Small feedback vertex sets in planar digraphs

Let $G$ be a directed planar graph on $n$ vertices, with no directed cycle of length less than $g\ge 4$. We prove that $G$ contains a set $X$ of vertices such that $G-X$ has no directed cycle, and $|X|\le \tfrac{5n-5}9$ if $g=4$, $|X|\le \tfrac{2n-5}4$ if $g=5$, and $|X|\le \tfrac{2n-6}{g}$ if $g\ge 6$. This improves recent results of Golowich and Rolnick.

if g = 4, |X| 2n− 5 4 if g = 5, and |X| 2n−6 g if g 6. This improves recent results of Golowich and Rolnick. A directed graph G (or digraph, in short) is said to be acyclic if it does not contain any directed cycle. The digirth of a digraph G is the minimum length of a directed cycle in G (if G is acyclic, we set its digirth to +∞). A feedback vertex set in a digraph G is a set X of vertices such that G−X is acyclic, and the minimum size of such a set is denoted by τ (G). In this short note, we study the maximum f g (n) of τ (G) over all planar digraphs G on n vertices with digirth g. Harutyunyan [1,4] conjectured that f 3 (n) 2n 5 for all n. This conjecture was recently refuted by Knauer, Valicov and Wenger [5] who showed that f g (n) n−1 g−1 for all g 3 and infinitely many values of n. On the other hand, Golowich and Rolnick [3] recently proved that f 4 (n) 7n 12 , f 5 (n) 8n 15 , and f g (n) 3n−6 g for all g 6 and n. Harutyunyan and Mohar [4] proved that the vertex set of every planar digraph of digirth at least 5 can be partitioned into two acyclic subgraphs. This result was very recently extended to planar digraphs of digirth 4 by Li and Mohar [6], and therefore f 4 (n) n 2 . This short note is devoted to the following result, which improves all the previous upper bounds for g 5 (although the improvement for g = 5 is rather minor). Due to the very recent result of Li and Mohar [6], our result for g = 4 is not best possible (however its proof is of independent interest and might lead to further improvements).
and for all g 6, f g (n) 2n−6 g . In a planar graph, the degree of a face F , denoted by d(F ), is the sum of the lengths (number of edges) of the boundary walks of F . In the proof of Theorem 1, we will need the following two simple lemmas. Proof. Assume that H has n vertices, m edges, f faces, and f 6 faces of degree at least 6. Let N be the sum of the degrees of the faces of H, plus twice the sum of the degrees of the vertices of V . Observe that N = 4m, so, by Euler's formula, N 4n + 4f − 8. The sum of degrees of the faces of H is at least 4(f − f 6 ) + 6f 6 = 4f + 2f 6 , and since each vertex of V has degree at least 2, the sum of the degrees of the vertices of V is at least 2|V |. Therefore, 4f + 2f 6 + 4|V | 4n + 4f − 8. It follows that f 6 2|U | − 4, as desired.
Lemma 3. Let G be a connected planar graph, and let S = {F 1 , . . . , F k } be a set of k faces of G, such that each F i is bounded by a cycle, and these cycles are pairwise vertex-disjoint.
where the first sum varies over faces F of G not contained in S.
Proof. Let n, m, and f denote the number of vertices, edges, and faces of G, respectively. It follows from Euler's formula that the sum of 3d(F ) − 6 over all faces of G is equal to 6m − 6f = 6n − 12 We are now able to prove Theorem 1.
Proof of Theorem 1. We prove the result by induction on n 3. Let G be a planar digraph with n vertices and digirth g 4. We can assume without loss of generality that G has no multiple arcs, since g 4 and removing one arc from a collection of multiple arcs with the same orientation does not change the value of τ (G). We can also assume that G is connected, since otherwise we can consider each connected component of G separately and the result clearly follows from the induction (since g 4, connected components of at most 2 vertices are acyclic and can thus be left aside). Finally, we can assume that G contains a directed cycle, since otherwise τ (G) = 0 min{ 5n−5 9 , 2n−5 4 , 2n−6 g } (since n 3). Let C be a maximum collection of arc-disjoint directed cycles in G. Note that C is non-empty. Fix a planar embedding of G. For a given directed cycle C of C, we denote by C the closed region bounded by C, and byC the interior of C. It follows from classical uncrossing techniques (see [2] for instance), that we can assume without loss of generality that the directed cycles of C are pairwise non-crossing, i.e. for any two elements C 1 , C 2 ∈ C, eitherC 1 andC 2 are disjoint, or one is contained in the other. We define the partial order on C as follows: C 1 C 2 if and only ifC 1 ⊆C 2 . Note that naturally defines a rooted forest F with vertex set C: the roots of each of the components of F are the maximal elements of , and the children of any given node C ∈ F are the maximal elements C C distinct from C (the fact that F is indeed a forest follows from the non-crossing property of the elements of C).
Consider a node C of F, and the children C 1 , . . . , C k of C in F. We define the closed region R C = C − 1 i kC i . Let φ C be the sum of 3d(F ) − 6, over all faces F of G lying in R C . Claim 4. Let C 0 be a node of F with children C 1 , . . . , C k . Then φ C 0 3 2 (g − 2)k + 3 2 g. Moreover, if g 6, then φ C 0 3 2 (g − 2)k + 3 2 g + 3.
Consider the following planar bipartite graph H: the vertices of the first partite set of H are the directed cycles C 0 , C 1 , . . . , C k , the vertices of the second partite set of H are the vertices of G lying in at least two cycles among C 0 , C 1 , . . . , C k , and there is an edge in H between some cycle C i and some vertex v if and only if v ∈ C i in G (see Figure 1). Observe that H has a natural planar embedding in which all internal faces have degree at least 4. Since k 1 and at least two of the cycles C 0 , . . . , C k intersect, the outerface also has degree at least 4. Note that the faces F 1 , . . . , F t of H are in one-to-one correspondence with the maximal subsets D 1 , . . . , D t of R C 0 whose interior is connected. Also note that each face of G ∩ R C 0 is in precisely one region D i and each arc of k i=0 C i (i.e. each arc on the boundary of R C 0 ) is on the boundary of precisely one region D i . For each region D i , let i be the number of arcs on the boundary of D i , and observe Figure 1. The region R C 0 (in gray) and the planar bipartite graph H.
A region D i with i 4 is said to be of type 1, and we set T 1 = {1 i t | D i is of type 1}. Since for any 4 we have 3 − 6 3 2 , it follows from the paragraph above that the regions D i of type 1 satisfy φ D i 3 i

.
Let D i be a region that is not of type 1. Since G is simple, i = 3. Assume first that D i is bounded by (parts of) two directed cycles of C (in other words, D i corresponds to a face of degree four in the graph H). In this case we say that D i is of type 2 and we set T 2 = {1 i t | D i is of type 2}. Then the boundary of D i consists in two consecutive arcs e 1 , e 2 of some directed cycle C + of C, and one arc e 3 of some directed cycle C − of C. Since g 4, these three arcs do not form a directed cycle, and therefore their orientation is transitive. It follows that |C + | g + 1, since otherwise the directed cycle obtained from C + by replacing e 1 , e 2 with e 3 would have length g − 1, contradicting that G has digirth at least g. Consequently, k i=0 |C i | (k + 1)g + |T 2 |. If a region D i is not of type 1 or 2, then i = 3 and each of the 3 arcs on the boundary of D i belongs to a different directed cycle of C. In other words, D i corresponds to some face of degree 6 in the graph H. Such a region D i is said to be of type 3, and we set T 3 = {1 i t | D i is of type 3}. It follows from Lemma 2 that the number of faces of degree at least 6 in H is at most 2(k + 1) − 4. Hence, we have Using these bounds on |T 2 | and |T 3 |, together with the fact that for any , we obtain: k + 3 2 g + 3, as desired. This concludes the proof of Claim 4.
Let C 1 , . . . , C k∞ be the k ∞ maximal elements of . We denote by R ∞ the closed region obtained from the plane by removing k∞ i=1C i . Note that each face of G lies in precisely one of the regions R C (C ∈ C) or R ∞ . Let φ ∞ be the sum of 3d(F ) − 6, over all faces F of G lying in R ∞ . A proof similar to that of Claim 4 shows that φ ∞ 3 2 k ∞ (g − 2) + 3, and if g 6, then φ ∞ 3 2 k ∞ (g − 2) + 6. We now compute the sum φ of 3d(F )−6 over all faces F of G. By Claim 4, If g 6, a similar computation gives φ 3g|C| + 6. On the other hand, it easily follows from Euler's formula that φ = 6n − 12. Therefore, |C| 2n−5 g−1 , and if g 6, then |C| 2n−6 g . Let A be a set of arcs of G of minimum size such that G − A is acyclic. It follows from the Lucchesi-Younger theorem [7] (see also [3]) that |A| = |C|. Let X be a set of vertices covering the arcs of A, such that X has minimum size. Then G − X is acyclic. If g = 5 we have |X| |A| = |C| 2n−5 4 and if g 6, we have |X| |A| = |C| 2n−6 g , as desired. Assume now that g = 4. In this case |A| = |C| 2n−5 3 . It was observed by Golowich and Rolnick [3] that |X| 1 3 (n + |A|) (which easily follows from the fact that any graph on n vertices and m edges contains an independent set of size at least 2n 3 − m 3 ), and thus, |X| 5n−5 9 . This concludes the proof of Theorem 1.

Final remark
A natural problem is to determine the precise value of f g (n), or at least its asymptotical value as g tends to infinity. We believe that f g (n) should be closer to the lower bound of n−1 g , than to our upper bound of 2n−6 g . For a digraph G, let τ * (G) denote the the infimum real number x for which there are weights in [0, 1] on each vertex of G, summing up to x, such that for each directed cycle C, the sum of the weights of the vertices lying on C is at least 1. Goemans and Williamson [2] conjectured that for any planar digraph G, τ (G) 3 2 τ * (G). If a planar digraph G on n vertices has digirth at least g, then clearly τ * (G) n g (this can be seen by assigning weight 1/g to each vertex). Therefore, a direct consequence of the conjecture of Goemans and Williamson would be that f g (n) 3n 2g .