G,) - V9(G3)) - - = { a , b , c , d , e , f } . We know from Theorem VI.63 that Gi has an A-trail Ti, i = 1,2. We want to show that no choice of T, and T2 permits extension of these A-trails to an A-trail T of G in a way similar to the one in the proof of Theorem VI.88. For this purpose, it suffices to show that T, does not induce the same type of splitting in V, := { a , b, c, d, e , f } as T,. To be more precise, let G , and G, have their 2-face-colorings carried over from a 2-face-coloring of D. Among G , and G, let G , be defined as the graph whose outer face F, is the unbounded region of H (where H is viewed as a simple, closed, plane curve), and suppose w.1.o.g. the 2-face-coloring of D is chosen in such a way that F, in G , is a 1-face. Whence we may conclude that the “outer” face int H of G, is a 2-face. As in the proof of Theorem VI.88, let {Vl,V j } be the A-partition of the arbitrarily chosen A-trail Tiof G;, i = 1,2. We want to show that the equations V, n V: = V, n V:, V, n V; = V, n Vz cannot hold. Suppose a E V t . Since d, e , f are boundary vertices of 1-faces of G , each of which contains a as boundary vertex as well, { d , e, f} C V$ follows by Theorem VI.87. By the same theorem however, { e , f} C V$ is impossible since {e,f} c bd(F,) for some 2-face F2 of G, and because the “outer”

VI. 152

VI. Various Types of Euierian Trails

Figure VI.38. An eulerian triangulation of the plane D with hamiltonian cycle H defining two 2-connected7 outerplane, eulerian graphs G,,G2 with D = G , U G,, H = G , n G,. D has no A-partition which induces A-partitions in both G , and G, (note that (V(G,)- V,(G,)) n (V(G,) -

V,(G,)) # 0). face of G, is a 2-face as well. Therefore, the above equations cannot hold if a E V:. Now suppose a E V l nV;. Since b and c are boundary vertices of 2-faces of G, each of which also has a as a boundary vertex, we have { b , c} c Vf by Theorem VI.87. On the other hand, { b , c} E V(bd(F,))for some 1-face of G,; thus, by Theorem VI.87 we must have { b , c } V:. Consequently, also in the case a E V: we may conclude that the above equations cannot hold. However, we leave it as an exercise to check that D has an A-trail indeed (we note, though, that n(D) = 3).


So, while in general it is not possible to deduce from some A-trail of G A-trails Ti in Gi, i = 1 , 2 (where Gi is defined as above with respect to a hamiltonian cycle H ) , even if H has the A-property, we do achieve the following interesting result, [REGN76a, Satz 4.2.21. We present a proof differing entirely from Regner's. Theorem VI.89. Suppose for the 2-face-colored, plane, eulerian graph G that it has a hamiltonian cycle H defining two outerplane eulerian graphs G,, G, satisfying G, U G, = G , G, n G, = H (see above). If G has an A-trail T inducing an A-trail Tiof Gi, T induces an A-trail T j in G, as well, { i , j } = {1,2}. Proof. We assume w.1.o.g. that F,, the unbounded face of G, (with bd(F,) = H ) , is a l-face in the 2-face-coloring of G, induced by the 2-face-coloring of G, and that G has an A-trail T inducing an A-trail T9 in G,. Let {V,, V,} be the A-partition of V ( G )= V(G,) corresponding to T ,T2respectively. We have to show that {V,, V,} is an A-partition of V(G,) as well. For this purpose we first observe that G1 := GLT1,, is a connected outerplane graph whose outer face F&, is a l-face with E(bd(F,)) E(bd(Fk))and such that E(bd(F&,))n E ( B ) # 8 for every cycle B of G1 (Lemma VI.5S, Theorem VI.59). If we can show that every non-trivial block of G1 - E(G,) is an end-block of GI, then G1 - (G, E ( H ) )has precisely one non-trivial component, namely (Gl)vl ,,, which satisfies the hypothesis of Lemma VI.58. Consider the graph

Go is well defined because the "outer" face F" of G, is a 2-face in the induced 2-face-coloring of G, (since F, is a l-face), and therefore E ( H ) = E(bd(F")); so, the construction of Go from (G2)vl,1amounts to adding diagonals of H lying in G,, i.e., adding the elements of E(G,) E ( H ) . Also, since T, appears as a run through the boundary of the uniquely determined l-face of (G2)v1,,, it follows by definition that Go is connected. Moreover, E(bd(F")) C E((G,)"] ,,) implies that 1) (G2)V1,1contains (E(bd(F,))) as a block;

therefore

2) Go has a block B" 5) (E(bd(F"))) with B"

N

G,;

VI. 154


3) every block C of Go - B" is a cycle with E ( C ) E E(G,) - E ( H ) and therefore, C is an end-block of Go (see 2)). We relabel the vertices of B" according to their original labels in G; then we obtain B" = G, and the isomorphism

For every block C of Go other than B" (see 3)) we have C f l B" C V,. Hence, the unique vc E CnB" is not affected by the transition from Go to Since every z E V(C - vc) is 2-valent in Go and therefore also in (GO),, ,,, it follows that every block C of Go - B" is an end-block of Finally, if we view G, and as unlabeled graphs, then the isomorphism (*) becomes an identity (in (*), the isomorphism can be chosen so as to act as the identity on E(G1) = E((Go)v,,l). Consequently, every non-trivial block of G1 - E(G,) is an end-block of G1 whose edge set belongs to G, - H . So,

which satisfies the hypothesis of Lemma VI.58 because G1 also does (the left side of the last equation only expresses the deletion of certain endblocks of G' which renders a connected graph and leaves unchanged the embedding properties espressed in that lemma). It follows that (Gl)vl has an A-trail. Consequently G, has an A-trail. The theorem now follows.

,,

Maybe, the above considerations on A-trails in plane, hamiltonian, eulerian graphs will be of help in eventually proving or disproving Conjecture VI.86. I do not know, however, which of the plane, 4-connected, eulerian graphs admit a hamiltonian cycle H having the A-property or at least an H such that the graphs G,, G, defined by int H U H and ext H U H , are eulerian.

So, let us consider Conjecture VI.86 for the case of eulerian triangulations D.The constructions performed before (see also the considerations preceding Conjecture VI.86) do not permit one to draw the conclusion that A-trails and non-separating A-trails are equivalent concepts in 4connected, eulerian triangulations of the plane (compare this with Conjectures VI.73, VI.74 and the discussion leading up to Lemma VI.75). Modify Conjecture VI.86 for triangulations of the plane to the extent


that one replaces ‘A-trail’ with ‘non-separating A-trail’. It follows from Theorem VI.71 and the relation &(D(G,))= XJG,) for connected plane cubic G, # K4, that this modified conjecture is equivalent to the BTC (Conjecture VI.72) for the case of cyclically 4-edge-connected, planar, bipartite, cubic graphs. But what does the existence of a (not necessarily non-separating) A-trail in D = D(G,) imply for G, ? The answer to this question is given in the next theorem. Recall that a dominating cycle in a graph G (or Tutte cycle if G is cubic) is a cycle C such that E(G - V ( C ) )= 0.

Theorem VI.90. Let G3 be a connected, plane, bipartite, cubic graph with bipartition {V.,V,}, and let D(G,) denote its dual. The following statements are equivalent. 1) D(G,) has an A-trail. 2) G, has a Tutte cycle C such that V’ := G,

- V ( C )satisfies

a) V, n V‘ C int C if and only if V, n V’ c ezt C ;

b) V l n V ’ n e z t C = O o r V l n V ’ n i n t C = 8 . The proof of Theorem VI.90 can be obtained by modifying Tutte’s proof of Theorem VI.71; it is therefore left as an exercise. Note that in Theorem VI.90, the equivalence between Tutte cycles of a special type in G, and A-trails in D(G,) does not require X,(G,) = 4. On the other hand, Tutte’s ‘Bridge Theorem’ guarantees the existence of a dominating cycle in planar, cyclically 4-edge-connected, cubic graphs = 4, the equivalence ex(Corollary 111.73). So, in the case of X,(G,) pressed in Theorem VI.90 boils down to the question whether in the non-empty set of Tutte cycles of G, one can find an element satisfying a) and b). We also note that regarding non-separating A-trails, it suffices to prove Conjecture VI.73 for 4-connected triangulations; this follows from Proposition VI.83, and it is tantamount to saying that it suffices to prove the BTC for the corresponding cyclically 4-edge-connected graphs (see Exercises VI.25 and VI.30).

VI. 156


VI.3.3. How to Find A-Trails: Some Complexity Considerations and Proposals for Some Algorithms Complexity considerations, although a very important part of today’s development of graph theory (in particular with regard to applications to various problems in operations research, for example), do not constitute an essential part of this book. Nevertheless, here and in other chapters such considerations will be performed depending on the importance I attribute to a (solved or unsolved) problem; however, I have restricted my considerations on this topic to such questions as whether a given problem is (at least) NP-complete or whether it can be solved/decided in polynomial time.l) In what follows, we basically restrict ourselves to 3-connected, planar, eulerian graphs. For, as we have seen before, a 2-connected, planar, eulerian graph G may have plane embeddings G , and G, such that G, has no A-trail while G, does (see Figures VI.26 - VI.28). Thus, we would have t o consider embedding problems as well, respectively we would have to check all plane embeddings concerning the existence or non-existence of A-trails. On the other hand, 3-connected, planar graphs G are uniquely embeddable on the plane (see Theorem 111,52); hence (see the discussion preceding Lemma VI.64), the existence of an A-trail in such a G which is also eulerian, is independent of any concrete embedding of G on the plane (or equivalently, 3-dimensional sphere). Nevertheless, for practical purposes we shall always consider some embedding of G. On the grounds of Regner’s construction of a planar, 3-connected, eulerian graph without A-trails (respectively statement (V1.A) preceding Lemma VI.76), and because of [GARE76a], we can formulate and prove the following result.

Theorem VI.91. The problem of determining whether a given connected, plane, eulerian graph has an A-trail, is an NP-complete problem. ’) I know that this is a very subjective point of view, and I foresee the disappointment and criticism of some of my colleagues. However, given the material I had to choose from, a thorough treatment of complexity and/or algorithms would have occupied a large part of the book. Moreover, I frankly admit that I am not a specialist on these topics. Hence I prefer to leave them to more competent researchers.

VI.3.3. A-Trails: Complexity Considerations, Algorithms

VI. 13'7

This is true even if the problem is restricted to 3-connected graphs.l)

Proof. By statement (VI.A), if G, is a planar, 3-connected, cubic graph, a planar, 3-connected, eulerian graph G with D(G,) c G exists such that no face boundary of D(G,) is a face boundary of G , and G has an A-trail if and only if G, has a hamiltonian cycle. Since the hamiltonian problem for G, is NP-complete, [GARE76a], it is (more than) sufficient to show that the A-trail problem for G 3 D(G,), is polynomially equivalent to the hamiltonian problem for G,, and that G can be constructed from G, in polynomial time. Since it follows from Theorems 111.89 and 111.90 that deciding whether a given graph is planar, and if so, finding an actual plane embedding, can be done in polynomial time, we may assume that G,, G respectively, are actually plane graphs. The problem of finding a 2-face-coloring of G can also be solved in polynomial time (Theorem 111.91). Hence, we may assume that the 2-face-coloring of G is given as well. Assume an A-trail T given in G 3 D(G,). A single run through T tells us at every vertex w which is reached (be it that 21 is reached in this run for the first time or not), whether 21 E V, or 21 E V, where {Vl,U,} is the A-partition of T. For, together with the embedding of G we can determine the face boundaries of D(G,) in polynomial time (Theorem 111.90~)~ so we know whether edges of E, consecutive in T belong to the boundary of a 1-face or 2-face of G. Thus, {V,,V,} is obtained in polynomial time from T . Having stored V ( D ( G , ) ) we obtain 5':= K n V ( D ( G , ) ) , such that (y')is a tree, i = 1,2, in polynomial time P 2 ( P G ) . Knowing the bijection wF E V ( D ( G , ) ) t+ bd(F) C G , for every face F of G, (which requires polynomial space and time only), we form Go = UVFE"{ bd(F) c G,. This and deleting those edges of Go which belong to bd(F') and bd(F") for F' # F", vF,uF,, E E((V:)), can be done and results in H , a hamiltonian cycle of G,. in polynomial time P3(pG3), Since, p G L C0nSt.pG3 (see below) P ( p G 3 ) := CpG) P 2 ( P G ) P 3 ( p G 3 ) is the polynomial in pG3 =I V(G,) I representing the time required to obtain the hamiltonian cycle H of G, from the A-trail T of G.

+

+

The first part of this theorem has been proved in [BENT87a], where the authors speak of 'non-intersecting eulerian trails' what we call A-trails. Also, their constructions involve multiple edges and cut vertices in most instances. It is also worth noting that the motivation for this paper is derived from a problem in flame cutting.


VI.158

Conversely, suppose a hamiltonian cycle H C G , be given. Then it takes to reach the vertex partition {V;,Vi} of D(G,) polynomial time Pi(pG3) such that (V;’) is a tree, i = 1 , 2 (V,l corresponds to the set of faces F of G, with F C int H ) . In order to obtain an A-partition {V,,V,} of V ( G ) such that V;‘ = V; n V ( D ( G , ) ) ,i = 1,2, we choose a particular way of constructing a plane, 3-connected7eulerian graph G with D(G,) C G and such that no face boundary of D(G,) is a face boundary of G. It has been noted in the discussion preceding Lemma VI.76 that the construction of G is not unique, and that it can be done in such a way that every face of D(G,) contains precisely one copy of either Hi of Figure VI.32 or of H,, of Figure VI.31. This construction will be performed as the last step of the proof; hence assume first that G with this property is given. Then - fG3 +5pG 3 -< 6pG, , where the second inequality follows from Euler’s PG < polyhedron formula, and the first inequality is due to the fact that a face of D(G,) (which corresponds to a vertex of G,, and vice versa) contains either 3 or 5 vertices of V ( G )- V(D(G,)). Having stored D(G,) and {V,l,V,l}, we can extend this partition to { v,,V,} as follows. a) The face F of D(G,) contains precisely three vertices; then it contains Ho. Following the notation of Figure VI.31, we may by symmetry assume w.1.o.g. that a , b E V,l,c E V,l. In this case we define { a l , b , } c V,, c1 E V,. Note that there are altogether only 6 choices for the distribution of a, b, c in V,l and V,l.

b) The face F of D(G,) contains precisely five vertices, so it contains Hi. We follow the notation of Figure VI.32, and having again six choices for the distribution of xi-l, xi,yi in V; and V,l,we assume by symmetry w.1.o.g. that either { Z ~ - ~ , Z ~c } V,l and yi E V,l, or {xjal,gi} c V[ and xi E Vi. In any case, let the vertex of G, uniquely determined by N ( z i - , ) n N(y,) n F , belong to V,; let the vertex ti of G , uniquely determined by N ( x i - , ) n N ( x i ) n F , belong to V, unless A = (xi-l, ti,xi) is the boundary of a 2-face and yi E V,l in which case we let ti belong to V,; and let the corresponding vertex of N ( z i ) n N(yi) n F belong t o V,. As for the remaining two vertices ui-, E N ( z i - , ) and ui E N ( z i )of V ( H J n F , define ui-,,ui E V,. Since

(V;’)

is a tree, i = 1 , 2 , and because of the way we extended { V:, V,l} to { V,,V,}, it follows from these considerations that { V,,V,} is an A-partition which can be obtained from {V;,V,l} in polynomial


VI.159

time Pi(pG3). So, together with the 2-face-coloring of G we can perform the transition from {V,, V2}to a cycle T which stands for an Atrail of G, by a step by step procedure. Namely: define GO := G, and Gi := (Gi-l)jviI,6, i = 1,...,p G , where V ( G ) = {zl, ,...)upG}, and the value of 6 E {1,2} is determined by zli E V,. Since this transition from Gi-l to Gi can be done in polynomial time, we reach T = GPG from {V,, V2}in polynomial time Pi(pG). Again, since pG = const.pG3, we reach the A-trail T of G from the given hamiltonian cycle H of G, in polynomial time P ' ( p G 3 ) = P i ( p G 3 ) Pi(pG3) Pi(pG). Thus, the problem of finding an A-trail in G 3 D(G,) is polynomially equivalent to the problem of finding a hamiltonian cycle in G,.

+

+

Finally, let us show that the construction of G with the property that every face of D(G,) contains precisely one copy of either Hi or H,,can be done in polynomial time. First we observe that the transition from G, to D(G,) can be done in polynomial time Ql(pG3). We anticipate now some of the theory on the Chinese Postman Problem (CPP) which we shall develop in Volume 2, Chapter VIII. Namely: given any connected graph H , there exists an eulerian supergraph H I with the following properties: 1) V ( H , ) = V ( H ) , E( H , )

2 E(H);

2) if x and y are not adjacent in H, they will not be adjacent in H,;

3) for every cycle C, of H , we have

Expressed more descriptively, 1)- 3) says that H I is obtained from H by replacing certain edges of H with an edge of multiplicity 2 in such a way that H , is eulerian and such that for every cycle C of H , the number of edges of C which are replaced in H , with a multiple edge is at most +Z(C). As we shall see in the discussion on the CPP, the construction of H , from H can be done in polynomial time. Note that in this construction,

=

d H I - - E ( H ) ( u ) dH(2r)(mod2) for every

zl

E V ( H )= V ( H , ) .

Applied to the construction of G from H = D(G,) the above amounts to saying that we can label the edges of D(G,) in polynomial time


VI. 160

with 0 and 1 in such a way that in D(G,) for E(D(G,))/X(e)= 1) (with X denoting the label function)

Q2(I)GJ)

I E, n El 1-

d D ( G 3 ) ((mod2) ~) for every

and such that for every cycle

El :=

E V(D(G,))

{e E

(1)

C of D(G,)

In particular, (2) holds for the triangular face boundaries A of D(G,). That is, for every such A, I E ( A )nEl {0,1}. Following the notation of Figures VI.31 and VI.32 we write V ( A ):= { a , b, c } if E ( A )n El = 8, and V(A) := { Z ~ - ~ , X ~if ,E~( A ~ )} n El # 0 with the notation chosen in such a way that w.1.o.g. { Z ~ - ~ Z=~E} ( A )n El. Among the two face boundaries containing Z ~ - ~ Z ; ,choose one and fix it; denote it as above with A. For each of the A chosen with E ( A ) n E, # 8 embed the graph Hiof Figure VI.32 in the corresponding face. For every other face boundary A = ( a ,b, c ) embed H, of Figure VI.31 in the corresponding face. Denote by G the graph resulting from D(G,) this way. This requires polynomial time Q , ( p G 3 )only, because for each of the pG3 face boundaries of D(G,) the embedding of H,, Hi respectively, can be done in constant time. Thus, the entire transition from G, to G can be done in polynomial time Q(PG,) = Qi( P G ~ ) Q,(PG~ 9 3 6 3 ).~ ~ Because of (1) and the very construction of G from D(G,) it follows that G is plane and eulerian; and it is 3-connected because G, is 3-connected. This finishes the proof of the theorem.

+

+

We note that in the second step of the proof of Theorem VI.91 it is not necessary to extend explicitly {V;,V,l)to {Vl, V2}. It suffices to construct G = (Gv:,1)v;,2 which is connected and 4-regular, and then find algorithmically an A-trail of G in polynomial time by successively applying the Splitting Lemma, say (see Volume 2, Chapter X). As for the complexity of the A-trail problem in planar, 4-connected graphs, it can very well be that it parallels the phenomenon concerning the complexity of the hamiltonian problem. For as we have noted before, in 4-connected planar graphs, a hamiltonian cycle can be found in polynomial time. This implies that a Tutte cycle in a planar cubic


VI. 161

graph G, with X,(G,) 2 4 can be found in polynomial time; in this case, L(G,) is a planar 4-connected graph (see Corollary 111.73). Of course, this does not guarantee that a Tutte cycle of the type described in Theorem VI.90, can be found in polynomial time.,) In fact, there is some evidence of the likelihood of the above parallelism. In addition to the facts quoted in the discussion following Conjecture VI.86 we may make the following observations (we restrict ourselves to outlines of proofs, the details are left as exercises). 1) An A-trail of a 2-connected outerplane simple graph can be found in polynomial time.

We use Theorem VI.87 to develop a procedure for constructing an A-trail in a 2-connected outerplane simple graph G. We assume G to be 2-facecolored with the outer face F, of G being a l-face. The procedure (an algorithm, so to say) amounts to successively marking vertices of valency exceeding 2 in such a way that ultimately every l-face F, # F, contains exactly one marked vertex. Denoting by V, the set of marked vertices and V, := V ( G )- V,, the partition {V,, V,} will then satisfy Theorem VI.87, and therefore, (Gvl,l)v,,2 will be a cycle representing an A-trail of ‘1 On the other hand, it has been claimed that the hamiltonian cycle problem is solvable in polynomial time for arbitrary edge graphs (see [GARE79a, p.199, [GT37]], where the authors refer to [LIUC68a]); hence the Tutte cycle problem would lie in P (see Lemma 111.69). This claim must be erroneous. In order t o see this, we first consider L(G,) for a 2-connected cubic graph G,. Next we introduce a new vertex vP+ifor every vertex vi of G, and add the new edge v ~ v ~ + i ~=, 1,.. . , p . Denote by H the graph thus obtained. Finally we consider L ( N ) and S(G,)UL(G,). The latter graph can be viewed as obtained from S(G,) by letting V(L(G,)) = V,(S(G,)) and then adding E(L(G3)). If we proceed thus L ( H ) and S(G3)UL(G3) are isomorphic graphs. With this visualization of L ( H ) ,it now follows by a straightforward argument that there is a l-l-correspondence between hamiltonian cycles of G, and such cycles in L ( H ) . Moreover, the above construction, the transformation of a hamiltonian cycle of G, into the corresponding hamiltonian cycle of L ( H ) and the inverse transformation can be performed in polynomial time. Whence we may conclude that the hamiltonian cycle problem for L ( H ) is NP-complete even if G, is planar and 3-connected (cf. [GARE76a]). We note that these arguments are a slight modification of Bertossi’s considerations in [BERTSla] who dealt with the analogous error concerning the hamiltonian path problem [GARE79a, p.199, [GT 3911. - Incidentally, I could not find in Liu’s book the claim quoted in [GARE79a].

VI. 162


G. The starting point, however, is the construction of the weak dual of G

D"(G) := D(G) - f, where f, is the vertex of D ( G ) ,the dual of G, corresponding to F,. It has been shown, [FLEI74d, Theorem 1 and Corollary], that D"(G) is a tree for 2-connected7outerplane graphs. Hence, in our case T" := D"(G) is a tree whose vertex bipartition {Vlw,VT}following the 2-face-coloring of G is such that the end-vertices of T" belong to V2". Choose vo E l.;U and orient the edges of "'2 in such a way that T" is transformed into an out-tree T," with V(T,") = V(T") and root vo. For every v E Vlw let F, # F, denote the l-face of G corresponding to v, and for any l-face F # F, of G let v F denote the corresponding vertex of Vlw.We observe that the following initial steps can be done altogether in polynomial time Po(n)where n =I V ( G )I: a) Establishing the face boundaries of G and classifying them according to whether they belong to a l-face or to a 2-face.

b) Establishing for every v E V ( G ) the set L, of l-faces Fl with v E V(bd(F1)). c ) Constructing the plane out-tree T," with root vo and vertex bipartition { Vlw,V?}. We first mark the root vo E V(T,"). Then we mark an arbitrary vertes v1 E bd(FVo),and let v1 belong to V, (note that Fvo is a l-face by the choice of vo). Consider next L,,,and mark v F E Vy for every F E Lvl - { F v 0 } . At this stage as well as at any later stage of this marking process, the following property is fulfilled precisely because G is a 2connected outerplane graph:

If P(vo,w ) is a path in T O W joining the root vo with w E V?, and if w has been marked, all elements of V ( P ( v o,w))nVlwhave been marked. Next we choose an unmarked vertex

(*> E Vlw subject to the condition that

all elements of V(P(vo,v) - v) n Vlwhave been marked already. (**) We mark this chosen v. Precisely because G is outerplane and simple it follows that there exists v2 E V(bd(F,)) such that


VI. 163

d(v2)2 4 and for every F E Lv2 - { F u } , vF E V(T,") has n o t been marked yet. (* * *I We mark v2. Then we mark in T," every vertex vF # v for F E Lv2-{ F,}. We observe that because of (*) the elements of Vlwalready marked together with the elements of V2wadjacent from these marked vertices, define a subtree T' of T," with root v,,. Therefore, if TI # T,",we choose z E V;U subject to condition (**) (with z in place of v) in order to mark x and to find v3 E V(bd(F,))satisfying (* * *) (with v3 in place of v2), and mark us. Then we mark in T," the vertex v F # z for every F E Lvj - {F,}. We observe that the subtree T" of T," defined after this new step of the marking procedure in the same way that TI was defined above, satisfies TI c TI' T,"

c

Therefore, choosing repeatedly a vertex of Vlwsubject to (**) and a vertex of G satisfying (* * *) and marking the latter vertex, eventually leads to an out-tree T ( i )satisfying

T'

c TI' c . . .

After this i-th step, the set V, C V ( G )consisting precisely of the marked vertices of V ( G ) ,and V, := V ( G )- V, defines a vertex partition. We note that it is precisely because of (* * *) that statement 2) of Theorem VI.87 is fulfilled. Hence {V,, V,} is an A-partition of V(G). We note in passing that the preceding algorithmic considerations together with Theorem VI.87 amount to another proof of Theorem VI.63 independent from the proof established earlier. As for the complexity of the above marking procedure we observe that after the j-th step, 1 5 j < i, the choice for v E VlWsubject to (**) is arbitrary in the set of sources of T," - V ( T ( j ) ) .So, constructing T ( j ) and choosing and marking such v among the sources of T," - V ( T ( j ) ) can be done in polynomial time Pl(n). Since the marking procedure in Tow can be combined with marking the corresponding elements in the sets Lu, the discovery of a vertex vj+2 E V ( G )satisfying (* * *) in the ( j l)-th step can be done in polynomial time P2(n).Hence, the time required to construct the A-partition {V,,V,} is at most P-,(n)+n(P,(n)+ P2(n)).Since the transition from {V,,V,} to (GVl,1)V2,2 can be done in polynomial time as well, it follows that finding an A-trail in G can be done in polynomial time.

+

VI. 164


On the grounds of 1) we arrive at the following statement.

2) If G i s a simple, plane, eulerian graph with a given hamiltonian cycle H having the A-property, a n A-trail of G can be found in polynomial time. By using the notation of the proof of Theorem VI.88 we may conclude from 1) that A-trails T, and T2 of GI and G,, respectively, can be found in polynomial time (where G, U G, = G, G, n G, = H ) . Hence the corresponding A-partitions {Vj,V i } of Ti,i = 1 , 2 , can be found in polynomial time. Since the four vertex sets

W; := vj‘ n ( V ( G i )- V2(Gi)), i , j E {1,2} can be found in polynomial time, we therefore arrive at Wj := ryj’ U Wj”, j = 1,2, in polynomial time. Since (Gw,,1)w2,2can then be obtained in polynomial time, and since this graph is a cycle (see the proof of Theorem VI.88), it corresponds to an A-trail of G obtained in polynomial time. We observe, however, that the above considerations are based on a given hamiltonian cycle H of G. Hence the question arises: how complex is the problem of deciding whether a given 4-connected, planar, eulerian graph has a hamiltonian cycle with the A-property ? Of course, in 2) we did not need any assumptions on K ( G )(however, K ( G )2 2 since G is hamiltonian), but in view of Conjecture VI.86 and Tutte’s theorem on hamiltonian cycles (Corollary 111.71) we restrict ourselves in the above question to 4-connected graphs. It should be noted that if G is any eulerian triangulation of the plane (4-connected or not, simple or not), this problem belongs to P ; for, in such G, if e, f E E ( G )are incident with v E V(G)and O+(v) = (. . . ,e(v), f(v), . . .), e and f either determine a unique hamiltonian cycle H having the A-property with e, f E E ( H ) , or else H does not exist at all (Exercise VI.32). Thus, it can very well be that the above question has a ‘polynomial’ answer as well. Note that for a triangulation D of the plane, a hamiltonian cycle having the A-property is a special type of a left-right path as defined in [SHAN75a]. We leave it to the reader to check the complexity of finding an A-trail in a graph covered by either of the Theorems VI.77.a and VI.79 (Exercise VI.33), and now consider the general case.

An A-Trail Algorithm For Arbitrary Plane Eulerian Graphs

VI.165

An A-Trail Algorithm for Arbitrary Plane Eulerian Graphs Our starting point is a 2-connected, plane, eulerian graph Go other than a cycle and, indirectly, Theorems VI.59 and VI.67. However, instead of considering unicolored face-rings we combine a vertex-marking procedure with a sequence of 1-splittings. '1 Step 0. Consider Go together with a 2-face-coloring such that the outer face Fomof Go is a 1-face. Denote V< = V(bd(F,")) - V,(Go), and let x1 E V< be arbitrarily chosen. Set i = 1. Step 1. Form G i := ( G j - l ) { z i l , l . Denote by Fi" the outer face of Gi, and let the 2-face-coloring of Gi be induced by the 2-face-coloring of

Gi-1Step 2. Mark those so far unmarked elements of V(bd(Fim))which are cut vertices of Gi,and let denote the set of unmarked vertices in V(bd(Fim))- V,(Gi).

v

Step 3. If Step 1.

If

V;.. # 8, then choose any xi+l

E

v. Set i = i + 1 and go to

v = 0, then continue.

Step 4. Set V,.* = {z/x E ( E ( G i )- E(bd(F;"))) U ( V ( G i )- V2(Gi)); x is unmarked if it is a vertex }. If V,.* = 8, then go to Step 7.

If

v*# 8, then continue.

Step 5. Search for the largest integer j

< i such that v;" - { z ~ + #~ }8.

If such j does not exist, then go to Step 8. If j exists, continue. Step 6. Mark xj+l and set

= v;" - {x~+~}. Set i = j and go to Step

3. Step 7. Go has an A-trail. Step 8. Go has no A-trail. '1 For the case of eulerian triangulations of the plane, this algorithm has been outlined already in [FLEI88a].


VI. 166

In fact, this algorithm does more than decide whether Go has an Atrail: for, if = 0 = V,.* for some i > 0 (which is the necessary and sufficient condition for reaching Step 7), then Gi satisfies the hypothesis of Lemma VI.58. Hence a run through bd(F;”) describes an A-trail To of Go. Moreover, an A-partition {Vf, V;} in Go corresponding to To is, in terms of the algorithm’s notation, defined by

v:

= {xl, ...,Xi} , V : = {x E V(G)/z has been marked or x E V,(Go)}

.

We note that the construction of G iand the determination of V,*,Ty* respectively, can be done very fast. This is also true for the determination of the integer j < i in Step 5. However, the algorithm on the whole is very slow. This becomes clear by studying the Steps 4,5 and 6: V;. = 0 and # 0 means that, for the time being, one cannot continue the ‘greedy approach’ expressed in Steps 1, 2 and 3; i.e., one has reached a deadlock. At this point, each block B of G isatisfying BnV,**# 0 defines a unicolored face-ring R (whose elements are 2-faces) with a complete set of links L, := V(bd(F,”)) - V2(B), where F r is the unbounded face of B (whose embedding is induced by that of G i ) and b d ( F p ) c bd(F;”). For, = 0 means that all elements of V(bd(F,“)) - V2(Gi)have been already marked; since this and B nV;,.* # 0 imply int bd(F,”) n V,.* # 0, we conclude that there are (at least two) 2-faces F2 with E(bd(F2))n E(bd(F,”)) # 0, whence R is defined by the set of these 2-faces. We note, in addition, that a marked vertex of G, need not be a cut vertex of Gi (see Step 6). Steps 5 and 6 are means of getting out of this deadlock. That is, one has to go back to some G j , j < i (for practical reasons one chooses the largest j ) and choose some vertex E - {zj+l} to which one applies a 1-splitting to obtain G[i+l. At this point has become a marked vertex, and all the algorithmic work in establishing Gj+l,. . .,G,is to no avail. That is, the marking and splitting procedures in these graphs are undone. But having shifted xj+l from to the set of marked vertices of G, does not say anything about the ‘final fate’ of xj+l in an A-partition {Vf,V;} of Go that might exist. For, at a later stage one might reach j‘ < j ; and continuing with Step 1 for i = j‘ 1 might result in the formation of a Vc, k > j’ containing the above again. We note however, that xj+l remains a marked vertex as long as the above Vj* has not been eshausted by Step 6.

v*

+

The working of the above algorithm (and its slowness) becomes even clearer if Go = D is an eulerian triangulation of the plane. For, in this

An A-Trail Algorithm For Arbitrary Plane Eulerian Graphs

VI.167

case the algorithm amounts to constructing connected graphs D, (induced by the vertices to which one applies a 1-splitting) and D, (induced by the marked vertices) such that

c

c

and for 6 = 1,2, D6 has a cycle only if = bd(I"6) where F6 is a &face (see Corollaries VI.68 and VI.69 and the subsequent discussion). And at every stage of the algorithm, the graph induced by the vertices to which one applies a 1-splitting, has this property. We leave it as an exercise to translate this algorithm into an algorithm for finding hamiltonian cycles in plane cubic bipartite graphs (it's the obvious algorithm !). However, in view of the equivalence between the Conjectures VI.73 and VI.74 as expressed by Lemma VI.75 one can modify the above algorithm in order for it to decide whether a given plane eulerian graph Go has a non-separating A-trail (Exercise VI.34.b)). If Go has such an A-trail, the outcome of this modified algorithm is a connected outerplane eulerian graph G i(for some i > 0) whose blocks are the boundaries of the 2faces F2 of Go and such that bd(F2) has more edges than it contains cut vertices of Gi.Restricting ourselves to eulerian triangulations Do of the plane this means that each block of Dicontains at most two cut vertices o f D i (*). Statement (*) is trivially true for i = 0, and for i = 1 as well; hence one is tempted to suspect that (*) can be preserved in all the graphs Do,D,, . .. ,Diof the modified A-trail algorithm. This "suspicion" leads us to the following more general conjecture where we consider simple eulerian triangulations of an n-gon in the plane. In this 0 (mod 3) must hold; and for every such n, triangulations of case, TI this type do exist (see e.g. [FLEI74b] and note that Lemma VI.76 is a consequence of this fact).

Conjecture VI.92. Let a positive integer n G O(mod3) be chosen, and let Do be a simple eulerian triangulation of an n-gon in the plane together with a 2-face-coloring of Do such that the outer face Fom of Do is a 1-face. Then (distinct) vertices xk E V(D,) - V2(D0)exist such that D, := (Dk-l)~zkl,l, k = 1,.. .,i, has the following properties: a)

xk

E V(bd(F,"_,))- V2(D,-1) (F,"_, is the outer face of D,-l);

b) every block of D, contains at most two cut vertices of D,; c) if k = i, then every block of D, is a (triangular) boundary of a 2-face.


VI. 168

Thus, because of property b) Conjecture VI.92 is a generalization of Conjecture VI. 73.

VI.3.4. Final Remarks on Non-Intersecting Eulerian Trails and A-Trails, and another Problem As we have noticed at the beginning of this section) non-intersecting eulerian trails and A-trails are in principle different concepts; they coincide however, for G with A(G) 5 4. This basic difference between these two types of eulerian trails can also be demonstrated topologically. Recall that an eulerian trail T of the connected eulerian graph can be determined by a sequence of splitting operations on vertices of valeiicy exceeding 2 such that the graph obtained after the last splitting operation in that sequence, is a cycle (see also Corollary V.13). On the other hand, a connected eulerian graph G on q edges can be viewed as obtained from a q-gon Cq by a homomorphism which acts bijectively on the edge sets of G and Cq (see Remark V.12). We apply these considerations to an arbitrary connected) plane, eulerian graph G and a non-intersecting eulerian trail T of G. Denote by C, the plane cycle corresponding to T. Moreover, similar to the concept of a &splitting, 5 = 1,2, we consider, in the construction of C, from G, the k-fold application of the splitting procedure to obtain k 2-valent vertices vl, . . . z'k from the 2k-valent vertex v, k > 1, as a simultaneous procedure to replace v with these k 2-valent vertices. Note that this replacement procedure can be done in such a way that vl,. . . vlclie in a (geometrical) circle K ( v ,E ) with center v and radius E , and x ( v , E ) contains no other vertices of G, C, respectively) and E ( v , E ) n ( E ( G )- E,) = 0. Note that C, is a simple closed curve since T is non-intersecting. Thus G can be viewed as obtained from a regular q-gon in the plane Cq( q =I E(G)I) as follows: 1L cq CT 4 G )

)

where cp is a homotopic deformation and

$ is a continuous mapping of the plane

&2

onto itself

VI.3.4. Final Remarks

VI. 169

such that

$(x) = v for x E r ( v , E ) and v E V ( G )- V2(G), $ : €2 -

u

K(v, E)

--+

€2

- V ( G )is bijective.

E V(G )-Vz ( G )

Since an A-trail is a special type of non-intersecting eulerian trail, the above is true, of course, if 5" is an A-trail. However, if T is an A-trail, we can assume CT as having the additional property that for every v E V ( G )- V2(G)

there exists 6 < E such that

r ( w , 6)n CT = {q,. . .,vk}.

(A41

Using planarity we can replace (A) with

precisely one of K ( v , E ) n int CT and K ( v ,E ) n e x t C, is a region. (A')

If we consider a 2-face-coloring of G (with F" being a l-face as usual) and the induced coloring of CT,K ( v ,E ) n int CT is a region if and only if T induces a 2-splitting in v. In the case of a non-intersecting eulerian trail which is not an A-trail, then for at least one v E V ( G )- V2(G),neither of these two sets defined in (A') is connected.

So we have a way of topologically constructing all connected, plane, eulerian graphs (compare this with [ABRH79a, Theorem 31) as well as those graphs among them which admit an A-trail. The A-trail problem can thus be re-interpreted as the problem of describing various classes of these graphs in terms of graph theoretical parameters. We observe that the above considerations can, in principle, be extended to arbitrary graphs embedded in some (orientable or non-orientable) surface F ,except where 2-face-colorings are involved. In particular, property (A) remains valid as the topological means for distinguishing A-trails from the other non-intersecting eulerian trails. Note that the concept of an A-trail was originally introduced via O+(v), O-(v) respectively, and did not require the concept of 6-splittings. However, for different nonintersecting eulerian trails TIand T2 of the same G , the cycles C, and CTz may not be homotopic.

As for A-trails in connected, plane, eulerian digraphs D and/or mixed graphs M it makes sense to consider this problem only if D is canonically oriented, and/or if M can be canonically oriented (it can easily be

VI. 170


decided, whether M admits such orientation; for, a canonical orientation is uniquely determined by just one given arc). To consider the A-trail problem in a canonically oriented digraph D is, however, tantamount to considering the problem in the underlying graph G. For, if G has an Atrail T , either T or T-l corresponds to an A-trail in D; and an A-trail of D corresponds to an A-trail of G. This follows directly from the relation between an A-trail and 6-splittings, 6 = 1,2. However, it might be of interest to study (in the contest of Theorem VI.33) those in-trees of D which correspond to A-trails of D. Another problem (which has not been treated explicitly in this chapter) considers arbitrary connected eulerian graphs G drawn in the plane in such a way that open edges do not contain points corresponding to vertices, two open edges intersect in at most one point at which they must cross each other, and no three open edges intersect in the same point. Such a drawing of G defines a certain O+(G) from which one deduces a special system of transitions X , by defining t E X,(v) f-t t = {ei, e i + i } , 1 5 i 5 k = k, = $d(v), and v E V(G) is arbitrarily chosen. However, X , = X , for some unique trail decomposition S of G. Whence the following questions arise: 1) For a given graph G, does there exist a drawing of G in the plane such that X , = X , for some eulerian trail T of G ? 2 ) How many drawings of G exist such that some eulerian trail T of G satisfies X , = X , (where X , relates to a particular drawing) ? 3) Which trail decompositions S of G, satisfying X , = X , can be obtained this way ? These questions have been raised in [HARB89a] where question 1) is answered in the affirmative for G = Ii2n+l, n EN.

VI.4. Exercises Exercise VI.1. Prove Corollary VI.2, part 2), by applying the Splitting Lemma (i.e., without using Theorem VI.l). Exercise VI.2. Construct a graph G and define a partition system P(G) satisfying Corollary VI.4.2) such that G has no P(G)-compatible path/cycle decomposition S in which the number of path elements of S equals half the number of odd vertices of G. Exercise VI.3. Prove Corollaries VI.5 and VI.6 by using the Splitting Lemma only.

VI.4. Exercises

VI.171

Exercise VI.4. Prove Lemma VI.7 and Lemma VI.8 by using the Splitting Lemma. Exercise VI.5. Prove Corollaries VI.9 and VI.10. Exercise VI.6. Show that the digraph of Figure VI.2 has no A*compatible eulerian trail. Exercise VI.7. Prove Corollary VI.13 (hint: for v with d,(v) > 6, shorn first that if I X * ( v )I= i d D ( v ) + l , then there exists CiE X * ( v ) ,i = 1,2,3, with I CiI= 2 and (At)+ 3 Ci (A:)-. Then apply the Splitting Lemma in a manner similar to the proof of Theorem VI. 11). Exercise VIA. Show that D, of Figure VI.4. has an X(D,)-compatible eulerian trail if and only if D has an X(D)-compatible eulerian trail, where X ( D ) is given and X ( D , ) = X ( D ) U X ( s , ) U X ( s z ) . Exercise VI.9. a) Construct connected eulerian digraphs D with the property that V ( D ) consists of 4-valent vertices and an odd number of 6-valent vertices. b) Taking two graphs of the type described in a), construct from it an eulerian digraph with an even number of arcs and a 4-valent cut vertex. Exercise VI.10. Prove Corollary VI.19. Exercise V I . l l . Prove Lemmas VI.27 and VI.28. Determine those graphs G whose valencies are multiples of four such that these two lemmas are also true for such G. Exercise VI.12. Show that the subdigraphs D, and Di = D* - DA of the digraph D* of Figure VI.7 satisfy statement 4)a) of Theorem VI.35, while statement 4)b) holds for Di only. Exercise VI.13. Prove Corollary VI.43. Exercise VI.14. Construct a 4-regular simple digraph which has two but not three pairwise compatible eulerian trails (hint: proceed similar to the discussion centering around Figure VI.14). Exercise VI.15. Prove Lemma VI.58. Exercise VI.16. Prove: if G satisfies the hypothesis of Lemma VI.58, then there exists a connected, outerplane, simple graph H having a subgraph H‘ homeomorphic to G, such that H has no A-trail. If one drops the property “simple”, then H can be constructed in such a way that H’ spans H .

VI.172


Exercise VI.17. Construct a 2-connected, outerplane, eulerian graph having no A-trail. Exercise VI.18. a) Construct an A-trail in the graph GT of Figure VI.28. b) Show that GT has no A-trail inducing a 1-splitting in v1 if the outer face of GT is a 1-face in the 2-face-coloring of GT. c ) Find other plane embeddings of the abstract graph underlying Go and GT, which admit A-trails. Exercise VI.19. Prove Lemma VI.65 and show by example that the converse of this Lemma is not true. Exercise VI.20. Show that the plane graph Go of Figure VI.26 (which has no A-trail) does have a vertex partition {Vl, V,} such that (V,) is acyclic, S = 1,2. Exercise VI.21. Prove that the graph of Figure VI.30 (a) is the smallest planar, 2-connected, cubic, bipartite non-hamiltonian graph, and that it is uniquely determined. Exercise VI.22. Prove: the following statements are equivalent. 1) Every plane, 3-connected, cubic, bipartite graph has a hamiltonian cycle (BTC). 2) Every plane, 3-connected, cubic, bipartite graph has a dominating cycle. Exercise VI.23. a) Prove: if G, (D) is a plane, bipartite, 3-connected, cubic graph (simple eulerian triangulation of the plane) whose face boundaries are 4-, 6-, or 8-gonal (whose vertices are 4-, 6-, or 8-valent), then G,(D) has precisely six 4-gonal face boundaries (Cvalent vertices) if it has no 8-gonal face boundary (8-valent vertex); it has precisely seven 4-gonal face boundaries (4-valent vertices) if it has precisely one 8-gonal face boundary (8-valent vertex). b) Construct an eulerian triangulation of the plane with 4-, 6-, and 8valent vertices only such that it has precisely one 8-valent vertex v; and v is not adjacent to any 4-valent vertex. Exercise VI.24. a) Prove Corollary VI.81. b) Starting from the octahedron, construct an eulerian triangulation of the plane D by a sequence of weak W4-extensions such that D has precisely one pair of adjacent 4-valent vertices. Exercise VI.25. Translate Conjecture VI.82 into the theory of cubic graphs to obtain a stronger (but equivalent) version of Conjecture VI.72.

VI.4. Exercises

VI.173

Exercise VI.26. Prove Lemma VI.84. Exercise V1.27. Prove: if G, is a connected 3-regular graph having a 2-factor Q4 consisting of 4-gons only, then G, has a hamiltonian cycle H 3 E(G,)- Q4 (hint: use Corollary VI.6 and suppress those 4-gons in

Q4 which either have diagonals or which contain an edge of multiplicity 2).

Exercise VI.28. Show that the graph D of Figure VI.3S has an A-trail. Exercise VI.29. Prove Theorem VI.90. Exercise VI.30. Show that it suffices to prove Conjecture VI.73 for the corresponding 4-connected graphs (hint: apply Proposition VI.83). Exercise VI.31. a) Concerning finding an A-trail in a 2-connected outerplane simple graph, work out the details of the proposed algorithm (in particular, show that properties (*) and (* * *) are valid). b) Check in detail the complexity considerations on this algorithm (how large is the exponent of Pi(,) for i = 0,1,2 ?).

Exercise VI.32. Show: a) If G is a(n) (eulerian) triangulation of the plane and e,f E E ( G ) are adjacent in a face boundary of G, then there is at most one hamiltonian cycle H of G having the A-property with e, f E E ( H ) ;and determining whether this H exists or not, can be done in linear time (note that G is already embedded in the plane). b) Using a) show that the problem of deciding the existence of a hamiltonian cycle with the A-property in a(.) (eulerian) triangulation of the plane, lies in P. Exercise VI.33. Study the complexity of finding an A-trail in D , where D is an eulerian triangulation of the plane as described in Theorems VI.77.a and VI.79. Exercise VI.34. a) Translate the A-trail algorithm for arbitrary plane eulerian graphs into an algorithm for finding a hamiltonian cycle, a Tutte cycle respectively, with the properties expressed in Theorem VI.90, in a plane cubic bipartite graph (hint: use Lemma VI.75, Theorem VI.90 and their respective proofs). b) Modify this A-trail algorithm in such a way that it decides whether a given connected plane eulerian graph has a non-separating A-trail.


VII. 1

Chapter VII

TRANSFORMATIONS OF EULERIAN TRAILS

In the preceding chapter, we have studied various types of eulerian trails in arbitrary eulerian graphs and digraphs, as well as in certain classes of eulerian graphs. We shall consider all eulerian trails in a fixed eulerian graph G. Is there a way of transforming an eulerian trail Tlof G into another eulerian trail T2 of G ? For this purpose, we must first explain under which conditions TI and T2 are considered to differ.

Definition VII.l. Two eulerian trails Tland T2 of the connected eulerian (di)graph G are considered to differ precisely if X , # X,. In other words, for an eulerian trail T of G, should the eulerian trail T' arise from T by reversing or cyclically rotating the edge sequence representing T , T and T' are considered to be equal. Thus, Definition VII.l is in line with the approach used by most authors to date to distinguish between eulerian trails. Moreover, this definition is suggested by Corollary v.10.

Of course, in the case of a digraph D,the reversal of the arc sequence representing an eulerian trail T' of D , to obtain an eulerian trail T' of D is out of the question, since traversing an arc (v,w) from w to v is not permitted in an eulerian trail. However, there are other ways of considering two eulerian trails as different which should be mentioned at this juncture. For example, W.T. Tutte considers T and T' to be different if the edge sequences (and/or arc sequences) representing T, T' respectively are different. More precisely, T # T' if T and T' either start at different vertices or if for some i E (1,. . . ,q } , the i-th edge (arc) traversed by T differs from the i-th edge (arc) traversed by T'. Comparing this view with the one expressed by Definition VII.l, it follows that this definition implies an equivalence relation on the set of eulerian trails of a (di)graph; each equivalence class

vII.2

VII. Transformations of Eulerian Trails

contains precisely 2q eulerian trails which define the same system of transitions and which are considered different in the sense of Tutte. For the purposes of this book, however, Tutte’s distinction between eulerian trails of a graph has no essential impact other than the above factor 2q. However, in the case of digraphs, Tutte’s distinction is of indirect relevance. There is yet another way of distinguishing between eulerian trails of a (di)graph. Suppose the edges (arcs) of a (di)graph H are arbitrarily labeled beforehand by f l , . . . , fq. An eulerian trail T of H , expressed as a sequence of edges (arcs),

T = f;.(1)7

f?r(2),

* *

.,f+)

,

then defines a permutation z of the elements of {1,2,. .. ,q } . Eulerian trails Tland T2 of H are then considered to differ if and only if the corresponding permutations 7il and 7r2 differ. This distinction between eulerian trails (which is in line with Tutte’s point of view) will play an essential role in securing a parity result on eulerian trails in digraphs which in turn yields an interesting application in linear algebra.

VII.l. Transforming Arbitrary Eulerian Trails in Graphs Let G be a connected eulerian graph other than a cycle, and consider in G a system of transitions X , defined by an eulerian trail T . Let t1,t2 E X ( v ) C X , for TJ E V ( G ) - V,(G) # 8 be arbitrarily chosen, f;}. Let H be the connected eulerian and denote t , = {fi,gi}, t , = graph obtained from G by splitting away those pairs of edges which correspond to the elements of X, - {tl,t2}.Consequently, d H ( u ) = 4 and V,(H) = V ( H )- {v}. Moreover, T corresponds t o an eulerian trail Tl of H with t,, t, E X T l . It now follows by the very structure of H that f;, f.& f{ cannot be half-edges within the same block of H . The Splitting or is connected. AssumLemma implies, therefore, that either ing w.1.o.g. that H1,, is connected, we have V2(H1,2)= V(Hl,,); i.e., is a cycle. Thus, a run through describes an eulerian trail T. of H . By construction and Definition VII.1, TI# T,.Moreover, T2 corresponds to an eulerian trail T’ # T of G whose corresponding system of transitions X‘ := X,, satisfies the equation

{fi,

VII. 1. Transforming Arbitrary Eulerian Trails in Graphs

VII.3

Summarizing these considerations, we arrive at the following result (see also [KOTZ68c, Theorem 11 and [ABFtH80a, Lemma 11). L e m m a VII.2. Let T be an eulerian trail of the connected eulerian graph G , where V ( G )- V,(G) # 0 , and let t,, t , be two different transitions of T at a vertex of valency exceeding 2. Then there exists an eulerian trail of G whose transitions coincide with those of T except for t , and t,. Re-interpreting the above equivalently in terms of representations of eulerian trails by edge sequences, we can describe the eulerian trails T and T' as follows = e l , ... 2

T' = el

7

---

7

Sm,n, en+l, * ' - 7 eq -1 em-, S,,n, en+l, - - 9 eq 7

(2)

*

where E ( G ) = { e l , . . . , e q } , Sm,n is a closed trail starting with em and ending with e , at the same vertex 2r E V ( G )- V,(G), and ern-, f,, em = g,, en = f2, en+, = f3. Hence (1) and (2) describe the same situation. In accordance with [ABRH80a, SKIL83aI (see also [KOTZ68c] which, historically, is the starting point for this chapter), we are led to the following. We retain the notation introduced hitherto.

Definition VII.3. Let T and T' be two eulerian trails of the connected eulerian graph G , and let 21 E V ( G )- V,(G) # 0. If the corresponding systems of transitions X,,X' respectively, satisfy (I), we can say that X' (TI)has been obtained from X, (T) by a 6-transformation which in itself is given by transforming t , and t, into ti and t;. We can also say, by analogy, that TI has been obtained from T by the reversal of the segment S ,, (or segment reversal for short), which starts and ends at the same vertex 21 (see (2)). We note that the inverse operation of a rc-transformation is also a 6transformation. This follows from (l),but also from (2) by observing = Sm,n. If we extend the validity of (1) for the case that (S;$)-' { t l , t 2 } = {ti,tb} (i.e., T' = T), then Definition VII.3 permits one to consider the identical transformation (i.e., X, = X ' ) a 6-transformation. For the sake of brevity we can write K(T)= T', if TI is obtained from T by a r;-transformation.') Since this formal equation does not specify by which 6-transformation TI has been obtained from T, we have 6(T) = T',n(T)= T" and T' # T" for different 6-transformations.

VII.4


Definition VII.4. Let I ( G ) denote the set of all eulerian trails of the connected eulerian graph G. For T,T' E I ( G ) we call T and T' +associates,') if To,.. .,T, E I ( G ) exist for some n E N so that and Ti= K ( T ~ - , )for1 5 i 5 n

T = To,T' = T,,

.

Thus, if T,T',T'' E I ( G ) are such that K ( T )= T' and such that T' and T" are K-associates, then T and T" are K-associates. From Definition VII.4 and the discussion immediately preceding it, we can now conclude that K-association defines an equivalence relation on I(G ) . That is, I ( G ) has a partition P(G, K ) whose classes are the equivalence classes under this relation. This leads to the question of the number of classes that P ( G ,K ) has. The answer to this question is given by the nest theorem. Theorem VII.5. IP(G,rc)

I=

1 for every connected eulerian graph G.

Proof. If G is a cycle, then I I ( G )I= 1 by Definition VII.1. The validity of the theorem follows in this case ( I ( G )= { T } )since T = K ( T )with the K-transformation being the identical transformation, whence we assume V ( G )- V,(G) # 8, so I'T(G)I> 1 by Lemma VII.2. Let us assume that we have Tl,T2 E I(G)which are not K-associates. Define X , := X,, X , := XT2 and choose T, and T . in such a way that I X , n X , I is as large as possible. Next consider 21 E V ( G )- V2(G)with X , ( v ) # X,(v), where Xi(.) E X i , i = 1,2, is the set of transitions at v induced by Ti. Such 2) must exist because of T,# T2. Since d(v) = 2k > 2, it follows from the choice of 21 that z ~ , ~ :=IXl(v) ( ~ I ) n X 2 ( v )I< k - 2 (note that z,,~(TJ) = k - 1 implies z,,,(v) = k which is impossible by the choice of 21). Consequently, there are r := k - z , , ~ ( u ) 2 2 transitions of X,(v)which do not belong to X,, and vice versa. We may denote these transitions in such a way that

{e:,fiIY 7 N . 7 f;} E X,(V) - X.(v), { 4 7 fil17-- - fiJ E X , ( 4 - X,(v>, * *

where

{i, ,... ,iT} = (1, ..., t} and ij ')

#j

for

1 5j 5 r .

Note that rc-associates are the same as associates in [SI) z , , ~ ( v )and therefore, IX,nX2 I>lX, nX2I. By the choice of T, and T', it follows that T3 and T' are K-associates. This and K(T,)= T3 imply that T, and T2 are K-associates, thus contradicting the choice of T, and T2,whence we can conclude that X, satisfies the equation (3). Now we apply Lemma VII.2 for T = T, and t , = t2+,s = 1,2. Using a symmetric argument we can conclude that there is

and X4 must satisfy the equation

'1 Note that these 2r transitions correspond to a 2-factor Q in K,, where fij/j = 1,. . . ,r} ,E(IC2,) consists of all conceivable tran-

V ( K 2 , ) = {e;,

sitions, and where Q consists only of even cycles because of the definition of these 2r transitions (see also the discussion after Conjecture VI.36 and before Theorem VI.37). Hence, we could have been even more specific concerning the choice of the indices ij, j = 1,. . ,r. This we shall do later on.

.

VII.6


Comparing (3) and (4)we can confirm the validity of the equation

S is the Kronecker delta. By the choice of TI and T z ,it follows where , that T' and T4 are rc-associates. This and the formal equations &(T,)= T', = T4 classify, however, Tl and T2 as being &-associates. This final contradiction implies that the theorem is valid. Note that the above proof differs from those of [ABRH80a, Theorem] and [SKIL83a, Theorem 11. Knowing that any connected graph G with 2k > 0 vertices of odd degree has a decomposition S into k open trails such that an odd vertex of G appears as an end-vertex of precisely one trail in S (see Corollary V.2), one is tempted to generalize Theorem VII.5 in the following way. Join a new vertex vo # V(G)to each of the odd vertices of G with precisely one edge in order to obtain a connected eulerian graph H . Each trail decomposition S of G as described above can then be extended to an eulerian trail of H . If we denote S = {T(ui,vi)/i = 1,.. . ,k}, where T ( u i , v i ) stands for the element of S joining the odd vertices ui,zti E

V ( G )7

is an eulerian trail of H . In fact, if k > 1, the same decomposition S of E ( G ) defines more than one such trail T H .Moreover, every eulerian trail TH of H defines a decomposition S of E(G)into open trails. Hence by Theorem VII.5, any two such decompositions S,, S, of E ( G ) can be transformed into each other by following a sequence of &-transformations applied to the respective eulerian trails T I ,T2 of H . Interpreting these 6transformations in terms of segment reversals and noting that a segment may start and end at vo $! V ( G ) ,it immediately follows that in general, we cannot conclude that S, and S2 can be transformed into each other by a sequence of segment reversals (respectively rc-transformations) performed only in G. Figure VII.l demonstrates this fact. Of course, we can extend the concept of segment reversal (respectively rc-transformation) to arbitrary graphs G by restricting ourselves in H to segments that contain no element of E,,, (and/or by restricting rc-transformations t o the elements of XS1- XS1(vo)).

VII.1. Transforming Arbitrary Eulerian Trails in Graphs

U

U

VII.7

V

2

2

V

I

I

Figure VII.l. The decomposition S, of E ( G ) into two , where the elements of X,, open trails T ( u , , v l ) , T ( u 2 v2), are described by the small arcs, cannot be transformed by segment reversals into a path decomposition S, := P ( u 1 ,u2),9(v1,21211 of W ) . However, if k = 1 (i.e., if G contains an open covering trail), any two such trails can be transformed into each other by a sequence of segment reversals. To see this form firstly H from G as above, and denote { u , v} = N H ( v o ) .Precisely, because vo is 2-valent in H one can define a correspondence TH E 7 ( H ) T i E %(G) ---f

where %(G)is the set of all open covering trails of G joining the odd vertices u, v E V ( G ) ,and where TH is defined by the edge sequence

T H = vou,T$,vv0

.

By Definition VII.l it is no loss of generality if the elements of 7 ( H ) are written in such a way that they start and end in vo. Hence, the above relation between THand TZ classifies the above correspondence as a bijection between 7 ( H ) and 70(G). Secondly, consider T,T' E I ( H ) such that K ( T )= T'. Assume G is a connected graph other than a path, hence we may consider

z E V ( H )- V2(H)= V ( G )- (V,(G) U v l ( G ) )# 8

.

VII.8


Now we can write T as an edge sequence of the form

where Sl,m--l and S,,, are segments starting and ending at z; hence T starts and ends at 2. Moreover, m = 2 (rn = q ) , if and only if the first (last) edge of T is a loop. Next suppose T # T'. Since K ( T )= T', T' is being obtained from T by the reversal of a segment S,,, starting and ending at y E V ( H )- V,(H). Because of (*) and Definition VII.1 we may suppose w.1.o.g. that y = z, r = 1, s = m - 1. Hence

(**) Following the notation of (1) and (2), we then have

However, for

,s;;, transitions X" satisfies (1)

T" = Sl,,-l

the corresponding system of as well, where t , ,t,, ti,t&are the transitions described in (* * *). Hence,

X" = X'

,

i.e.,

TI' = TI .

That is, T' can be obtained from T by reversing the segment Sm,q instead of Sl,m-l. In other words, if K ( T )= T' and T # T I , then T' can be obtained f r o m T by reversing a segment not containing a prescribed edge (see also [SKIL83a, Lemma 11). Generalizing this for any two K associates T,T'E 7 ( H ) we can therefore say that T' can be obtained f r o m T b y a sequence of segment reversals, such that none of the reversed segments contains any element of E ( H )-E(G). This and Theorem VII.5 are equivalent, however, to the following result ([SI 2 odd vertices) and in the light of the discussion following Theorem VII.5, it is not clear how to handle the classification problem concerning S(G), where S(G) is the set of all decompositions S of E(G) i n t o k open trails. Should one simply adhere to the concept of segment reversal, as has been done for the cases k = 0 (Theorem 141.5) and k = 1 (Corollary VII.6), or should one use certain segments of the eulerian trails of H as well for segment reversals ? The discussion thus far suggests that one should adhere to segment reversals as in the cases k = 0 , l . The question we are then faced with is (see also the discussion of Figure VII.1): if G is a connected graph with precisely 21; > 2 odd vertices, how large can I P(G, K ) I be ? Here, P( G ,rc) denotes the partition of S(G) induced by rc-association which remains an equivalence relation, provided one considers an open trail T ( u , v )and its inverse T ( u , v ) - l to be the scme object (see Definition VII.l). Is I P(G,rc) I a function f ( k ) depending on k only ? Maybe f ( k ) = k ? In my view, these are questions worth answering; it would seem that nobody has undertaken that task.

VII.2. Transforming Eulerian Trails of a Special Type Instead of generalizations, let us now turn to specialization. We should consider in a connected eulerian graph the set I*of all eulerian trails of a special type and ask whether any two of them are K-associates within I*. More precisely, if T ,TI E 7*are chosen arbitrarily, does a sequence of objects Tl, . . . ,T, E I*arise such that TI = T , T, = T' and ~(2'~)= Ti+1for i = 1,.. . ,n - 1 ? Turning first to the concept of compatibility, we are soon faced with some disappointment. If G is a connected eulerian graph together with a system of transitions X and eulerian trails T , TI compatible with X, it is generally not true that G contains eulerian trails Tl, . . . ,T, compatible with X and satisfying

T = T I ,TI = T,,

and

Ti+l = ~(2';) for 1 5 i 5 n - 1 .

This can be seen from Figure VII.2, where we consider 1

Then

x, := (X' - { t i ,ti})lj {t,, t,}

,

(1")

v11.12


where t,, t , E X,(v), defines an eulerian trail T of G. This follows from (*) and the discussion following (l‘),which in turn classify the transformation of X , into X‘, and/or of X’ into X T , as expressed by (1’) and (l”),as being inverse operations to one another. As a summary of this discussion we introduce the following terminology.

Definition VII.7. Let G be a connected eulerian graph with V ( G )V,(G) # 0 , let T E I ( G ) and let S’ be a trail decomposition of E ( G ) with IS‘I= 2; denote by X T and X’ the respective systems of transitions. If for some v E V ( G )- V,(G) there exist t,,t, E X,(v), ti,ti E X ’ ( v ) which satisfy equation (l’),we can say that X‘ has been obtained from X, by a K-detachment; it can be formally described as S = d ( T ) . If the same objects X,, S’, ti,t:, i = 1,2, satisfy equation (l”),we can say that X , has been obtained from X’ by a K-absorption, which is T‘ E I ( G ) is said to be expressed by the formal equation T = ~’’(s’). obtained from T by a rc*-transformationif a trail decomposition S‘ of E ( G ) exists such that S‘ = ~’(2“)and T’ = K”(S’). We then adapt the cursory description K* = K”K‘ and T’ = K * ( T (which ) is justified by the equation T‘ = K”(K’(T))). Remark. Note that in Definition VII.7 I S‘ I= 2 by the very definition of a K-detachment. Moreover, applying a K*-transformation means that the corresponding K-absorption may or may not be applied at the same vertex where the corresponding +detachment has been applied. We note, however, that in Chapter X (see Volume 2) we shall use the terms Kdetachment and K-absorption in the more general sense which is induced by Definition VII.7 (just delete in this definition the requirement I S’ I= 2 and view T as another trail decomposition). One can define K*-associationin the same way as &-associationhas been defined (simply replacing K with K* in Definition VII.4). We leave it to the reader to show (Exercise VII.2) that n*-association also defines an equivalence relation on ‘T(G). However, for the next result we need to combine tions.

K-

and K*-transforma-

Definition VII.8. For T,T’ E I ( G ) , where G is a connected eulerian graph, we may say that T’ can be obtained from T by a K ~ transformation, in short T‘ = K ~ ( Tif)either , T‘ = K ( T )or T‘ = K*(T). We define Kc,-associationin the same way as Ic-association has been defined (see Definition VII.4), and leave it to the reader to show that K , -

VII.2. Transforming Eulerian Trails of a Special Type

VII.13

association also defines an equivalence relation on 7 ( G ) . In the case under discussion, however, we are interested in transforming eulerian trails compatible with a given system of transitions X in such a way that one does not use any other eulerian trail in such transformations. So we are ultimately led to the following terminology.

Definition VII.9. Let 7 ( G , X ) denote the set of all eulerian trails (of the connected eulerian graph G) compatible with a given system of transitions X = X ( G ) . Furthermore, denote by S2(G) the set of all trail decompositions S of G with I S I= 2, and by S2(G,X ) all those elements of S2(G) which are compatible with X. Call eulerian trails T and T‘ Kx-associates if To,. . .,T, E 7 ( G ,X ) exist such that for 1 5 i 5 n , 1) T = To, T‘ = T, and Ti= I E , ( T ~ -and ~ ) , 2) if Ti= K ~ ( T ~ =- ~ ) I E * ( T ~for - ~some ) i, then

Similarly, nX-association defines an equivalence relation on 7 ( G ,X) (see Exercise VII.2). This induces a partition P ( G , K ~ of) 7 ( G , X ) whose elements are classes of nx-associates. The next theorem can be viewed as the analogue to Theorem V11.5 for eulerian trails compatible with X .

Theorem VII.10. I P ( G , K ~ I=) 1 for every connected eulerian graph G and every prescribed system of transitions X of G. Proof. If G is a cycle, the theorem follows by using the same arguments as in the proof of Theorem VII.5 (note that X = 0 in this case), whence we have V(G) - V2(G) # 0. The case 7 ( G , X ) = { T }is trivial because it implies IE,(T) = K(T) = T,where IE denotes the identical transformation in this case. Hence, assume I I ( G , X ) I> 1 and suppose Tl,T2E I ( G , X ) exist which are not Icx-associates. We proceed along the lines of the proof of Theorem VII.5, retaining the notation used there. Again we choose TIand T2 in such a way that X , n X , is as large as possible for their respective systems of transitions X , and X,. Subject to this condition, choose G with minimal E(G) 1; V,(G) = 0 follows. In the sequel, if e is a loop, its two half-edges will be distinguished by different subscripts rather than el,”’ (el refers to the fact that we deal with a half-edge).

I

I

I

Again, Tl # T2 must hold (the identical transformation is a IE,transformation), so v E V(G) exists with d(v) = 2k and x,,,(v) := p , ( v ) n X2(v)I 5 k - 2. Moreover, for T := k - z , , ~ ( v )we again have the

VII.14


relations

and because of the footnote in the proof of Theorem VII.5 we may assume w.1.o.g. that { e ; ,fi-l} E X2(v)- X,(v), i = 1 , . . . ,m, putting fh = f&. Consequently, n = 2 (see the proof of Theorem VII.5 concerning the index n ) , whence we have

We consider four systems of transitions (see - in the proof of Theorem V11.5 - (3), (4) and the equations immediately preceding them):

We observe that for i E {3,4), these definitions imply

X l = XTi if and only if Xr = Xsi Xr = XTi if and only if Xl = XSi

(**I

where TiE 7 ( G )and SjE S2(G); moreover,

Ti# 7 ( G ,X ) implies SiE S2(G,X)

{ e ; , f;)E X

.

Xi = X, for T3 E 7 ( G ) {eh, f;} E X if Xi = XT4 for T4E 7 ( G ) .

(* * *)

if

(5)


VII.15

Otherwise, one obtains a contradiction as in the proof of Theorem VII.5. Consequently, by (5) and (**), {e;,e;} E Xonly i f X r = X , for TiE I ( G ) , i = 3,4.

(6)

Also, {ei, ek} E Xfl, i = 3,4, and the choice of T,, T2E 7 ( G 7X ) yield:

X y = X,,

i = 3,4,

implies {T3,T4} I ( G , X )

.

(7)

Otherwise, one proceeds as in the proof of Theorem V11.5 to obtain a contradiction. Now we can consider three main cases. Case 1. Suppose X i = X,, X y = X T j for Tj,Tj E 7 ( G ) . By (**), i # j , hence { i , j } = {3,4}; w.1.o.g. i = 3 and j = 4. By (5)

and (6), {ei,f;} E X implying rn > 2, and {ei,ek},{fi,f;} 6 X . Since T3 $ 7 ( G ,X ) , it follows from (***) that S, E S,(G, X ) where Xs3 = X;. Depending on whether T4 E ‘ T ( G , X ) or not, we may distinguish two cases.

Subcase 1.1. T4E 7 ( G , X ) , Denote S, = { S , , S f } such that el,e2 E E(S,) and therefore, f,,f, E E ( S f ) . Note first that X , ( v ) - {t,,,, t,,,} # 0 since m > 2. Next consider a run through T4starting at v along el. Since T4E 7 ( G ) and {ei ,e;} E XT4 = X:, one arrives at along e, only after having passed all elements of E(G). Consequently, since this is not the case with ) which a respect to S,, there must be a vertex w E V ( ( S , ) )n V ( ( S f ) at transition { h i , g ; } E X t satisfies h, E E(S,),g, E E ( S / ) . Hence, we find h, ,g2 E E, such that {hi,ha} E X , (w) in any case, and where (9;,g i } E X,(w) if w # v,whereas we denote g: = fi , i = 1,2, if w = v. Observe that {f;,f;> E X?. In any case, { { h ~ , h ~ ] , { g ~ , gnx, ~ } } nX, = 8 by construction and because of m > 2 and {ei, ek} E X < n X t . Moreover, for

{T,

s} = { 1,2},

{ { h : , g : } , { h ; , g b } } n x # 0 i~P~ies,{{h’l7gb},(h;,9~}}nX =0

*

So, one obtains the validity of this equation for some choice of T and s , where { T , s } = {1,2}. We can write S, for some j and k, where { j , k} = { 1,2}, in the form S, = ej,. . . ,h,, h,,

. . . ,e,

VII.16


(possibly e j = h, or ek = h, but {ej,ek} # {h,, h,}), and

(note that h, E E(S,) because of h, E E(S,) and {hi,hL} E X1(zu),and that, similarly, g2 E E ( S f ) ) . Hence we obtain

T. = e j , . . . ,h,,S,, h,, . . .,el, . By the very construction of

Tj,k it follows that Tj,kE I ( G , X ) .

Consequently, since K'(T,) = S, E S,(G,X) and ~ " ( s ,=) Tj,k E I ( G , X ) it follows that Tland Tj,k are KX-associates. Moreover, T7and T4are Kx-associates because of T4E I ( G , X ) and T4= K(T*) = q(T2). On the other hand, since { { h i ,hh}, {gi, g : } } n X , n X , = 0

The choice of T, and T,,and T4,Tj,k E I ( G , X ) imply that T4 and Tj,k are Kx-associates. So we can finally conclude that T, and T2are /cx-associates. This contradiction settles Subcase 1.1. Subcase 1.2. T4# I ( G , X ) . By (* * *), S4 E S , ( G , X ) , and Xs4 = X i by (**). That is, fk} E X since {ei, eh} # X , and thus {eh, fk} # X. Moreover, retaining the notation of Subcase 1.1 we define S, = {S,, S}, as above and

{fi,

In this case, however, S; assumes the role played by S, in Subcase 1.1, and S, assumes the role played there by T4. We can consider a run through S, starting at 21 along el; following the elements of X, and belonging to S, one finally arrives at 21 after having passed the elements of E(S,), with e2 being the last edge in this run. This run, re-interpreted a s a run through certain elements of E(S,*)and E ( S j ) starting with el E E(S,*)and ending with e2 E E ( S j ) , must therefore pass at some w E V ((Sj)) n V ((Sz)) from S,* over to S*. Hence, we may conclude in a manner similar to Subcase 1.1 that { h f, , g i } E X f ( w ) exists such that h, E E(S,*), g1 E E ( S j ) ; moreover there are h,,g, E E, such that { h i ,h:} E X,(w)in any case; and {gi, gi} E X,(w)for w # 21, whereas


VII.1'7

fh}

for w = v we denote gi = fk,91 = e; ({e;, E X i ) . Since we have in any case {hl, h2} C E(SH) and {gl,9,) C E ( S j ) ,we may conclude as in Subcase 1.1 that for

S,* = e l , . . . , h j ,h,, and

. . . ,fl , where

Si = g s , . .. , g r , where T;,k

= e l , . . ., hj,

{j,k} = {1,2}

{ s , r } = {1,2}

,

,

s;,h k , . . ., f i

satisfies T*k E 7 ( G ,X ) because { { h i ,gt}, { h i ,9:)) n X = 0 for fixed {j,Ic} = (1,2} and one of the two possible choices of {s,r } = { 1,2}. Note that S j can be expressed in the above manner for both choices of s and r . Since, by construction,

{h; 7

E X,(4

in any case (note that rn

- Xl(4

and

(9:7 9 3

> 2), while

{ e ; 7 f ; } E X Tj,k * nX,-X,nX2 and because X , n X ,

# XI (4

7

c X T . , we therefore conclude that 3*k

IXlnX,. a s k I>lxlnx,I

-

Consequently, by the choice of T', and T2, T, and TAk are tcx-associates, and by construction, T'lkand 57, are tcx-associates. Hence Tl and T, are tcX-associates. This contradiction settles Subcase 1.2 and therefore Case 1. We note, however, that in both Subcases 1.1 and 1.2, the special consideration of the case v = w for defining g i = fi,i = 1 , 2 , respectively g1 = f, and g2 = e2, is not really necessary. It shows, though, how in the case II = w one c m construct T'j,k.,Tj,krespectively, by changing a minimum number of elements of the original X,, X,. With Case 1 being settled we thus are left with the cases where X i = XTi, i = 3,4, or X r = XTi, i = 3,4.

Case 2. X i = X,, i = 3,4. By ( 5 ) , { { e i 7 f 4 } , { e $ , f & }C } X , so rn > 2; moreover, Xfl = XSi by (**), and Si E S 2 ( G , X ) ,i = 3 , 4 by (* * *). Retaining most of the notation previously used, we write

s, = { S e ,Sf}, s, = {S,*,s;>


VII.18

where

{e1,e2}

Subcase 2.1.

E E ( S e ) n E(S,*),{fly f 2 ) E {f17 fm} c E ( S j ) * E(S,) # E(S,*).This assumption plus the fact {el, e,} C E(Sf)7

E(S,) n E(S,*)implies that starting a run through S,* at v along el, one finds {h;,gi} E X2(w) for some w E V ( ( E ( S , * ) )such ) that h, E E(S,), g1 E E ( S f ) . As before we then also find h,, g2 E E, such that {hi,hh} EXl(w)inanycase, {9:,94} E X l ( w ) i f v # w ( s o 9 2 E E ( S f ) ) , and where we denote 9: = f:, i = 1,2, if v = w (see Subcase 1.1). We retain the notation used in the preceding considerations for the sake of a uniform approach and repeat them for a run through S, starting at v along e l : we find correspondingly {h;,gi} E X,(w) for some w E V ( ( E ( S , ) ) such ) that h, E E(S,*),g1 E E(S;), and we find h,,g., E E, such that { h i ,hb} E X2(w) in any case, and {gi, gi} E X2(w) if # w , whereas for 21 = w we define gi = g; = f;. Note the difference to Subcase 1.2 which is due to the fact that there X, = Xi.

fi,

As in Case 1we conclude that by construction

and that Tj,k, TlkE 7 ( G ,X) for the eulerian trails constructed in a way analogous to Subcase 1.1,1.2 respectively. We have

This and the choice of TI and T2 classifies Tj,k and T;k as KX-associates. Moreover, by construction Tland Tj,k are KX-associates, and so are T2 and T'Lk. Whence we conclude that Tland T2 are Kx-associates, a contradiction which settles Subcase 2.1.

Subcase 2.2. E(S,) = E(Sz). Consequently, E ( S f ) = E(S;). This and f, E E ( S j ) as well as {e;, E Xf imply em, f, E E ( S f ) = E ( S j )which, in turn, together with {e;, f&-l} E X: implies em,fm-l E E ( S j ) = E ( S f ) (observe that rn > 2), a.s.0. Consequently, since fl E E ( S f )= E ( S j ) we have E, - {e1,e2} E E ( S f )= E ( S j ) . We consider two cases.

fh}

> 3. Then

{eL,f;} E Xf, {eL,f;} E X:, and because of {ei, f;}, {ei, f;} E X and rn > 3 we have

(2.2.1). rn


VII.19

By the preceding arguments and because we can write S, = f i , . ..,e3, f3 , . . .,f j for some choice of

{i, j} = { 1,2}

S; = fk,. . . , e 3 , f 2 , .. . ,fi for some choice of

{k,Z} = {l,m},

and

it follows that both

T = fi,

-

,e3, S,,

f3,.

- - ,f j

and T*= fk,

- - - ,e3,S,*,

f2,.

- .,fi

(where both S, and S: are considered to start with el and end with e2) belong to 7 ( G ,X).By construction, {e$,

ei} u (X,n X,)= X, n X,.

which classifies T and T* as tcx-associates because of the choice of Tl and T,. This and the construction of T from T,,respectively T*from T', imply that TIand T2 are Kx-associates, again a contradiction.

(2.2.2) m = 3. Then X(v) 2 {{ei, f;}, {e;, f;}, {e;, f;}}.') Because of {e1,e2} E: E(S,) = E(SZ) there is a total of three possibilities for S, and Si to pass through the edges of E, - {e1,e2}. Depending on these possibilities we may construct T,T* E I(G, X). (a) S, = fl,. . ., f 3 , e 3 , .. . ,f2. Note that Si cannot have a segment S3,, = e,, . . . ,f2 because {ei, E XT,which implies that X y IS3,* C Xg. This and S, c S; would contradict the fact that S; is a closed trail. Anyway, with S, satisfying the above equation we conclude that TI is of the form

,

fi}

Observe that rn = 3 does not mean that d ( v ) = 6. This is immediately apparent from the footnote in the proof of Theorem VII.5, where 2m is the length of a cycle in the 2-factor Q of K2,.. Moreover, {e;, E X ( v ) follows f;}, {eh, f;} E X ( v ) , as well as from a combination of from m = 3 and Case 1 and (5) if we set tlg = {e;, f;}.

{ei,

fi}


VII.20

(b) Si = f,,. . .,f,, e 3 , .. . ,f3. We conclude in a manner similar to case (a) that S, cannot have a segment of the form e 3 , .. .,f3. Here T2 and T* E I ( G ,X) obtained from T2 by a n-transformation are of the form

( c ) Either S, = f l , . . . ,e3,f 3 , . . .,f2 or Then we have for Tl and T2 the equations

si = f,,- .. ,e3,f 2 , . . . ,f3.

This yields for T E I ( G ,X) obtained from T, by a n-transformation,

while one obtains T* E I ( G ,X)from S, = {S:,

Si} by a n-absorption,

In all cases (a),(b),(c), it follows that T and Tl, respectively T* and T,, are nx-associates, and that one always has

X,

n X, X, , X,n X, C_ X,,

respectively

.

On the other hand, if one compares any of the constructed T with any of the constructed T * ,one can conclude either

{fi, fi}

{ei, e ; } E X, n X,. , or E X , n X,. or T is as in case (a) andT* is as in case ( c ) .

,

Thus, for the first two instances we may conclude I X,nX,, />I X, nX, 1, classifying T and T* as nx-associates. T and T* are, however, xXassociates in the third instance as well: for in case ( c ) we have {eb, f;} E X,, which implies I X,nX,. ]>I 1, so T* and Tlare nx-associates. This and T, = n(T) finally proves that T and T* are nx-associates in any case. Consequently, since we have T = K ~ ( T ~T* ) ,= n1(T2) in all cases under consideration, it follows that TIand T2 are nx-associates.

xlnx,


VII.21

This contradiction finally settles the case (2.2.2) and thus Subcase 2.2. Hence, Case 2 has been settled.

Case 3. X r = X,, i = 3,4. By (7), Tj # ' T ( G , X ) if TkE I ( G , X ) , for any choice of { j , k} = {3,4}, hence w.1.o.g. j = 3, k = 4. Subcase 3.1. T4E I ( G , X ) . Then Xi = XsJ, where S3 E S , ( G , X ) by (* * *). Moreover, {el,, eb} # X, {fi ,fl} E X . Observe that m > 2 and T4= e l , . . . ,f,, f,, . . .,e,. Retaining the notation used before we write

Now we proceed similarly to Subcase 1.1. We start a run through T4at 21 along el E E ( S f )and note that this run will end with e2 after all other edges of G have been passed. Consequently, since f,,e2 E E(S,) but E(S,) c E ( G ) , at some vertex w there exists { h i , g i } E X,(w) - X , ( w ) such that h, E E(S,), g, E E ( S f ) . Correspondingly, let h,,g, E E, be chosen in such a manner that { h i , h i } E X , (and thus h, E E(S,)) and {gi,g;} E X , if w # v, while for v = w we denote gi = el,, gi = fi (thus g1,g2 E E ( S f ) in any case). Continuing now as in Subcase 1.1, we obtain Tj,k E ' T ( G , X ) from S, by a 6-absorption; i.e., Tj,k and T, are KX-associates. Moreover, {{fi, e ; } } U X , n X , C X , n XTj,k by construction and due to the structure of T' and S,, implying that Tj,k and T2 are 6X-associates because of the choice of T, and T,. Hence T, and T2 are K;x-associates. This contradiction to the choice of T, and T2 settles Subcase 3.1. Subcase 3.2. T4 # I ( G ,X ) . Since the case T3 E 7 ( G ,X ) is symmetric to Subcase 3.1, we only have to consider the case T3 $ I ( G , X ) . It follows from (* * *) that S,, S, E S,(G, X ) , where Xg = Xs3 and X i = Xs4. Retaining the notation previously used we obtain the following expressions:

Now we have E(S,*)# E(S,) # E ( S j ) . Hence, we may proceed in a manner similar to the one employed in Subcase 2.1, by starting a run through S,* at along f,. Since f, E E(S,) and el E E ( S j ) one finds {hl,,gi} E X,(w)for some w E V((S,*)) such that h, E E(S,), g1 E E ( S f ) ;and similarly one finds h,, 9, E E , such that { h i ,h i } E X , ( w )

vII.22


in any case, and where {gi,gi} E X,(w) if w # v, while we denote gi = ei, gi = fl if w = v. Note that {g1,g2} C E ( S f ) ,and {gi,gi} E X i - X 2 since either w # v or rn > 2; moreover, { h i ,hb} # { f;, ek} since

{ f L e 3 E x2. The eulerian trail T’,!E I ( G , X ) constructed in a manner similar to Subcase 1.1 now satisfies the relation

Because of the choice of T, and T2 this relation implies that T2 and

Tj,k are Kx-associates. This and K,(T,) = Tj,k imply that TI and T, are Kx-associates, again a contradiction. This settles Subcase 3.2 and therefore Case 3 (Note here that the construction of T’,k as in Subcase 2.1 is not necessary simply because Tj+contains a transition of T2 not yet accounted for in X, n X,, whereas in Subcase 2.1 the construction and Tikwas necessary because of {ei, ea} # X,U X,). of both Tj,k Having obtained a contradiction in each of the three main cases (which cover all possibilities concerning the structure of T, and T2),we may conclude that 7 ( G , X )cannot contain two elements which are not I C ~ associates. Theorem VII.10 now follows. A closer look at the proof of Theorem VII.10 reveals that we have, in fact, proved a stronger result. In order to see this, we must consider for the connected eulerian graph G a partition system P ( G ) a special case of which is a system of transitions X = X ( G ) . Accordingly, we denote by 7 ( G ,P ( G ) )the set of all P(G) -compatible eulerian trails of G , and introduce correspondingly the set S2(G,P(G)). Moreover, we estend Definition VII.9 to obtain the concept of +(,)-association which again defines an equivalence relation on 7 ( G ,P(G))whose equivalence classes define the corresponding partition P(G,K ~ ( ~ ) For ) . these more general concepts, one then deduces the following lemma whose proof is left to the reader (Exercise VII.3).

Lemma VII.ll. Let G be a connected eulerian graph together with an arbitrary partition system P(G). Consider an arbitrary T E 7 ( G , W )(S ) E S,(G, W ) )and ) let { e ; , f;},{ e ; , f;} E (X,)be arbitrarily chosen. Then, for all K E P ( G ) ,

x,

VII.2.1. Special Types of tcl-Transformations

VII.23

This lemma guarantees the validity of the equations analogous to (w) and (* * *), and the relations corresponding to (5), (6),(7) also remain valid (simply replace X with P(G); see the beginning of the proof of Theorem VII.10). These equations and relations, however, were of central importance to the proof of Theorem VII.10. Hence one arrives at the next result which is a generalization of Theorem VII.10, and which can be proven the same way as Theorem VII.10 (one only has to change symbolic expressions such as {e’,f’} E X to {e’,f’} K E P ( G ) ) . It is therefore left as an exercise to translate the proof of Theorem VII.10 into a proof of the next result.

Theorem VII.12. IP(G,sp(Gl) I= 1 for every connected eulerian graph G and every partition system P(G) satisfying I K 15 i d ( v ) for every K E P(u) P(G) and every v E V ( G ) . Note that in Theorem VII.12, restrictions on the size of the elements of P ( G ) are necessary and sufficient in order to have 7 ( G ,P ( G ) )# 8 (see Theorem VI.l). ‘1

VII.2.1. Applications to Special Types of Eulerian Trails and tc,-Transformations Some of the next results are direct consequences of Theorem VII.10. We consider connected graphs G which are 4-regular7 or such eulerian G which are embedded in a surface, together with three types of eulerian trails. In the first two cases, we may consider G as embedded in a surface and denote the set of all A-trails, non-intersecting eulerian trails respectively, of G by 7 A ( G ) I, N l ( G )respectively. In the third case we consider a 2-factorization {Ql, Q2}of a 4-regular G (see Corollary VI.3) and denote by 7a(G;Q1,Q2) the set of all eulerian trails of G in which edges of Q1 and Q 2 alternate (i.e., eulerian trails compatible with the cycle decomposition S = Q1U 0,). Moreover, for TI, T2 E I decided to prove Theorem VII.10 and to leave the proof of Theorem VII.12 as an exercise (instead of obtaining the former result as a corollary of the latter) because a system of transitions can be visualized more easily than a general partition system (and, of course, because there is no real difference between the respective proofs). Moreover, in the ensuing discussion Theorem VII.10 is of more direct relevance than Theorem VII.12.

VII.24


7 A ( G )(respectively T l ,T2 E 7N1(G),Tl, T, E x ( G ;Q1,Q 2 ) ) satisfying T2 = K ~ ( Twe ~ )call , this K1-transformation a KA-transformation, (respectively KNI-transformation,rc,-transformation), and define K A association (respectively KlvI-association,K, -association) correspondingly. We require, however, that if the rcA-transformation is a K * transformation (see Definition VII.8), the trail decomposition S’ satisfying ~’(2’~) = 5’‘ (see Definition V11.7) must result from Tlby choosing at v e V,(G) the pair of non-intersecting transitions different from X , (v). Similarly, if a KNI-transformation (K,-transformation) is a K * transformation, K’(T~) = S’ implies that the corresponding system of transitions X,, is non-intersecting (compatible with the cycle decomposition Q 1U Q,). We note that if G is a plane graph, a KA-transformation is necessarily a K*-transformation, while this is not always so if G is embedded in a surface other than the plane or sphere (Exercise VII.5a), b)). Moreover, in the case of 4-regular graphs, the concepts of K ~ and KNItransformation coincide. Again we leave it as an exercise to show that each of the concepts of K ~ - K, N I - , and rc,-association defines an equivalence relation on the respective set of eulerian trails (Exercise VII.5.c)).

Corollary VII.13. Let G be a connected 4-regular graph embedded in a surface. Then any two elements of 7,(G) are nA-associates. Proof. Since G is embedded in a surface, O+(v) is given by this embedding for every w € V ( G ) . Hence it is possible to speak of intersecting and non-intersecting transitions. Since G is 4-regular, the intersecting transitions are determined solely by the embedding of G. Let X be the system of these intersecting transitions. Then we have

and KX-association is equivalent to KA-association in this case. Corollary V11.13 now follows from Theorem VII.10. Another way of constructing all elements of I,(G) and of describing a KA-transformation has been developed in [KOTZ68c] for the case where the above G is plane (and for our purposes, this is the relevant case when we speak of A-trails). Namely, consider a 2-face-coloring of the plane connected 4-regular graph G (using the colors 1 and 2) and construct for i = 1,2 the graph M ias follows: every element x € V ( M i ) represents bijectively an i-face F ( x ) of G , and z,y E V ( M i )are joined ) V ( b d ( F ( y ) ) )I edges; i.e. every vertex v E V ( G ) by I V ( b d ( F ( z ) ) n

VII.2.1. Special Types of tcl-Transformations

VII.25

corresponds to an edge ei(u) E E(Mi). Let 7 denote the set of spanning trees of Mi.The nest result relates 'T;. to IA(G) and describes tcA-transformations in I A ( G ) (see [KOTZ68c7Theorems 14-17]).

Theorem VII.14. Let G be a plane connected 4-regular graph, and let Mi, i = 1,2, be the graphs constructed above in accordance with a fixed 2-face-coloring of G. Then the following is true. 1) There exist bijections ai between IA(G) and 'T;., i = 1 , 2 , and a bijection between I1and I2 which are defined by the following properties:

(i) If {Vl, V2}is the A-partition of T A E TA(G), then ai(TA)= ({ei(u)E B(n/r,)/~ E y}) E ?;, i = 1,2.

(ii) For Tl E 5 and TA E TA(G)satisfying Tl = a1(TA), cy1,2(T1) = ({e2(u) E E(M2)/el(v) E ( T l ) } ) = a2(TA) E 3 (iii) a2 = a1,2~1. 2 ) The following statements are equivalent for TA,TA E IA(G) . (i)

nA(TA)

= T i and TA

*

# Ti.

(ii) ai(TA)U cyi(Ti) has precisely one cycle, i = 1,2. Proof. 1) Consider for T A E TA(G) its A-partition {Vl,V2} and construct the plane graphs Hi+1 := G,,i, i = 1 , 2 (putting 3 = 1). By Theorem VI.59, Hiis a connected outerplane graph1) whose blocks are its cycles which in turn are the boundaries of the i-faces of G, i = 1,2. Hence, one can define Ti C M iby

V(Ti) = V(Mi), E(Ti)= {XY E E(Mi)/x:,Y E '(Ti) A (bd(FH;(2)) n bd(FH;(Y))# 0)) (where the subscripts Hiindicate that one considers the corresponding faces in H i ) . Because of the structure of Hi, I bd(F,;(x)) n bcl(F,;(y)) I< 1 for every x,y E V(T,). Because of the definition of Hi and because of u E V(G) w ei(u) E E ( M i ) ,we conclude for i = 1 , 2 that xy = e i ( u ) E E(Ti) if and only if b d ( F H i ( z )n ) b d ( F H i ( y ) )= {u} E

(i +

V;.

More precisely: Hiis a planar graph embedded in such a way that some 1)-face contains in its boundary an edge of every block.

VII.26


(see Lemma VI.58). Observing further that

d H j ( v )= 4 if and only if

ei(v) E E(Ti)

(4

and considering the subdivision graph S(T,), it follows from the above that E(Ti)= { e i ( v )E E ( M i ) / v E V,} ,

S(T,) 11 bc( Hi)

.

Since Hiis connected, bc(Hi) is a tree. This and V(Ti)= V(Mi) implies

Ti= ({ei(v) E E(M,)/V E y})E

.

(**)

Noting that Hi is already characterized by an A-partition {Vl, V,},it = Tidefines a bijection between follows from (*) and (**) that ai(TA) as expressed in (i). 7,(G) and In order to see that (ii) also holds true, it suffices to compare H , with H., and T, with T2 (with Ti defined as above, i = 1,2). Observe first that HI and H2 describe the same TA E TA(G). Hence al(TA) = T, , a2(TA) = T,. By (*) and the very definition of H , and H2we can therefore conclude that e1(2r) E E(T,) if and only if e 2 ( u ) 6 E(T,). Hence a1,2(T1)= T2 defines the bijection between TIand 3 as expressed by (ii) . The validity of (iii) now derives from the preceding considerations.

2) Suppose KA(TA) = T i for TA,TiE TA(G). As has been noted before (see also Exercise VII.5), 'cA(TA) = lc*(TA);i.e., K* = K " K ' , where d is a K-detachment and K" is a 6-absorption. There thus exists S E S,(G) with X, containing non-intersecting transitions only, such that 6'(TA)= S and d'(S) = Ti. Denote by v ( w ) the vertex of G at which the 6-detachment (+absorption) takes place, and denote further Ti := a,(TA),T;' := a i ( T i ) , i = 1,2. Let {V,, V,} ({V:,V;})denote the A-partition of G corresponding to T A (Ti). Note that TJ # w if and only if T A # TA. W.1.o.g. i = 1 (the case i = 2 can be handled in an analogous manner), The transition from TA to S, i.e. the application of a 6-detachment at TJ, is equivalent to changing {V,,V,} to either {Vl U {TJ}, V, - {v}} or {V,-{TJ>, V,U{v}} depending on whether TJ E V. or TJ E V,. Consequently

VII.2.1. Special Types of K1-Transformations

VII.27

and equivalently, T, is transformed either into the unicyclic graph T, := TlU el(v) with e,(v) necessarily belonging to the uniquely determined cycle C of T,, or into the forest TF := T, - el(v) which has precisely two components K, and K2 (see l)(i)). However, since the transition from S to T i , i.e. the application of a /c-absorption at w, must result in the A-partition {V;, Vl} of T i , we can conclude that either

(i.e., v E V, if and only if w E V,.Observe that qTl = qT, = IV,l = IV,ll), or = (V,- {w}) u {w}, = (V, u {w}) - {w} .

v;

v;

Correspondingly and equivalently, Ti = a, (Ti)results from T, by deleting el(w), respectively from TFby adding el(w). In the first case el(w) E E ( C ) ,in the second case el(w) = x1x2,where x j E V(Kj), j = 1,2; this follows from the fact that Ti E 7,. The above construction of Ti from T, shows that in any case,

Tln TI

= T, u 5": - {e,(v), el(w)}

is a forest which has precisely two components, such that each of el(w) and el(w) joins vertices in different components. Hence,

has precisely one cycle (which necessarily contains el(.), el(w)). Thus 2) (i) implies 2) (ii) . Conversely, suppose that CY~(T~)UCY;(T~) , i = 1 , 2 , contains precisely one cycle; call it C. TA# TA follo~vs.Denote Ti := ai(TA),Ti := ai(T;). Since Ti,T{ E ?;., and because l ) ( i ) has already been proven we can conclude that

I E(T;)- E(T,')I=I E(T{)- E(T;)I=

1 ;

(otherwise, either TiUT: had more than one cycle or Ti = Ti contradicting T A # Ti).Denote

VII.28


Consequently, Ti can be obtained from Tiby deleting e i ( v ) first and then adding e i ( w ) (or, likewise, by adding ei(w) first and then deleting e i ( v ) ) . It follows from the first part of the proof of 2 ) that the transition from Tito Ti- e i ( v ) corresponds to the transformation of TA to S, where S E S,(G) and X,(v) is a pair of non-intersecting transitions. Hence, this transformation is a n-detachment, i.e., d(TA)= S. Similarly, since Ti’= (Ti - e i ( v ) )U { e i ( w ) }is a tree it follows from the bijection expressed by 1)(i) and by construction, that the construction of Ti from S is a tiabsorption involving non-intersecting transitions only; i.e. , d’(S) = T’. Thus T i = KI”(K’(T~)) = R ~ ( T i.e., ~ ) 2)(ii) , implies 2)(i). This finishes the proof of the theorem. Note that every spanning tree of a connected graph G can be obtained from any other spanning tree of G by a sequence of the ’edge out-edge in’ procedure described in the above proof. This and Theorem VII.14 yield a proof of Corollary VII.13 (restricted to plane graphs) independent of Theorem VII.10. In fact, one can produce a result analogous to Theorem VII.14 for arbitrary connected plane eulerian graphs G. There are two ways, though, to approach this problem. If one wants to adhere to graphs, then (in constructing the corresponding Mi) V(Mi) not only consists of the “capitals” of the i-faces, but also of V(G). In this case, z y E E ( M i ) if and only if I {z, y} n V(G) I= 1, where x E V(G), and y corresponds to an i-face F ( y ) with z E b d ( F ( y ) ) . Note that this way of defining M icorresponds to ,”(Mi) in the case of a 4-regular G (see the discussion immediately preceding Theorem VII.14). In the present (more general) situation, however, there is no longer a bijection aibetween I A ( G ) and the set ?;. of spanning trees of Mi. Rather, TA E TA(G) with A-partition {V,,V,} corresponds to any TiE 7 in which d T i ( v j ) = 1 for every vj E vj and dTi(vi) = $dG(vi) for every v i E q ,where { i , j } = {1,2}. So, every TA E I A ( G ) is associated with a set of trees ?;(TA)C ?;.. Nevertheless, the duality between 7,and 3 as expressed by equation l ) ( i i ) in Theorem VII.14 remains valid in a modified form; the same can be said of part 2 ) of that theorem. However, the concept of hypergraphs might be a more appropriate tool to describe an analogy to Theorem VII.14 for arbitrary plane connected eulerian graphs G. In that case V(Mi) is defined as in the case of 4regular graphs, and correspondence between v E V(G) and e i ( v ) E E ( M i )

VII.2.1. Special Types of Kl-Transformations

VII.29

is preserved as well. But now

ei(v) = {z E V ( M i ) / F ( z )is ani-face with v E V ( b d ( F ( z ) ) ) } . Defining a spanning hypertree Tiof M iin a way similar to the case of graphs, one can show that Theorem VII.14 as stated above remains valid even in this more general sense. We leave it as an exercise to check the details of the above discussion on the two possible extensions of Theorem VII.14 to arbitrary plane coiinected eulerian graphs (Exercise VII.6). We note, however, that because of the bijection cyi between I,(G) and the set ?;. of spanning hypertrees (in the general case), deciding the existence of a spanning hypertree in a general hypergraph is an NP-complete problem. For, the construction of M ifrom G can be done in polynomial time and vice versa, provided G is a plane graph. So, if we restrict our considerations to planar 3-connected eulerian graphs G, the face boundaries of such G are uniquely determined independent of any actual embedding of G (see Theorem 111.52). Consequently, in this case the hypergraph M iis determined solely by G and vice versa. Now, since for such G deciding the existence of an A-trail is an NP-complete problem (see Theorem VI.91), it follows that the problem of deciding the existence of a spanning hypertree Tic M iis also NP-complete. We leave it as an exercise to check the details of these complexity considerations (Exercise VII.7). One should not overlook, though, that just as Theorem VII.14 as such does not imply Corollary VII.13 (see the paragraph immediately following the proof of Theorem V11.14), the extension of Theorem VII.14 to arbitrary plane connected eulerian graphs G as such does not yield any information about the equivalence classes of TA(G ) under KA-association. Of course, a KA-transformation can only involve 4-valent vertices since we also require that S induces a 1- or 2-splitting at every vertex of G. In fact, in this case T,T’ E TA(G)belong to the same equivalence class if and only if their respective A-partitions {Vl, V,} and {V;, V,l} satisfy the equations

It is left as an exercise to check that in the general situation as well, IcA-association is in fact an equivalence relation on TA(G)and that the above equations determine the equivalence classes of TA(G)under K ~ association (Exercise VII.8).

-


VII.30

However, if we consider the set ‘TN1(G) of an arbitrary connected eulerian graph embedded in a surface, we can derive a result analogous to Theorem VII.10. The proof of this result will be simpler than that of Theorem VII.10, but it will show at the same time that one does not have the same degree of freedom in applying the corresponding Ic,-transformation. For this proof we shall need the following lemmas.

Lemma VII.15. Let G be a graph embedded in a surface and having an even vertex v. Let X ( v ) be a system of non-intersecting transitions at v. Then for at least one t E X ( v ) , t = {e’, f’} say, el and f’ are consecutive in O+(v). Moreover, if v @ V,(G) then there are at least two such transitions. Proof. Consider O+(v) := (ei , . . . , e h k ) . If k = 1, then X ( v ) = { {ei, ek}}, and the lemma follows in this case simply from the fact that G is an embedded graph. If k = 2 , then either X ( v ) = { {ei, eh}, {ei, ei}} or X ( v ) = {{ei, ei}, { e ; , e ; } } because X ( v ) is a set of non-intersecting transitions, whence we can conclude that the lemma is also true for k = 2 . So consider the case Ic > 2 , and suppose there exists { e i , e>}

E X ( v ) such that i - 1 # j # i + 1 ,

where we put i - 1 = 2k if i = 1, and i

(*I

+ 1 = 1 if i = 2 k .

Among all possible choices for i and j satisfying (*) take one for which li-jl is as small as possible. W.1.o.g. suppose that in O+(v) the notation has been chosen in such a way that i = 1. Hence 2 < j < 2 k . Precisely because X ( v ) is a set of non-intersecting transitions it follows that {ei,e&} $! X ( v ) if 2 5 I < j < m 5 2 k . Consider the open integer intervals

(**I

:= (1,j) and

and choose I , n E IZ-nl

I’j,zk := ( j , 2 k )

Il,jsuch that

isassmallaspossibleandt:={ei,e~}E X ( v )

.

(***)

Then t has the property as described in the lemma with e‘ = ei , f‘ = e:; otherwise, we obtain a contradiction either to (*) and the choice of i and j , or to (w).

VII.2.1. Special Types of rcl-Transformations

VII.31

As for finding a second transition t' E X(v)having the property stated in the lemma, either t' = {ekk-, ,ekk} E X ( v ) (in which case nothing has to be proved any more), or else { e i , e b k } E X(v), where j < p < 2k - 1. In the latter case, consider the corresponding open integer interval IPlzk:= ( p , 2 k ) and consider Z',n' E Ip,2ksatisfying (* * *). Using the same argument as above and because rnax{Z,n} < min{Z',n'}, we can conclude that t' = {ei,,eL,} # t and that t' also has the property cited in the lemma (with e' = e ; , , f' = el,,). Lemma V11.15 now follows. It follows from Lemma VII.15 that if X, is a system of non-intersecting transitions corresponding to a (non-intersecting) eulerian trail T of the connected eulerian graph G which is embedded in a surface then at every vertex, T behaves in at least one transition like an A-trail.

Lemma VII.16. Let G be a connected eulerian graph embedded in a surface. Suppose X, is a system of non-intersecting transitions for some S = {Se,Sg} E S2(G). Then for every w E V ( ( E ( S , ) ) n ) V((E(S,))) there exists T, E 7 N 1 ( G such ) that T, results from S by a 6-absorption at w. Proof. Choose w E V ( ( E ( S , ) ) n ) V ( ( E ( S , ) ) )arbitrarily and consider X,(w) together with O+(w)= ( h i , . . . ,hLk); k > 1 because of the choice of w. Suppose w.1.o.g. that the notation for O+(w)has been chosen in such a way that h, E E(S,). For t , = { h i ,h i } E X,(w) we therefore have h j E E(S,). Let t, = {hh,h;} E X,(w)be chosen such that h,, h, E E(S,); w.1.o.g. rn < n. Moreover, j < rn or j > n since X, is a system of non-intersecting transitions; w.1.o.g. j < rn. Suppose {h:, h:} fZ

X,(w) if

r and s satisfy j

< r < m < n < s 5 2k .

consists of non-intersecting transitions only. Hence

x*= (X,- X,(w)) u XJW) is a system of non-intersecting transitions of G. Moreover, it follows from the definition of X,(w)that the transformation of X, into X * defines a Kabsorption at w (see equations (*) and (1") in the discussion immediately

VII.32


preceding Definition VII.7). That is, X ” = X,, and T, results from S by a r;-absorption.

where

T,,, E IN1(G);

If, however,

then define

ti := {h:, h:}, tk := t ,

if h,, h, E E(S,) ti := t,, ti := { h i ,h:}, otherwise,

and repeat the above argument with ti,ta in place of t l , t , (note that {h:, h:} E X s ( w ) implies {h,, h,} n E(Se) # 0 if and only if { h r ,h,} C

E ( S e ) ) .Since d(w) < 00 it follows that for some {ty),t:)) X s ( w ) , the above 6-absorption transforming S into T, E I N I ( G ) can , be performed , ,t (9 , instead of t,, t,. Lemma VII.16 now follows. by using t (9 In accordance with the notation previously used, P(G,‘ c N I ) denotes the induced by rcNI-association. partition of lNI(G) Theorem VII.17. I P ( G ,tcIvI) I= 1 for every connected eulerian graph G embedded in a surface.

Proof. The theorem being trivially true if

I INI(G)I=

1 (since the identical transformation is a KNI-transformation), we proceed indirectly by assuming that 17NI(G)I> 1 and that eulerian trails T,,T’. E 7jNI(G) exist which are not sNI-associates. Denoting their respective system of transitions by X , and X , (we retain the notation of the proof of Theorem VII.10 whenever possible), we can again assume that Tland T2have been chosen in such a manner that I X , n X , I is as large as possible, but this - 2) is as small time subject to the condition that a(G) := CVEV(G)(d(v) as possible. a ( G ) > 0 follows from ( I N 1 ( GI> ) 1. Let v be any vertex for which X,(v) # X,(v). By Lemma VII.15 a transition t = {e’, f’} E X,(v) exists such that O+(v) = (. . . ,e’, f’, . . .). Suppose t E X,(v) holds as well. One can then form G‘ by splitting e and f away from 21 in such a way that V(G’) = V(G)u{velf} and e , f are the edges incident with v,,~,and where G‘ can be viewed as embedded in the same surface as G. Since t E X , n X , it follows that Ti corresponds where Xi is the system of to an eulerian trail Ti of G’ with Xi = Xi,

VII.2.1. Special Types of Icl-Transformations

VII.33

transitions corresponding to Ti’, i = 1,2. Observe that because of the position of t in G and because of

it follows that any system of non-intersecting transitions of G‘ corresponds to a system of non-intersecting transitions in G. Since a(G’) < a(G),it follows from the choice of G that Ti and Ti are KNI-associates. Viewing the various KNI-transformations used in transforming 2’: into T: as operations performed in G, and noting (*), it follows that T, and T ! are also nN1-associates. This contradicts the choice of Tl and T2;hence

t # Xl(V). So, consider t , = {e’,g:} and tf = {f’,ga}, where t,,tf E X,(v), and define for t* = {gi, ga}

x,. = ( X , - { t , , t f } ) u { t , t * } , x; = x,

.

Precisely because of the structure of t E X,(v)= X;(v)it follows that X; is a system of non-intersecting transitions. Moreover,

since t E X; n Xl whereas t # X,. Consequently, if X; = X,, for T* E 7 N I ( G )it, follows from the choice of Tl and T2 that T* and T2 are KNI-associates. This and T* = /cN1(Tl) implies, though, that Tl and T2 are rCNI-associates, a contradiction to the choice of T, and T2 which implies that X; = X, for some S E S,(G). In accordance with the notation used earlier, denote S = {S,, Sg},where w.1.o.g. e, f E E(S,) and, therefore, gl,g2 E E ( S g )(this follows from S E S,(G)).

As in the proof of Theorem VII.10, consider now a run through T,starting at v along e . Since t E X, this run will end at v right after passing f . Sincee, f E E(S,), gl,g2 # E(S,), and therefore E(S,) c E(T2)= E ( G ) , it follows that there is a vertex w at which T2 passes from S, into Sg, and that d(v) > 4 if w = v. In any case, w E V ( ( E ( S , ) ) n ) V((E(Sg))). For t E X ; n Xz, let G’ be obtained (as in the first part of the proof) by splitting away e and f from v (note t = {el, f’}), and redefine Xi = Xi. ,i = 1 , 2 , correspondingly; but now Xi = X,,, for S’ E S2(G’), where S‘ = (Sd, SL} and E(S:) = E(S,), E(S;) = E ( S g ) .We distinguish between two cases.

VII.34


Case 1). Xi(w) n X;(w) = 8. In this case, Lemma VII.16 can be applied directly to G' and S' so as to obtain T'k E I N I ( G ' ) ,where Tk results from S' by a K-absorption at w. It follows that Tk and T . are IcNz-associates (where Ti E TN1(G')corresponds to T 2 ) ,for the definition of Xf together with X i ( w ) n Xi(w) = 0 and (**) ensure the validity of I X,, w fl Xi 1>1 X, n X, I. Thus, the corresponding eulerian trails T,,T2 E I N z ( G )are K-associates. This and K'(T,)= S, K"(S)= T,, and because X; is a system of non-intersecting transitions, classify TIand T2 as KNI-associates. This contradiction settles Case 1). Case 2). X ~ ( w ) n X ~ ( w # )0. In this case we consider first O+(w) in

G'

and write

O+(w) = ( h i ,h;, . . . ,hLk) , k 2 2 (note that dG,(w) 2 4 regardless of w = or w # u). Consider an arbitrary element t' = {hk,hk} E X i ( w ) n Xi(w), where 1 5 m < n 5 2 k ; w.1.o.g. rn = 1. By the very definition of non-intersecting transitions it follows that t" = {hi,h:} # X((w)U Xi(w)for i and j satisfying 1 < i < n < j 5 2k. Consequently, any element of (Xi(w)UXi(w)) -{t'} lies on either of the two sides of the open curve, ,C , defined by hk U h;. Thus C,, defines a partition of Xf(w), i = 1,2,

(possibly Xf,l(w) = 0 or Xi,r(w) = 0, but Xf,,(w) U Xf,,(w) # 0); Xf,r(w) (Xf,T(w))contains precisely those transitions in Xf(w) which lie on the Zeft (right) side of, ,C, (where Zeft side and right side denote the opposite sides of C,,,). This partition (* * *) gives rise to the replacement of w by three vertices w,, w,,w t fZ V ( G )in such a way that

It follows from the definition of the partition (* * *) that this replacement procedure can be performed in such a way that the resulting graph G{ can be viewed as embedded in the same surface as G. Moreover, it also follows from (* * *) that S' and Tl are transformed into corresponding elements Si E S2(G',) and Ti,t E 7 N z ( G i ) .Finally, it follows from (* * *) and the construction of G', that every system of non-intersecting transitions of G{ corresponds to a system of non-intersecting transitions of G' and G.

VII.2.1. Special Types of tc1-Transformations

VII.35

We repeat the above procedure if necessary until we arrive at the connected eulerian graph H with the following properties:

1) V ( H )2 V ( G ’ -

W)

;

ti of X : ( W )n X i ( w ) are represented in H by 2-valent vertices w j , and E:j = ti ; 2 ) the elements

3) dH(w,) 2 4 for every

w, E V ( H )- (V(G’ - W ) u { W j / t ; E X ; ( W )n x ; ( ~ ) ,} ) and

Xi(W,)

c Xil(W),

i = 1,2

,

where these X i ( w a ) satisfy

whereas the vertices of G’

- w remain unaltered in H .

4) H can be viewed as embedded in the same surface as G‘ and obtained from G‘ in such a way that X , ( H ) is a system of non-intersecting transitions which can be written as

X , ( H ) = x;, i = 1 , 2 (see 2 ) and 3), and note that a transition is a pair of half-edges independent of incidences with actual vertices). Moreover, every system of non-intersecting transitions in H corresponds to such a system in G’ as well. Because of 4), X l ( H ) = X s H for SH = { S e , H , S g , HE} S 2 ( H ) corresponding to the above S’ E S2(G’),and X 2 ( H ) = X , for T H E I N I ( H ) corresponding to T!E 7 N I ( G ’ )We . note that since T2passes at w from S, into Sg we can therefore find some

at which T H passes from Se,H into S g , H .Consequently, since X 1 ( w H )n x 2(wH ) = 8 by 3) we have reduced Case 2 ) to Case 1) with

VII.36


H,w H ,S H ,T H in place of G', w ,S', Ti respectively. Thus, the application of Lemma VII.16 yields 2': E INI(H) which results from S , by a K-absorption at w H . Arguing now along the same lines as in Case 1)we can conclude that 2': and T H are KNI-associates which is tantamount to saying that T,, the eulerian trail of G corresponding to T z , and T2 are tcNI-associates. Moreover, by the same token we can conclude that 2' and T, are tcNI-associates. Hence Tland T2 are IcNI-associates, a contradiction which settles Case 2. Having obtained a contradiction in all possible cases, the theorem now follows. It should be noted that surface embeddings of G and the various derived graphs were not necessary for the formulation and proof of Theorem VII.17. What really mattered was that O+(v) was given for every v E V ( G ) which is sufficient for defining non-intersecting transitions. On the other hand, the proof presented here shows that the constructions of various graphs which are needed to find the appropriate / c N I transformations for transforming Tl into T,, permit adherence to the same surface all the way. Moreover, the proof of Theorem VI.17 indicates that in general more (algorithmic) work is required to establish the appropriate rcNI-transformations than in the case of KX-association. It is possible that in the case of S E S,(G) without intersecting transition, no &-absorption at w yields 57, E I N I ( G )satisfying I X , n X,[>I X,n X 2 I (see Figure VII.3). Therefore, one is forced to produce S from TI in such a way that already S satisfies I X , n X , ]>I X , n X , 1, and thereafter to perform a %-absorption at w without involving any element of

X l ( 4 nX 2 ( 4 Finally we turn to &,-transformations of eulerian trails in connected 4regular graphs G. It should be recalled that for a 2-factorization {Q, ,Q3} of G, X ( G ;Q,,Q,)denotes the set of eulerian trails of G in which ed&s of Q , and Q2 alternate (see the discussion preceding Corollary VII.13). As another application of Theorem VII.10 we obtain the following result Q1,Q 2 )into equivalence in which P ( G , IC,) denotes the partition of 7,(G; classes under rc,-association.

Corollary VII.18. Let G be a connected 4-regular graph, and let {Ql, Q2}be a 2-factorization of G (which exists by Corollary VI.3). Then I "(G, 6,) I= 1. Proof. S = Q1U Q2 is a special type of cycle decomposition of G; its induced system of transitions is denoted by X , . By the very definition

VII.3i


G Figure VII.3. The connected plane eulerian graph G with whose transitions are marked with eulerian trail TI (T2) small (dotted) arcs. A K-detachment of TI at v involving the edges e, f yields S E S2(G)without intersecting transitions. i = 1,2, the subsequent Because of the structure of Xi(w), n-absorption at w yielding T, E 7 , 1 ( G )implies IX, ( w ) n W

X d w ) I=IX,(w)n X2(4I= 0. of S and

Xs respectively, it follows that

Also, 6,-association and case. Thus,

“x’s -association

P(G7 K a ) = P(G7 K

are identical concepts in this X ~ ).

These observations and the validity of Theorem VII.10 imply the validity of Corollary VII.18. In the case of K,-transformations, one can generalize this concept and Corollary V11.18 in a way similar to the generalization of Theorem VII.10 which led to Theorem VII.12. Namely: suppose G is a connected eulerian

VII.38


graph with an even number of edges. Then G can be decomposed into two edge-disjoint eulerian subgraphs G, and G, with d,, ( u ) = dGz( u ) for every Y E V ( G ) (see Corollary VI.2). Denote by 7a(G;G,, G,) the set of eulerian trails which alternately pass through edges of G, and G,. Generalize the concept of &,-transformation and K,-association correspondingly. Then the following result can be derived from Theorem VII.12 in much the same way as we derived Corollary VII.18 from Theorem VII.10. We therefore leave its proof to the reader (Exercise VII.9).

Corollary VII.19. Let G be a connected eulerian graph with an even number of edges, and let {G,,G,} be a decomposition of G into two subgraphs having the same degree sequence (d(v,), . . .,d(u,)}. Then Tl and T2 are Ka-associates for every T,, T2 E 7=(G; GI, G,).

In the case of 4-regular graphs G, however, one obtains for a fixed T E 7,(G;Q,, Q,) a classification of the vertices of G into ‘odd’ and ‘even’ vertices as follows. Let

T be written in the form

and suppose ui = uj+,. Consider the segment

S; . = ui, ei, . ..,ej, Y. 3+1 93

7

and the eulerian trail T* obtained by reversing the segment Sij;i.e.,

T*= u1,el , . . . ,e i - , ,Si;i ,ej+l , . . .,e q 7u1 . It follows from the very definition of T and T* that T* E 7,(G;Q1, Q,) if and only if I E(Si,j) 1(mod 2). Consider the case where I E(S;,,) 0 (mod2). Then a Ka-transformation + ~ be a &-transformation. On the other involving Y := vui= Y ~ cannot hand,

lE(Si,j)I= 0 (mod2) implies V((E(Si,j)))nV((E(G)-E(S,,j))) # {u} . Otherwise, (E(Si,j))would have, except for the 2-valent Y, only 4-valent vertices, i.e. ( E ( S i , j ) would ) be homeomorphic to a 4-regular graph with


VII.39

an odd number of edges, something that is impossible. So, a 2-valent vertex w E V((E(Si,,.))) - (21) exists and T is of the form

where w = urn = vn, and

We now show that these equations imply the existence of a teatransformation involving 21 and w. For such K , = IC.' I K I and Sj+l,i-l.21.J + l , ej+l,.

. . ,eq,v1,e1,. . . , e i - l , ~ iwe can define 4 T ):= S'

= {S+

s,.+l,i-l}

which is compatible with S := Q1 U Q2 precisely because of I E ( S i .) 0 (mod2). If we denote by Sm-l,m and S n - l , n the segments of T obtained from Silj,S,.+l,i-lrespectively, by cyclic rotations so as to start and end 1 at w ,then for any S:-l,n E {Sn-l,n,S~-l,n} 9'.

since S' is compatible with S , it follows that

That is, there is a 6-absorption involving w such that n"(S') = T* ; i.e.,

K,(T)= T* for

K,

= K .' I K .I

.

Observe, in general, that if one fixes an initial vertex u1 and an initial edge el for T , T defines for every 21 E V(G)a unique trail decomposition Sh = { S l , u , S 2 , u }(each of whose elements starts and ends at u). Moreover, I E(G)= 0 I (mod 2) since G is 4-regular; hence we have

This congruence leads to the following more general definition.

VII.40


Definition VII.20. Let T be an eulerian trail of the connected 4-regular graph G starting at v,, E V ( G ) along e E Eu0,and consider for every v E V ( G )the unique trail decomposition S, = (S,, S,,,} which consists of the two segments of T starting and ending at v. Then we can call T-even or T-odd, depending on whether I E(S,,,) I (and therefore, by (l),I E(S,,,) I) is even or odd.

In fact, because of (l),the classification of 2'-even and T-odd vertices is, for a given eulerian trail T , independent of the choice of an initial vertex v1 and initial edge el (Exercise VI1.10). Hence every eulerian trail T of G defines a partition of V(G)into 2'-even and T-odd vertices, where this partition depends on 2' only. It follows from this definition and the preceding discussion that riatransformations are &-transformations precisely at T-odd vertices of G, while a &,-transformation which is a &*-transformation,is composed of a &-detachment at a 2'-even vertex and of a &-absorption at some (T-even or T-odd) vertex w E V((E(S,,,)))n V((E(S,,,))). It is not true, however, that this vertex w must be 2'-even (or 2'-odd) in any case. This fact is demonstrated by the graphs G, and G, of Figure VII.4. There, the eulerian trails Ti,j E X ( G i ;Qj,,, Qi,2), i,j = 1,2, satisfy the equations

d ( T i , J= ) s;,,,

K"(s:,,)

= Tj,,

where the &-detachment 12 is performed at the Tj,,-even vertex v, while the &-absorption IC" is performed at w which is T,,,-odd in GI and T2,?even in G,. Consequently, in G, a &,-transformation involving w is possible such that

However, is T*-odd indeed. Hence, there exists a &,-transformation That is, the above &*-transformationtransformsuch that n(T*) = TI,,. ing T,,, into TI,, can be replaced with two &,-transformations which are &-transformations. Such a replacement is not possible, however, in the case of T2,,and T,,,. As we shall see in a moment, this phenomenon is a general one.

Definition VII.21. Let G be a connected 4-regular graph, and let T,, T, E 7a(G; Q,, Q 2). Suppose for S = Q1 U Q, and for some v, w E

VII.2.1. Special Types of Rl-Transformatioiis

VII.41

Figure VII.4. Two 4-regular graphs G , with eulerian E 7a(Gi;Qi,l, Q i , 2 ) i,j , = 1,2, where the edges trails Ti,j of Qi,k are marked by k, k = 1,2, and the transitions of Ti,, are marked by small arcs, while the transitions of Ti,, are marked by small dotted arcs, i = 1 , 2 . Since T, and TI,, define the same transitions at x E V ( G , ) , x is represented by two 2-valent vertices. v is Ti,,-even in G,, i = 1 , 2 , while w is T,,,-odd in G , and T2,1-even in G,.

~"(s')

V ( G ) that ~'(2"~) = S' E S,(G;X,) , = T', where K' is a Rdetachment at v and K" is a R-absorption at w..Call := K"R' reducible or irreducible (regarding v, w) depending on whether w is T,-odd or T,even (note that v is Tl-even in any case since K~ is a K*-transformation). L e m m a VII.22. Let G ,T,, T,, S , S', = R'R'', v and w be the objects used in Definition VII.21. Then the following holds: 1) tea is reducible if and only if there exist T' E 7 a ( G ; Q 1 , Q 2and ) Rtransformations K , and K , involving UT, v respectively, such that K , ( T ~=) T* and K ~ ( T= * )T,.

VII.42


2) K , is irreducible if and only if there exists S“ E S2(G,X,) such that R’(T,) = S”, E’’(S”) = T2, where iif (2’)is a K-detachment (&-absorption) at w (v).

Proof. 1) Suppose K , is reducible. By Definition VII.21, w is Tl-odd. Hence, by the discussion following Corollary VII.19 and Definition VII.20, there exist T* E 7,(G;Q1, Q 3 ) and a rc,-transformation tcW which is a ICtransformation such that K . ~ ( T =~T* ) and X T . - X,,(w) = X , -

x, (4We claim that v is T*-odd. This becomes visible when we express Tl in the following way:

where vo = u j = v m = v and v i = v k = w. Moreover, we assume w.1.o.g. that e,,ej+, E E(Q,), f,,fj7 fm E E(Q,). This is possible since v is Tl-even. Since w is T,-odd we must have for one of the two choices for T , s satisfying { T , s} = { 1 , 2 }

Hence T* is of the form

T* =vO,e,, vl, f,,.. . ,h:,vj,h i , . . . ,ej+,, vj7 fj, . . .

. . .,hy, vi,h i , . . . f,, vm . That is, starting in v = vo along el E E(Q,) we reach v = v j again after passing ej+l E E(Q,), i.e. v is 7”-odd. Consequently, there exists a K,-transformation K , which is a K transformation such that K,(T*) = T$ E h* i a

. . . , h i , 211,h:, . . . ,f m , T J ,

*

,

fj,Vj,ej+l,

*.

.

Thus, changing the transitions at w = zli = v k so as not to violate compatibility with X,, we obtain two edge-disjoint subtrails covering It

It

Sl,, = T J ~ ,hi St,, = T

. ., fj, T J ~ e, j + l ,

7

h:,

J ~ ,

*

* 2

* * * 7

f m , v m , e l , 211,

That is, S‘ := {Sy,,,Sl,w} E S,(G,X,) detachment iz’ at w.

hi,T

f1, * *

J

7

~ 9

hi, T J ~

-

and ilt(Tl)= S” for some

K-

However, by a rotation we can rewrite the elements of S“ so as to obtain

Si,, = vj, ej+,, . . . , h&,vk,hy,. . . ,fj, v j Sh,v =

el >

> f1,

T J ~

.

*

>

hi, T J ~ ,h;,

* * *

> fm,TJm

*

That is, S’ = {S;,,, Si,v} yields ~’(7’;)= S’ for some 7’: E 7 , ( G ; Q1, Q,). Since S’ results from S” by a cyclic rotation of the sequences describing its elements, and since the inverse operation of a rc-detachment is a Kabsorption, we have i?’(S”) = T i for a Ic-absorption ill‘ at TJ. We can now conclude as in the first part of the proof of 1) that = T2. ’ R”iZ’ satisfy Conversely, suppose two Ka-transformations K ~ ~ I Cand n”(n’(Tl>) = ilfl(il’(Tl)) = T,,where K‘ is a K-detachment at TJ and ill is a n-detachment at w. Then both TJ and w must be Tl-even (see the discussion following Definition VII.20), hence n, is irreducible. This finishes the proof of 2). Lemma VII.22 now follows.

VII.44


Lemma VII.22 says that a 6,-transformation K”K’ is reducible if and only if one can avoid taking a ‘detour’ via S2(G,X,), by applying two n-transformations K , and K , , each of which yields an element of 7=(G;Q1, Q,). This fact, i.e. staying with rc-transformations instead of combining a 6-detachment with a K-absorption, justifies the term Teducible. On the other hand, if d‘~‘ is irreducible, it is irrelevant at which of the two vertices w and w one starts with a K-detachment, i.e. in this case, the product K”K is commutative in the sense that the transformations 6’ and K”, viewed as operations on systems of transitions, yield d’d(T1)= dd’(T1)= T2 where d(T1),d’(T1)E S,(G, X,) and in general, d(T1)# 6”(T1). Summarizing the discussion following Corollary VII.19 we are thus led to the following result which is based on Corollary VII.18.

Theorem VII.23. Let G be a connected 4-regular graph G with 2factorization {Ql, Q,}. For any two eulerian trails T ,T’ E %(G;Q1, Q2) a sequence of eulerian trails To,. . . ,T, exists for some m 2 1 such that To = T , T, = T’, TiE 7,(G; Ql,Q,), i = 0,. . .,m, and such that for i = 1, . . . ,rn, Tiresults from Ti-1 by the application of a &,-transformation. If this K,-transformation is a 6-tramformation, it is applied at a Ti-1odd vertex. If, however, it is a rc*-transformation, it is an irreducible Ka-transformation applied at two Ti-,-even vertices w and w, where w belongs to both segments of Ti-1 starting and ending at w. In this case it is irrelevant whether one applies the &-detachment at w and the Kabsorption at w, or vice versa. It should already be noted at this point that Theorem VII.23 will play a central role in G. Sabidussi’s approach to what has become known as the Compatibility Problem [SABI84a]. In fact, Theorem VII.23 can be translated into [SABI84a, (7.7) Theorem], and a large part of the discussion leading to Theorem VII.23 is contained in that paper. For a short description of Sabidussi’s approach to the Compatibility Problem as well as for a discussion of various other approaches and various aspects of the Compatibility,Problem, the intested reader might resort to [FLEI88a] at this point.

VII.3. Transformation of Eulerian Trails in Digraphs

VII.45

VII.3. Transformation of Eulerian Trails in Digraphs Just as we used Theorems VII.10 and VII.12 to derive various results concerning the transformations of special types of eulerian trails in graphs, we can apply Theorem V11.12 to obtain a result for digraphs analogous to Theorem VII.5. For this purpose denote, by analogy, by 7 ( D )the set of all eulerian trails of the connected eulerian digraph D, and define K - , K*-, rcl-transformations and 6-detachments and K-absorptions. Of course, these operations have to be defined in such a way that every transition at v E V ( D )always consists of an incoming and an outgoing halfarc incident with v. This is why, for a digraph D , K-transformations are not a sufficiently general tool to transform T E 7 ( D ) into every other element of 7 ( D ) (see Figure VII.5). However, as we shall see below, Kl-transformations suffice to establish a result for digraphs hnalogous to Theorem VII.5.

As in the case of graphs one can define the concepts of K - , K * - , K ~ association, and one can show that each of these concepts defines an equivalence relation on 7 ( D ) ;it is left as an exercise to do this (Exercise VII.ll). Such an equivalence relation on I ( D ) defines again, of course, a partition of 7 ( D ) into equivalence classes as before. For our purposes we shall consider the partition P ( D , K ~ ) . Theorem VII.24.

Iw ,K 1 ) I=

Let D be a connected eulerian digraph.

Then

1.

Proof. Our aim is to deduce Theorem VII.24 from Theorem VII.12. For this purpose consider G, the connected eulerian graph underlying a given connected eulerian digraph D. For every v E V ( D ) ,let

be the partition of A: into the classes of half-arcs incident from v, to v respectively. Since V ( D )= V ( G )one then defines for the same ZI

P,(G):= {(E:)+ by letting, for e = v z E E(G),e' belong to (Ez)+((Ez)- respectively) if and only if for the corresponding arc a E A ( D ) ,a' E (A;)+ (u' E (A:)respectively), i.e. if and only if u = ( z I , ~(u ) = (z,v) respectively). This

-

VII.46


D Figure VII.5. D has precisely two eulerian trails, TI and T2 (marked by small and dotted small arcs). K(T,)# T2 for any of the two conceivable K-transformations, but K*(T,)= T2 for each of the two possible K*-transformations K* = ~ " d (compare this with Exercise VII.5). definition is meant to include multiple arcs (in which case each arc of the form (v, x), (z, u ) respectively, has to be considered separately in the above correspondence), as well as loops (v,v)in D (in which case one half-loop belongs to (E:)+ and the other half-loop belongs to ( E z ) - ) . Finally define P ( G ) := Pu(G) .

U

uEV(G)

P(G) is, by definition, a partition system of G such that I K I= i d ( v ) for every I< E P,(G) and every v E V(G)= V(D).Hence, I ( G , P ( G ) )# 0 by Theorem VI.l, and I P ( G ,K ~ ( ~I=) 1 ) by Theorem VII.12. However, precisely because of the definition of P(G) there is a 1-1-correspondence between P(G)-compatible eulerian trails of G and eulerian trails of D , and by the same token, TG,TA E I ( G , P ( G ) ) are rip(C)-associates if and only if the corresponding elements T,, Tb E 7(D)arc K,-associatcs. That is, I P ( D , tcI) I= 1. The theorem now follows.


VII.47

Observe that the above application of Theorem VI.l and the bijection between 7 ( G ,P ( G ) )and 7 ( D )yields a proof of the first part of Theorem IV.8. Another approach to transform an arbitrary T E 7 ( D ) into any other T’ E 7 ( D ) has been chosen in [XIAX84a].l) In this approach one does not have to deal with trail decompositions as in the case of I C ~ transformations; rather, at every step one transforms an element of 7 ( D ) into another element of 7 ( D ) .The tool for this approach is given by the next definition.

Definition VII.25. Let D be a connected eulerian digraph and suppose some T E 7 ( D )can be expressed in the following way:

T = . . . ,u , T ( u ,v), 21,. . .,u,T’(u,v), 21,. . .

,

where T ( u ,v) and T’(u,v) are arc-disjoint trails joining u and v (possibly u = v), and suppose A(T(u,v))UA(T’(u,v)) # A ( D ) (which holds in any case if u # v). Then

T’ = . . . ,u,T’(u,v), v, . . .,u,T(u,v), v, . . . is said to be obtained from T by a .r-transformation which exchanges T ( u ,v) and T’(u,v). We express this transformation by the formal equation T ( T )= TI.

Remark VII.26. 1) Xia speaks of a T-transformation instead of 7transformation; but since we used the greek letter K for various transformations and since T usually stands for an eulerian trail, the use of the letter T instead of T in naming the above transformation seems appropriate. 2) It follows from Definition VII.25 that not only does T‘ E 7 ( D )hold for

T E 7 ( D )and T’ = .r(T),but we even have T‘ # 7’.This is guaranteed by the condition A(T(u,v)) U A(T’(u,v)) # A ( D ) . That is, contrary to ‘1 This paper has been published in Chinese, the English abstract of which, however, is insufficient t o permit a full understanding of the essentials. Doz. Wolfgang Ruppert of Vienna, has translated the most important passages of the paper, and I hope t o reproduce them correctly in the ensuing discussion. In any case, I wish t o express my thanks to Doz. Wolfgang Ruppert for having taken time t o translate major parts of the paper.

VII.48


various previous transformations, a 7-transformation can never be the identical transformation. So the question arises: which are the digraphs having precisely one eulerian trail ? The answer is simple enough: a connected eulerian digraph D has precisely one eulerian trail if and only if every cycle of D is a block of D and A(D) 5 4. The proof of this statement is left as an exercise. 3) However, if T‘ = T ( T )then , for the same pair of trails T ( u ,v): T’(u, u) used in transforming T into T’(see Definition VII.25) we obtain T(T’)= T , i.e. T , in a certain sense, satisfies the formal equation .r2(T)= r(T’) = T . Consequently, it makes sense to call T,T’ E 7 ( D ) .r-associates if either T = T’ or there exists a sequence To,.. .,T,, rn > 0, where Ti E 7 ( D ) for i = 0,. . . ,rn such that To = T , T , = T‘ and Ti = T ( T ~ - ~ ) for i = 1,.. .,m. Again, 7-association defines an equivalence relation on 7 ( D )which in turn defines a partition P ( D ,7) (see Exercise VII.13). The main result of [XIAX84a] is the following. The proof presented here differs, however, from the proof given in the paper cited.

Theorem VII.27. I P ( D , 7)I= 1for every connected eulerian digraph D . Proof. If I 7 ( D )I= 1, the theorem is trivially true since T E 7 ( D )is the only 7-associate of itself, whence we can assume 1 7 ( D )I> 1.

Suppose the theorem fails for D. Then T,T’ E 7 ( D ) exist, which are not 7-associates. Suppose T and T’ have been chosen such that I X , fl XTr I is as large as possible subject to the condition that a ( D ) := C , E v ~ D ) ( o d ( v-) 1) =I A ( D ) I - I V ( D )I is as small as possible. For practical reasons, we can deduce some property D must have attributable to those restrictions.

FOTevery v E V ( D ) ,X,(v)

nX

p (v)= 0

OT

d(v) = 2

.

(1)

Otherwise, let v be a vertex with X,(v) n X,r(v) # 0 and d ( v ) > 2. Let t = {e’, f’} E X T ( v )n X,,(v) and split away the arcs e, f to form a new eulerian digraph D, (note that { e , f } n A: # 0 # { e , f} n A ; ) . The eulerian trails T and T‘ correspond to eulerian trails T,, Ti of D, , hence D, is connected. Moreover, X , = X,, X,; = X p , hence IX , n XTr I= I X , nX,; I. However, a ( D l ) < a ( D ) since I A ( D , ) /=I A ( D ) I, but I V(D,)I=I V ( D ) I +l. Thus, T, and T‘; are -r-associates. We note that for every Tl(z, y) in D, used in one of the .r-transformations which have been employed for the transformation of T, into Ti, we cannot have


VII.49

v E {x,y}, where ve,f E V ( D , ) - V ( D ) , for there are no two arce4 &sjoint open (or closed) trails in D, starting (or ending) at ve,f. Hence T,(x, y) corresponds to T ( z ,y) in D which is an open trail if and only if T,(z,y) is an open trail. Consequently, T, and Ti being 7-associates is tantamount to saying T and T‘ are 7-associates, a contradiction to our assumption. Consider now an arbitrary v E V ( D )-V,(D), and let t E X,,(v) be arbitrarily chosen. Denote t = {e(v)-,e(v)+}where e- E A;, e+ E A: and let t,, t, E X,(v) be chosen in such a way that t , = {f(v)-, e(v)+},t , = { f ( v ) + , e ( v ) - } .Because of (l),t , # t,; and f- E A;, f+ E A:. Now, a run through T starting at v along e+ reaches v along e- before f+ or f- have been passed. Hence T contains a segment S(e+,e-) c 2’. On the other hand, T’, also viewed as starting at v along e+, has e- as its last arc. Hence there exist transitions t 3 , t , E X,,(w) for some

w E v ( ( A ( S ( e + , e - ) ) )n ) V ( ( A ( T) A(S(e+,e-)))) such that t3

where g+ E A$

= (s(4-7 h(W>+} 9

t, = {9(w)+, h ( 4 - 1

n A(S(e+,e - ) ) , 9-

E A;

n A(S(e+,e-)),

h+ E A$ n ( A ( D )- A(S(e+,e - ) ) ) , h- E A, n ( A ( D )- A(S(e+,e - ) ) )

,

and t, = {9(w)+, 9(w>-} E X,(w> (see Figure VII.6 and the proof of Theorem VII.17). Possibly we have v = w. In any case, though, T can be expressed in the form T = v,S ( e + ,e-), v,f + ,. ..,h-, w,. ..,f-,v . That is, there is a proper subtrail of T joining v and w and having f+ as its first and h- as its last arc; denote this subtrail by T’(v,w). Observe that T’(v, w)contains none of the arcs of S(e+, e-). Similarly, S(e+,e-) contains a subtrail T ( v , w ) having e+ as its first and 9- as its last arc. Writing S(e+, e-) = T ( v ,w),w ,T ( w ,u ) we can express T in a more explicit form,

T = v,T(v,W),W,T(W,v),v,T’(ZI,W),W,. ..,j-,v

.

VII.50

VII. Transforinations of Eulerian Trails

D Figure VII.6. Eulerian trails T,T' E 7 ( D ) whose transitions are marked with small (respectively, small dotted) arcs. Possibly v = w. The segment S ( e + , e - ) C T contains g- and g+, but none of f - , f + ,h-h+. Consequently, we obtain T* E 7 ( D ) ,T # T * ,by the .r-transformation which exchanges T ( v ,w ) and T'(v,w). That is,

T* = v,T/(v,w), w,T(w,v),v,T(v,w), w,. . . ,f-,v

.

Possibly T* = T'. In any case, however, we have d ( v ) > 2, d(w) > 2 and t , t , E X,. f l X,, 2 X , n X,, . This and (1) together with the choice of T and TI imply that T* and T' are 7-associates (which is true also if T* = T I ) . Therefore, and because 7 ( T )= T * ,we can conclude that T and T' axe 7-associates. This contradiction to the choice of T and T' finally implies the validity of the theorem. For reasons of time and space, we have restricted ourselves to the above study of transforming 'unrestricted' eulerian trails in digraphs and leave it to the interested reader to find suitable transformations by which one can relate any P(D)-compatible (respectively, (D-D, )-favoring) eulerian trail to any other such trail. Although I have not occupied myself with the search for such transformations, I believe that K- and K*-transformations suffice in order to handle this problem for digraphs in a way similar to

VII.4. Find Remarks and Some Open Problems

VII.51

the case of graphs (note that a 7-transformation can be viewed as the combination of a 6-detachment with a 6-absorption). As for transforming aneulerian trails in a digraph D it suffices to consider 07.The graph GT underlying D+ is a connected eulerian graph because an aneulerian trail in D corresponds to an aneulerian trail in 07 and thus to an eulerian trail in GT, and vice versa. Thus, transforming aneulerian trails in D is equivalent to transforming eulerian trails in GT which can be done by the exclusive use of tc-transformations (Theorem VII.5).

VII.4. Final Remarks and Some Open Problems In the preceding sections of this chapter we developed various types of transformations in order to be able to construct every eulerian trail of a special type, provided one such eulerian trail was given. This approach to handling a whole set 7 of eulerian trails gives rise to the following question: Given T,TI E 7, how often does one have to apply a certain type of in order to transform T into TI ? (*I

K~ -transformations

In other words, what is the complexity of transforming T into T’?We note that a single 6-transformation can be performed in polynomial time, even if one has to search for a vertex u at which a 6-transformation should be performed. Hence the construction of TI, T[ E 7 ( G ) ,satisfying TI= rc(T),TI = rc(T’)(with the 6-transformations performed at u ) , and

can be performed in polynomial time (Exercise VII.15). Hence the original question (*) is the real key to the question about the complexity of &-association. This question has been answered by A. Bouchet for connected 4-regular graphs G. He showed that for any T ,T’ € 7 ( G )it takes at most 6-transformations to transform T into T‘, [BOUCSOa]. Together with this result, however, he presents a conjecture which says that I +l. can be replaced with (V(G) However, analogous complexity studies for the various types of tcltransformations (which have been developed for the various types of

VII.52


eulerian trails in graphs and digraphs treated so far), have yet to be conducted. I suspect that they all will amount to finding polynomial time algorithms. Finally, we note an interesting structural feature concerning rc-transformations. Define the eulerian trail graph G ( 7 , n ) as the graph whose vertices are the elements of 7 := 7 ( G ) ,and T1T2E E ( G ( 7 ,I C ) ) if and only if IC(T,)= T2.Note that @,)' = T2implies 4 T 2 ) = Tl.Hence G ( 7 , IC) is a well defined graph. It follows from Theorem VII.5 that G ( 7 , IC) is connected, since any two eulerian trails of G can be obtained from each other by a sequence of Ic-transformations. However, Zhang and Guo verified a much stronger result which we present without proof (for details, see [ZHAF86a]).') Theorem VII.28. If I7 ( G )I> 2, G ( 7 , IC) is edge-hamiltonian.

Again (similar to Bouchet's result [BOUCSOa]) one is led to the question whether there are results analogous to Theorem VII.28, if one restricts the considerations t o special classes of eulerian trails (such as IX:= 7 ( G ,X), a.s.0.) and the appropriate rcl-transformations, and defines correspondingly a restricted eulerian trail graph (e.g., G('TX,f i x ) ) . Because of Theorems VII.10, VII.12, VII.17, VII.23, VII.24, VII.27 and Corollaries VII.13, VII.18, VII.19, these restricted eulerian trail graphs are connected in any case. But are they hamiltonian ? How can they be characterized, how can one type of restricted eulerian trail graph be distinguished from another ? We note that in [ZHAFSSa] the authors announce that they found a result similar to Theorem VII.28 for the directed euler tour graph, but they do not specify which type of transformation serves as the basis for defining the edges of that graph. In view of Theorem VII.14 and the paragraph following its proof, the following is of relevance to the above questions. Construct the tree graph G, of a graph G, where V ( G T )= (T & G / T is a spanning tree of G}, and T1T2E E(G,),if and only if 1 E(T, UT,)- E(T, nT,)I= 2 (compare this equation with tcA-transformationsand the set '7;. of spanning trees of M i in the discussion leading to the formulation of Theorem VII.14). In fact, tree graphs G, have been already considered implicitly in [OREO62a], explicitly in [CUMM66a]. In the latter reference it has been shown that GT is edge-hamiltonian (a short proof of this result has been given in '1 Zhang and Guo speak of the euler tour graph Eu(G) that we have termed the eulerian trail graph G ( ~ , I c ) .

VII.5. Exercises

VII.53

[SHAN68a]). In view of Theorem VII.14 it is conceivable that the affinity between this result on tree graphs and Theorem VII.28 is not a coincidence. For, the tree graph ( M i ) T of the graph M iis nothing but G(TA(G), I C ~ ) Concerning . other, earlier, studies of tree graphs we refer to [BAR068a, BAR068bl. As for another structural insight regarding tree graphs, we mention [SHAN8la]; there H. Shank proves that if G,, (GT)Tand ( ( G T ) T ) , are all eulerian, ( ( ( G , ) T ) T ) T is not eulerian unless G, E {K,, K 3 } . Finally, we note that for a digraph D, the in-tree graph DT concerning a fixed root q, E V ( D ) can be defined in a way similar to the above GT. This has been done in [DORF74a] where it is shown that DT is always connected. We conclude our considerations by remarking that graphs similar to eulerian trail graphs or tree graphs can be found in linear algebra; there one deals with interchange graphs whose vertices are certain m x n matrices (see [BRUA82a] for details). Also here one is faced with the question of hamiltonicity (regarding interchange graphs). According to Li Qiao, this is a ’big question’.

VII.5. Exercises Exercise VII.1. Show that in Figure VII.2 there is no sequence T I ,. . .,T,, of X‘-compatible eulerian trails of K2 with T = T,, T’ = T, and rc(Ti)= Ti+1,1

Eulerian Graphs and Related Topics: Part 1, Volume 1 (Annals of Discrete Mathematics, Volume 45)

Eulerian Graphs and Related Topics: Part 1, Volume 1 (Annals of Discrete Mathematics, Volume 45)

Eulerian Graphs and Related Topics: Part 1, Volume 2 (Annals of Discrete Mathematics, Volume 50)