On Evenly-Equitable , Balanced Edge-Colorings and Related Notions

A graph G is said to be even if all vertices of G have even degree. Given a k-edge-coloring of a graph G, for each color i ∈ Zk = {0, 1, . . . , k − 1} let G(i) denote the spanning subgraph of G in which the edge-set contains precisely the edges colored i. A k-edgecoloring of G is said to be an even k-edge-coloring if for each color i ∈ Zk, G(i) is an even graph. A k-edge-coloring of G is said to be evenly-equitable if for each color i ∈ Zk, G(i) is an even graph, and for each vertex V ∈ V(G) and for any pair of colors i, j ∈ Zk, |deg G(i) (V) − deg G(j) (V)| ∈ {0, 2}. For any pair of vertices {V, w} let mG({V, w}) be the number of edges between V and w in G (we allow V = w, where {V, V} denotes a loop incident with V). A k-edge-coloring of G is said to be balanced if for all pairs of colors i and j and all pairs of vertices V and w (possibly V = w), |mG(i)({V, w}) − mG(j)({V, w})| ≤ 1. Hilton proved that each even graph has an evenly-equitable k-edge-coloring for each k ∈ N. In this paper we extend this result by finding a characterization for graphs that have an evenly-equitable, balanced k-edge-coloring for each k ∈ N. Correspondingly we find a characterization for even graphs to have an evenly-equitable, balanced 2-edge-coloring. Then we give an instance of how evenly-equitable, balanced edge-colorings can be used to determine if a certain fairness property of factorizations of some regular graphs is satisfied. Finally we indicate how different fairness notions on edge-colorings interact with each other.


Introduction
When considering edge-colorings of graphs it is usually desired to have some fairness properties imposed on the number of edges colored by each color.Below we define some important such notions and then explore the existence of edge-colorings satisfying combinations of these conditions.
In what follows, a graph  is called even if all vertices of  have even degree.Given a -edge-coloring of a graph , for each color  ∈ Z  = {0, 1, . . .,  − 1} let () denote the spanning subgraph of  in which the edge-set contains precisely the edges colored .Then a -edge-coloring of  is called an even -edge-coloring if for each color  ∈ Z  , () is an even graph.A -edge-coloring of  is said to be equitable if for each vertex V ∈ () and for each pair of colors ,  ∈ Z  , |deg () (V) − deg () (V)| ∈ {0, 1}.Moreover, a -edge-coloring of  is said to be evenly-equitable if (i) for each color  ∈ Z  , () is an even graph, (ii) for each vertex V ∈ () and for any pair of colors ,  ∈ Z  , |deg () (V) − deg () (V)| ∈ {0, 2}.
For any pair of vertices {V, }, let   ({V, }) be the number of edges between V and  in  (we allow V = , where {V, V} denotes a loop incident with V).A -edge-coloring of  is said to be balanced if for all pairs of colors  and  and all pairs of vertices V and  (possibly V = ), | () ({V, }) −  () ({V, })| ≤ 1.A -edge-coloring of  is said to be equalized if ||(())| − |(())|| ≤ 1 for each pair of colors ,  ∈ Z  .
Due to de Werra's work in [1][2][3][4] it has been known since the 1970s that for each  ∈ N every bipartite graph has a -edge-coloring that is balanced, equitable, and equalized at the same time.One important result for more general graphs is by Hilton, who proved in [5] that each even graph has an evenly-equitable -edge-coloring for each  ∈ N, thereby completely settling this problem (see Theorem 9).The existence of equitable -edge-colorings is much more 2 International Journal of Combinatorics problematic and very unlikely to be completely solved, since, for example, settling the existence of equitable Δ-edgecolorings is equivalent to classifying Class 1 graphs (see [6,7] for some results on this topic).One general result on equitable -edge-colorings was found by Hilton and de Werra [8] who proved that if  ≥ 2 and  is a simple graph such that no vertex in  has degree equal to a multiple of , then  has an equitable -edge-coloring.More recently, Zhang and Liu [9] extended this result by proving that for each  ≥ 2, if the subgraph of  induced by the vertices which have degree equal to a multiple of  is a forest, then  has an equitable edge-coloring, thereby verifying a conjecture made by Hilton in [10].
In Section 2 we extend Hilton's result [5] by finding a characterization for graphs that have an evenly-equitable, balanced -edge-coloring for each  ∈ N (see Theorem 1).We then use this result to find a different kind of characterization for even graphs to have an evenly-equitable, balanced 2-edgecoloring (see Theorem 2).
In Section 3 we prove Theorems 5 and 6, the latter of which uses results from the previous section.The proof of Theorem 6 provides an instance of how evenly-equitable, balanced edge-colorings can be used to ensure that a certain fairness property of factorizations of some regular graphs is satisfied.This particular notion of fairness is defined as follows.A -factorization of a graph in which the vertices have been partitioned into parts is said to be fair if for each two parts (possibly they are the same) the number of edges between these two parts in each factor differs from the number in each other factor by at most one.
For completeness, in Section 4 we address the existence of all other combinations of the three edge-coloring properties (namely, evenly-equitable, balanced, and equalized), finding weakest subsets of conditions that will guarantee (if possible) that a graph  has a -edge-coloring which has the following properties in turn: ( 1 ) evenly-equitable, balanced, and equalized, ( 2 ) evenly-equitable and equalized, ( 3 ) balanced and equalized, ( 4 ) evenly-equitable, ( 5 ) balanced, and ( 6 ) equalized.

Coloring Results
The following characterization can be used to find evenlyequitable, balanced -edge-colorings.The proof has the flavor of Hilton's proof in [5] of the case where the additional property of being balanced was not required but is modified to deal with extra complications that arise in this new setting.Theorem 1.For each positive integer , a graph  (possibly with loops) has an evenly-equitable, balanced -edge-coloring if and only if it has an even, balanced -edge-coloring.
Proof.Proving the "only if " result is trivial.To show the "if " result, we first prove the assertion for the case when  is connected and loopless.Let  be an even, balanced -edgecoloring of .Among all pairs of colors ,  ∈ Z  and all vertices V ∈ () suppose that |deg () (V) − deg () (V)| = 2 is as large as possible (where  ∈ N).If  ∈ {0, 1}, then this edge-coloring is evenly-equitable, so assume  ≥ 2. Let   be the spanning subgraph of  induced by the edges colored  and .From   form a new graph   by adding an uncolored loop at each vertex V satisfying deg   (V) ≡ 2 (mod 4).Then For each pair of vertices {V, } with V,  ∈ Z  and for any color ℎ ∈ Z  , let  (,) ({V, }) = min{ () ({V, }),  () ({V, })}, and let  , ({V, }) be a set of size 2 (,) ({V, }) containing precisely  (,) ({V, }) edges of each of the colors  and  joining vertices V and .So | , ({V, })| is even.Let  , (V) = ⋃ ∈()\{V}  , ({V, }) and  , = ⋃ 0≤V<<  , ({V, }).Define   =   −  , .Since | , (V)| is even for each V ∈ (), and since the original edge-coloring is even, each component of   is an eulerian graph and has no multiple edges since  is balanced (possibly it has an uncolored loop at some of the vertices).The following argument establishes property (4); namely, each component of   has an even number of edges.First note that by the assumption of this theorem for all ℎ ∈ Z  each component of (ℎ) is eulerian, so the size of each edge-cut in  (ℎ) is even (so it is also even in   (ℎ)) . (2) Let  be any component of   and let  = [ , ].Let  1 = ([()]); so | 1 | is even (since there are an even number of edges in  , between each pair of vertices).Let  2 be the edge-cut [(), () \ ()], which by the definition of ≡ 0 (mod 4) by (1) . ( So, Let   be a new 2-edge-coloring of   formed as follows.For each component  of   , alternately color the edges of an eulerian circuit of  with  and .This yields a balanced 2-edge-coloring of   (  is simple) where by ( 4), for each vertex V ∈ (), Now add the edges in  , with their original colors to   and remove the uncolored loops that were added when forming   .Then clearly the resulting graph is   and this new 2-edge-coloring   satisfies |deg To show that   is also even, consider the following cases (in which deg   () (V) refers to edge-coloring   with   ).
Case 1.One has deg   (V) ≡ 0 (mod 4).Note that in this case we are not adding a loop at V when forming   .Now look at the following subcases.
Subcase 1.1.∑ ∈(  )\{V}  (,) ({V, }) is odd.So, an odd number of edges incident with V of each of the colors  and  were removed when forming   from   .So, deg   (V) ≡ 2 (mod 4) and hence by (5) . Putting back the removed edges shows that V is incident with an even number of edges of each color in the edge-coloring   of   .Subcase 1.2.∑ ∈(  )\{V}  (,) ({V, }) is even.So, an even number of edges incident with V of each of the colors  and  were removed when forming   .So, deg   (V) ≡ 0 (mod 4) and hence deg   () (V) ≡ deg   () (V) ≡ 0 (mod 2).Putting back the removed edges shows that V is incident with an even number of edges of each color in the edge-coloring   of   .Case 2. One has deg   (V) ≡ 2 (mod 4).Note that in this case an uncolored loop is added to V when forming   .Now look at the following subcases.Subcase 2.1.∑ ∈(  )\{V}  (,) ({V, }) is odd.So, after adding an uncolored loop at V, an odd number of edges incident with V of each of the colors  and  were removed when forming   .Then deg   (V) ≡ 2 (mod 4), so by (5) Subcase 2.2.∑ ∈(  )\{V}  (,) ({V, }) is even.So, after adding an uncolored loop at V, an even number of edges incident with V of each of the colors  and  were removed when forming   .Then deg   (V) ≡ 0 (mod 4), so by (5) in the new edge-coloring deg   (V) = deg   () ≡ 0 (mod 2).So, for each  ∈ {V, } and each  ∈ {, }, deg Repetition of this procedure yields an evenly-equitable, balanced -edge-coloring of .
For the case when  has loops and is possibly disconnected, simply remove all the loops from  and apply this procedure to each component of the resulting loopless graph to get an evenly-equitable, balanced -edge-coloring of each component.Then put back the loops; it is easy to color them in a balanced way without destroying the evenly-equitable property at each vertex.
Note that in the statement of Theorem 1 we cannot replace the condition on the existence of an even, balanced -edgecoloring by a weaker set of conditions, as is illustrated by the next two examples.A cycle of length 3 with a cycle of length 2 intersecting in one of its vertices is an even graph and clearly has a balanced (and equalized) 2-edgecoloring, but no 2-edge-coloring that is evenly-equitable and balanced.The graph 2 2 (the graph with two vertices and two edges joining these two vertices) has an even (actually evenly-equitable) 2-edge-coloring, but no 2-edge-coloring that is evenly-equitable and balanced.While these two graphs are trivial, they can be generalized to more complicated examples.
Theorem 1 leads to the problem of finding conditions guaranteeing that a graph has an even, balanced -edgecoloring.The following result addresses that problem.Recall that our unusual definitions of even and odd vertices and of  (2) are given at the end of Section 1.
Theorem 2.  has an even, balanced 2-edge-coloring if and only if  is even and  (2) has no components with an odd number of odd vertices.Proof.To prove the necessity, suppose that an even, balanced 2-edge-coloring of  is given.Since the given 2-edge-coloring is balanced, for each pair of vertices V and , the   ({V, }) −  ,2 ({V, }) edges between V and  that are to be deleted when forming  (2) from  can be chosen so that they are shared evenly among the two color classes.Let  be a component in  (2) .Now since the given 2-edge-coloring of  is even, for each color  ∈ Z 2 , an odd vertex in  contributes an odd number to the degree sum of the graph  (2) (), and an even vertex in  contributes an even number to the degree sum of the graph  (2) ().Hence the number of odd vertices in  must be even.
To show the sufficiency, color the edges in  as follows.To satisfy the balanced property, for each pair of vertices {V, } ⊆ () color (  ({V, }) −  ,2 ({V, }))/2 (note that by definition of  ,2 this is an integer) of the edges between V and  with each color  ∈ Z 2 .Let  * be the graph induced by the edges that have been colored so far, and note that the graph induced by the uncolored edges is  (2) .Also note that by the definition of odd and even vertices, for each  ∈ Z 2 , if and only if V is an odd (even) vertex.

( * )
Since  is an even graph and since   ({V, }) −  ,2 ({V, }) is even for each {V, } ⊆ (),  (2) is also an even graph.For each component  in  (2) color the edges of an eulerian tour of  as follows.Start by coloring the first edge in the eulerian tour with  ∈ Z 2 and then switch to +1 (modulo 2) whenever the eulerian tour reaches an odd vertex for the first time.Note International Journal of Combinatorics that if the first vertex in the eulerian tour is even, then the first and last edges in the eulerian tour will have the same color because an even number of color switches will occur (by assumption there are an even number of odd vertices).Similarly, if the first vertex, say V, is odd, then the first and the last edges will have different colors if deg  (2) (V) = 2 (since no color switch is made at V) and they will have the same color if deg  (2) (V) > 2 (since then the eulerian tour will pass through V, so a color switch will occur at V).This coloring of the edges in  (2) has the property that for each V ∈ () and for each , by ( * ), (i), and (ii) each vertex in () has even degree and hence the given 2-edge-coloring has the desired properties.
It appears to us that a generalization of Theorem 2 for three or more colors may be difficult to obtain.
The following result characterizes graphs which have an evenly-equitable, balanced 2-edge-coloring.
Corollary 3. Suppose that  is an even graph.Then  has an evenly-equitable, balanced 2-edge-coloring if and only if  (2)  has no components with an odd number of odd vertices.
Proof.This follows immediately by Theorems 1 and 2.

An Application Using Amalgamations
In this section edge-colorings that satisfy another notion of equally distributing edges across color classes are considered, namely, that of fairness.Not only are the edge-colorings equitable, but also for any given partition  of the vertices, for each two parts in  (possibly they are the same) the edges between vertices in the two parts are equally divided among the color classes.While the results here (Theorems 5 and 6) address general partitions, these types of questions naturally arise when edge-coloring the complete multipartite graph   1 ,...,  , in which the partition is chosen to be the parts of the graph.For example, it has been shown when there exist fair equitable edge-colorings of   1 ,...,  in which each color class induces a hamilton cycle [11] or a 1-factor [12].
To prove Theorem 5, the method of amalgamations is used.A graph  is said to be the -amalgamation of a graph  if  is a function from () onto () such that  = {,V} ∈ () if and only if {(), (V)} ∈ ().The function  is called an amalgamation function.We say that  is a detachment of , where each vertex V of  splits into the vertices of  −1 ({V}).An -detachment of  is a detachment in which each vertex V of  splits into (V) vertices.Amalgamating a finite graph  to form the corresponding amalgamated graph  can be thought of as grouping the vertices of  and forming one supervertex for each such group by squashing together the original vertices in the same group.An edge incident with a vertex in  is then incident with the corresponding new vertex in ; in particular an edge joining two vertices from the same group becomes a loop on the corresponding new vertex in .
In what follows, [] denotes the subgraph of  induced by the edges colored  (so unlike (), [] is not necessarily a spanning subgraph), and   () denotes the number of loops at  in .The following theorem was proved in much more generality by Bahmanian and Rodger in [13], but this is sufficient for our purposes.The following theorem provides a necessary condition for the existence of fair 2-factorizations of 4-regular graphs ( ≥ 1).For any graph  and any partition  of (), let () be the -amalgamation of , where  maps two vertices in  to the same vertex in () if and only if they are in the same element of .
Proof.Suppose that  has a fair 2-factorization.Let  1 and  2 be the subgraphs of  induced by the edges corresponding to the 2-factors of .Since at each vertex in  the number of edge-ends incident with a vertex is a multiple of 4 and since these edge-ends are shared evenly among  1 and  2 , the number of edge-ends incident with each vertex in  in each of  1 and  2 is even.So, by the definition of odd and even vertices, in  (2) an odd vertex is incident with an odd number of edge-ends in each of  1 and  2 , and an even vertex is incident with an even number of edge-ends in each of  1 and  2 .Let  be a component of  (2) .Clearly ∑ V∈() deg  (V) is an even number and where ∑ V∈() is even deg  (V) is an even number and each term in the summation ∑ V∈() is odd deg  (V) is an odd number by the above observation.Hence the number of odd vertices in () must be even.
To investigate whether the necessary condition given in Theorem 5 is also sufficient for a graph to have a fair 2factorization, we introduce the notion of -equivalence.
Let  1 and  2 be two graphs with ( 1 ) = ( 2 ) = , and let  be a partition of .Then  1 is said to be -equivalent to  2 if for all   ,   ∈  (possibly  = ) ( 1 (  ,   )) = ( 2 (  ,   )), where (  (  ,   )) denotes the number of edges in   (for  = 1, 2) between the parts   and   .So if  1 and  2 are -equivalent, then  = ( 1 ) = ( 2 ).If either  1 or  2 has a fair 2-factorization, then Theorem 5 shows that (1) must be satisfied.To investigate the strength of (1), Theorem 6 shows that if  is a 4-regular graph for which  (2) = () (2) satisfies (1), then  is -equivalent to some graph (which is simple if a certain necessary condition is met) with a fair 2-factorization.Conjecture 7 goes on to make a much stronger claim that if  1 is -equivalent to  2 , then  1 has a fair 2-factorization if and only if  2 does.Theorem 6.Let  1 be a 4-regular graph.Let  be any partition of ( 1 ).Let  = ( 1 ).Suppose  (2) has no components with an odd number of odd vertices.Then there exists a graph  2 such that (iii)  2 has a fair 2-factorization (with respect to the given partition ), (iv)  2 can be chosen to be simple if and only if for all Note that it is long known by Petersen's 2-factor theorem (see, e.g., [14]) that every 2-regular graph has a 2factorization.The importance of Theorem 6 is that if the condition of the theorem is satisfied, then regardless of the partition  that is chosen, the resulting factorization of  2 (formed with  in mind) is fair.
Proof.By the supposition  (2) has no components with an odd number of odd vertices.Clearly  is even since  1 is even.So  satisfies the conditions of Corollary 3 and hence it has an evenly-equitable, balanced 2-edge-coloring.By the evenly-equitable property of this 2-edge-coloring, each color appears on exactly half of the edge-ends incident with each vertex of  (a loop contributes two edge-ends to the incident vertex).Notice that  is the -amalgamation of  1 where (V 1 ) = (V 2 ) if and only if V 1 and V 2 are in the same element of .For each V ∈ () define (V) = deg  (V)/4 = | −1 (V)|.By (i) of Theorem 4, there exists an -detachment  2 of  such that (1)  2 is -equivalent to  1 , (2) for each vertex V of  the edges of each color incident with V are shared as evenly as possible among the vertices in  −1 (V) (i.e., the vertices in the corresponding part of  2 ).
Note that, by (ii) and (iii) of Theorem By (2), in  2 each color is on two edges incident with each vertex.So, in  2 the subgraph induced by the edges of each color is a 2-factor, and hence this 2-edge-coloring is a 2-factorization of  2 .The fairness of this 2-factorization follows from the following observation: There is a one-to-one correspondence between the edges colored  joining any pair of vertices  and  in  and the edges colored  between the two corresponding parts  −1 () and  −1 () of  2 .So, the balanced property of this 2-edge-coloring implies the required fairness property of the 2-factorization.
In the light of Theorems 5 and 6 we make the following conjecture.

Other Combinations of Requirements
As described in the introduction we now consider other combinations of edge-coloring properties in turn.The results in this section are straight forward to obtain but are reported here for completeness.
( 1 ) Evenly-equitable, balanced, and equalized: as is discussed below, the examples in Figure 1 show that there are graphs which have an even, balanced, equalized 2-edgecoloring, but no 2-edge-coloring that is evenly-equitable and equalized.So, for each positive integer , no matter which combination of the conditions on the existence of an even -edge-coloring, balanced -edge-coloring and equalized edge-coloring of a graph  is used, it is not possible to guarantee that  has a -edge-coloring which is evenlyequitable, balanced, and equalized.
A graph is said to be of color-type 1 if it is connected and simple and has an even, equalized 2-edge-coloring but has no evenly-equitable, equalized 2-edge-coloring.Note that any edge-coloring of a color-type 1 graph is balanced because it is simple.In  1 there are two 3-cycles that intersect in just the top vertex; color the six edges in these 3-cycles with color 0 and color the remaining edges with color 1 to produce an even, balanced, equalized 2-edge-coloring. 1 does not have an evenly-equitable, equalized 2-edge-coloring, since in every evenly-equitable 2-edge-coloring one color class must be 2-regular and spanning and so has 7 edges.So,  1 is of color-type 1.In fact, a basic search shows that there is no color-type 1 graph with fewer vertices nor one on 7 vertices with less than 12 edges.In  2 the six edges of the two 3-cycles can be colored with color 0 and the edges of the 5-cycle with color 1, thereby producing an even, balanced, equalized 2-edge-coloring. 2 does not have an evenly-equitable, equalized 2-edgecoloring, since the only evenly-equitable 2-edge-coloring has one color class consisting of the three edges in the middle 3cycle.So,  2 is of color-type 1.In fact, another basic search shows that there is no color-type 1 graph with fewer edges nor one with 11 edges on less than 9 vertices.
Note that  2 suggests a way to construct infinitely many color-type 1 graphs: Take any cycle of length  as the middle cycle, attach to it a cycle of length  on the left and a cycle of length  on the right where  ∈ { +  − 1,  + ,  +  + 1}, and , ,  ≥ 3.
Since we cannot guarantee the existence of an evenlyequitable, balanced, and equalized -edge-coloring of a graph , even with the strong assumption that  has a -edgecoloring which is even, balanced, and equalized, we focus our attention on conditions implying the existence of edge-colorings that are ( 2 ) evenly-equitable and equalized, ( 3 ) balanced and equalized, ( 4 ) evenly-equitable, ( 5 ) balanced, and ( 6 ) equalized; evenly-equitable, balanced edgecolorings are the focus of Section 2.
( 2 ) Evenly-equitable and equalized: the examples in Figure 1 show that even with the strong assumption that a graph  has an even, balanced, equalized -edge-coloring,  does not necessarily have an evenly-equitable, equalized -edge-coloring; characterizations of graphs with such edgecolorings would seem to be difficult to find.
( 3 ) Balanced and equalized: such edge-colorings are always easy to find as is stated in the following theorem.Theorem 8.For each positive integer , each graph has a balanced, equalized -edge-coloring.
Proof.Let  be a graph with  edges (loops, being special types of edges, are also included in this count).Form an ordering ( 1 ,  2 , . . .,   ) of the edges of  where loops incident with the same vertex appear consecutively in the list, as do the edges joining the same pair of vertices.For 1 ≤  ≤  color   with  (modulo ).This -edge-coloring is clearly balanced and equalized.
Note that the condition that  is even is clearly necessary.( 5 ) Balanced: by Theorem 8 for each positive integer , any graph  has a balanced -edge-coloring.
( 6 ) Equalized: by Theorem 8 for each positive integer , any graph  has an equalized -edge-coloring.
The discussion above leads to the chart in Table 1.

Figure 1 :
Figure 1: Examples of graphs that are not of color-type 1.