# (m,n)-cycle covers

 Importance: High ✭✭✭
 Author(s): Celmins, Uldis A. Preissmann, Myriam
 Subject: Graph Theory » Basic Graph Theory » » Cycles
 Keywords: cover cycle
 Posted by: mdevos on: March 7th, 2007

\begin{conjecture} Every \Def[bridgeless]{bridge (graph theory)} graph has a (5,2)-cycle-cover. \end{conjecture}

Definition: If $G=(V,E)$ is a graph, a binary cycle of $G$ is a set $C \subseteq E$ such that every vertex of the graph $(V,C)$ has even degree. An $(m,n)$-cycle-cover of $G$ is a list $L$ consisting of $m$ cycles so that every edge of $G$ is contained in exactly $n$ of these cycles.

Since every binary cycle can be written as a disjoint union of edge sets of ordinary cycles, the above conjecture is a strengthening of the cycle double cover conjecture. For positive integers $m,n$ it is natural to ask what family of graphs have $(m,n)$-cycle-covers. The following chart gives some information about this question for small values of $m$ and $n$. A "yes" in the $(m,n)$ box indicates that every graph with no cut-edge has an $(m,n)$-cycle-cover. A "no" indicates that no graph has an $(m,n)$-cycle-cover. A more detailed explanation of the entries in this chart appears below it.

m
n

 2 3 4 5 6 7 8 9 10 11 2 Eulerian NZ 4-flow NZ 4-flow 5CDC conj open 4 no Eulerian 5 post. sets B-F conj yes [BJJ] yes 6 no Eulerian 7 post. sets ? ? yes [F] yes

We did not include odd values of n, since any graph with an $(m,n)$-cycle-cover for an odd integer $n$ must be \Def[Eulerian]{eulerian graph}. The entry "NZ 4-flow" is short for \Def[nowhere-zero]{nowhere-zero flows} 4-flow. Thus, our chart indicates that ($G$ has a nowhere-zero 4-flow) if and only if ($G$ has a $(3,2)$-cycle-cover) if and only if ($G$ has a $(4,2)$-cycle-cover). These equivalences were discovered by Tutte [Tu].

Two of the $(m,n)$ boxes are conjectures. The 5CDC conj is the 5 cycle double cover conjecture and the B-F conjecture is the \OPref[Berge-Fulkerson conjecture]{the_berge_fulkerson_conjecture}. In both of these cases, the conjecture is equivalent to the assertion that every graph with no cut-edge has an $(m,n)$-cycle-cover (i.e. it would be accurate to put a "yes" in the corresponding. box). For emphasis, we state the Berge-Fulkerson conjecture again below in this new form.

\begin{conjecture}[The Berge-Fulkerson conjecture] Every graph with no cut-edge has a (6,4)-cycle-cover. \end{conjecture}

The fact that the above conjecture is equivalent to the usual statement of the Berge-Fulkerson conjecture was discovered by Jaeger [J]. For cubic graphs this equivalence is easy to see, since $M_1,\ldots,M_6$ satisfy the Berge-Fulkerson conjecture if and only if $E\M_1,\ldots,E\M_6$ is a $(6,4)$-cycle-cover. By Jaeger's argument, the weak Berge-Fulkerson conjecture is equivalent to the statement that there exists a fixed integer $k$ so that every bridgeless graph has a $(3k,2k)$-cycle-cover.

A postman set is a set of edges $J$ such that $E(G)\J$ is a cycle. The entry "k post. sets" in the $(k,k-1)$ box of the above chart indicates that a graph G has a $(k,k-1)$-cycle-cover if and only if it is possible to partition the edges of $G$ into $k$ postman sets. This equivalence follows immediately from the definition. Rizzi's Packing postman sets conjecture is thus equivalent to the following conjecture on cycle-covers.

\begin{conjecture}[the packing postman sets conjecture] If every odd edge-cut of $G$ has size $\ge 2k+1$, then $G$ has a $(2k+1,2k)$-cycle-cover. \end{conjecture}

Next we turn our attention to orientable cycle covers. If $H$ is a directed graph a map $\phi:E(H) \rightarrow \{-1,0,1\}$ is a 2-flow or an oriented cycle if at every vertex of $H$, the sum of $\phi$ on the incoming edges is equal to the sum of $\phi$ on the outgoing edges. It is easy to see that the support of a 2-flow is always a cycle. Furthermore, for any oriented cycle, there is a list $L$ of edge-disjoint circuits with directions so that an edge $e$ is forward (backward) in a circuit of $L$ if and only if $\phi(e)=1$ ($\phi(e)=-1$). So as in the unoriented case, an oriented cycle may be viewed as the edge-disjoint union of oriented circuits. For an even integer $n$, a $(m,n)$-oriented-cycle-cover of a graph $G$ is a list of $m$ oriented cycles so that every edge of $G$ appears as a forward edge $n/2$ times and a backward edge $n/2$ times. The following conjecture is the common generalization of the orientable cycle double cover conjecture and the five cycle double cover conjecture. It is due to Archdeacon and Jaeger.

\begin{conjecture}[The orientable five cycle double cover conjecture] Every graph without a cut-edge has a (5,2)-oriented-cycle-cover. \end{conjecture}

Considerably less is known about $(m,n)$-oriented-cycle-covers. We sumarize some of what is known for small values of $m$ and $n$ below.

m
n

 2 3 4 5 6 7 8 9 10 11 2 Eulerian NZ 3-flow NZ 4-flow O5CDC conj open 4 no Eulerian ? ? ? conj. open 6 no Eulerian ? ? ? ? yes [DG]

Every graph with an $(m,n)$-cycle-cover also has a $(2m,2n)$-oriented-cycle-cover obtained by taking two copies of each cycle with opposite orientations. Thus, by Bermond, Jackson, and Jaeger's $(7,4)$-cycle-cover theorem, every bridgeless graph with no has a $(14,8)$-oriented-cycle-cover. DeVos and Goddyn have observed that Seymour's 6-flow theorem can be used to construct an $(11,6)$-oriented-cycle-cover for every bridgeless graph. By combining these, we find that for every even integer $n \ge 10$ there exists an $m$ so that every bridgeless graph has an $(m,n)$-oriented-cycle-cover. This question is still open for $n=2,4,10$.

The following conjecture appears in the above chart. \begin{conjecture}[The orientable eight cycle four cover conjecture] Every graph with no cut-edge has a (8,4)-oriented-cycle-cover. \end{conjecture}

This conjecture may be viewed as a sort of oriented version of the Berge-Fulkerson conjecture. To see this analogy, note that ($G$ has a nowhere-zero 4-flow) if and only if ($G$ has a $(3,2)$-cycle-cover) if and only if ($G$ has a $(4,2)$-oriented-cycle-cover). The Berge-Fulkerson conjecture and the above conjecture assert respectively that every bridgeless graph has a $(6,4)$-cycle-cover and a $(8,4)$-oriented-cycle-cover (i.e. a cover with double the parameters which are equivalent to a nowhere-zero 4-flow). As with most of the conjectures in this area, the above conjecture is trivially true for graphs with nowhere-zero 4-flows and it holds for the Petersen graph.

## Bibliography

[A] D. Archdeacon, Face coloring of embedded graphs. J. Graph Theory, 8(1984), 387-398.

[BJJ] J.C. Bermond, B. Jackson, and F. Jaeger, Shortest covering of graphs with cycles, J. Combinatorial Theory Ser. B 35 (1983), 297-308. \MRhref{0735197}

*[C] A. U. Celmins, On cubic graphs that do not have an edge-3-colouring, Ph.D. Thesis, Department of Combinatorics and Optimization, University of Waterloo, Waterloo, Canada, 1984.

[F] G. Fan, Integer flows and cycle covers, J. Combinatorial Theory Ser. B 54 (1992), 113-122. \MRhref{1142267}

[J] F. Jaeger, Flows and Generalized Coloring Theorems in Graphs, J. Combinatorial Theory Ser. B 26 (1979) 205-216. \MRhref{0532588}

[J88] F. Jaeger, Nowhere zero flow problems. Selected Topics in Graph Theory 3 (L.W.Beineke and R.J.Wilson eds.), Academic Press, London (1988), 71-95.

*[P] M. Preissmann, Sur les colorations des arêtes des graphes cubiques, Thèse de 3ème cycle, Grenoble (1981) .

[T54] W.T. Tutte, A Contribution on the Theory of Chromatic Polynomials, Canad. J. Math. 6 (1954) 80-91. \MRhref{0061366}

[T66] W.T. Tutte, On the Algebraic Theory of Graph Colorings, J. Combinatorial Theory 1 (1966) 15-50. \MRhref{0194363}

* indicates original appearance(s) of problem.