On Acyclic Edge-Coloring of Complete Bipartite Graphs

# On Acyclic Edge-Coloring of Complete Bipartite Graphs

Ayineedi Venkateswarlu e-mail: venku@isichennai.res.in; Corresponding author Computer Science Unit, Indian Statistical Institute - Chennai Centre, MGR Knowledge City Road, Taramani, Chennai – 600113, INDIA. Santanu Sarkar e-mail: santanu@iitm.ac.in Department of Mathematics
Indian Institute of Technology Madras, Chennai – 600036, INDIA.
A. Sai Mali e-mail: sai.mali.mail@gmail.com Computer Science Unit, Indian Statistical Institute - Chennai Centre, MGR Knowledge City Road, Taramani, Chennai – 600113, INDIA.
###### Abstract

An acyclic edge-coloring of a graph is a proper edge-coloring without bichromatic (-colored) cycles. The acyclic chromatic index of a graph , denoted by , is the least integer such that admits an acyclic edge-coloring using colors. Let denote the maximum degree of a vertex in a graph . A complete bipartite graph with vertices on each side is denoted by . Basavaraju, Chandran and Kummini proved that when is odd. Basavaraju and Chandran provided an acyclic edge-coloring of using colors and thus establishing when is an odd prime. The main tool in their approach is perfect -factorization of . Recently, following their approach, Venkateswarlu and Sarkar have shown that admits an acyclic edge-coloring using colors which implies that , where is an odd prime. In this paper, we generalize this approach and present a general framework to possibly get an acyclic edge-coloring of which possess a perfect -factorization using colors. In this general framework, we show that admits an acyclic edge-coloring using colors and thus establishing when is an odd prime.

Keywords: Acyclic edge-coloring, Acyclic chromatic index, Perfect -factorization, Complete bipartite graphs

## 1 Introduction

Let be a finite and simple graph. A proper edge-coloring of is an assignment of colors to the edges so that no two adjacent edges have same color. So it is a map with for any adjacent edges , where is the set of colors. The chromatic index, denoted by , is the least integer such that admits a proper edge-coloring using colors. A proper coloring of is acyclic if there is no two colored cycle in . The acyclic edge chromatic number (also called acyclic chromatic index), denoted by , is the least integer such that admits an acyclic edge-coloring using colors. The notion of acyclic coloring was first introduced by Grünbaum [11] in , and the concept of acyclic edge-coloring was first studied by Fiamc̆ík [10]. Let be the maximum degree of a vertex in . It is obvious that any proper edge-coloring requires at least colors. Vizing [22] proved that there always exists a proper edge-coloring with colors. Since any acyclic edge coloring is proper, we must have . On the other hand, in , Fiamc̆ík [10] (also independently by Alon, Sudakov and Zaks [2]) posed the following conjecture.

###### Conjecture 1.1

for any graph .

In [2] it was proved that there exists a constant such that for any graph with girth is at least . It was also proved in [2] that for almost all -regular graphs. Later Něsetřil and Wormald [18] improved this bound and showed that for a random regular graph . In another direction, there have been many results giving upper bounds on for an arbitrary graph . For example, Alon, McDiarmid and Reed [1] proved that . Molloy and Reed [13] improved this bound and showed that . Recently, Ndreca et. al. obtained  [19] which is currently the best upper bound for an arbitrary graph . Muthu, Narayanan and Subramanian [14, 17] obtained better bounds: for graphs with girth at least 220; for graphs with girth at least . The acyclic edge-coloring of planar graphs has been deeply studied in recent years. See [23, Section 3.3] for a nice account of recent results.

The Conjecture 1.1 was shown to be true for some special classes of graphs. Burnstein [8] showed that when . Hence the conjecture is true when . Muthu, Narayanan and Subramanian proved that the conjecture holds true for grid-like graphs [15] and outerplanner graphs [16]. It has been observed that determining is a hard problem from both theoretical and algorithmic points of view [23, p. 2119]. In fact, we do not yet know the values of for some simple and highly structured graphs like complete graphs and complete bipartite graphs in general. Fortunately, we can get the exact value of for some cases of complete bipartite graphs, thanks to the perfect -factorization.

Let be the complete bipartite graph with vertices on each side. The complete bipartite graph is said to have a perfect 1-factorization if the edges of can be decomposed into disjoint perfect matchings such that the union of any two perfect matchings gives a Hamiltonian cycle and it is of length (see Section 2 for more details). It is known that when , where is an odd prime, or and odd, then has a perfect 1-factorization (see [7]). One can easily see that if has a perfect 1-factorization then . And also we have the following result due to Basavaraju, Chandran and Kummini [4].

###### Theorem 1.1

, when is odd.

Hence when . By a result of Guldan [12, Corollary 1], we can also get when .

The main idea here is to give different colors to the edges in different -factors in , and removal of (one) two vertices on each side and their associated edges gives the required edge-coloring of with colors. But a different approach is needed to deal with when . In 2009, Basavaraju and Chandran [5] proved that for any odd prime . We can view their approach as follows: suitably pick one edge from each -factor and partition these edges into two groups and each group can possibly be assigned a different color to get the required result. Following this approach, Venkateswarlu and Sarkar have recently shown that for any odd prime  [21]. In this paper we view this approach in a more general setting and propose a general framework for the proof. The only remaining infinite class of complete bipartite graphs that are known to have perfect -factorization is , where is odd prime. In this general framework we provide an acyclic edge-coloring of using colors when . Therefore we state our main result as follows.

###### Theorem 1.2

, where is an odd prime.

Therefore the acyclic chromatic index is equal to for all the three known infinite classes of complete bipartite graphs having a perfect -factorization, and the Conjecture 1.1 holds true for such graphs.

In the next section we discuss some preliminaries and in Section 3 we present a general framework to possibly get an acyclic edge-coloring of which possess a perfect -factorization using colors. Then we present the proof of Theorem 1.2 in this framework in Section 4.

## 2 Preliminaries

Let be an integer. We treat elements of the ring as integers in the range . We denote the complete bipartite graph as with and . We use to define edges though our graph is undirected. This is only for ease of presentation in associating a perfect matching in with a permutation of the label set , which we discuss below. Accordingly, the use of arrows in the Figures 1 and 2 below is to explicitly emphasize the correspondence between a perfect matching and its associated permutation map. We use the terms ‘composition’ and ‘product’ of permutations interchangeably. Note also that a permutation can be decomposed as a product of disjoint cycles uniquely (upto reorder of the cycles and cyclic rotation of the elements within a cycle) and it is called disjoint cycle decomposition. We use (instead of the usual union notation ) to signify union of ‘disjoint’ sets.

### 2.1 Perfect matching and Perfect 1-factorization

A matching in a graph is a set of edges without common vertices, and a perfect matching is a matching which matches all vertices of the graph. In the case of complete bipartite graph , a perfect matching is a set of edges satisfying:

• for each vertex there exists a vertex such that .

• if and are in then .

So by labelling the vertices in both and with elements of (or an appropriate label set of size ), we can interpret a perfect matching in as a permutation of the label set , say . For convenience, let us illustrate this through an example. Let and consider the graph with the same labels from to for the vertices on the top and the bottom , as depicted in the figure below. Let then .

Note that the union of any two perfect matchings of forms a collection of disjoint cycles. These cycles can also be seen from the disjoint cycle decomposition of the composition of their corresponding permutations. Let be as mentioned above and . Then .

We have and the arrows are placed accordingly in the figure above. The fixed element corresponds to the common edge represented by the normal line in Figure 2, and the two cycles and correspond to the cycles and respectively in . By a careful observation of the example, we can see the following general result.

###### Lemma 2.1

Let and be two perfect matchings of and let be a cycle of length in the disjoint cycle decomposition of , i.e., , where the subscripts are taken modulo . Then the corresponding cycle in is of length and the participating edges are given by and which appear alternatively in as depicted in Figure 3 : the edges of and are represented by lines (without arrows) with colors red and green respectively.

A perfect matching is also called a -factor, and a partitioning of the edges of a graph into -factors is a -factorization. A -factorization is perfect if the union of any two of its -factors (perfect matchings) is a Hamiltonian cycle. As pointed out in the introduction, there are three infinite classes of complete bipartite graphs known to have perfect -factorization, namely, with an odd prime. Let us illustrate it by considering the complete bipartite graph . As discussed above the -factors can be described by permutations of . Consider the -factors given by the permutations for . In Figure 4 the edges of are the green colored lines and the edges of are the red colored lines and they correspond to the permutations and respectively. The induced subgraph formed by the edges is depicted below (without arrows), and the corresponding permutation is equal to .

Now we can see from the above lemma that is a perfect 1-factorization of the complete bipartite graph if and only if the permutation is a full cycle (of length ) for any with , where is the permutation corresponding to . Next we present a general framework to possibly get an acyclic edge-coloring of which possess a perfect -factorization using colors, where is odd.

## 3 A general framework

Let be an odd integer. Suppose that the complete bipartite graph has a perfect 1-factorization. Then there exist disjoint perfect matchings covering all the edges of , say , such that the union of any two perfect matchings gives a Hamiltonian cycle (which is of length ). Let be a proper subset of and consider the following partial coloring:

 assign\ color ci to the\ edges\ in M∗i. (1)

The remaining edges to be colored is given by

 M=n−1⨆i=0(Mi∖M∗i),

and these edges will be assigned some other colors different from the colors in .

###### Lemma 3.1

There can not be a cycle in the induced subgraph formed by the edges from the union of two color classes and for any with .

Ideally, to get an acyclic edge-coloring of , our aim should be to use only two more colors for coloring the (uncolored) edges in , and thus attaining the lower bound on . In other words, partition into and , if possible, in such a way that the induced subgraph of does not contain a cycle for any and . If such a partition of exists, then the edges of and can be assigned distinct colors, and one can easily see that the proposed edge-coloring is proper and acyclic. These observations and our intuition suggest that to minimize the number of edges (size of ) that are yet to be assigned colors. This can be done by taking to be a (proper) maximal subset of , i.e., take for some . Thus the size of the set will be (near)111One can take for exactly one and other ’s as mentioned, and observe that Lemma 3.1 holds true in such a case as well. minimal. Additionally, we have to make sure that a suitable partition of exists satisfying the other requirements mentioned above. Below we present a strategy to choose (to be a perfect matching), so that the other requirements on can possibly be worked out using permutations of the label set .

As discussed in the previous section, we can interpret each perfect matching in as a permutation of the label set and let us denote it by . Now, if possible, select one edge from each such that the set of edges gives a perfect matching in . Let be its corresponding permutation of . Then we have for .

###### Remark 3.1

A perfect -factorization of complete bipartite graph is equivalent to a Hamiltonian latin square of order . Our choice of perfect matching with for is equivalent to a transversal (of length ). These concepts are well studied in literature (see [9, p. 135–151]). It was conjectured by Ryser that every latin square of odd order has a transversal [6]. Such a transversal is also said to be rainbow matching (see [3] for details). In fact, it is enough to take a transversal of length as pointed out in the footnote below. It was conjectured by Brualdi and Stein independently that every latin square of odd order has a transversal of length  [6, 20]. As far as we know it is not known that in general a Hamiltonian latin square of order has a transversal (of length or ). In fact, we need a transversal which satisfies an additional property as discussed below.

Now set and assign colors as in (1). As discussed our aim is to use two additional colors for coloring the remaining edges (given by ) and obtain an acyclic edge-coloring of . For this purpose, we need to partition into and , in such a way that the induced subgraph of does not contain a cycle for any and . In other words, we must have edges from both and in any cycle in the induced subgraph of . Note that by Lemma 2.1 the cycles of the induced subgraph of can be obtained from the cycles of the permutation in its disjoint cycle decomposition. So in order to see such a partition of exists or not, we analyse cycle structure of the permutations for . We now see that such a partition of , if exists, can be obtained from a partition of the label set , and it is due to the one-to-one correspondence between the label set and a perfect matching.

Suppose as a product of disjoint cycles. Observe that there is exactly one common edge , and so is the only fixed element in the permutation , i.e., and for any . Let us take and let be the length of the cycle . Then we must have for and . Note that for a cycle in the disjoint cycle decomposition of these permutations, its corresponding cycle in is of length ; half of the edges are from and the other half are from (see Lemma 2.1). We now try to partition into and by analysing all the cycles , such that at least one element from both and appear in the representations of all those cycles . If such a partition of exists, then the corresponding partition of is given by and , and by Lemma 2.1 we can see that the partition of into and gives the required result. In general, if possesses a perfect -factorization, the difficulty is to identify a suitable perfect matching that can help to get an acyclic edge-coloring of using only colors. Let us illustrate the technique by considering the case , where is an odd prime.

### 3.1 The case of Kp,p for an odd prime p

This case was studied in [5] and we present here a slight variant of it. Take to be the perfect matching corresponding to the permutation for . We can see that the decomposition gives a perfect -factorization of . Now consider to be the perfect matching given by the permutation , where is a generator of . Let be the multiplicative inverse of in . Note that in . We can easily check that . We also have the following.

• , where represents the common edge and is a cycle of length containing .

• for , where represents the common edge and is a cycle of length containing .

Therefore we can get the required result with a partition of into and . Then the corresponding partition of is given by and . Observe that the cycle of corresponding to contains exactly one edge which belong to and contains the edge . The other edges of the cycle belong to . Now the final assignment of the colors is as follows:

• the edges in are colored with for ;

• the edges in are colored with ;

• the edges in are colored with .

From the above discussion and by Lemma 3.1, it is clear that the proposed edge-coloring (with colors) of is proper and acyclic.

Note that the proposal in [5] is and and one can easily see that the result is still valid with such a choice as well.

## 4 The case of Kp2,p2 for an odd prime p≥5

In this section we provide an acyclic edge-coloring of with colors, where is an odd prime . We follow the general framework described in the previous section. Accordingly we now summarize the set-up in this case. We use elements of for labelling the vertices of on both sides. Let be a generator of and let be its inverse. Observe that in . We consider the following.

• Let be the perfect matching corresponding to the permutation of the label set defined by

 π(a,b)((c,d))=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩(a,a+b+d)if c=0 and a+b+d≠0(a+xb,0)if c=0 and a+b+d=0(a+c+xb,0)if c≠0 and b+d=0(a+c,b+d)if c≠0 and b+d≠0

Then from [7] (with and ), we can see that the perfect matchings form a perfect -factorization of .

• We choose the perfect matching corresponding to the permutation defined by

 π((c,d))=(yc,xd);
• We choose the following partition of the label set :

 I(2)={(0,1),(1,0),(z,z),(z,zx) for z∈Z∗p} and I(1)=I∖I(2)

Then the corresponding partition of is given by

 M(1) ={(c,d)↦(yc,xd):(c,d)∈I(1)}, M(2) ={(c,d)↦(yc,xd):(c,d)∈I(2)}.

We have and .

Let for . The edge-coloring of that we consider is as follows:

• the edges in are colored with for ;

• the edges in are colored with ;

• the edges in are colored with .

According to the general framework discussed in the previous section, the two requirements that need to be satisfied to establish the above edge-coloring of is proper and acyclic are as follows:

• for , there is exactly one fixed element in the permutation . That is there is exactly one edge common to both and ;

• for , elements from both and must appear in the representation of the cycles of length in the disjoint cycle decomposition of the permutation .

Let us now prove that the above two requirements are satisfied in our set-up. For brevity of expression, we sometimes use the following notation.

 x′ =1x−1=y1−y and xi=x+x2+⋯+xi for i=1,2,…,p−2, y′ =1y−1=x1−x and yi=y+y2+⋯+yi for i=1,2,…,p−2.
###### Proposition 4.1

For , we have .

Proof : An edge is common to both and if and only if . Therefore by checking the four cases

 (a,a+b+d) =(0,xd) if c=0 and a+b+d≠0 (a+xb,0) =(0,xd) if c=0 and a+b+d=0 (a+c+xb,0) =(yc,xd)if c≠0 and b+d=0 (a+c,b+d) =(yc,xd)if c≠0 and b+d≠0

for , we get

 M∩M(a,b)={(ay′,bx′)↦(−ax′,−by′)}

and hence the proof.

The above proposition shows that the first requirement is satisfied in our set-up. In what follows we prove that the other requirement is also satisfied. For this purpose, we now analyse cycle structure of the permutations for . The inverse permutation of is given by . So we get

 π−1∘π(a,b)((c,d))=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩(xa,y(a+b+d))if c=0 and a+b+d≠0(xa+x2b,0)if c=0 and a+b+d=0(x(a+c)+x2b,0)if c≠0 and b+d=0(x(a+c),y(b+d))if c≠0 and b+d≠0

The above permutation can be decomposed into the following three permutations.

 π(0)(a,b)((c,d))=(x(a+c),y(b+d))
 π(1)(a,b)((c,d))={(c,ya+d)% if c=xa(c,d)otherwise
 π(2)(a,b)((c,d))={(c+x2b,d)if d=0(c,d)otherwise
###### Proposition 4.2

For , we have

 π−1∘π(a,b)=π(2)(a,b)∘π(1)(a,b)∘π(0)(a,b)

Proof : Note that

 π(1)(a,b)(π(0)(a,b)((c,d)))={(x(a+c),ya+y(b+d))if x(a+c)=xa (⇔c=0)(x(a+c),y(b+d))otherwise

and splits each of above two cases into two subcases depending on whether the second component is zero or not. Thus we get

 π(2)(a,b)(π(1)(a,b)∘π(0)(a,b)((c,d)))=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩(xa,ya+y(b+d))if c=0 % and a+b+d≠0(xa+x2b,0)if c=0 and a+b+d=0(x(a+c)+x2b,0)if c≠0 and b+d=0(x(a+c),y(b+d))if c≠0 and b+d≠0

and hence the proof.

With the above decomposition and the following result, analysis of the cycle structures can be simplified which we will see later.

###### Proposition 4.3

Let , then we have

 π−1∘π(xa,xb)=σ−1∘(π−1∘π(a,b))∘σ.

Proof : We have

and so we get

Now one can check that this is equal to .

Thus for , the permutations are all conjugates of each other, and so they all have same cycle structure. In fact we get the disjoint cycle decomposition of by the symbol transformation , i.e., replacing the symbols by in the disjoint cycle decomposition of . Therefore it is enough to study the cycle structure of for .

We now analyse the cycle structure of for by dividing into four cases: , where represents nonzero elements of . We discuss these cases one by one and we show that elements from both and appear in the representation of the cycles (of length ) in the disjoint cycle decomposition of for elements in each of these four cases.

In Section 3 we have mentioned cycles with single element explicitly to emphasize that there is exactly one fixed element. As we have already proved it in Proposition 4.1, for convenience we follow the convention and in the discussion below we do not explicitly mention cycles with single element in the disjoint cycle decomposition of permutations. Accordingly we count only the cycles of length in the disjoint cycle decomposition. For simplicity, we also use some common notation in presenting the disjoint cycle decomposition in each of these cases.

Case:

The permutation is the identity map, and so . We can see from the definition that the disjoint cycle decomposition of can be given by

 π−1=C0C1⋯Cp−1Cp,

where the cycles are given by

 Cj Cp =((1,0)(x,0)(x2,0)⋯(xp−2,0))

and it is evident that they are of length . The missing element is fixed by .

Note that the cycle contains exactly one element and other elements belong to , and also the cycle contains exactly one element and other elements belong to .

For , the elements of the cycle are of the form for . Since is also a generator of we have either or for some . Note that .

• If (is a square), then we have and , and observe that and are the only elements of which belong to and they are of the form . The remaining elements of belong to .

• If (is a non-square), then we have and