~ u l s a Graphs n m
and
'ela6ed rn Topics I
EULERIAN GRAPHS AND RELATED TOPICS Part 1, Volume 1
ANNALS OF DISCRET...
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~ u l s a Graphs n m
and
'ela6ed rn Topics I
EULERIAN GRAPHS AND RELATED TOPICS Part 1, Volume 1
ANNALS OF DISCRETE MATHEMATICS
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 LIN7; 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
NORTHHOLLAND 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
O 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 B.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
To those who cannot be refrained from asking questions. Therefore, also to my son.
Portrait of Leonard Euler (17071783) by the Alsatian painter Franz Bernhard Frey (17161806). The portrait must have been painted during Euler's years in Berlin (17411766); for he had lost one eye in 1735 (during his first stay in St. Petersbug), whereas he lost the second eye in 1766.
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 1976177 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 cooperation with Bill Jackson, the 'Chinese Postman' Guan Meigu, and  recently  Andriis Frank). In the course of these discoveries, it transpired that in various proofs, the application of the FourColor Theorem (or the assumption of the validity of the FourColor Conjecture, if you wish) could be avoided by using the compatibility result. Unfortunately, there is no simple analogy to the compatibility result in the case of nonplanar graphs, however, in recent joint work with Bill Jackson we have put forward a conjecture which contains as a special At this point a remark seems appropriate: I do not have any opinion as to whether the AppelHakenproof of the FourColor Conjecture is correct (thus establishing the FourColor Theorem) or not. I would prefer t o see a proof of that conjecture/theorem which is not computeraided (but which graph theorist wouldn't prefer that ?), or a t least it might be desirable for various independent teams t o try and produce computeraided proofs along the lines adopted by K. Appel and W. Haken (but who has the time and money for it ?).
...
VIII
Preface
case an equivalent formulation of the cycledoublecover 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 1977178 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 threehour 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 first 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 90page version of Eulerian Graphs (originally named Towards Theories o n 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 booli (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 Wenger7s) on arbitrarily traceable graphs, and enumeration in eulerian graph theory, respectively. My colleagues Mr. L. Dirnitrov 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 undertooli joint research, contributed  directly or indirectly  to this bool;: 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, F'ranqois 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
CONTENTS
PREFACE
. . . . . . . . . . . . . . . . . . . . . . . . ziii
I.
INTRODUCTION . . . . . . . . . . . . . . . . . . 1.1
I1.
THREE PILLARS OF EULERIAN GRAPH THEORY Solution of a Problem Concerning the Geometry of Position . . . . . . . . . . . . . . . . . . .
. 11.1
. . 11.12
On the Possibility of Traversing a Line Complex Without Repetition or Interruption . . . . . . . . . . . . . 11.23 From 0. Veblen's "Analysis situs" I11.
. . . . . . . . . . 11.26 BASIC CONCEPTS AND PRELIMINARY RESULTS . 111.1
111.1.
Mixed Graphs and Their Basic Parts . . . . . . . . . 111.2
111.2.
Some Relations Between Graphs and (Mixed) (Di)graphs. Subgraphs . . . . . . . . . . . . . . . . . . . . 111.9
111.3.
Graphs Derived from a Given Graph
111.4.
Walks, Trails. Paths. Cycles. Trees; Connectivity
111.5.
Compatibility. Cyclic Order of Ihim called G e o m e t r y of position. This part of geometry is concerned call be determined by position alone, ancl nit11 precisely with that ~vl~ich the investigation of properties of position; in this matter one should neither be concerl~edwith quantities nor their calculation. However, the lcinds of problems belonging to 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 cluite well known and relates to the following. In Iiijnigsberg 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. Fro~n 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 I { t ~ ) assumption. Consider again the components C' > { v ~ , ~ of  eZ. We claim that e3 E E(Ctt) is not a bridge of C": otherwise, by analogy to the preceding argument (but with e, in place of e, ) we may conclude again that v is a cut vertex. Therefore, since e3 is not a bridge and G. of C", a cycle C3 > {e3) exists in C"; it is also a cycle in However, GI := (E(C'))G and G" := (E(C"))G satisfy G' n G" = { L * ) ; c GI and C3 c G" satisfy Cl n C3 = {v). NOWwe thus C1 := proceed as in case I). The lemma follows since case I ) has been settled and because the last part of the lemma is a direct consequence of 2 ) . It is not true, however, that if H = D is strongly connected then Dl ,,or Dl,, is also strongly connected (and thus weakly connected and bridgeless). This becomes clear if {al, a2,a3) C A t ( g A;) in which case But even if a l E A: ul,? is a source (sink) in both D I Z and may be strongly connected. nor and a2,a, E A;, then neither On the other hand, since a weakly connected eulerian digraph is in fact strongly connected, the above consideration will not have any negative impact. We shall also need another result when splitting arcs away in digraphs. There, the symbol Di,j will mean that we have split away a; E A; and a: E A t . L e m m a 111.2'7 (Double Splitting Lemma). Let D be a weakly connected bridgeless digraph having a vertex v with dl := id(v) > 3 and d2 := od(u) > 3. Let A; = {a;, . . . ,a;,), A t = {a:, . . . ,a i 2 ) , and suppose D l and Dl ,4 are both bridgeless and weakly connected. ') If both (D,,,),,; and (D1,3)2,4are disconnected, then at least one of (D1,4)2,3 and (D1,2)4,3is weakly connected and bridgeless. Proof. To shorten the proof as much as possible we apply parts of the proof of Lemma 111.26. To this end we consider G := G D , GI,, := (GD),,, , ei := e , f j = ea+, a.s.0. Whence assume that both (G1,3)4,2 a. 3 are disconnected. Arguing as in the proof of Lemma 111.26 and we may assume w.1.o.g. that G is loopless, and it follows from conclusion By the Splitting Lemma we may assume that if any Dl,j is disconnected or has bridges, then j = 2.
111.32
111. Basic Concepts and Preliminary Results
2) of this proof that {e4,f,) and {e,, f 4 ) are 2cuts in GI,, . We may even conclude that the blocks B4,2 > {e4, f,) and B2,4 > {e,, f 4 ) of GI,, are different: namely, they correspond to blocks Bi := (E(B4,,)), B;,, := (E(B2,,))in both (G1,3)4,2and (GI,3)2 . Just rkcitll Definition 111.17.3) and observe that c((G1,,),,,) = c ( ( G ~ , ~ = ) ~2,implies ~) B;,, nB;,, = 0 in both (G1,3)4,, and (G1,3)2,4and thus B4,, n B2,4 = {u) in G1 . However, the last equation and Lemma 111.19.3) secure B4,2# B2,4and classify u as a cut vertex. Moreover, is bridgeless and loopless. This and Lemma 111.19.2) imply that B4,, (B2,4) contains a cycle C4,, > {e4,f2) (C2,4> {e,, f,}) such that C4,, n C2,4= {u) holds true in GI,, .
,
l
,
Now consider in the block B1,, > {el, f3) and compare its poB2,4 and v. To this end form F := sition in bc(G1,,) with that of B4., b c ( ~ ~ ,3IUB,,~} ) where ~ ~ 2 E, E4 ( ~ C ( G ~joins , ~ ) )u, B 2 , 4 E v ( ~ c ( G ~ , ~ ) ) . By Proposition 111.23, bc(G,,) is a tree which is an acychc graph implying that uB2,4 is a bridge a& therefore, F is a forest with precisely two components TI and T,. W.1.o.g. B2,4 E V(T2); B4,, E V(Tl) follo\vs since vB4,, E E ( F ) while v and B2,4 belong to different components of F (cf. Theorem 111.18.2)). However, we could have considered vB,,, instead of vB2,4 since B2,4 and B4,, played symmetrical roles so far. In any case, we may assume w.1.o.g. that B1,3 E V(Tl). B1,, # B2,4follows, but B1,, = B4,, may hold. In any case, B1,. contains a cycle Clz > {el, f,} such that C1,, n C2,4 C {u) with equallty holding only if v E V(B1,3). We also note that dBz,4(u)= dB4,2(v)= dB,,, ( ~ 1 , = ~ 2) since {e,, f4) and {e4,f2) are 2cuts. For = {v2,41u1,3) v((G1,3)2,4) define := ((G1,3)2,4)~o( ~ 2 ~is4 the new vertex obtained by splitting away e, and f,). In fact, bc(G,) satisfies the equation bc(Go) = F U { V ~ , U ~ B ~ where ~ , U vo ~ is B the ~ ~ ~ , vertex obtained by contracting Vo, and the notation concerning the blocks is retained in the transition from to G o . By construction, TI, T2 bc(Go); thus every vertex of T1 can be joined to v by a path. This is also true for vo since adjoining vo,v,B1,, to a path P(B1,,, V) yields a required path. Since B2,4E V(T2) every vertex of T2can be joined in T, to B2,, by a path which can be extended via vo and P(B1,,, u) to a path ending in u. Thus bc(Go) is connected (cf. Lemma III.15), and the edge sets of the blocks are the same in and Go (only the position of the blocks has been changed). However, forming (G1,4)2,3is now tantamount to splitting vo appropriately into v1,4 and v2,3 such that EvIs4= {el, f,), Ev2,3= { e , , f3). For
111.4. Walks, Trails, Paths, Cycles, Trees; Connectivity
111.33
the sake of simplicity we retain the symbols for the objects considered so far. Thus, in Go we have C2 n C1,3 = {vO) since B2,4r l B 1 , = {v,). Therefore, forming (G0)i,4 = (G1,4)2,3 (the splitting operation being persuch formed a t vo) transforms C2, and Clz into two paths P2,?and PlY3 that P 2 , 4 n Piy3= { v ~ ,v~~, , ~ )It. now follows from 2) in the proof of the Splitting Lemma that (G1,4)2,3 is connected. This, the last equation and conclusion 4) of the proof of the Splitting Lemma imply that (G0)1,4= (G1,4)2,3is bridgeless. This finishes the proof. We define another mixed graph in connection with cut sets.
Definition 111.28. Let H = V U K be a weakly connected mised graph and let be a cut set of the form Ic0 = I
>
+
Definition 111.34. 1 ) Two paths PI and P2are called internallydisjoint or simply disjoint (or independent) if Pl n P2 C V, (PI) n V, (P,). They are called totally disjoint if Pl n P, = 0. 2) Let x, y be two nonadjacent vertices of the mixed graph H (i.e., $ GH for p = pH). The local connectivity ~ ( xy), is defined as the smallest vertex cut V, such that x and y belong to different components of H  V,. We also say V, separates x f r o m y. The next theorem plays a central role throughout graph theory. T h e o r e m 111.35. The following statements are true for a simple rnised graph H and k E N, k 5 p~  1. 1) For every pair of nonadjacent vertices x, y E V(H), K(X,y) is the maximum number of pairwise disjoint paths in GH joining x and y, and ~ ( xy), 2 n(H) (Menger's T h e o r e m [MENG27a]). 2) H is kconnected if and only if for every x, y E V(H), x # y, there are k pairwise internallydisjoint paths in G joining x
111. Basic Concepts and Preliminary Results
111.36
and y , [WHIT32a]. '1 Quite a few proofs of Menger's theorem exist; we refer the interested reader to [DIRA66a] and [McCU84a]. We note that the former reference lists several early proofs of this famous theorem and that the latter reference contains a very short proof of this theorem for digraphs from which the original theorem can be deduced easily (in the context of digraphs, however, the definition of a set of vertices separating x and y as given in [McCU84a] differs from Definition 111.34.2)). However, we also shall use other types of connectivity.
Definition 111.36. 1) Let H be a mixed graph and let I, be the set of edge/arc cuts of H. We define
and call this number the edgeconnectivity of H. Moreover, we say that H is kedgeconnected if X(H) 2 k. Moreover, X(x, y)  the local edgeconnectivity  is the size of a smallest Ii, E E, separating x from y (i.e. x and y belong to different components of H  I 1. Thus C 'absorbs' at least two edges of E,U E,. 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 Zcut 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 Zcut E, such that x and y belong to different components of G E,. 2 = K(G) = X(G) follows. However, no component Co of G Eois a tree, where Eois an arbitrary 2cut; otherwise, we may conclude as before that C, is joined to G  V(Co) by at least four edges: again a contradiction. Thus X(G) 5 X,(G) 2 from which 2 = K(G) = X(G) = X,(G) follows. This proves the lemma.
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 0+and 0isomorphisms. However, if we have an embedding GI of a planar 3connected graph G on the torus, the corresponding O+(G1) may satisfy O+(G) # O+(Gl) # 0(G), where O+(G) and 0  ( 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). T h e o r e m 111.53. Let a nontrivial connected graph G be given together with O+(G). There esists a surface 3 and a 2cell embedding GI of G on 3 such that G = $ G I . We will frequently use the following result regarding plane graphs. L e m m a 111.54. Let G be a plane graph. The following is true. 1) If G is 2connected 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 Fo such that every covering walk Wo of bd(Fo) passes a vertex at least twice. Such vertex is a cut vertex of G. If Wo passes e E E(bd(Fo)) at least twice then e is a bridge of G. Conversely, if v (e) is a cut vertex (bridge) of G then there is some face Fo, such that v E V(bd(Fo)) (e E E(bd(Fo))) and every covering walk of bd(Fo) passes v (e) at least twice.
3) K(G) = 0 if and only if a face Fo exists such that bd(Fo) 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 (nonorientable 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 y(G), the nonorientable 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 y(G) = y(G) = 0, are characterized as follows.
111.54
111. Basic Concepts and Preliminary Results
Theorem 111.55. (Kuratowski's Theorem). A graph G is planar if and only if it has no subgraph homeomorphic to K3,3or I&. 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 I 0, it follows that P ( F ( G ) ) is uniquely determined up to permutation of the color classes of F(F(G)). Define the natural 1factorization of G , L := P ( E ( G ) ) := {L,, L,, L3) by letting e E Li if and only if the two faces F' and F" for which e E bd(F1)flbd(F"), satisfy {F',F") F j u F k and {i,j, k) = {1,2,3). Thus, if F E Fj, then E(bd(F)) is a cycle in which edges of Liand Lk alternate. Since the 3facecoloring P ( F ( G ) ) can be defined via a natural 1factorization of G, we arrive at the next theorem. Theorem 111.67. The following statements are equivalent for a plane, connected, cubic bipartite graph G.
1) G has a 3facecoloring (it does have one by Theorem 111.66). 2) G has a natural 1factorization. 3) G has a 2factor Q such that every C E Q is a face boundary and l(C) is even.
4) D(G) has a 3coloring.
We also characterize the plane graphs G with xF(G) = 2.
111.62
111. Basic Concepts and Preliminary Results
T h e o r e m 111.68 (TwoFaceColoring Theorem). A plane graph is 2facecolorable if and only if it is eulerian. Theorem 111.68 follows, for example, from the facts that for a plane graph G, D(G) is bipartite if and only if G is eulerian, and that 2vertescolorings are equivalent to being bipartite. Moreover, Theorem 111.68 and Theorem 111.66 are equivalent. This follows from Theorem 111.67 and the construction of G/Q which is a plane eulerian graph for plane bipartite cubic G and Q = LiULj, i # j , where Li, L j E C. = {L,, L,, L3) and L is the natural 1factorization of G (note that the inverse operation yielding G from G/Q yields a plane bipartite cubic graph provided G/Q is a plane eulerian graph). In this context, G/Q or G/L (where G is a plane graph and Q is a 2factor of G whose elements are face boundaries or L is a 1factor of G) is defined in the obvious way suggested by the topology of the plane and the definition of O+(v), v E V(G). Instead of giving a formal definition, we refer to Figure 111.17. In fact, Theorem 111.68 admits the construction of a special type of eulerian orientation in plane eulerian graphs. Let G be a plane eulerian graph, and suppose a 2facecoloring is given such that the unbounded face has color 1. The faces F belonging to the second color class F2(G) are thus all bounded faces and S = {bd(F)/F E F,(G)) is a trail decomposition of G since every e E E(G)belongs to precisely one bd(F), F E F2(G). Replace e with an arc a, such that when passing a, from the tail of a, to the head of a, the face F lies on the lefthand side. Doing this for every e E E(G) yields a counterclockwise orientation of every bd(F), F E F,(G) and thus an orientation DG of G. DGis eulerian because for every t = {e', f')E X(v) C X s , v is the head of a , if and only if v is the tail of a j . Moreover, it follows from the definition of DG that in O+(v) = (el(v), ez(v), . . .,ed(v)), d = d(v), v is the head of ei if and only if v is the tail of ei+,(v), i = 1,.. .,d (setting d 1 = 1). Thus in O+(v) E O+(DG) ingoing and outgoing halfarcs alternate. We shall refer to DG , D g respectively, as the canonical orientation of G.
+
111.9. Harniltonian Cycles We start with a relation between hamiltonian cycles and dominating cycles. For this purpose we call the graph K,,, the nstar (or just s t a r ) .
111.9. Hamiltoniall Cycles
Figure 111.17. Contracting a) a face boundary C, b) an edge e, in a plane graph G. Lemma 111.69 [HARA69a, p.831. A loopless graph G has a nontrivial dominating trail if and only if L(G) is hamiltonian and G is not an nstar for n = pc  1. Hamiltonian cycles are another central theme in graph theoretical re
111.64
111. Basic Concepts and Preliminary Results
search. Owing to an erroneous assumption on the part of Tait the search for hamiltonian cycles in planar graphs has been greatly influenced by the search for a proof of the Four Color Conjecture. In fact, Tait believed that every planar, 3connected cubic graph G had a hamiltonian cycle H (Tait's Conjecture). If true, Tait's conjecture would have solved the Four Color Problem. For if H is a hamiltonian cycle of the planar cubic graph G, then G has a 1factorization C = {L1, L,, LJ) such that L1 U L, = E ( H ) and L, = E(G)  E(H) (note that pG must be even by Lemma 111.4 and because G is 3regular). Thus, by Tait's Equivalence Theorem (Theorem 111.65) any embedding of G on the plane is 4facecolorable. Note that the existence of hamiltonian cycles and of 1factorizations is independent of any embedding, whereas a facecoloring depends on an actual embedding. However, Tait's conjecture was disproved by Tutte in 1946 who subsequently proved that every 4connected planar graph is hami1tonian.l) This result is a direct consequence of a more general theorem for which we need some additional notation. Consider a graph G other than a forest, and let C be a cycle of G. Form G1 := G  V(C) and for every component B1 C G, let
That is, B arises from B, by adding all vertices of C adjacent to a vertex of B1, and all edges joining these additional vertices to V(Bl). We call v E V(B) n V(C) = V(B)  V(Bl) a v e r t e x of Cattachment (or just v e r t e x of a t t a c h m e n t if no confusion arises) and we call B a C bridge (or just bridge).2) Note that for two Cbridges B1 and B2 we have E(B1) f l E(B,) = 0, and every v E B1 f l B2 (if such v exists) is a vertex of attachment. A chord or diagonal of C, i.e., an edge e $ E ( C ) such that h ( e ) V ( C ) ( h is the incidence function of G), is also called Since then many planar, 3connected, cubic graphs without hamiltonian cycles have been developed by various authors. We quote [ZAKS82a] concerning references on this topic and note that in this paper a graph of the said type has been developed such that it contains only face boundaries of a length not exceeding 7. 2, It will become clear from the context that Cbridges = bridges have nothing in common with bridges as defined earlier. Tutte's terminology has caused some graph theorists (myself included) headaches. The best cure for these headaches, however, is to enjoy his wonderful results.
111.9. Hamiltonian Cycles
111.6.5
a Cbridge. Finally, considering e E E(G) and the faces F', F" uith e E bd(F1) n bd(FU), we call a cycle Co a terminal cycle of e with respect to F' (Ffl) if a) e E E(C0), b) F' E intCo and F" E extCo, or F" E intCo and F' E extCo, c) the 'side7 of Co containing F' (F") i.e. intCo or extCo  contains as few faces as possible. Note that if G is 2connected, then such Co is nothing but bd(F1) (bd(Fl1)). The next result is the essence of what is quite commonly called Tutte's "Bridge Theorem" (see [TUTT77a7Theorem I]. This paper is an updated version of [TUTT56b] which already contained Corollary 111.71 below).
Theorem 111.70. For every edge e contained in some cycle of a planar graph, there is a cycle C such that 1 ) C contains e; 2) every Cbridge has at most three vertices of attachment; 3) if E(B)
n E(Co) # 0 for a Cbridge B
and a terminal cycle Co of e, then B has exactly two vertices of attachment.
Moreover, from this theorem we can easily deduce the following results which will be of relevance in the course of this book.
Corollary 111.71. Every (simple) planar 4connected graph G has a hamiltonian cycle H. Proof. Let C be a cycle as described in Theorem 111.70 and assume that some Cbridge B is not a chord of C. Then B n C is a vertex cut V, with I V, 1 < 4 unless B n C = V(C) and every e E E(G)  (E(C) U E ( B ) ) is a chord of C (Theorem 111.70.2)). By hypothesis, C is a triangle, the two face boundaries C' and C" of G containing e are terminal cycles and satisfy (E(C1) n E(B)) U (E(Cl1) n E(B)) # 0 . By Theorem III.70.3), B has exactly two vertices of attachment implying that C contains a digon which contradicts the fact that G is simple. It follows that every e E E ( G  C) is a chord of C; whence H := C is a harniltonian cycle of G. Corollary 111.71 generalizes the next result due to H. Whitney [WHIT3la].
Corollary 111.72. Let D be a simple triangulation of the plane such that every triangle of D is a face boundary. Then D is hamiltonian.
111.66
111. Basic Concepts and Preliminary Results
Noting that the hypothesis of this corollary amounts to saying that K(D)2 4 for D # I
It should be noted that the preceding results (Tutte's Bridge Theorem included) also follow from a more general result of C. Thomassen which in turn generalizes Corollary 111.71 (see [THOMS3a] and [CHIB86a]; the latter paper corrects a flaw in the former). The following fol1;lore result relates hamiltonian cycles in a plane cubic graph to trees in D(G).
Lemma 111.74. The following statements are equivalent for a plane cubic graph G. 1) G is hamiltonian.
2) V(D(G)) has a partition {V,, V,) such that both (V,) and (I/,) are trees.
Moreover, given a partition as described in 2) and F' := {F, E F(G)/u E Vl), the edges of G belonging to the boundary of precisely one element of F', define a hamiltonian cycle of G.
Proof. 1) implies 2). Consider a hamiltonian cycle H c G and view it as a Jordan curve. Thus H defines a partition of the set of its chords,
such that every e E Eb lies in i n t H , whereas every f E E, lies in extH (here, e and f are viewed as open edges). Analogously, H defines a partition of the set of faces,
where the subscripts have the same meaning as above. Let V, C V(D(G)) be the vertex set corresponding to Fb C F(G), let V2 be defined analogously with respect to F, and consider (V,), (V2) C D(G). W.l.o.g., if
111.9. Hamiltonian Cycles
111.67
one of these graphs contains a cycle C , then it is (Vl). However, viewing D(G) constructed in the plane (see Definition III.59), it follows that C can be viewed as a Jordan curve composed of capitals of certain elements of Fband (topological) arcs eFp1 corresponding to certain elements of Eb . By definition, eFFl n e # 8 if and only if e E E(bd(F)) n E(bd(Ff))n Eb , in which case I eFFl n e I=[ ~ F F n , E I= 1. This equation and Theorem 111.61 imply, however, that the vertices x and y incident with e lie on different sides of C (note that V(G) n D(G) = 8 by definition). Assume w.1.o.g. that x E intC. However, V(H) = V(G) > {x, y) implies that there is a path P joining x and y in H. Consequently, H r l C # 8 by Theorem 111.61. On the other hand, the definition of D(G) and (Vl) implies C C intH. This and C n H # 8 cannot be true simultanously. It follows that (V,) is acyclic. By analogy, (V,) is acyclic. Since G is cubic, D(G) is a triangulation of the plane. It now follows from Proposition 111.63.1) that (V,) and (V,) are trees. The implication now follows. 2) implies 1). Suppose (Vl), (V2) C D(G) are trees for some partition {V,, V,) of V(D(G)). Let F' C F ( G ) be as defined in the statement of the lemma. By a straightforward inductive argument (which we omit for the sake of brevity) it follo~vsthat F' defines a cycle H C G such that e E E ( H ) if and only if e E E(bd(F,)) for precisely one F, E F', v E V, . View H again as a Jordan curve and assume w.1.o.g. that (V,) C i n t H (note that for every F, E F', Fvn i n t H # 0 if and only if F, f l (extH u H ) = 8, in which case F, c i n t H follows). Suppose H is not a hamiltonian cycle. It follows from the construction of H and by assumption that x E (V(G)  V ( H ) )n extH exists (note: Go := (UFyEFI E(bd(Fv))) is a 2connected outerplane graph). Therefore, x E V(bd(Fw)) for some w E V2. In fact, since F, c i n t H for every v E Vl , it now follows that every face Fwsatisfying x E V(bd(Fw)) also satisfies Fw C extH and w E Vz . This means, however, that ({w E V2/x E V(bd(Fw)))) contains a cycle which is a cycle of (12).This contradicts the hypothesis that (V2) is a tree. Hence H is a harniltonian cycle. The lemma now follows.
We now turn to hamiltonian cycles in arbitrary simple graphs. First, we state a famous theorem by 0. Ore which generalized a result established by G.A. Dirac and greatly influenced hamiltonian graph theory. Theorem 111.75 (Ore's Theorem). If G is a simple graph of order p such that d(x) d(y) 2 p whenever x and y are nonadjacent, then G is hamiltonian.
+
Proof. The theorem is true if G e K p . Proceeding indirectly we consider
111. Basic Concepts and Preliminary Results
111.68
G with maximal qG (pG = p is fixed) for which the theorem fails. Hence x, y E V(G) exist such that Ez n E, = 0 implying that G, = G U {xy) is a simple graph; it has a hamiltonian cycle HI > {xy) by assumption and the choice of G. Writing P := HI  {xy) as a vertex sequence (in which x = vl, y = up),
1. Assume a E N U (0) exists such that k > 2a and b(G) 2 k  a . Then G is hamiltonian. Proof. We proceed indirectly. Since the corollary holds if G 11. I i k , k , let G be chosen such that qG is maximal and the corollary does not hold for G. Consider nonadjacent vertices vl E Vl , v2 E V2 and form G1 := G U {vlv2). It follows from the choice of G that G1 is hamiltonian and every hamiltonian cycle of G1 contains vlv2. Thus, G has a hamiltonian path which can be written as a vertex sequence
where v, = v1,1,v2 = and vi,j E 5 , 1 5 j 5 k, i = 1,2. It follows that { V ~ U ~ V, ~~ , U ~ , ~ E(G); ) otherwise, (E(P)  { ~ , , ~ v , , ~U) ) {vlv2,j, v ~ v ~ defines , ~ } a hamiltonian cycle of G contradicting the choice of G. Consequently,
That is, k
< 2a2 which contradicts the hypothesis. The corollary follows.
Concerning hamiltonian cycles in digraphs there is an analogy to Theorem 111.75, [MEYN73a]; a short proof of this result is given in [BOND77a].
111.10. Incidence and Adjacency Matrices, Flows and Tensions 111.69
Theorem 111.77 (Meyniel's Theorem). If D is a strongly connected simple digraph of order p, and if d(x) d(y) 2 21,  1 for every pair of nonadjacent vertices x, y E V(D), then D is hamiltonian.
+
+
Corollary 111.78. If D is a simple digraph of order p such that d(x) d(y) 2 2p3 whenever x and y are nonadjacent, then D has a hamiltonian path. For a survey on hamiltonian graph theory we refer the interested reader to JC. Bermond's article in [BEIN78a, Chapter 61.
111.10. Incidence and Adjacency Matrices, Flows and Tensions There are various connections between graph theory and other branches of mathematics as well as other sciences (see e.g. [WILS79a]). For the purposes of this monograph (especially Chapters VIII and IX, which are contained in Volume 2), certain links with linear algebra and linear programming (LP) are of particular interest, this being the case not only for the sake of implementing graph algorithms, but also for proving graph theoretical results. We restrict our considerations to graphs and digraphs and begin with the representation of graphs by matrices.
Definition 111.79. 1) Let G be a graph whose sets of vertices and edges are given as lists, V(G) = {v,, . . .,v,), E(G) = {el,. . . ,e,}. Let B(G) be a p x q matrix such that the rth row represents v, , 1 5 r 5 p, the sth column represents e , , 1 5 s 5 q, and b,,, = 1 if and only if e, E Evv ; otherwise b,,, = 0. Similarly, if D is a loopless digraph, V(D) = {v,, . . . ,up}, A(D) = {a,, . . . ,a,}, then the p x q matrix B(D) has b,, = 1 if and only if a, E A t r , b , , = 1 if and only if a, E A; ; otherwise b , , = 0. We call B(G) (B(D)) the incidence matrix of G (D). 2) Let G, D be given as I), but now D need not be loopless. Let A(G) and A(D) be p x pmatrices in each of which the ith row and ith column represent vi ; the respective entries satisfy a,,,(G) =I E,, n E, I and a i j ( D ) =I A: n A; I if i # j, 1 5 i, j 5 p, whereas ai,,(G) = ai,i(D) = A,;. We call A(G) (A(D)) the adjacency matrix of G (D). We malie the following elementary observations.
111.70
111. Basic Concepts and Preliminary Results
1) The ith row sum of B(G) (B(D)) equals d(vi) (od(v,)  id(vi)), 1 5 i 5 p, whereas the jth column sum of B(G) (B(D)) equals 2 (0) provided G (D) is loopless, 1 j q.
<
Readers interested in flows and tensions (particularly with respect to operations research) are referred to [BERG62a7 FORD62a, EVEN79al and AvondoBodino's survey article in [WILS79a]. We note, however, that all these maximum flow problems can be reduced to LP problems. In various instances, the latter will be formulated but not discussed in detail.
111.11. Algorithms and Their Complexity Over the past 25 years, graph algorithms and the study of their complexity have been an evergrowing subject. Quite a few books have been published on this topic (or, at least, they have this topic one of its main themes). However, although we develop algorithms in various chapters Again, for such arcs a one usually assumes infinite capacity. The lower bound for P(a) given here does not imply any restriction on any flow.
111.11. Algorithms and Their Complexity
111.73
(not only in chapter X) and study (or, at least, mention) their complesity, this is not to be considered one of the main tasks of this monograph. Therefore, for the sake of brevity of this chapter, we refer the interested reader to R.C. Read's survey article in [WILS79a] and to [EVEN79a, PAPA82a, JUNG87al. These references also contain more or less explicit complexity considerations  including an introduction to the theory of NPcompleteness. Nonetheless, we refer readers interested in a thorough study of NPcompleteness to the Lclassic' [GARE79a]. Hence, although no clear answer to the Shakespearian query ("To P or to N P ? That is the question !") may have been obtained upon concluding this monograph, we assume from now on that the reader is acquainted with the class N P and its subclasses P (the good guys) and NPC (at present the not so good guys, for whom there's little hope of ever being good guys), as defined in [GARE79a, EVEN79a, PAPA82al. We note, however, that S. Even also quotes two other concepts of NPcompleteness, due to Cook, and Aho, Hopcroft and Ullman. In the following, we summarize (without proof) some graph theoretical decision problems, all of which are in P. As for problems in NPC we quote them as they 'come along' in the context of this monograph. The explicit complexity of polynomial algorithms will also be quoted in the relevant places. Theorem 111.87. Let H be a mixed graph. Polynomial time algorithms exist for a) determining K(H),X(H);
b) determining whether H is strongly connected (if H is a digraph, say); c) finding a spanning tree;
d) finding an intree or outtree rooted at v,, E V(H) if H = D is a strongly connected digraph. From now on, whenever Q is a set of real numbers, we set Q+ := { T E QIr > 01. Theorem 111.88. Let G be a graph, and let f : E(G) t R+ U (0) be given. For Eo E(G), define f (Eo) = CeEEo f ( e ) . Polynomial time algorithms exist for determining
111.74
111. Basic Concepts and Preliminary Results
a) the distances between all pairs u, v E V(G) (i.e., paths P = P ( u , v) such that f ( E ( P ) )is minimum);')
b) a minimum cost spanning tree (i.e., a tree T such that f ( E ( T ) ) is minimum); c) a minimum cost perfect matching if G has one (i.e., a 1factor
L such that f (L) is m i n i m ~ r n ) . ~ ) We note that problems such as determining whether a given set of edgeslarcs and vertices is separating (thus, in particular, whether k E I 2. L,et el, e,, e, be three edges incident with v. By the Splitting Lemma, at least one of the graphs Gig and is connected; say, GI,, is connected. Since a(G1,,) < o(G), and Gl,z is also eulerian, we conclude by induction that contains an eulerian trail Since E(Gl,,) = E(G) and any pair of adjacent edges of is also such a pair of G, therefore, TI,, induces an eulerian trail T of G. Hence, 1)implies 2). 1) implies 3). We proceed by induction on a(G) as in the above proof until we reach GI., (in the case G = K,, S = 0 is the unique cycle
IV. 2
IV. Characterization Theorems and Corollaries
,,
decomposition of G). By induction, has a cycle decomposition S, . Every C E S1,2not containing vl,2 is a cycle in G as well. Let Cl , E Slz denote the cycle with v , , ~ E V(C1,2). If v $ V(Clt), then c,,? corresponds in G to a cycle Ci,2 with v E V(C;,,). Otherwise, V(C1,2) > { v ~ ,v) ~ ,and C1 corresponds to a subgraph H of G in which v is 4valent and the other virtices are 2valent. Hence H can be decomposed into two cycles C1,C2 with ei E E(Ci), i = 1,2 (just start a run in v along el and stop after reaching v again; this gives C1, and C2 = H  C1). Consequently, we obtain a cycle decomposition of G by defining in the case u $ V(C1,2)
and in the case v E V(C1,2)
2) implies 1). Consider a run through a fixed eulerian trail T starting at v E V(G). For T = v, we have G = Iil,for which the implication is true. Hence we assume q > 0. For every x E V(G), x # v (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 v we have a pairing of the edges incident with v: 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 = {C, , . . .,C,) be a fixed cycle decomposition of G (possibly S = 0). For every v E V(G) and every i = 1, . . .,t , we V(C,) or v E V(C,). have dci(v) = 0 or 2 depending on whether u Consequently, since 1
we have
t
This finishes the proof of Theorem IV.1.
IV.l. Graphs
IV.3
We note in passing that in the proofs of the preceding two implications, a loop vv or xx was counted twice which is in accordance with the convention that a loop contributes two units to the degree of a vertes. We also note that if G is disconnected, then it is eulerian if and only if it has a cycle decomposition (see Exercise IV.l). 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 blocli 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 [I 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, (x), E, (y), E, (xy) where E,(x) consists of the edges of G incident with x 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.l). 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,(x) 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 = IE,(x)l, let C(ei, ej) denote a cycle of G, containing the two different edges ei, e j E E, (x), and let yG(e) (YG,(ei)) denote the number of cycles of G (G,) containing the edge e (ei E E, (x)). By hypothesis we have
and by induction, yG, (ei)

1 (mod 2)
.
IV.l. Graphs
Figure IV.l.The graph G,
Therefore we obtain (by observing that IE,(xy)l = X  1)
EL,
(note that in yGe(ei) every cycle C(ei, ej) is counted once in rGe (ei) 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 3.41). 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 a t 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 C1, . . .,C, be the cycles containing e. By Theorem IV.5, r 1 (mod 2). We proceed by induction on q = IE(G)I.
IV.6
IV. Characterization Theorems and Corollaries
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 r = 1 and G is a cycle with S = {C1) as its unique cycle decomposition. Assume, therefore, that Gi # 0, 1 5 i 5 T . By induction, Gi has an odd number of cycle decompositions (note that Gi need 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,
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, McI(E84aj. 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 TG denote the number defined in the statement of the theorem, and let ~ ~ (denote e ) 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 =(V(G) (= 1 , then G = K l , q = 0, and E(G) = 0 is the only subset contained in the unique spanning tree G = IC1. Hence we assume p > 1, q > 0 (Note that K, is eulerian). Let e = xy be an arbitrary edge. Form G, by contracting e with z E 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 Ge is disconnect,ed if and only if e is a bridge of G. 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.1. 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 11correspondence between spanning trees of G containing e and spanning trees of G,. Consequently, we have by the above and by induction
with T
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 TG = 0 (mod 2), 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 G, is eulerian as well, but G  e is not eulerian. By induction and (2), (3) we then have rG
= 1 + 0 = 1(mod 2) ,
in accordance with the statement of the theorem.
IV.8
IV. Characterization Theorems and Corollaries
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 x and y are the only odd vertices of G. Consequently,
lee.,
rG

0 (mod 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 [BOND86a] 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 i f 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 [ K O N I ~p.29 ~ ~ 1. , Theorem IV.8. For a weakly connected digraph D , any two of the following statements are equivalent:
1) id(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
Proof. 1) implies 2). Proceeding by induction on
we first note that if a(D) 5 0, then D = I 0; consider any v E V(D) with id(v) = od(v) > 1, and let a, = (v,, v) be any arc incident to v, and let a2 = (v, v2),a 3 = (v, v3) be any two arcs incident from v. Both Dl,: and are eulerian, and by the Splitting Lemma at least one of them is weakly connected; say, is weakly connected. By induction, since a(D1,?) < a(D), Dl,z has an eulerian trail TI,, which 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 a ( D ) > 0. As in the proof of the preceding implication we apply the Splitting Lemma and assume w.1.o.g. that is weakly connected. Let S1,2be a cycle decomposition of Arguing as in the proof of Theorem IV.l we conclude that the cycle Ci,:! E S1,2 corresponds in D either to a cycle C1,2or to a subdigraph of D which is the arcdisjoint union of two cycles C,, C,. Hence, also for digraphs we obtain a cycle decomposition S by defining
depending on whether Cit2 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 4edgeconnected 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
IV. Characterization Theorems and Corollaries
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 Vl, V2 contains as many arcs incident from Vl as it has arcs incident from V2 (in other words: i = 1,2).
a+(q) = a(5)
for
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, = { u ) or V, = {v) for arbitrary v E V(G). Say, V, = {v). By hypothesis, there are as many arcs of A,  A, incident from u as there are arcs of A, incident from V(D)  {v), i.e., incident to u. That is, od(v) = id(v) for arbitrary v 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(Ao), the tieup of D with respect to A, (see Figure IV.2.).
Figure IV.2. The eulerian digraph D and its tieup D(A,).
For i = 1,2, we conclude from the fact that odDi(v) = odD(v) = idD(v) = idDi(v) for every v E y,that odD,(2;) = idDi(z;) holds as well (see Corollary III.5a). Since an arc of A, is incident from 5 , to 5 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 K, namely idDi(zi) = odD,(zi),i = 1 , 2 (note that this equation is independent of i because idD,(z,) = odDz(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 I$,4,, r 2 1. The same digraphs sho~v that there is no analogue of Corollary IV.6 to digraphs either; for they have an even number of cycle decompositions (see Exercises IV.7 and IV.8). As for an analogue of Theorem IV.7 to digraphs we discuss the digraphs of Figure IV.3.
Figure N . 3 . Two eulerian digraphs.
For both D, and D2we look at the spanning outtrees rooted at v, w, x respectively, and count the number . r ~ ~ of ( t subsets ) of A ( D i ) contained in some spanning outtree rooted at t E {v,w,x),i = 1,2. For t = v we
IV. Characterization Theorems and Corollaries
IV.12
have in Dl the corresponding subsets
and in D2,
That is, rD,(v)
3 rD2 (v)
+ 1= 0 (mod 2) .
Hence, by symmetry we have for arbitrary t E (u,w,x) and i = 1,2 rDi(t) = i
+ 1 (mod 2)
.
To determine for i = 1,2 the parity of rDi, the number of subsets contained in some spanning outtree, we observe that no subset of size at least three can be counted in rDi,and that no subset of size two counted in rDi(t), t E {v, w, x), is counted in rD.(t'), t' E {v, w, x), t' # t. Therefore we count in TD1 nine subsets of size two and in rD1twelve such sets. Since every oneelement subset and the empty set must be counted in rD, we obtain for i = 1,2
= + 1(mod 2) .
r~~ i
Whence we conclude that there is no simple analogy (if there is any) of Theorem IV.7 for digraphs. It is interesting to note however, that the number of subsets (including the empty set) of A(Di), i = 1,2, each of which is contained in either a spanning outtree or spanning intree of Di, is  3) 7 = 19. But, under the same condition, the same number is achieved for any noneulerian digraph obtained from Di by arc reversals.  We note already now that the digraphs of Figure IV.3 are distinguished in another respect as well: namely, the number of spanning outtrees rooted at t E {v, w, x) is 3 for Dl and 4 for D2. This fact will play an important role in the question of how to enumerate eulerian trails in graphs.
((i)
+
IV.3. Mixed Graphs
IV. 13
IV.3. Mixed Graphs Turning now to mixed graphs we have the following result (in [FLEI83b] I attributed it to Batagelj and Pisanski, [BATA77a]; they as well as myself were not aware at that time, however, that this result already appears in the book Flows in Networks by Ford and Fulkerson, [FORDGZa, Theorem 7.1, p.601.  This mistake of quotation was pointed out to me after [FLEI83b] had been published in [BEIN83a]).
Theorem N.ll. Let H be a weakly connected mixed graph. Then any two of the following statements are equivalent. 1) For every X C V(H), fH (X) := e(X)a nonnegative even integer.
1 a + ( X )  a(X) I is
2) H has an eulerian trail T. 3) H has a cycle decomposition S.
Proof. Let T be an eulerian trail of H starting at v, say, or let S be a cycle decomposition of H. A run through T or any C E S induces an orientation on the elements of E ( H ) ,E(C) respectively (in the case where A(C) = 8, choose arbitrarily one of the two possible orientations). Doing this for every C E S, we obtain by Theorem IV.8 from either T or S an eulerian digraph Dl with A(H) c A(Dl). In Dl we have for every X c V(Dl) = V ( H ) ,a t ( X ) = a r ( X ) (the subscript 1 refers to Dl in order to avoid confusion with a+(X), a(X) respectively). Thus we have
where et(X)
+ eff(X)= e(,Y);
consequently
implying 0 5 2eIf(X) = e(X)  (a(X) and 0
 a+(X))
< 2ef(X) = e(X)  (a+(X)  a(X))
.
IV. 14
IV. Characterization Theorems and Corollaries
Therefore, e(X) I a+(X)  a(X) I is a nonnegative even integer. Thus we have shown that 2) implies 1) and 3 ) implies 1 ) . On the other hand, if we have in H an eulerian trail T (or a cycle decomposition S), then we obtain from T (S) as above an eulerian digraph DT (DS) in which the original T (S) induces an eulerian trail TD (cycle decomposition SD). By Theorem IV.8., TD(SD) is equivalent to the existence of a cycle decomposition S ( D ) (an eulerian trail T(D)), which in turn induces in H a cycle decomposition ST (an eulerian trail T s ) , simply because A(H) C A(D) for D E {DT,DS). Hence we have shown the equivalence of 2) and 3). To finish the proof of the theorem it suffices to show that starting from I), one can produce an eulerian digraph D by orienting the elements of E(H) in such a way that A(H) C A(D). The application of Theorem IV.8 to D then yields an eulerian trail T or a cycle decomposition S in H the same way as in the preceding paragraph. Suppose there is a mixed graph H (not necessarily connected) satisfying the condition stated in 1) which cannot be transformed into an eulerian digraph D with A(H) C A(D). Among all possible counterexamples let H be chosen in such a way that I V(H) I I E(H)I is as small as possible.
+
In what follows we call a cut set Cx corresponding to X C V(H) critical if fH(X) = 0; we call it trivial if m i n { l X I, I V(H)  X 1) = 1. Also, we call statement 1 ) of Theorem IV.ll the cut condition. W.1.o.g. we may assume H to be loopless since by definition a cut set cannot contain loops, and by the choice of H, loops lie in A(H); hence having obtained an eulerian orientation Do of H  A(H), Do u A(H) is an eulerian orientation of H. Thus, if p =)V(H) I= 1, then E ( H ) = A ( H ) = 0 and H = I 1 can be assumed. Next we observe the following facts concerning an arbitrary mixed graph H*: a ) F o r every X C V(H*), fH*(X) = f H I (V(H*)  X); this follows from the cut condition. Hence, in what follows it is no loss of generality if we assume in the case of a critical cut set Cx that X does not contain a prescribed vertex. b)If H* satisfies the cut condition and has a critical cut set Cx,then the orientation of the elements of E ( X ) is uniquely determined in every eulerian orientation D of H (if such D exists). For, E ( X ) # 0 implies
IV.3. Mixed Graphs
IV.15
a + ( X ) # a(X), and if a+(X) > a(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 tieup of H* (see also Figure IV.2) with respect to a critical cut set Cx of H*, and if Hi* has an eulerian orientation Df ,i = 1,2, then an eulerian orientation of H* is induced in a natural way by D; and Dg (see also Exercise IV.9). Note that for {zf } = V(H:)  V(H), fHf (zf) = fH. (X) (even if CX is not critical). c)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 f H r (Y) is even for any Y C V(H;*). i
Now we prove
H has n o nontrivial critical cut sets.
(1
Suppose it does. Then there exists X C V(H) with I X I> 1, I V(H) I> 1, and f H ( X ) = 0. Let Hi with {zi} = V(Hi)  V(H) be the (mixed) (di)graphs of the tieup with respect to Cx, i = 1,2.
X
Suppose fH.(Y) < 0 for some Y C V(Hi) with zi 6 Y (see a ) , c ) ) , i E {I,2 ) . +hen Y C V(H) and Cy is also a cut set of H by the definition of Cx and a tieup; and fH(Y) = fHi (Y) 0. Th'is contradiction shows that Hl and H2 satisfy the cut condition. Since Cx is nontrivial, I V(Hi) I
Our next step is to show that for every v E V(H), fH(v) = 0.
(2)
Suppose there is x E V(H) with fH(x) # 0; by the cut condition, fH(x) 2 2. That is, there is el = us,e, = wx E E,. Form HI,, by splitting el, e2 away from x with {xl,,} = V(H1,2)  V(H) if I A, U EZI> 2; otherwise let Hl,z= H with x1,2 = x. Let Ho = (HI,,  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 and hence to an eulerian orientation of H; and V(Ho) 5 V(H), I E(H,) I=I E ( H ) I 1.
IV. 16
IV. Characterization Theorems and Corollaries
fH(x) >_ 2 implies fHo (x) 2 0 if x E V(Ho); but suppose there exists X c V(Ho) with x @ X such that f H o ( X ) < 0 (see a), and note that fHo(X) is even anyway). Let Cx denote the corresponding cut set in
Ho If uw @ Cx, then Cx = Cx, is also a cut set of < 0, with fHI,P(X1) where XI = X or XI = X U { x ~ , depending ~) on whether I {u, w) n X I= 0 or = 2. If uw E CX, then w.1.o.g. u E X, w $ X , and C i = (Cx{uw))~{uxl,2) is a cut set of with f H l V z(X) < 0 (in this case x1,2 $Z XI = X ) . Consequently, in both cases we obtain XI C V(H) defining a cut set with f H(XI) < 0 unless uw @ CXI and x , , ~E XI. We now coilclude that u, w E X, XI  { x ~ , defines ~) in H a cut set C = Cx U {ux, wx), and thus f H (XI 0. Consequently f H (XI  { x ~ , ~= ) )0. Therefore, 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).
2 satisfies dD(v) 2 8 and I X*(v) 12 idD(v) 1,
+
then we obtain a true statement. The proof of Corollary VI.13 can be reduced to an application of Theorem VI.ll (if one properly applies the Splitting Lemma at every v with dD(v) > 2), and is therefore left to the reader (Exercise VI.7). Moreover, by the same argument used in Exercise VI.7, one can lower the lower bound in 2') to ?jdD (v) if every vertex v satisfies dD(v) > 8 unless it is 2valent. Thus we obtain another corollary. Corollary VI.14 If we replace in Corollary VI.13 dD(v) 2 8 with dD(v) > 8 and I X*(v) 12 $dD(v) 1 with I X*(v) 12 gdD(v), then we obtain a true statement.
+
By a further specialization concerning dD(v) and P ( D ) we obtain another result. Corollary VI.15. Let D be a connected eulerian digraph, and let a system of transitions X (D) be given. If S(D) > 6, then D has an eulerian trail compatible with X(D). Proof. By Corollary VI.14, it suffices to replace every 8valent vertex in an appropriate manner with four 2valent vertices. The application of Corollary VI.14 to the digraph thus obtained then finishes the proof of Corollary VI. 15. If D has no 8valent vertices, then D and X(D) satisfy the hypothesis of Corollary VI.14, and Corollary VI.15 follows. If however, D has an 8valent vertex v indeed, then let a+ E (At)+ and af E (At) such , 1,2,3,4), and and are that w.1.o.g. X(v) = { { a ~ , a ~i) = connected. If (D1,3)2,4 and (D1,3)4,2are both disconnected, then by the Double Splitting Lemma (Lemma III.27), at least one of (D1,4)2,3 and (D1,2)4,3 is connected; say, (D1,4)2,3is connected. Moreover, by the Splitting Lemma either ((D1,4)2,3)3,1or ((D1,4)2,3)3,2is connected; say, H = ((D1,4)2,3)3,1is connected. But then H is a connected eulerian digraph obtained from D by replacing v with four 2valent vertices, and
VI.l.l. P(D)Compatible Eulerian Trails in Digraphs
VI. 13
such that every ( X ( D )X(U))compatible eulerian trail of H corresponds to an X(D)compatible eulerian trail of D. If, however, (D1,3)2,4or (D1,3)4,, is connected then a further application of the Splitting Lemma yields the same conclusion (note that eulerian digraphs are bridgeless in any case by Corollary IV.9). Applying this construction to every 8  d e n t vertex of D we obtain an eulerian digraph to which Corollary VI.14 can be applied. Corollary VI.13 now follows. Figure VI.2 shows that Corollary VI.15 is best possible concerning the restriction on the minimum valency esceeding 2; and there are infinitely many examples demonstrating this fact. To see this, take any 3regular digraph D and subdivide any two arcs a l , a,. Then join the two subdivision vertices s l , s2 by two arcs of the form (sl, s2) and two arcs of the form (s2,s l ) , and define x(sl) and X (s,) as indicated by Figure VI.4.
Figure VI.4. The 3regular digraph D, constructed from the 3regular digraph D.
We leave it a s an exercise to check that Dl has an eulerian trail compatible with X ( D l ) = X ( D ) U X ( s l ) U X(s2) if and only if D has an eulerian trail compatible with a given X(D). Consequently, if D is the digraph of Figure VI.2 or any 3regular digraph obtained from this digraph by repeated application of the extension indicated by Figure VI.4,
VI.14
VI. Various Types of Eulerian Trails
and if X ( D ) is defined correspondingly, then D has no X(D)compatible eulerian trail. However, if X(D) = XT is the transition system of an eulerian trail T of D , then we obtain a somewhat stronger result which can be viewed as the analogue to Corollary VI.5 for digraphs. Corollary VI.16. If D is a connected eulerian digraph with S(D) > 4 and T is an eulerian trail of D, then a Tcompatible eulerian trail Tf of D exists.
Proof. In view of Corollary VI.15 it suffices to construct from D a connected eulerian digraph D t without 4 and 6valent vertices with X ( D t ) being induced by X T such that no transition of T at a 6valent vertex of D defines A: where x is 2valent in Dl. In fact we can do better by constructing Dt in such a way that X(D1) defines an eulerian trail of Dl. For this purpose we assume D to have a 6valent vertex v (otherwise, by Corollary VI.15, nothing has to be proved). Write T in the form
where X(v) = {{ei(v),ei(v)+)li = 1,2,3) e,(v)+ E (At)+, i = l,2,3. Then
XT with e,(v)
E (A:),
is an eulerian trail of D which follows T everywhere except in v where it behaves like a Tcompatible eulerian trail of D. Doing this step by step for every 6valent vertex of D we finally obtain an eulerian trail T1which coincides with T in all but the 6valent vertices; and in these vertices it behaves like a Tcompatible eulerian trail of D. Denoting by Dl the digraph obtained from D by replacing every 6valent vertex with three 2valent vertices such that TI corresponds to an eulerian trail of Dl, we have arrived at an eulerian digraph without 4 and 6valent vertices with a system of transitions X(Df) induced by T and defining an eulerian trail of Dl. By the construction of Df,X(Dt) respectively, any eulerian trail of D f resulting from the application of Corollary VI.15 to Dl, corresponds to a Tcompatible eulerian trail of D. Corollary VI.16 now follows. It is clear that Corollary VI.16 cannot be extended to include eulerian digraphs with 4valent vertices (see the discussion preceding Definition
VI.l.l. P(D)Compatible Eulerian Trails in Digraphs
VI.15
VI.12). It is not clear which 2regular 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 Db (with D, and Db being arcdisjoint), while the degree condition is being transformed into
(any 'red' arc incident to v is matched by a 'blue' arc incident from v, 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 4valent cut vertex x (such digraphs exist !  see Exercise VI.9). Then we can write
D = D, U D2, Dl n D, = x, and Di is connected for i = 1,2. Thus idDl (x) = idD, (x) = odDl (x) = odD2(2) = 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 in such a way that I A(Di) 1 s 0 (mod 2), i = 1,2, and w.1.o.g. the arc incident to x in Dl has been colored red. Then 1') (see above) is still fulfilled for this new bluered 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 D2. 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 halfarcs incident to and from v forming the first, the corresponding red halfarcs the second; moreover, P(v) satisfies 1'). Let P ( D ) = UP(v). 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. Various Types of Eulerian Trails
VI. 16
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(v) be written as Pi(.)
= P i , l ( ~ ) U P i , 2 ( v ) with
P i , l ( ~ ) n P i , 2 (=~0)
where Pi,,(v) contains precisely the halfarcs of Pi(v) which are incident to v (hence Pi,2contains only halfarcs incident from v). L,et P(D) = P(v), and let a new digraph D p be derived from D in the following way: every v E V(D) is replaced by two vertices vl,2 and v2,,, and the incidences in Dp are defined by
uv~vw)
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 p 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 halfarc belonging to Pi(v), then this halfarc belongs to Pi,,(v), and T must leave v via an halfarc belonging to Pi+l,2 (v) where we put i 1 = 1 if i = 2. Consequently, because of the definition of the incidences in Dp, T defines a unique eulerian trail of Dp.
+
Conversely, if D p has an eulerian trail Tp, then Tp defines a unique eulerian trail T of D by definition of Dp. Moreover, by this definition it follows that the halfarcs defining a transition of Tp in v , , ~or vZI1 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 kregular digraph having an even number of arcs. Then D can be decomposed into two arcdisjoint subdigraphs Dl and D2 such that dDi(v) = k for every v E V(D) and i = 1,2. In particular, if k = 2, then D is the arcdisjoint union of two 1factors.
VI.1.2. Aneulerian Trails, Bieulerian Digraphs and Orientations V1.17
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 Zvalent vertices v and v+ such that odDo(v) = idDo(v+) = 0 for the digraph Do thus obtained. By this definition, Do is 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 2colored 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 v E V(D) idGb(v)= odGb(v)= idG,(v) = odG,(v) = 1. That is, each of Gb and G, is a 1factor 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 O+(A:). We state the most general result only, leaving its proof a s an exercise (Exercise VI.lO). Corollary VI.19. Let D be a connected eulerian digraph with a prescribed order O+(At) of the halfarcs incident with v, for every v E V(D), such that two consecutive elements in O+(A:) are not both incident to (from, respectively) v. Then D has a nonintersecting 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 pathlcycle 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 TD in D corresponding to TG is called an antidirected euler i a n trail, in short an aneulerian trail, if for every section ai, v ~ +ai+l ~ , of TD,a; and are both either incident to vi+, 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. Various Types of Eulerian Trails
VI. 18
The following is an immediate consequence of Definition VI.20.
Corollary VI.21. If D is a bieulerian digraph, then for every v E V(D), idD(v) = odD(v) r 0 (mod 2) and, consequently, ) A(D) 10 (mod 2). 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, ) A(D) (r 0 (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 0 7 as obtained from D by splitting every v E V(D) into v,v+ E V(D7) and by letting v, vS respectively, be incident with the elements of ( A t )  , (A:)+ respectively.
Theorem VI.22. Let D be an eulerian digraph. The following statements are equivalent. 1) dD(v)
0 (mod 4) for every v E V(D), and DT is connected.
2) D is bieulerian.
Proof. If dD(v) 0 (mod 4), then idD(v) = odD(v) r 0 (mod 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 DT 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 dD(v) s 0 (mod4) for every v E V(D) by Corollary VI.21. By definition of DT, an aneulerian trail of D corresponds to an aneulerian trail of DT; hence DT is connected. This finishes the proof of the theorem. The tworegular 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 dD(v) 0 (mod 4) for every v E V(D). Then D has a unique decomposition S,, into maximal aneulerian subdigraphs of D , and ) s,, ) = c(D7). Proof. Consider 07, and let its components be denoted by C1,. . . ,Clc, 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 dD(v) G 0 (mod 4) for every v E V(D), we have d(v+) = d(v) r 0 (mod2) in D; for every v+, v E V(D;). Consequently, Gi is eulerian, and an eulerian trail of Gi corresponds to an aneulerian trail of Ci by definition of Dr. Consequently, if Dl, . . . ,Dk are the subdigraphs of D corresponding to Cl, . . . ,Ck respectively, then S ,, = {Dl,. . . ,Dk) is a decomposition of D into aneulerian subdigraphs (note that there is a 1  1 correspondence between aneulerian trails of Ci and aneulerian trails of Di, 1 _< i 5 k). Moreover, precisely because of the definition of D+, and because C1, . . . ,Ck are its components, S,, is uniquely determined and Di is a maximal aneulerian subdigraph of D ,, (=c(D7). This finishes the proof of for i = 1,.. .,k. Consequently, I S the theorem. We observe that if D is 2regular, then the aneulerian subdigraphs of D are necessarily the maximal aneulerian subdigraphs of D simply because the vertices of D; are 2valent. Which 4regular 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 4regular graphs and 3regular 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]). T h e o r e m VI.24. A 4regular graph G4 has a bieulerian orientation if and only if there is a hamiltonian, bipartite, 3regular graph G3 with hamiltonian cycle H such that G4 N G3/L where L = E(G3)  E ( H ) . Proof. If G, has a bieulerian orientation, then fix such an orientation, call it D4, and construct the bipartite digraph DT a s before. The aneulerian trail of D, guarantees that D r is connected. Therefore, for H being the cycle of the underlying graph Go, G3 = GOU {v+v/v E V(D4)) is a 3regular graph with a hamiltonian cycle H; by the construction of D+ and G3, the latter is bipartite as well, and G, z G,/L where L = {v+v/v E V(D,)) = E(G,)  E(H). Conversely, if G, G3/L, and if H = G3  E(L) is a hamiltonian cycle of the 3regular 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(Do) defined by the sources, sinks respectively, coincides with the bipartition of V(G3) since Do is connected. Thus, every e E L joins a source with a sink since G3
VI.20
VI. Various Types of Eulerian Trails
is bipartite. Hence, D4 obtained from Do by identifying the same pairs of vertices as in forming G3/L from G3, is 2regular, and a run along the aneulerian trail of Do corresponds to an aneulerian trail of D,. Consequently, since G4 is underlying D4 by hypothesis and by construction, G4 has a bieulerian orientation. The theorem now follows. Thus, G3 being bipartite is an essential property in order that G4 has a bieulerian orientation. However, G4 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(v) for every v E V(G), then it has no bieulerian orientation.
 0 ( m o d 4 )
Proof. Suppose an orientation D of G has an aneulerian trail TDstarting w.1.o.g. a t v E V, along the arc (v, w) where w E V2, and {V,, V2} is the bipartition of V(G) = V(D). Then the next arc in TD is of the form (x, w), with x E Vl. Then the arc following (x, w) in TD is of the form (x, y) with y E V2, a.s.0. Consequently, the first component of an arc of D always belongs to Vl, while the second component lies in V2. 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 dG(v) 0 (mod 4) 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 endblock B, of G and its only vertex v (G cannot have 2valent vertices by hypothesis, so every endblock of G is a loop). Depending on the degree of v we distinguish between two cases.
VI.1.2. Aneulerian Trails, Bieulerian Digraphs and Orie~ltations VI.21
1) dG(v) = 4. Then we form GI obtained from G  B, by suppressing in G  B1 the 2valent vertes v (see Figure VI.5).
Figure VI.5. G1 obtained from G by deleting the loop B1 and suppressing the 2valent vertes v of G  B1; possibly x = y.
GI satisfies the hypothesis of the theorem and has fewer blocks than G. Let D be an arbitrary eulerian orientation of G, and let Dl be the eulerian orientation of G1 induced by D. By induction, Dl is bieulerian. Assuming w.1.o.g. (x,u) E A(D), we have (u, y) E A(D) and (x,y) E A(D1). Considering now any aneulerian trail Tl of D l , we see that TI can be extended to an aneulerian trail T of D by replacing the arc (x,y) with (x, v), u, A(B1), v, (v, y), where A(B1) means that we pass through B1 against its orientation (w.1.o.g. Tl passes (x, y) from x to y. Observe that for an aneulerian trail, the inverse sequence is also an aneulerian trail). 2) dc(u) > 4. Because of 1) handled already, we assume w.1.o.g. that no 4valent vertex of G is incident with a loop of G. Consequently, if we look a t the blockcut vertex graph bc(G), then no endvertes of bc(G) is adjacent to a 2valent 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 dC(u) 0 (mod 4), u is incident with two loops, B, and B, say. Now form G, = G  (B, U B,) .
VI.22
VI. Various Types of Eulerian Trails
Gl satisfies the hypothesis of the theorem because 4 5 dG(v)  4 = d,, (v). Let D be an eulerian orientation of G as in case 1)with (x, v) E A(D). The orientation Dl of G, induced by D is eulerian; by induction, D l is even bieulerian. Replace in any aneulerian trail TIof Dl the arc (x, v) with (x, v), v, A(B,) ,v, A(B2), thus obtaining an aneulerian trail of D (A(B1) means, as above, that we pass through B, against its orientation, while A(B2) stands for passing through B2 according to its orientation; also here we assume w.1.o.g that Tl passes (x, v) from x to v). 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 Cregular graphs without bieulerian orientation. For example, if G is obtained from two 4regular graphs G; by subdividing an edge ei with one vertex vi,i = 1 , 2 , and then identifying the two 2valent vertices v, and v2, then G is a 4regular graph with cut vertex v arising from identifying v, and v2. It is straightforward to see that G has a bieulerian orientation if and only if G, and G2 have such orientations. In fact, one can prove the validity of the following statement. Lemma VI.27. A connected 4regular graph G has a bieulerian orientation if and only if for each block B C G which is homeomorphic to a 4regular graph B*, B* has a bieulerian orientation.
One proceeds similarly to show the validity of the next lemma. Lemma VI.28. For a connected 4regular graph G, every eulerian orientation is bieulerian if and only if for every block B of G which is homeomorphic to a 4regular 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 4regular graph.
Because of the Lemmas VI.27 and VI.28, it suffices to deal with 2connected graphs if one wants to determine all 4regular graphs which have bieulerian orientations, and/or the property that every eulerian
VI.1.2. Aneulerian Trails, Bieulerian Digraphs and Orientations VI.23
orientation is bieulerian. The example above which has no bieulerian orientation a t all, has, however, a cut vertex; but in [BERM79d], a 2connected, Cregular 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 D5of 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 vertexsequence 1,3,2,4,3,5,4,1,5,2,1determines the aneulerian trail corresponding to these orientations. Thus the I(5 has an eulerian and thus a bieulerian orientation which is unique up to isomorphisms. However, if we label the vertices of the I& beforehand, then the I(5 has several eulerian orientations which are, however, bieulerian as well. In fact, the number of bieulerian orientations of any 4regular graph containing a prescribed arc (x, y) is even, [BERM79d, Theorem 6.21.
Figure VI.6. For a given cyclic orientation of the outer pentagon of the the two possible choices of orienting the inner pentagram yield isomorphic eulerian orientations of the IC5, with the isomorphism indicated by the vertex labeling.
VI.24
VI. Various Types of Eulerian Trails
As we have seen in Corollary VI.18, every 2regular digraph can be decomposed into two 1factors. In general, there will be more than one such decomposition. Hence the question: what is a necessary and sufficient condition for a 2regular digraph to have a unique 1factorization ? This question is answered as a consequence of the following proposition. Proposition VI.29. Let D be a 2regular digraph. Then there is a 11correspondence between the 1factors of D and the maximal independent arc sets of D r . Proof. If F is a 1factor of D l then we have odF(v) = idF(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 D 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. VCTehave IA(C)I r 0 (mod 2) for every cycle C of G because DT is bipartite D; by construction. Moreover, since D is 2regular, the vertices of DT are 2valent. It follows that M covers all vertices of DT. 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 idF(v) = odF(v) = 1; i.e., F is a 1factor of D. Proposition VI.29 now follows. For a 2regular digraph D with 1factor F, D  F is a 1factor F' of D as well; and M' = A(DT)  M is a maximal independent arc set of D; 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 (MI n (D;
 C))
is a maximal independent arc set of DT different from M and MI. These considerations together with Proposition VI.29 yield the following relation between the concept of a 1factorization and that of being bieulerian. Corollary VI.30. A 2regular digraph has a unique 1factorization if and only if it is bieulerian.
VI.1.3. DoFavoring 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 follo~vsfrom [BERM78b, Theorems 2.3 and 2.41. These are summarized as follo~vs. Corollary VI.31. Let D be a 2regular digraph. Then the number of 1factors of D equals 2C(D;) and, consequently, the number of 1factorizations of D equals 2c(D;)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 4regular 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,, Gb,G , with just one Zvalent vertex a, b, c respectively, and all the other vertices are 4valent. Consequently, G+ = G, U Gb U Gc U {ab, bc, cn} is a 4valent 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 Dofavoring if and only if for every v E V(D), T traverses every arc of Do incident from v before it traverses any arc of Dl := D  Do incident from v. Of course, every connected eulerian digraph D can be written as D = DoUDl with A(Do)nA(D,) = 0; just take Dl = Dl V(Do) = V(D)
VI.26
VI. Various Types of Eulerian Trails
and A(Do) = 0. In this case any eulerian trail of D is favorin^ in^.') Also, Definition VI.32 indicates that the existence of a Dofavoring eulerian trail T may depend on the choice of an initial vertex for T. We prove two theorems on the existence of Dofavoring eulerian trails depending on the structure of Dl := 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 Dl = D  Do is a spanning intree of D with root v. Then a Dofavoring eulerian trail starting (and ending) at v exists. Conversely, if T is an eulerian trail of D starting (and ending) at v, and if we mark at every w E V(D), w # v, the last arc of T incident from w, then D l , the subdigraph of D induced by the marlied arcs, is a spanning intree with root v (and hence T is a (D Dl)favoring eulerian trail of D). Proof. Let Do C D be chosen such that Dl = D  Do is a spanning intree 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 l 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 Dl, there is a path P(z,v) C_ Dljoining z to v. Write
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,) is not contained in T; therefore, also (u, ,u,) is not contained in T (note that (ul, u,) can be contained in T only if all arcs incident to ul are contained in T), a.s.0. In particular, (uk,v) is not contained in T, contradicting the fact that idT(v) = odT(v) = idD(v) = odD(v). Thus, T contains all arcs of D. This observation is contained in [ B E R K ~where ~ ~ ] the author attributes Theorem IV.8 to I.J. Good, an error apparently inherited from [I
This question has a positive answer if X(G) m  1; and if X(G) 2 2m then one can even prescribe the direction in which these m edges are traversed by T. These and more general results are contained in [CAIM89a]; they are derived by using [JACK88d, Theorem 11.
VI.2. Pairwise Compatible Eulerian Trails As we have seen in Corollary VI.5, if G is an arbitrary connected eulerian graph without 2valent vertices and X any system of transitions of G, then there exists an eulerian trail T of G compatible with X. This is true in particular if X = X T , is the system of transitions induced by an eulerian trail T' of G. Whence the following question arises: what is the m a x i m u m number of pairwise compatible eulerian trails in a connected eulerian graph G ? This question was first asked by A.J.W. Hilton and B. Jackson with the latter proving that if S(G) = 2k > 2, then there are a t least max(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 [FLEI86a] for the class of eulerian graphs G where each block of G is a cycle. Such graphs demonstrate that the bound 2k  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 if f is the only other edge incident with v and belonging to the same block a s 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 .XT 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(v) to form a transition with other halfedges incident with v. 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 = v, l , w , 2 , ~ , 3 , w , 4 , vT2 ; = v,1,w,3,v,4,w,2,v;T3 = v , ~ , w , ~ , v , ~ , w , ~ , v .
In fact, one might conjecture that if one strengthens the hypothesis of Conjecture VI.36 by additionally assuming G to be 2connected, then one can even find 21;  1 pairwise compatible eulerian trails. Figure VI.8 points into this direction. In the case where G = 1 L, and vice versa. (*I For, by the above considerations, an eulerian trail of GI (written as an edge sequence) must be of the form
where we assume T to start at v and with the notation chosen in such 6 j Sj+l a way that {eij ,eij ) = {e;, ,e.:}' for j = 1, . . .,m, and {il, . . .,im} = (1,. . . ,m). This sequence, reinterpreted as a sequence of vertices of L(G1) defines nothing but a hamiltonian cycle H of L(G1) with L C E(H). By the same token, a hamiltonian cycle H of L(G1) with L C E ( H ) defines an eulerian trail of GI. Now, if Tl and T2 are two compatible eulerian trails of GI, then the corresponding hamiltonian cycles HI and H2 of L(G1) necessarily satisfy
This follows from the preceding considerations and the fact that XT1fl XT, = 0 by the very definition of compatibility. Observing now that if H is a hamiltonian cycle of L(G1) corresponding to an eulerian trail of GI, then (*) implies that E ( H )  L is a 1factor of L(G1), and considering (**), we are led to the following theorem which is therefore equivalent to the validity of Conjecture VI.36 in the case where G consists of just one vertex and m 2 2 loops, [FLEI86a].
Theorem VI.37. Let n = 2m > 2 be a positive even integer, and let L be a 1factor of the complete graph K,. Then there exists a 1factorization L = {L1,. ..,L,l) of K , such that L = L, and L U Li is a hamiltonian cycle of K,, for i = 2, . . .,n  1. Proof. To obtain a 1factorization as stated in the theorem we consider K , obtained from K, by contracting the edges of the 1factor L = (el,. . .,em) C K,. Let the notation be chosen in such a way that e j = eie;, i = 1,.. . , m , thus obtaining V(I(,) = {e:, ey/i = 1,.. .,m);
VI. Various Types of Eulerian Trails
VI.38
and for the sake of simplicity denote V(Ir',) = {e,, . . . , e m ) (Observe that this notation is in accordance with the discussion preceding the statement of Theorem VI.37). Let e = eilei2 be a fixed edge of I i m . By the above choice of notation we have in K , the identities e;, = e!t l e!', I , %  e!%?el!1 2 and eil ,ei2 E L. Thus, e corresponds to a set He of four edges in Ii , , He = {e:le:2, e: e: ,eil e: ,e: ei2) (see Figure VI.lO); and this correspondence between e E E(Iir,) and He C E(I L of I(, as before (see Figure VI.ll). G* is then obtained from PE; by adding for every edge e = eiej E L3 the elements of He = (eie;, eye;, e:ey, eye;). Assuming w.1.o.g. elez E L, we classify the edges of G*  L as follows: For i = 1,2,3, if ere, E Li, then
By definition, {Li, Ly, N i / i = 1,2,3) is a partition of E(G*)  L. This partition is the basis for constructing the 1factors L;, 1 5 j 5 6, as required. We distinguish between two cases. a) m = 0 (mod 4). We consider the hamiltonian cycle H' of I(, induced by L2 and L3 and the corresponding 5regular subgraph PE;, of I
VI.2. Pairwise Compatible Eulerian Trails
VI.51
hamiltonian cycles containing a given lfactor of a related graph. Let G be a connected eulerian graph with 6(G) = 2k > 2, and let r systems of transitions of G, X,, . . . ,X,, be given. Consider an arbitrary vertex v E V(G), and let Xo(v) be any system of transitions at v if v is not a cut vertex; otherwise, let the system of transitions at v, Xo(v), be defined in such a way that it induces a system of transitions at v in every block containing v (since G is eulerian, every block containing v has an even number of halfedges incident with v; hence Xo(v) can X/O(v), and let be defined in the above manner). Let Xo = UvEV(Gl Xi(v) C Xi, i = 1,.. . , r , be the corresponding systems of transitions at v. Now let ei, . . .,e&,d = d(v) G O(mod 2), be the halfedges incident with v, and consider the complete graph Icdwith V(ICd) = {eill = 1 , . . . ,d). Analogous to what has been said in the discussion preceding Theorem VI.37, we conclude that a system of transitions at v corresponds to a lfactor L(v) of ICd. We mark in K d the edges of Lj(v), the lfactor corresponding to X j (v), j = 0, . . . ,r . We know that constructing an eulerian trail T of G is tantamount to replacing step by step, every vertex v of G with t = ;d(v) 2valent vertices vl, . . . , v, such that, at each step, the resulting graph G, is connected. Now, if T is compatible with Xi, then XT(v) n Xi(v) = 0, i = 1,. . .,r. Consequently, XT(v) corresponds to a lfactor LT(v) in ICd with LT(v) n Li(v) = 0, i = 1,.. . , r .
(o>
This is a necessary condition for the existence of T compatible with X i , i = 1,.. ., r . But it is not sufficient because it does not take care of the fact that G, has to be connected (observe that the elements of XT(v) are precisely the sets EGj, j = 1 , . . . ,t). On the other hand, a consideration of the statement (*) preceding Theorem VI.37, plus the observation that Xo(v) is uniquely defined if and only if every block of G containing v contains precisely two halfedges incident with v, yields the following conclusion: If LT(v) U LO(v)is a hamiltonian cycle of ICd, then G, is connected. (00) Summarizing (0) and (oo), we seem to be faced with the following ques0 (mod 2)) contain a lfactor L tion: does Kd (with 4 5 d = d(v) such that Li(v) n L = 0, i = 0,. . .,r, and Lo(v) U L is a hamiltonian
VI. Various Types of Eulerian Trails
VI.52
cycle ? Well, let us consider the case r = 1. If d = 4 then I 4, then d 2 6, and by Theorem VI.37, Kd has a 1factorization {L, ,. . ., Ldl) such that Lo(v) = L1 and Lo(v) U Li is a hamiltonian cycle for i = 2,. . . ,d  1. d 2 6 implies $ d < d  2; i.e., the 1factor Ll(v) (which has $ d edges) satisfies L,(v) n Li= 0 for at least one i E (2,. . . ,d  1). Whence we conclude that the above question has a positive answer if r = 1 (we note that the same conclusion could have been reached without using Theorem VI.37, by employing a) compatibility results of section VI.l, b) the Splitting Lemma, and c) the structure of Lo(v)). So, the problem we are really faced with is this (we put Li = Li(v)):
+
Let r 1 1factors Lo, L,, . . .,L, of the I&be given, where d > 2 is a n even integer. How large can r be such that Kd has a 1 factor L with L n Li = 0, i = 0,. . . ,r , and L U Lo is a hamiltonian cycle ? (o o o) The answer to (o o o) follows from the next result (see [ H A G G ~ ~ ~ ] ) . Observe that a graph of even order satisfying Ore's Theorem (Theorem 111.75) contains a 1factor because it has a hamiltonian cycle (see [BERM83a] for a generalization of the next theorem). Theorem VI.40. Let H be a simple graph of even order d 2 4 such that d H (x) d H (y) 2 d 1 for every pair of nonadjacent vertices of H , and let L be an arbitrary 1factor of H. Then L is contained in some hamiltonian cycle of H.
+
+
Proof. We proceed indirectly. Since K d satisfies vacuously the hypothesis of the theorem and since any prescribed 1factor L c E(Iid) is contained in some harniltonian cycle of Kd by Theorem VI.37, therefore we may choose a graph H having the following properties: a) H is a proper subgraph of K d satisfying the hypothesis of the
theorem, b) H has a 1factor L not contained in any hamiltonian cycle of H, c)
/ E ( H ) I is as large as possible.
Consequently, if x and y are any two nonadjacent vertices of H, then H U (xy) has a harniltonian cycle C containing L. xy E E ( C ) , otherwise
VI.2. Pairwise Compatible Eulerian Trails
VI.53
H is not a counterexample to the theorem. Writing C as a sequence of vertices with x = xl, y = xd,
we conclude that
By the same argument used in the proof of Ore's Theorem we conclude that d(x) d(y) 2 d + 1 implies the existence of vertices xi and xi+, such that x1xi+,,xdxi € E ( H ) , 2 5 i 5 d  2. If xixi+, 4 L, then the cycle C, with
+
(see the proof of Ore's Theorem) is a hamiltonian cycle of H containing L, contradicting the choice of H. Thus xixi+, E L, and i is odd
.
(4
To obtain a contradiction concerning the choice of H satisfying (*), we decompose H into two subgraphs H1 and H2 where H, is bipartite with L C E(H,) and, subject to this property, IE(H1) I is as large as possible, and H2 = H  HI. Consequently, V(H1) = V(H). Furthermore, let Hi = H1  L. Observe that H1 is connected because of the hypothesis of the theorem; hence the vertex bipartition {Al, B1) of V(Hl) is uniquely determined, and I A, I=I B1 I= $ d because of L C E(Hl). W.1.o.g. 2, E A,. We deduce some properties of H,. Consider an arbitrary e = xjxj+, E L ; w.1.o.g. x j E A, (otherwise, replace the index j 1 with j  1). Now define a new bipartite graph H;* with bipartition V(H;*) = A; U B; as follows:
+
A;1 = ' (  {xj)) U {xj+l), B; = (B1  ( x ~ + ~u) ){xj), E(H,') = {uv E E ( H ) / u E A;, v E B;) . Observe that
VI. Various Types of Eulerian Trails
VI. 54
Using the fact that L c E(H;) by the definition of H;, we conclude from the choice of H, that IE(H;) 151E ( H l ) I. This and the above equation yield dH;(e) dH2(e) for arbitrary e E L . (**I
>
Consequently, if e =
f = xkxk+l E L, then
+
This inequality together with dH(e) = dH, (e) dH2(e) yields 2 d ~(e) ; 2 d ~(f) i dH(e) dH(f) 
+
>
+
+ 2 for any e E L .
Now, since e = X ~ X ? + ~f , = xkxkcl and w.1.o.g. xj,xk E A,, we obtain from the preceding Inequality and because of the choice of HI
(note: the maximality of I E(H1) I then implies x .xk+,,xkxj+, whence dH(xj) dH( x ~ + ~ d) 1 by hypothesisj.
+
> +
6 E(H)
We construct from H1 a digraph D as follows: replace every edge xy E E(Hl) with the arc (y,x) where x E Al; from the resulting digraph DHl produce D := DH1/L (where L stands for the arc set in DH, corresponding to L in H,). For practical reasons we label the vertices of D with the labels of the elements of L C E ( H l ) in accordance with the contraction procedure. This gives
Since e, f E V(D) are not adjacent for the above e, f E L if and only if xj X k + , ,xkxj+l fi! E ( H l ) (remember that H, is bipartite !), and because d = 2 I V(D) I, we obtain from (* * *) the inequality dD(e)+dD(f)
> 2 1V(D)I 1
if
e, f E V(D)are nonadjacent.
(****)
However, since we do not know at this point whether D is strongly connected or not, Meyniel's Theorem cannot be applied to D (note that D has no multiple arcs because of the definition of DH1). But for the application of Corollary 111.78 strong connectedness is not required. Whence we conclude that D contains a hamiltonian path PD which we write as a vertex sequence, PD = eil, ei2,. . . ,ejm
VI.2. Pairwise Compatible Eulerian Trails
VI.55
where m = i d =I V(D) I and {eil ,. . . ,ei, ) = V(D) = L. Because of the orientation DH1 of Hl it follows that PDcorresponds to an antidirected hamiltonian path in DHl; therefore, PDcorresponds to a hamiltonian path P of H1 containing L. If the endvertices of P are joined by an edge e in H, then P U {e) is a hamiltonian cycle of H containing L, contradicting the choice of H. W.1.o.g. we can relabel the vertices of P such that E(P) = E(C){xdxl) and thus PD= el,e2,. . . , e m for ei = X , ~  ~ X , ~=, ~1,.. ., m. It follows from the hypothesis (see the arguments preceding (*)) that X ~ X ~ xdxi + ~ , E E(H); and by (*), i is odd. Also, since P C Hl and x1 E Al by assumption, we have Al = { x , ~  ~j , = 1,..., fd}, B1 = {xZj/j = I , . . . , i d ) . This yields xlxi+,, xjxd E E(H1) since i is odd. Thus, (xi+l, x,), (xd,xi) E A(DHl) by definition of DHl, and these two arcs correspond to the arcs (ej, el), (em,ej) E A(D), where ej = xixi+, and j = whence we conclude that PDU {(ej, el), (em,ej)) is a strongly connected spanning subdigraph of D ; i.e. D is strongly connected. By Meyniel's Theorem, D has a hamiltonian cycle CD. We conclude as above concerning the relation between PDand P that CD corresponds to a hamiltonian cycle Co of Hl with L C E(Co). Since V(Hl) = V(H), we obtain the final contradiction to the choice of H. The theorem now follows.
q;
We are now in a position to answer the question (o o o) raised before stating Theorem VI.40. This question is equivalent to asking whether the graph H = (ICd  U:='=Li) o U Lo has a hamiltonian cycle containing Lo. By definition of H and because Li n L j # 0 may hold for some i # j, 0 i, j r , we have dH(u) 2 (d 1)  ( r 1) 1 = d  T  1 for arbitrary v E V(H). Therefore,
2,; and NT = d  1 for odd d, while NT = d  2 for d = 4 (see Figure VI.15). But we have not obtained the best possible result for NT simply because we lost some information by utilizing the same approach as in the undirected case.
3, and let Lo,L1,. . .,L, be 1factors of ICd d , where r Then (ICd  U:=O Li) U LO has a hamiltonian cycle containing Lo.
< 9.
Proof. Denote L = Lo and H =
 U:=O Li) U Lo. We have
Consequently,
Corollary V1.49 now follows from Corollary VI.48. We note in passing that if we replace in Corollary VI.49 the restrictions on d and r with d 3 and r 5 max{l, 9 1 , then the conclusion is false precisely for d = 3. One only needs to choose L1 with I Lo n Ll I= 1. If, however, LonLl = 8, these weaker restrictions admit the same conclusion as in Corollary VI.49.
>
On the grounds of the preceding discussion we deduce our next result from Corollary VI.49 in the same way we reached Theorem VI.42 by applying Corollary VI.41.
Theorem VI.50. Let D be a connected eulerian digraph with S(D) > 6, 6( 0 )4 and let X I , . . .,X, be any r systems of transitions, where r 5 7. Then D has an eulerian trail compatible with Xi, i = 1,.. . ,r . This theorem together with the remark following Corollary VI.49 yields the analogue to Corollary VI.43.
>
Corollary VI.51. Let D be a connected eulerian digraph with S(D) 6. Then D has at least max(2, pairwise compatible eulerian trails.
y}
Again, this lower bound for the maximum number of pairwise compatible eulerian trails is, in general, best possible (see Exercise VI.4). However, in view of Theorem VI.47 we put forward the following conjecture (we do not bother, however, to formulate the conjecture on 1factorizations
VI.68
VI. Various Types of Eulerian Trails
in the Kd corresponding to Conjecture VI.39 with the former's validity implying the validity of Conjecture VI.52). Conjecture VI.52. The maximum number of pairwise compatible eulerian trails in a connected eulerian digraph D with S(D) > 4, is a t least 16(D) 2  2.
We note in passing that most of the above results appear in [FLEISOa].
Of course, one can try to apply the discussion performed in this section so far to mixed graphs. However, in the case where every block is a cycle, one is no longer faced with seeking special types of 1factorizations of the or I 2 has an Atrail T. Let v E V(G)with d = d(v) > 2 be arbitrarily chosen and suppose {ei, e;) is the first transition defined by T in v. W.1.o.g. T = . . . ,el, v, e 2 , .. .. Then
where
VI.3. ATrails in Plane Graphs
VI. 71
and e2; = e2;+, if and only if eZi is a loop, i E {I,.. .,i d ) (where we put ed+l = el)Proof. Suppose T contains an extended segment T' of the form
where e2il,e,i, f , g are the only edges of T' incident with v, and f # e2i+l, f # el respectively if 2i = d. By rotating, if necessary, the indices in O+(v), we may assume w.1.o.g. that i = 1; hence f = e j where j E (4, . . . ,d). Then T' contains a subsequence C which is a cycle starting and ending in v and containing e2 and ej. Consequently, if s(ek) is a subdivision point of el,, k = 1,3, s(el) lies in the exterior of C if and only if s(e3) lies in the interior of C. Now, because of the assumption i = 1 and because of the possibility of cyclically rotating the initial vertex in the sequence T, we may, w.l.o.g., express T in the form
where T" is the subsequence of T' starting and ending in v and not containing e, and g, while T"' is the remainder of T, i.e. the subsequence of T having g as its first and el as its last edge. By construction, e3 E T"'. Hence, viewing T"' as a closed curve starting and ending in v we conclude from the above that T"' contains an open curve To,whose ends are s(e,) and s(e3). We conclude from the Jordan Curve Theorem (Theorem III.61), that there is a vertex w on C such that at least one section fi, w, fi+l of T is a section of T"', and the subdivision points s(fi), s(fi+,) lie on different sides of C. But then fi and fi+, do not belong to the same face boundary of G (see Corollary III.6la). This contradiction implies the validity of Lemma VI.53. Lemma VI.53 implies that if we know one transition of an Atrail T at v E V(G), we know the other transitions of T at v as well; i.e. if {ei, e;) E XT(v) and O+(v) = (ei, ek, . . .,e&)or 0(v) = (e',, eh, . . . ,e&),XT(v) = I {{ei, eh}, {ej, ei), . . ., {ed, , e&)}. Consequently, we can say more generally:
If T is an Atrail of the plane eulerian graph G with 6(G) > 2, and if O+(v) = (e',, e;, . . .,e&,, e&)for an arbitrarily chosen v E V(G), either XT(v) = {{ei, ei), 0.
XT(v) = {{eb,ej),
. . . ,{e&,,,
 . {ei, e:,>  7
,
e&))
.
(1)
VI.72
VI. Various Types of Eulerian Trails
Observe that the validity of (I)rests (implicitly) on the Jordan Curve Theorem which does not hold for surfaces of genus other than 0. However, the concept of an Atrail can be defined for graphs G embedded on arbitrary surfaces. In fact, an Atrail in G yields the same local information expressed by (I). On the other hand, ( I ) is a weaker statement than Lemma VI.53; for, the conclusion of Lemma VI.53 says that the Atrail T 'picks up' the edges incident with v either in their clockwise or in their counterclockwise cyclic ordering induced by the embedding of G. Which of these two 'pick up' procedures is followed by T, depends, of course, on the direction in which one passes through T. In any case, once an orientation of T has been chosen by fixing an initial vertex and an initial edge, T defines a partition of V = V(G), V = V+ U V, where V+ (V) contains precisely those vertices v at which T 'picks up' the edges incident with v according to O+(v) (0(v)). If T is passed in the opposite direction, these two sets V+ and V naturally interchange their roles. That is, the above partition {V+, V  ) of V(G) is uniquely determined by T with the meaning of V+, V respectively, depending on the direction in which T is passed. So, the question arises whether it is possible to obtain this partition of V(G) independent of an orientation of T and without using any O+(v) and/or 0(v). To see that this is possible, we make use of Theorem 111.68. Consider for the plane eulerian graph G with A(G) > 2 a 2facecoloring (with 1 and 2 denoting the two colors), and let T be an Atrail of G. Choose an arbitrary v E V(G) with d = d(v) > 2 and suppose the halfedges incident with v are labeled in such a way as e:, ek, . . . ,e> that O+(v) = (ei, e;, . . . ,e&). We further assume that the face whose boundary contains el and e, ,is colored with color 1. Then the faces F2,1 (FZi) with {e2;,,e2;) E(bd(F2;1)) ({e2;,e2;+11 C E(bd(F2;))) are colored 1 (2), i = 1, . . . , +d. Now, if we split v into k = i d 2valent vertices dl),. . . ,v(*) following XT(v) (i.e., {E:,;)/i = 1,. . . , k) = XT(v)) such that the resulting graph G, is plane, then by ( I ) , the faces F2i2+s, i = 1,.. . , k, appear (homotopically) unchanged in G, while the faces F2i+l6, i = 1,.. . , k, become one face in G,, whereby 6 = 1 or 6 = 2 depending on the pattern of XT(v) (see Figure VI.18). Thus we are led to the following definition (recall that a 1face, 2face respectively, is a face colored 1, respectively 2, in a 2facecoloring).
c
Definition VI.54. Let G be a plane, eulerian, 2facecolored graph
VI.3. ATrails in Plane Graphs
VI.73
with A(G) > 2, and let v C V(G) with 2k = d(v) > 2 be arbitrarily chosen. Suppose that the notation is chosen in such a way that O+(v) = (e',, eb, . . . ,e&) and ei, eb belong to the boundary of a 1face. For S E {I,2 ) 7 define X ~ ( V= ) ((eki2+6, ekil+6)/i =  7 k) = e',) and form the plane graph G, such that Ez,, = +6), i = 1,.. . ,k, and bd(F) is a face boundary both in G and G,, for every 6face F of G. Then we say that G, is the result of a (6 1)splitting applied to u E V(G), where we put 6 1 = 1 if 6 = 2 (see Figure VI.18) '1.
+
+
In view of Definition VI.54 and the paragraph preceding it we can say that an Atrail T of the plane eulerian graph G without 2valent vertices induces a partition {Vl, V2) of V(G) such that v E Vg if and only if XT(v) = X 6 + l ( ~ (putting ) 6 1 = 1 for 6 = 2). That is, V6 contains precisely those vertices a t which T induces a 6splitting, 6 = 1,2. Of course, Vl = 0 or V2 = 0 may hold; the same can be said concerning the above partition {V+, V) (see, e.g., G2 in Figure VI.17).
+
Observe that in Definition VI.54, O+(v) was used for practical purposes only. As Figure VI.18 indicates, a 6splitting can be defined solely on the grounds of G being a plane eulerian graph and of a fixed 2facecoloring of G (see [FLEI74a, Definition 2 I). Corollary VI.55. Let G be a plane eulerian 2facecolored graph without 2valent vertices which has an Atrail T, and let {V+,V), {V,, V2) be the two partitions of V(G) induced by T (see above). Then {Vl, V2) = {V+, v).
Proof. Choose for T an initial vertex x and an initial edge e = xy, Combining Definition VI.54 with the discussion preceding it we can say that if v is a cut vertex of G, bd(F2i2+s) can be a cycle in G, while it is not a cycle in G, but there is a 11correspondence between the walks through the components of bd(F2i2+6) in G, and the corresponding walks in G; i.e., ~ ~ ( b d ( F ~ ~ =  ~~+~ ~, () b) d ( F ~ ~  ~Similarly, + ~ ) ) . if v is a cut vertex of G, F2i+l6 might be the same face in G for i = 1 , . . . ,k, and so this face becomes one face in G, as well; but in this case ~ ~ ( b d ( F ~ ~ + 2. Then H has the following properties: 1)
If F, # Fo is a 1face of G with v E V(bd(Fl)), then every w E V(bd(F,))  {v) is a cut vertex of H.
2)
The cut vertices of H are precisely the vertices defined in 1).
3) H is a nontrivial blockchain.
Proof. To prove property 1) we observe that the outerplanarity of G yields v, w E V(bd(Fl)) n V(bd(Fo)). That is, there exist two simple open curves CoyC, in the plane such that a) Ci n G = {v, w), i = 0 , l ;
b) Ci  {v, w) lies in the interior of the face F,, i = 0 , l ; and, consequently, c)
C := Co U C, is a simple closed curve in the plane satisfying C n G = {v,w).
VI.3. ATrails in Plane Graphs
VI.81
By construction and the Jordan Curve Theorem, C divides F, C, i = 0,1, into two parts which lie on different sides of C. From this and the fact that Fo and Fl are 1faces, we can conclude that each of the two sides of C contains a positive even number of edges incident with v and w, respectively. Since G is a simple graph, each of the two sides of C must contain at least one vertex of G. That is, v and w separate G. More precisely, we can express G a s G = G'
u G" ,
where G' n G" = {v, w)
and dGl(v)r dG,(w) E dG,,(v)
= dG,,(w) = O(mod2) .
(*)
Consequently, GI and GI1 lie on different sides of C (except for v and w, of course, which lie on C). Hence, if we form the plane graph HI by splitting v into two vertices v' and v" in such a way that E,, = E, n E(Gt) and Evil = E, n E(GU) and such that v' (v") lies on the same side of C as GI (G"), then we can write
H, = Hi u H:
,
where
Hi n H: = {w)
and Hi, Hr and Hl are all connected and eulerian by (*). Moreover, w is the only element of HI which lies on C (hence Hi and Hi' lie on different sides of C except for w); and by construction, w is a cut vertex of H1. Also, by the above construction,
and H can be written as
H = HI u HI1, where H'
f l
H" = {w}
and
H" = (H:),,,,, . That is, H' and H" lie on different sides of C except for w. This classifies w as a cut vertex of H as well.
'
= ( H V l,
Now we turn to the proof of property 2). First of all, the 2valent vertices ~ ( ~i 1=, 1,.. .,k, k = idG("), arising from the 1splitting of v, cannot be cut vertices of H; otherwise, E,(;, is a set of two bridges of H, thus contradicting the fact that H is eulerian.
VI. 82
VI. Various Types of Eulerian Trails
Now consider the path
and let x E V(Po) be chosen subject to the condition that x g' V(bd(Fl)) if F, # Fo is a 1face of G and v E V(bd(F,)). By the above and 1) it suffices to show that x is not a cut vertex of H. In any case, the choice of x implies that vx @ E,  {el, ed), where {el,ed) = E, n E(bd(Fo)). Let the paths PI and P2 be defined by
(possibly E(P,) = 0 or E(P2) = 0). We can consider two cases. a) Suppose there exists xi E V(P;)  {x), i = 1,2, such that either x1x2 E E ( H ) , or for some j E (1,. . ., k), xlv(j), x2v(j) E E(H). Then let P3= P3(xl, x2) be the path joining xl and x2 in Po; x E V(P3)  {x, , x2) follows. Define the cycle C j of H by
C j = (E(P3) u {xlx2)) if xlx2 E E ( H ) , Cj = (E(P3) U E,(,,) otherwise ) (note that {x,v(j), x2v(j)) = E,(jl if xlx2 @ E(H)). In any case, ~ ( j:= (V(Cj)) is a %connected subgraph of H. In fact, Cj C bd(F,), where F, is the unbounded (outer) face of H. Since H and ~ ( jC)H are also outerplane with Cj being the boundary of the outer face of ~ ( j )we, can ) every u E V(Cj)  {xl, x2, v(j)} (observe conclude that E, c E ( H ( ~ ) for that the elements of E, nE(P,) are consecutive in O+(u)). Consequently, u cannot be a cut vertex of H since ~ ( jis)a 2connected subgraph of H and thus a subgraph of a block of H. Observing that
we can conclude that x is one of these vertices u and, therefore, x is not a cut vertex of H. b) Assuming now that xi E V(Pi)  {x), i = 1,2, with the properties stated in a) does not exist, we immediately deduce the validity of the following equations:
VI.3. ATrails in Plane Graphs
VI.83
Consequently, assuming first that E ( P l ) # 0 # E(P2), it follows that v and x separate G. Moreover, these two equations imply even that in G there exists a face F # Fo with v, x E ~ ( b d ( ~ ) ) By . ~ the ) choice of x, F cannot be a 1face; hence it is a 2face both in G and H. Thus, v(j) E V(bd(F)) n V(H) for some j E 1 . .,k . Similar to case a ) ~ )denote the path containing x and defined by let P3 = P ~ ( X ~ ,CXPO {v(j)xl ,v(j)x2) = EvO).Using the same symbols, define Cj and ~ ( j as in case a); again Cj C bd(F,), and Cj is a cycle both in G and H. Continuing the argument as in case a), we can conclude that x is not a cut vertex of H if x = u for some u E V(P3)  {xl,x2). By this and because vx $ E,  {el, ed) by the choice of x, we are led to the following conclusions: x E {x,, x,), hence vx E E, C E(G) which implies vx Z: {el, ed). So, either E(Pl)= 0 or E(P2)= 0. W.1.o.g. vx = el. Let x' denote the element of V ( G ) {v) incident with e2 which is defined by O+(v) = (e',,eh,. . . ,e&),and let in this case P3 = P3(x, 2') c PObe the path joining x and x' in Po. Again, by using the same symbols as before and by observing that in this case j = 1, we define the same way as before the cycle C, which is the boundary of the outer face of H(l) where, as before, ~ ( l:=) (V(C1)) is a Zconnected outerplane subgraph of H. But also in this case we can conclude E, C E ( H ( ~ ) ) ;for we can express C1 as
C, = Pi u {e,) for Pi = P3U {v, vx)
.
This gives the situation of case a ) with v and x' assuming the roles played by xl and x,. That is, x is not an end vertex of Pi; so, as before, E, c E(H(')) follows, and therefore, x is not a cut vertex of H. This finishes the proof of property 2). To see that H is a nontrivial blockchain once again consider Po as defined a t the beginning of the proof of property 2), and let x,, x, x2 be three different vertices lying in this order on Po.We claim that x is not a cut vertex of H if xl and x2 belong to the same block B c H.
B is 2connected since H is eulerian and simple. Hence there is a cycle C C B with xl, x2 E V(C). For the path P3= P3(x1, x2) E Po,form the Observe that in the definition of PI it is not said whether Pl is the path in Po'to the left7 or 'to the right' of X; hence it is no loss of generality, if we write the same index i on both sides of the last equation. 2, This implication is true, however, for every separating pair of vertices in a 2connected plane graph.
)
VI. Various Types of Eulerian Trails
VI.84
2connected subgraph B1 of H defined by
(possibly B, = C ; B1 # B if B, # C). In any case, B1 C B because C C B and K ( B ~2) 2. Also, x E V(B1) since x E V(P3); hence x E V ( B ) . Moreover, P3 C B implies that P3C bd(F*), where F*is the unbounded (outer) face of the outerplane graph B . Now, we observe as before that the elements of E, n E(P3) are consecutive in O+ (x) ; hence E, c E (B), i.e., x is not a cut vertex of H. That is, since j = 1,.. . ,k , is not a cut vertex of H and since V ( P o )= V(H)  {v(j)/j= 1,.. . , k), a block of H contains at most two cut vertices of H. Consequently, since H is connected but not 2connected and since H is a simple graph, it follo~vs that a cut vertex of H belongs to two blocks of H; that is bc(H) is a nontrivial path. This implies the validity of property 3) and finishes the proof of Lemma VI.61. Finally, we prove a property of 2connected outerplane graphs. Lemma VI.62. Let B be a 2connected outerplane simple graph whose outer face is F,, and let B := B  e for some edge e = x y E E(bd(F,)). Denote by F2 # F, the other face of B with e E E(bd(F2)). Then Bhas the following properties: 1) V(bd(F2))  {x, y) is the set of cut vertices of B. 2)
B is a (nontrivial) blockchain with x and y belonging to different endblocks of B, and they are not cut vertices of B.
Proof. The lemma is trivially true if B is a cycle, i.e., if B is a path. Hence suppose that B is not a cycle. Any w E V(bd(F2)) {x, y) is a cut vertex of B: this follows from the fact that w E V(bd(F2))nV(bd(F,)) and xy E E(bd(F2)) n E(bd(F,)) which implies that there is a plane simple closed curve C containing nothing in B except w and precisely one point of the edge x y (with x y viewed as a topological image of the open unit interval (0,l)). Thus x and y lie on different sides of C; hence, every path from x to y in B passes through w; i.e., w is a cut vertex of B. Observe that V(bd(F2))  {x, y) # 0 because G is a simple graph. Now let z be any cut vertex of B. Since B is 2connected while B is not 2connected, it follows from Theorem 111.32 that B is a nontrivial blockchain with x and y belonging to different endblocks of B, and they are not cutvertices of B. Hence every path P(x, y) C B contains z , and
VI.3. ATrails in Plane Graphs
VI.85
so every cycle of B containing xy also contains z. Since bd(F,) is such a cycle, and since x and y are not cut vertices of B , z E V(bd(F,))  {x, y } follows. This proves the lemma. Now we turn to Regner's result on Atrails in outerplane eulerian graphs.
Theorem VI.63. Let G be a 2connected outerplane simple eulerian graph, and consider a 2facecoloring of G such that the outer face Fois a 1face. Let v E V(bd(Fo)) with d(v) > 2 be arbitrarily chosen (if such v exists). Then G has an Atrail inducing a 1splitting in v. Proof. If G has only 2valent vertices it is a cycle; a run through this cycle represents an Atrail, whence we can assume that G has at least one vertex whose valency exceeds 2. We observe that G being a 2connected 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 2valent 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 1splitting in v renders the graph GO,,, having precisely two 4valent 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 Atrail To inducing a 1splitting in v and a 2splitting 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 Atrail T inducing a 1splitting in the chosen vertex v and a 2splitting in the other two 4valent 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 el, ed E E(bd(Fo)) where Fo is the outer face of G. Now form H = G,,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 x,, . . .,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 x j 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. Various Types of Eulerian Trails
VI.86
Figure VI.22. The 2connected outerplane eulerian graph Go having an Atrail Towhich induces a 1splitting in v. B,, . . . ,B,+, such that xl E V(Bl), x, E V(B,+l) and xi,,xi E V(Bi) for 2 5 i 5 r. As for the distribution of the j = 1,.. . ,k, d(v) = 2k, in these blocks Bi, i = 1,.. . , r 1, we have
+
since G is 2connected and because of the definition of v(j), j = 1,.. . ,k. W.l.o.g., v(') E V(B,), v(" 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(Bi, ). Of course, there may be blocks of H not containing any v(j), j = 1, . . .,k; but for j # m we have i # i, because among the four vertices of Poincident 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 of v(j), j = 1,. . ., k, that if j < m, then ij < i,. In addition to the xi, i = 1,. . .,r , we introduce zo and x,+, defined in G by vxo = el and U X , + ~ = ed. Thus, Bi n {xj/j = 0,.. .,r 1) = {xi,, xi). Figure VI.23
+
+
VI.3. ATrails in Plane Graphs
VI.87
illustrates the structure of H which follows from Lemma VI.61 and the above argument.
G
H
Figure VI.23. The graphs G and H = G,,l.
The outer face F, of the blockchain H is a 1face; and Po C bd(F,). Moreover, bd(F,) n E(Bi) # 8 for i = 1,.. . , r 1 because E(Bi) n E(Po)# 0. This and the chosen notation allow the application of Lemma VI.60. Therefore, an Atrail of G inducing a 1splitting in v necessarily induces a 2splitting in each of xl, .. .,x,. On the other hand, if we can and xi, i = show that Bi has an Atrail inducing a 2splitting in 1,. . .,r 1, then H has an Atrail by Lemma VI.60, which is equivalent to saying that G has an Atrail inducing a 1splitting in v. Hence, to finish the proof of the theorem, it suffices to show that (Bi)(zil,zil,2 has an Atrail for i = 1,.. . ,r 1. For this purpose, fix an arbitrary i E (1,. . .,r 1) and consider Bi . Depending on the position of Bi in H and its structure, we can consider the following cases.
+
+
+
+
(I) V ( B j ) n { v ( j ) / j = l , . . . , k } = 0
.
Denote for short B = Bi, x = xil, and y = xi. Let A , = (x, v, tl), A 2 = (y, v, t2) be two triangles with t l , t2 $ V ( G ) ,and construct a plane embedding of the planar graph
VI.8S
VI. Various Types of Eulerian Trails
such that t, and t2 are 2valent 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(bd(Fk)) = V(P(x, y ) ) , where
P(x, y) = bd(Fk) n Po ;
otherwise, Po U {el, e d ) = 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 f as defined above and illustrated by Figure VI.24 is outerplane indeed. Moreover, since B is 2connected, we can conclude by the same reasoning that P' (x, y) := bd(FL)  P ( x , y) contains no edge other than xy (see Figure VI.24).
Figure VI.24. Forming the outerplane graph B+ from the outerplane graph B by adding the triangles Al 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 Atrail T+ of B+ inducing a 1splitting in v (the 2facecoloring of B + is induced by the 2facecoloring of G via the induced 2facecoloring of B). Observing that T+, read as an edge sequence, can also be regarded as an Atrail of B:, we can conclude that by Lemma VI.60, B has an Atrail T such that the vertex splittings defined by X T ( x ) and XT(y), are Zsplittings.
VI.3. ATrails in Plane Graphs
VI.89
However, if B has precisely n 1 vertices of valency exceeding 2, induction cannot be applied to B+ as constructed above. So in this case we have to proceed differently. Consider the 2face F2of B (which, in fact, is also a 2face of G) with xy E bd(F2), and form B = B  {xy). By Lemma VI.62 we can write the blockchain B as the union of its blocks,
... U B , , where t =I V(bd(F2))I 1, and B i n B3T # 0 if and only if I i  j I= Furthermore, we can denote for j = 1 , . . ., t  1, B=B,U
1.
wj E V(bd(&))  {x, y) by Lemma VI.62. In addition, denote wo = x and w, = y (w.1.o.g. x E B,, y E B,). Note that since dB ( z )
 1(mod 2 ) , if and only if
z E {x, y)
,
we have for j E J := { I , . . . , t ) dB (wj,) zz dBr (wj) 3 1(mod 2), 3
dg7(u) 3
3
= O(rnod2)
if
u
#W~~,W~.
B being outerplane and V(bd(F2)) = {wj/j = 0,. . . ,t ) yields
Hence wjl wj E E(B3T) for j E J, and by the above congruences, Bi := Bj  {wj,wj) is eulerian. Of course, E(BS) = 0 may hold true for some (but not all) j E J. Consider Jo:= { j E J / E ( B i ) # 0) and observe that since B3T is a block of B, it follows that BIT is 2connected for j E Jo;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 nontrivial blockchain; thus it has a cut vertex, and wj,,wj belong to different endblocks 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 B;
U
j E Jo
VI.90
VI. Various Types of Eulerian Trails
Figure VI.25. A block B = Bi of H, the noneulerian blocks B; of B  {xy), i E J, and the eulerian blockchains B; of By  { W ~  ~ W ~ j) ,E Jo.
where Bj is a nontrivial eulerian blockchain for j E Jo; any two of these blockchains have at most one vertex in common, and that vertex belongs to V(bd(F2)). Figure VI.25 illustrates this structure of B. Our aim now is to find for every j E Jo two subtrails covering B; and combine them with the cycle bd(F,) so that it results in an Atrail 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 2connected outerplane eulerian graph B? by attaching
AY)
to Bi two triangles A? = ( W ~  ~ , P tl), , = (wj,p, t,) at wj,, wj respectively, where p, tl, t2 g' V(G) and t l , t2 are 2valent in BF. We now have wjl in place of x, wj in place of y, p in place of v, and Bj in place of B, if we compare this construction with the one illustrated by Figure VI.24. The fact that B is 2connected while B; is a blockchain (and therefore, cannot contain wjlwj) does not matter. In any case, Bf has nf < n vertices of valency exceeding 2; this derives from the fact that B is a nontrivial blockchain. Now, by induction we can conclude that B: has an Atrail T; inducing in cj a 1splitting. By
VI.3. ATrails in Plane Graphs
VI.91
Lemma V1.60, T : induces a 2splitting in wj,, wj and p (observe that (B:)~,,~ is a blockchain with wjWl,wj7p as three of its cut vertices). Hence we can write T : as an edge sequence as follows:
where Ti') and Ti') are edgedisjoint closed trails covering B;. T:)' duces a 2splitting in wj,
in
, as does Ti') with respect to wj. Since Ti')
and T J ~are ) Atrails of the respective components of (B;)c, ,1, it follows 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 Ti'), Ti'). By analogy to the proof of Lemma VI.60, we can now choose for every j E J, edges ej, fj,gj, hj with ej, f j E E,.1  1 nE(B(i),gj, hj E E w j n E ( B j ) so that the trails Ti') and Ti') (viewed as edge sequences) can be written in the form  e j , T l j , f j 7 and Ti') = hj,Tzj7gj
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,, contains edges of E,, while T!")' does not exist; if j # Jo,j 1 E J,, Ti'+') contains edges of E,, while Ti') does exist and contain not exist; and if j, j 1 E J,, both Ti') and T!")' edges of E,,. Moreover, if we look at O+(wj) in B, either
+
TA')
+
depending on which of the above four cases concerning j, j+ 1 E J prevails (wjl = wi if j =O). Defining for i = 1,2 and j E J
~T/"=T/') if j E J,,
and 6 T / j 1 = 0
if j E J  J,
,
VI.92
VI. Various Types of Eulerian Trails
and taking into account the above equations concerning T!)' and O+(wj), we conclude that
is an Atrail of B inducing in wj, j E J U (0) and, therefore, especially in wo = xil and wt = xi a 2splitting. In other words, (Bi)(ri, , r i l , 2 has an Atrail. This settles the case (I). Then there exists precisely one j E { I , . . . ,k) so that v(j) E V(Bi) (see the discussion preceding Figure VI.23). Again denote for short B = B i , x = xil, y = xi, set v = v(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 xy 51 E(B), we suppress v to obtain B* which is of the type treated in (I);and any Atrail of B* corresponds to an Atrail of B, and vice versa. Thus, for xy 6 E ( B ) induction applied to B* yields an Atrail of B as required. Whence we assume xy E E(B): B contains a triangular 2face A = (v, 2, Y). If A = B, a run through A is an Atrail of B as required because of our additional definition of a 6splitting in Zvalent vertices. Assume, therefore, A # B. In this case we have precisely the situation illustrated by Figure VI.25 with t = 2, Jo= (21, and v in place of wo, x in place of wl, and y having the same meaning as in Figure VI.25. Note that B  v is a 2connected outerplane simple graph; hence B := (B  v)  {xy) = B  A is a nontrivial blockchain by Lemma VI.62. As in case (I) we find c E P(x, y)  {x, y) which is a cut vertex of B, and by induction an Atrail TB of B which induces a 1splitting in c and, therefore, a 2splitting 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 2connected eulerian plane graph has an Atrail. In fact, when I first considered the problem of finding Atrails in plane
VI.3. ATrails in Plane Graphs
VI.93
eulerian graphs, I thought that 2connectedness would be a sufficient condition for a plane eulerian graph G to admit a partition {V;, V2/) of V(G)  {v) for some v E V(G), such that (Gvl,1)V;,2 does not contain a subgraph homeomorphic to the graph G1 of Figure VI.17. For, if one considers the problem of finding an Atrail 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 Atrail, if and only if for every u E V(G) there is a partition {V., V2/) of V(G)  {u) with the property just described. Therefore, the graph G1 of Figure VI.17 characterizes the forbidden stage that one has to avoid in trying to produce an Atrail by a sequence of 6splittings, 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 2connected plane eulerian simple graphs having only 4 and 6valent vertices and not having any Atrail. 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 2facecolored with the outer face being a 1face. Suppose now that Go has an Atrail T. Observing that (Go)(V1,V61,2 is a disconnected graph and applying Corollary VI.57, it follows from the symmetry of Go that this graph also has an Atrail inducing a 1splitting in v,. W.1.o.g. suppose T has been chosen with this property.

This implies, however, with necessity that the 6splitting induced by T in vi, 2 i 6, satisfies the congruence i 6 (mod 2). That is, we arrive a t the graph H = , v ~ , v ~,{ ~ ,,,,, , I ,l,2 as illustrated by Figure VI.27, where T, viewed as an edge sequence, is an Atrail of H as well. Now, if one applies the corresponding 6splitting to every 4valent vertex of H, one obtains a graph homeomorphic to the graph G1 of Figure VI.17.') No matter which 6splitting 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 Atrail.
<
1, be m distinct faces of the 2facecolored plane eulerian graph such that, for a fixed 6 E {1,2}, Fiis a 6face for i = 1,.. . ,m. Let us assume distinct vertices v , , ~v, ~ ,. .~.,,v,,~ exist with v;,++~E bd(Fi) n bd(F;+,), i = 1,...,m (with the subscripts read mod m). Then we call R = {F,, . . . ,F,) a unicolored facering of G and LR := {vl,*, . . ., a complete set of links of R (if m = 2, then v1,2 # v2,1). The following result relates Apartitions to facerings and complete sets of links (see [FLEI74a, Theorem 1 and Corollary I]). Theorem VI.67. Let G be a 2facecolored Zconnected plane eulerian graph without 2valent vertices, and let {Vl, V2} be a partition of V(G). The following statements are equivalent.
VI.3. ATrails in Plane Graphs
VI.101
(1) G has an Atrail T whose Apartition is {Vl, V2). (2) For every unicolored facering R of G and every complete set of links LR of R, if the elements of R are 6faces then LR V6, 6 E {I?2).
Moreover, if we replace in (1) "Apartition" with "perfect Apartition" and in (2) " L R V6" with "Vl 2 LR Sf V2", then these stronger statements are equivalent as well. Proof. (1) implies (2). Suppose for some unicolored facering R of G, whose elements are 6faces, 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 facering of G and 6 = 1, distinct 1faces Fl, . . . ,Fmof G exist such that
and by definition,
Now let Ci c Fi U bd(Fi) be a simple open curve with Ci n bd(Fi) = {vil ,i7 v ~ , ; + ~ ) .Then C = Ci is a simple closed curve. Precisely because Fi is a 1face, 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 E,i,i+l,i = 1,.. .,m (otherwise, Ci and Ci+, would lie in faces of different colors, or else Ci = Ci+, which implies Fi = Fi+l). Moreover, C n G = LR S Vl by assumption.
ULl
W.1.o.g. suppose the notation chosen in such a way that for O + ( V ~ , ~ +=, ) (ei ,e;, . . . ,e&,),di = d(viVi+,) > 2, precisely the open edges el, e2, . . .,en; lie in the interior of C. Then 2 5 ni = 0 (mod 2) and di  ni 2 follows from the preceding paragraph. Then we can perform the 1splitting in vi,++,, i = 1,. . . ,m, in such a way that v(l), . . .,v ( ~ ; )lie entirely in the interior of C, while v("'+') ,. . . ,v(") lie entirely in the exterior of C, where mi = $ni and ki = i d i (see Figure VI.18). Consequently, since C n G = LR c Vl, we can conclude that H is disconnected.
>
,,
Since H is disconnected and LR C Vl, it follows that Gvl is disconnected, and further that (Gvl ,l)v2,2is disconnected. On the other hand,
VI.102
VI. Various Types of Eulerian Trails
{V,, V2) being the Apartition of an Atrail T of G with V6 containing precisely those vertices at which T induces a 6splitting, 6 = 1,2, implies that (GVl,?)V2,2is a cycle (compare this with Corollary V1.57). This contradiction proves the implication. ( 2 ) implies ( 1 ) . Consider H = (Gvl,l)v,,2 which is a 2regular graph. If
H is connected, then by Corollary VI.57, {V,, V2) is an Apartition obtained from some Atrail T inducing 6splittings precisely in the elements of V6, 6 = 1,2. In this case, the implication follows. Hence suppose H to be disconnected. Choose Vo V(G) as small as possible such that Ho := (GVd,l)Vdf,2 is disconnected, where Vd = Vo n V,, V,' = Vo n V2. It follows that Ho has a face Fo such that bd(Fo) is disconnected; w.1.o.g. Fois the unbounded face of Ho. Furthermore, since we did not make any assumptions concerning the 2facecoloring of G, we may now assume that Fo is a 1face. : Now, if V', # 0, consider w E .V We claim that already HI := (GV;,I)V;~{~},~ has a disconnected face boundary (whence HI is disconnected): for, Ho = (H1)Iw),2, and by definition of a 6splitting, 6 E {1,2), the application of the 2splitting to a vertex of H, leaves the 1faces of H, (homotopically) invariant (see the footnote at the end of Definition VI.54); i.e., bd(Fo) is disconnected already in H,. This contradicts the choice of Vo; hence V : = 0 and, therefore Vo & V,. Again by the choice of Vo, and since G is connected, we must have Vo C VFo for VFo := v ( ( E ( b ~ l ( F ~ ) ) )since ~ ) , Vo Vl and Ho being disconnected implies that GVonvF0 ,, has a disconnected outer face. Consider C,,a component of bd(Fo), define C2:= bd(Fo)  C,, and let C be a simple closed curve lying in Fo and such that int C > H,, e x t C > Ho HI, where H1is the component of Ho with HI > C1.The choice of Vo implies that for every v E Vo, some of the di)lie in int C (i.e., in HI) and some lie in e x t C (i.e., in H,  H,), i = 1,.. .,:d,(v); .  . . otherwise, one obtains C n GV0{,~,,= 0 and lntc n G V O  ~ , ~#, , 0 # e+t C n Gvo~,l,l for some v E V,. W.1.o.g. we may assume that C is drawn in such a way that the transition from Ho to G, viewed as a topological procedure, leaves C invariant (ie., w.1.o.g. C is a simple closed curve in G as well compare this with Figure VI.18) and C nG = Vo while (C  Vo)nF ( ~ ) = 0 for every 2face F ( ~of) G.
C n G = Vo implies that we can write Vo as
VI.3. ATrails in Plane Graphs
VI. 103
where and vi,++, divide C into two simple open curves C: and Cy such that C: n 5 = {u;,,~, C:  {vi,,;, u;,~+,) lies in a uniquely determined 1face, call it Fi, i = 1,. . .,m. It follows from the choice of Vo that Fi#Fj for i # j , l < i , j < m , (i2 and that bd(F+)n bd(Fj)
#0
if and only if
i j 1 1 m  1
,
(i3)
where bd(Fi) n bd(Fi+,) = {u,,++,) (putting m
+ 1= I)
.
(i4
Otherwise, one could draw a simple closed curve C in the plane such that C n F(,) = 0 for every 2face of G, CnG= V; c Vo, and (C V,) n Fi = 0 for at least one but not all i E {I,.. .,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 facering and Vo as a complete set of links of R. Since R is a set of 1faces and Vo C V,, R and Vo violate the validity of (2); hence H must be connected in any case.
,
To finish the proof of the theorem suppose now that the Apartition in (1) is a perfect Apartition, and suppose in (2) that LR fulfills the stronger relation V, 2 LR 5 ;call the corresponding statements (1') and (2'). If {V,, V2) is a perfect Apartition, then, by definition, (GVl,1)V2,2is a cycle (which corresponds to T), and so is (Gv,,l)v, ,. Denote by T* the Atrail of G corresponding to the second cycle and define V;' := V,, V; := V,. Since (1)implies (2) a twofold application of this implication yields Vz 2 LR Jf Va for every unicolored facering R and every complete set of links LR; i.e., LR V6, 6 = 1,2. Whence we conclude that (1') implies (2'). Observe that {V6, V ): = {V, ,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 facering R. Suppose (2') holds. Define V$ as above. Since (2) implies (1) we obtain from the proof of this implication that both (Gvl ,l)v2a and (Gv; , = (Gv2,1)v,,2 are cycles; i.e. {V,, V,) is a perfect Apartition. Theorem VI.67 now follows.
VI.104
VI. Various Types of Eulerian Trails
G, 6 = The theorem just proved indicates that if we look at G, = (V,) 1,2, where (Vl, V2) is an Apartition and G satisfies the hypothesis of the theorem , then G, must have a very special structure. In order to 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 i n t C and ext C contain (open) edges of G (where we view C as a simple closed curve in the plane). Furthermore, both i n t C and extC contain 1faces and 2faces as well. In particular, every e E E(C) belongs to some bd(F,) where F6 is a 6face, 6 = 1,2. Hence, considering
we conclude that R, is a unicolored facering having a complete set of links LR V(C), 6 = 1,2. By Theorem VI.67, V(C) If V,, and therefore, C G6, 6 = 1,2. Hence,
if G6 contains a cycle I 2) as a complete set of links of R6,5 E {1,2). In this case, LR If V,, S E {1,2), by Theorem VI.67; hence ( L R ) C G6, 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 Atrail of the connected 2facecolored plane eulerian graph, G, and if {V,, V2) is a corresponding Apartition of G, G, := (V6) contains a cycle C, only if C is the boundary of a 6face, 6 E (1,2).
We point out, however, that neither GI nor G2 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 {V,, V2) such that G, = (V,) is acyclic, although G does not have an Atrail. This can be seen by studying the graph Go of Figure VI.26; we leave this as an exercise.
VI.3. ATrails in Plane Graphs
VI. 105
Figure VI.29. (a) The foursided antiprism H, (b) an Atrail T of H, and (c) the subgraphs induced by the Apartition of T. Observe that H is 4regular and has, except for two 4gonal faces, triangular faces only.
The next result, however, shows that the converse of Corollary VI.68 is true provided G6 is connected. For this result, however, we need some considerations on unicolored facerings. Let us consider a unicolored facering R = {Fly F2,.. .,F,), m 2 2, of the 2connected, 2facecolored plane eulerian graph G. Suppose bd(Fi)nbd(Fj)=O
for
lijl>l,l 2 holds. That is, at least one 1face Fl # F, exists. a) Let T be an Atrail of G, and let Fl # F, be arbitrarily chosen. Moreover, let {V,, V,) denote any Apartition defined by T. Since Fl is a 1face we must have I V(bd(Fl)) n Vl I> 0; this follows exclusively from the fact A(G) > 2; for in this case, V(bd(Fl)) n Vl = 0 implies that Gv2,2 is disconnected with (E(bd(Fl))) as a component. Suppose now V(bd(Fl))nVl > {v, w). Then R = {F,,F,) is a unicolored facering with LR = {v, W ) a s a complete set of links. The elements of R are 1faces, and LR C Vl follows from the assumption. By Theorem VI.67, however, this contradicts the fact that {Vl, V2} is an Apartition. b) Suppose the partition {V,, V2) of V(G) satisfies I V(bd(F,)) n V, I= 1 for every 1face F, # F,. Consider an arbitrary unicolored facering R
VI. Various Types of Eulerian Trails
VI.148
whose elements are 6faces 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.
I I>
Suppose first 6 = 1. Since R 1 it follows that Fl E R for some F1 # F,, and by definition I V(bd(Fl)) n L R I= 2. Consequently, V(bd(Fl)) r l V, 1 which violates 2).
I
I>
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 facering R1 having L as a complete set of links. We want to show that L R = V(bd(F*)) for some 1face F* # F,.
I
IE
It follows from the minimality of LR that bd(F1) n bd(Ftl) n LR {O,1} for different elements F1,Fl1E R. Hence we can espress R and LR in the following form:
where
LR n bd(F,)
n bd(Fi+l) = { v ~ , + + ~for) i = 1,.. .,m
bd(Fi)nbd(Fj) = 0 for j  i
> 1, 15 i < j 5 m, { i , j ) # {l,m} .
As in the first part of the proof of Theorem VI.67, we construct a simple closed curve C with
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 Pi,i+,joining vil,i a d v ~ , ; +in~ bd(Fi) whose open edges lie in int C must satisfy
and therefore,
VI.3.2. ATrails and Hamiltonian Cycles in Eulerian Graphs VI.149
Hence, (LR) C int C U LR and (LR) is a cycle. This, X(e) = l ( e E E(G)), and the rninimality of LR imply that (int C  E((LR)))n G = 0
.
That is, adjacent edges in (LR) are consecutive in O+(v), where v is their common endvertex. Thus (LR) is a face boundary bd(F*). Since E((LR))fl E(bd(F,)) # 0 for i = 1 , . . . ,m, it follows of necessity that F* is a 1face; and F* # F, because (LR)n ext C = 0. Thus, we have found a 1face F*with
This contradiction to statement 2) finishes the case 6 = 2; whence we conclude that LR V6, 6 E {1,2}, for every unicolored facering R whose elements are 6faces, with LR being a complete set of links of R. By Theorem VI.67, G has an Atrail whose Apartition is {V,, 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 Aproperty. Then G has an Atrail. Proof. We can write G as the union of two 2connected, outerplane, eulerian graphs GI and G2 with GlnG2 = H (see the discussion preceding Theorem VI.87). Let T, and T2 be Atrails of GI, G2 respectively; by Theorem VI.63, T, and T2 exist (we start with a 2facecoloring of G and apply Theorem VI.63 to Gi whose 2facecoloring is induced by that of G, i = 1,2. Thus, the face of G, 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 Aproperty of H we have for arbitrary v E V(G) ~ G ~ (>U2,) only if dGj(v) = 2 and dci(v) = dG(v), {i,j) = {I, 2). (*) Consider now {V;', V;), an Apartition corresponding to Ti in G,, i = 1,2, and define
W: :=
n (V(Gi)  V.(G,))
for arbitrary i, j E {I,2)
.
VI. Various Types of Eulerian Trails
VI. 150
This definition together with V:
w' n W: (observe that W j
= 0 for
n Vj = 0 and (*) yield
( j i) # ( k 1 )
2 V2(G,) for {i, k)
{ j k 1)
{
l
(+r)
= {l,2)).
Consequently, {l.V;/i, j = 1,2) is a partition of V(G)  V2(G); therefore, Wj := W j U Wf, j = 1,2, defines a partition of V(G)  V2(G). Now we observe that a 2valent vertex v2 cannot belong to any complete set of links LR of any unicolored facering R; for v2 belongs to precisely one 6face, 6 = 1,2. Hence, if we can show for such R and L that LR g W6, provided the elements of R are 6faces, 6 E {1,2), then for V2(G) an Apartition {V,, V,) of V(G) is obtained by defining any V," V, := W, U V,", V2 := l.V2 U (V2(G)  V,") (see Theorem VI.67).
s
So, let R and LR be as above. If all elements of R are 6faces of Gi, i E {1,2), then the existence of Tiimplies LR g Wi by Theorem VI.67, 6 E {1,2}. Since Wi 2 V,(G,) for {i,j) = {1,2}, LR g Wb follows. Hence we have to assume that some elements of R are faces of G, and some are faces of G2; and all of them are 6faces, 6 E {1,2}. We further specify the situation by assuming w.1.o.g. that the "outer" face F1( 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(Fl) 2 G I , bd(F,) 2 G2. Define
+
i2 = min{i/l
< i 5 m A bd(Fj) zG,}
+
and
il = i2  1 .
Then we obtain in G, a unicolored facering R' whose elements are 6faces, and a corresponding complete set of links L' = LRt by defining
We have L' E LR and L' V(G1). If we had LR C W6, this and W i C V2(G1) would imply L' E Wi C VJ. But Theorem VI.67, applied to G,, says that the existence of TIand its Apartition {V:, Vi) implies L' V,'. Thus, we must have LR g Ws. Again by Theorem VI.67 and the above definition of V, and V2, this means that (V,, V2) is an Apartition of G; i.e., G has an Atrail. Theorem VI.88 now follows.
VI.3.2. ATrails and Hamiltonian Cycles in Eulerian Graphs VI.1.51
However, not every Atrail T of G is obtainable from Atrails T, and T, of G, and G2 respectively, as described in the proof of Theorem ~ 1 . 8 g That is, some Apartition {Vl, V2) of V(G) may not be an Apartition of V(Gi) = V(G) for i = 1,2. This can be seen already in the case of O, which we can view as obtained from the cycle C6 = (v,, . . . , v6) by adding A, = (vl ,v3, v5) and A 2 = (v2, v4,us). Embed A, in the bounded region of C6; thus A 2 lies in the unbounded region of C6. We assume the outer face of the 2facecolored O6 to be a 1face. Defining H := C6, GI := H U A,, G, := H u A,, and Vl := {u2, v3, v6), V2 := {v,, v4, v5) we conclude that although {Vl , V2) is a (perfect) Apartition of V(G), {Vl , V,) is not an Apartition of V(G1) = V(G), nor of V(G2) = V(G) (note that H has the A  ~ r o ~ e r tFor, ~ ) . (G1){u121.51,2 is disconnected, and so is (G2){vz,V6 However, if an Apartition {V,, V2) of V(G) is an Apartition of V(Gi) as well, i E {1,2), we may say that the latter is an induced Apartition of the former. Correspondingly, we can say that the Atrail Tiof Gi is induced by the Atrail T of G. The graph of Figure VI.38 sho~vsan eulerian triangulation D of the plane together with a hamiltonian cycle H such that the corresponding graphs G,, G2 satisfying G1 U G2 = D and Gl n G2 = H (see the discussion preceding Theorem VI.87), are eulerian. We discuss this graph D: H does not have the Aproperty since (V(G1)  V,(G,)) n (V(G2)  V,(G,))   = {a, b, c, d, e, f ) . We know from Theorem VI.63 that Gi has an Atrail Ti, i = 1,2. We want to show that no choice of TI and T, permits extension of these Atrails to an Atrail T of G in a waysimilar to the one in the proof of Theorem VI.88. For this purpose, it suffices to show that TI does not induce the same type of splitting in Vo := {a, b, c, d, e, f ) as T2. To be more precise, let G, and G2 have their 2facecolorings carried over from a 2facecoloring of D. Among G, and G2 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 2facecoloring of D is chosen in such a way that F, in G, is a 1face. Whence we may conclude that the "outer7' face int H of G2 is a 2face. As in the proof of Theorem VI.88, let {V;', Vj) be the Apartition of the arbitrarily chosen Atrail Tiof Gi, i = 1,2. We want to show that the equations V, n V: = Vo n V ,: Vg n V2' = Vo fl V22 cannot hold. Suppose a E Vll. Since d, e, f are boundary vertices of 1faces 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 Vj2is impossible since {e, f ) c bd(F2) for some 2face F, of G2 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 2connected, outerplane, eulerian graphs G,,G2 with D = G1 U G2, H = G1 n G,. D has no Apartition which induces Apartitions in both G1 and G, (note that (V(Gl)  V2(Gl)) n (V(G,) Q(G2)) # 0). face of G, is a 2face as well. Therefore, the above equations cannot hold if a E V.: Now suppose a E Vl nV;. Since b and c are boundary vertices of 2faces of G2 each of which also has a as a boundary vertex, we have {b, c ) c V,' by Theorem VI.87. On the other hand, {b, c ) E V(bd(Fl)) for some lface of Gl ; thus, by Theorem VI.87 we must have {b, c) V: . Consequently, also in the case a E V2' we may conclude that the above equations cannot hold. However, we leave it as an exercise to check that D has an Atrail indeed (we note, though, that n(D) = 3).
VI.3.2. ATrails and Hamiltonian Cycles in Eulerian Graphs VI.153
So, while in general it is not possible to deduce from some Atrail of G Atrails Ti in Gi, i = 1,2 (where Gi is defined as above with respect to a harniltonian cycle H ) , even if H has the Aproperty, 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 2facecolored, plane, eulerian graph G that it has a hamiltonian cycle H defining two outerplane eulerian graphs G I , G2 satisfying G1 U G2 = G, G1 n G, = H (see above). If G has an Atrail T inducing an Atrail Ti of Gi, T induces an Atrail Tj in G j as well, {i,j) = {1,2). Proof. We assume w.1.o.g. that F,, the unbounded face of G1 (with bd(F,) = H), is a lface in the 2facecoloring of G1 induced by the 2facecoloring of G, and that G has an Atrail T inducing an Atrail T9 in G,. Let {Vl, V,) be the Apartition of V(G) = V(G,) correspondillg to T, T2respectively. We have to show that {V,, V,) is an Apartition of V(Gl) as well. For this purpose we first observe that G1 := GL,l,l is a connected outerplane graph whose outer face F&, is a lface with E(bd(F,)) E(bd(Fk)) and such that E(bd(F&,))n E ( B ) # 0 for every cycle B of G1 (Lemma VI.5S, Theorem VI.59). If we can show that every nontrivial block of G1  E(G1) is an endblock of G1, then G1  (G2 E ( H ) ) has precisely one nontrivial component, namely (G, ) vl ,1, which satisfies the hypothesis of Lemma VI.58. Consider the graph
Go is well defined because the "outer" face FM of G2 is a 2face in the induced 2facecoloring of G2 (since F, is a lface), and therefore E ( H ) = E(bd(Fw)); so, the construction of Go from (G,),l,, amounts to adding diagonals of H lying in GI, i.e., adding the elements of E(G1) E ( H ) . Also, since T2 appears as a run through the boundary of the uniquely determined lface of (G,) vl ,1, it follows by definition that Go is implies that connected. Moreover, E(bd(FM))C E((G2)", 1) (G,),] ,, contains (E(bd(Fm))) as a block;
therefore 2) Go has a block B w 3 (E(bd(Fw))) with Bw E GI;
VI. 154
VI. Various Types of Eulerian Trails
3 ) every block C of Go  Bm is a cycle with E(C) E E(G,)  E ( H ) and therefore, C is an endblock of Go (see 2)). We relabel the vertices of Bm according to their original labels in G; then we obtain Bm = G1 and the isomorphism
For every block C of Go other than Bm (see 3)) we have C f l Bm C V,. Hence, the unique vc E C n Bm is not affected by the transition from Go to Since every x E V(C  vc) is 2valent in Go and therefore ,1, it follows that every block C of Go  Bm is an endblock also in ,l. Finally, if we view G1 and as unlabeled graphs, of then the isomorphism (*) becomes an identity ( ~ n (*), the isomorphisnl can be chosen so as to act as the identity on E(G1) = E((GO)V,,l). Consequently, every nontrivial block of G1  E(Gl) is an endblock of G1 whose edge set belongs to G2  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 G1 which renders a connected graph and leaves unchanged the ,l embedding properties espressed in that lemma). It follows that has an Atrail. Consequently G1 has an Atrail. The theorem now follows. Maybe, the above considerations on Atrails 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, dconnected, eulerian graphs admit a hamiltonian cycle H having the Aproperty or at least an H such that the graphs G I , G2 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 Atrails and nonseparating Atrails 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
VI.3.2. ATrails axtd Hamiltonian Cycles in Eulerian Graphs VI.155
that one replaces 'Atrail' with 'nonseparating Atrail7. It follows from Theorem VI.71 and the relation &(D(G3))= X,(G3) for connected plane cubic G3 # Kq, that this modified conjecture is equivalent to the BTC (Conjecture VI.72) for the case of cyclically 4edgeconnected, planar, bipartite, cubic graphs. But what does the existence of a (not necessarily nonseparating) Atrail in D = D(G3) imply for G3 ? 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,, V2), and let D(G3) denote its dual. The following statements are equivalent. 1) D(G3) has an Atrail. 2) G3 has a Tutte cycle C such that V' := G3  V(C) satisfies
a) Vl n V' c int C if and only if V2 n V'
c ext C;
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 G3 and Atrails in D(G3) does not require Xc(G3) = 4. On the other hand, Tutte's 'Bridge Theorem' guarantees the existence of a dominating cycle in planar, cyclically 4edgeconnected, cubic graphs (Corollary 111.73). So, in the case of Xc(G3) = 4, the equivalence expressed in Theorem VI.90 boils down to the question whether in the nonempty set of Tutte cycles of G3 one can find an element satisfying a) and b). We also note that regarding nonseparating Atrails, it suffices to prove Conjecture VI.73 for Cconnected triangulations; this follows from Proposition VI.83, and it is tantamount to saying that it suffices to prove the BTC for the corresponding cyclically 4edgeconnected graphs (see Exercises VI.25 and VI.30).
VI. 156
VI. Various Types of Eulerian Trails
VI.3.3. How to Find ATrails: 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) NPcomplete or whether it can be solved/decided in polynomial time. l) In what follows, we basically restrict ourselves to 3connected, planar, eulerian graphs. For, as we have seen before, a 2connected, planar, eulerian graph G may have plane embeddings GI and G2 such that G , has no Atrail while G2 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 nonexistence of Atrails. On the other hand, 3connected, planar graphs G are uniquely embeddable on the plane (see Theorem 111.52); hence (see the discussion preceding Lemma VI.64), the esistence of an Atrail in such a G which is also eulerian, is independent of any concrete embedding of G on the plane (or equivalently, 3dimensional sphere). Nevertheless, for practical purposes we shall always consider some embedding of G. On the grounds of Regner's construction of a planar, 3connected, eulerian graph without Atrails (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 Atrail, is an NPcomplete 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. ATrails: Complexity Considerations, Algorithms
VI.137
This is true even if the problem is restricted to 3connected graphs.1)
Proof. By statement (V1.A),if G3 is a planar, 3connected, cubic graph, a planar, 3connected, eulerian graph G with D(G, ) c G exists such that no face boundary of D(G3) is a face boundary of G, and G has an Atrail if and only if G3 has a hamiltonian cycle. Since the hamiltonian problem for G3 is NPcomplete, [GARE76a], it is (more than) sufficient to show that the Atrail problem for G 3 D(G,), is polynomially equivalent to the hamiltonian problem for G,, and that G can be constructed from G3 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 2facecoloring of G can also be solved in polynomial time (Theorem 111.91). Hence, we may assume that the 2facecoloring of G is given as well. Assume an Atrail T given in G 3 D(G3). A single run through T tells us at every vertex v which is reached (be it that v is reached in this run for the first time or not), whether v E Vl or v E V2 where {Vl, C' ,)' is the Apartition 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 Ev consecutive in T belong to the boundary of a 1face or 2face of G. Thus, {V,,V,) is obtained in polynomial time Pl(pG) from T. Having stored V(D(G,)) we obtain v;l := V;: nV(D(G,)), such that (Kt)is a tree, i = 1,2, in polynomial time P2(pG). Knowing the bijection vF E V(D(G3)) tt bd(F) C G3 for every face F of G3 (which requires polynomial space and time only), we form Go = U V F c y bd(F) c G,. This and deleting those edges of Go which belong to bd(F1) and bd(F") for F' # F", 2]FlvFll E E((V:)), can be done in polynomial time P3(pG3),and results in H, a hamiltonian cycle of G,. Since, pG 5 const.pG3 (see below) P(pG3) := Pl (pG) P2 (pG) P3(pG3) is the polynomial in pG3 =I V(G3) I representing the time required to obtain the hamiltonian cycle H of G3 from the Atrail T of G.
+
+
The first part of this theorem has been proved in [BENT87a], where the authors speak of 'nonintersecting eulerian trails' what we call Atrails. 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
VI. Various Types of Eulerian Trails
Conversely, suppose a hamiltonian cycle H c G3 be given. Then it takes polynomial time Pi(pG3)to reach the vertex partition {V,', Vi) of D(G3) such that (V;') is a tree, i = 1 , 2 (V,' corresponds to the set of faces F of G3 with F C int H). In order to obtain an Apartition {V,, V2) of V ( G ) such that V;' = V; n V(D(G,)), i = 1,2, we choose a particular way of constructing a plane, 3connected, eulerian 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 PG <  fG3 +5pG3 <  6pG3,where the second inequality follows from Euler's polyhedron formula, and the first inequality is due to the fact that a face of D(G3) (which corresponds to a vertex of G3, and vice versa) contains either 3 or 5 vertices of V(G)  V(D(G,)). Having stored D(G3) and {V,', V,'), {V,, V2) as follows.
we can extend this partition to
a) The face F of D(G3) 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,', c E V2/. In this case we define {al, b , ) c V2, c1 E V,. Note that there are altogether only 6 choices for the distribution of a, b, c in V,' and V,'.
b) The face F of D(G3) 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 xil, xi, yi in V; and V2/, we assume by symmetry w.1.o.g. that either { z ~  ~ , xC~ V, ) ' and yi E V,', or { X ~  ~ , Y ~C) V[ and xi E Vi. In any case, let the vertex of G, uniquely determined by N(xi,) n N(yi) n F, belong to V2; let the vertex ti of G, uniquely determined by N ( X ;  ~ )nN(xi) nF, belong to V, unless A = (xil, ti, xi) is the boundary of a 2face and yi E V2/ in which case we let ti belong to V2; and let the corresponding vertex of N(xi) n N(yi) n F belong to V,. As for the remaining two vertices ui, E N(xi,) and ui E N(xi) of V(Hi) n F, define ui, ,ui E V2. Since ( K t ) is a tree, i = 1,2, and because of the way we extended {V:, V,') to {V, ,V2), it follows from these considerations that {V, ,V2) is an Apartition which can be obtained from {V,', V,') in polynomial
VI.3.3. ATrails: Complexity Considerations, Algorithms
VI.159
time Pi(pG3). So, together with the 2facecoloring of G we can perform the transition from {Vl, V2) to a cycle T which stands for an Atrail of G, by a step by step procedure. Namely: define Go := G, and G := ( G i l ) i , , i = I , . . .,pG, where V ( G ) = {v ,,.. .,upG), and the value of 6 E {1,2) is determined by vi E Vg. Since this transition from Gil to Gi can be done in polynomial time, we reach T = GPG from {Vl, V2) in polynomial time P;(pG). Again, since pG = const.pG3, we reach the Atrail T of G from the given hamiltonian cycle H of G3 in polynomial time Pi(pG3)= Pi (pG3) Pi(pG3) Pi (pG). Thus, the problem of finding an Atrail in G 3 D(G,) is polynomially equivalent to the problem of finding a hamiltonian cycle in G3.
+
+
Finally, let us show that the construction of G with the property that every face of D(G3) contains precisely one copy of either Hi or Ho, can be done in polynomial time. First we observe that the transition from G3 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 HI with the following properties:
1) V ( H , ) = V ( H ) ,
E(H,)2E(H);
2) if x and y are not adjacent in H, they will not be adjacent in HI; 3) for every cycle C1 of H1 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 Hl is eulerian and such that for every cycle C of H, the number of edges of C which are replaced in H1 with a multiple edge is at most l(C). As we shall see in the discussion on the CPP, the construction of H I from H can be done in polynomial time. Note that in this construction, dH1E(H)(~) dH(v) (mod2) for every v E V(H) = V(H1)
.
Applied to the construction of G from H = D(G3) the above amounts to saying that we can label the edges of D(G3) in polynomial time
VI. Various Types of Eulerian Trails
VI. 160
Q 2 k G 3 )with 0 and 1 in such a way that in D(G,) for El := {e E E(D(G3))/X(e) = 1) (with X denoting the label function)
I E, n El 1
dD,G3,(v)(mod 2) for every u E V(D(G3))
(1)
and such that for every cycle C of D(G3)
In particular, (2) holds for the triangular face boundaries A of D(G3). That is, for every such A , I E(A) n El (0, 1). Following the notation of Figures VI.31 and VI.32 we write V(A) := {a, b, c) if E ( A ) n El = 0, and V(A) := { X ~  ~ , Xyi) ~ , if E(A) n El # 0 with the notation chosen in such a way that w.1.o.g. {xi,xi) = E ( A ) n El. Among the two face boundaries containing xi,xi, choose one and fix it; denote it as above with A.
IE
For each of the A chosen with E(A) n El # 8 embed the graph Hiof Figure VI.32 in the corresponding face. For every other face boundary A = (a, b, c) embed Ho of Figure VI.31 in the corresponding face. Denote by G the graph resulting from D(G3) this way. This requires polynomial time Q3(pG3)only, because for each of the pG3face boundaries of D(G,) the embedding of Ho,Hi respectively, can be done in constant time. Thus, the entire transition from G3 to G can be done in polynomial time Q(PG~ ) = Q1 ( P G)~4 QZ ( 1 ) )~ ~Q3 ( 1 ) ).~ ~Because of (1) and the very construction of G from D(G3) it follows that G is plane and eulerian; and it is 3connected because G3 is 3connected. 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,') to {Vl, V2). It suffices to construct G = (Gv;,l)v;t which is connected and Cregular, and then find algorithmically an Atrail of G in polynomial time by successively applying the Splitting Lemma, say (see Volume 2, Chapter X). As for the complexity of the Atrail problem in planar, 4connected 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 Qconnected planar graphs, a harniltonian cycle can be found in polynomial time. This implies that a Tutte cycle in a planar cubic
VI.3.3. ATrails: Complexity Considerations, Algorithms
VI.161
graph G3 with X,(G,) 2 4 can be found in polynomial time; in this case, L(G3) is a planar 4connected 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.l) 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 malie the following observations (we restrict ourselves to outlines of proofs, the details are left as exercises). 1) An Atrail of a 2connected outerplane simple graph can be found in
polynomial time. We use Theorem VI.87 to develop a procedure for constructing an Atrail in a 2connected outerplane simple graph G. We assume G to be 2facecolored with the outer face F, of G being a 1face. The procedure (an algorithm, so to say) amounts to successively marking vertices of valency exceeding 2 in such a way that ultimately every 1face Fl # F, contains exactly one marked vertex. Denoting by Vl the set of marked vertices and V2 := V(G)  Vl, the partition (V,, V,) will then satisfy Theorem VI.87, and therefore, (Gvl ,l)v, ,2 will be a cycle representing an Atrail of 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, [ G ~ 3 7 ] ] 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(G3) for a 2connected cubic graph G3. Next we introduce a new vertex vp+; for every vertex ui of G3 and add the new edge V ; V ~ + ~ , i = 1,.. . , p . Denote by H the graph thus obtained. Finally we consider L(H) and S(G3)uL(G3). The latter graph can be viewed as obtained from S(G3) by letting V(L(G3)) = V2(S(G3)) and then adding E(L(G3)). If we proceed thus L(H) and S(G3)U L(G3) are isomorphic graphs. With this visualization of L (.H,) ,it now follows by a straightforward argument that there is a 11correspondence between hamiltonian cycles of G3 and such cycles in L(H). Moreover, the above construction, the transformation of a hamiltonian cycle of G3 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 NPcomplete even if G3 is planar and 3connected (cf. [GARE76a]). We note that these arguments are a slight modification of Bertossi's considerations in BERT^^^] 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 [ G A R E ~ ~ ~ ] .
VI. 162
VI. Various Types of Eulerian Trails
The starting point, however, is the construction of the weak dual of G
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 Dw( G ) is a tree for 2connected, outerplane graphs. Hence, in our case T W := D W ( G ) is a tree whose vertex bipartition {Vlw,V2W)following the 2facecoloring of G is such that the endvertices of T w belong to V,ZU. Choose vo E 1';" and orient the edges of T w in such a way that T w is transformed into an outtree TT with V(T?) = V ( T W )and root vo. For every v E Vlw let Fv # F, denote the 1face of G corresponding to v , and for any 1face F # F, of G let v F denote the corresponding vertex of V;". 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 1face or to a Zface.
b) Establishing for every v E V ( G ) the set Lv of 1faces Fl with v E V(bd(F1))c) Constructing the plane outtree TT with root vo and vertex bipartition {Vlw,V2W). We first mark the root vo E V ( T T ) . Then we mark an arbitrary vertes v1 E bd(Fvo), and let vl belong to V , (note that FVo is a 1face by the choice of v,). Consider next Lv, and mark vF E V;U for every F E Lvl  {Fvo). 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 ( v o ,w ) is a path in Tow joining the root vo with w E V;U, and if w has been marked, all elements of V ( P ( v o w)) , nVlw have been marked. (*> Next we choose an unmarked vertex v E Vlwsubject to the condition that all elements of V ( P ( v 0v, )  v ) r l Vlw have been marked already. (**) We mark this chosen v. Precisely because G is outerplane and simple it follows that there exists v2 E V(bd(Fv))such that
VI.3.3. ATrails: Complexity Considerations, Algorithms
VI.163
d(v2) 2 4 and for every F E Lv2 {F,), vF E V(T,") has n o t been (* * *> marked yet. 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 VlW already marked together with the elements of V2w adjacent from these marked vertices, define a subtree T' of T," with root v,. Therefore, if Ti # Tow, we choose x E V;U subject to condition (**) (with x in place of v) in order to mark x and to find v3 E V(bd(F,)) satisfying (* * *) (with v3 in place of v,), and mark u3. Then we mark in T," the vertex vF # x 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 Therefore, choosing repeatedly a vertex of Vlw subject to (**) and a vertex of G satisfying (* * *) and marking the latter vertex, eventually leads to an outtree T ( i )satisfying
After this ith step, the set Vl C V(G) consisting precisely of the marked vertices of V(G), and V2 := V(G)  Vl defines a vertex partition. We note that it is precisely because of (* * *) that statement 2) of Theorem VI.87 is fulfilled. Hence {Vl, V2) is an Apartition 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 jth step, 1 5 j < i, the choice for v E Vlw subject to (**) is arbitrary in the set of sources of T,"  ~ ( ~ (  1 ' 1So, ) . constructing T(j) and choosing and marking such v among the sources of T,"  v ( T ( ~ ) ) 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 L,, the discovery of a vertex vj+2 E V(G) satisfying (* * *) in the ( j 1)th step can be done in polynomial time P2(n). Hence, the time required to construct the Apartition {Vl, V2) is at most P,(n)+n(P, (n)+ P2(n)). Since the transition from {V,, V2) to (GVl,1)V2,2can be done in polynomial time as well, it follows that finding an Atrail in G can be done in polynomial time.
+
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VI. Various Types of Eulerian Trails
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 Aproperty, a n Atrail 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 Atrails TI and T2 of GI and G2, respectively, can be found in polynomial time (where G, U G2 = G, G, n G, = H). Hence the corresponding Apartitions {V;',Vi) of Ti, i = 1,2, can be found in polynomial time. Since the four vertex sets
can be found in polynomial time, we therefore arrive at Wj := Wj' U W f , j = 1,2, in polynomial time. Since (Gw, ,l)w, ,2 can then be obtained in polynomial time, and since this graph is a cycle (see the proof of Theorem VI.88), it corresponds to an Atrail of G obtained in polynomial time. We observe, however, that the above considerations are based on a given harniltonian cycle H of G. Hence the question arises: how complex is the problem of deciding whether a given 4connected, planar, eulerian graph has a hamiltonian cycle with the Aproperty ? Of course, in 2) we did not need any assumptions on n(G) (however, n(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 4connected graphs. It should be noted that if G is any eulerian triangulation of the plane (4connected 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 Aproperty 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 Aproperty is a special type of a leftright path as defined in [SHAN75a]. We leave it to the reader to check the complexity of finding an Atrail 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 ATrail Algorithm For Arbitrary Plane Eulerian Graphs
VI.165
An ATrail Algorithm for Arbitrary Plane Eulerian Graphs Our starting point is a 2connected, plane, eulerian graph Go other than a cycle and, indirectly, Theorems VI.59 and VI.67. However, instead of considering unicolored facerings we combine a vertexmarking procedure with a sequence of 1splittings. '1 Step 0. Consider Go together with a 2facecoloring such that the outer face Fomof Go is a 1face. Denote Vo* = V(bd(F,"))  V,(Go), and let x1 E V< be arbitrarily chosen. Set i = 1. Step 1. Form Gi := (Gil)(zil,l. Denote by Fim the outer face of Gi, and let the 2facecoloring of Gi be induced by the 2facecoloring of
G,,. Step 2. Mark those so far unmarked elements of V(bd(Fim)) which are cut vertices of Gi, and let V1:* denote the set of unmarked vertices in V(bd(Fim))  V2(G;). Step 3. If Step 1.
If
V;.* # 0, then choose any xi+, E V. Set i = i + 1 and go to
v = 0, then continue.
Step 4. Set v,** = {x/x E (E(Gi)  E(bd(F;"))) U (V(Gi)  V2(Gi));x is unmarked if it is a vertex ). If v,** = 0, then go to Step 7.
If
T*# 0, then continue.
Step 5. Search for the largest integer j
< i such that V;  {xj+,) # 0.
If such j does not exist, then go to Step 8. If j exists, continue. Step 6. Mark xj+, and set
=
V; {aj+,). Set i = j and go to Step
3. Step 7. Go has an Atrail. Step 8. Go has no Atrail.
'1 For the case of eulerian triangulations of the plane, this algorithm has been outlined already in [ F L E I ~ ~ ~ ] .
VI. Various Types of Eulerian Trails
VI. 166
In fact, this algorithm does more than decide whether Go has an Atrail: for, if = 0 = T*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;DO) describes an Atrail To of Go. Moreover, an Apartition {Vf, V;) in Go corresponding to To is, in terms of the algorithm's notation, defined by
v:
= {xl, ...,X i ) , V2' = {x E V(G)/x has been marked or x E V,(Go)} We note that the construction of Gi and the determination of V,*,T.7' 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 V;.** # 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 Gi satisfying B n y * # 0 defines a unicolored facering R (whose elements are 2faces) with a complete set of links LR := V(bd(FgOO))  V2(B), where Fp is the unbounded face of B (whose embedding is induced by that of G,) and bd(Fp) c bd(F,OO). For, = 0 means that all elements of V(bd(Fp))  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) 2faces F2 with E(bd(F2)) n E(bd(Fg)) # 0, whence R is defined by the set of these 2faces. We note, in addition, that a marked vertex of Gi need not be a cut vertex of G, (see Step 6). Steps 5 and 6 are means of getting out of this deadlock. That is, one has to go back to some Gj, j < i (for practical reasons one chooses the largest j ) and choose some vertex E  {xj+l} to which one applies a 1splitting to obtain G;+l. At this point x ? + ~ has become a marked vertex, and all the algorithmic work in establishing Gj+l,. . .,Gi is to no avail. That is, the marking and splitting procedures in these graphs are undone. But having shifted xj+, from to the set of marked vertices of Gj does not say anything about the 'final fate' of xj+l in an Apartition {Vf ,V2') 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 Vz, k > j' containing the above xi+l again. We note however, that xj+l remains a marked vertex as long as the above Vj* has not been eshausted by Step 6.
+
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 ATrail Algorithm For Arbitrary Plane Eulerian Graphs VI. 167 case the algorithm amounts to constructing connected graphs Dl (induced by the vertices to which one applies a 1splitting) and D2 (induced by the marked vertices) such that
and for 5 = 1,2, D6 has a cycle C only if C = bd(F6) where F6is 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 1splitting, 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 nonseparating Atrail (Exercise VI.34.b)). If Go has such an Atrail, the outcome of this modified algorithm is a connected outerplane eulerian graph Gi(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 Di contains at most two cut vertices of Di (*). 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,, . . .,Di of the modified Atrail algorithm. This "suspicion7' leads us to the following more general conjecture where we consider simple eulerian triangulations of an ngon in the plane. In this case, n 0 (mod 3) must hold; and for every such n, triangulations of 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 r 0 (mod 3) be chosen, and let Do be a simple eulerian triangulation of an ngon in the plane together with a 2facecoloring of Do such that the outer face Fooo of Do is a 1face. Then (distinct) vertices xk E V(Do)  V2(Do) exist such that Dk := (Dkl)~z,l,l, k = 1,. . .,i, has the following properties: a) x k E V(bd(F,OO_,)) V2(Dk,) ( F Z , is the outer face of b) every block of Dk contains at most two cut vertices of Dk;
c) if k = i, then every block of Dk is a (triangular) boundary of a 2face.
VI. 168
VI. Various Types of Eulerian Trails
Thus, because of property b) Conjecture VI.92 is a generalization of Conjecture VI.73.
VI.3.4. Final Remarks on NonIntersecting Eulerian Trails and ATrails, and another Problem As we have noticed at the beginning of this section, nonintersecting eulerian trails and Atrails 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 valeilcy 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 qgon 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 nonintersecting eulerian trail T of G. Denote by CT the plane cycle corresponding to T. Moreover, similar to the concept of a 6splitting, 6 = 1,2, we consider, in the construction of CT from G, the kfold application of the splitting procedure to obtain k 2valent vertices v,, . . . ,vk from the 2kvalent vertex v, k > 1, as a simultaneous procedure to replace v with these k 2valent vertices. Note that this replacement procedure can be done in such a way that v,, . . .,v k lie in a (geometrical) circle K(v, E) with center v and radius E, and x ( v , E) contains no other vertices of G, CT respectively, and E ( v , e) n (E(G)  E,) = 0. Note that CT is a simple closed curve since T is nonintersecting. Thus G can be viewed as obtained from a regular qgon in the plane Cq(q =I E(G) I) as follows: 1L cT4 G
c,
where cp is a homotopic deformation
and $Jis a continuous mapping of the plane E2 onto itself
VI.3.4. Final Remarks
VI. 169
such that $(x) = v for x E ((v, E) and v E V(G)  V2(G),
$: £ 
U
Ii(u, E )
+ E2
 V(G) is bijective.
v€V(G)Vz(G)
Since an Atrail is a special type of nonintersecting eulerian trail, the above is true, of course, if T is an Atrail. However, if T is an Atrail, we can assume CT as having the additional property that for every v E V(G)  V2(G) there exists 5 < E such that E ( v , 6)n CT = {vl,. . .,vk). (~4) Using planarity we can replace (A) with
precisely one of K(v, E) n int CT and K(v, E) n ext CT is a region. (A') If we consider a 2facecoloring of G (with Foo being a 1face as usual) and the induced coloring of CT, K(v, E) n int CT is a region if and only if T induces a 2splitting in v. In the case of a nonintersecting eulerian trail which is not an Atrail, then for at least one u 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 Atrail. The Atrail problem can thus be reinterpreted 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 nonorientable) surface 3, except where 2facecolorings are involved. In particular, property (A) remains valid as the topological means for distinguishing Atrails from the other nonintersecting eulerian trails. Note that the concept of an Atrail was originally introduced via O+(v), 0(v) respectively, and did not require the concept of 5splittings. However, for different nonintersecting eulerian trails TIand T2 of the same G, the cycles CTl and CT, may not be homotopic. As for Atrails 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
VI. Various Types of Eulerian Trails
decided, whether M admits such orientation; for, a canonical orientation is uniquely determined by just one given arc). To consider the Atrail 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 Tl corresponds to an Atrail in D ; and an Atrail of D corresponds to an Atrail of G. This follows directly from the relation between an Atrail and 5splittings, 5 = 1,2. However, it might be of interest to study (in the contest of Theorem VI.33) those intrees of D which correspond to Atrails 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 Xo by defining t E XO(v)+t t = {e:,ei+i), 1 5 i 5 k = k, = $d(v), and v E V(G) is arbitrarily chosen. However, Xo = Xs for some unique trail decomposition S of G. Whence the following questions arise: 1)For a given graph G, does there exist a drawing of G in the plane such that Xo = XT for some eulerian trail T of G ? 2) How many drawings of G exist such that some eulerian trail T of G satisfies XT = Xo (where Xo relates to a particular drawing) ? 3) Which trail decompositions S of G, satisfying Xo = Xs can be obtained this way ? These questions have been raised in [HARB89a] where question 1) is answered in the affirmative for G = I$,+,,n E N.
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 pathlcycle 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.lO. 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 dD(v) > 6, show first that if IX*(v) I= ;dD(v)+1, 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.ll).
u!=,
Exercise VI.8. Show that Dl of Figure VI.4. has an X(D1)compatible eulerian trail if and only if D has an X(D)compatible eulerian trail, where X ( D ) is given and X(D1) = X ( D ) U X ( s l ) U X(s,). Exercise VI.9. a) Construct connected eulerian digraphs D with the property that V ( D ) consists of 4valent vertices and an odd number of 6valent 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 4valent cut vertex. Exercise VI.lO. 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 Dl and D i = D *  DL 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 4regular 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 Atrail. If one drops the property "simple", then H can be constructed in such a way that H' spans H.
VI.172
VI. Various Types of Eulerian Trails
Exercise VI.17. Construct a 2connected, outerplane, eulerian graph having no Atrail. Exercise VI.18. a) Construct an Atrail in the graph GT of Figure VI.28. b) Show that GT has no Atrail inducing a 1splitting in vl if the outer face of G; is a 1face in the 2facecoloring of G r . c) Find other plane embeddings of the abstract graph underlying Go and G,: which admit Atrails. 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 Atrail) does have a vertex partition {Vl, V2)such that (V6)is acyclic, 6 = 1,2. Exercise VI.2 1. Prove that the graph of Figure VI.30 (a) is the smallest planar, Zconnected, cubic, bipartite nonhamiltonian graph, and that it is uniquely determined. Exercise VI.22. Prove: the following statements are equivalent. 1) Every plane, 3connected, cubic, bipartite graph has a hamiltonian cycle (BTC). 2) Every plane, 3connected, cubic, bipartite graph has a dominating cycle. Exercise VI.23. a) Prove: if G3 (D) is a plane, bipartite, 3connected, cubic graph (simple eulerian triangulation of the plane) whose face boundaries are 4, 6, or 8gonal (whose vertices are 4, 6, or 8valent), then G, (D) has precisely six Cgonal face boundaries (Cvalent vertices) if it has no 8gonal face boundary (8valent vertex); it has precisely seven 4gonal face boundaries (4valent vertices) if it has precisely one 8gonal face boundary (8valent vertex). b) Construct an eulerian triangulation of the plane with 4, 6, and 8valent vertices only such that it has precisely one 8valent vertex v; and v is not adjacent to any 4valent 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 W4extensions such that D has precisely one pair of adjacent 4valent 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 G3 is a connected 3regular graph having a 2factor Q4 consisting of 4gons only, then Gg has a hamiltonian cycle H > E(G3)  Q4 (hint: use Corollary VI.6 and suppress those 4gons 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.38 has an Atrail. Exercise VI.29. Prove Theorem VI.90. Exercise VI.30. Show that it suffices to prove Conjecture VI.73 for the corresponding 4connected graphs (hint: apply Proposition VI.83). Exercise VI.31. a) Concerning finding an Atrail in a 2connected 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 (n) 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 Aproperty 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 harniltonian cycle with the Aproperty in a(n) (eulerian) triangulation of the plane, lies in P. Exercise VI.33. Study the complexity of finding an Atrail 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 Atrail 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 ancl their respective proofs). b) Modify this Atrail algorithm in such a way that it decides whether a given connected plane eulerian graph has a nonseparating Atrail.
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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 TI of 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 TI and T, of the connected eulerian (di)graph G are considered to differ precisely if XTl # XT2. 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.lO.
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 ith edge (arc) traversed by T differs from the ith edge (arc) traversed by TI.Comparing this view with the one expressed by Definition VII. 1, 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 29 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,, . . . , f,. An eulerian trail T of H, espressed as a sequence of edges (arcs), then defines a permutation .r; of the elements of {1,2,. . . ,q ) . BAerian trails TI and 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 XT defined by an eulerian trail T. Let tl,t2 E X ( v ) C XT for v E V(G)  V2(G) # 8 be arbitrarily chosen, and denote tl = {f;,gi), t2 = {fi,f;). Let H be the connected eulerian graph obtained from G by splitting away those pairs of edges which correspond to the elements of X T  {t,, t 2 } . Consequently, d H ( v ) = 4 and V2(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 halfedges within the same block of H. The Splitting Lemma implies, therefore, that either or is connected. Assuming w.1.o.g. that HI,, is connected, we have V2(H1,2) = V(H1,,); i.e., is a cycle. Thus, a run through HI,z 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' := XT, satisfies the equation
x t = (X,{t,,t,))~(t',,t;) where t', := {f;, f;),th := {f;, 9:).
,
(1)
VII.l. Transforming Arbitrary Eulerian Trails in Graphs
VII.3
Summarizing these considerations, we arrive a t 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)  V2(G) # 0, and let t,, t2 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 tl and t2. Reinterpreting the above equivalently in terms of representations of eulerian trails by edge sequences, we can describe the eulerian trails T and TI as follows
where E(G) = {el,. . . ,eq), Sm,,is a closed trail starting with em and ending with e, at the same vertex v E V(G)  V2(G), and em, f,, em = g,, en = f 2 , en+, = f3. Hence (1) and (2) describe the same situation. In accordance with [ABRH80a, SKIL83al (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 TI be two eulerian trails of the connected eulerian graph G, and let v E V(G)  V2(G) # 0. If the corresponding systems of transitions X T , X 1 respectively, satisfy (I), we can say that XI (TI) has been obtained from XT (T) by a Ktransformation which in itself is given by transforming tl and t2 into t i and t;. We can also say, by analogy, that TI has been obtained from T by the reversal ,, (or segment reversal for short), which starts of t h e segment S and ends at the same vertex v (see (2)). We note that the inverse operation of a Ktransformation is also a Ktransformation. This follows from (I), but also from (2) by observing that (S;$)l = S,,. If we extend the validity of (1) for the case {tl,t2) = {ti,tb) (i.e., T' = T), then Definition VII.3 permits one to consider the identical transformation (i.e., XT = X I ) a Ktransformation. For the sake of brevity we can write K(T) = TI, if TI is obtained from T by a r;transformation.') Since this formal equation does not specify by which Ktransformation TI has been obtained from T,we have K(T) = T',n(T) = T" and T' # TI1 for different Ktransformations.
VII.4
VII. Transformations of Eulerian Trails
Definition VII.4. Let I ( G ) denote the s e t of all eulerian trails of the connected eulerian graph G. For T,T1 E I ( G ) we call T and TI K associate^,^) if To, . . .,Tn E I ( G ) exist for some n E N so that
T = T o , T 1 = T n , and T i = ~ ( T i  , ) f o r l I i < n
.
Thus, if T, TI, T" E I ( G ) are such that K(T) = TI and such that TI and T" are Kassociates, then T and TI1are Kassociates. From Definition VII.4 and the discussion immediately preceding it, we can now conclude that Kassociation 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. T h e o r e m VII.5. IP ( G , K)I= 1 for every connected eulerian graph G.
Proof. If G is a cycle, then I7(G) I= 1 by Definition VII.1. The validity of the theorem follows in this case ( I ( G ) = {TI) since T = K(T) with the Ktransformation being the identical transformation, whence we assume V(G)  V2(G) # 0, so IT(G) I> 1 by Lemma VII.2. Let us assume that we have T I ,T2 E I(G)which are not Kassociates. Define X, := XTl, X2 := XT2and choose TI and T2 in such a way that I X, n X2 I is as large as possible. Next consider v E V(G)  V2(G) with X,(v) # X2(v), where Xi(v) Xi, i = 1,2, is the set of transitions at v induced by Ti. Such v must exist because of TI# T2. Since d(v) = 2k > 2, it follows from the choice of v that x1,2(v) :=I X1(v) n X2(v) I k  2 (note that x , , ~ ( v )= k  1 implies x , , ~ ( v )= k which is impossible by the choice of v).
Consequently, there are T := k  x , , ~ ( v ) 2 transitions of Xl(v) which do not belong to X2, and vice versa. We may denote these transitions in such a way that
where {i, ,... ,iT}={l,..., r } and
ij#j
for
l s j s ~.
Note that Kassociatesare the same as associates in [SIv ) X ~ , ~ and ( V )therefore, I X 3 nX 2 I>I X 1 n X 2 I. By the choice of Tl and T2, it follows that T3 and T, are Kassociates. This and K ( T ~=)T3 imply that Tl and T2 are Kassociates, thus contradicting the choice of Tl and T2,whence we can conclude that X 3 satisfies the equation (3). Now we apply Lemma VII.2 for T = T2 and t , = t2,,, s = 1,2. Using a symmetric argument we can conclude that there is
and X4 must satisfy the equation
Note that these 2r transitions correspond to a 2factor Q in K 2 , where V ( K 2 , ) = {e; ,f;, /j = 1, . . . ,r ) ,E(IC2,) consists of all conceivable transitions, 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 i j ,j = 1,. . ,r. This we shall do later on.
.
VII.6
VII. Transformations of Eulerian Trails
Comparing (3) and (4) we can confirm the validity of the equation
where dm,, is the Kronecker delta. By the choice of TI and Tz7it follows that T3 and T4 are Kassociates. This and the formal equations &(TI) = T,, n(T,) = T4 classify, however, Tl and T2 as being Kassociates. This final contradiction implies that the theorem is valid. Note that the above proof differs from those of [ABRH80a7Theorem] 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 endvertex 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(ui, vi) stands for the element of S joining the odd vertices ui7vi 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 S1,S, of E(G) can be transformed into each other by following a sequence of Ktransformations applied to the respective eulerian trails TI,T2of H. Interpreting these Ktransformations in terms of segment reversals and noting that a segment may start and end at vo 6 V(G), it immediately follows that in general, we cannot conclude that S1 and S2 can be transformed into each other by a sequence of segment reversals (respectively Ktransformations) performed only in G. Figure VII. 1 demonstrates this fact. Of course, we can extend the concept of segment reversal (respectively Ktransformation) to arbitrary graphs G by restricting ourselves in H to segments that contain no element of EVo (and/or by restricting Ktransformations t o the elements of XS1 XSl(v,)).
VII.1. Transforming Arbitrary Euleriaxl Trails in Graphs
VII.7
Figure VII.l. The decomposition S1 of E(G) into two open trails T(ul, vl), T(u,, v,), where the elements of Xs, are described by the small arcs, cannot be transformed by segment reversals into a path decomposition S, := (P(u1, 212 1, q(v1, v2 I of E(G).
>
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) = NH(vO).Precisely, because vo is 2valent in H one can define a correspondence TH E 7 ( H ) t E % (G) 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
By Definition VII.l it is no loss of generality if the elements of T ( H ) are written in such a way that they start and end in v,. Hence, the above relation between TH and T$ classifies the above correspondence as a bijection between I ( H ) and T0(G). Secondly, consider T,T1 E I ( H ) such that K(T)= TI. Assume G is a connected graph other than a path, hence we may consider
VII.8
VII. Transformations of Eulerian Trails
Now we can write T as an edge sequence of the form
where S1,mland S,,, are segments starting and ending at x; hence T starts and ends at x. Moreover, m = 2 (m = q), if and only if the first (last) edge of T is a loop. Next suppose T # TI. Since K(T)= TI, T' is being obtained from T by the reversal of a segment S,,, starting and ending at y E V(H)  V2(H). Because of (*) and Definition VII.l we may suppose w.1.o.g. that y = x, r = 1, s = m  1. Hence
Following the notation of (1) and (2), we then have
tl = {ei, eb}, ti = {ei, el,),
t2
= {e',,,e',),
t; = {el,l, eb}.
However, for
T" = S1,,, ,s;;, the corresponding system of transitions X" satisfies (1) as well, where t,, t2,t',,t; are the transitions described in (* * *). Hence,
That is, T' can be obtained from T by reversing the segment Sm,,instead of S1,ml. In other words, if K(T) = T' and T # TI,then T' can be obtained from 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 T ( H ) we can therefore say that T' can be obtained f r o m T by 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) into k open trails. Should one simply adhere to the concept of segment reversal, as has been done for the cases k = 0 (Theorem VII.5) and k = 1 (Corollary VII.G), 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.l): if G is a connected graph with precisely 21; > 2 odd vertices, how large can I P ( G , r i ) I be ? Here, P ( G , K ) denotes the partition of S(G) induced by Kassociation which remains an equivalence relation, provided one considers an open trail T(u, v) and its inverse T(u, v)l to be the s~uneobject (see Definition VII.l). Is I P ( G , K) I a function f(k) depending on k only ? Maybe f (k) = k ? In my view, these are questions worth answering; it ~vould 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 Kassociates within I*. More precisely, if T, TI E I* are chosen arbitrarily, does a sequence = of objects T I , . . . ,Tn E I*arise such that T, = T, Tn = T' and K(T,.) Ti+l for i = I , . . .,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 TI,. . . ,Tn compatible with X and satisfying T=T,,T1=Tn,
and
T i + l = ~ ( T i ) for
l 2, and {e', ,eh), { fi ,f;) 6 X. Since T3 $ I ( G , X ) , it follows from (***) that S3 E S2(G,X ) where XSS= X;. Depending on whether T4 E 7 ( G , X ) or not, we may distinguish two cases. Subcase 1.1. T4 E 7 ( G , X). Denote S3 = {S,, Sf) such that el, e2 E E(S,) and therefore, f17 f2 E E ( S f ) Note first that Xl (v)  {t,,, ,tl,,) # 8 since m > 2. Next consider a run through T4 starting at v along el. Since T4 E I ( G ) and {e', ,eh) E XT4 = Xt, one arrives at v along e2 only after having passed all elements of E(G). Consequently, since this is not the case with respect to S,, there must be a vertex w E V((S,)) n V((Sf)) at which a transition {hi, 9:) E X t satisfies hl E E(S,), gl E E ( S I ) . Hence, we find h2,g2 E E, such that {hi, ha) E X1(w) in any case, and where {g: , g i ) E Xl(w) if w # v, whereas we denote g: = fi , i = 1,2, if w = V. Observe that {fi ,f;) E X?. In any case, {{hi, ha), { g ; , gh)) n Xl n X2 = 0 by construction and because of m > 2 and {e',, eh) E X< n Xt. Moreover, for {r, s) = {l,2),
So, one obtains the validity of this equation for some choice of r and s, where {r, s) = {1,2). We can write S, for some j and k, where {j,k) = {1,2), in the form
VII.16
VII. Transformations of Eulerian Trails
(possibly e j = hl or ek = h2 but {ej, ek) # {hl, h2)), and
(note that h2 E E(S,) because of hl E E(S,) and {hi, h;) E X1(w), and that, similarly, g2 E E(Sf)). Hence we obtain
By the very construction of Tj,,, it follows that Tj,kE T(G, X ) . Consequently, since /cl(Tl) = S3 E S2(G,X) and /c1I(S3) = Tj,lcE I ( G ,X ) it follows that T, and TjVk are /cxassociates. Moreover, T, and T4 are /cxassociates because of T4 E I ( G , X ) and T4 = /c(T2) = q ( T 2 ) . On the other hand, since {{hi, hh), jg;, n Xl n X, = 0 {ei,e;) E X:nx,,,
implies
I X : ~ I X ~ ,I>IX1 ,~ n X 2I
.
The choice of Tl and T2, and T4,Tj,, E I ( G , X) imply that T4 and Tj,,, are /cxassociates. So we can finally conclude that Tl and T2 are /cxassociates. This contradiction settles Subcase 1.1. Subcase 1.2. T4 6T(G, X ) . By (* * *), S4 E S2(G,X ) , and XS4 = X i by (**). That is, {fi, fk) E X since {ei, eh) 6X, and thus {eh, fk) 6X. Moreover, retaining the notation of Subcase 1.1we define S3= {S,, Sf) as above and
In this case, however, S; assumes the role played by S3 in Subcase 1.1, and S, assumes the role played there by T,. We can consider a run through S, starting a t v along el; following the elements of Xl and belonging to S,, one finally arrives at u after having passed the elements of E(S,), with e2 being the last edge in this run. This run, reinterpreted as 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((S3) from S,t over to S*. Hence, we may conclude in a f manner similar to Subcase 1.1 that {hl,gi) E Xf(w) exists such that h1 E E(S,*), g, E E ( S j ) ; moreover there are h2,g2 E E, such that {hi, hi) E X2(w) in any case; and {gi,gi) E X2(w) for w # v, whereas
VII.2. Transforming Eulerian Trails of a Special Type
VII.17
for w = v we denote gi = fk , gi = e; ({e;, fh) E Xi). Since we have in any case {h,, h2} C E(S:) and {g,, g2} C E ( S j ) , we may conclude as in Subcase 1.1 that for
S: = e l , . . . ,hj, h k , . . . ,f, , where {j,k} = {I, 2}
,
and
Sj = g s,...,
g,,, where
{ s , ~ } = {1,2}
,
T;,k = e l , . .., h j , S j 7 h k , .. . , f l satisfies T*k E I ( G ,X ) because {{hj,g;}, fl X = 0 for fixed {j,k) = (1,2} and one of the two possible choices of {s,r} = { 1,2}. Note that Sj can be expressed in the above manner for both choices of s and r . Since, by construction, {h', 7
E X2(w)  X l (w) and (9: 7 9;)
in any case (note that m
6 X, (w)
> 2), while
and because Xl n X2 c XTt , we therefore conclude that 1.k
Consequently, by the choice of Tl and T2,Tl and TAk are tcxassociates, and by construction, Tlk and T2 are tcxassociates. Hence T, and T2 are tcxassociates. 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 gi = fi, i = 1,2, respectively g, = f, and g2 = e2, is not really necessary. It shows, though, how in the case v = w one can construct Tj,k, T;,k respectively, by changing a minimum number of elements of the original XI, X2. With Case 1 being settled we thus are left with the cases where Xf = XTi,i = 3 , 4 , or X;"= XTi, i = 3,4.
Case 2. X f = X Z , i = 3,4. By ( 5 ) , { { 4 , f;), {e;, fk}} C X , so m > 2; moreover, X;" = Xsi by (**), and Si E S2(G,X), i = 3 , 4 by (* * *). Retaining most of the notation previously used, we write
VII.18
VII. Transformations of Eulerian Trails
Subcase 2.1. E(Se) # E(S:). This assumption plus the fact {el, e,} C E(Se) n E(S,*) implies that starting a run through S,*at v along el, one finds {hi,g;) E X2(w) for some w E V((E(S,*))) such that h1 E E(Se), 9, E E(Sf). As before we then also find h,, g2 E E, such that ,hh} E Xl(w) in any case, {gi, 9;) E Xl (w)if v # w (so 92 E E(S/)), and where we denote g: = 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 el : we find correspondingly {hi, 9;) E Xl(w) for some w E V((E(Se))) such that hl E E(S:), gl E E(S;), and we find h2,9., E E, such that {hi, h;} E X2(w) in any case, and {g;, 9;) E X2(w) if # w , whereas for v = w we define g; = fi, g; = f;. Note the difference to Subcase 1.2 which is due to the fact that there Xs4 = Xi. As in Case 1 we conclude that by construction
and that Tj,k,Tlk E T ( 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 Ti,* and T;* as K~associates. Moreover, by construction T, and Tj,k are K~associates, and so are T2 and Ttk. Whence we conclude that T, and T2 are nxassociates, a contradiction which settles Subcase 2.1.
Subcase 2.2. E(Se) = E(S:). Consequently, E ( S f ) = E(S;). This and f, E E ( S j ) as well as {e',, fh} € Xf imply em, f, E E ( S f ) = E ( S j ) which, in turn, together with {e',, f&l} E X: implies em,fm, E E ( S j ) = E ( S f ) (observe that m > 2), a.s.0. Consequently, since fl E E(Sf) = E ( S j ) we have E,  {el,e2} E(Sf) = E(Sj). We consider two cases. (2.2.1). m > 3. Then {e:, f;} E X f , {e;, f;} E X:, and because of {e;, f;},{ei, f;) E X and m > 3 we have
VII.2. Transforming Eulerian Trails of a Special Type
VII.19
By the preceding arguments and because we can write
Sf = fi, ...,e3, f3,. . .,f j for some choice of {i, j} = {I, 2) and
S; = f k , . . . ,e3, f,, . . . ,fi for some choice of {k, Z} = {I, m}, it follows that both
T = f i ,...,e3,Se,f3 ,...,f j and T* =fk,...,e3,S,*,f,,...,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 (X1 n X,)
= X,
n X,.
which classifies T and T* as tcxassociates because of the choice of Tl and T,. This and the construction of T from TI, respectively T* from T2,imply that TIand T2 are K~associates, again a contradiction.
>
{{e:, f;}, {ei, fi},{ei, f:}).') Because of {el,e2} E: E(S,) = E(SZ) there is a total of three possibilities for Sf and S; to pass through the edges of E,  {el, e,). Depending on these X). possibilities we may construct T, T* E I(G, (2.2.2) rn = 3. Then X(v)
(a) Sf = f l , . . ., f3,e3,..., f,. Note that S; cannot have a segment S3,, = e3,. . . ,f2 because {ej, fi} E XT, which implies that Xy c Xg. This and S3 c Sj would contradict the fact that Sj is a closed trail. Anyway, with Sf satisfying the above equation we conclude that TI is of the form
,
consequently, we obtain T E 7 ( G , X) by a 6transformation,




'I Observe that m = 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 Zfactor Q of Kz,. Moreover, {ek, E X(v) follows from m = 3 and {e:, f;), {ek, fi) E X ( v ) , as well as from a combination of Case 1 and (5) if we set tl,Z= {ei, fi).
fi)
VII. Transformations of Eulerian Trails
VII.20
(b) S j = f,, . . .,f 2 , e3, . . . ,f,. We conclude in a manner similar to case (a) that Sf cannot have a segment of the form e3, . . .,f3. Here T2 and T* E I ( G , X ) obtained from T2 by a Ktransformation are of the form
(c) Either Sf = f l ,..., e3,f3,..., f 2 or Then we have for Tl and T2 the equations
S) = f l ,  . . , e 3 , f z , . . . , f 3 .
This yields for T E I ( G , X ) obtained from T, by a ntransformation:
while one obtains T* E I ( G ,X ) from S4= {S:, Sj) by a nabsorption,
In all cases (a),(b),(c), it follows that T and TI, respectively T* and T2, are nxassociates, and that one always has X l n X2
X,
, X, n X2 X,,
respectively
.
On the other hand, if one compares any of the constructed T with any of the constructed T*,one can conclude either
{ei, e ; ) E X T n X,. , or {fi,fi) E X T n X,. or T is as in case (a) and T*is as in case (c).
,
Thus, for the first two instances we may conclude I XT nXT, / > I X, nX,I, classifying T and T* as nxassociates. T and T* are, however, t c x associates in the third instance as well: for in case (c) we have (eb, f;) E XTI which implies I XlnXT. I >1 XlnX2 1, so T* and Tl are nxassociates. This and TI = n(T) finally proves that T and T* are nxassociates in any case. Consequently, since we have T = tc1(T1), T* = n1(T2) in all cases under consideration, it follows that TIand T2 are nXassociates.
VII.2. Transforming Eulerian Trails of a Special Type
VII.21
This contradiction finally settles the case (2.2.2) and thus Subcase 2.2. Hence, Case 2 has been settled.
Case 3. XI' = XTj, i = 3,4. By (7), Tj $ 7 ( G , X ) if Tk E I ( G , X), for any choice of {j,k} = {3,4), hence w.1.o.g. j = 3, k = 4. Subcase 3.1. T4 E I ( G ,X). Then Xi = XsJ, where S, E &(G, X) by (* * *). hloreover, {e',, eb} $ X, {fi, fl) E X. Observe that m > 2 and T4 = el, . . . ,f, , f m ,. . . ,e2. Retaining the notation used before we write
Now we proceed similarly to Subcase 1.1. We start a run through T, at v 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 fl,e2 E E(Se) but E(Se)C E(G), at some vertex w there exists {hi,gi} E X2(w)  X1(w) such that h, E E(S,), gl E E(Sf). Correspondingly, let h2,g? E Ewbe chosen in such a manner that {hi, h;} E X1 (and thus h2 E E(s,)) and {gi, 9;) E X, if w # v, while for v = w we denote gi = e',, gi = fi (thus g,, g, E E ( S f ) in any case). Continuing now as in Subcase 1.1, we obtain Tj,k E I ( G , X ) from S3 by a Kabsorption; i.e., Tj,, and TI are nxassociates. Moreover, {{fi, e ; } } U Xl n X2 X2 n XTjekby construction and due to the structure of T, and S3,implying that TjTk and T2 are rtxassociates because of the choice of TI and T2. Hence T, and T2 are K~associates.This contradiction to the choice of TI 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,, S4 E S2(G,X ) , where Xj = Xs3 and X i = XS4. Retaining the notation previously used we obtain the following expressions:
Now we have E(S,*) # E(S,) # E(Sj). Hence, we may proceed in a manner similar to the one employed in Subcase 2.1, by starting a run through S,*at u along f,. Since fl E E(Se) and el E E ( S f ) one finds {hi, gi) E X,(w) for some w E V((S,*)) such that h, E E(Se), g, E E(Sf); and similarly one finds h2,g2 E Ewsuch that {hi, h i ) E Xl(w)
VII. Transformations of Eulerian Trails
VII.22
in any case, and where {gi, gi) E Xl(w) if w # v, while we denote gi = e i , gi = if w = v. Note that {gl, g,) C E ( S f ) , and {g:, gi) E Xi  X2 since either w # v or m > 2; moreover, {hi, hb) # {f;, ek) since
fl
{f:,
E X2.
The eulerian trail Tj,* E I ( G , X ) constructed in a manner similar to Subcase 1.1 now satisfies the relation
Because of the choice of Tl and T2 this relation implies that T2 and Tj,k are K~associates.This and /c1(T1) = Tj,k imply that TI and T2 are nxassociates, 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 Tjr contains a transition of T2 not yet accounted for in X1 n X2, whereas in Subcase 2.1 the construction of both Tj,kand Tik was necessary because of {ei, eb) X1 U X2). Having obtained a contradiction in each of the three main cases (which cover all possibilities concerning the structure of Tl and T,), we may conclude that I ( G , X ) cannot contain two elements which are not tcxassociates. Theorem VII. 10 now follows. A closer look a t 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 I ( 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 s association which again defines an equivalence relation on I ( G , P(G)) whose equivalence classes 1 )these . more general define the corresponding partition P(G,~ ~ ( ~For 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 P(G)) (S E S2(G, P(G))) and let je:, f i ) , {e:, f;) E X~ (XS) be arbitrarily chosen. Then, for all K E P(G),
VII.2.1. Special Types of 6,Transformations
VII.23
This lemma guarantees the validity of the equations analogous to (**) 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.lO). These equations and relations, however, were of central importance to the proof of Theorem VII.lO. 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 {el, f l ) E X to {el, 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 , rcp(G)) I= 1 for every connected eulerian graph G and every partition system P(G) satisfying I K 15 i d ( v ) for every I< E P(v) 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 T(G, P(G)) # 8 (see Theorem VI.l).
VII.2.1. Applications to Special Types of Eulerian Trails and 6,Transformat ions Some of the next results are direct consequences of Theorem VII.lO. We consider connected graphs G which are Cregular, 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 Atrails, nonintersecting eulerian trails respectively, of G by TA(G), TNI(G) respectively. In the third case we consider a 2factorization {Q1, Q2) of a Cregular G (see Corollary VI.3) and denote by 7,(G;Q1,Q2) the set of all eulerian trails of G in which edges of Q, and Q, alternate (i.e., eulerian trails compatiT2 E ble with the cycle decomposition S = Q,U Q2). Moreover, for TI, 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
VII. Transformations of Eulerian Trails
7 ' , (G) (respectively TI,T2 E INr (G), TI,T2 E I,(G; Ql ,Q,)) satisfying T, = nl(Tl), we call this K~transformationa nAtransformation, (respectively nNItransformation, n,transformation), and define K A association (respectively association, n, association) correspondingly. We require, however, that if the nAtransformation is a K*transformation (see Definition VII.8), the trail decomposition S' sat isfying nl(Tl) = St (see Definition VII.7) must result from Tl by choosing a t v e V4(G) the pair of nonintersecting transitions different from XTl(v). Similarly, if a nNItransformation (natransformation) is a K * transformation, nl(Tl) = St implies that the corresponding system of transitions X s , is nonintersecting (compatible with the cycle decomposition Q1 U Q2). We note that if G is a plane graph, a %Atransformation 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 4regular graphs, the concepts of KA and KNItransformation coincide. Again we leave it as an exercise to show that each of the concepts of KA,KNI, and K,association defines an equivalence relation on the respective set of eulerian trails (Exercise VII.5.c)). Corollary VII.13. Let G be a connected 4regular graph embedded in a surface. Then any two elements of TA(G) are nAassociates. Proof. Since G is embedded in a surface, O+(v) is given by this embedding for every v € V(G). Hence it is possible to speak of intersecting and nonintersecting transitions. Since G is 4regular, the intersecting transitions are determined solely by the embedding of G. Let X be the system of these intersecting transitions. Then we have
and K~association is equivalent to nAassociation in this case. Corollary V11.13 now follows from Theorem VII.10. Another way of constructing all elements of TA(G) and of describing a K~transformationhas 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 Atrails). Namely, consider a 2facecoloring of the plane connected 4regular graph G (using the colors 1 and 2) and construct for i = 1 , 2 the graph Mias follows: every element x € V(Mi) represents bijectively an iface F(x) of G, and x, y E V(Mi) are joined by I V(bd(F(x))) n V(bd(F(y))) I edges; i.e. every vertex v E V(G)
VII.2.1. Special Types of K~Transformations
VII.25
corresponds to an edge ei(v) E E(Mi). Let 7 denote the set of spanning trees of Mi. The nest result relates 'T;. to TA(G) and describes K~transformations in TA(G) (see [KOTZ68c7Theorems 1417]).
Theorem VII.14. Let G be a plane connected 4regular graph, and let Mi, i = 1,2, be the graphs constructed above in accordance with a fixed 2facecoloring of G. Then the following is true.
1) There exist bijections cri between TA(G) and 'T;., i = 1,2, and a between TI and T2 which are defined by the following bijection properties: (i) If {Vl, V2) is the Apartition of TAE TA(G), then a;(TA) = ({ei(v) E E(Mi)/v E y)) E ?;, i = 1,2. (ii) For Tl E I, and TA E TA(G) satisfying Tl = al(TA), ~ 1 , 2 ( T l= ) ({e2(v) E E(M2)/e1(v) E(T1))) = a2(T,) E 3 . (iii) a, = al,2cr1.
e
2) The following statements are equivalent for TA7 TA E TA(G).
(i)
K A (TA) =
T i and TA# Ti.
(ii) ai( TA) U oi(TA) has precisely one cycle, i = 1,2. Proof. 1) Consider for TA E TA(G) its Apartition {V,, V2) and construct the plane graphs Hi+l := Gv,,i, i = 1 , 2 (putting 3 = 1). By Theorem VI.59, Hi is a connected outerplane g a p h l ) whose blocks are its cycles which in turn are the boundaries of the ifaces of G, i = 1,2. Hence, one can define Ti C Mi by
(where the subscripts Hi indicate that one considers the corresponding faces in Hi). Because of the structure of Hi, I bd(FHi(x)) f l bd(FH,(y)) I< 1 for every x, y E V(Ti). Because of the definition of Hi and because of v E V(G) w ei(v) E E(Mi), we conclude for i = 1 , 2 that xy = ei(v) E E(Ti) if and only if bd(FHi (x)) n bd(FHi (y)) = {v) E
(i
+
V;.
More precisely: Hi is a planar graph embedded in such a way that some 1)face contains in its boundary an edge of every block.
VII.26
VII. Transformations of Eulerian Trails
(see Lemma VI.58). Observing further that d H i (v) = 4 if and only if
e ,(v) E E (Ti)
(4
and considering the subdivision graph S(Tj), it follows from the above that
E ( T i ) = { e i ( v ) E E ( M i ) / v ~ V , ), Since Hi is connected, bc(Hi) is a tree. This and V(Ti) = V(&li) implies
Noting that Hi is already characterized by an Apartition {I 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 h1 E E(Se). For tl = {hi, hi) E Xs(w) we therefore have hj E E(S,). Let t2 = {h',, h',) E Xs(w) be chosen such that h,, h, E E(S,); w.1.o.g. m < n. Moreover, j < m or j > n since Xs is a system of nonintersecting transitions; w.1.o.g. j < m. Suppose {h',,hL) @Xs(w) if r and s satisfy j
< r < m < n < s 5 2k .
Then Xa(w) := (Xs(w)  j t 1 7 tz)) U {{hi, h 3 7 {hi, h:,)) consists of nonintersecting transitions only. Hence
x*= (X,
 X,(w)) U X, (w)
is a system of nonintersecting transitions of G. Moreover, it follows from the definition of X,(w) that the transformation of Xs into X * defines a Kabsorption at w (see equations (*) and (1") in the discussion immediately
VII. Transformations of Eulerian Trails
VII.32
E TNI(G); preceding Definition VII.7). That is, X * = XT,, where T,,, and Tw results from S by a Kabsorption.
If, however, {h:, h',) E Xs(w)
and j
< T < m < n < s < 2k for some r and s ,
then define t', := {h:, h:), tk := t2 if h,, h, E E(S,) otherwise, t', := t,, t i := {hi, h:), and repeat the above argument with t i , ta in place of tl, t2 (note that {h',, h:) E Xs(w) implies {h,, h,) n E(Se) # 0 if and only if {h,, h,) C E(Se)). Since d(w) < oo it follows that for some {ty), tyl) Xs(w), the above Kabsorption transforming S into Tw E TNI(G), can be performed (9 instead of t,, t,. Lemma VII.16 now follows. by using tl(9,t2 In accordance with the notation previously used, P(G, "NI) denotes the partition of IN1(G) induced by rcNIassociation.
Theorem VII.17. I P(G, tclvI) G embedded in a surface.
I=
1 for every connected eulerian graph
Proof. The theorem being trivially true if ) TNI(G)
I=
1 (since the identical transformation is a tcNItransformation), we proceed indirectly by assuming that 1 TNI(G) I> 1 and that eulerian trails TI,T2 E TNI(G) exist which are not sNIassociates. Denoting their respective system of transitions by X1 and X2 (we retain the notation of the proof of Theorem VII.10 whenever possible), we can again assume that Tl and T2 have been chosen in such a manner that I X1 n X2 I is as large as possible, but this time subject to the condition that a(G) := CVEV(G) (d(v)  2) is as small as possible. a(G) > 0 follows from (TN1(G)(> 1. Let v be any vertex for which Xl(v) # X2(v). By Lemma VII.15 a transition t = ( e l , f') E X2(v) exists such that O+(v) = (. . . ,e', f', . . .). Suppose t E Xl(v) holds as well. One can then form G' by splitting e and f away from v in such a way that V(G1) = V(G) u {v,, f ) and e, f are the edges incident with v , , ~ ,and where G' can be viewed as embedded in the same surface a s G. Since t E X1 n X2 it follows that Ti corresponds to an eulerian trail T;'of G' with Xf = Xi, where Xf is the system of
VII.2.1. Special Types of nlTransformations
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 nonintersecting transitions of G' corresponds to a system of nonintersecting transitions in G. Since a(Gt) < a(G),it follows from the choice of G that Ti and Ti are nNIassociates. Viewing the various nNItransformations used in transforming T: into T: as operations performed in G, and noting (*), it follows that Tl and T; are also nNIassociates. This contradicts the choice of Tl and T2; hence t # X1(v). SO, consider t, = {el,g:} and tf = {ft,ga}, where te,tf E Xl(v), and define for t* = {gi ,ga}
x; = (XI  {t,, tf)) U {t, t*), x;
= X2
.
Precisely because of the structure of t E X2(v) = X;(v) it follows that X; is a system of nonintersecting transitions. Moreover,
since t E X; n Xl whereas t # X1. Consequently, if X; = XT. for T* E TNI(G), it follows from the choice of TI and T2 that T* and T, are nNIassociates. This and T* = nNI(T1) implies, though, that Tl and T2 are nNIassociates, a contradiction to the choice of Tl and T2 which implies that X; = Xs for some S E S2(G). In accordance with the notation used earlier, denote S = {Se,Sg}, where w.1.o.g. e, f E E(Se) and, therefore, gl, g2 E E(Sg)(this follows from S E S2(G)). As in the proof of Theorem VII.lO, consider now a run through T2starting at v along e . Since t E X2 this run will end at v right after passing f . Since e, f E E(S,), gl ,g2 # E(S,), and therefore E(Se) c E(T2)= E(G), it follows that there is a vertex w at which T2 passes from Se into S,, 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 X,*, let G' be obtained (as in the first part of the proof) by splitting away e and f from v (note t = {el, fl}), and redefine Xi = Xi* ,i = 1,2, correspondingly; but now X i = X,, , for St E S2(G1), where St = (S:, SL) and E(S:) = E(Se), E(S;) = E(Sg). We distinguish between two cases.
VII.34
VII. Transformations of Eulerian Trails
Case 1). Xi(w) n X;(w) = 0. In this case, Lemma VII.16 can be applied directly to G' and S' so as to obtain Tk E TN1(Gf), where Tk results from S' by a Kabsorption at w. It follows that Tk and T; are associates (where Ti E INI(Gt) corresponds to T,), for the definition of Xf together with Xi(w) n Xi(w) = 0 and (**) ensure the validity of I XT,w f l Xi 1>1 X, n X, I. Thus, the corresponding eulerian trails Tw,T2 E TNz(G) are Kassociates. This and K'(T,) = S, K"(S) = Tw,and because X; is a system of nonintersecting transitions, classify TI and T2 as K ~ ~  a s s o c i a t eThis s . contradiction settles Case 1). Case 2). X i (w) nX;(w)
# 0. In this case we consider first O+(w) in G'
and write O+(w) = (hi, h;, . . . ,hLk) , k
>2
(note that dG,(w) 2 4 regardless of w = v or w # v). Consider an arbitrary element t' = {h',, h',} E Xi(w) n Xi(w), where 1 5 m < n 5 2 k ; w.1.o.g. m = 1. By the very definition of nonintersecting transitions it follows that t" = {hi, hi) $ X((w) U Xi(w) for i and j satisfying 1 < i < n < j 5 2k. Consequently, any element of ( X i ( w ) ~ X i ( w )) {t') lies on either of the two sides of the open curve, ,C , defined by h k U h;. Thus, ,C, defines a partition of Xf(w), i = 1,2,
(possibly Xi,, (w) = 0 or Xf,r(w) = 0, but Xi,, (w) U Xf,r(w) # 0); Xi,,(w) (Xi,r(w)) contains precisely those transitions in Xf(w) which lie (where left side and right side denote the on the left (right) side of, ,C, opposite sides of.),C ,, This partition (* * *) gives rise to the replacement of w by three vertices w,, w,, w, $ 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,, E INz(G;). Finally, it follows from (* * *) and the construction of G', that every system of nonintersecting transitions of G{ corresponds to a system of nonintersecting transitions of G' and G.
VII.2.1. Special Types of &ITransformations
VII.35
We repeat the above procedure if necessary until we arrive at the connected eulerian graph H with the following properties: 2) the elements ti of Xi(w) n Xi(w) are represented in H by 2valent vertices wj, and E:j = ti ; 3) dH(w,)
2 4 for every
and Xi(wa) C X;(w),
i = 1,2
,
where these Xi(w,) satisfy
whereas the vertices of G'
 w remain unaltered in H.
4) H can be viewed as embedded in the same surface as GI and obtained from GI in such a way that Xi(H) is a system of nonintersecting transitions which can be written as
(see 2) and 3), and note that a transition is a pair of halfedges independent of incidences with actual vertices). Moreover, every system of nonintersecting transitions in H corresponds to such a system in G' as well. Because of 4), Xl(H) = XSH for SH = {Se,H7 Sg,H) E S2(H) correI (H) sponding to the above S1E S2(GI), and X2(H) = XTH for TH E IN corresponding to T!E 'T;VI(G1). We note that since T2passes a t w from S, into Sgwe can therefore find some
at which TH passes from Se,H into SS,H. Consequently, since X1(wH) n X2(wH) = 0 by 3) we have reduced Case 2) to Case 1) with
VII.36
VII. Transformations of Eulerian Trails
H, WH, SH,TH in place of G', w, Sf,Ti respectively. Thus, the application of Lemma VII.16 yields T t E TNI(H) which results from SH by a Kabsorptionat wH. Arguing now along the same lines as in Case 1)we : and TH are associates which is tantamount can conclude that T to saying that Tw,the eulerian trail of G corresponding to T z , and T2 are tcNIassociates. Moreover, by the same token we can conclude that TI and Tw are tcNIassociates. Hence TI and T2 are tcNIassociates, 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 nonintersecting transitions. On the other hand, the proof presented here shows that the constructions of various graphs which are needed to find the appropriate /cNItransformations for transforming T, 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 rcNItransformations than in the case of K~association.It is possible that in the case of S E S2(G) without intersecting transition, no &absorptionat w yields Tw E TNI(G) satisfying I XTwn X2 I>I Xl n X2 I (see Figure VII.3). Therefore, one is forced to produce S from TI in such a way that already S satisfies I Xs n X, ]>I X, n X , 1, and thereafter to perform a Kabsorption at w without involving any element of X,(W>n x2 (4Finally we turn to K,transformations of eulerian trails in connected 4regular graphs G. It should be recalled that for a 2factorization {Q, ,Q,) of G, I,(G; Q,, Q,) denotes the set of eulerian trails of G in which edges of Q, and Q2 alternate (see the discussion preceding Corollary VII.13). As another application of Theorem VII.10 we obtain the following result in which P(G, IC,) denotes the partition of 7, (G; Q,, Q,) into equivalence classes under K,association.
Corollary VII.18. Let G be a connected Cregular graph, and let {Q,, Q,) be a 2factorization of G (which exists by Corollary VI.3). Then I P(G, 6,) I= 1. Proof. S = Q1 U Q2 is a special type of cycle decomposition of G; its induced system of transitions is denoted by Xs. By the very definition
VII.2.1. Special Types of K~Transformations
VII.37
Figure VII.3. The connected plane eulerian graph G with eulerian trail TI (T2) whose transitions are marked with small (dotted) arcs. A Kdetachment of TI at v involving the edges e, f yields S E S2(G) without intersecting transitions. Because of the structure of Xi(w), i = 1,2, the subsequent nabsorption a t w yielding T, E T,,(G) implies IXT ( w ) n X*(w) l=IX,(w) n X2(w) I= 0. ut
of S and Xs respectively, it follows that
Also, 6,association and nxsassociation are identical concepts in this case. Thus, p(G7 "a) = p(G7 KX, ) . 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
VII. Transformations of Eulerian Trails
graph with an even number of edges. Then G can be decomposed into two edgedisjoint eulerian subgraphs G1 and G2 with dG,(v) = dG2(v) for every v E V(G) (see Corollary VI.2). Denote by Ta(G; GI, G2) the set of eulerian trails which alternately pass through edges of G, and G2. 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.lO. 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 {GI, be a decomposition of G into two subgraphs having the same degree sequence (d(vl ), . . .,d(v,)}. Then Tl and T2 are K,associates for every TI,T2 E I,(G; GI, G2).
In the case of Cregular graphs G, however, one obtains for a fixed T E Ta(G; Q,, Q2) a classification of the vertices of G into 'odd7 and 'even7 vertices as follows. Let T be written in the form
and suppose vi = vj+,. Consider the segment
and the eulerian trail T* obtained by reversing the segment Si,j; i.e.,
T*= vl, el, . . . ,ei,,, ;s
ej+l, . . .,e,, v,
.
It follows from the very definition of T and T*that T* E T,(G; Q,, Q2) 1(mod 2). if and only if I E(Si,j)
(=
Consider the case where I E ( S i j ) 0 (mod 2). Then a &,transformation involving v := vi = vj+, cannot be a &transformation. On the other hand, IE(Si,,.)I= O(mod2) implies V((E(Si,j)))nV((E(G)E(Si,j))) # {v} . Otherwise, (E(Sij)) would have, except for the Zvalent v, only Cvalent vertices, i.e. (E(Sij)) would be homeomorphic to a 4regular graph with
VII.2.1. Special Types of K*Transformations
VII.39
an odd number of edges, something that is impossible. So, a 2valent vertex w E V((E(Sij)))  {v} exists and T is of the form
where w = vm = v,, and
We now show that these equations imply the existence of a K,transformation involving v and w. For such K, = K If K I and Sj+l,il..vJ.+ l , ej+l,. . . ,eq,vl, e l , . . . ,eil, vi we can define
which is compatible with S := Q1 U Q2 precisely because of I E(Si .) Ir 0 (mod 2). If we denote by Sm,,, and S n 1,n the segments of T obtalned from S i j,Sj+l,ilrespectively, by cyclic rotations so as to start and end 1 at w, then for any S:l ,, E {Snl,n, SL1 ,) 9'.
since Sf is compatible with S, it follows that
That is, there is a Kabsorption involving w such that nff(S') = T* ; i.e.,
K,(T) = T* for
K,
= K I1K I
.
Observe, in general, that if one fixes an initial vertex vl and an initial edge e , for T, T defines for every v E V(G) a unique trail decomposition Sh = {S1,,,S2,,) (each of whose elements starts and ends at v). Moreover, I E(G) 0 I (mod 2) since G is 4regular; hence we have

This congruence leads to the following more general definition.
VII.40
VII. Transformations of Eulerian Trails
Definition VII.20. Let T be an eulerian trail of the connected 4regular graph G starting at v, E V(G) along e E Eva, and consider for every v E V(G) the unique trail decomposition S, = (S,,,, S2,,) which consists of the two segments of T starting and ending at v. Then we can call v Teven or Todd, depending on whether ( E(S1,,) ( (and therefore, by (I), I E(S2,,) I) is even or odd. In fact, because of (I), the classification of Teven and Todd vertices is, for a given eulerian trail T, independent of the choice of an initial vertex vl and initial edge el (Exercise VII.10). Hence every eulerian trail T of G defines a partition of V(G) into Teven and Todd vertices, where this partition depends on T only. It follows from this definition and the preceding discussion that r;,transformations are Ktransformations precisely at Todd vertices of G, while a 6,transformation which is a K*transformation,is composed of a &detachmentat a Teven vertex and of a Kabsorption at some (Teven or Todd) vertex w E V((E(S,,,))) n V((E(S2,,))). It is not true, however, that this vertex w must be Teven (or Todd) in any case. This fact is demonstrated by the graphs G1 and G2 of Figure VII.4. There, the Qi,,, Qi,2), i, j = 1,2, satisfy the equations eulerian trails Ti,j E I,(Gi;
where the &detachment 6' is performed at the Ti,,even vertex v, while the 6absorption 6" is performed at w which is TI,,odd in GI and T,,?even in G2. Consequently, in GI a 6,transformation involving w is possible such that
However, v is T*odd indeed. Hence, there exists a 6,transformation such that n(T*) = That is, the above K*transformationtransforming TI,,into can be replaced with two 6,transformations which are Ktransformations. Such a replacement is not possible, however, in the case of T2,, and T2,2.AS we shall see in a moment, this phenomenon is a general one.
Definition VII.21. Let G be a connected Cregular graph, and let T1,T2 E T,(G; Q,, 9,). Suppose for S = Q1 U Q2 and for some v, w E
VII.2.1. Special Types of %ITransformations
Figure VII.4. Two 4regular graphs Gi with eulerian Qi,2),i, j = 1,2, where the edges trails T i j E 7,(Gi; Qi,,, of Qi,,, are marked by k, k = 1,2, and the transitions of Ti,l are marked by small arcs, while the transitions of Ti,, are marked by small dotted arcs, i = 1,2. Since TI and TI,, define the same transitions at x E V(G1), x is represented by two 2valent vertices. v is Ti,1even in Gi, i = 1,2, while w is TI,,odd in GI and T2,,even in G,.
V(G) that &'(TI)= St E S2(G;X S ) , K"(S')= T 2 , where K' is a Kdetachment a t v and 6'' is a Kabsorption at w: Call t ~ , := K"K' reducible or irreducible (regarding v, w) depending on whether w is TIodd or TIeven (note that v is T,even in any case since K , is a K*transformation).
Lemma VII.22. Let G, Tl, T,, S, St,K, = K'K", v and w be the objects used in Definition VII.21. Then the following holds: 1) K, is reducible if and only if there exist T* E ?,(G; Q,, Q,) and Ktransformations K, and K, involving w, v respectively, such that K,(T~)= T* and 6,(T*) = T,.
VII.42
VII. Transformations of Eulerian Trails
2) n, is irreducible if and only if there exists S" E S,(G, Xs) such that z1(T1) = S", it"(S1') = T2, where E' (2') is a Kdetachment
(&absorption) at w
(v).
Proof. 1) Suppose K, is reducible. By Definition VII.21, w is TIodd. Hence, by the discussion following Corollary VII.19 and Definition VII.20, there exist T* E Ta(G; Q1, Q,) and a nutransformation K , which is a Ktransformation such that nw(Tl) = T* and XT*  XT*(w) = XTl 
x, (4We claim that v is T*odd. This becomes visible when we express TI in the following way:
where uo = uj = vm = v and vi = vk = w. Moreover, we assume w.1.o.g. that e1,ej+, E E(Ql), f l , f j , f m E E ( Q 2 ) . This is possible since v is TIeven. Since w is TIodd we must have for one of the two choices for r , s satisfying {r, s) = { l , 2 )
Hence T* is of the form
T* =vO,el, v,, f l , . . . ,hi, vj, h i , . . . ,ej+l, uj7fj, . . . ,
. . .,hy, vi, h i , . . . ,f,, urn . That is, starting in u = vo along el E E(Ql) we reach v = vj again after passing ej+l E E(Q,), i.e. v is T*odd. Consequently, there exists a natransformation nu which is a Ktransformation such that nv(T*) = T$ c I,(G; Q,, Q,) and XTc XT. (v) = XT;  XT; (u) . Consequently,
and
VII.2.1. Special Types of K~Transformations
These equations and TI,T2,T,' E
VII.43
I,(G; Q1, Q2) yield T2 = Ti.
Conversely, suppose for TI, T2 E 7, (G; Q1, Q2) that there exists T* E 7, (G; Q1, Q2) and Ktransformations K, and K, (involving w, v respectively) such that K,(T~)= T* and K,(T*) = T2. Definition VII.20 and the subsequent discussion imply that w is TIodd. On the other hand, since K, = K"K' is a K*transformation,v must be TIeven (see Definition VII.21). Hence, K, = r c t t ~ 'is reducible by Definition VII.21. This finishes the proof of statement 1). 2) If
is irreducible, then w is also TIeven. Hence, by using the same symbols with the same meaning as in the first part of the proof of I ) , and such that (*) is fulfilled we obtain K,
Thus, changing the transitions at w = vi = v, so as not to violate compatibility with Xs , we obtain two edgedisjoint subtrails covering 'I
'I
Sl,, = vi,hi
f j , v j , e j + l , . . . , h ~ , ~, ~
St,, =~k,h:,...,fm,~m,el,vl,fl,...,h:,viThat is, St' := {S:,,, Sl,,) E S2(G,XS) and E1(Tl) = S" for some Kdetachment izt at w. However, by a rotation we can rewrite the elements of S" so as to obtain
That is, St = {Si,,,Si,,) yields K'(T;) = Stfor some T: E 7, (G; Q1, Q2). Since St results from S" by a cyclic rotation of the sequences describing its elements, and since the inverse operation of a Kdetachment is a Kabsorption, we have ii"(S") = T; for a Kabsorption Elt at v. We can now conclude as in the first part of the proof of 1 ) that T; = T2. Conversely, suppose two K,transformations K"K' and R"Z' satisfy K"(K'(T~))= E"(Et(Tl)) = T2, where nt is a Kdetachment at v and E' is a Kdetachment at w. Then both v and w must be TIeven (see the discussion following Definition VII.20), hence n, is irreducible. This finishes the proof of 2). Lemma VII.22 now follows.
VII.44
VII. Transformations of Eulerian Trails
Lemma VII.22 says that a K,transformation K"K' is reducible if and only if one can avoid taking a 'detour' via S2(G,Xs), by applying two Ktransformations K, and K, , each of which yields an element of Ta(G; Q,, Q,). This fact, i.e. staying with Ktransformations instead of combining a Kdetachment with a Kabsorption, justifies the term reducible. On the other hand, if K"K' is irreducible, it is irrelevant at which of the two vertices v and w one starts with a Kdetachment, i.e. in this case, the product K"K is commutative in the sense that the transformations K' and K", viewed as operations on systems of transitions, yield K"K'(T,)= K'K"(T,) = T2 where K'(T,), K"(T,) E S,(G, Xs) and in general, K'(T,) # rctt(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 4regular graph G with 2factorization {Q,, Q,). For any two eulerian trails T, T' E I,(G; Q1, QP) a sequence of eulerian trails To,.. . ,T, exists for some m 2 1 such that To = T, T, = T',Ti E Ta(G;Q,,Q,), i = 0,. . .,m, and such that for i = 1,. . . ,m, Ti results from Ti, by the application of a K,transformation. If this K,transformation is a Ktraasformation, it is applied at a Ti,odd vertex. If, however, it is a K*transformation, it is an irreducible K,transformation applied at two Ti,even vertices v and w, where w belongs to both segments of Ti, starting and ending at v. In this case it is irrelevant whether one applies the Kdetachment at v 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 VII.12 to obtain a result for digraphs analogous to Theorem VII.5. For this purpose denote, by analogy, by I ( D ) the s e t of all eulerian trails of t h e connected eulerian d i g r a p h D, and define K, K*, rcl transformations and Kdetachmentsand Kabsorptions. 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 ~vhy,for a digraph D , Ktransformations are not a sufficiently general tool to transform T E I ( D ) into every other element of I ( D ) (see Figure VII.5). However, as we shall see below, K~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*, 6,association, and one can show that each of these concepts defines an equivalence relation on I ( 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 I ( D ) into equivalence classes as before. For our purposes we shall consider the partition P ( D , K,). T h e o r e m VII.24. IP(D, K l ) I= 1
Let D be a connected eulerian digraph.
Then
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 halfarcs incident from v, to v respectively. Since V(D) = V(G) one then defines for the same v
by letting, for e = vx E E(G), e' belong to (Ez)+ ((Ez) respectively) if and only if for the corresponding arc a E A(D), a' E (A;)+ (af E (A:)respectively), i.e. if and only if a = (v, x) (a = (x, v) respectively). This
VII.46
VII. Transformations of Eulerian Trails
Figure VII.5. D has precisely two eulerian trails, TI and
T2 (marked by small and dotted small arcs). &(TI) # T2 for any of the two conceivable Ktransformations, but ri*(Tl) =
T, for each of the two possible K*transformationsK* = ~ " r i ' (compare this with Exercise VII.5).
definition is meant t o include multiple arcs (in which case each arc of the form (v, x), (x, v) respectively, has to be considered separately in the above correspondence), as well as loops (v, v) in D (in which case one halfloop belongs to (E:)+ and the other halfloop belongs to (E:)). Finally define P ( G ) := Pv(G) .
U
vEV(G)
P(G) is, by definition, a partition system of G such that I I< I= + d ( u ) 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 , I= 1 by Theorem VII.12. However, precisely because of the definition of P(G) there is a 11correspondence 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 associates if and only if the corresponding elements TD,Tb E 7 ( D ) arc 6,associates. That is, I P ( D , K , ) I= 1. The theorem now follows.
VII.3. Transformation of Eulerian Trails in Digraphs
VII.47
Observe that the above application of Theorem VI.l and the bijection between I ( G , P(G)) and I ( D ) yields a proof of the first part of Theorem IV.8. Another approach to transform an arbitrary T E I ( D ) into any other Tr E I ( D ) has been chosen in XIA AX^^^].^) In this approach one does not have to deal with trail decompositions as in the case of K , transformations; rather, at every step one transforms an element of I ( D ) into another element of I ( 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 I ( D ) can be expressed in the following way:
where T(u, v) and Tr(u, v) are arcdisjoint trails joining u and v (possibly u = v), and suppose A(T(u, v)) u A(Tr(u, v)) # A(D) (which holds in any case if u # v). Then
is said to be obtained from T by a Ttransformation which exchanges T(u, v) and Tr(u,v). We express this transformation by the formal equation T(T) = TI. R e m a r k VII.26. 1) Xia speaks of a Ttransformation instead of Ttransformation; 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 Tr E I ( D ) hold for T E I ( D ) and Tr = T(T), but we even have Tr # T. This is guaranteed
by the condition A(T(u, v)) U A(Tr(u,v))
# A(D).
That is, contrary to
This paper has been published in Chinese, the English abstract of which, however, is insufficient to permit a full understanding of the essentials. Doz. Wolfgang Ruppert of Vienna, has translated the most important passages of the paper, and I hope to reproduce them correctly in the ensuing discussion. In any case, I wish to express my thanks to Doz. Wolfgang Ruppert for having taken time to translate major parts of the paper.
VII.48
VII. Transformations of Eulerian Trails
various previous transformations, a rtransformation 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 i f and only i f every cycle of D i s a block of D and A(D) 5 4. The proof of this statement is left as an exercise. 3) However, if T' = r(T), then for the same pair of trails T(u, v): T1(u,u ) used in transforming T into T' (see Definition VII.25) we obtain r(T1) = T, i.e. T , in a certain sense, satisfies the formal equation r2(T) = r(Tr)= T . Consequently, it makes sense to call T, T' E I ( D ) rassociates if either T = T' or there exists a sequence To,. . .,Tm, m > 0, where Ti E I ( D ) for i = 0 , . . .,m such that To = T, Tm = T' and Ti = T ( T ~  ~ ) for i = 1,. . .,m. Again, rassociation defines an equivalence relation on I ( D ) which in turn defines a partition P(D, r ) (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 , r) I= 1for every connected eulerian digraph D . Proof. If 1 I ( D ) I= 1, the theorem is trivially true since T E I ( D ) is the only rassociate of itself, whence we can assume 17(D) I> 1.
Suppose the theorem fails for D. Then T,T' E I ( D ) exist, which are not rassociates. Suppose T and T' have been chosen such that I X, n X,, I is as large as possible subject to the condition that C ( D := ~ C , E V ( D ) ( ~ d (~ )1) =I A(D) I  1 V(D) I is as small as possible. For practical reasons, we can deduce some property D must have attributable to those restrictions.
For every v E V(D), XT(v) n X T r ( v ) = 0 ord(v) = 2 Otherwise, let v be a vertex with XT(v) n XTr(v)
.
# 0 and d(v) > 2.
(1) Let
t = {e', f') E XT(v) n XT,(v) and split away the arcs e, f to form a new eulerian digraph Dl (note that {e,f} n A: # 0 # { e , f} n A;). The eulerian trails T and T' correspond to eulerian trails TI,Ti of Dl, hence Dl is connected. Moreover, XTl = XT, XT; = XTr , hence I XT n XT. I= IX , nXT; I. However, o(D1) < o(D) since I A(Dl) /=I A(D) 1, but I V(D,) I=I V(D) I +l. Thus, Tl and T,'are Tassociates. We note that for every Tl(x, y) in Dl used in one of the Ttransformations which have been employed for the transformation of Tl into Ti, we cannot have
VII.3. Transformation of Eulerian Trails in Digraphs
VII.49
v ~ E, {x, ~ y), where vePf E V(D1)  V(D), for there are no two arcdisjoint open (or closed) trails in Dl starting (or ending) at v , , ~ . Hence Tl(x, y) corresponds to T(x, y) in D which is an open trail if and only if T,(x, y) is an open trail. Consequently, Tl and Ti being Tassociates is tantamount to saying T and T' are Tassociates, a contradiction to our assumption. Consider now an arbitrary v E V(D)  V2(D), and let t E XTl(v) be arbitrarily chosen. Denote t = {e(v), e(v)+) where e E A;, e+ E A: ancl let t,, t2 E XT(v) be chosen in such a way that tl = {f (v), e(v)+), t2 = {f(v)+,e(v)). Because of (I), t, # t2; 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 T. On the other hand, TI,also viewed as starting a t v along e+, has e as its last arc. Hence there exist transitions t3, t4 E XT1(w) for Some
such that
and t, = {9(w)+, 9(w)1 E XT(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 T1(v,w). Observe that T1(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 g as its last arc. Writing S(e+, e) = T(v, w), w, T(w, v) we can express T in a more explicit form,
VII.50
VII. Transforlllations of Eulerian Trails
Figure VII.6. Eulerian trails T, TI E I ( 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 + ,hh+. Consequently, we obtain T* E I ( D ) , T # T*, by the Ttransformation which exchanges T(v, w) and T1(v,w). That is,
Possibly T* = TI. In any case, however, we have d ( v ) > 2, d(w) > 2 and t, t, E X T . n X T , 1 XT n X,,. This and (1) together with the choice of T and TI imply that T* and TI are Tassociates (which is true also if T* = TI). Therefore, and because r(T) = T*, we can conclude that T and TI are Tassociates. This contradiction to the choice of T and TI 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 Dl )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. Final Remarks and Some Open Problems
VII.51
the case of graphs (note that a Ttransformation can be viewed as the combination of a 6detachment with a 6absorption). As for transforming aneulerian trails in a digraph D it suffices to consider Dr.The graph GT underlying D+ is a connected eulerian graph because an aneulerian trail in D corresponds to an aneulerian trail in DT 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 Ktransformations (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 K~ transformations in order to transform T into TI ? (*I In other words, what is the complexity of transforming T into TI? We note that a single 6transformation can be performed in polynomial time, even if one has to search for a vertex v at which a 6transformation should be performed. Hence the construction of TI, T[ E T(G), satisfying TI = rc(T), Ti = rc(T1) (with the 6transformations performed at v), 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 Kassociation. This question has been answered by A. Bouchet for connected Cregular graphs G. He showed that for any T, T' € 7(G) it takes at most 6transformations to transform T into T', [BOUCSOa]. Together with this result, however, he presents a conjecture which says that can be replaced with ( V(G) I +l. However, analogous complexity studies for the various types of nltransformations (which have been developed for the various types of
VII.52
VII. Transformations of Eulerian Trails
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 ntransformations. Define the eulerian trail graph G ( 7 , n) as the graph whose vertices are the elements of 7 := 7(G), and TIT2 E E ( G ( 7 , n)) if and only if n(T1) = T2. Note that &(TI) = T2 implies 4 T 2 ) = TI. Hence G ( 7 , n) is a well defined graph. It follows from Theorem VII.5 that G ( 7 , n) is connected, since any two eulerian trails of G can be obtained from each other by a sequence of ntransformations. However, Zhang and Guo verified a much stronger result which we present without proof (for details, see [ z H A F ~ ~ ~ ] ) . ' ) Theorem VII.28. If I 7(G) I> 2, G ( 7 , n ) is edgehamiltonian.
Again (similar to Bouchet's result [BOUC9Oa]) one is led to the question whether there are results analogous to Theorem VII.28, if one restricts the considerations to special classes of eulerian trails (such as Tx := 7(G, X), a.s.0.) and the appropriate nltransformations, and defines correspondingly a restricted eulerian trail graph (e.g., G(Tx, fix)). 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 [ZHAF86a] 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 GT of a graph G, where V(GT) = (T & G / T is a spanning tree of G), and TIT2 E E(G,),if and only if / E(Tl uT2)  E(Tl nT2) I= 2 (compare this equation with tcAtransformationsand the set '7;. of spanning trees of M i in the discussion leading to the formulation of Theorem VII.14). In fact, tree graphs GT have been already considered implicitly in [ORE062a], explicitly in [CUMM66a]. In the latter reference it has been shown that GT is edgehamiltonian (a short proof of this result has been given in Zhang and Guo speak of the euler tour graph Eu(G) that we have termed the eulerian trail graph G(7,n).
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 (Mi)T of the graph Mi is nothing but G(TA(G),K ~ ) 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 GT, (GT)T and ((GT)T)T are all eulerian, (((GT)T)T)T is not eulerian unless GT E {K1,K3). Finally, we note that for a digraph D , the intree graph DT concerning a fixed root vo E V(D) can be defined in a way similar to the above GT. This has been done in [ D O R F ~ where ~ ~ ] it is s h o w 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 TI, . . .,T,, of XIcompatible eulerian trails of K2 with T = TI, T' = T, and n(Ti) = Ti+1,1 < i 5 n  1. Describe the segment reversals by which one obtains T2 from T and TI1from T2. Exercise VII.2. Show that each of the concepts of K* and K ~ association defines an equivalence relation on T(G). Show the same for the concept of K*association with respect to T(G, X). Exercise VII.3. Prove Lemma VII.ll. Exercise VII.4. Translate the proof of Theorem VII.10 into a proof of Theorem VII.12. Exercise VII.5. a) Show that if K ~ ( T=~T2 ) for Tl ,T2 E TA(G), where G is a connected plane 4regular graph, then this rcAtransformation is a rc*transformation. b) Construct an example of a nonplanar G embedded in a surface where this conclusion does not hold. c) Prove: each of the concepts of KA, ICNI and K,association defines an equivalence relation on the respective set of eulerian trails.

VII.54
VII. Transformations of Eulerian Trails
Exercise VII.6. Extend Theorem VII.14 to arbitrary connected plane eulerian graphs, in two ways such that a) Mi is a (bipartite) graph, b) Miis a hypergraph (see the first part of the discussion following the proof of Theorem VII.14). Exercise VII.7. Show that deciding the existence of a spanning hypertree in Mi (see Exercise VII.6.b)) is an NPcomplete problem if one restricts oneself to the consideration of those Mi which correspond to planar 3connected eulerian graphs (see the second part of the discussion following the proof of Theorem VII.14). Exercise VII.8. a) Show that the concept of K~association, extended to arbitrary connected plane eulerian graphs, defines an equivalence relation on 7,(G). b) Show that two elements of TA(G)belong to the same equivalence class under K~association if and only if their respective Apartitions coincide on the vertices whose valency exceeds 4. Exercise VII.9. Generalize the concept of K,transformation for connected eulerian graphs having an even number of edges and prove Corollary VII.19 (Hint: apply Theorem VI.l, respectively Corollary VI.2 and Theorem VII.12). Exercise VII.lO. Let T be an eulerian trail of the connected 4regular graph G. Show that in Definition VII.20, v E V(G) is Teven (Todd) independent of the choice of an initial vertex and initial edge of T. Exercise VII.ll. Define K, n* and rclassociation for digraphs and show that each of these concepts defines an equivalence relation on 7 ( D ) . Exercise VII.12. Prove: the following two statements are equivalent. 1) D is a connected eulerian digraph having precisely one eulerian trail. 2) Every cycle of D is a block of D and A(D) 5 4. Exercise VII.13. Prove that Tassociation defines an equivalence relation on T(D). Exercise VII.14. Consider P(D)compatible eulerian trails in a connected eulerian digraph D and find transformations by which any such trail can be transformed into any other such trail. Study the same problem with respect to (D  Do)favoring eulerian trails. Exercise VII.15. Let G be a connected eulerian graph, and let T,T1E 7(G). Show:
VII.5. Exercises
VII.55
1)The problem of finding a vertex u a t which XT # X T , ( v ) can be solved in polynomial time. 2) The construction of TI, Ti E I ( G ) satisfying xT(v)=  XTl ( v ),XTt  XTt ( v ) = XT;  XT; ( v ), K ( T ) = TI, r(T1)= Ti and I XTl ( v ) n XT; ( v ) I>I X T ( u ) n X T , ( v ) 1, can be done in polynomial time.
xT
xTl
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