Variations on the Petersen colouring conjecture

TL;DR

This paper investigates a variation of the Petersen colouring conjecture, providing a near-solution by showing that most edges in a bridgeless cubic graph can be coloured with four colours to nearly satisfy the conjecture's conditions.

Contribution

It introduces a new partial colouring approach for bridgeless cubic graphs that nearly satisfies the Petersen colouring conjecture, with a tight bound proven.

Findings

A 4-colour edge-colouring where at most 80% of edges violate the conjecture's condition.

The bound is tight and only the Petersen graph reaches this limit.

A simple discharging method is used for the proof.

Abstract

The Petersen colouring conjecture states that every bridgeless cubic graph admits an edge-colouring with $5$ colours such that for every edge $e$, the set of colours assigned to the edges adjacent to $e$ has cardinality either $2$ or $4$, but not $3$. We prove that every bridgeless cubic graph $G$ admits an edge-colouring with $4$ colours such that at most $\frac45\cdot|V(G)|$ edges do not satisfy the above condition. This bound is tight and the Petersen graph is the only connected graph for which the bound cannot be decreased. We obtain such a $4$-edge-colouring by using a carefully chosen subset of edges of a perfect matching, and the analysis relies on a simple discharging procedure with essentially no reductions and very few rules.

Equations

$$3t+1+\sum _{i=1}^t\frac{{\ell }_i-2}{5}=1+\frac{1}{5}\left(13t+\sum _{i=1}^t{\ell }_i\right).$$

$$5+13t<3\cdot \sum _{i=1}^t{\ell }_i.$$

Full text

This work was partially supported by P.H.C. Proteus [37455VB]; ARRS [BI-FR-PROTEUS/17-18-009, P1-0383].

1. Introduction

At the ninth British Combinatorial Conference in 1983, Fouquet and Jolivet [FoJo83] introduced strong edge-colourings of cubic graphs. This notion was further studied by Jaeger, who formulated a conjecture which is, arguably, one of the most challenging conjecture in graph theory. Proving Jaeger’s conjecture to be true would have tremendous consequences, such as confirming the Cycle double cover conjecture, the Berge-Fulkerson conjecture and the nowhere-zero $5$-flow conjecture.

For any integer $k\geqslant 3$, consider a $k$-edge-colouring of a cubic graph $G=(V,E)$, that is, a mapping $f\colon E\to\mathopen{\{}1,\dotsc,k\mathclose{\}}$ such that $f(e)\neq f(e^{\prime})$ for every two edges $e$ and $e^{\prime}$ that share a vertex. For an edge $e\in E$, let $\mathscr{E}(e)$ be the set of four edges adjacent to $e$. The edge $e$ is rich if $\left\lvert f(\mathscr{E}(e))\right\rvert=4$, while it is poor if $\left\lvert f\left(\mathscr{E}(e)\right)\right\rvert=2$. The edge-colouring $f$ is normal if every edge is either rich or poor. The Petersen colouring conjecture reads as follows.

Conjecture 1.1** (The Petersen coloring conjecture—Jaeger, 1985).**

Every cubic bridgeless graph admits a normal $5$-edge-colouring.

Now it is maybe a good time to explain the links with the ubiquitous Petersen graph $\mathscr{P}$. A Petersen colouring of a cubic graph $G=(V,E)$ is a mapping $g$ that associates to each edge of $G$ an edge of $\mathscr{P}$ such that if two edges $e$ and $e^{\prime}$ of $G$ share a vertex, then so do the edges $g(e)$ and $g(e^{\prime})$ of $\mathscr{P}$. As observed by Jaeger [Jae85], normal colourings and Petersen colourings of cubic graphs are in one-to-one correspondence.

Indeed, as is well known, the Petersen graph $\mathscr{P}$ can be seen as the Kneser graph with parameters $5$ and $2$, defined as follows: the vertices are in one-to-one correspondence with the $2$-element subsets of $\mathopen{\{}1,\dotsc,5\mathclose{\}}$ and two vertices are adjacent if and only if the corresponding subset are disjoint. With this definition in mind, we can label every edge $uv$ of $\mathscr{P}$ by the unique integer $\ell(uv)\in\mathopen{\{}1,\dotsc,5\mathclose{\}}$ that does not belong to $X_{u}\cup X_{v}$, where $X_{w}$ is the $2$-element subset of $\mathopen{\{}1,\dotsc,5\mathclose{\}}$ that corresponds to $w$, for every vertex $w$ of $\mathscr{P}$. Notice that if $g$ is a Petersen colouring of a cubic graph $G=(V,E)$, then $f\colon E\to\mathopen{\{}1,\dotsc,5\mathclose{\}}$ defined by $f(e)\coloneqq\ell(g(e))$ is a normal colouring of $G$.

Conversely, assume that $f$ is a normal $5$-edge-colouring of a cubic graph $G=(V,E)$. Keeping in mind the labelling $\ell$ of the edges of $\mathscr{P}$ given above, we define the mapping $g\colon E\to E(\mathscr{P})$ as follows. For each edge $e=uv\in E$, we define $g(e)$ to be the edge $e^{\prime}$ of $\mathscr{P}$ such that first $\ell(e^{\prime})=f(e)$, and second $e^{\prime}$ is incident to the vertex $w\in V(\mathscr{P})$ such that the three colours assigned by $f$ to the edges of $G$ incident to $u$ are the elements of $X_{w}\cup\{f(e)\}$. A straightforward checking ensures that $g$ is a Petersen colouring of $G$. We just proved the following equivalence, which was first established by Jaeger [Jae85].

Proposition 1.2**.**

Let $G$ be a cubic graph. Then, $G$ admits a normal colouring if and only if $G$ admits a Petersen colouring.

Notice that a $3$-edge-colouring of a (connected) cubic graph $G$ is precisely a normal colouring in which every edge is poor. Fouquet and Jolivet [FoJo83] coined the term strong colouring to define an edge-colouring in which every edge is rich. This corresponds precisely to an edge-colouring “at distance $2$” or, in other words, to a vertex-colouring of the square of the line graph of $G$. The unique normal colouring of the Petersen graph is a strong colouring.

Despite original approaches [MaSk05, NeSa08, Sam11], few progress has been made on the Petersen colouring conjecture: ways to infirm it remain elusive as possible counter-examples must be snarks, that is bridgeless cubic graphs that are not $3$-edge-colourable (the ones we know are usually obtained from well-structured graph operations, for which the Petersen colouring conjecture can be verified [Bil15, HaSt14]), and confirming the conjecture is expected to be a difficult task since as reported earlier this would confirm several difficult and most researched graph conjectures.

In view of the difficulty of the question, it is natural to ask for weaker versions of the conjecture. Because a strong colouring is normal, we know that every cubic graph admits a normal colouring using at most $10$ colours: indeed Andersen [And92] and, independently, Horák, He and Totter [HHT93] established this statement (which confirms, for the particular case of graphs with maximum degree $3$, a conjecture of Erdős and Nešetřil formulated in 1985 during a seminar in Prague). Further, it has been noted before [Bil15] that Seymour’s $8$-flow theorem provides, for any bridgeless cubic graph, a normal $7$-edge-colouring. To the best of our knowledge, whether a normal $6$-edge-colouring can be found for any such graph is still an open question. A line of study is then to find an edge-colouring that is normal on a large proportion of the graph, which we formalise in the next subsection. Before that, we end this part with a remark.

As mentioned earlier, a $3$-edge-colourable graph $G$ always has a Petersen colouring: let $e_{1}$, $e_{2}$ and $e_{3}$ be the three edges incident to an arbitrary vertex $v$ of the Petersen graph. Label $e_{i}$ by $\ell(e_{i})\coloneqq i$ for each $i\in\{1,2,3\}$. If $c$ is a $3$-edge-colouring of $G$, then defining $f\colon E(G)\to E(\mathscr{P})$ by $f(e^{\prime})\coloneqq e_{c(e^{\prime})}$ yields a Petersen colouring of $G$. This Petersen colouring is, in some sense, trivial: it only uses the incidences at a single vertex of the Petersen graph. This is equivalent to saying that the original graph admits a $3$-edge-colouring. A fact that seems worth pointing out, however, is that Petersen colourings of bridgeless cubic graphs are either “trivial” (and hence the graph admits a $3$-edge-colouring) or surjective.

1.1. The rich, the poor and the medium.

Given an edge-colouring of a cubic graph $G$, define an edge $e$ to be medium if it is neither rich nor poor. Since the Petersen colouring conjecture states that every bridgeless cubic graphs admits a $5$-edge-colouring such that no edge is medium, it seems interesting to investigate the minimum number of medium edges in edge-colourings of bridgeless cubic graphs. As observed by Bílková [Bil15]*p. 9, Petersen’s perfect matching theorem combined with Vizing’s edge-colouring theorem (and some further analysis if the graph has cycles of length less than $5$) directly yield for every bridgeless cubic graph a $5$-edge-colouring such that at least one third of the edges are rich or poor. This lower bound was improved [Bil15]*Theorem 3.2 to two thirds of the edges for cubic graphs having a $2$-factor consisting of two cycles of the same length — the class of “generalised prisms” — and to roughly half the edges in graphs with no short cycles [Bil15]*Theorem 3.6.

Ways how poor, rich and medium edges combine in edge-colourings looks intriguing. We already pointed out that an edge-colouring with poor edges only is actually a $3$-edge-colouring. Oppositely, an edge-colouring with rich edges only is a strong colouring. Notice that every normal $5$-edge-colouring of the Petersen graph actually contains no poor edge: it is thus a strong colouring.

We consider $4$-edge-colourings and prove the following result.

Theorem 1.3**.**

Every connected cubic bridgeless graph $G$ admits a $4$-edge-colouring such that at most $\frac{4}{5}\cdot\left\lvert V(G)\right\rvert$ edges are neither rich nor poor. Furthermore, no $4$-edge-colouring of $G$ yields less medium edges if and only if $G$ is the Petersen graph.

Since an $n$-vertex cubic graph has $\frac{3n}{2}$ edges, Theorem 1.3 ensures that every bridgeless cubic graph $G$ admits a $4$-edge-colouring containing at most $\frac{8}{15}\cdot\left\lvert E(G)\right\rvert$ medium edges. This bound cannot be improved in general, since the Petersen graph has $15$ edges and each of its $4$-edge-colourings yields at least $8$ medium edges.

2. Proof of Theorem 1.3

We demonstrate the upper bound: every bridgeless cubic graph $G$ admits a $4$-edge-colouring such that at most $\frac{4}{5}\cdot\left\lvert V(G)\right\rvert$ edges are medium. While developing the proof, we shall see that the only case where the bound must be attained is if $G$ is the Petersen graph. We proceed by induction on the number of vertices of $G$. The induction yields the conclusion in a standard way if $G$ has a triangle, and we thus first deal with this case. To make the argument smoother, we actually prove the result for (loopless) bridgeless cubic multi-graphs. For instance any $4$-edge-colouring of a triple edge between two vertices yields three poor edges. Similarly, if a cubic graph $G$ contains two vertices with exactly two parallel edges between them, then in any $4$-edge-colouring of $G$ either both edges are poor or both edges are medium.

As reported earlier, every $3$-edge-colouring of a cubic graph contains only poor edges, and hence the statement of Theorem 1.3 is correct if $G$ admits a $3$-edge-colouring, and hence in particular if $\left\lvert V(G)\right\rvert=2$. We hence consider a connected bridgeless cubic multi-graph $G$ that admits no $3$-edge-colouring, and we set $n\coloneqq\left\lvert V(G)\right\rvert$ (so $n\geqslant 10$). Our first two arguments are standard and well known to people used to graph colouring but they are included for completeness.

We use induction to prove the statement if $G$ contains a multi-edge. Indeed, suppose that $e_{1}$ and $e_{2}$ are two different edges with end-vertices $v_{1}$ and $v_{2}$. For each $i\in\{1,2\}$, let $u_{i}$ be the neighbour of $v_{i}$ different from $v_{3-i}$. Since $n>2$ and because $G$ is bridgeless, $u_{1}\neq u_{2}$. Let $G^{\prime}$ be the bridgeless cubic multi-graph obtained from $G-\{v_{1},v_{2}\}$ by adding a new edge $e^{\prime}$ between $u_{1}$ and $u_{2}$. The induction hypothesis ensures that $G^{\prime}$ admits a $4$-edge-colouring $c^{\prime}$ yielding at most $\frac{4}{5}\cdot(n-2)$ medium edges. It is straightforward to deduce from $c^{\prime}$ a $4$-edge-colouring $c$ of $G$ with no more medium edges, which thus prove the statement (including “the furthermore part”) if $G$ contains a multi-edge. One obtains a $4$-edge-colouring $c$ of $G$ by setting $c(e)=c^{\prime}(e)$ if $e\in E(G)\cap E(G^{\prime})$ and $c(v_{i}u_{i})\coloneqq c^{\prime}(e^{\prime})$ for each $i\in\{1,2\}$, and letting $c(e_{1})$ and $c(e_{2})$ be the two colours of the two edges of $G^{\prime}$ incident to $u_{1}$ that are different from $e^{\prime}$. The obtained $4$-edge-colouring $c$ of $G$ yields at most $\frac{4}{5}\cdot(n-2)$ medium edges, which is less than $\frac{4}{5}\cdot n$. Consequently, we may assume that $G$ is simple.

We now use induction to prove the statement if $G$ contains a triangle $v_{0}v_{1}v_{2}$. Indeed, we then define $G^{\prime}$ to be the multi-graph obtained from $G$ by contracting $v_{0}$, $v_{1}$ and $v_{2}$ into a single vertex $x$. It follows that $G^{\prime}$ is a bridgeless and cubic multi-graph with $n-2$ vertices, and therefore the induction hypothesis yields that $G^{\prime}$ admits a $4$-edge-colouring $c^{\prime}$ yielding at most $\frac{4}{5}\cdot(n-2)$ medium edges. This colouring $c^{\prime}$ can be extended into a $4$-edge-colouring $c$ of $G$ by setting $c(e)\coloneqq c^{\prime}(e)$ if $e\in E(G)\cap E(G^{\prime})$ and $c(v_{i}v_{i+1})\coloneqq c^{\prime}(v_{i+2}u_{i+2})$ where $i\in\{0,1,2\}$ is considered modulo $3$ and $u_{i}$ is the neighbour of $v_{i}$ not in $\{v_{i+1},v_{i+2}\}$. The obtained $4$-edge-colouring $c$ yields no more medium edges in $G$ than $c^{\prime}$ does in $G^{\prime}$. Consequently, we may assume that $G$ is a simple bridgeless cubic graph with no triangle.

It remains to deal with the case where $G$ is a bridgeless cubic graph with no triangles. Let $F$ a $2$-factor of $G$ and $M$ the perfect matching such that $F=G-M$. Because $G$ admits no $3$-edge-colouring, we know that $F$ contains at least two odd cycles. In particular, there exists an edge in $M$ that is not a chord of a cycle in $F$. Note also that cycles of $G$ with length (at most) $5$ have no chord, as $G$ has no triangle.

If $v\in V(G)$, we define $C_{v}$ to be the cycle in $F$ to which $v$ belongs and $v^{\prime}$ to be the unique neighbour of $v$ in $G$ such that $vv^{\prime}\in M$. For every cycle $C\in F$, there is a cyclic ordering $\varphi_{C}$ of the vertices in $V(C)$, which we extend to the edges in $M$ that are incident to a vertex in $C$: two edges $e$ and $e^{\prime}$ in $M$ that have each exactly one end-vertex on a given cycle $C\in F$ are consecutive if their end-vertices in $C$ are consecutive with respect to $\varphi_{C}$. (Notice that chords are purposely excluded from this definition.)

An edge-selection is a subset $S$ of $M$ with the following properties:

1. (1)

every edge in $S$ is incident to two different odd cycles in $F$ (in particular, no edge in $S$ is a chord of a cycle in $F$); 2. (2)

every cycle in $F$ is incident to at most two edges in $S$; and 3. (3)

if a cycle $C\in F$ is incident to two edges in $S$, then these two edges are consecutive.

Given an edge-selection $S$ and a cycle $C\in F$, the degree $\deg_{S}(C)$ of $C$ in $S$ is the number of edges incident to $C$ that belong to $S$, and hence $\deg_{S}(C)\in\{0,1,2\}$. If in addition $C^{\prime}\in F$, then $C$ and $C^{\prime}$ are $S$-adjacent if $S$ contains an edge incident to both $C$ and $C^{\prime}$. An $S$-component of $G$ is an inclusion-wise maximal subset $K$ of $F$ such that for every two distinct cycles $C$ and $C^{\prime}$ in $K$, there exists a sequence $(C_{i})_{0\leqslant i\leqslant t}$ of cycles in $K$ such that $C_{0}=C$, $C_{t}=C^{\prime}$ and for every $i\in\{0,\dotsc,t-1\}$, the cycles $C_{i}$ and $C_{i+1}$ are $S$-adjacent. The edges in $S$ joining two cycles in $K$ are said to be associated with $K$. This relation is a surjective mapping from $S$ to the set of all $S$-components of $G$. We need a last definition. If $K$ is an $S$-component, then we let $G_{K}$ be the multi-graph with vertex set $K$ and $x$ edges between $C$ and $C^{\prime}$ where $x\in\{0,1,2\}$ is the number of edges in $S$ that are incident to both $C$ and $C^{\prime}$ in $G$. It follows from the definitions that $G_{K}$ is either a single vertex, or a path, or a cycle, or two vertices joined by two parallel edges.

Among all edge-selections of maximum order, we choose one such that the number of $S$-degree-$2$ cycles is as large as possible. We construct a $4$-edge-colouring of $G$ with the following properties:

• •

every edge in $M$ is coloured $4$ and no other edge is coloured $4$;

• •

an edge is coloured $3$ only if it belongs to an odd cycle in $F$;

• •

every odd cycle in $F$ has exactly one edge coloured $3$;

• •

every edge in $S$ is adjacent to two edges coloured $3$; and

• •

if an edge $e$ in $S$ is medium, then it is associated with an $S$-component $K$ such that $G_{K}$ is an odd cycle and $e$ is the only medium edge associated with $K$.

To see why such a $4$-edge-colouring exists, start by colouring the edges of $G$ that belong to $M$ with $4$. Next colour the edges of every even cycle in $F$ using $\{1,2\}$. By (2) and (3) every odd cycle $C\in F$ has an edge that is incident to all the edges in $S$ incident to $C$: colour this edge with $3$. The remaining uncoloured edges span a vertex-disjoint collection of paths, and we colour them using $\{1,2\}$, independently for each $S$-component $K$. If $G_{K}$ is not an odd cycle, then one can ensure that no edge associated with $K$ is medium. If $G_{K}$ is an odd cycle, then one can ensure that exactly one edge $S$ associated with $K$ is medium.

Our goal is to demonstrate that the obtained $4$-edge-colouring of $G$ contains at most $4n/5$ medium edges, where $n$ is the number of vertices of $G$. We use a discharging argument to count the medium edges: we start by assigning a charge of $1$ to each medium edge, and thus throughout all the process the total charge in the graph $G$ is precisely the number of medium edges.

We shall define a number of discharging rules: in the first ones, medium edges send charge to cycles in $F$ to which they are incident. Later, some cycles in $F$ will send some charge to other cycles in $F$. We apply the rules in order and analyse the global state of the charge in the graph after one or more rules have been applied. At the end, we prove that for each $S$-component $K$ the sum of the charges of the cycles in $K$ is at most $\frac{4}{5}$ times the number of vertices belonging to cycles in $K$, which implies the sought upper bound on the number of medium edges.

Fact 1**.**

A cycle in $F$ contains zero medium edge if it is even and $3$ medium edges if it is odd.

Proof.

Indeed, if the edge $e$ belongs to an even cycle, then its colour $c$ belongs to $\{1,2\}$, and each of its end-vertices is incident to an edge coloured $4$, which belongs to $M$, and an edge coloured $3-c$, which belongs to the even cycle. So $e$ is poor. Let $C=v_{0}\dotso v_{2k}$ is an odd cycle in $F$ with $v_{0}v_{1}$ being its only edge coloured $3$. For each $i\in\{2,\dotsc,2k-1\}$, let $c_{i}\in\{1,2\}$ be the colour of the edge $v_{i}v_{i+1}$. Then the edge $v_{i}v_{i+1}$ is incident to two edges coloured $4$ and two edges coloured $3-c_{i}$, and hence $v_{i}v_{i+1}$ is poor. Consequently, the only medium edges on $C$ are $v_{0}v_{1}$, $v_{1}v_{2}$ and $v_{2k}v_{0}$. ∎

Our first rule reads as follows.

(R0) Every medium edge that belongs to a cycle $C$ in $F$ sends $1$ to $C$.

After applying rule (R0), an edge has charge [math] except if it is a medium edge that belongs to $M$, in which case it has charge $1$. In addition, a cycle in $F$ has charge [math] if it is even and charge $3$ if it is odd. For our next rule, notice that if $e$ is a medium edge that belongs to $M$, then $e$ is adjacent to an edge coloured $3$, which must belong to an odd cycle in $F$.

(R1) Let $e$ be a medium edge that belongs to $M$, and let $C$ and $C^{\prime}$ be the two cycles in $F$ to which $e$ is incident, such that $C$ is odd and the edge coloured $3$ on $C$ is adjacent to $e$. If $C^{\prime}$ is even, then $e$ sends $1/2$ to $C^{\prime}$ and $1/2$ to $C$. If $C^{\prime}$ is odd its edge coloured $3$ is adjacent to $e$, then $e$ sends $1/2$ to each of $C$ and $C^{\prime}$. If $C^{\prime}$ is odd and its edge coloured $3$ is not adjacent to $e$, then $e$ sends $1$ to $C$ and nothing to $C^{\prime}$.

After applying rule (R1), every edge has charge [math]. If $C\in F$ is an even cycle, then its charge is at most $1/2\cdot|V(C)|$, which is less than $4|V(C)|/5$. In addition, if $C\in F$ is an odd cycle, then one of the following occurs:

• •

$C$ has $S$-degree [math] and charge at most $5$;

• •

$C$ has $S$-degree $1$ and charge at most $4$; or

• •

$C$ has $S$-degree $2$ and charge at most $7/2$.

Indeed, if $C$ is incident to exactly one edge in $S$, then this edge cannot be medium by the construction of the $4$-edge-colouring. If $C$ is incident to exactly two edges in $S$, then at most one of them is medium, in which case it sends $1/2$ to $C$. It follows that the charge of $C$ is at most than $\frac{4}{5}\cdot|V(C)|$ unless $C$ has length $5$ and $S$-degree [math].

The next step is to apply the three following rules, the third one being illustrated in Figure 2.

(R2) If $C$ is a cycle of length $5$ in $F$ with $S$-degree [math], then $C$ sends $1/5$ to $C_{v^{\prime}}$ for each $v\in V(C)$.

(R3) Let $C=v_{0}\dotso v_{4}$ be a cycle of length $5$ in $F$ of $S$-degree $1$, with $v_{1}v^{\prime}_{1}$ being the unique edge in $S$ incident to $C$. For each $i\in\{0,2\}$, if $C_{v^{\prime}_{i}}$ is not a cycle of length $5$ with $S$-degree $1$, then $C$ sends $1/5$ to $C_{v^{\prime}_{i}}$ through $v_{i}$.

(R4) Let $C=v_{0}\dotso v_{4}$ be a cycle of length $5$ in $F$ of $S$-degree $1$, with $v_{1}v^{\prime}_{1}$ being the unique edge in $S$ incident to $C$. Let $i\in\{0,2\}$. Suppose that $C_{v^{\prime}_{i}}$ is a cycle of length $5$ of $S$-degree $1$, written $v^{\prime}_{i}u_{1}u_{2}u_{3}u_{4}$ such that $u_{2}$ is its unique vertex incident to an edge in $S$. If $C_{u^{\prime}_{2}}$ is a cycle of $S$-degree $2$ then $C$ sends $1/5$ to $C_{u^{\prime}_{2}}$ through $v_{i}$.

We now check that after applying (R2)–(R4), for every $S$-component $K$, the sum of the charges of the cycles in the component is at most $4/5$ times the number of vertices belonging to cycles in $K$ with equality only if $K$ contains only cycles of length $5$. We first analyse the current charge of every cycle $C\in F$. If $C$ is even, then it has charge at most $(1/2+1/5)\cdot\left\lvert V(C)\right\rvert<\frac{4}{5}\cdot\left\lvert V(C)\right\rvert$.

If $C$ is a cycle with $S$-degree [math] and length $2k+1$, where $k\geqslant 2$, then $C$ does not receive any charge. Indeed, there cannot be an edge $e$ in $M$ incident to $C$ and a cycle of length $5$ and $S$-degree [math], as $S\cup\{e\}$ would then contradict the maximality of $S$. This shows that $C$ does not receive charge by (R2). Moreover, if $C$ would receive charge by (R3) then the definition of (R3) would imply the existence of a cycle $C^{\prime}=v_{0}v_{1}v_{2}v_{3}v_{4}$ in $F$ of length $5$ and $S$-degree one such that $v_{1}v^{\prime}_{1}\in S$ and $C=C_{v^{\prime}_{0}}$ (or $C=C_{v^{\prime}_{2}}$). Consequently, $S\cup\{v_{0}v^{\prime}_{0}\}$ would contradict the maximality of $S$. Therefore, the final charge of $C$ is at most $5$ if $k\geqslant 3$, which is less than $4\cdot(2k+1)/5$. If $k=2$ then by rule (R2) the final charge of $C$ is at most $5-5\cdot\frac{1}{5}=4=4\cdot(2k+1)/5$.

Let $C$ be a cycle with $S$-degree $1$. Note that $C$ can receive charge only because of (R2) and (R3). Moreover, $C$ does not receive charge through its vertex incident to an edge in $S$. If $C$ has length $2k+1$ with $k\geqslant 3$, then before applying (R2)–(R4) the charge of $C$ was at most $4$ since the edge in $S$ incident to $C$ is not medium, and hence its final charge is at most $4+2k/5$, which is less than $4\cdot(2k+1)/5$ since $k\geqslant 3$.

If $C$ has length $5$, then let us write $C=v_{0}\dotso v_{4}$ with $v_{1}v^{\prime}_{1}\in S$. Recall that $v_{1}v^{\prime}_{1}$ cannot be medium because the $S$-component to which $C$ belongs is not an odd cycle. Furthermore, $C$ can receive charge only by (R2). Observe that none of $C_{v^{\prime}_{0}}$ and $C_{v^{\prime}_{2}}$ is an odd cycle with $S$-degree [math] for otherwise adding the edge $v_{0}v^{\prime}_{0}$ or $v_{2}v^{\prime}_{2}$ to $S$ would contradict its choice. Therefore $C$ can receive charge only through $v_{3}$ or $v_{4}$, for a total of at most $2/5$. It follows that if $C$ sends $1/5$ through each of $v_{0}$ and $v_{2}$, due to rules (R3) and (R4), then its final charge surely is at most $\frac{4}{5}\cdot\left\lvert V(G)\right\rvert$. Let us identify precisely when $C$ sends charge through $v_{2}$, the case for $v_{0}$ being identical. If $C_{v^{\prime}_{2}}$ is not a cycle of length $5$ with $S$-degree $1$, then (R3) applies. So assume that $C_{v^{\prime}_{2}}$ is a cycle of length $5$ with $S$-degree $1$, written $v^{\prime}_{2}u_{1}u_{2}u_{3}u_{4}$. None of $u_{1}$ and $u_{4}$ is incident to an edge in $S$, for otherwise adding the edge $v_{2}v^{\prime}_{2}$ to $S$ would contradict its maximality. So we can assume without loss of generality that $u_{2}u^{\prime}_{2}\in S$. Now we observe that if $C_{u^{\prime}_{2}}\neq C$, then $C_{u^{\prime}_{2}}$ cannot be a cycle of $S$-degree $1$, for otherwise $(S\setminus\{u_{2}u^{\prime}_{2}\})\cup\{v_{2}v^{\prime}_{2}\}$ would contradict the fact that $S$, among all edge-selections of maximum order, creates the maximum number of cycles with $S$-degree $2$. Therefore in this case $C_{u^{\prime}_{2}}$ is a cycle of $S$-degree $2$ and hence $C$ sends $1/5$ to $C_{u^{\prime}_{2}}$ by (R4). It follows that the only case where $C$ does not send charge through $v_{2}$ is when $C_{u^{\prime}_{2}}=C$ and hence $v^{\prime}_{1}=u_{2}$ (or $v^{\prime}_{1}=u_{3}$). In this situation, we argue that $C$ cannot receive charge through $v_{3}$. Indeed $C$ can receive charge through $v_{3}$ only by (R2), which applies if and only if $C_{v^{\prime}_{3}}$ is a cycle of length $5$ and $S$-degree [math], as illustrated in Figure 3. In this case, $(S\setminus\{v_{1}v^{\prime}_{1}\})\cup\{v_{2}v^{\prime}_{2},v_{3}v^{\prime}_{3}\}$ would contradict the maximality of $S$. Consequently, we proved that either $C$ sends $1/5$ through $v_{2}$ or $C$ receives nothing through $v_{3}$. By symmetry of the roles played by $v_{2}$ and $v_{0}$, either $C$ sends $1/5$ through $v_{0}$ or $C$ receives nothing through $v_{4}$. Since $C$ can receive charge only through $v_{3}$ and $v_{4}$, we therefore conclude that the final charge of $C$ is not greater than its charge before applying (R2)–(R4), that is $4$.

It remains to deal with cycles with $S$-degree $2$. Let us write $C=v_{0}\dotso v_{2k}$, where $k\geqslant 2$, with $v_{1}v^{\prime}_{1}\in S$ and $v_{2}v^{\prime}_{2}\in S$. Suppose first that $G_{K}$ is not an odd cycle, where $K$ is the $S$-component to which $C$ belongs. In particular, none of $v_{1}v^{\prime}_{1}$ and $v_{2}v^{\prime}_{2}$ is medium. In this case, we show that the charge of $C$ is at most $\frac{4}{5}\cdot(2k+1)$, with equality only if $k=2$. Observe that, for each $i\in\{1,2\}$, the cycle $C$ can receive some charge through $v_{i}$ only if $C_{v^{\prime}_{i}}$ is a cycle of length $5$ and $S$-degree $1$. Further, according to (R4), the cycle $C$ can receive at most $2/5$ through $v_{i}$, because only the two vertices at distance two from $v^{\prime}_{i}$ on $C_{v^{\prime}_{i}}$ can be involved in an application of (R4). As a result, it is enough to prove that if $C$ receives $2/5$ through $v_{i}$, then $C$ does not receive any charge through $v_{6-3i}$. More explicitly, if $C$ receives $2/5$ through $v_{1}$ then $C$ receives nothing through $v_{3}$; and if $C$ receives $2/5$ through $v_{2}$ then $C$ receives nothing through $v_{0}$. In total, the final charge of $C$ would then be at most $3+(2k+1)/5$, which is at most $\frac{4}{5}\cdot(2k+1)$ since $k\geqslant 2$, with equality if and only if $k=2$.

Let us establish the assertion above: assume without loss of generality that $C$ receives $2/5$ through $v_{2}$ because of (R4). Writing $C_{v^{\prime}_{2}}=v^{\prime}_{2}u_{1}u_{2}u_{3}u_{4}$, we deduce that each of $C_{u^{\prime}_{2}}$ and $C_{u^{\prime}_{3}}$ is a cycle of length $5$ and $S$-degree $1$. In addition, by the definition of (R4) for each $i\in\{2,3\}$ the vertex of $C_{u^{\prime}_{i}}$ incident to an edge in $S$ is a neighbour of $u^{\prime}_{i}$ on $C_{u^{\prime}_{i}}$. Suppose first that $C_{u^{\prime}_{2}}\neq C_{u^{\prime}_{3}}$, as illustrated in Figure 4. Then $(S\setminus\{v_{2}v^{\prime}_{2}\})\cup\{u_{2}u^{\prime}_{2},u_{3}u^{\prime}_{3}\}$ contradicts the maximality of $S$. If, on the contrary, $C_{u^{\prime}_{2}}=C_{u^{\prime}_{3}}$, then without loss of generality we may write $C_{u^{\prime}_{2}}=w_{0}u^{\prime}_{2}w_{1}w_{2}u^{\prime}_{3}$ with $w_{0}w^{\prime}_{0}\in S$, as illustrated in Figure 5. Now, if $C$ receives charge through $v_{0}$ then $C_{v^{\prime}_{0}}$ is a cycle of length $5$ and $S$-degree [math] or $1$. In the latter case, we notice that the vertex of $C_{v^{\prime}_{0}}$ incident to an edge in $S$ is consecutive to $v^{\prime}_{0}$ on $C_{v^{\prime}_{0}}$ and, consequently in both cases $C_{v^{\prime}_{0}}\neq C_{u^{\prime}_{2}}$. From this and the fact that $v^{\prime}_{0}\notin\{u^{\prime}_{2},u^{\prime}_{3}\}$, we deduce that in any case $(S\setminus\{v_{2}v^{\prime}_{2}\})\cup\{v_{0}v^{\prime}_{0},u_{2}u^{\prime}_{2}\}$ contradicts the choice of $S$.

It remains to deal with the case where $C$ belongs to an $S$-component $K$ such that $G_{K}$ is an odd cycle. In this case, we prove the sum of the charges of all cycles in $K$ to be less than $\frac{4}{5}$ times the number of vertices belonging to cycles in $K$. Every cycle in $K$ is incident to exactly two edges in $S$. It follows that none of these cycles receive charge by (R4). Moreover, exactly one edge in $S$ associated with $K$ is medium. Consequently, if $C\in K$ is not incident to a medium edge that belongs to $S$, then the final charge of $C$ is at most $3+\frac{1}{5}\cdot(\left\lvert V(C)\right\rvert-2)$. If $C$ is one of the two cycles incident to the medium edge in $S$ associated with $K$, then the final charge of $C$ is at most $\frac{7}{2}+\frac{1}{5}\cdot(\left\lvert V(C)\right\rvert-2)$. It now suffices to sum these quantities over the cycles in $K$: let us write $K=\{C_{1},\dotsc,C_{t}\}$ where $t$ is an odd number at least $3$. Setting $\ell_{i}\coloneqq\left\lvert V(C_{i})\right\rvert$ for each $i\in\{1,\dotsc,t\}$, the sum of the final charges of the cycles in $K$ is at most

[TABLE]

It only remains to show that this last quantity is less than $\frac{4}{5}\cdot\sum_{i=1}^{t}\ell_{i}$, that is,

[TABLE]

Since each cycle in $F$ has length at least $5$, one has $\sum_{i=1}^{t}\ell_{i}\geqslant 5t$, and hence (1) holds because $t\geqslant 3$.

Looking at the above inequalities, we observe that as soon as $F$ contains a cycle $C$ of length different from $5$, then the number of medium edges is less than $\frac{4}{5}\cdot\left\lvert V(G)\right\rvert$. Therefore, the number of medium edges obtained is strictly less than $\frac{4}{5}\cdot\left\lvert V(G)\right\rvert$ unless $F$ contains only cycles of length $5$. As it turns out, it has been proved [DMP08] that every connected bridgeless cubic graph different from the Petersen graph admits a $2$-factor containing a cycle of length different from $5$.111See also [DK] for a different and short argument. This concludes the proof of Theorem 1.3.

3. Further work

We point out that, using more involved discharging rules and a lengthier analysis, one can show that there exists a positive $\varepsilon$ (which we did not try to optimise) such that for every connected bridgeless cubic graph $G$ different from the Petersen graph, there exists a $4$-edge-colouring yielding at most $(4/5-\varepsilon)\left\lvert V(G)\right\rvert$ medium edges.

It seems stimulating to try and obtain upper bounds for the least possible number of medium edges in a $k$-edge-colouring of a bridgeless cubic graph. As we saw, this number is [math] if $k\geqslant 7$ and at most $\frac{4}{5}\cdot\left\lvert V(G)\right\rvert$ if $k=4$. Since the Petersen colouring conjecture states that this number should be [math] when $k=5$, can one obtain at least a sub-linear (in the number of vertices) upper bound in this case? What can be proved when $k=6$?