Optimal acyclic edge-coloring of cubic graphs

Optimal AcyclicEdge-Coloring of CubicGraphs

Lars Døvling Andersen,1 Edita Macajova,2 and Jan Mazak2

1DEPARTMENT OF MATHEMATICAL SCIENCES

AALBORG UNIVERSITY, 9220 Aalborg SØ, DENMARKE-mail: [email protected]

2DEPARTMENT OF COMPUTER SCIENCE

FACULTY OF MATHEMATICS, PHYSICS AND INFORMATICS

COMENIUS UNIVERSITY, 842 48 BRATISLAVA, SLOVAKIAE-mails: [email protected], [email protected]

Received May 30, 2010; Revised September 23, 2011

Published online in Wiley Online Library (wileyonlinelibrary.com).DOI 10.1002/jgt.20650

Abstract: An acyclic edge-coloring of a graph is a proper edge-coloringsuch that the subgraph induced by the edges of any two colors is acyclic.The acyclic chromatic index of a graph G is the smallest number of colors inan acyclic edge-coloring of G. We prove that the acyclic chromatic index ofa connected cubic graph G is 4, unlessG is K4 or K3,3; the acyclic chromaticindex of K4 and K3,3 is 5. This result has previously been published byFiamcIk, but his published proof was erroneous. � 2011 Wiley Periodicals, Inc. J

Graph Theory

Keywords: acyclic edge-coloring; cubic graphs

1. INTRODUCTION

Various types of edge-colorings of graphs have occurred in graph theory for morethan a century. Among these colorings the most important are proper colorings—those

Journal of Graph Theory� 2011 Wiley Periodicals, Inc.

1

2 JOURNAL OF GRAPH THEORY

that do not allow two adjacent edges to have the same color. Considerable effort hasbeen devoted to proper edge-colorings where cycles that contain only two colors areforbidden.

A proper k-edge-coloring of G such that there are no two-colored cycles in G iscalled an acyclic k-edge-coloring of G. The concept of acyclic colorings was introducedby Grunbaum in [9].

In this article all the considered graphs are finite and simple, i.e. without multipleedges and loops. Let �=�(G) denote the maximum degree of a vertex in a graph G.Throughout the article, any proper k-edge-coloring uses the colors denoted by 1,2, . . . ,k.The chromatic index of a graph G, denoted by �′(G), is the least number of colorsneeded to color the edges of G by a proper edge-coloring. Similarly, we define theacyclic chromatic index (also called acyclic edge chromatic number) of a graph G,denoted by a′(G), to be the least k such that G has an acyclic k-edge-coloring. Obviously,a′(G)≥�′(G). An alternative definition of the acyclic chromatic index gives a slightlydifferent point of view: the acyclic chromatic index of a graph G is the minimumnumber of matchings which suffice to cover all edges of G in such a way that the unionof any two matchings does not contain a cycle.

It is known that a simple graph G with maximal degree � has chromatic indexeither � or �+1. How many new colors do we have to use if we do not want toallow two-colored cycles? It is known that a′(G)≤16� (see [2, 10]), but recent results(e.g. [1, 3, 11]) suggest that this bound is far from tight. The conjecture of Fiamcık [8]and later by Alon et al. [1] says that a′(G)≤�+2 for every graph G. Basavaraju andChandran [4] proved that a′(G)≤7 for graphs with �≤4.

In this article we focus on graphs with �=3. Burstein’s result in [5] implies that forcubic graphs a′(G)≤5; therefore, the conjecture is true for �=3. In 1980, Fiamcık [6]published an article claiming that K4 is the only cubic graph requiring five colors in anacyclic edge-coloring; nevertheless, both the result and the proof were incorrect. Fouryears later, in [7], he corrected one of the errors of the previous article, and stated acorrect result that there are two exceptional cubic graphs requiring five colors in anacyclic edge-coloring. However, the proof still contains a big gap: usage of Lemma 2in [6] eliminates the cycle of colors 1 and 2, but it does not ensure that no two-coloredcycles of other pairs of colors are created. This may be the reason why the result ofFiamcık has fallen into obscurity.

In [3], Basavaraju and Chandran proved that a′(G)≤4 for all subcubic graphs (graphswith maximal degree at most 3) containing a vertex of degree at most 2.

In every proper 3-edge-coloring of a cubic graph G the edges colored by any of thecolors form a perfect matching. Hence the subgraph of G induced by the edges of anytwo colors is a 2-factor, which is a nonempty set of two-colored cycles. Therefore,to have an acyclic edge-coloring of a cubic graph we need at least four colors. Infact this argument holds for any d-regular graph, so such a graph always requiresat least d+1 colors for an acyclic edge-coloring, as was also observed in [1]. Weprove that four colors are optimal for all connected cubic graphs with the exceptionof K4 and K3,3, the two graphs mentioned by Fiamcık in [7], for which five colors areoptimal.

Our main result is captured in the following theorem. The proof in Section 2 isalgorithmic and does not rely on any probabilistic arguments. As a simple corollarywe are able to determine the acyclic chromatic index of all cubic graphs.

Journal of Graph Theory DOI 10.1002/jgt

OPTIMAL ACYCLIC EDGE-COLORING OF CUBIC GRAPHS 3

FIGURE 1. An acyclic 5-edge-coloring of K4 and K3,3.

Theorem 1.1. Let G be a connected graph with �(G)≤3 different from K4 and K3,3.Then a′(G)≤4.

Corollary 1.2. The acyclic chromatic index of a connected cubic graph G is 4unless G is K4 or K3,3, for which a′(K4)=a′(K3,3)=5.

Proof. For a cubic graph G we have proved that a′(G)≥4. An acyclic 5-edge-coloring of the graphs K4 and K3,3 is shown in Figure 1.

It remains to prove that the acyclic chromatic index is at least 5 for both ofthese two graphs. Suppose that we have an acyclic 4-edge-coloring of K4. The graphK4 has six edges, and the edges colored by any of the colors form a matching, henceno color is used on more than two edges. Therefore, there are two colors such thatboth of them color exactly two edges, and edges colored by these colors form a2-factor, which is a two-colored cycle of length 4 in our case. We have derived acontradiction.

Suppose that we have an acyclic 4-edge-coloring of K3,3. The graph K3,3 has nineedges. If there are two colors such that both of them color three edges, the edgescolored by these colors form a 2-factor, which is a two-colored cycle of length 6 in ourcase. Otherwise exactly one color, say 1, colors three edges and any other color colorstwo edges. Choose a cycle C of length 4 in K3,3 such that two of its edges are coloredby the color 1. The other edges in this cycle must be of different colors, say 2 and 3.Look at one of the edges not belonging to C and adjacent with the edge of C that iscolored by color 2. Its color can be either 3 or 4. Since our coloring is proper and eachcolor different from 1 is used twice, the colors of all the other edges are determined.It is easy to check that in both cases there is a two-colored cycle, hence the graph K3,3cannot have an acyclic 4-edge-coloring. �

2. ACYCLIC EDGE-COLORING OF SUBCUBIC GRAPHS

In this section we prove Theorem 1.1. We proceed by induction on the number ofvertices of G. If G is a 1-vertex graph, then the assertion is trivial. We therefore assumethat G is a connected graph on at least two vertices with �(G)≤3, G �=K4, G �=K3,3,and for every connected graph H of smaller order, with �(H)≤3 and different fromK4 and K3,3, a′(H)≤4.

If G contains a vertex of degree 1 or 2 then the proof can be found in [3]. Therefore,we will assume that G is cubic.



FIGURE 2. Creating G′ from G in Case 1, k=4.

First, assume that G contains a bridge. Let e be a bridge. Let V1 and V2 be thevertex-sets of the two components created by removing e and let Gi be the graphinduced by Vi, i=1,2.

Both G1 and G2 contain a vertex of degree 2, thus by [3] have an acyclic4-edge-coloring �1 and �2, respectively. We can obtain an acyclic 4-edge-coloring �of G by letting �(e) be any color and permuting the colors of �1 and �2 on G1 andG2 so that the edges of G adjacent to e have colors different from �(e). Thus to theend of the proof we will assume that G is bridgeless.

The proof splits into several cases depending on the girth g(G) of G.

Case 1. The girth of G is even and g(G)≥6.Let C be any of the shortest cycles in G. Let the vertices of C be u0,v0,u1,v1, . . . ,

uk−1,vk−1 where k=g(G) /2≥3 (the indices will be taken mod k in what follows). Letu′

i and v′i be the neighbors of ui and vi, respectively, which do not belong to C (the

vertices u′0,v′

0,u′1,v′

1, . . . are distinct, because if a vertex outside C had two edges to Cthere would be a cycle of shorter length than C). Let G′ be the graph constructed inthe following way: we remove from G the cycle C together with all the edges adjacentto this cycle and join u′

i to v′i for i=0,1,2, . . . ,k−1 (Fig. 2 for the case k=4). The

resulting graph G′ is cubic and simple because g(G)>4: a multiple edge in G′ yieldsa cycle of length 4 in G. Note that G′ might not be connected but it contains nocomponent isomorphic to K4 or K3,3. Indeed, such a component would yield a cycleshorter than C in G; we prove this assertion in the next paragraphs.

Assume that G′ contains a component K isomorphic to K4 or K3,3. Since G does notcontain cycles of length 4, any such cycle in K must contain at least one of the edgesadded in the construction of G′ (we call such edges added in what follows). If K =K4,one can easily check that K must contain two adjacent added edges, contradicting thatu′

0,v′0,u′

1,v′1, . . . are distinct.

If K =K3,3, then since K does not contain two adjacent added edges, it must containthree added edges forming a 1-factor (otherwise there is a 4-cycle in G ). Since no other



edges in K are added, the remaining six edges of K forming a cycle of length 6 belongto G. Thus the girth of G is 6, the same is the number of vertices of C. Therefore, anyvertex of C is adjacent to a vertex of K. It is easy to check that in any case G containsa cycle of length 4 or 5.

Now we return to the main argument of Case 1. The induction hypothesis usedon each component of G′ yields an acyclic 4-edge-coloring �′ of G′; we extend thiscoloring to an acyclic 4-edge-coloring � of G.

Let �(e)=�′(e) for all edges e common for G and G′ and �(uiu′i)=�(viv′

i)=�′(u′iv

′i).

It remains to define � on the edges of C. We do this in several steps. The directionu0,v0,u1, . . . along C will be referred to as clockwise. Let D be a cycle in G otherthan C. We say that D enters C at a vertex w and leaves C at a vertex w′, if the pathw–w′ belongs to both C and D, goes along C in the clockwise direction and cannotbe extended in any direction. We say that D is of intersection �, if the path w–w′ haslength �. The cycle D can have more than one common path with C, in such a casewe choose w and w′ so that their clockwise distance along C is maximal.

Now we define � on the remaining edges of G; after this step � will be a proper4-edge-coloring of G such that any cycle different from C will contain at least threedistinct colors. Consider the edge viui+1. If the colors �(viv′

i) and �(ui+1u′i+1) are

different, we choose any of the two remaining colors for �(viui+1). Otherwise we set�(viui+1) to be the color different from the colors of the three edges incident with v′

i.We thus color all k edges viui+1.

Consider the edge uivi. It is adjacent to four edges, but two of them (uiu′i and viv′

i)are colored by the same color, so there is at least one free color to color the edgeuivi. If �(vi−1ui) and �(viui+1) are different, we choose for uivi the only possible colorthat is not already used on its neighboring edges. Otherwise, if �(vi−1ui)=�(viui+1)we color uivi with color different not only from �(vi−1ui) and �(uiu′

i) but also from�(vi−1v′

i−1).Now we verify that all the cycles of G, except possibly C, contain at least three

different colors. All cycles disjoint with C are not two-colored by the induction hypoth-esis. Let D be a cycle of intersection 1. Clearly, C and D can intersect on severalintervals of length 1. If at least one of these intervals is an edge viui+1, then D containsat least three colors, as �(viui+1) is different from the colors of three edges incidentwith v′

i. If, on the other hand, all these intervals are edges uivi then, by the inductionhypothesis, D has a corresponding cycle in G′ colored by at least three colors in �′and therefore there are at least three colors used on D itself.

Now let D be a cycle of intersection at least 2. Such a cycle containing only two colorscannot enter C in ui: the colors of uiu′

i and viv′i are the same, hence the colors �(uiu′

i),�(uivi), and �(viui+1) are three different colors contained in this cycle. Therefore,assume that D enters C in vi. Then D cannot leave C in vi+1, because the colors ofviui+1 and vi+1v′

i+1 cannot be the same. If �(viui+1) and �(vi+1ui+2) are different,then clearly D contains at least three colors. It only remains to consider the case that�(viui+1)=�(vi+1ui+2). In that case we have chosen �(ui+1vi+1) to be different from�(viv′

i), hence at least three different colors are used on D.At this moment, the only possibly two-colored cycle is C. Assume that it contains

only two colors, say 1 and 2. The edges incident with vertices of C but not belongingto C are colored by colors 3 and 4. If �(u1u′

1)=�(u2u′2)=3, we can set �(u2v2)=4

to obtain an acyclic 4-edge-coloring of G. Otherwise we may assume that the edges



uiu′i for i=1,2, . . . ,k are colored alternately by colors 3 and 4 (this assumption

allows us to replace color 1 on the edges of C by color 2 and vice versa withoutintroducing two-colored cycles of intersection 1). Moreover, we may assume that�(u1u′

1)=�(v1v′1)=3, �(u2u′

2)=�(v2v′2)=4, �(u1v1)=2 and �(v1u2)=1. There are

three cases for the colors f1 and f2 of the two edges incident with u′2 and not incident

with u2:

• if {f1, f2}={2,3} we set �(v1u2)=4 and �(u2u′2)=1,

• if {f1, f2}={1,2} we set �(v1u2)=4 and �(u2u′2)=3,

• if {f1, f2}={1,3} we set �(v1u2)=4 and �(u2u′2)=2 and color the edges of C

different from v1u2 by colors 1 and 2 alternatively to obtain a proper 4-edge-coloring.

The resulting coloring � is an acyclic 4-edge-coloring of G.

Case 2. The girth of G is at least 5 and is odd.Let C be one of the shortest cycles of G. Let u0,v0,u1,v1, . . . ,uk−1,vk−1,w be the

vertices of this cycle in the order in which they lie on the cycle, and let v−1 =uk =w.Let u′

i, v′i, and w′ be the neighbors of ui, vi, and w, respectively, which do not lie on

the cycle C.From the graph G we construct a graph G′ by removing the cycle C with all the

edges adjacent to this cycle and joining u′i to v′

i for i=0,1,2, . . . ,k−1. As the girth ofG is at least 5 and C is a shortest cycle, the resulting graph G′ contains no loops and nomultiple edges. Moreover G′ has �=2, �=3 and no component of G is isomorphic toK4 or K3,3. The last assertion can be proved in almost the same way as we have doneit in Case 1, the difference occurs only for K =K3,3, where we have three added edgesforming a 1-factor: since the graph G in this case has girth at most 6 and the girth isodd, it is equal to 5, hence C has only five vertices. This contradicts the fact that K hassix distinct vertices that have six pairwise distinct neighbors lying on C. The inductionhypothesis applied to each component of G′ gives an acyclic 4-edge-coloring �′ of G′.We extend this coloring to an acyclic 4-edge-coloring � of G.

First, set �(e)=�′(e) for each edge e of G that have no vertex in common with Cand set �(uiu′

i)=�(viv′i)=�′(u′

iv′i).

Next, we describe the coloring of the three edges incident with w. We may assumethat the colors used in �′ are 1,2,3,4, the colors of the edges incident with w′ but notincident with w are 1, and 2 and �(vk−1v′

k−1)≤�(u0u′0) in what follows. There are

several possibilities for the colors of u0u′0 and vk−1v′

k−1, all of them with the desiredcoloring of the edges incident with w are in Figure 3. As in the previous case, thedirection u0,v0,u1, . . . along C is referred to as clockwise. Recall the definitions of theterms enter, leave and to be of intersection for a cycle that has an edge in commonwith C, they apply also in this case. Note that from Figure 3 there is no two-coloredcycle entering C at vk−1, and no two-colored cycle entering at w and leaving at u0.

We continue by coloring the edges viui+1, i=1,2, . . . ,k−2, doing this in exactlythe same manner as we did in Case 1. As the next step we color the edge u0v0 sothat there will be no two-colored cycle that enters C at w. Any two-colored cycle ina proper edge-coloring has its edges colored alternately by two colors. We utilize thisfact and choose the color of one of the edges of C so that this edge together with thealready colored edge ww′ will exclude the possibility of a two-colored cycle enteringC in w. If �(wu0) �=�(v0u1) we set �(u0v0) to be different from �(wu0), �(v0u1), and



FIGURE 3. Coloring of the edges incident with w in Case 2.

�(u0u′0), otherwise, if �(wu0)=�(v0u1), we set �(u0v0) to be different from �(wu0),

�(u0u′0) and �(ww′).

Then we color the edges uivi, i=1,2, . . . ,k−1, exactly as in Case 1 to get a proper4-edge-coloring � of G. The induction hypothesis assures that any two-colored cyclein G has to contain at least one edge from C. The cycle C itself has odd length, henceit cannot be two-colored in a proper edge-coloring. From the construction of � wealready know that no two-colored cycle enters C at vk−1 or w, and it follows as in Case1 that no cycle of intersection 1 is two-colored. It also follows as in the previous casethat no two-colored cycle of intersection at least 2 can enter C in ui. If such a cycleenters C at vj, then by the above j<k−1, and it again follows from the constructionthat the cycle cannot be two-colored. So � is an acyclic 4-edge-coloring of G, hencea′(G)≤4.

Case 3. The girth of G is 4.Let C4 =v1v2v3v4 be a 4-cycle in G. First, assume that C4 contains two non-adjacent

edges such that neither of them is contained in a 4-cycle different from C4. Fori=1,2,3,4, let v′

i denote the neighbor of vi not belonging to C4. There are no trianglesin G and thus no two vertices of v′

1, v′2, v′

3, and v′4 coincide. Moreover, since G is

bridgeless, the graph G−V(C4) has either one or two components.If G−V(C4) is disconnected, there are two vertices in any of the two compo-

nents of G−V(C4) joined by an edge to C4 in G. Without loss of generality wecan assume that v′

1 and v′2 are in different components of G−V(C4). We create the

graph G′ by removing the vertices of C4 together with the incident edges and adding



edges joining v′1 with v′

2 and v′3 with v′

4. The graph G′ is a cubic, connected and ofsmaller order than G and it contains a 2-edge cut, therefore G′ �=K4 and G′ �=K3,3. Bythe induction hypothesis, G′ has an acyclic 4-edge-coloring �′. We define an acyclic4-edge-coloring � of G by setting �(e)=�′(e) for every edge e not incident with avertex of C4, �(v1v′

1)=�(v2v′2)=�′(v′

1v′2), and �(v3v′

3)=�(v4v′4)=�′(v′

3v′4). Further,

if �′(v′1v′

2)=�′(v′3v′

4), we set �(v2v3) to be the color not assigned to an edge inci-dent with v′

2 in G, �(v1v4) to be the color not assigned to an edge incident with v′1

in G and if �(v2v3) �=�(v1v4) we set �(v1v2) and �(v3v4) to be the color differentfrom �(v1v′

1), �(v2v3), and �(v1v4); otherwise, if �(v2v3)=�(v1v4),we set �(v1v2) and�(v3v4) to be the two different colors, both different from �(v1v′

1) and �(v2v3). If, onthe other hand, the colors �′(v′

1v′2) and �′(v′

3v′4) are different, we set �(v1v2)=�(v3v′

3),�(v3v4)=�(v1v′

1) and set �(v2v3) and �(v1v4) to be two different colors, both differentfrom �(v1v′

1) and �(v3v′3). In both cases it can be easily checked that the resulting

coloring is acyclic.Now suppose that G−V(C4) is connected and the edges v1v2 and v3v4 are in no

other 4-cycle of G than C4. We form the graph G′ in the same way as if G−V(C4)was disconnected. If G′ �=K4 and G′ �=K3,3 we find an acyclic 4-edge-coloring of Ganalogously. If G′ =K4 or G′ =K3,3 the graph G together with an acyclic 4-edge-coloring is depicted in Figure 8(a)–(c).

Now we proceed to the case that C4 does not contain a pair of nonadjacent edgessuch that neither of them is contained in a 4-cycle different from C4. Then C4 has tocontain two adjacent edges, each of them contained in a 4-cycle different from C4. Letv1v2 and v2v3 be two such edges.

We first deal with the case that G contains a 4-cycle D different from C4 whichcontains both v1v2 and v2v3. Since G is simple, neither v3v4 nor v4v1 is contained inD. Therefore, there is a vertex v5 different from v2 and v4, adjacent to both v1 and v3.Let v′

2, v′4, and v′

5 be the neighbor of v2, v4, and v5, respectively, that is not containedin {v1,v2, . . . ,v5}. Since G is bridgeless, the vertices v′

2, v′4, and v′

5 exist but they neednot be pairwise distinct. Moreover, it could not happen v′

2 =v′4 =v′

5, for otherwiseG=K3,3.

Assume that the vertices v′2, v′

4, and v′5 are pairwise distinct. We form a new graph G′

from G in the following way. We remove from G the vertices v1,v2, . . . ,v5 and add anew vertex u linked by an edge with the vertices v′

2, v′4, and v′

5 (Fig. 4). Clearly, G′ issimple, cubic and connected and is distinct from K4 since G does not contain triangles.If G′ is distinct from K3,3, then a′(G′)≤4 by the induction hypothesis. Let �′ be anacyclic 4-edge-coloring of G′. An acyclic 4-edge-coloring � of G can be constructed bysetting �(e)=�′(e) for every edge e in G which is not incident with any of v1,v2, . . . ,v5and �(v1v2)=�(v3v5)=�(v4v′

4)=�′(uv′4), �(v1v5)=�(v2v′

2)=�(v3v4)=�′(uv′2), and

�(v1v4)=�(v2v3)=�(v5v′5)=�′(uv′

5). If G′ =K3,3, an acyclic 4-edge-coloring of G isdepicted in Figure 8(d).

Assume that v′2 �=v′

4 =v′5. Let v6 be the neighbor of v′

4 distinct from v4 and v5; as Gis bridgeless, v6 �=v′

2. We create a new graph G′ by deleting the vertices v1,v2, . . . ,v5,v′4

and adding a new vertex v joined to v′2 and v6. Then G′ is connected and contains

a vertex of degree 2 and has maximum degree 3. Hence a′(G′)≤4 by [3]. Let �′ bean acyclic 4-edge-coloring of G′. We extend �′ to an acyclic 4-edge-coloring of Gin the following way. Without loss of generality we can assume that �′(vv′

2)=1 and�′(vv6)=4.



FIGURE 4. Creating G′ from G in Case 3 if there is a 4-cycle D containing bothv1v2 and v2v3 and v ′

2, v′4, and v ′

5 are pairwise distinct.

FIGURE 5. Creating G′ from G in Case 3 if there is a 4-cycle D containing bothv1v2 and v2v3 and v ′

2 �=v ′4=v ′

5.

We set �(e)=�′(e) for every edge e that is not incident with any of v1,v2, . . . ,v5,v′4

and let �(v1v5)=�(v3v4)=�(v2v′2)=1, �(v3v5)=�(v4v′

4)=2, �(v1v4)=�(v2v3)=�(v5v′

4)=3, and �(v1v2)=�(v′4v6)=4, see Figure 5. If v′

4 �=v′2 =v′

5 or v′5 �=v′

2 =v′4 the

proof can be done in a similar way.Now assume that there is no 4-cycle different from C4 containing two adjacent edges

of C4, and C4 contains two adjacent edges, say v1v2 and v2v3, each contained in a4-cycle different from C4. Let D1 =v1v2v5v6 be a 4-cycle such that {v3,v4}∩{v5,v6}=∅and let D2 be a 4-cycle different from C4 containing v2v3. Since G is cubic and simple,D1 and D2 share exactly the edge v2v3. Then there exists a vertex v7 different fromv1,v2, . . . ,v6 and adjacent to both v3 and v5. Note that no two vertices of {v4,v6,v7}are joined by an edge, for otherwise G would contain a triangle. Let v′

4, v′6, and v′

7 bethe neighbor of v4, v6, and v7, respectively, different from all the vertices v1,v2, . . . ,v7;the vertices v′

4, v′6, and v′

7 are not necessarily distinct. If v′4, v′

6, and v′7 are pairwise

distinct, we obtain G′ from G by removing the vertices v1,v2, . . . ,v7 together with theincident edges and adding a new vertex u joined by an edge to v′

4, v′6, and v′

7. Since Gis triangle-free, G′ is different from K4. If G′ is also different from K3,3, then since it isconnected and cubic and has fewer vertices than G, G′ has an acyclic 4-edge-coloring �′by the induction hypothesis. We construct an acyclic 4-edge-coloring � of G by setting�(e)=�′(e) for every edge e which is incident with no vertex from {v1,v2, . . . ,v7}and further set �(v1v2)=�(v6v′

6)=�(v5v7)=�′(uv′6), �(v1v4)=�(v2v3)=�(v5v6)=

�(v7v′7)=�′(uv′

7), �(v1v6)=�(v2v5)=�(v3v7)=�(v4v′4)=�′(uv′

4), and �(v3v4) be thecolor different from the colors of edges incident with u in G′ (Fig. 6). If G′ is K3,3, anacyclic 4-edge-coloring of G can be found in Figure 8(e).

Now we deal with the case that two of the vertices v′4, v′

6, and v′7 are equal and

different from the third. Assume that v′4 =v′

6 �=v′7, the proof for other pairs can be done

analogously (in fact the situations are completely symmetric). Let u be the neighbor



FIGURE 6. Creating G′ from G in Case 3 if there is a 4-cycle D1 containing v1v2and a 4-cycle D2 containing v2v3, and v ′

4, v′6, and v ′

7 are pairwise distinct.

FIGURE 7. Creating G′ from G in Case 3 if there is a 4-cycle D1 containing v1v2and a 4-cycle D2 containing v2v3, and v ′

4=v ′6 �=v ′

7.

of v′4 different from v4 and v6. Since G is bridgeless, u is different from v′

7. We formthe graph G′ by removing the vertices v1,v2, . . . ,v7 and v′

4 from G and adding a newvertex v joined to each of v′

7 and u by an edge. The graph G′ is different from K4 andK3,3 as it has a vertex of degree 2, and it is connected, has maximum degree 3 and hasfewer vertices than G. We take an acyclic 4-edge-coloring �′ of G′ and extend it toan acyclic coloring � of G. Without loss of generality we may assume that �′(vv′

7)=1and �′(vu)=4. We set �(e)=�′(e) for every edge not incident with any of v1,v2, . . . ,v7and �(v1v6)=�(v2v5)=�(v3v4)=�(v7v′

7)=1, �(v1v4) = �(v2v3) = �(v5v7) = �(v6v′4)

=2, �(v1v2)=�(v3v7)=�(v4v′4)=3 and �(v5v6)=�(v′

4u)=4 (Fig. 7).If all three vertices v′

4, v′6, and v′

7 coincide, graph G is the 3-dimensional cube Q3.An acyclic 4-edge-coloring of Q3 is shown in Figure 8(f).

Case 4. The girth of G is 3.

First, suppose that G contains a triangle with no edge included in another triangle;denote one such triangle C3 and let C3 =v1v2v3. Let v′

i be the neighbor of vi notcontained in C3. Create a new graph G′ by removing the vertices of C3 together withthe incident edges and introducing a new vertex u connected to v′

1, v′2, and v′

3. Sinceno edge of C3 was included in another triangle, G′ is simple.

If G′ =K4 or G′ =K3,3 then G is isomorphic to one of the graphs in Figure 9; anacyclic 4-edge-coloring of G is given in the figure. Otherwise, G′ is connected andhas fewer vertices than G, thus by the induction hypothesis, a′(G′)≤4. Let �′ be anacyclic 4-edge-coloring of G′. We extend �′ to an acyclic 4-edge-coloring of G: we



(a) (b) (c)

(d) (e) (f)

FIGURE 8. Acyclic 4-edge-colorings of small graphs from Case 3.

FIGURE 9. Acyclic 4-edge-coloring of K4 and K3,3 with a vertex expanded to atriangle.

FIGURE 10. Creating G′ from G in Case 4 if every triangle shares an edge withanother triangle.

set �(e)=�′(e) for every edge e not incident with a vertex of C3, set �(v′ivi)=�′(v′

iu),�(v1v2)=�′(v3v′

3), �(v1v3)=�′(v2v′2), and �(v2v3)=�′(v1v′

1).Now suppose that each triangle in G has an edge in common with another triangle.

Fix a triangle C3 =v1v2v3 in G. Since G �=K4, C3 has exactly one edge, say v1v2,included in a triangle different from C3. Let v4 be the common neighbor of v1 and v2different from v3. Let v′

3 be the neighbor of v3 different from v1 and v2 and let v′4 be

the neighbor of v4 different from v1 and v2. Since G does not contain a bridge, we havev′

3 �=v′4. Construct a new graph G′ by removing the vertices v1, v2, v3, and v4 from G

and introducing a new vertex u of degree 2 adjacent to v′3 and v′

4. Let �′ be an acyclic4-edge-coloring of G′; G′ admits such a coloring by Basavaraju and Chandru [3].



We construct a coloring � of G as follows: we set �(e)=�′(e) for every edge ewhich is not incident with either of vi for i=1,2,3,4 and �(v′

3v3)=�(v2v4)=�′(v′3u),

�(v′4v4)=�(v1v3)=�′(v′

4u). Further, let c1 and c2 be the two colors different from�′(uv′

3) and �′(uv′4). To finish the definition of the coloring set �(v2v3)=�(v1v4)=c1,

and, finally �(v1v2)=c2 (Fig. 10). Clearly, � is acyclic. This concludes the proof.

ACKNOWLEDGMENTS

The authors thank Mickael Montassier for valuable discussions on the subject andMartin Skoviera for his comments leading toward improvement of this article.The first author acknowledges the support of The Danish Council for IndependentResearch/Natural Sciences. The second and the third author acknowledge partialsupport from the grants VEGA 1/0634/09, APVV-0111-07, APVV-0223-10, and fromAPVV-ESF-EC-0009-10 within the EUROCORES Programme EUROGIGA (projectGReGAS) of the European Science Foundation.

REFERENCES

[1] N. Alon, B. Sudakov, and A. Zaks, Acyclic edge-colorings of graphs, J GraphTheory 37 (2001), 157–167.

[2] N. Alon, C. J. H. McDiarmid, and B. A. Reed, Acyclic coloring of graphs,Random Struct Algorithms 2 (1991), 277–288.

[3] M. Basavaraju and L. S. Chandran, Acyclic edge coloring of subcubic graphs,Discrete Math 308 (2008), 6650–6653.

[4] M. Basavaraju and L. S. Chandran, Acyclic edge coloring of graphs withmaximum degree 4, J Graph Theory 61 (2009), 192–209.

[5] M. I. Burstein, Every 4-valent graph has an acyclic 5-coloring (in Russian),Soobsc Akad Nauk Gruzin SSR 93 (1979), 21–24 (Georgian and Englishsummaries).

[6] J. Fiamcık, Acyclic chromatic index of a graph with maximum valency three,Arch Math (Brno) 16 (1980), 81–88.

[7] J. Fiamcık, Acyclic chromatic index of subdivided graph, Arch Math (Brno)20 (1984), 69–82.

[8] J. Fiamcık, The acyclic chromatic class of a graph (in Russian), Math Slovaca28 (1978), 139–145.

[9] B. Grunbaum, Acyclic colorings of planar graphs, Isr J Math (1973),390–408.

[10] M. Molloy and B. Reed, Further algorithmic aspects of the local lemma,Proceedings of the 30th Annual ACM Symposium on the Theory ofComputing, Dallas, TX, USA, New York, NY, 1998, 524–529.

[11] J. Nesetril and N. C. Wormald, The acyclic edge chromatic number of arandom d-regular graph is d+1, J Graph Theory 49 (2005), 69–74.


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