Coloring dense digraphs

The \emph{chromatic number} of a digraph $D$ is the minimum number of acyclic subgraphs covering the vertex set of $D$. A tournament $H$ is a \emph{hero} if every $H$-free tournament $T$ has chromatic number bounded by a function of $H$. Inspired by the celebrated Erd\H{o}s--Hajnal conjecture, Berger et al. fully characterized the class of heroes in 2013. Motivated by a question of the first author and Colin McDiarmid, we study digraphs which we call \emph{superheroes}. A digraph $H$ is a \emph{superhero} if every $H$-free digraph $D$ has chromatic number bounded by a function of $H$ and $\alpha(D)$, the independence number of the underlying graph of $D$. We prove here that a digraph is a superhero if and only if it is a hero, and hence characterize all superheroes. This answers a question of Pierre Aboulker.


Introduction
Every digraph in this paper is simple, loopless and finite, where a digraph D is simple if for every two vertices u and v of D, there is at most one arc with endpoints {u, v}. Given a digraph D, we denote by V (D) the vertex set of D. The independence number α(D) of a digraph D is the independence number of the underlying graph of D. A subset X of V (D) is acyclic if the subgraph (i.e. sub-digraph) of D induced by X contains no directed cycle. A k-coloring of a digraph D is a partition of V (D) into k acyclic sets, and the chromatic number χ(D) is the minimum number k for which D admits a k-coloring. This digraph invariant was introduced by Neumann-Lara [13], and naturally generalizes many results on the graph chromatic number (see, for example, [3], [9] [10], [11], [12]).
Given digraphs D and H, we say that D is H-free if there is no induced subgraph of D isomorphic to H. A digraph T is a tournament if there is an arc between every pair Theorem 1.5. If H is a superhero, then ∆(H, T 1 , T k ) and ∆(H, T k , T 1 ) are superheroes for any k ≥ 1.
One may inquire how large g(H, α) needs to be for particular digraphs H. For digraphs not containing an oriented triangle, we believe that the following statement may be true. Conjecture 1.6. There is an integer ℓ such that if D is a C 3 -free digraph with α(D) ≤ α, then χ(D) ≤ α ℓ .
Indeed, if Conjecture 1.6 is true, then every C 3 -free digraph D either has an independent (hence, acyclic) set of size n 1/2ℓ or has chromatic number at most (n 1/2ℓ ) ℓ = √ n, and hence has an acyclic set of size √ n. Consequently, Conjecture 1.1 in the special case of H = C 3 would hold for ε := 1/2ℓ.
While targeting a polynomial bound for chromatic number of C 3 -free digraphs, we could not even achieve an exponential bound. However, by an algorithmic approach, we are able to obtain a factorial bound. Theorem 1.7. If D is a C 3 -free digraph with α(D) ≤ α, then χ(D) ≤ 35 α−1 α! and such a coloring can be found in polynomial time.
On another front, one may be interested in the chromatic number of digraphs with simple local structure. It was conjectured in [2] (Conjecture 2.6) that there is a function g such that if T is a tournament in which the set of out-neighbors of each vertex has chromatic number at most k, then χ(T ) ≤ g(k). The conjectured was verified for k = 2 in [6] and for all k in [7]. Here, we prove a generalization to digraphs with bounded independence number. Theorem 1.8. There is a function g such that if D is a digraph with α(D) ≤ α and that the set of out-neighbors of each vertex has chromatic number at most k, then χ(D) ≤ g(k, α).
The structure of the paper is as follows. We prove Theorem 1.8 in Section 2, which is the main tool to prove Theorem 1.3. Section 3 and 4 are devoted to proving Theorems 1.4 and 1.5, and hence complete the proof of Theorem 1.3. In Section 5, we will prove Theorem 1.7 to support Conjecture 1.6.

Notation and remarks
Given a digraph D, we say that u sees v and v is seen by u if uv is an arc in D. to at least one vertex of X. When it is clear in the context (most of the time), we omit the subscript D in this notation. We will use throughout the paper the fact that V (D)\X = M + D (X) ∪ N − D (X) ∪ N o D (X) for any X ⊆ V (D). Given a digraph D and a set X ⊆ V (D), we denote by D[X] the subgraph of D induced by X. When the context is clear, we often use X to denote D[X], and say chromatic number of X to refer to the chromatic number of D [X]. Given a digraph D we say that a set X ⊆ V (D) is a dominating set of D if every vertex v ∈ V (D)\X is seen by at least one vertex of X. A subset Y of V (D) is independent (or stable) if any two distinct vertices in Y are non-adjacent. Given a digraph D and two disjoint sets X, Y ⊆ V (D), we denote X → D Y (or just X → Y ) if there is no arc from Y to X in D.
A side remark is that some proofs in this paper that proceed by induction on α use the fact that if α(D) ≤ α, then α(N o (v)) ≤ α − 1 for every v, and thus χ(N o (v)) is bounded. In these inductive proofs, we often cite known results on tournaments for the base case α = 1. However, our proofs are indeed self-contained since to prove the base case α = 1, we just repeat the same arguments and use the fact that in a tournament, N o (v) = ∅ for any vertex v. Hence for example, the proof of Theorem 1.5 can serve as an alternative proof for Theorem 4.1 in [2] (that if H is a hero, then so are ∆(H, T k , T 1 ) and ∆(H, T 1 , T k )).

From local to global
We start with some observations regarding the size of a dominating set in an acyclic digraph.
Proposition 2.1. An acyclic digraph D has a dominating set which is also a stable set, and hence has size at most α(D).
Proof. We proceed by induction on |D| to show that every acyclic digraph D has a dominating set S which is stable. The statement clearly holds for |D| = 1. For |D| > 1, since D is acyclic, there is a vertex v with no in-neighbors.
C be a class of digraphs closed under taking subdigraphs. We say that C is tamed if for every integer k there exists K and ℓ such that every digraph T ∈ C with χ(T ) ≥ K contains a set A of ℓ vertices such that χ(A) ≥ k. Note that a class of digraphs with bounded chromatic number is indeed tamed.
The following proposition is straightforward.
Let us restate Theorem 1.8. Theorem 2.3. For every positive integers α and t, there is a function f α (t) such that every t-local digraph D with α(D) ≤ α has chromatic number at most f α (t).
Claim 2.4. For every t, the class of t-local digraphs with independence number at most α is tamed.
Let D be a t-local digraph with vertex set V such that α(D) ≤ α and χ(D) ≥ K ′ . Let B be a dominating set of D of minimum size b. By Proposition 2.2, we have In particular, b ≥ K ′ /(t+1) ≥ 2k(αs+1). Consider a subset W of B of size k(αs+1). By Proposition 2.2, we have χ(N + (W )) ≤ kt(αs + 1). By induction hypothesis on α − 1, for In particular, by the tamed property applied to k, one can find a set A ⊂ M − (W ) such that A has ℓ vertices and χ(A) ≥ k.
Consider now a subset S of W of size αs+1. We claim that χ(N + (S)) ≥ s. If not, we can cover N + (S) by at most s − 1 acyclic sets. Since every acyclic set has independence number at most α, it has a dominating set of size at most α by Proposition 2.1. Hence N + (S) has a dominating set, say S ′ of size at most α(s − 1) ≤ |S| − 2. But this yields a contradiction since the set (B \ S) ∪ S ′ ∪ {x}, where x is an arbitrary vertex in A, would be a dominating set of T of size less than |B|. Therefore, χ(N + (S)) ≥ s. By Proposition 2.2 applied to N + (A), we have χ(N + (A)) ≤ ℓt. Hence Thus, by the tamed property applied to k, there is a subset A S of N + (S) ∩ M − (A) such that |A S | = ℓ and χ(A S ) ≥ k.
We now construct our subset of V with chromatic number at least k + 1. For this we consider the set A ′ consisting of vertices A ∪ W to which we add the collection of A S , for all subsets S ⊆ W of size αs + 1. Observe that the number of vertices of A ′ is at most To conclude, it is sufficient to show that χ(A ′ ) ≥ k + 1. Suppose not, and for contradiction, take a k-coloring of A ′ . Since |W | = k(αs + 1) there is a monochromatic set S in W of size αs + 1 (say, colored 1). Recall that A S ⊆ M − (A) and A ⊆ M − (W ) ⊆ M − (S), so we have all arcs from A S to A and all arcs from A to S, and note that since χ(A) ≥ k and χ(A S ) ≥ k, both A and A S have a vertex of each of the k colors. Hence there are u ∈ A and w ∈ A S colored 1. Since A S ⊆ N + (S), there is v ∈ S such that vw is an arc. We then obtain the monochromatic cycle uvw of color 1, a contradiction. Thus, χ(A ′ ) ≥ k + 1, completing the proof of the claim. ♦ We now can finish the proof of the theorem. Since the class of t-local digraphs with independence number at most α is tamed, by applying tamed property for k = t + f α−1 (t) + 1, we have that there exist (K, ℓ) such that every t-local digraph D with α(D) ≤ α and χ(D) ≥ K contains a set A of ℓ vertices and Consequently, t-local digraphs have chromatic number at most f (t) := max(K, ℓ+ℓt).
By reversing direction of all arcs, we have the following theorem. 3 Proof of Theorem 1.4 Theorem 1.4 states that if H 1 and H 2 are superheroes, then so is H 1 ⇒ H 2 . We will reuse the notions of r-mountains and (r, s)-cliques introduced in [2]. Let us first give the idea of the proof of Theorem 1.4 for a special case: (C 3 ⇒ C 3 )-free tournaments have bounded chromatic number. Given a (C 3 ⇒ C 3 )-free tournament T , suppose that there is a a small set Q in T with chromatic number 3. Then for any partition of Q into Q 1 , Q 2 , at least one partition has chromatic number at least 2, and so contains a copy of C 3 . Let Y Q 1 ,Q 2 ⊆ V (D)\Q be the set of vertices seeing all vertices of Q 1 and seen by all vertices of Q 2 . Observe that Y Q 1 ,Q 2 is C 3 -free, otherwise a copy of C 3 in Y Q 1 ,Q 2 together with a copy of C 3 in either Q 1 or Q 2 forms a copy of C 3 ⇒ C 3 . Note that V (T )\Q is covered by only 2 |Q| such sets Y Q 1 ,Q 2 , and hence χ(T ) is bounded.
Hence we wish to find such a small set of vertices Q with chromatic number 3. To this end, we call an arc uv of T thick if N − (u) ∩ N + (v) contains a copy of C 3 . If T has no thick arcs, then intuitively T should have simple structure, and thus, bounded chromatic number. Suppose that T contains a (not necessarily oriented) triangle uvw where all of the three arcs are thick. Then for each of the three thick arcs, we take its thickness-certificate (i.e., a copy of C 3 ) and together with u, v, w we obtain a set Q of at most 12 vertices. It is straightforward to verify that Q has chromatic number at least 3, and thus, by the argument above χ(T ) is bounded. If T contains no triangle of thick arcs, then for any vertex v, the set of vertices adjacent to v by a thick arc induces a thick-arc-free tournament, which, intuitively, should have bounded chromatic number. We then easily bound the chromatic number of the sets of non-thick in-neighbors and non-thick out-neighbors of v, and hence bound the chromatic number of T .
The proof of the general case is in the same vein. Intuitively, we search for a small set Q with large chromatic number as described above. We will capture the notion of the set Q with the definition of an object called an r-mountain, and the notion of a triangle of thick arcs with objects called (r, s)-cliques. Given a digraph D, the formal definitions (which are borrowed from [2]) of r-thick-arc, (r, s)-clique, and r-mountain in D are defined inductively on r as follows. Every vertex of D is a 1-mountain. For every r, s ≥ 1, • An (r, s)-clique of D is a set S ⊆ V (D) such that |S| = s, and for every distinct vertices u, v ∈ S, either uv or vu is an arc that is r-thick.
• Given an (r, r + 1)-clique S and a certificate C u,v for every distinct u, v ∈ S, then the tournament induced on S ∪ ( u,v∈S C u,v ) is an (r + 1)-mountain of D.
Note that if a digraph D contains an (r, r + 1)-clique, then D contains an (r + 1)mountain, which is the (r, r + 1)-clique together with certificates of all r-thick arcs of that (r, r + 1)-clique. Hence, if D contains no (r + 1)-mountain, then D contains no (r, r + 1)-clique.
Proof. A small remark is that the second hypothesis seems redundant since if D contains no (r, 2)-clique, then D contains no r-thick arc, and so contains no r-mountain. However, the second hypothesis is necessary for the case s=1.
Since H 1 and H 2 are superheroes, there is b 4 such that every H 1 -free (or H 2 -free) digraph D with α(D) ≤ α has χ(D) ≤ b 4 . We first identify all r-thick arcs of D. Fix an arbitrary vertex v, then V (D)\{v} can be partitioned into four sets: N * the set of neighbors of v that are connected to by an r-thick arc; The crucial fact is that the digraph induced by the set N * does not contain an (r, s)-clique; indeed, an (r, s)-clique together with v would form an (r, s + 1)-clique, a contradiction to the fact that D has no (r, s + 1)-cliques. Hence by the second hypothesis, and we haveĤ 1 ⇒Ĥ 2 forming a copy of H 1 ⇒ H 2 , a contradiction. Hence, This means that uv is an r-thick arc, contradicting u / ∈ N * . Hence Thus we have, Recall that if D contains no (r + 1)-mountain, then D contains no (r, r + 1)-clique. We are now ready to show that digraphs which do not contain a mountain have bounded chromatic number.
Proof. We proceed by induction on r. If D contains no 1-mountain, then D has no vertices, and we can set g α (r) := 0. Now suppose by induction that g α (r) exists. We will show that g α (r + 1) exists. First, we claim the following.
To prove Theorem 1.4, it suffices to prove the following lemma.
Proof. We proceed by induction on α. Since H 1 ⇒ H 2 is a hero (see [2], Theorem 3.2), Lemma 3.5 is true for α = 1. Suppose that Lemma 3.5 is true for α − 1 and let c 0 = f (α − 1). Since both H 1 and H 2 are superheroes, there exists c 1 such that if D is any H 1 -free or H 2 -free digraph, then χ(D) ≤ c 1 . Let D be a (H 1 ⇒ H 2 )-free digraph with vertex set V and α(D) ≤ α. If D does not contain a (c 0 + 2c 1 )-mountain, then by applying Lemma 3.4 to D with b 0 = c 0 , there is c 2 such that χ(D) ≤ c 2 . Thus, it remains to consider the case that D contains a (c 0 + 2c 1 )-mountain.
Note that for every vertex v ∈ V \Q, there always exists a partition of Q into some sets Q 0 , Q 1 , Q 2 such that v is non-adjacent with every vertex in Q 0 , sees every vertex in Q 1 and is seen by every vertex in Q 2 . Hence, V \Q can be written as the union of all possible Hence

Proof of Theorem 1.5
It is proved in [14] that for each integer k ≥ 1, every tournament with at least 2 k−1 vertices contains a copy of T k . Let R(a, b) be the Ramsey number of (a, b), i.e., the smallest n such that any graph on n vertices either contains an independent set of order a or a clique of order b.
Proof. Suppose for a contradiction that there is a T k -free digraph D with at least R(α + 1, 2 k−1 ) vertices. Then the underlying graph of D contains either an independent set of size α + 1 or a clique of size 2 k−1 . The former case is impossible since α(D) ≤ α. Thus D contains a tournament of size 2 k−1 , and hence contains a copy of T k , a contradiction.
Recall that Theorem 1.5 states that if H is a superhero, then so are ∆(H, T k , T 1 ) and ∆(H, T 1 , T k ) for any integer k ≥ 1. We will prove that if H is a superhero, then so is ∆(H, T k , T 1 ) for any k ≥ 1. This is sufficient. Indeed, if H is a superhero, then so is H rev , the digraph obtained from H by reversing all its arcs. Thus, ∆(H rev , T k , T 1 ) rev = ∆(H, T 1 , T k ) is also a superhero.

Theorem 4.2. For every superhero H and every pair of integers
The idea of the proof of Theorem 4.2 is as follows. Fix a large number c and call a subset B of V (D) with χ(B) = c a bag. We aim at finding a longest chain of disjoint bags B 1 , ..., B t in V (D), together with a partition V (D)\ B i into sets we call zones Z 0 , ..., Z t such that there is no backward arc in D (where uv is a backward arc if the bag or zone containing u has higher index than the bag or zone containing v). Then, using maximality of t, we show that the chromatic number of every zone is bounded. Observe that all B i and Z i have bounded chromatic number, while D has no backward arc, and so D has bounded chromatic number. However, the requirement that there is no backward arc in D is too strong and so we will have to slightly relax it. In doing so, we will need to allow some backward arcs, but we will want to do so in a very controlled manner. This leads us to the following definitions.
Given an integer c ≥ 1 and a ∆(H, Since H is a superhero, such c 1 clearly exists. Note that every T k -free digraph D with α(D) ≤ α has at most c 1 vertices by Proposition 4.1, and so has chromatic number at most c 1 .

Proof of Theorem 4.2.
We proceed by induction on α. Since ∆(H, T k , T 1 ) is a hero (see [2], Theorem 4.1), the theorem holds for α = 1. Suppose that Theorem 4.2 is true for α − 1 and let c 0 = f (H, k, α − 1). Let D be a ∆(H, T k , T 1 )-free digraph with vertex set V and α(D) ≤ α. An important observation is that for every v ∈ V , the digraph N o (v) is a ∆(H, T k , T 1 )-free and has independence number at most α − 1, and hence We also would like to recall some useful formulas. For every v ∈ X ⊆ V , and if Y ⊆ V and Y ∩ X = ∅, then (recalling that M + (X) is the set of vertices seen by all vertices of X) Let |H| = h and set c := 2(c 0 + c 1 )(h + k). Let us assume that B 1 , ..., B t is a c-bagchain of D with t as large as possible. In the proof of this theorem, we drop prefix cof c-bag and c-bag-chain for convenience. By definition of bag-chain, every bag behaves well with bags proceeding or succeeding it. In the following claim, we show that every bag in fact behaves well with any other bag.
Proof. We proceed by induction on r. For r = 1, both (a) and (b) holds by definition of bag-chain. Suppose that both statements are true for r − 1. We now prove (a) for r. Suppose for a contradiction that there is v in some B i such that and

(4.4)
For each x ∈Ĥ, by (4.1) we have χ(N o (x) ∩ B i−1 ) ≤ c 0 , and by induction hypothesis of (b) applied to x and r − 1, we have Then there exists a copy of T k in M + (Ĥ) ∩ N − (v) ∩ B i−1 , sayT k . Note that by construction, we have all arcs fromĤ toT k , fromT k to v and from v toĤ. Then ∆(Ĥ,T k , v) forms a copy of ∆(H, T k , T 1 ), a contradiction.
The proof of (b) for r is similar but not symmetric. In order to obtain a copy of ∆(H, T k , T 1 ), we first get a copy of T k in B i+r , and then a copy of H in B i+1 . This proves the claim. ♦ We next prove a stronger statement that every bag behaves well with the union of all other bags proceeding or succeeding it. Let B i,j = j s=i B s for any 1 ≤ i ≤ j ≤ t (if i < 1 or j > t or j < i, we set B i,j = ∅).
Proof. We repeat the same argument as in the proof of Claim 4.3. Suppose for a contradiction that the first statement is false, i.e., there is v in some B i such that We now show that the union of all bags has bounded chromatic number. We note that in the following proof, we will use only two hypotheses: Claim 4.5 and that χ(B i ) is bounded for every i. The reason for our remark is that we will re-use the arguments of this proof for for subsequent claims. Proof. An arc uv with u ∈ B j , v ∈ B i , and j > i is called a backarc with span j − i.

For every i and every
We have a formula similar to (4.4): Now let G be the undirected graph with vertex set B 1,t and uv ∈ E(G) if u ∈ F v or v ∈ F u . Then B i is a stable set in G for every i. We now color the vertices of G by c 1 colors as follows. First, color all B t properly by color 1. Suppose that we have already colored B i+1 , ..., B t . Every vertex v in B i is incident (in G) with at most c 1 − 1 vertices in B i+1,t (those belong to F v ) and independent (in G) with all other vertices in B i , so we can always properly color v, and so properly color B i . When the process of coloring ends, we obtain a partition of B 1,t into c 1 sets of colors, say X 1 , ..., X c 1 , where each X s is a stable set in G. We now claim that the chromatic number of the graph induced by X s is small.
To prove (B), we define a sequence of indices i 1 , i 2 , ... inductively as follows. Let i 1 = 1, and for every r ≥ 1, let i r+1 > i r be the smallest index such that χ(B ir,i r+1 ) > 4c. The sequence ends by i ℓ with χ(B i ℓ ,t ) ≤ 4c (i.e. there is no i ℓ+1 satisfying the condition). Set A r := B ir,i r+1 −1 for every 1 ≤ r ≤ ℓ − 1 and A ℓ := B i ℓ ,t . Then B 1,t = ℓ r=1 A r , and by definition of the sequence, χ(B ir,i r+1 −1 ) ≤ 4c for every 1 ≤ r ≤ ℓ − 1. In other words, for every 1 ≤ r ≤ ℓ, χ(A r ) ≤ 4c. (4.5) Suppose that there is a backarc uv with u / ∈ F v and u ∈ A r , v ∈ A r ′ , where r ≥ r ′ + 2. Suppose that u ∈ B j and v ∈ B j ′ . Then j ≥ i r since B j ⊆ A r = B ir,i r+1 −1 , and j ′ < i r ′ +1 since B j ′ ⊆ A r ′ = B i r ′ ,i r ′ +1 −1 , and so j ′ < i r−1 since r ′ + 1 ≤ r − 1. Also note that χ(B i r−1 ,ir ) > 4c by definition of the sequence. Thus we have This deduces an observation that for any r ≥ r ′ + 2, there is no backarc uv with u ∈ A r , v ∈ A r ′ and u / ∈ F v . Now fix an arbitrary X s and let X s,r = X s ∩ A r for every 1 ≤ r ≤ ℓ. Observe that if uv is an backarc with u, v ∈ X s , then u / ∈ F v (otherwise, u ∈ F v , so uv ∈ G and so X s is not stable in G, a contradiction). Hence combining with the observation in the paragraph above, we have that there is no backarc from X s,r to X s,r ′ for any r, r ′ with r ≥ r ′ + 2; in other words, X s,r ′ → X s,r for any r, r ′ with r ≥ r ′ + 2. Recall from (4.5) that χ(X s,r ) ≤ 4c. Thus χ r≥1 X s,2r−1 ≤ 4c and χ r≥1 X s,2r ≤ 4c. This gives χ(X s ) = χ r≥1 X s,r ≤ 8c, which proves (B). Hence This completes the proof of Claim 4.6. ♦ We now turn our attention to the vertices not in the bags. We partition V \B 1,t into sets Z 0 , ..., Z t called zones as follows. For every x ∈ V \B 1,t , if i is the largest index such that χ(N − (x) ∩ B i ) ≥ c 1 , then x ∈ Z i ; otherwise, x ∈ Z 0 . We first show that every Z i behaves well with any B j sufficiently far from it.
Proof. The former inequality is obvious by the partition criterion. For the latter, the proof follows the same idea as that of Claim 4.3, but is a bit more involved. Suppose for a contradiction that χ(N + (v) ∩ B i−r ) ≥ c 1 for some r ≥ 2. Then there is a copy of H in

Then by applying (4.3) we have
. Note that R = R 1 ∪ R 2 , and so either χ(R 1 ) ≥ c 1 or χ(R 2 ) ≥ c 1 . If χ(R 1 ) ≥ c 1 , there is a copy of H in R 1 , sayĤ ′ . Then ∆(Ĥ ′ ,T k , v) forms a copy of ∆(H, T k , T 1 ), a contradiction. If χ(R 2 ) ≥ c 1 , there is a copy of T k in R 2 , sayT ′ k . Then ∆(Ĥ,T ′ k , v) forms a copy of ∆(H, T k , T 1 ), a contradiction again. This proves the claim. ♦ While Claim 4.7 shows that every zone behaves well with every sufficiently far enough bag. The next claim shows the converse, i.e., every bag behaves well with every sufficiently far enough zone.

Claim 4.8. For every i and
Proof. We repeat the argument in the proof of Claim 4.3. Suppose for a contradiction that the first statement is false, i.e., there is v in some B i and r ≥ 2 such that Proof. Suppose for a contradiction that In order to bound the chromatic number of the union of zones, we also need that each zone has bounded chromatic number. We will prove this by employing the assumption that the bag-chain B 1 , ..., B t is of maximum length. Proof. Suppose that some Z i contains a bag-chain of length 6, say Y 1 , ..., Y 6 . Note that we have every v ∈ B i−3 by Claim 4.8; and Thus by definition of bag-chain, B 1 , ..., B i−3 , Y 1 , Y 2 , ..., Y 6 , B i+3 , ..., B t is a bag-chain of length t + 1, which contradicts the maximality of t. Hence no Z i contains a bag-chain of length 6. ♦ Lemma 4.11. There is c ′ such that if a ∆(H, T k , T 1 )-free digraph D ′ with α(D ′ ) ≤ α does not contain any c-bag-chain of length 6, then χ(D ′ ) ≤ c ′ .
We defer the proof of Lemma 4.11 for now.
We now show that this is sufficient to prove the theorem. To do so, we group zones by indices modulo 3 and follow a similar argument as in Claim 4.6. Fixed 0 ≤ s ≤ 2, and for every i ≡ s mod 3, 1 ≤ i ≤ t, let Z s ⌊i/3⌋ := Z i . Then χ(Z s j ) ≤ c ′ for every j, and for every v ∈ Z s j , by Claim 4.9, we have By applying Lemma 4.11 for D ′ = Z i and using Claim 4.10, it is immediate that χ(Z i ) ≤ c ′ for every i. Thus, χ(Z s j ) ≤ c ′ for every j. We can repeat exactly the argument of Claim 4.6 (with c replaced by c ′ and by using Claim 4.9 instead of Claim 4.5) to deduce for every s = 0, 1, 2. (Remark: we may suppose that c ′ ≥ c, which is necessary for the argument at the end of the proof of Property (A) that 3(c 1 + c 0 )(k + 1) ≤ 2c ≤ 2c ′ ). Hence From Claims 4.6 and the above inequality, we have which proves Theorem 4.2.

Proof of Lemma 4.11
Lemma 4.11 asserts that no bag-chain of length 6 is enough to force a digraph (with bounded independence number and ∆(H, T k , T 1 )-free) to have bounded chromatic number. It is trivial by definition that no bag-chain of length 1 forces bounded chromatic number. In the next lemma, we show that no bag-chain of length 2 can force bounded chromatic number, which contains all difficulty in the proof of Lemma 4.11. In the following lemma, c, c 0 , c 1 , h, k are the values as in the proof of Theorem 4.2. Recall that c = 2(c 0 + c 1 )(h + k). Proof. Let J be the tournament H ⇒ T k . By Theorem 1.4, J is a superhero and contains both H and T k as subtournaments. Let D be a digraph with vertex set V and satisfying the hypotheses of the lemma. A copy of J in D is called a ball. A ballĴ is colored red if χ(M + (Ĵ)) ≤ d and blue if χ(M − (Ĵ)) ≤ d (note that we do not color vertices of a ball but color the ball as a single object). A ball certainly can be both red and blue, in which case we color it arbitrarily with one of the colors.
Claim 4.13. Every ball is either red or blue.
Proof. Suppose for a contradiction that a ballĴ is neither red nor blue. Then there are LetĤ be a copy of H in N + (v) ∩ B 1 . LetT k be a copy of T k inĴ. Then ∆(Ĥ,T k , v) forms a copy of ∆(H, T k , T 1 ), a contradiction. Hence for otherwise a copy of T k in N − (v) ∩ B 2 would yield a contradiction. Hence B 1 , B 2 is a d-bag-chain of length 2, a contradiction. ♦ For every vertex v of V , we color v as follows. If there are c 1 + 1 vertex-disjoint red ballsĴ 1 , ...,Ĵ c 1 +1 such thatĴ i has complete arcs toĴ j for every i < j, and v ∈Ĵ c 1 +1 , then we color v red. If there are c 1 + 1 blue ballsĴ 1 , ...,Ĵ c 1 +1 such thatĴ i has complete arcs toĴ j for every i < j, and v ∈Ĵ 1 , then we color v blue. If v satisfies both conditions, we color v arbitrarily. After the process of coloring, we obtain a partition of V into R the set of red vertices, B the set of blue vertices, and U the set of uncolored vertices.
Proof. Let K be the tournament of J ⇒ J ⇒ ... ⇒ J (2c 1 + 2 times J). Since J is a superhero, then so is K by Theorem 1.4. Hence there is d 1 such that every K-free digraph D ′ with α(D ′ ) ≤ α has chromatic number at most d 1 . Suppose that U contains a copy of K, sayK. Since every ball is either red or blue, we can find c 1 + 1 vertex-disjoint monochromatic ballsĴ 1 , ...,Ĵ c 1 inK such thatĴ i has complete arcs toĴ j for every i < j. Then either vertices ofĴ 1 are blue or vertices ofĴ c 1 +1 are red, a contradiction with the fact that all vertices of U are uncolored. Hence U is K-free, and so χ(U ) ≤ d 1 .
♦ It remains to show that R and B have bounded chromatic number, which can be done by applying Theorems 1.8. To do so, we need to prove that N + (v) has bounded chromatic number for every v ∈ R. Claim 4.15. There is d 2 such that χ(N + (v)) ≤ d 2 for every v ∈ R.
Proof. Fix v ∈ R. Then there are vertex-disjoint red ballsĴ 1 , ...,Ĵ c 1 +1 where v ∈Ĵ c 1 +1 andĴ i has complete arcs toĴ j for every i < j. Let L = c 1 i=1Ĵ 1 . Note that v is seen by all vertices of L. For every u ∈ L, we have For every partition of L into L 1 , L 2 (L 1 or L 2 may be empty), let Y L 1 , there is a partition of L into some L 1 , L 2 such that x sees all vertices of L 1 and is seen by all vertices of L 2 , and so x ∈ Y L 1 ,L 2 . Hence We now show that • Otherwise, if there is noĴ i ⊆ L 2 , then L 1 contains at least one vertex of eachĴ i , 1 ≤ i ≤ c 1 , and so |L 1 | ≥ c 1 . Hence L 1 has a copy of T k , sayT k . Note that all vertices ofT k see v since v ∈Ĵ c 1 +1 . If χ(Y L 1 ,L 2 ) ≥ c 1 , then Y L 1 ,L 2 contains a copy of H, sayĤ, then ∆(Ĥ,T k , v) forms a copy of ∆(H, T k , T 1 ), a contradiction. Hence Note that |L| = |J|c 1 = (h + k)c 1 . Thus, there are 2 (h+k)c 1 possible ways to partition L into L 1 , L 2 . Hence This completes the proof of the claim. ♦ Then χ(N + (v)∩R) ≤ χ(N + (v)) ≤ d 2 for every v ∈ R. Then by applying Theorem 2.3 for digraph R with t = d 2 , we have χ(R) ≤ d 3 for some d 3 .
We now prove that B has bounded chromatic number. The proof is slightly different from that of Claim 4.15 due to asymmetry of ∆(H, T k , T 1 ).
Proof. Fix v ∈ B. Then there are vertex-disjoint blues ballsĴ 1 , ...,Ĵ c 1 +1 where v ∈Ĵ 1 andĴ i has complete arcs toĴ j for every i < j. Let L = c 1 +1 i=2Ĵ i . Note that v sees all vertices of L. For every u ∈ L, we have For every partition of L into L 1 , L 2 (L 1 or L 2 may be empty), let Y L 1 , , there is a partition of L into some L 1 , L 2 such that x sees all vertices of L 1 and is seen by all vertices of L 2 , and so x ∈ Y L 1 ,L 2 . Hence We now show that χ(Y L 1 ,L 2 ) ≤ d for every Y L 1 ,L 2 .
• If there isĴ i ⊆ L 2 , thenĴ i contains a copy of H, sayĤ. Note that v is in the first ball, so v sees all vertices ofĤ. If χ(Y L 1 ,L 2 ) ≥ c 1 , then Y L 1 ,L 2 contains a copy of T k , sayT k , then ∆(Ĥ,T k , v) forms a copy of ∆(H, T k , T 1 ), a contradiction. Hence • Otherwise, we have two remarks: (1)Ĵ c 1 +1 must have a vertex in L 2 , say z.
(2) EveryĴ i , 2 ≤ i ≤ c 1 must have a vertex in L 1 . Then |L 1 ∪ {v}| ≥ c 1 , and so L 1 ∪ {v} contains a copy of T k , sayT ′ k , such that all vertices ofT ′ k are in one of theĴ i , 1 ≤ i ≤ c 1 (note that v ∈Ĵ 1 ). Observe that z is seen by all vertices ofT ′ k since z ∈Ĵ c 1 +1 . If χ(Y L 1 ,L 2 ) ≥ c 1 , then Y L 1 ,L 2 contains a copy of H, sayĤ ′ , then ∆(Ĥ ′ ,T ′ k , z) forms a copy of ∆(H, T k , T 1 ), a contradiction. Hence This completes the proof of Lemma 4.12.
We are now ready to prove Lemma 4.11. Proof. Suppose for a contradiction that χ(Z i ) ≥ g(g(g(c))). We will show that D contains a c-bag-chain of length 8. By applying Lemma 4.12 to D with d := g(g(c)), we have that D contains a g(g(c))-bag-chain of length two, say X 1 , X 2 . Hence χ(X 1 ) = χ(X 2 ) = g(g(c)) and • χ(N + (v) ∩ X 1 ) ≤ c 1 for every v ∈ X 2 , and Apply Lemma 4.12 again to X 1 (res. X 2 ) with d := g(c), we obtain a g(c)-bag-chain of length two, say Hence by definition, Y 1 , ..., Y 4 forms a g(c)-bag-chain of length 4. Note that χ(Y s ) = g(c) for every 1 ≤ s ≤ 4. Repeating the argument we obtain a c-bag-chain of length 2 inside each Y s , and hence obtain a c-bag-chain B 1 , ..., B 8 of length 8 inside D. This contradicts the fact that D has no c-bag-chain of length 6, and so completes the proof.

Triangle-free digraphs
In this section, we prove Theorem 1.7. We present an efficient algorithm to color a C 3 -free digraph whose independence number is α(D) ≤ α with at most 35 α−1 α! colors. For a digraph D, let n = |V (D)| denote the size of its vertex set. Let poly(n) denote the function n k for some rational number k > 0. With respect to the time complexity of our algorithm, our main goal is to show that it is a polynomial depending only on n, i.e. poly(n). We therefore do not optimize the running time nor do we provide a tight analysis. Moreover, note that each time we argue that a subroutine can be performed in poly(n) time, the value of k may be different.
For each integer α ≥ 1, define h(α) to be the minimum number such that every C 3 -free digraph D with α(D) ≤ α has chromatic number at most h(α). Conjecture 1.6 asserts that h(α) ≤ α ℓ for some ℓ. Clearly h(1) = 1 since every C 3 -free tournament is acyclic. However, h(α) is still unknown for all α ≥ 2. We believe that h(2) = 2 or 3 even though the best bound we have is h(2) ≤ 18 (by tweaking the proof of Theorem 1.7). Theorem 1.7 gives an exponential upper bound for h(α) by the function g(α) := 35 α−1 α!. Since our proof of Theorem 1.7 will use induction, we will assume that a C 3 -free digraph with independence number can be colored with at most g(α − 1) colors. This is true for α = 1, and we will prove it for α. Let us restate Theorem 1.7. The rest of this section is devoted to proving Theorem 5.1. We begin with some observations regarding the size of a dominating set in a digraph.  We now show that Y is a set with the desired properties. Suppose for a contradiction that there is and v / ∈ S since Y dominates all vertices of S. There is u ∈ S seeing v since S is a dominating set of D. Note that u / ∈ Y ; otherwise this contradicts Y ⊆ N + (v). There is y ∈ Y seeing u since Y is a dominating set of S. Then u, v, y are distinct vertices where u sees v, v sees y, and y sees u. Hence we obtain a copy of C 3 in D, a contradiction.
We now present some definitions and useful lemmas. First, we re-define a bag so that it inherits all properties of a bag as defined in Section 4 and so that it can be tested efficiently. Recall that u and v are neighbors if either uv or vu is an arc. We can check in poly(n) time (e.g. O(n 4 )) whether or not a set B is a bag of D by exhaustively checking all triples in V (D) \ B. Suppose that the n vertices of a digraph can be partitioned into disjoint sets such that the subgraph induced on each set has independence number at most α − 1. Let t(α − 1, n) denote the maximum (over all such possible partitions of the vertices) total time required by our algorithm (the algorithm Color-Digraph, which we will define shortly) to color all of the subgraphs, each with at most g(α − 1) colors. The following claim follows from this definition.
We now fix an arbitrary C 3 -free digraph D such that α(D) ≤ α, and we omit the subscript D from the relevant notation when the context is clear.
is not a bag of D, then: (a) We can partition S into three disjoint sets, S 1 , S 2 and S 3 , such that α(D[S i ]) ≤ α−1 for i ∈ {1, 2, 3}. This procedure takes time poly(n).
Proof. If S is not a bag of D, then we can, in time poly(n), find a triple {x, y, z} ∈ V (D) \ S such that every v ∈ S is not incident to at least one of x, y or z. Thus, each vertex v ∈ S belongs to either N o (x) ∩ S, N o (y) ∩ S or N o (z) ∩ S. Each of these sets has independence number at most α − 1. The total time for this procedure is poly(n). The second assertion follows from the (a) and from the definition of the function t(α, n). ♦ We can check in poly(n) time (e.g. O(n 5 )) if a bag B is poor by testing whether or not N − (v) ∩ B and N + (v) ∩ B are bags for every v ∈ B.
Proof. Since B is poor, then for each v ∈ B, either N − (v) ∩ B or N + (v) ∩ B is not a bag of D. Then we can partition B into two sets, L and R, where N − (v) ∩ B is not a bag for every v ∈ L, and N + (v) ∩ B is not a bag for every v ∈ R.
Applying Proposition 5.3 to R, we can find a set . And so: For v ∈ Y R , note that N + (v) ∩ R is not a bag, and so by (a) from Claim 5.6, we can partition N + (v) ∩ R into three sets, each with independence number at most α − 1. Additionally, we have the set N o (v) ∩ R, which also has independence number at most α − 1. Overall, we can partition R \ Y R into 4α sets, each with independence number α − 1. The same argument can be applied to Y L to obtain 8α disjoint sets. Therefore, in time poly(n), we can partition B \ Y R ∪ Y L , and in time t(α − 1, |B|) we can color B \ {Y r ∪ Y L } with 8α · g(α − 1) colors. We can then color each v ∈ Y R (respectively, v ∈ Y L ) with an arbitrary color used to color the set In general, we do not know how to color a bag efficiently, and a bag may be very large (e.g. V (D) is a bag). Our aim is therefore to find poor bags, since these can be colored using Claim 5.8. The first step of our algorithm is to find a chain of poor bags. Recall that B i → B i+1 means there is no arc from B i to B i+1 . Moreover, if each B i is a poor bag, then this sequence is a chain of poor bags. Given a chain of bags We can partition the vertices in V (D) \ C into sets Z 0 , . . . , Z t , which we call zones, as follows. For every v ∈ V (D) \ C, let i be the largest index such that v is seen by at least one vertex in B i . Then vertex v is assigned to zone Z i . Otherwise, we assign v to zone Z 0 . This partition is unique and can be done in time poly(n). As in the case of the bags and zones used in Section 4, these bags and zones exhibit useful properties. The proofs we present here are similar, but much simpler.
Proof. Property (a) holds for r = 1 by definition of a chain of bags. Suppose that (a) holds for r − 1 > 1, and suppose that there is an arc uv with u ∈ B i+r and v ∈ B i . Since B i+1 is a bag, there is x ∈ B i+1 such that x is a common neighbor of u and v. Then by induction hypothesis, vx, xu are arcs, and so vxu is a copy of C 3 , a contradiction. Hence (a) holds for r.
Property (b) holds for all r ≥ 1 by the partitioning criterion of vertices into zones.
To prove property (c), suppose that there is an arc zv with z ∈ Z i+r and v ∈ B i for some r ≥ 2. Then there is u ∈ B i+r such that uz is an arc by the partitioning criterion of vertices into zones. Since B i+1 is a bag, there is x ∈ B i+1 such that x is a common neighbor of u, v, z. By property (a), vx and xu are arcs. If xz is an arc, then vxz is a copy of C 3 . Otherwise, zx is an arc, and so xuz is a copy of C 3 . Either way, we reach the contradiction, and so (c) holds for every r ≥ 2.
To prove property (d), suppose that there is an arc uv with u ∈ Z i+r and v ∈ Z i for some r ≥ 3. Since B i+1 is a bag, there is x ∈ B i+1 such that x is a common neighbor of all u and v. By property (b), vx is an arc, and by property (c), xu is an arc. Hence vxu is a copy of C 3 , a contradiction. Hence (d) holds for every r ≥ 3. ♦ We now show how to find a chain of poor bags, which can be colored using Claim 5.8. Proof. Let B 1 , . . . , B t be the chain of poor bags output by Find- Chain(D, B). The bags in this chain are pairwise disjoint. From Step 1 of Find-Chain, it follows that each B i is a bag. Furthermore, each B i must be poor; otherwise, Step 1 would be applied to B i to return poor bags inside B i . Observe that for every pair of consecutive bags B i , B i+1 in the chain, B i+1 ⊆ N + (v i ) and B i ⊆ N − (v i ). If there is an arc xy with x ∈ B i+1 and y ∈ B i , then v i xy is a copy of C 3 , a contradiction. Hence B 1 , . . . , B t is a chain of poor bags by definition. The procedure Find-Chain(D, B) runs in time poly(n). ♦ Claim 5.12. If Find-Chain(D, B) returns a chain of t poor bags, then we can color B with 8α · g(α − 1) + (t − 1) · g(α − 1) colors in time poly(n) + t(α − 1, |B|).
Proof. Suppose that Find-Chain(D, B) returns B 1 , v 1 , B 2 , . . . , v t−1 , B t and that C = {B 1 , . . . , B t } is the chain of poor bags output by the procedure. Since each B i is a poor bag, it can be colored using Claim 5.8. By Claim 5.10, B i → B j for every i < j, and so we can color the vertices in C using 8α · g(α − 1) colors in time poly(n) + t(α − 1, |C|) time.
Observe that each v ∈ B \ C is either (i) some v i in the output sequence returned by Find-Chain(D, B), or (ii) belongs to N o (v i ) for some v i in this output sequence. For any v ∈ V (D), the set N o (v) has independence number α − 1. Therefore, the vertices in t−1 i=1 N o (v i ) can be colored with (t − 1) · g(α − 1) colors. Note that v i can be colored with an arbitrary color from the color palette used for N o (v i ). This coloring can be found in time poly(n) + t(α − 1, |B \ C|). ♦ Claim 5.16. Color-Digraph(D) runs in time poly(n).
Proof. We now analyze the running time. Finding a maximal chain C takes poly(n) time, as does the procedure of partitioning the vertices in V (D) \ C into zones. Once we have found this partitioning, we can color the vertices in C in time poly(n)+t(α−1, |C|) using Claim 5.8, and we can color each zone Z i in time poly(n) + (α − 1, |Z i |) using Claim 5.12 and Corollary 5.13. So applying Claim 5.5, the total running time of Color-Digraph(D) is at most poly(n) + t(α − 1, n). This leads to the following recurrence relation: t(α, n) = poly(n) + t(α − 1, n) = α · poly(n).
We remark that our algorithm is actually just partitioning V (D) into disjoint subsets, where each subset has independence number at most α − 1. In other words, it first partitions the vertices in V (D) into sets where each set is a poor bag or not a bag (for example, a zone Z i is either a poor bag or not a bag). Then, it further partitions these sets into sets with independence number at most most α−1. (For example, by Claim 5.6, a set that is not a bag can be partitioned into three disjoint sets, each with independence number at most α − 1. Similarly, in the proof of Claim 5.8, a poor bag is partitioned into 8α disjoint sets, each with independence number at most α − 1.) Once the subgraphs on these induced subsets are colored (recursively) using g(α − 1) colors, then certain subsets are allowed to use the same color palette, and these color palettes can be coordinated in time poly(n). The initial partitioning procedure and the final coordinating procedure require poly(n) time, while the recursive coloring requires t(α − 1, n) time. ♦ Theorem 5.1 follows from Claims 5.14, 5.15 and 5.16.