eulerian colorings and the bipartizing matchings conjecture of fleischner

European Journal of Combinatorics 27 (2006) 1088–1101www.elsevier.com/locate/ejc

Eulerian colorings and the bipartizing matchingsconjecture of Fleischner

Zdenek Dvoraka,1, Tomas Kaiserb,1, Daniel Kral’ a,1

aDepartment of Applied Mathematics, Charles University, Malostransk´e nam. 25, 118 00 Praha 1, Czech Republicb Department of Mathematics, University of West Bohemia, Univerzitn´ı 8, 306 14 Plzeˇn, Czech Republic

Available online 18 July 2006

Abstract

A trigraph is a multigraph with half-edges colored in three colors. We introduce the notion of an Euleriancoloring of a trigraph and show that the existence of two orthogonal Eulerian colorings in a special classof trigraphs is closely related to the bipartizing matchings conjecture of Fleischner, and hence to the cycledouble cover conjecture and Tutte’s 5-flow conjecture. We prove that every trigraph has an Eulerian coloringand that a rainbow cubic trigraph has a pair of orthogonal Eulerian colorings if and only if it has a perfectmatching.c© 2006 Elsevier Ltd. All rights reserved.

1. Introduction

Fleischner [3] has proposed a novel approach to several deep graph-theoretic conjectures, suchas the cycle double cover conjecture [9,10] or the5-flow conjecture of tutte [12]. A key ingredientin this approach is the bipartizing matchings conjecture (Conjecture 1.2below, abbreviated BMConjecture). In this paper, we introduce the notion of an Eulerian coloring of a trigraph, providinga setting that allows a generalization of the bipartizing matchings conjecture and inspires newquestions. We also prove the counterparts of major known facts about bipartizing matchings inthe more general context.

E-mail addresses:[email protected](Z. Dvorak),[email protected](T. Kaiser),[email protected](D. Kral’).

1 Institute for Theoretical Computer Science (ITI), Charles University, Praha, Czech Republic. Supported by projectLN00A056 of the Czech Ministry of Education.

0195-6698/$ - see front matterc© 2006 Elsevier Ltd. All rights reserved.doi:10.1016/j.ejc.2006.06.002

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http://dx.doi.org/10.1016/j.ejc.2006.06.002

Z. Dvorak et al. / European Journal of Combinatorics 27 (2006) 1088–1101 1089

Fig. 1. (a) A bipartizing matchingM (bold) in the Petersen graphP. Thedominating cycle is shown dashed. (b) ThegraphP − M is a subdivision ofK3,3.

Let us describe in detail the connection between the BM conjecture and the cycle doublecover conjecture which asserts that any bridgeless graph has acycle double cover (CDC), i.e., a(multi)set of circuits which covers each edge exactly twice. It is well known [13, Theorem 7.2.4]that it is enough to prove the CDC conjecture for 3-connected cubic graphs, in fact, even only forsnarks. (Recall that asnark is a cubic cyclically 4-edge-connected graph of girth at least 5 whichis not 3-edge-colorable. See [6] or [11] for moreinformation.)

It is easy to find a CDC of a Hamiltonian cubic graph. Although no snark has a Hamiltoncycle, they are all conjectured to be rather close to being Hamiltonian:

Conjecture 1.1 (Dominating Cycle Conjecture [2] ). Every snark has adominating cycle, i.e., acycle incident with every edge.

If a cubic graph has a dominating cycle, does it help in finding a CDC? Not in an obviousway. However, with the help of the BMConjecture of Fleischner [3] (Conjecture 1.2below),it is indeed possible to construct a CDC. LetC be a dominating cycle in a cubic graphG. Abipartizing matchingin G (with respectto the dominating cycleC) is a matching M that is edge-disjoint fromC, covers all the vertices not onC, andhas the property thatG− M is a subdivisionof a cubic bipartite graph.Fig. 1 shows a bipartizing matchingM in the Petersen graph withrespect to the “outside” dominating cycle. Indeed, the removal ofM yields a subdivision ofK3,3.

Conjecture 1.2 (Bipartizing Matchings Conjecture [3] ). Given any dominating cycle C in asnark, there is a pair of (edge-)disjoint bipartizing matchings with respect to C.

Fleischner [3,4] showed thatConjecture 1.2implies the existence of a CDC and a nowhere-zero 5-flow for any snark with a dominating cycle. (For information on nowhere-zero flows,consult [8].) Thus, if Conjectures 1.1and 1.2 hold, then the CDC conjecture and the 5-flowconjecture follow. Speaking about two disjoint bipartizing matchings, it is natural to ask whetherat least one bipartizing matching (with respect to a given dominating cycle) always exists.Fleischner and Stiebitz [5] showed that this is indeed the case:

Theorem 1.3. Any cubic graph has a bipartizing matching with respect to any dominating cycle.

As shown in [3], an analogue ofConjecture 1.2holds for Hamiltonian cubic graphs. Moreprecisely, any cubic graph with a Hamilton cycleC has a pair of disjoint bipartizing matchingswith respect toC. The following alternative proof of this fact is the starting point of our work.

Two chordsof a Hamilton cycleC in G intersect if one of them separates the end-vertices ofthe other. Letus define thecircle graphCG(G) of G (with respect toC) to be the graph whosevertices are the chords of the cycleC in G, and whose edges join intersecting pairs of chords.An example of a circle graph is given inFig. 2.

1090 Z. Dvorak et al. / European Journal of Combinatorics 27 (2006) 1088–1101

Fig. 2. (a) A graphG with a Hamilton cycleC (dashed). (b) The circle graph CG(G) of G with respect toC.

Throughout this paper, anEuleriangraph is a graph all of whose vertices have even degrees(the graph may or may not be connected).

Proposition 1.4. Let C be a Hamilton cycle in a cubic graph G. A matching M⊂ E(G)− E(C)

is bipartizing ifand only if the set V(CG(G)) − M induces an Eulerian graph.

Proof. Consider a matchingM ⊂ E(G) − E(C). For each chordf �∈ M of the cycleC, fix acycleC f ⊂ G − M consisting of f and one of the two parts ofC delimited by the endverticesof f . We prove the following:

Claim. The degree ofany chord f , as a vertex ofCG(G)−M, has the same parity as the numberof vertices on Cf whose degree in G− M is 3.

A vertex ofC f has degree 3 inG − M if andonly if it is an endvertex of a chord not inM. Achordh �∈ M contributes 1 such vertex if it crossesf , and it contributes 0 or 2 such vertices if itdoes not crossf . Sincethe degree off in CG(G) − M is precisely thenumber of chordsh �∈ Mcrossing f , the claim follows.

To prove the equivalence in the proposition, note thatM is bipartizing if and only if each cyclein G − M contains an even number of degree 3 vertices. In particular, ifM is bipartizing, theneach cycleC f contains an even number of degree 3 vertices. By the claim, the degree of eachf ∈ V(CG(G)) is even.

For the converse, letG′ be the graph obtained fromG−M by suppressing all degree 2 vertices,and letC′ andC′

f denote the cycles corresponding toC andC f in G′, respectively. By the claim,all of these cycles are of even length. If the number of chordsf �∈ M of C is t , then the set

{C′} ∪ {C′f : f �∈ M is a chord ofC}

is a set oft + 1 independent vectors in the cycle spaceZ(G′) of G′ (see, e.g., [1, p. 52]). ThegraphG′ has 2t vertices and 3t edges, so the dimension ofZ(G′) is 3t − 2t + 1 = t + 1.Hence, we have a basis ofZ(G′) consisting of even cycles. Clearly,G′ is bipartite, and soM isbipartizing. �

Thus,G has two disjoint bipartizing matchings with respect to the Hamilton cycleC if andonly if CG(G) contains two induced Eulerian subgraphsH1 and H2, such that V(CG(G)) =V(H1) ∪ V(H2). Not only is the latter condition always true. Somewhat surprisingly, there isa much stronger result due to T. Gallai (see Problem 5.17(a) in [7]). We include a proof sincewe use similar ideas later, in the more complicated proof ofTheorem 4.1. As usual, the inducedsubgraph of a graphH on a setX ⊂ V(H ) is denoted byH [X].Theorem 1.5. The vertices ofany graph H can bepartitioned into two sets, each of whichinduces an Eulerian subgraph of H .


Proof. If H itself is Eulerian, then the trivial partition intoV(H ) and∅ does the job. Otherwise,let v be a vertex of odd degree and let its neighborhood be denoted byN. Remove v andperform thecomplementation on N(delete all edges onN and replace all former non-edgeson N with edges) toobtain a graphH ′. SinceH wasnot the one-vertex graph,H ′ is nonempty.By induction,V(H ′) has apartition V ′

1 ∪ V ′2 into sets inducing Eulerian subgraphs. Since|N| is

odd,|V ′i ∩ N| is even for precisely onei , sayi = 1. If we reverse the complementation onN,

the degree of eachw ∈ V ′1 ∩ N in H [V ′

1] changes from even to odd, while all other degrees inH [V ′

1] and all degrees inH [V ′2] stay even. Thus, the partition ofV(H ) into V1 = V ′

1 ∪ {v} andV2 = V ′

2 has the desired property.�

By Proposition 1.4, thepartition in Theorem 1.5yields a partition of the chords ofC in G intotwo bipartizing matchings, thus providing a new proof of an analogue ofConjecture 1.2in theHamiltonian case. This brings up the following question: is there a generalization of the circlegraph that can be used ifC is just a dominating cycle? We describe such a generalization in thefollowing section.

One more comment on the proof ofTheorem 1.5is in order. An alternative argument involvesa system oflinear equations overZ2 where eachv ∈ V(H ) hasits own variablexv and definesthe following equation:

∑

w∈N(v)

(xv + xw + 1) = 0.

This system of equations always has a solution, and the solutions are in one-to-onecorrespondence with the partitions fromTheorem 1.5. The interesting thing is that the twoarguments, despite their apparent dissimilarity, are in fact very close to each other. Roughlyspeaking, the complementation on the neighborhood of a vertex (as in the proof ofTheorem 1.5)corresponds to a step of the Gauss elimination on the above system of equations.

2. Trigraphs

We shall now define trigraphs and Eulerian colorings. These concepts will enable us to transferquestions about bipartizing matchings to a more general level, in the direction indicated byProposition 1.4.

Throughout,C will always be adominating cycle of a cubic graphG. Any vertexnot onCwill be calledinternal. Forour purposes, it may be assumed thatC has no chords (so every edgenot onC is adjacent to an internal vertex). This is ensured by the followingchord eliminationprocess (originally from [3]) whose outcome is illustrated inFig. 3. Let c1c2 be a chord ofC,with the vertexc1 adjacent to a vertexd onC. Subdivide the edgesc1c2 andc1d, and join the twonew vertices by an edge. The resulting graphG′ is cubic and the dominating cycleC′, arisingfrom C, hasone chord fewer. Furthermore,G′ has a bipartizing matching with respect toC′ ifand only if G has one with respect toC. A similar claim holds for a pair ofdisjoint bipartizingmatchings.

We now define an analogue of the circle graph from Section1. In contrast to the previouscase, there are essentially 3 possible configurations of two internal verticesx andy, as shown inFig. 4. To tell them apart,one can form a cyclic sequenceσ that contains the symbolx for eachneighbor ofx, and the symbol y for each neighbor ofy, as they appear inthe clockwise orderaroundC. Then, internal verticesx andy form adisjoint pair if σ = xxxyyy(as inFig. 4(a)), acrossingpair if σ = xxyxyy(Fig. 4(b)), or analternatingpair if σ = xyxyxy(Fig. 4(c)).


Fig. 3. A graph obtained by repeated chord elimination from the graph inFig. 2(a). The added vertices and edges areshown in grey.

(a) A disjointpair.

(b) A crossingpair.

(c) Analternatingpair.

Fig. 4. The three configurations of two internal vertices.

We now define our analogue of the circle graph. Atrigraph is a pair(T, c), whereT is aloopless multigraph whose edges are viewed as composed of two half-edges, andc is a coloringof the half-edges by the colors 1, 2, 3. If an edgee joins verticesx and y, then the half-edgeincident withx is denoted byex. The colorc(ex) of ex will be referred to as thecolor of e at x.In cases where the coloringc is clear from the context, we just speak of a trigraphT .

The trigraph TG(G, C) of a cubic graphG, with respect to a dominating cycleC, is definedas follows. Its vertices are the internal vertices ofG. Two vertices are joined by 0, 1 or 3 edgesaccording to whether they forma disjoint, crossing or alternating pair, respectively.

The coloring of the half-edges of TG(G, C) reflects the special role of certain edges, e.g., ina crossing pair of internal vertices. Choose a labelling� of V(C) by 1, 2 and 3 such that theneighbors of each internal vertex get all three labels. For a pair of verticesx, y ∈ V(TG(G, C)),let Gxy be the graph obtained fromG by removing all internal vertices exceptx and y, andsuppressing all vertices of degree 2 this creates. Thus,Gxy has 8 vertices, and is isomorphic toone of the graphs inFig. 4. We shall now define the coloring in each of the three cases. At thesame time, we shall define the notion of arelatedpair of vertices.

For a disjoint pair x, y, there isno edge betweenx andy to color, and there will also be norelated neighbors ofx andy.

If x andy form a crossing pair, letx′ andy′ be the unique neighbors ofx andy, respectively,that are not contained in any triangle inGxy. The (single) edgexy in TG(G, C) will be assignedthe color �(x′) at the vertexx, and the color �(y′) at y. The verticesx′ andy′ form a related pair(written asx′ ∼ y′), and there are no other related pairs among the neighbors ofx andy.

If x andy form an alternating pair, then for each neighborx′ of x, there isaunique vertex,y′,suchthatx′ andy′ are not contained in any 4-cycle inGxy. Correspondingly, one of the 3 edgesjoining x to y in TG(G, C) will have the color �(x′) at x and�(y′) at y. Theother choices ofx′ yield the colorings of the remaining two edges joiningx andy in TG(G, C). We have 3 pairsof related vertices in this case: eachx′ is related to the associatedy′. This completes both thedefinition of TG(G, C) and of the related vertices.


Fig. 5. (a) A cubic graphG. (b) The trigraph TG(G,C) with respect to the dashed dominating cycleC of G.

Note that TG(G, C) dependsonthe choice of the labelling�. However, trigraphs obtained fordifferent choices of� are equivalent in the following sense: trigraphsT1 andT2 areisomorphic ifthere is an isomorphismϕ : T1 → T2 of uncolored multigraphs, and for each vertexv ∈ V(T1)

and edgese, e′ incident withv, the colors of e, e′ at v are the same if and only if the colors ofϕ(e), ϕ(e′) atϕ(v) are the same. Thus, isomorphic trigraphs only differ by color permutations ateach vertex. Their combinatorial properties arethe same, and we shall regard such trigraphs asidentical. With this provision, TG(G, C) is well-defined.

An example of acubic graphG with a dominating cycleC, along with the correspondingtrigraph TG(G, C), is shown inFig. 5. We representthe colors 1, 2, 3 by shades of grey (fromlighter todarker).

We remark that not all trigraphs arise as TG(G, C) for a cubic graphG and a dominatingcycleC. A trivial obstruction is thepresence of more than 3 parallel edges joining two vertices.It is, however, not difficult to construct such examples without any multiple edges as well.

3. Eulerian colorings of trigraphs

Fix a cubic graphG and a labelling� of the verticesof the dominating cycleC by {1, 2, 3}.Let xi denote the neighbor of an internal vertexx with �(xi ) = i . Define the(x, i )-segmentSx,i to be the arc ofC delimited by the neighbors ofx different from xi . More formally,Sx,i is the component ofC − {x1, x2, x3} containing no neighbor ofxi . Further, let Cx,i bethe cycle formed by the segmentSx,i , the two neighbors ofx adjacent to it, and the vertexxitself.

The notion of a bipartizing matching has a natural counterpart in the trigraph setting. Let afullmatching in G be a matching disjoint from C and covering all internal vertices. Since we haveassumed thatC has no chords, all bipartizing matchings are full. Any full matchingM inducesa (not necessarily proper) vertex coloringfM of TG(G, C) by the colors 1, 2, 3: a vertexx ofTG(G, C) gets color i in fM if the edgexxi is contained inM. In fact, this defines a bijectivecorrespondence between full matchings inG and vertex colorings of TG(G, C) by 1, 2, 3. For atrigraphT and a coloringf : V(T) → {1, 2, 3}, Tf is the multigraph onV(T) obtained fromT by the removal of all the edgese such that for at least one endvertex ofe, c(ex) = f (x). Thecoloring f is Eulerian if Tf is an Eulerian graph. Eulerian colorings are related to bipartizingmatchings by the following analogue ofProposition 1.4.

Proposition 3.1. Let M be a full matching in a cubic graph G with a dominating cycle C, andlet fM be the corresponding vertex coloring ofTG(G, C). The matching M is bipartizing if andonly if fM is an Eulerian coloring.


Proof. Let x be an internal vertex of G with neighborsx1, x2, x3 and assume thatxx1 ∈ M. Forconvenience, we shall writeT for TG(G, C) and f for fM .

Let us call a vertexz ∈ V(C) survivingif z is not incident with an edge inM (and hence isnot suppressed after the removal ofM). We show:

Claim. The number s of surviving vertices in Sx,1 has the same parity as the degree d(x) of xin Tf .

To prove this claim, we consider each internal vertexy ∈ V(T) in turn and evaluate itscontribution tos andd(x) based on the type of the pair(x, y).

If y forms a disjoint pair withx, then all the yi reside in a single(x, j )-segment, and exactly2 of them are surviving. Thus, the contribution ofy to s is 0 or 2. This is fine because there areno edges betweeny andx in Tf .

Next, assumey forms a crossing pair withx. If any neighbor ofy is related tox1, then noyi

is contained inSx,1 — azero contribution to the numbers. On theother hand, the contribution tod(x) is also zero, since any edge colored 1 atx gets deleted on theway toTf . If someyi is relatedto x2, then the neighbors ofy other thanyi are either both inSx,1 or both inSx,3, andyi is in theother one of these two segments. It follows that the contribution tos is 0 or 2 if yyi ∈ M, and 1otherwise. Nowy contributes 0 tod(x) if yyi ∈ M (because edges coloredi at y are deleted),and 1 if yyi �∈ M. Hence, the contribution tod(x) is as required. The case that someyi is relatedto x3 is symmetric.

Finally, consider the case thatx and y form an alternating pair. The only neighbor ofy inSx,1 (say,yj ) is related tox1. If yj is surviving, then the contribution ofy to s is 1. So is thecontribution tod(x), because the edge coloredj at y gets deleted because of its color atx,exactly one other edge will be deleted because of its color aty, andone edge remains. Thus,assume thatyj is not surviving, i.e.yyj ∈ M. Then twoof the three edges betweenx andy arein Tf , soy contributes an even number tod(x) and 0 tos as desired.

We have shown that the numbers of surviving vertices inSx,1 is the same, modulo 2, asd(x).Note that the number of surviving vertices inCx,i is s + 2 (becausex is not surviving, but twoof its neighbors are). This implies that ifM is bipartizing, thenfM is Eulerian.

To see the converse, proceed exactly as in the proof ofProposition 1.4. Assume thatfM isEulerian and that there are a total oft internal vertices. LetG′ be the graph obtained fromG− Mby suppressing all degree 2 vertices, and letC′ andC′

x,i denote the cycles corresponding toCandCx,i in G′. Theset + 1 even cycles form a basis of the cycle spaceZ(G′) of G′. Hence,G′is bipartite andM is bipartizing. �

4. The existence of Eulerian colorings

The main result of this section is a generalization ofTheorem 1.3which justifies the hope thattrigraphs provide a reasonable setting for the study ofConjecture 1.2:

Theorem 4.1. Every trigraph has an Eulerian coloring.

For the time being, we postpone the proof ofTheorem 4.1, andobserve first that it sufficesto prove the theorem for a special class of trigraphs. To begin with, we may restrict ourselves torainbowtrigraphs, i.e. trigraphs in which no two edgese, e′ with the same endverticesx andyhave the same color atx. Suppose a trigraphT contains such edges and form a trigraphT ′ asfollows. If c(ey) = c(e′

y), then removeeande′ from T . Otherwise, replaceeande′ with a single


edgee′′ colored so thatc(e′′x) = c(ex) andc(ey) �= c(e′′

y) �= c(e′y). In either case, it is easy to

check that any Eulerian coloring ofT is an Eulerian coloring ofT ′ and vice versa. Since the newtrigraphT ′ has fewer edges thanT , repeating the above process yields a rainbow trigraph. Theresult is well-defined. Note that rainbow trigraphs contain at most 3 edges between any pair ofvertices.

Next, we eliminate all the triple edges. Letx, y be vertices ofT joined by 3 parallel edgese1, e2, e3. Permutethe colors of all half-edges incident withy so as to make eachei have thesame color atx andy. (Notethat this permutation applies to all edges incident withy, not justthe edgesei .) Now the trigraphT ′′ is obtained by replacing the edgesei with 2 parallel edgesh, h′, colored as follows:

c(hx) = c(h′y) = 1 and c(hy) = c(h′

x) = 2.

Observe that iff is any coloring of the vertices ofT (with 1, 2, 3), then the number of edgesbetweenx andy in Tf has the same parity as inT ′′

f . It follows thatT has an Eulerian coloring ifand only if T ′′ does. Eventually, this process yields a rainbow trigraph with no triple edges.However, since the result of this reduction depends on the order in which triple edges areeliminated, we prefer to allow triple edges in the following definitions.

We are now going to define the ‘flip’, an operation that corresponds to the neighborhoodcomplementation of Section1. Fix a vertex x of a rainbow trigraph(T, c) and a colork ∈{1, 2, 3}. Let Xk = Xx,k be the set of verticeswhich are joined tox by at least one edgewhose color atx differs from k. For each vertexz ∈ Xk, we define thedistinguished colorγk(z) ∈ {1, 2, 3} as the uniqueγ such that the number of edgesebetweenx andz with c(ex) �= kandc(ez) �= γ is even. It is easy to check that the choice is indeed unique in each of the possiblesituations:

• if x andz are joined by a single edgee with c(ex) �= k, setγk(z) = c(ez),• if x andz are joined by two edgese, e′ with c(ex) = k, setγk(z) = c(e′

z),• if x andz are joined by two edgese, e′ with k �∈ {c(ex), c(e′x)}, thenγk(z) is the colordifferent

from bothc(ez) andc(e′z),• if x andz are joined by three edgese, e′, e′′ with c(ex) = k, setγk(z) = c(ez).

The (x, k)-flip is an operation whose result is a rainbow trigraphT∗x,k on V(T) − {x}. To

perform the(x, k)-flip in T , start with the trigraphT − x, add a new edge between every twovertices inXk and set the color of each new edge at an endvertexz to beγk(z). Finally, reducethe result to a rainbow trigraph.

For an example, considerthe(x, 1)-flip in the trigraphT shown inFig. 6(a). The edges addedto T − x are shown inFig. 6(b), and the resulting rainbow trigraph is inFig. 6(c). Recall thatcolor 1 corresponds to the lightest shade of grey. We haveX1 = {z, w, t}. The distinguishedcolors areγ1(z) = 1 andγ1(w) = γ1(t) = 2. Note how the addition of a new edge betweenw

andt (colored 2 on both ends) and the subsequent reduction to a rainbow trigraph eliminates anedge ofT .

Fix a coloring f of the trigraphT andk ∈ {1, 2, 3}. Set

Xk( f ) = {z ∈ Xk : f (z) �= γk(z)}.We make severalobservations used later in the proof ofTheorem 4.1.

Lemma 4.2. With respect to degree parity, the(x, k)-flip behaves like the complementation onXk( f ). More precisely, letw andw′ be two vertices of T− x. If m denotes the number of edges


Fig. 6. (a) A trigraphT. (b) The intermediate step of the(x, 1)-flip in T. (c) The result of the(x, 1)-flip.

betweenw andw′ in Tf and m∗ denotes the number of edges between them in(T∗x,k) f , then

m �≡ m∗ (mod 2) if and only if bothw andw′ are in Xk( f ).

Proof. Performing the (x, k)-flip, one begins withT − x and adds a suitably colored edgebetween each pair of vertices inXk. Let us call the resulting intermediate trigraphT . Assumethatw,w′ ∈ V(T − x) and letm denote the number of edges betweenw andw′ in Tf . We provethe following:

Claim. Thenumbers m andm have different parity if and only if bothw andw′ are in Xk( f ).

If either of the two vertices is not contained inXk, then the claim is triviallytrue, since the edgesbetweenw andw′ in T are, in this case, the same as inT − x. If bothw andw′ are inXk, thenT contains a new edge between them, coloredγk(w) at w andγk(w

′) at w′. This edge isin Tf

if andonly if f (w) �= γk(w) and f (w′) �= γk(w′), i.e., if and only if w,w′ ∈ Xk( f ). The claim

follows.To obtainT∗

x,k, one finally reducesT to a rainbow trigraph. It is easy to check this does notchange the parity in question, i.e., thatm ≡ m∗ (mod 2). Thus, the statement of the lemma isimplied by the claim. �

For a coloringf of T − x, let f x →k be theextension of f to T obtained by assigning colorkto x. We have the following easyobservation.

Lemma 4.3. If f is a coloring of T− x, then fx →k is an Eulerian coloring of T if and only ifall of the following conditions hold:

(i) all z �∈ Xk( f ) have even degree in(T − x) f ,(ii) all z ∈ Xk( f ) have odd degree in(T − x) f , and(iii) the size of Xk( f ) is even. �

Another observation shows that the behavior of colorings under flips is somewhat similar asin Lemma 4.3. In particular, condition (i) is the same in both statements.

Lemma 4.4. A coloring f of T − x is Eulerian for T∗x,k if and only if all of the following hold:

(i) all z �∈ Xk( f ) have even degree in(T − x) f , and(ii) all z ∈ Xk( f ) have degree (in(T − x) f ) different from|Xk( f )| modulo 2. �

Proof of Theorem 4.1. Let T be a trigraph. As noted above,T may be assumed to be rainbow.We proceed similarly as in the Fleischner–Stiebitz proof [5] of Theorem 1.3. Let x ∈ V(T). Weshow:


Claim.

e(T) ≡ e(T∗x,1) + e(T∗

x,2) + e(T∗x,3) (mod 2). (1)

We now prove this congruence. Fix a coloringf of T − x. We show that the numberA ofEulerian extensions of this coloringf to T has the same parity as the numberB of distincti suchthat f is an Eulerian coloring ofT∗

x,i . We mayassume thatA �= B (otherwise there is nothing toprove).

If k is a color such thatf x →k is Eulerian, then we claim thatf is Eulerian forT∗x,k. To begin

with, conditions (i)–(iii) ofLemma 4.3are satisfied. These conditions imply the two conditionsof Lemma 4.4. Indeed, condition (i) is identical in these two lemmas, while (ii) ofLemma 4.4follows from (ii) and (iii) of Lemma 4.3. Thus,Lemma 4.4shows thatf is Eulerian forT∗

x,k asclaimed.

By our assumption thatA �= B, there must be somej suchthat f is Eulerian forT∗x, j , but

f x → j is not Eulerian. ByLemma 4.4, all z �∈ X j ( f ) have even degrees (all degrees are taken in(T −x) f in this paragraph) and allz ∈ X j ( f ) have degrees different from|X j ( f )| modulo 2. Onthe other hand (sincef x →i is not Eulerian), either (ii) or (iii) ofLemma 4.3fail to hold. Thus,either somez ∈ X j ( f ) has even degree, or|X j ( f )| is odd. Each of thesepossibilities leads tothe same conclusion: allz ∈ X j ( f ) have even degrees, and|X j ( f )| is odd.

An important consequence is thatf must be Eulerian forT − x. To prove this, recall thatf isEulerian forT∗

x, j and note that the degree of anyz �∈ X j ( f ) is the same in(T∗x, j ) f as in(T −x) f ,

i.e., even. Since we have shown that allz ∈ X j ( f ) also have even degrees in(T − x) f , it followsthat(T − x) f is Eulerian.

This has implications for colorsi �= j . Consider any colori ∈ {1, 2, 3}. By Lemmas 4.3and4.4,

f is Eulerian forT∗x,i if and only if eitherXi ( f ) = ∅ or |Xi ( f )| is odd, while

f x →i is Eulerian if and only ifXi ( f ) = ∅. (2)

In particular, the differenceB − A is the number of colorsi suchthat|Xi ( f )| is odd.It is not hard to see from the definition of theXi ( f ) (checking the several cases that may

occur) that each neighbor ofx is contained in 0 or 2 of the setsXi ( f ). Thus X3( f ) =X1( f ) ⊕ X2( f ), where⊕ is the symmetric difference. Hence, either 0 or 2 of the setsXi ( f )are of odd size. By the above, it follows thatB − A is even. Thus, we managed to show thatA ≡ B(mod 2) and to prove the congruence(1).

With (1) established, we prove, by induction on|V(T)|, that the numbere(T) of Euleriancolorings ofT is odd. This is clear ifT has a single vertex. Otherwise, the induction hypothesisimplies that each ofe(T∗

x,i ) (wherei = 1, 2, 3) is odd. By(1), e(T) is odd as well. The proof isfinished. �

5. Orthogonal Eulerian colorings

Two Eulerian colorings of a trigraphT areorthogonalif they differ at each vertex. Observethat a pair of disjoint bipartizing matchings in a cubic graphG (with a dominating cycleC)corresponds to a pair of orthogonal Eulerian colorings of TG(G, C). In view of Conjecture 1.2,one might ask for conditions implying that a trigraph has a pair of orthogonal Eulerian colorings.

There are trigraphs with no pair of orthogonal Eulerian colorings, since there are examplesof cubic graphs with no pair of disjoint bipartizing matchings (for a given dominating cycle).


Fig. 7. Trigraphs with no orthogonal pair of Eulerian colorings.

For instance, an example due to Fleischner yields the trigraph shown inFig. 7(a). There are also2-connected such trigraphs, seeFig. 7(b) and (c).

We have investigated in detail the existence of two orthogonal Eulerian colorings incubictrigraphs, i.e., ones in which each vertex has degree 3. (We define thedegreeof a vertexx in amultigraph to be the total number of edges incident withx; thus, any multiple edges are countedwith their multiplicities.) A cubic trigraph isbalancedif the three half-edges incident with anyvertex have all three colors. Rather unexpectedly, it turns out that the existence of a perfectmatching in such a trigraph plays a key role:

Theorem 5.1. A balanced cubic trigraph T has a pair of orthogonal Eulerian colorings if andonly if T has a perfect matching.

Proof. We begin with a reformulation of the problem. Asuitablecoloring of a cubic multigraphH is a coloring of its half-edges by colors from{B, Y, G} (blue, yellow and green) with theproperty that

(1) the 3 edges incident with any vertex get all 3 colors,(2) the edgese whose color on both ends differs fromB form an Eulerian subgraph, and(3) the same property holds forY in place ofB, i.e., the edgesewhose color on both ends differs

from Y form an Eulerian subgraph.

One way to interpret this definition is to regard the half-edges of colorG as being ‘blue’ and‘yellow’ at the same time. Property (2), for instance, then states that each vertex is incident withan even number of edges with both ends ‘yellow’.

A balanced cubic trigraphT has a pair of orthogonal Eulerian colorings if and only if itsunderlying multigraphH has a suitable coloring. This can be seen as follows. Given the Euleriancolorings f1 and f2 of T , assign colorB to all half-edgesh of H satisfyingc(hy) = f1(y),wherey is the endvertex ofh. Similarly, color h with Y if c(hy) = f2(y), and assign colorGto the remaininghalf-edges. The result is a suitable coloring ofH , and the correspondence isbijective.

We define thetypeof an edge to be the unordered pair of the colors of its ends. Assume nowthatH has a suitable coloringa. Consider a vertexx ∈ V(H ) and the possible color assignmentson edges incident withx. We shalluse the notationa(ex) just like in the case of trigraphs. Letax,G denote the color of the edgee with a(ex) = G on the end opposite tox. Defineax,B andax,Y in the analogous way. The properties (1)–(3) above restrict the possible values ofax,G, ax,B

andax,Y. In fact, it is straightforward to check that the following is the complete list ofadmissibleconfigurations:

(a) ax,G = G, ax,Y �= B andax,B �= Y,(b) ax,G = B, ax,Y = B andax,B �= Y,(c) ax,G = Y, ax,Y �= B andax,B = Y.


Fig. 8. An auxiliary half-edge coloringa. The solid edges form the 2-factorF , theordinary edges are dotted and theparity edges are dashed. The root cycle ofF is the cycle of length 4. Changing the circled entries toG, one obtains asuitable coloring.

Conversely, let a coloringa of the half-edges ofH in {B, Y, G} be given. Assume that theconfiguration at each vertexx is admissible. Then it is easy to see thata is suitable.

Note that in each of the 3 cases, exactly one edge incident with eachx is of type GG orBY. Consequently, the edges of these two types form a perfect matching inH . This proves oneimplication from the theorem.

We now prove the other implication. LetF be a perfect matching inH . ThenF = H − E(F)

is a 2-factor of H . Contract each component ofF to a single point (preserving any multipleedges but discarding loops) to obtain a multigraphH ′. We claim thatH ′ has an acyclic spanningsubgraphP such that for each vertexx, thedegree ofx in H ′ has the same parity as its degreein P. (Such a subgraph is usually called aparity subgraphof H ′.) Indeed, to getP, it suffices toremove cycles (more precisely, their edge sets) fromH ′, as long as there are any.

Let us return to the multigraphH . Each edge ofP may be naturally identified with an edgeof F ⊂ H ; let uscall theseparity edges. Furthermore, other edges ofF correspond to edges inE(H ′) − E(P); thesewill be calledordinary edges.

We now construct an auxiliary coloringa of the half-edges ofH that will later be modifiedto a suitable coloring. (SeeFig. 8.) Assign colorG to both ends of each ordinary edge ofH .Consider a cycleZ of F . We have the following requirements on the coloring ofZ:

(i) each edge ofZ is of typeB B or Y Y,(ii) if two incident edges ofZ share an endvertex of an ordinary edge, then their types differ,

while if they share an endvertex of a parity edge, their types are the same.

Conditions (i) and (ii) can be ensured because by the construction ofP, the cycleZ is incidentwith an even number of ordinary edges. Subject to these restrictions, any prescribed coloring ofa half-edge onZ can be uniquely extended toZ.

The freedom in the choice of the coloring ofZ enables us to impose another requirement foreach parity edgee:


(iii) if one end of a parity edge is incident with edges of typeY Y, then the other end is incidentwith edges of type B B.

This can be achieved as follows. Consider a componentR of P. Choose a root vertexr ofthe treeR and fix a coloring satisfying (i) and (ii) on the cycle ofF corresponding tor (therootcyclefor R). TraverseR from the root to the leaves; at each vertex, fix the (unique) coloring ofthe corresponding cycle ofF satisfying (i), (ii) and (iii). Performed for all componentsR, thisoperation produces a coloring meeting all of our restrictions. Furthermore, color the half-edgesof each parity edge byB or Y soas to mismatch, at each end, the color of the incident half-edges.

The resulting coloringa is certainly not suitable: at each end of a parity edge, the two incidentedges are either both of typeB B or both of typeY Y. However, this can be fixed. Orient eachcycle Z of F . If a half-edgeh on Z is situated clockwise from theendvertex of a parity edge,then changethe color of h to G. After this modification, each vertex is incident with half-edgesof all 3 colors.

Let us determine the configurations of colors around a vertexx of H , to seeif theyare admissible (i.e., if they match the set of admissible configurations (a)–(c)). The symbolsax,G, ax,R andax,Y have the same meaning as before. The values are as in the following table:

ax,G ax,Y ax,B Condition

G Y/G B/G If x is an endvertex of an ordinary edge,B B B/G If x is incident with the ‘Y’ endof a parity edge,Y Y/G Y If x is incident with the ‘B’ endof a parity edge.

A comparison with the list of admissible configurations shows that indeed, all of theseconfigurations are admissible. This means that we have constructed a suitable coloring.�

As shown by the trigraph inFig. 7(c), Theorem 5.1cannot be extended to imbalanced cubictrigraphs. We conclude this paper with an open problem. Call a vertexx of a trigraphbad if alledges incident withx, except for one edgee, have the same color atx, andehas a different color.Each of the examples inFig. 7contains a bad vertex. This suggests the following question. It canbe shown that an affirmative answer would implyConjecture 1.2.

Problem 5.2. Does every 2-connected rainbow trigraph with no bad vertices admit a pair oforthogonal Eulerian colorings?

Acknowledgment

The second author wishes to thank Herbert Fleischner for interesting conversations onbipartizing matchings, and also for bringing the term ‘circle graph’ to the authors’ attention.

References

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eulerian colorings and the bipartizing matchings conjecture of fleischner

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