On the existence of factors intersecting sets of cycles in regular graphs

A recent result by Kardoš, Máčajová and Zerafa [J. Comb. Theory, Ser. B. 160 (2023) 1–14] related to the famous Berge-Fulkerson conjecture implies that given an arbitrary set of odd pairwise edge-disjoint cycles, say $\mathcal O$, in a bridgeless cubic graph, there exists a $1$-factor intersecting all cycles in $\mathcal O$ in at least one edge. This remarkable result opens up natural generalizations in the case of an $r$-regular graph $G$ and a $t$-factor $F$, with $r$ and $t$ being positive integers. In this paper, we start the study of this problem by proving necessary and sufficient conditions on $G$, $t$ and $r$ to assure the existence of a suitable $F$ for any possible choice of the set $\mathcal O$. First of all, we show that $G$ needs to be $2$-connected. Under this additional assumption, we highlight how the ratio $\frac{t}{r}$ seems to play a crucial role in assuring the existence of a $t$-factor $F$ with the required properties by proving that $\frac{t}{r} \geq \frac{1}{3}$ is a further necessary condition. We suspect that this condition is also sufficient, and we confirm it in the case $\frac{t}{r}=\frac{1}{3}$, generalizing the case $t=1$ and $r=3$ proved by Kardoš, Máčajová, Zerafa, and in the case $\frac{t}{r}=\frac{1}{2}$ with $t$ even. Finally, we provide further results for the case where even cycles are included.

💡 Research Summary

The paper investigates a natural generalisation of a recent result by Kardoš, Máčajová and Zerafa (2023) concerning the Berge–Fulkerson conjecture. Their theorem states that in any bridgeless cubic (3‑regular) graph, for an arbitrary collection 𝒪 of pairwise edge‑disjoint odd cycles there exists a 1‑factor intersecting every cycle of 𝒪 in at least one edge. The authors ask whether an analogous statement holds for an r‑regular graph G and a t‑factor F (1 ≤ t ≤ r − 2).

Main question (Question 1.6).
Given a 2‑connected r‑regular graph G and any set 𝒪 of pairwise edge‑disjoint odd cycles, does there always exist a t‑factor F such that F is edge‑disjoint from every cycle in 𝒪?

The paper first establishes two necessary conditions.

1. 2‑connectivity. Using explicit constructions (Theorem 2.1), the authors show that if G has a cut‑vertex, one can embed a set 𝒪 for which every t‑factor must intersect at least one cycle of 𝒪. Hence the graph must be 2‑connected.

2. Ratio t⁄r ≥ 1⁄3. Theorem 3.1 builds an r‑connected, r‑regular graph consisting of three copies of a complete bipartite graph K_{r‑2,r} together with a family of r triangles (the set 𝒪). By counting edges of a hypothetical t‑factor that avoids 𝒪, the authors derive the inequality 3(r‑2)t + r ≤ t(3r‑3), which simplifies to t⁄r ≥ 1⁄3. Thus any positive answer to Question 1.6 requires t⁄r ≥ 1⁄3.

Having identified the necessary constraints, the authors turn to sufficient conditions. They prove the conjectured condition is sufficient in two important families.

Theorem 1.7 (t⁄r = 1⁄3).
Let t be any positive integer and let G be a 2‑connected 3t‑regular graph. For any set 𝒪 of pairwise edge‑disjoint odd cycles and any prescribed edge e∈E(G), there exists a t‑factor F containing e such that F meets each cycle of 𝒪 in a non‑empty matching. The proof extends the cubic case (t = 1) proved in