Jones' Conjecture in subcubic graphs

We confirm Jones' Conjecture for subcubic graphs. Namely, if a subcubic planar graph does not contain $k+1$ vertex-disjoint cycles, then it suffices to delete $2k$ vertices to obtain a forest.


Introduction
We investigate the connection between the maximum number of vertex-disjoint cycles in a graph and the minimum number of vertices whose deletion results in a cycle-free graph, i.e. a forest. A cycle packing of a (multi)graph G is a set of vertex disjoint cycles that appear in G as subgraphs. We denote the maximum size of a cycle packing of G by cp(G).
A feedback vertex set of a (multi)graph G is a set S of vertices such that G − S is a forest. We denote an arbitrary minimum feedback vertex set of G as F V S(G) and denote its size by fvs(G).
Erdős and Pósa [4] showed that there is a constant c such that for any graph G, fvs(G) c · cp(G) log cp(G), and that this upper-bound is tight for some graphs. This seminal result led to rich developments. Two main directions: determining which structures can replace "cycles" there (free to increase the bounding function), and improving the bounding functions. Most notably, in the first line of research, Cames van Banteburg et al. [1] proved that "cycle" can directly be replaced with "minor of a given planar graph", which was earlier showed by Robertson and Seymour [9] with a worse bounding function.
We focus on the second line of research, and are interested in the best possible bound in the original theorem, when restricted to the specific case of planar graphs. A drastic improvement is then possible, and we were motivated by the following elusive conjecture.
Note that Conjecture 1 is tight for wheels or for the dodecahedron. Currently, the best known bound is that every planar graph G satisfies fvs(G) 3 · cp(G), as proved independently by Chappel et al. [2], Chen et al. [3], and Ma et al. [7].
In his PhD Thesis, Munaro [8] considered the case of subcubic graphs and made significant progress. Here we complete the case, and prove that Jones' Conjecture holds for subcubic graphs.
2 Proof of Theorem 2

Notation
A multigraph is simple if it has no loops or multi-edges. In this case, we simply refer to it as a graph.
Let G = (V, E) be a (multi)graph. For W ⊆ V , we denote by G[W ] the subgraph of G induced by W , and by G − W the subgraph of G induced by Graph is called cubic if all of its vertices have degree exactly 3. Graph is called subcubic all of its vertices have degree at most 3.
A (multi)graph is k-connected if it has at least k + 1 vertices and the removal of at most k − 1 vertices leaves the graph connected. A (multi)graph is k-edge-connected if the removal of at most k − 1 edges leaves the graph connected. Note that a subcubic (multi)graph with at least k+1 vertices is k-connected if and only if it is k-edge-connected. In a connected (multi)graph, a separating edge or bridge is an edge the removal of which disconnects the graph.
A (multi)graph is essentially 4-edge-connected if the removal of at most three edges does not yield two components with at least two vertices each. A (multi)graph is cyclically 4-edge-connected if the removal of at most three edges does not yield two components that both contain a cycle. For a cubic (multi)graph, these last two notions are equivalent.

Proof
We proceed by contradiction. Let G be a counter-example to Theorem 2 with the minimum number of vertices. To obtain a contradiction, we first argue that G is a simple graph that is essentially 4-edge-connected, as follows. Proof. While Claims 4, 5, 6 are known and easy properties of a minimum counter-example to Jones' conjecture on subcubic graphs (see e.g. [8]), we include their proofs because we believe they may constitute a useful warm-up. The uninterested reader may skip them.
First the result is easily checked on small multigraphs, so we can assume that G has at least 4 vertices. Hence, for k 3, G is k-connected if and only if it is k-edge connected. Proof. Suppose G has a vertex v with degree at most 1. Then G − v satisfies Jones' Conjecture by minimality of G. As no cycle of G contains v, fvs(G − v) = fvs(G) and cp(G − v) = cp(G), therefore G also satisfies Jones' Conjecture, a contradiction.
Suppose G has a vertex v with degree 2, and let u and w be the two neighbors of v. Then G = G − v + uw satisfies Jones' Conjecture, so fvs(G ) 2 · cp(G ). The cycles of G are in bijection with the cycles of G , by exchanging the edges uv and vw and an edge uw when appropriate. Hence cp(G) = cp(G ). Moreover, if S is a feedback vertex set of G that does not contain v, then S is a feedback vertex set of G , and if S is a feedback vertex set of G that contains v, then (S \ {v}) ∪ {u} is a feedback vertex set of G . Thus fvs(G) fvs(G ) 2 · cp(G ) = cp(G), and G satisfies Jones' Conjecture, a contradiction. Hence G is cubic.
Proof. Suppose that G is not 2-connected. As G is cubic, that means that G is not 2edge-connected. Let e be a separating edge of G. Both components G 1 and G 2 of G − e verify Jones' Conjecture by minimality of G. Since e is separating, it is not in any cycle of G. The union of any feedback vertex set of G 1 and any feedback vertex set of G 2 is a feedback vertex set of G, so fvs(G) fvs(G 1 ) + fvs(G 2 ). The union of any cycle packing of G 1 and any cycle packing of G 2 is a cycle packing of G, so cp(G) cp(G 1 ) + cp(G 2 ). Therefore G satisfies Jones' Conjecture, a contradiction.
Proof. Assume that it is not 3-connected, and thus not 3-edge-connected. Let e = u 1 u 2 and e = v 1 v 2 be a 2-edge-cut, where u 1 and v 1 are in the same connected component of Note that this may lead to a double (or even triple) edge. See Figure 1 for an illustration. By minimality of G, we know that G 1 , G 2 , G 1 and G 2 all satisfy Jones' Conjecture. Figure 1: The graphs G, G 1 , and G 2 in Claim 6.
By symmetry, cp(G 2 ) = cp(G 2 ) + 1. Therefore, every cycle packing of G 1 contains the edge u 1 v 1 and every cycle packing of G 2 contains the edge u 2 v 2 . We can thus combine a cycle packing of G 1 and a cycle packing of G 2 by making a single cycle out of those two cycles. So cp(G) = cp(G 1 ) + cp(G 2 ) + 1. However, if S 1 is a feedback vertex set of G 1 and S 2 is a feedback vertex set of G 2 , then S 1 ∪ S 2 ∪ {u 1 } is a feedback vertex set of G. Therefore fvs(G) fvs(G 1 ) + fvs(G 2 ) + 1 2 cp(G 1 ) + 2 cp(G 2 ) + 1 < 2 cp(G), a contradiction. Therefore G is 3-connected.
In particular since G is cubic, Claim 6 implies that G is a simple graph.
Proof. Assume that G is not essentially 4-edge-connected, and thus not cyclically 4-edgeconnected. Consider a non-trivial 3-edge-cut {e A , e B , e C }. Let G 1 and G 2 be the two components of G \ {e A , e B , e C }. For i ∈ {1, 2}, we define G ABC i as the graph obtained from G by contracting G 3−i into a single vertex x. See Figure 2 for an illustration. We define G AB i (resp. G AC i , G BC i ) as the graph obtained from G i by connecting with an edge vertices from G i incident to e A and e B (resp. to e A and e C for G AC i or to e B and e C for the electronic journal of combinatorics 28(4) (2021), #P4.5 G BC i ). Again, this may lead to a double edge. Note that for both values of i, all of G i , have fewer vertices than G, and thus satisfy Jones' Conjecture. First note that for both values of i, fvs(G) fvs(G ABC i ) + fvs(G 3−i ). In order to prove that let us assume without loss of generality that i = 1. Then remove F V S(G 2 ) from G. What remains from G 2 after deleting F V S(G 2 ) is a forest that could hypothetically create connections between vertices from G 1 incident to e A , e B , e C . However if we are given any tree T and its three vertices u, v, w ∈ V (T ) then P uv ∩ P vw ∩ P wu = ∅ (in fact it is always a single vertex), where P uv , P vw , P wu are sets of vertices on unique paths between corresponding vertices, so it is possible to break the connections between all three pairs of these vertices by removing a single vertex of T . Because of that we see that ) what leads to fvs(G) fvs(G ABC 1 )+fvs(G 2 ). Therefore we see that fvs(G) fvs(G ABC i )+fvs(G 3−i ) fvs(G 1 )+fvs(G 2 )+1 2 cp(G 1 )+2 cp(G 2 )+1. We also have cp(G) cp(G 1 ) + cp(G 2 ), yet fvs(G) > 2 cp(G).
It follows that for both values of i: And that: fvs(G) = fvs(G 1 ) + fvs(G 2 ) + 1 We are now ready for a closer analysis. Indeed, take without loss of generality i = 1 and x = C. Let us consider a minimum feedback vertex set S of G AB 1 . Note that in G 1 \ S, there is no path between the vertex incident to e A and the vertex incident to e B or at least one of them is in S. As a consequence, either vertex incident to e C is in S or one of them, say the vertex incident to e A , is either in S or is not in the same component as the vertex incident to e C . For any minimum feedback vertex set S of G BC 2 , we observe that S ∪ S is a feedback vertex set of G, hence the conclusion. In particular, by combining with We assume without loss of generality that z 1 = C. Note that from (iii), cp(G AB 2 ) = cp(G 2 ), hence z 2 = C, and fvs(G AB 2 ) = fvs(G 2 ) by (ii). We assume without loss of generality z 2 = A. By symmetry, cp(G BC 1 ) = cp(G 1 ) and fvs(G BC 1 ) = fvs(G 1 ). From (i) applied with i = 1 and x = A, there is y ∈ {B, C} such that fvs(G ABC−y 2 ) = fvs(G 2 ) + 1. Note that y = C, so y = B and fvs(G AC 2 ) = fvs(G 2 ) + 1. We derive from (ii) that cp(G AC 2 ) = cp(G 2 ) + 1, hence cp(G AC 1 ) = cp(G 1 ) by (iii). Again from (ii), we obtain fvs(G AC 1 ) = fvs(G 1 ). Therefore fvs(G AB 2 ) = fvs(G 2 ), fvs(G BC 1 ) = fvs(G 1 ), and fvs(G AC 1 ) = fvs(G 1 ). Now, (i) applied with i = 2 and x = C yields a contradiction.
The conclusion follows directly from Claim 7.
To obtain the desired contradiction, we combine Lemma 3 with the following very convenient theorem from Munaro [8]: Theorem 8 (Theorem 3.4.10 in [8]). If G is a simple subcubic graph which is a counterexample to Jones' Conjecture and which has the minimum number of vertices, then G is not cyclically 4-edge-connected.
The main ingredient of the proof of Theorem 8 is an explicit formula for the minimum size of a feedback vertex set in a simple cubic cyclically 4-edge-connected graph. In [8], this formula is deduced using properties of the maximum genus of such a graph. In the next section, we include a more direct proof that, despite relying on the same combinatorial backbone -the matroid matching problem -is more self-contained and highlights the key properties of highly connected cubic graphs that allows us to compute the minimum size of a feedback vertex set.
3 An alternative proof of Theorem 8 As argued with Claim 4 in the proof of Lemma 3, to prove Theorem 8 it suffices to handle simple cubic graphs that are cyclically 4-edge-connected. That is, we henceforth assume G is a simple planar cyclically 4-edge-connected cubic graph and our goal is to prove that fvs(G) 2 cp(G).
Instead of directly comparing fvs(G) and cp(G), we bound the value of each of them by a function of the number of vertices n. Note that the dual graph of G is also planar, hence it admits an independent set of size f 4 , where f is the number of faces in the plane embedding of G. Note that an independent set in the dual of G is a cycle packing in G. (In fact, it is a face packing, that is, a cycle packing where every cycle is a face.) Therefore, G satisfies cp(G) Equivalently, we need to argue that in such a graph, the maximum number of vertices in an induced forest is 3n−1 4 . To look for a largest induced forest in G, we rephrase the question as a matroid matching problem in the line graph of G. To this end, we need a few definitions.
Let H be a multigraph. A ν-pair in H is a pair of edges of H with a common endpoint. A ν-graph is a pair (H, V) where H is a multigraph and V is a partition of E(H) into ν-pairs. A cactus in (H, V) is a set X ⊆ V such that F ∈X F is a forest. Let β(H, V) be the maximum size of a cactus in (H, V).
Let G be a graph as in Theorem 9 and let L(G) be the line graph of G. Construct a ν-graph (H, V) as follows. Take V (H) = V (L(G)) = E(G). For every v ∈ V (G), let e 1 , e 2 , e 3 be the three edges of G incident with v; the line graph L(G) features a triangle e 1 e 2 e 3 . Pick arbitrarily two edges of this triangle, say e 1 e 2 and e 2 e 3 , add them to E(H) and add them as a ν-pair F v to V. The following observation is straightforward. Hence, the question of determining the maximum size of an induced forest in G is equivalent to the question of computing β(H, V).
To achieve this goal, we rely on a min-max formula by Lovász [6], following an exposition of Szigeti [10]. Let (H, V) be a ν-graph. Let P = {V 1 , . . . , V } be a partition of V (H) and let Q = {U 1 , . . . , U k } be a partition of V. For every 1 i k, let P (U i ) be the set of those indices 1 j for which there is a ν-pair in U i where one or both of the edges have an endpoint in V j . Define It is not difficult to observe that, for fixed P and Q, val(P, Q) is an upper bound on β(H, V). Lovász proved that some choice of P and Q actually yields a tight upper bound [6,10].
Recall that in our setting V (H) = E(G), so P = {V 1 , V 2 , . . . , V } is actually a partition of E(G). Furthermore, every ν-pair F v ∈ V corresponds to a vertex v ∈ V (G). By somewhat abusing the notation, we henceforth treat Q = {U 1 , U 2 , . . . , U k } as a partition of V (G). With this notation, P (U i ) is the set of indices j such that V j contains an edge (of G) incident with a vertex of U i (which is a subset of V (G)).
Consider now the following pair of partitions (P 0 , Q 0 ): • P 0 is the finest partition of E(G): each V i is a singleton set and = |E(G)| = 3n 2 . • Q 0 is the coarsest partition of V (G): k = 1 and U 1 = V (G).
Since n is even due to G being cubic, We conclude that β(H, ν) 3n− 1 4 and it remains to argue that the minimum of val(·, ·) is attained for the pair (P 0 , Q 0 ).
To this end, pick a pair (P, Q) of partitions that: • minimizes val(P, Q), • subject to the above, maximizes , and • subject to the above, minimizes k.
It remains to show that P = P 0 and Q = Q 0 . We prove this in two steps.
Let P (U i ) be the set of those indices j ∈ [ ] for which V j contains an edge incident with a vertex in U i . Similarly, let P (U i ) be the set of those indices j ∈ [ + 1] for which V j contains an edge incident with a vertex in U i . Clearly, P (U i ) P (U i ) ⊆ { , + 1}.
Let uv = e, and let a, b ∈ [k] be such that u ∈ U a and v ∈ U b , where a and b are not required to be distinct. Observe that if i ∈ [k] \ {a, b}, then P (U i ) = P (U i ). For i ∈ {a, b}, it holds that + 1 ∈ P (U i ), ∈ P (U i ) and, as P has sets V j , ∈ P (U i ). From the above observations it in particular follows that Observe that (6) and (7) imply that the only possibility for (5) to hold is that a = b and both of the following equalities hold: The above implies that |P (U a )| is even and |P (U a )| = |P (U a )| + 1, hence ∈ P (U a ). Similarly we infer that |P (U b )| is even, |P (U b )| = |P (U b )| + 1, and ∈ P (U b ). Consider now a partition Q of V (G) that is created from Q by merging U a and U b into one set U . Let P (U ) be the set of those indices j ∈ [ + 1] for which there is an edge of V j that is incident with a vertex in U ; that is, P (U ) = P (U a ) ∪ P (U b ). Recall that both and + 1 are elements of P (U a ) and P (U b ), and thus Consequently, as both |P (U a )| and |P (U b )| are even, We infer that val(P , Q ) val(P, Q) while |P | > |P|, a contradiction to the choice of P and Q. This finishes the proof of the lemma.
Lemma 13. If P = P 0 , then Q = Q 0 , that is, Q partitions V (G) into one set.
Proof. Assume the contrary, k 2. By the assumption P = P 0 , we can treat P (U i ) as the set of edges of E(G) incident with at least one vertex of U i . By reordering the sets U i if necessary, we can assume that for some 0 r k it holds that |P (U i )| is even if and only if i r. Furthermore, let δ(U i ) be the set of edges of G having exactly one endpoint in U i . Since G is cyclically-4-edge-connected, δ(U i ) 3 for every i ∈ [k] and, furthermore, δ(U i ) 4 for every i ∈ [r]. Then, as G is cubic, Hence val(P, Q) 3n 4 val(P 0 , Q 0 ) while |Q| > |Q 0 |, a contradiction.
As we argued, Lemmas 12 and 13 finish the proof of Theorem 9, which in turn implies Theorem 8.

Conclusion
Through a non-trivial combination of elementary tricks and using a nice preliminary result of [8], we were able to close the case of Jones' Conjecture for subcubic graphs.
The obvious question is whether this can be at all used to solve the whole conjecture. The reduction we have for subcubic graphs extends easily to the general setting, in the sense that a smallest counter-example to Jones' Conjecture is essentially 4-edge-connected. It is not difficult to argue in a similar way that such a graph is 3-vertex-connected. However, a much harder question is whether it is essentially 4-vertex-connected. While it still seems possible, such a result using our approach would require additional tricks. Note that being in the general setting also gives us more leeway regarding possible reductions (no need to shy away from increasing the maximum degree, as long as there are fewer vertices).
A second obstacle to generalization is that even assuming that a smallest counterexample is essentially 4-vertex-connected, Theorem 8 only deals with the subcubic case. Another argument must then be devised.
A different approach would be not to aim for the conjectured bound of 2 but simply for any bound better than the existing one of 3. Unfortunately, this does not seem conceptually much easier. Let us emphasize this: a simple discharging argument yields fvs(G) 3 cp(G) for every planar graph G, while even significant effort fails to grant a factor of (3 − ) instead of 3.
To highlight how little we understand around Jones' Conjecture, we conclude by posing the following stronger conjecture. Note that the example of many nested disjoint cycles shows that the embedding cannot be fixed. Also note that the simple discharging argument mentioned above does not imply the following conjecture with a factor of 3 instead of 2.
Conjecture 14. For every planar graph G, where fp(G) is the maximum size of a face-packing of G, i.e., a cycle-packing where, for some embedding of G, every cycle bounds a face.