Cycle decompositions of pathwidth-6 graphs

# Cycle decompositions of pathwidth-6 graphs

Elke Fuchs, Laura Gellert and Irene Heinrich
###### Abstract

Hajós’ conjecture asserts that a simple Eulerian graph on vertices can be decomposed into at most cycles. The conjecture is only proved for graph classes in which every element contains vertices of degree or . We develop new techniques to construct cycle decompositions. They work on the common neighbourhood of two degree- vertices. With these techniques we find structures that cannot occur in a minimal counterexample to Hajós’ conjecture and verify the conjecture for Eulerian graphs of pathwidth at most . This implies that these graphs satisfy the small cycle double cover conjecture.

## 1 Introduction

It is well-known that the edge set of an Eulerian graph can be decomposed into cycles. In this context, a natural question arises: How many cycles are needed to decompose the edge set of an Eulerian graph? Clearly, a graph with a vertex of degree cannot be decomposed into less than many cycles. Thus, for a general graph , we cannot expect to find a cycle decomposition with less than many cycles. Hajós’ conjectured that this number of cycles will always suffice.111Originally, Hajós’ conjectured a bound of . Dean [4] showed that Hajós’ conjecture is equivalent to the conjecture with bound .

###### Conjecture 1 (Hajós’ conjecture (see [11])).

Every simple Eulerian graph has a cycle decomposition with at most many cycles.

Granville and Moisiadis [7] showed that for every and every , there exists a connected graph with vertices and maximum degree at most whose minimal cycle decomposition consists of exactly  cycles. This shows that — even if the maximal degree is restricted to — the bound is best possible.

A simple lower bound on the minimal number of necessary cycles is the maximum degree divided by . This bound is achieved by the complete bipartite graph that can be decomposed into Hamiltonian cycles (see [10]). In general, all graphs with a Hamilton decomposition (for example complete graphs  [1]) trivially satisfy Hajós’ conjecture.

Hajós’ conjecture remains wide open for most classes. Heinrich, Natale and Streicher [9] verified Hajós’ conjecture for small graphs by exploiting Lemma 6, 8, 10, and 11 of this paper as well as random heuristics and integer programming techniques:

###### Theorem 2 (Heinrich, Natale and Streicher [9]).

Every simple Eulerian graph with at most vertices satisfies Hajós’ Conjecture.

Apart from Hamilton decomposable (and small) graphs, the conjecture has (to our knowledge) only been shown for graph classes in which every element contains vertices of degree at most . Granville and Moisiadis [7] showed that Hajós’ conjecture is satisfied for all Eulerian graphs with maximum degree at most . Fan and Xu [6] showed that all Eulerian graphs that are embeddable in the projective plane or do not contain the minor satisfy Hajós’ conjecture. To show this, they provided four operations involving vertices of degree less than that transform an Eulerian graph not satisfying Hajós’ conjecture into another Eulerian graph not satisfying the conjecture that contains at most one vertex of degree less than . This statement generalises the work of Granville and Moisiadis [7]. As all four operations preserve planarity, the statement further implies that planar graphs satisfy Hajós’ conjecture. This was shown by Seyffarth [12] before. The conjecture is still open for toroidal graphs. Xu and Wang [13] showed that the edge set of each Eulerian graph that can be embedded on the torus can be decomposed into at most cycles. Heinrich and Krumke [8] introduced a linear time procedure that computes minimum cycle decompositions in treewidth- graphs of maximum degree .

We contribute to the sparse list of graph classes satisfying Hajós’ conjecture. Our class contains graphs without any vertex of degree or — in contrast to the above mentioned graph classes.

###### Theorem 3.

Every Eulerian graph of pathwidth at most satisfies Hajós’ conjecture.

As graphs of pathwidth at most contain two vertices of degree less than , it suffices to concentrate on graphs of pathwidth exactly . All such graphs with at most one vertex of degree or contain two degree- vertices that are either non-adjacent with the same neighbourhood or adjacent with four or five common neighbours. We use these structures to construct cycle decompositions.

With similar ideas, it is possible attack graphs of treewidth . As more substructures may occur, we restrict ourselves to graphs of pathwidth .

A cycle double cover of a graph is a collection of cycles of such that each edge of is contained in exactly two elements of . The popular cycle double cover conjecture asserts that every -edge connected graph admits a cycle double cover. This conjecture is trivially satisfied for Eulerian graphs. Hajós’ conjecture implies a conjecture of Bondy regarding the Cycle double cover conjecture.

###### Conjecture 4 (Small Cycle Double Cover Conjecture (Bondy [3])).

Every simple -edge connected graph admits a cycle double cover of at most many cycles.

As a cycle double cover may contain a cycle twice, we can conclude the following directly from Theorem 3.

###### Corollary 5.

Every Eulerian graph of pathwidth at most satisfies the small cycle double cover conjecture.

## 2 Reducible structures

All graphs considered in this paper are finite, simple and Eulerian. We use standard graph theory notation as can be found in the book of Diestel [5].

In order to prove our main theorem, we consider a cycle decomposition of a graph as a colouring of the edges of where each colour class is a cycle. We define a legal colouring of a graph as a map

 c:E(G)↦{1,…,⌊\sfrac(|V(G)|−1)2⌋}

where each colour class for is the edge set of a cycle of . A legal colouring is thus associated to a cycle decomposition of that satisfies Hajós’ conjecture.

Using recolouring techniques, we show the following lemmas for two degree- vertices with common neighbourhood of size , or . All proofs can be found in Section 4.

###### Lemma 6.

Let be an Eulerian graph with two degree- vertices with

 N(u)=N∪{v}N(v)=N∪{u}.

Let all Eulerian graphs obtained from by addition or deletion of edges with both end vertices in have a legal colouring.
If contains at least one edge, or if contains a vertex that is adjacent to at least three vertices of then also has a legal colouring.

###### Lemma 7.

Let be an Eulerian graph with two degree- vertices with

 N(u)=N∪{u,xv}N(v)=N∪{v,xu}.

Let be an --path in . Further let all Eulerian graphs obtained from by addition and deletion of edges with both end vertices in and by optional deletion of have a legal colouring.
If contains at least one edge not equal to , or if contains a vertex that is adjacent to at least three vertices of then also has a legal colouring.

###### Lemma 8.

Let be an Eulerian graph with two degree- vertices with

 N(u)=N(v)=N.

Let all Eulerian graphs obtained from by addition or deletion of edges with both end vertices in have a legal colouring.
If contains at least one edge, or if contains a vertex that is adjacent to at least three vertices of then also has a legal colouring.

The next two results are not necessary for the proof of Theorem 3. We nevertheless state them here.

The first lemma is useful for graphs with an odd number of vertices.

###### Lemma 9.

Let be an Eulerian graph on an odd number of vertices that contains a vertex of degree or with neighbourhood . Let be obtained from by addition or deletion of arbitrary edges in . If has a legal colouring, then has a legal colouring.

If a graph contains a degree- vertex with independent neighbours , then it is clear that a legal colouring of can be transformed into a legal colouring of . Granville and Moisiadis [7] observed a similar relation for a degree- vertex.

###### Lemma 10 (Granville and Moisiadis [7]).

Let be an Eulerian graph containing a vertex with neighbourhood such that contains the edge but not the edge . If has a legal colouring, then also has a legal colouring.

Generalising this idea, we analyse the neighbourhood of a degree- vertex.

###### Lemma 11.

Let be an Eulerian graph that contains a degree- vertex with neighbourhood such that is a clique and . If has a legal colouring, then has a legal colouring.

## 3 Recolouring Techniques

In this section, we provide recolouring techniques that are necessary to prove Lemma 6, 7 and 8. For a path or a cycle we write or to express that all edges of respectively are coloured with colour . We start with a statement about monochromatic triangles.

###### Lemma 12.

Let be a graph with legal colouring that contains a clique . Then there is a legal colouring of in which the cycle is not monochromatic.

###### Proof.

Figure 1 illustrates the recolourings described in this proof. Assume that is monochromatic of colour in . First assume that

 an edge of colour j\coloneqqc(y1y) is adjacent to y2 (1)

for two distinct vertices in . Without loss of generality, the path of colour between and along the path does not contain the vertex (where ). Flip the colours of the monochromatic paths and , ie set , and for all other edges . The obtained colouring is legal: By construction, all colour classes are cycles and at most many colours are used. Further, the cycle is not monochromatic.

If (1) does not hold, we can get rid of one colour. Set , , , and for all other edges . By construction, all colour classes are cycles and is not monochromatic. ∎

Figure 2 illustrates the following simple observation.

###### Observation 13.

Let be an --path that is vertex-disjoint from an --path . Then there are three possibilities to connect and by two vertex-disjoint paths that do not intersect for . Two of the possibilities yield a cycle — the third way leads to two cycles.

Lemma 14, 15 and 16 are all based on the same elementary fact: Let and be graphs with . If allows for a cycle decomposition with at most cycles, then any cycle decomposition of that uses at most one cycle more than the cycle decomposition of shows that is not a counterexample to Hajós’ conjecture.

This fact leads us to the following inductive approach: Given a graph with two vertices and of degree , we remove and from and might remove or add edges to obtain a graph . If has a cycle decomposition with at most cycles we construct a cycle decomposition of from it. We reroute some of the cycles in an appropriate way such that and are each touched by two cycles. Now, there remain some edges in that are not covered. If those edges form a cycle, we have found a cycle decomposition of . If a cycle is not rerouted to or twice, the cycle decomposition of satisfies Hajós’ conjecture.

To describe this inductive approach in a coherent way, we regard the cycle decomposition of as a legal colouring. Then we regard the above reroutings as recolourings where we have to make sure that no colour appears twice at or . If the edges that have not yet received a colour form a cycle, we associate the new colour to this cycle. The obtained colouring of the edges then uses at most many colours and each colour class is a cycle. Thus, we have constructed a legal colouring.

###### Lemma 14.

Let be an Eulerian graph without legal colouring that contains two adjacent vertices and of degree with common neighbourhood . Define and let be a legal colouring of .

1. If contains a path of length then is monochromatic in .

2. Let contain an independent set of size . If is not an independent set or if there is a vertex in that is adjacent to , and , then does not have a legal colouring.

3. If contains an induced path of length then does not have a legal colouring.

4. If contains a triangle , a vertex that is not adjacent to and and a vertex adjacent to then does not have a legal colouring.

###### Proof of (i).

If has a colour different from and , then set

 c(y1uy2)=c′(y1y2)c(y2vy3)=c′(y2y3)c(y3uvy4)=c′(y3y4).

If has a colour different from and , then set

 c(y1uy2)=c′(y1y2)c(y2vuy3)=c′(y2y3)c(y3vy4)=c′(y3y4).

The case distinction makes sure that the modified colour classes remain cycles. By further setting and for all other edges we have constructed a legal colouring of . ∎

###### Proof of (ii).

Set and let be a legal colouring of .
First assume that . Then one can easily check that the following is a legal colouring of .

 c(y2uvy1)=c′′(y2y1)c(y2vy3)=c′′(y2y3)c(y3uy1)=c′′(y3y1) c(y4uy5vy4)=⌊\sfrac(|V(G)|−1)2⌋ c(e)=c′′(e) for all other edges e (2)

By symmetry, we are done unless the triangle is monochromatic in . By Lemma 12, we can suppose that there is no vertex in that is adjacent to , and . Suppose that is not independent. Without loss of generality, we can assume that contains an edge, say incident to one of the vertices of the independent -set. (Otherwise, we can choose another suitable independent -set in ). Then by construction the following is a legal colouring of .

 c(y1uvy4)=c′′(y1y4)c(y2uy3)=c′′(y2y3)c(y2vy3)=c′′(y2y1y3) c(y1y4uy5vy1)=⌊\sfrac(|V(G)|−1)2⌋ c(e)=c′′(e) for all other edges e\qed
###### Proof of (iii).

Let have a legal colouring and let be the unique vertex in .
If , set

 c(y1uy2)=c′′(y1y2)c(y2vuy3)=c′′(y2y4y1y3)c(y3vy4)=c′′(y3y4) c(uy5vy1y2y3y4u)=⌊\sfrac(|V(G)|−1)2⌋.

If is different from and , set

 c(y1uvy3)=c′′(y1y3)c(y4vy1)=c′′(y4y1)c(y2uy4)=c′′(y2y4) c(uy5vy2y3u)=⌊\sfrac(|V(G)|−1)2⌋.

If is different from and , the colouring is defined similarly by relabelling the vertices .
If is different from and , set

 c(y1vy3)=c′′(y1y3)c(y4vuy1)=c′′(y4y1)c(y2uy4)=c′′(y2y4) c(uy5vy2y3u)=⌊\sfrac(|V(G)|−1)2⌋.

Further set for all other edges in all cases. Again, the case distinction makes sure that all colour classes are cycles and we have constructed a legal colouring. ∎

###### Proof of (iv).

Let be a legal colouring of . First assume that . Then set

 c(y2vuy3)=c′′(y2y3)c(y1uy4)=c′′(y1y4)c(y3vy4)=c′′(y3y4) c(uy5vy1y3y2u)=⌊\sfrac(|V(G)|−1)2⌋.

If , the colouring is defined as above by interchanging the roles of and .

Now assume that . If , then the cycle is monochromatic. Set

 c(y4uvy5)=c′′(y4y5) c(y1vy3y2uy1)=c′′(y1y2y3y4y1)c(y1y3uy5y4vy2y1)=⌊\sfrac(|V(G)|−1)2⌋.

If , then either or . If , set

 c(y2uy3vy4)=c′′(y2y3y4)c(y1vuy4)=c′′(y1y4) c(y1uy5vy2y3y1)=⌊\sfrac(|V(G)|−1)2⌋.

If , set

 c(y2uy3)=c′′(y2y3)c(y1vy4)=c′′(y1y4)c(y3vuy4)=c′′(y3y4) c(y1uy5vy2y3y1)=⌊\sfrac(|V(G)|−1)2⌋.

By setting for all other edges we have constructed a legal colouring for in all cases. ∎

If and are adjacent degree- vertices that have a common neighbourhood  of size , we call the two vertices that are adjacent with exactly one of the private neighbours of and . Here, we denote them by and . If there is a --path in , it is possible to translate all techniques of Lemma 16. It suffices to delete , and to obtain another Eulerian graph: In all recolourings of Lemma 16, the edges for one vertex were contained in the new colour class . If we have two private neighbours and it suffices to replace the path by the path in this colour class. This means, we can regard as a single vertex .

###### Lemma 15.

Let be an Eulerian graph without legal colouring that contains two adjacent vertices and of degree with common neighbourhood and as well as . Let be an --path in . Define and let be a legal colouring of .

1. If contains a path with of length then is monochromatic in .

2. Let contain an independent set of size . If contains an edge or if there is a vertex in that is adjacent to , and then does not have a legal colouring.

3. If does not contain the edges for two vertices but contains an edge with end vertex or then does not have a legal colouring.

4. If contains the edges with and but not the edges then does not have a legal colouring.

5. If contains a triangle with , a vertex that is not adjacent to and and a vertex adjacent to then does not have a legal colouring.

###### Proof of (i).

The proof is very similar to the proof of Lemma 14.(i) if we regard as one single vertex. We will nevertheless give a detailed proof. By symmetry of and (and thus of and ), we can assume that is either contained in or is equal to . Suppose that is not monochromatic.

If , then set

 c(y1uvy2)=c′(y1y2)c(y2uy3)=c′(y2y3)c(y3vy4)=c′(y3y4).

If , then set

 c(y1uy2)=c′(y1y2)c(y2vuy3)=c′(y2y3)c(y3vy4)=c′(y3y4).

If , then set

 c(y1uy2)=c′(y1y2)c(y2vy3)=c′(y2y3)c(y3uvy4)=c′(y3y4).

If the following completes by construction a legal colouring of :

 c(y1y2y3y4uxuPxvvy1)=⌊\sfrac(|V(G)|−1)2⌋ c(e)=c′(e) for all other edges e

Now suppose that and that are not contained in the path . Then, the following completes by construction a legal colouring of :

 c(y1y2y3y4ux4vxvPy1)= ⌊\sfrac(|V(G)|−1)2⌋ c(e)=c′(e) for all other edges e\qed
###### Proof of (ii).

The proof is very similar to the proof of Lemma 14.(ii) if we regard as one single vertex. ∎

###### Proof of (iii).

Assume that is a legal colouring of and let . By symmetry of and (and thus of and ) we can suppose that .
If has a colour different from the colour of and , set

 c(xuuvy1)=c′′(xuy1)c(y1uy2)=c′′(y1y2)c(y2vxv)=c′′(y2xv) c(y3uy4vy3)=⌊\sfrac(|V(G)|−1)2⌋.

An analogous colouring can be defined if has a colour different from the colour of and .
If has a colour different from the colour of and , then set

 c(xuuy1)=c′′(xuy1)c(y1vuy2)=c′′(y1y2)c(y2vxv)=c′′(y2xv) c(y3uy4vy3)=⌊\sfrac(|V(G)|−1)2⌋.

Now suppose that all three edges have the same colour. Then, has a different colour. Set

 c(xuuy2)=c′′(xuy1y2)c(y1uvy4)=c′′(y1y4)c(y2vxv)=c′′(y2xv) c(uy3vy1y4u)=⌊\sfrac(|V(G)|−1)2⌋.

In all cases, set for all other edges . The case distinction now makes sure that we constructed a legal colouring for . ∎

###### Proof of (iv).

Assume that is a legal colouring of . Without loss of generality let .

First suppose that all three edges have the same colour. Then, has a different colour and the following gives by construction a legal colouring for :

 c(xuuy1)=c′′(xuy1)c(y1