On Compatible Normal Odd Partitions in Cubic Graphs

# On Compatible Normal Odd Partitions in Cubic Graphs

J.L. Fouquet
L.I.F.O., Faculté des Sciences, B.P. 6759
Université d’Orléans, 45067 Orléans Cedex 2, FR
email: jean-luc.fouquet@univ-orleans.fr
J.M. Vanherpe
L.I.F.O., Faculté des Sciences, B.P. 6759
Université d’Orléans, 45067 Orléans Cedex 2, FR
email: jean-marie.vanherpe@univ-orleans.fr
###### Abstract

A normal odd partition of the edges of a cubic graph is a partition into trails of odd length (no repeated edge) such that each vertex is the end vertex of exactly one trail of the partition and internal in some trail. For each vertex , we can distinguish the edge for which this vertex is pending. Three normal odd partitions are compatible whenever these distinguished edges are distinct for each vertex. We examine this notion and show that a cubic -edge-colorable graph can always be provided with three compatible normal odd partitions. The Petersen graph has this property and we can construct other cubic graphs with chromatic index four with the same property. Finally, we propose a new conjecture which, if true, would imply the well known Fan and Raspaud Conjecture.

Keywords:Cubic graph; Edge-partition

## 1 Introduction

For basic graph-theoretic terms, we refer the reader to Bondy and Murty . A walk in a graph is a sequence , where are vertices of , and are edges of and and are the ends of , . The vertices and are the end vertices and and are the end edges of this walk while are the internal vertices and are the internal edges. The length of is the number of edges (namely ). The walk is odd whenever is odd, even otherwise. The walk is a trail if its edges are distinct and a path if its vertices are distinct. If is a walk of , then () is a subwalk of (subtrails and subpaths are defined analogously) .

If is an internal vertex of a walk with ends and , then and are the subwalks of obtained by cutting at . Conversely if and have precisely one common end , then the concatenation of these two walks at gives rise to a new walk (denoted by ) with as an internal vertex. When there is no possible confusion as to the edges being used, it would be convenient to omit the edges in the description of a walk, i.e., can be shortened to .

In what follows, is a cubic graph on vertices where loops and multiple edges are allowed.

###### Definition 1

A partition of into trails is normal when every vertex is an internal vertex of some trail of , say , and an end vertex in , . The length of a normal partition is the maximum length of the trails in the partition, that is .

If is a normal partition, then . We can associate to each vertex the unique edge with end that is the end edge of a trail of . We shall denote this edge by and it will be convenient to say that is the marked edge associated to . When it is necessary to illustrate our purpose by a figure, we represent the marked edge associated to a vertex by a close to this vertex.

Let be a vertex such that is an internal vertex in and an end vertex in (as an end of ). We can associate to the set containing the end vertices of . Note that and are not necessarily distinct, in this case we have . When and are the ends of , one of these two vertices is certainly different from . Let us transform into a new normal partition by the so called switching operation (see Definition 2).

###### Definition 2

Let be a normal partition and be a vertex of the graph such that is an internal vertex in and an end vertex in . Let and be the ends of , ().

• When , let , and .

• When , let us write where , , and .
We set and (see Figure 1).

The normal partition is the result of the switch of on .

###### Definition 3

A normal partition of into trails is odd when every trail in is odd. For each trail of odd length , let us say that an edge of is odd whenever the subtrails of obtained by deleting have odd lengths. The edges of that are not odd are said to be even.

Given two normal partitions and , is the set of vertices such that . It must be clear that two normal partitions and are identical whenever

###### Definition 4

Two normal partitions and of into trails are compatible when for every vertex of (in other words ).

Given three normal partitions , and we let . We say that has three compatible normal odd partitions , and whenever these partitions are pairwise compatible, that is .

It is shown in  (see Theorem 5) that a cubic graph without loops can always be provided with three compatible normal partitions.

###### Theorem 5

 A cubic graph has three compatible normal partitions if and only if has no loop.

Normal odd partitions are directly associated to perfect matchings and it is natural to ask whether the problem of finding three compatible normal odd partitions is connected to the edge-coloring problem. We show that cubic graphs with chromatic index 3 can be provided with three compatible normal odd partitions. It turns out that the Petersen graph, the Flower snarks, and the Goldberg snarks have also three such partitions.

## 2 Preliminary results

### 2.1 Switching equivalence

In  we proved that if and are two normal partitions of a cubic graph then we can transform into by a sequence of at most switchings. In other words and are switching equivalent.

When is a normal odd partition, a switching leading to a new odd partition is said to be an odd switching. When we can transform a normal odd partition in by a sequence of odd switching operations, and are said to be odd switching equivalent.

###### Theorem 6

Any two normal odd partitions of a cubic graph are odd switching equivalent.

Proof Let denote the set of edges incident with a vertex for which there exists a normal odd partition of odd switching equivalent with , such that and for all , such that .

If is a normal odd partition of a cubic graph , then for every vertex of there exists a normal odd partition of such that and for all and such that , , and are odd switching equivalent. Therefore, for every .

Assume that and are normal odd partitions of that are not odd switching equivalent and such that has maximum cardinality. Then there is a vertex . Since and , we have . Therefore, there exist two normal odd partitions and of that are not odd switching equivalent and , a contradiction.

###### Theorem 7

Let be a cubic graph. Then has an odd normal partition if and only if has a perfect matching.

Proof If is a perfect matching of , then is a factor of . Let us give an orientation to this factor and for each vertex let us denote the outgoing edge . For each edge such that , let be the path of length obtained by concatenating , and . Then is a normal odd partition (of length ) of . Conversely let be a normal odd partition of . A vertex is internal in exactly one trail of . The edges of this trail being alternatly odd and even, is incident to exactly one odd edge. Hence the odd edges defined above induce a perfect matching of .

Given an odd normal partition of , we can define the associated perfect matching as the set of odd edges of . Conversely, given a perfect matching , we can say that a normal odd partition is conformal to whenever is the set of odd edges of . Let be a normal odd partition obtained from a normal odd partition by one operation of switching. If and are conformal to a perfect matching , then we can say that we have performed a conformal (to ) switching. This operation of conformal switching is not always possible on a vertex. Indeed, assuming that is an internal vertex in and an end vertex of this trail, then the conformal switching is not allowed since we would obtain a cycle in the transformation (see Figure 2, the edge of is the dashed edge).

###### Theorem 8

If is a cubic graph of order at least four and is a perfect matching in , then any two normal odd partitions and conformal to are conformal switching equivalent.

Proof Assume that and are two normal odd partitions conformal to a perfect matching that does not belong to the same equivalence class. Suppose that is maximum. In particular, we have .

Let and let , and be its neighbors. Put , , and . Without loss of generality we may assume that is an edge of and that , while . Since a conformal switching of on leads to a conformal normal partition where while nothing is changed elsewhere, we can suppose that this conformal switching is not allowed on . In the same way a conformal switching of on is not allowed as well. Hence is an internal vertex of and an end vertex of this trail. Symmetrically, is an internal vertex of and an end vertex of this trail (see Figure 3). We suppose that is the second end vertex of and the second end vertex of .

###### Claim 1

The vertices and are distinct.

Proof If , then we have (formally we need to distinguish between and ) and . Hence . Let us denote by the edge of incident to () on the subtrail of joining to .

We may assume that , otherwise would be a graph on two vertices, a contradiction.

But now, conformal switchings of on , , and lead to a normal partition conformal switching equivalent to . Whether belongs or not to , has more vertices than , a contradiction.

###### Claim 2

The vertices and are distinct.

Proof Assume not: thus . From Claim 1, we then have .

We have , otherwise we could transform to by using a conformal switching on followed by a conformal switching on and we would obtain , a contradiction.

But now, conformal switchings of on , , and lead to a normal partition conformal switching equivalent to . Whether belongs or not to , has more vertices than , a contradiction.

Similarly . Consider the subtrails . There is a certainly a vertex on that trail for which the associated marked edge . Assume that is the first such vertex when running from to on (let us remark that ). Hence . Let be the neighboring vertex of on (it may happen that , in which case ) and let be the trail of ending in with the marked edge .

Since and , we have .

###### Claim 3

.

Proof Assume that . Since , we can perform a conformal switching of on leading to the conformal partition :

 T′′=T−{T,Q}+{T(y,w)+Q,T(w,v)}

In the same way, we certainly have (take instead of ). Since by Claim 1 , we must have either or . By considering , we can decide without loss of generality that (if not, we consider where the roles of and are exchanged).

In , the vertex is an internal vertex of and an end vertex of a trail with . A conformal switching is allowed on and this switching leads to the conformal partition

 Q=T−{T,S}+{T(y,u2)+S,u2v}.

We have or . But now we can perform a conformal switching on followed by a conformal switching on . The first switching on leads to defined as follows:

 R=Q−{T(y,u2)+S,vu2}+{S+T(u2,v)+vu2,T(v,w)}.

The second switching on leads to defined by

 S=R−{S+T(u2,v)+vu2,T(v,w)}+{u2v+T(v,w)+T(w,v),T(w,u2)+S}.

We have now and in . Since this set has at least one vertex more than , we have a contradiction. Figure 4: Two non-equivalent conformal partitions for the cubic graph on two vertices (the dashed edges are odd).

It turns out that the cubic graph on two vertices depicted in Figure 4 has two non-equivalent conformal partitions (with respect to the dashed edge).

### 2.2 Miscellaneous

The following proposition will be essential in the next section.

###### Proposition 9

Let be a cubic graph having three normal partitions , and . If is an edge of such that and are not in , then one of the followings is true:

• is an internal edge in exactly one partition,

• is an internal edge in exactly two partitions.

Moreover, in the second case, the edge itself is a trail of the third partition.

Proof Assume that is an end edge in , in , and in . Then in or we would have two partitions (say and ) for which ( respectively), a contradiction.

If is an internal edge in , and , then let and be the two other neighbors of . We would then have

• or

• or

• or ,

which is impossible.

Assume now that is an internal edge of a trail in and in and let and be the two other neighbors of . Up to the names of the vertices, we have

• .

>From the third partition , we must have . In the same way we would obtain . Hence the trail containing is reduced to , as claimed.

Given a normal partition , the average length of the trails in is denoted while is the number of trails of length .

###### Proposition 10

 Let be a normal partition of a cubic graph on vertices. It follows that

• ,

• .

Hence a normal partition whose average length is has all its trails of length .

###### Proposition 11

If is a cubic graph with three compatible normal odd partitions, then is bridgeless.

Proof Assume that is a bridge of and let be the component of containing . Since has three compatible normal odd partitions, one of these partitions, say , is such that . Thus the edges of are partitioned into odd trails. We have

 m=|E(C)|=3(|C|−1)+22

and is even whenever while is odd whenever . The trace of on is a set of trails and this number is odd when and even otherwise. Hence when we must have an odd number of odd trails partitioning , but in that case is even, a contradiction. When , we must have an even number of odd trails partitioning , but in that case is odd, contradiction.

In Figure 5, we show provided with three compatible normal odd partitions. Let us remark that, following Theorem 12, we need to have trails of length in at least one partition.

## 3 On cubic graphs with chromatic index three

In this section the existence of three compatible normal odd partitions in cubic -edge-colorable graphs is considered.

###### Theorem 12

If is a cubic graph, then the following are equivalent:

• i) has three compatible normal odd partitions of length

• ii) has three compatible normal odd partitions, where each edge is an internal edge in exactly one partition

• iii) is bipartite.

Proof Assume first that can be provided with three compatible normal odd partitions of length , say , , and . Since the average length of each partition is (Proposition 10), each trail of each partition has length exactly . Thus , , and are three normal odd partitions and from Proposition 9, each edge is the internal edge of one trail in exactly one partition. Conversely suppose that can be provided with three compatible normal odd partitions where each edge is an internal edge in exactly one partition, the edge of each trail of length must be an internal edge of two partitions, thus there is no trail of length in any of these partitions. Since the average length of each partition is , that means that each trail in each partition has length exactly . Hence .

We prove now that . Let , and be three compatible normal odd partitions of length . Following the proof of Theorem 7, the set of internal edges of trails of ( and respectively) is a perfect matching, say ( and respectively).

Let be a trail of and let and be the third neighbors of and respectively. By definition, we have and .

Assume without loss of generality that is an internal edge of a trail of . The trail does not use : otherwise , a contradiction to since and are compatible. Hence uses and .

Assume now that is an internal edge of a trail of . Reasoning in the same way, we get that . These two results lead to the fact that must be a trail in , which is impossible since each trail has length exactly .

Hence is an internal edge in a trail of . Thus the two internal vertices of can be distinguished, as follows from the fact that the end edge to which they are incident is internal in (say white vertices) or (say black vertices). The same holds for each trail in (and incidently for each partition and ). We can now remark that is an end edge of a trail in . This end edge cannot be an internal edge in since the trail of length going through ends with . Hence is an internal edge in and is a black vertex. Considering now , this vertex is the internal vertex of a trail of length of . Since and is a perfect matching, cannot be incident to an other internal edge of a trail in and must be a black vertex. Hence is a white vertex and its neighbors are all black vertices. Since we can perform this reasoning for each vertex, is bipartite as claimed.

Conversely, suppose that is bipartite and let be the bipartition of its vertex set. In the following, a vertex in will be represented by a circle () while a vertex in will be represented by a bullet (). >From König’s Theorem , is a cubic -edge-colorable graph. Let us consider a coloring of its edge set with three colors . A trail of length that is obtained by considering an edge ( and ) colored with together with the edge colored incident with and the edge colored with incident with will be said to have the type .

It can be easily checked that the set of trails of type is a normal odd partition of length . We can define in the same way as the set of trails of type and as the set of trails of type .

Hence , , and are three normal odd partitions of length . We claim that these partitions are compatible. Indeed, let be a vertex and and be its neighbors. Assume that is colored with , is colored with and is colored with . Hence is internal in a trail of and . The edge is internal in a trail of and . The edge is internal in a trail of and . Since the same reasoning can be performed in each vertex of , the three normal partitions , and are compatible.

###### Theorem 13

Let be a cubic graph with three compatible normal odd partitions , , and . If has length then is a cubic -edge-colorable graph .

Proof Since has length , every trail of has length (see Proposition 10). Hence there is no edge which can be an internal edge of a trail of and a trail of , since by Proposition 9 such an edge would be a trail of length in . Thus the perfect matchings associated to and (see Theorem 7) would then be disjoint and induce an even -factor of , which means that is a cubic -edge-colorable graph as claimed.

Lemmas 14, 15, and 16, below, together with Theorem 17, will be useful in proving that a cubic graph with a -edge-coloring has three compatible normal odd partitions which are conformal with respect to this coloring (see Corollary 18).

###### Lemma 14

Let be a cubic -edge-colorable graph. Assume that has a proper 3-edge-coloring together with three compatible normal odd partitions , , which are, respectively, conformal to , and . Then the graph obtained from by subdividing an edge such that with two vertices and ( adjacent to and adjacent to ) and joining these two vertices by an additional edge has also this property.

Proof Assume that is colored with .

We get a proper -edge-coloring of as follows: let the edges and be colored while the two other edges incident to and are colored with the two remaining colors.

Since is colored , this edge is internal in . Moreover, by Proposition 9, we know that is an end edge in some other partition, say . For , we are going to transform the normal odd partition of , conformal to , into the normal odd partition of , conformal to .

Let us put and , while the edge incident to and colored with is denoted by and the edge incident to and colored is denoted by .

Let be the trail containing . We write where and are end vertices of . In order to get the normal odd partition the subtrail of is split into two subtrails, namely and (see Figure 6). Thus

 T′Red=TRed−{R}∪{R(r,x)+xe1ue3v,R(y,s)+ye2ve4u}.

Obviously all the trails of have odd edges of color .

The edge is an end edge of some trail of . This trail has and some other vertex as end vertices. We replace the subtrail of with and we consider the trail of length , . Hence we get a normal odd partition of conformal with as follows:

 T′Blue=TBlue−{B}∪{xe1u,ve4ue3ve2y+B(y,b)}.

We have now two cases to consider.

Case 1: is an end edge of some trail of .

Hence is one end vertex of while is the other one. We replace the subtrail of with and we add the trail of length . In other words:

 T′Yellow=TYellow−{Y}∪{ue3ve4ue1x+Y(x,c),ye2v}.

Case 2: is an internal edge of some trail of .

We write , where and are end vertices of . We replace the subtrail of with and we add the trail of length . Thus

 T′Yellow=TYellow−{Y}∪{Y(f,x)+xe1ue4ve3u,ve2y+TYellow(y,g)}.

In all cases, we get a normal odd partition of conformal with and we can check that these three normal odd partitions and of are compatible, as expected.

###### Lemma 15

Let be a 3-edge-colorable cubic graph with the proper 3-edge-coloring . If , , and are three compatible normal odd partitions conformal, respectively, to , , and , then the graph obtained from by expanding a vertex by a triangle also has this property.

Proof Let be a vertex of with neighbors , , . Let us expand the vertex by a triangle, say . We color the edges , , and in order to get a proper -edge-coloring in  (see Figure 7). Assume without loss of generality that the edge (respectively, , ) is colored (respectively, , ).

We suppose . The vertex is an internal vertex of some trail in , say , we replace in with and we add the trail of length , . Thus we get a normal odd partition of the edge set of all of whose trails have odd edges of color .

We proceed in a similar way for and for and we get three compatible normal odd partitions with the desired property.