On coloring digraphs with forbidden induced subgraphs

We prove a conjecture by Aboulker, Charbit, and Naserasr by showing that every oriented graph in which the out‐neighborhood of every vertex induces a transitive tournament can be partitioned into two acyclic induced subdigraphs. We prove multiple extensions of this result to larger classes of digraphs defined by a finite list of forbidden induced subdigraphs. We thereby resolve several special cases of an extension of the famous Gyárfás–Sumner conjecture to directed graphs stated by Aboulker et al.


| INTRODUCTION
All graphs and digraphs considered in this paper are simple. We say that a digraph is an oriented graph if it does not contain directed cycles of length two (digons) and we call it bioriented if each of its arcs belongs to a digon. Given a digraph D, an acyclic k-coloring of D, also referred to as a k-dicoloring, is an assignment → c V D S : ( ) of colors from a color set S of size k to the vertices such that every color class ∈ c i i S ( ), −1 induces an acyclic subdigraph of D. The dichromatic number → χ D ( ) as introduced by Erdős [6] and Neumann-Lara [8] is the smallest integer k for which an acyclic k-coloring of D exists. Given a set of (di)graphs F , we denote by F Forb ( ) ind the set of (di)graphs which do not contain an induced sub(di)graph isomorphic to a member of F . Given a class of (di)graphs , we denote by  → χ ( ) (or  χ ( ), respectively) the maximum (di) chromatic number of (di)graphs in the class (∞ if the latter is unbounded). We say that a finite set F of (di)graphs is heroic, if F Forb ( ) ind has bounded (di)chromatic number. For ∈ k , we denote by → K k the transitive tournament of order k. A classical research topic in the theory of graph coloring is to study the chromatic number of graphs with forbidden induced subgraphs, we refer to [9] for a recent survey article summarizing important results on this topic. Extending this area of research to digraphs, Aboulker, Charbit, and Naserasr [2] recently initiated the systematic study of the relation between excluded induced subdigraphs and the dichromatic number and asked the following intriguing question. Problem 1.1. Characterize the inclusionwise minimal heroic sets of digraphs.
In the following, let us mention a few related questions and results from the literature.
• Maybe the most important open problem for coloring undirected graphs with forbidden induced subgraphs, namely, the Gyárfás-Sumner Conjecture [7,10], can be restated in digraph terminology as follows: Conjecture 1.2. If a minimal heroic set F of digraphs includes → K 2 (the oriented edge), then F consists of at most three members, namely, → K 2 , a biorientation of a forest, and a biorientation of a clique.
Note that in the above setting, F Forb ( ) ind contains only bioriented graphs, whose chromatic number coincides with their dichromatic number. Hence, the above is the same as saying that the minimal heroic sets of graphs consist of a forest and a clique, which is the classical phrasing of the Gyárfás-Sumner Conjecture.
• In [4] Berger, Choromanski, Chudnovsky, Fox, Loebl, Scott, Seymour, and Thomassé studied the dichromatic number of tournaments which exclude a single fixed tournament H as an (induced) subdigraph. They defined a hero as a tournament H such that the tournaments excluding isomorphic copies of H have bounded dichromatic number. In other words, a digraph H is a hero if the set ↔ K K H { , , } 2 2 is heroic, where K 2 is the anticlique of order 2. The main result of Berger et al. in [4] is a recursive characterization of all heroes, it implies in particular that all tournaments on at most four vertices are heroes.
Motivated by Problem 1.1, Aboulker et al. [2] made the following conjecture for oriented graphs which exclude the oriented out-(or in-) star S 2 + (or S 2 − ) with two leaves, as well as the directed triangle → C 3 .  other oriented paths on four vertices. Two of them, which are called P (2, 1) + and P (2, 1) − in [2], consist of two oppositely oriented dipaths of length two and one, respectively.

| Structure of the paper
In Section 2 we investigate the structure of digraphs in the class ↔ ( ) , , ind 2 2 + 3 + and use these insights to prove Theorem 1. In Section 3 we prove Theorem 2. Finally, in Section 4 we prove Theorem 3 and we conclude with final comments in Section 5.

| Note
After the submission of this manuscript, independently discovered proofs both for Theorem 1 and Remark 5.2 (which appears in the conclusion) have appeared in the arXiv-preprint [1] by Aboulker, Aubian, and Charbit. Their proof of Theorem 1 is quite different, as they obtain and use a full structural characterization of the class ↔ ( ) , , ind 2 2 + 3 + of digraphs. In our work, however, the approach was not to obtain such a structural result, but instead to generate 2-colorings of these oriented graphs directly using local reductions.

| Notation
Given a digraph D, we denote by V D ( ) its vertex-set and by Arcs are denoted as u v ( , ), where u is the tail of the arc and v is its head. For ∈ v V D ( ) we denote by N u N u ( ), ( ) the sets of out-and in-neighbors of v in D, respectively. We generalize this notation to vertex subsets by putting k j are also arcs (indices ordered cyclically). For these more special digraphs, 2-colorability has been noted already in [2], we refer to Figure 1 for examples from this class.
We remark the following simple fact for later use.
Proof. To verify that is some newly added vertex representing U , and the following arcs: the arcs of D inside . In the following we prepare the proof of Theorem 1 with a set of useful lemmas, starting with two operations for digraphs which preserve the containment in the class ↔ ( ) Proof. Suppose towards a contradiction that ∕ D U contains a ↔ K 2 spanned by vertices x y , . Clearly we need to have Since U is an out-module, the vertices u y , 2 span a ↔ K 2 in D, which is impossible. Next, suppose there are vertices x y z , , spanning an , spans an S 2 + in D, in each case a contradiction. Finally, suppose that there is a copy of then since U is an out-module any vertex ∈ u U together with the remaining three vertices induces a W 3 + in D. If x U is not the source vertex of the W 3 + , then let x y z , , denote the other three vertices of the W 3 + , such that x is the source vertex, and Since U is an out-module, we must have ∈ u y A D ( , ) ( ). Then u y z , , are three out-neighbors of x in D and ∈ u y y z A D ( , ), ( , ) ( ). Hence, by transitivity (recall that  Proof. As z is not connected to x, adding x z ( , ) does not create a ↔ K 2 . Suppose it creates an S 2 + and let t be the vertex in the copy of S 2 ( ( ))) and thus y z t { , , } induces an S 2 + in D. Now suppose a W 3 + is created. First consider the case that x is the source vertex of this copy of W 3 + and let z a b , , be the other three vertices such that If x is not the source vertex of the copy of W 4 + , then the source vertex together with x and z induces an S 2 + . Obtaining a contradiction in each case, we conclude the proof. □ The next lemma shows the existence of out-modules with special properties.
Proof. Let M be the set of vertices of a strong component of  . 1 Then M satisfies the required properties: Consider any Then u v , are distinct outneighbors of t and thus have to be adjacent in D.
To see why M is an out-module, we use Observation 2.3. So let ∈ x y M , Then D T [ ] is a (possibly empty) transitive tournament. 1 One may obtain such a component by contracting all strong components and selecting a component corresponding to a sink in the resulting acyclic digraph.
Proof. We prove the assertion by showing that D T [ ] is a tournament and contains no directed triangle. First, suppose towards a contradiction there are nonadjacent members . This implies further that t u , 1 2 , as out-neighbors of u 1 , are adjacent. However, we cannot have ∈ t u A D ( , ) ( ) 1 2 , for then also ∈ t u A D ( , ) ( and t u , 2 2 would form a digon in D. Hence, and therefore u t t , , 2 1 2 induce an S 2 + in D, a contradiction.
Next, suppose towards a contradiction that some vertices ∈ t t t T , , {1, 2, 3}, for otherwise M being an out-module would imply that t j and u j span a ↔ K 2 in D. Next notice that not all of u u u , , 1 2 3 can be equal, for otherwise u t t t , , , 1 1 2 3 would induce a W 3 + in D. If exactly two of them are equal, , then t 1 and u 2 are adjacent and hence by our initial remark we have . However, now u u t t t = , , , , then u 1 must be adjacent to both t 2 and t 3 , hence u 1 sees all of t t t , , 1 2 3 , so u t t t , , , 1 1 2 3 span a W 3 + , again a contradiction.
Finally consider the case that u u u , , 1  . Then u 3 must be adjacent to both t 1 and t 2 , and by our initial remark this means that ∈ u t u t A D ( , ), ( , ) ( . Hence, u t t t , , , 3 1 2 3 form a W 3 + in this case, a final contradiction which concludes the proof. □ We are now sufficiently prepared to give the proof of Theorem 1. In fact, to make our inductive proof work we state a stronger version of the claim, which allows one to enforce a monochromatic coloring on the closed out-neighborhood of a vertex. Proof. Suppose towards a contradiction that the claim is wrong, and let D v ( , ) be a counterexample to the claim minimizing v D ( ). □ Claim 1. D is strongly connected.
Proof. Suppose not, then there is a partition of V D ( ) into nonempty parts X Y , such that no arc in D starts in X and ends in Y . This property together with the fact that induces a transitive tournament implies that there exist vertices ∈ ∈ x X y Y , such Possibly after permuting colors in c Y and putting these colorings together yields an acyclic 2-coloring of D in which the closed out-neigborhood of v is monochromatic, a contradiction (note that we do not create a monochromatic directed cycle in the process, as such a cycle would have to traverse an arc from X to Y ). □

Note that Claim 1 implies that
. Hence we may apply Lemma 2.6 to the vertex v of D and find an out-module , the definition of T here coincides with the one in Lemma 2.7. Now, Lemma 2.7 implies that D T [ ] is a (possibly empty) transitive tournament.
, so such a cycle has both colors. This is a contradiction, since we assumed that D does not admit an acyclic 2-coloring in which the closed out-neighborhood of v is monochromatic. □

Claim 3 in particular implies that
has an out-arc to every other out-neighbor of ⧹ x M T in D 0 , and this shows

. For the second claim, suppose towards a contradiction that there exists
and that there exists a vertex ∈ ⧹ m M T such that ∈ u m A D ( , ) ( ). By definition of T , this however shows that ∈ m T , a contradiction. □ In the following, let D be the digraph defined by Figure 2 for an illustration).   and suppose we know that the claim holds for D i−1 . Note that The number of vertices of D satisfies 3 by Claim 3. Hence, the minimality of D implies that the assertion of the theorem holds for D . Applying this assertion to the vertex ⧹ x M T in D , we find that there exists an acyclic . Hence, by the initial assumption on D, the coloring c cannot be acyclic, that is, there is a directed cycle C in D which is monochromatic in the coloring c. Since c M is an acyclic coloring, we must have , contradicting that c is an acyclic coloring. Hence there must be an arc ∈ x y A C ( , ) ( ) such that ∈ x M and ∉ y M. However, this means , and hence c y ( ) = 1. Thus C is a cycle in color 1, and since c t c t ( ) = ( ) = 2 M for every ∈ t T , it cannot intersect T. Let z be the first vertex of M we meet when traversing C in the forward direction, starting at y. Then ∈ ⧹ z M T. Let P be the subpath of C from x to z. Now

| ADDING A DOMINATING SINK TO A HERO
In this section our goal is to prove Theorem 2. Let us first prove the following lemma.
be the vertex-trace of P and consider the partition Then there exists no arc in D starting in A i and ending in A j .

Proof. Suppose towards a contradiction that there are vertices
Then x i and y as out-neighbors of x must be adjacent in D. By definition of However, now the directed path described by the vertices u x x x y x x v = , , …, , , , …, = for r = 0, 1, 2. Further note that the two sets ( ) implies that D is strongly connected, for the dichromatic number of D equals the maximum of the dichromatic numbers of its strong components.
Let ⊇ S Y denote a set of vertices in D defined as follows: . Let us note that in any case, Putting it all together, we find that Proof. We first note that it suffices to argue that there is no arc in D from S to The sets ≥ X ( ) i i 0 are by definition pairwise disjoint, and so there exists ≥ k 1 such that ≠ ∅ X X , …, k For an illustration see Figure 3. □ Claim. X i is an independent set of D for every ≥ i 0.
Proof. We prove the claim by induction on i. The claim trivially holds for i = 0 since X x = { } 0 , and since D does not contain a → K 3 , also X N x = ( ) 1 + must be an independent set in D. Now let ≥ i 2 and suppose that we already established that X X , …, i 0 −1 are independent. Now suppose that there is an arc and ∈ x y X , if i is even. In order for this path not to be an induced P (1, 1, 1) + two nonconsecutive vertices of the path must be adjacent. However, since D does not contain → K 3 , this is only possible if x and y 2 (i odd), respectively, x 2 and y (i even) are adjacents. .
Here, a tight lower bound would be provided by the following construction: Take a threefold blow-up of a directed four-cycle (every arc being replaced by an oriented K 3,3 ) and connect each of the three blow-up triples by a directed triangle. This oriented graph is Let us further remark at this point that there exists a very simple proof that if we exclude both S 2 + and S 2 − , that is, we consider locally complete oriented graphs (where the out-and in-neigborhood of every vertex induces a tournament), then we can show that the exclusion of any hero indeed bounds the dichromatic number as follows. .
Towards a contradiction suppose that → χ D C ( ) > 2 0 for some ∈ ↔ coloring, which will contradict our assumption → χ D C ( ) > 2 0 and thus conclude the proof. To verify this, note that D contains no S 2 + centered at an in-neighbor of v and no S 2 − centered at an out-neighbor of v, and hence there is no arc leaving and nor arc entering . Thus every directed cycle in D is either disjoint from . In all cases it cannot be monochromatic, since c′ is an acyclic coloring and since c + and c − are acyclic colorings with disjoint color sets. □ In the last section of this paper we investigated oriented graphs excluding the antidirected 4-vertex-path P (1, 1, 1 are heroic for all ≥ k 4.

ACKNOWLEDGMENTS
My sincerest thanks go to the anonymous reviewers for the careful reading of the manuscript and several helpful suggestions which greatly improved the presentation of the paper. In particular I am indebted to one of the reviewers for finding a simplified proof of Lemma 2.6. I would like to thank Lior Gishboliner and Tibor Szabó for stimulating and fruitful discussions