Coloring dense digraphs

TL;DR

This paper extends the concept of heroes in tournaments to dense digraphs, introducing superheroes, and proves that the classes of heroes and superheroes coincide, providing a complete characterization.

Contribution

It establishes that a digraph is a superhero if and only if it is a hero, thus fully characterizing superheroes in the context of dense digraphs.

Findings

Superheroes and heroes are equivalent classes of digraphs.

Complete characterization of superheroes in dense digraphs.

Answers a previously open question in the field.

Abstract

The chromatic number of a digraph $D$ is the minimum number of acyclic subgraphs covering the vertex set of $D$. A tournament $H$ is a 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. We extend this framework to dense digraphs: A digraph $H$ is a 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 Aboulker, Charbit and Naserasr.

Equations

$$K^{\prime }:=2k(\alpha s+1)(t+f_{\alpha -1}(t)+1),$$

$$\chi (D)=\chi (B\cup N^{+}(B))\le \chi (B)+\chi (N^{+}(B))\le (t+1)b.$$

$$\chi (M^{-}(W))\ge$$

$$\ge$$

$$\chi (N^{+}(S)\cap M^{-}(A))\ge$$

$$\ge$$

$$=$$

$${\ell }^{\prime }:=\ell +k(\alpha s+1)+\ell \left(\genfrac{}{}{0ex}{}{k(\alpha s+1)}{\alpha s+1}\right).$$

$$N^{+}=\big{(}M^{+}(\hat{H}_{1})\cap N^{+}\big{)}\cup\big{(}N^{o}(\hat{H}_{1})\cap N^{+}\big{)}\cup\big{(}N^{-}(\hat{H}_{1})\cap N^{+}\big{)}.$$

$$\chi (N^{+})\le$$

$$\le$$

$$\chi (D)\le$$

$$\le$$

$$\chi (N^o(v))\le f(H,k,\alpha -1)=c_0.$$

$$\chi (N^{+}(X))\le \sum _{v\in X}\chi (N^{+}(v))\text{ and }\chi (N^o(X))\le \sum _{v\in X}\chi (N^o(v)),$$

$$Y=(M^{+}(X)\cap Y)\cup\big{(}(N^{-}(X)\cup N^{o}(X))\cap Y\big{)}.$$

$$B_{i-1}=\Big{(}M^{+}(\hat{H})\cap B_{i-1}\Big{)}\cup\Big{(}\big{(}N^{-}(\hat{H})\cup N^{o}(\hat{H})\big{)}\cap B_{i-1}\Big{)},$$

$$B_{i-1}=\Big{(}N^{-}(v)\cap B_{i-1}\Big{)}\cup\Big{(}\big{(}N^{+}(v)\cup N^{o}(v)\big{)}\cap B_{i-1}\Big{)}.$$

$$\begin{split}B_{i-1}=&\Big{(}M^{+}(\hat{H})\cap N^{-}(v)\cap B_{i-1}\Big{)}\cup\Big{(}\big{(}N^{-}(\hat{H})\cup N^{o}(\hat{H})\big{)}\cap B_{i-1}\Big{)}\\ &\ \ \ \ \cup\Big{(}\big{(}N^{+}(v)\cup N^{o}(v)\big{)}\cap B_{i-1}\Big{)}.\end{split}$$

$$\chi (M^{+}(\hat{H})\cap N^{-}(v)\cap B_{i-1})\ge$$

$$-\sum _{x\in \hat{H}}\chi (N^{-}(x)\cap B_{i-1})-\chi (N^{+}(v)\cap B_{i-1}),$$

$$\ge$$

$$>$$

$$B_{i+1,j-1}=$$

$$\displaystyle\ \ \ \ \cup\Big{(}\big{(}N^{-}(v)\cup N^{o}(v)\big{)}\cap B_{i+1,j-1}\Big{)}.$$

$$\chi (B_{i+1,j-1})\le$$

$$\displaystyle\ +\sum_{x\in\hat{T_{k}}}\chi\Big{(}N^{+}(x)\cap B_{i+1,j-1}\Big{)}+\chi\Big{(}N^{-}(v)\cap B_{i+1,j-1}\Big{)}$$

$$\le$$

$$\chi (A_r)\le 4c.$$

$$\chi (B_{j^{\prime },j})\ge \chi (B_{i_{r-1},i_r})\ge \chi (B_{i_r,i_{r+1}})>4c,$$

$$\chi (B_{1,t})\le \sum _{s=1}^{c_1}\chi (X_s)\le 8c(c_1+1).$$

$$B_{i-1}=\Big{(}M^{+}(\hat{H})\cap B_{i-1}\Big{)}\cup\Big{(}\big{(}N^{-}(\hat{H})\cup N^{o}(\hat{H})\big{)}\cap B_{i-1}\Big{)},$$

Full text

1 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 $\alpha(D)$ 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., subdigraph) 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 $\chi(D)$ 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 of distinct vertices of $T$. A tournament $H$ is a hero if there is a number $f(H)$ such that $\chi(T)\leq f(H)$ for every $H$-free tournament $T$. The class of heroes was fully characterized in 2013 by Berger et al [2].

One may try to study the more general question for arbitrary digraphs: for which digraph $H$ does there exist a constant $\varepsilon(H)>0$ such that every $H$-free digraph $D$ satisfies $\chi(D)\leq|V(D)|^{1-\varepsilon(H)}$? In fact, it was conjectured in [8] that this holds for every $H$.

Conjecture 1.1**.**

For every digraph $H$, there is $\varepsilon>0$ such that if $D$ is a $H$-free digraph, then $\chi(D)\leq|V(D)|^{1-\varepsilon}$.

The conjecture is open even when $H$ is the oriented triangle, $C_{3}$. In fact, it can be viewed as a strengthening of the Erdős–Hajnal conjecture. For a comprehensive survey on the Erdős–Hajnal conjecture, we refer the reader to [5]. Given a digraph $D$, we denote by $\beta(D)$ the maximum number of vertices of an acyclic subset of $V(D)$. The following is one formulation of the Erdös–Hajnal conjecture.

Conjecture 1.2**.**

For every tournament $H$, there exists a constant $\varepsilon(H)>0$ such that every $H$-free tournament $T$ satisfies $\beta(T)\geq|V(T)|^{\varepsilon(H)}$.

Motivated by Conjecture 1.1, we say that a digraph $H$ is a superhero if for every integer $\alpha\geq 1$, there is a number $g(H,\alpha)$ such that $\chi(D)\leq g(H,\alpha)$ for every $H$-free digraph $D$ with $\alpha(D)\leq\alpha$. There are several remarks. First, if a tournament $H$ is a superhero, then $H$ is a hero by definition. Second, if $H$ is not a tournament, then $H$ is not a superhero. Indeed, one can easily construct a tournament $T$ with arbitrarily high chromatic number (see Lemma 3.1 for example). Note that $H$ contains two non-adjacent vertices. Hence $T$ contains no induced isomorphic copy of $H$, and so $H$ is not a superhero. From two remarks above, it follows that every superhero is a hero. Aboulker, Charbit and Naserasr [1] conjectured that the converse is also true. The main result of this paper is the affirmation of this fact.

Theorem 1.3**.**

Every hero is a superhero.

Given tournaments $H_{1},H_{2},H_{3}$, we denote by $H_{1}\Rightarrow H_{2}$ the vertex-disjoint union of $H_{1}$ and $H_{2}$ with complete arcs from $H_{1}$ to $H_{2}$, and $\Delta(H_{1},H_{2},H_{3})$ the vertex-disjoint union of $H_{1},H_{2},H_{3}$ with complete arcs from $H_{1}$ to $H_{2}$, from $H_{2}$ to $H_{3}$, and from $H_{3}$ to $H_{1}$. For every integer $k\geq 1$, let $T_{k}$ denote the transitive tournament on $k$ vertices, i.e. the acyclic tournament on $k$ vertices. The class of heroes are constructed in [2] as follows:

• •

The singleton $T_{1}$ is a hero.

• •

If $H_{1}$ and $H_{2}$ are heroes, then $H_{1}\Rightarrow H_{2}$ is a hero.

• •

If $H$ is a hero, then $\Delta(H,T_{k},T_{1})$ and $\Delta(H,T_{1},T_{k})$ are heroes for every $k\geq 1$.

The main result of [2] is that any tournament that cannot be constructed by this process is not a hero. Obviously $T_{1}$ is a superhero. Thus to prove Theorem 1.3, it suffices to prove the following two theorems.

Theorem 1.4**.**

If $H_{1}$ and $H_{2}$ are superheroes, then $H_{1}\Rightarrow H_{2}$ is a superhero.

Theorem 1.5**.**

If $H$ is a superhero, then $\Delta(H,T_{1},T_{k})$ and $\Delta(H,T_{k},T_{1})$ are superheroes for any $k\geq 1$.

One may inquire how large $g(H,\alpha)$ 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 $\ell$ such that if $D$ is a $C_{3}$-free digraph with $\alpha(D)\leq\alpha$, then $\chi(D)\leq\alpha^{\ell}$.

Indeed, if Conjecture 1.6 is true, then every $C_{3}$-free digraph $D$ either has an independent (hence, acyclic) set of size $n^{1/2\ell}$ or has chromatic number at most $(n^{1/2\ell})^{\ell}=\sqrt{n}$, and hence has an acyclic set of size $\sqrt{n}$. Consequently, the special case of Conjecture 1.2 when $H=C_{3}$ would hold for $\varepsilon:=1/2\ell$.

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 $\alpha(D)\leq\alpha$, then $\chi(D)\leq 35^{\alpha-1}\alpha!$ 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 $\chi(T)\leq g(k)$. 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 $\alpha(D)\leq\alpha$ and that the set of out-neighbors of each vertex has chromatic number at most $k$, then $\chi(D)\leq g(k,\alpha)$.

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. Sections 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.

1.1 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 $D$. For every $v\in V(D)$, we denote by $N_{D}^{+}(v)$ (respectively, $N_{D}^{-}(v)$) the set of out-neighbors (respectively, in-neighbors) of $v$ in $D$. Let $N(v)=N_{D}^{+}(v)\cup N_{D}^{-}(v)$. For every $X\subseteq V(D)$, let $N^{+}_{D}(X)=\bigcup_{v\in X}N^{+}_{D}(v)$ (respectively, $N^{-}_{D}(X)=\bigcup_{v\in X}N^{-}_{D}(v)$), the set of vertices seen by (respectively, seeing) at least one vertex of $X$, and let $M^{+}_{D}(X)=\bigcap_{v\in X}N^{+}_{D}(v)$ (respectively, $M^{-}_{D}(X)=\bigcap_{v\in X}N^{-}_{D}(v)$), the set of vertices seen by (respectively, seeing) all vertices of $X$. Let $N_{D}(X)$ denote $N_{D}^{+}(X)\cup N_{D}^{-}(X)$. We say that two vertices $u,v$ are non-adjacent if there is no arc with endpoints $\{u,v\}$. For every $v\in V(D)$, we denote by $N_{D}^{o}(v)$ the set non-adjacent vertices of $v$ in $D$. For a subset $X$ of $V(D)$, we denote by $N^{o}_{D}(X)$ the set of vertices of $V$ non-adjacent to at least one vertex of $X$. When it is clear from context (most of the time), we omit the subscript $D$ in this notation. We will use throughout the paper the fact that $V(D)\setminus{X}=M^{+}_{D}(X)\cup N^{-}_{D}(X)\cup N^{o}_{D}(X)$ for any $X\subseteq V(D)$.

Given a digraph $D$ and a set $X\subseteq V(D)$, 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 $X\subseteq V(D)$ is a dominating set of $D$ if every vertex $v\in V(D)\setminus{X}$ 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 $X,Y\subseteq V(D)$, we denote $X\to_{D}Y$ (or just $X\to Y$) if there is no arc from $Y$ to $X$ in $D$. A key observation is that if $X\to Y$, then \chi(X\cup Y)=\max\big{(}\chi(X),\chi(Y)\big{)}.

A side remark is that some proofs in this paper that proceed by induction on $\alpha$ use the fact that if $\alpha(D)\leq\alpha$, then $\alpha(N^{o}(v))\leq\alpha-1$ for every $v$, and thus $\chi(N^{o}(v))$ is bounded. In these inductive proofs, we often cite known results on tournaments for the base case $\alpha=1$. However, our proofs are indeed self-contained since to prove the base case $\alpha=1$, we just repeat the same arguments and use the fact that in a tournament, $N^{o}(v)=\emptyset$ 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 $\Delta(H,T_{k},T_{1})$ and $\Delta(H,T_{1},T_{k})$).

2 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 $\alpha(D)$.

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. Then $V(D)\setminus\{v\}=N^{+}(v)\cup N^{o}(v)$. Applying induction to $D[N^{o}(v)]$, we obtain a stable dominating set $S^{\prime}$ of $D[N^{o}(v)]$. Clearly $S:=S^{\prime}\cup\{v\}$ is a dominating set of $D$. Note that $S^{\prime}\subseteq N^{o}(v)$, and so $S$ is stable. ∎

Given $t\geq 1$, a digraph $D$ is $t$-local if for every vertex $v$ we have $\chi(N^{+}(v))\leq t$. Let $\cal C$ be a class of digraphs closed under taking subdigraphs. We say that $\cal C$ is tamed if for every integer $k$ there exists $K$ and $\ell$ such that every digraph $T\in\cal C$ with $\chi(T)\geq K$ contains a set $A$ of $\ell$ vertices such that $\chi(A)\geq k$. Note that a class of digraphs with bounded chromatic number is indeed tamed.

The following proposition is straightforward.

Proposition 2.2**.**

Let $D$ be a $t$-local digraph. Then for every subset $X$ of $V(D)$, \chi\big{(}N^{+}(X)\cup X\big{)}\leq t|X|.

Let us restate Theorem 1.8.

Theorem 2.3**.**

For every pair of positive integers $\alpha$ and $t$, there is a function $f_{\alpha}(t)$ such that every $t$-local digraph $D$ with $\alpha(D)\leq\alpha$ has chromatic number at most $f_{\alpha}(t)$.

Proof.

We proceed by induction on $\alpha$. The case $\alpha=1$ was proved in [7]. Suppose that Theorem 2.3 is true for $\alpha-1$ (i.e., $f_{\alpha-1}(t)$ exists for every $t$).

Claim 2.4**.**

For every $t$, the class of $t$-local digraphs with independence number at most $\alpha$ is tamed.

Proof..

We fix some arbitrary $t$ and show the property by induction on $k$. The claim is trivial for $k=1$. Assuming now that $(K,\ell)$ exists for $k$, we want to find $(K^{\prime},\ell^{\prime})$ for $k+1$. For this, we set $s:=K+\ell f_{\alpha-1}(t)+\ell t$ and

[TABLE]

and fix $\ell^{\prime}$ later.

Let $D$ be a $t$-local digraph with vertex set $V$ such that $\alpha(D)\leq\alpha$ and $\chi(D)\geq K^{\prime}$. Let $B$ be a dominating set of $D$ of minimum size $b$. By Proposition 2.2, we have

[TABLE]

In particular, $b\geq K^{\prime}/(t+1)\geq 2k(\alpha s+1)$. Consider a subset $W$ of $B$ of size $k(\alpha s+1)$. By Proposition 2.2, we have $\chi(N^{+}(W))\leq kt(\alpha s+1)$. By induction hypothesis on $\alpha-1$, for every $x\in W$, we have $\chi(N^{o}(x))\leq f_{\alpha-1}(t)$ since $N^{o}(x)$ is $t$-local and $\alpha(N^{o}(x))\leq\alpha-1$. Hence, (recalling that $M^{-}(X):=\bigcap_{x\in X}N^{-}(x)$ for any $X\subseteq D$)

[TABLE]

In particular, by the tamed property applied to $k$, one can find a set $A\subset M^{-}(W)$ such that $A$ has $\ell$ vertices and $\chi(A)\geq k$.

Consider now a subset $S$ of $W$ of size $\alpha s+1$. We claim that $\chi(N^{+}(S))\geq s$. If not, we can cover $N^{+}(S)$ by at most $s-1$ acyclic sets. Since every acyclic set has independence number at most $\alpha$, it has a dominating set of size at most $\alpha$ by Proposition 2.1. Hence $N^{+}(S)$ has a dominating set, say $S^{\prime}$ of size at most $\alpha(s-1)\leq|S|-2$. But this yields a contradiction since the set $(B\setminus S)\cup S^{\prime}\cup\{x\}$, where $x$ is an arbitrary vertex in $A$, would be a dominating set of $T$ of size less than $|B|$. Therefore, $\chi(N^{+}(S))\geq s$.

By Proposition 2.2 applied to $N^{+}(A)$, we have $\chi(N^{+}(A))\leq\ell t$. Hence

[TABLE]

Thus, by the tamed property applied to $k$, there is a subset $A_{S}$ of $N^{+}(S)\cap M^{-}(A)$ such that $|A_{S}|=\ell$ and $\chi(A_{S})\geq k$.

We now construct our subset of $V$ with chromatic number at least $k+1$. For this we consider the set $A^{\prime}$ consisting of vertices $A\cup W$ to which we add the collection of $A_{S}$, for all subsets $S\subseteq W$ of size $\alpha s+1$. Observe that the number of vertices of $A^{\prime}$ is at most

[TABLE]

To conclude, it is sufficient to show that $\chi(A^{\prime})\geq k+1$. Suppose not, and for contradiction, take a $k$-coloring of $A^{\prime}$. Since $|W|=k(\alpha s+1)$ there is a monochromatic set $S$ in $W$ of size $\alpha s+1$ (say, colored 1). Recall that $A_{S}\subseteq M^{-}(A)$ and $A\subseteq M^{-}(W)\subseteq M^{-}(S)$, so we have all arcs from $A_{S}$ to $A$ and all arcs from $A$ to $S$, and note that since $\chi(A)\geq k$ and $\chi(A_{S})\geq k$, both $A$ and $A_{S}$ have a vertex of each of the $k$ colors. Hence there are $u\in A$ and $w\in A_{S}$ colored 1. Since $A_{S}\subseteq N^{+}(S)$, there is $v\in S$ such that $vw$ is an arc. We then obtain the monochromatic cycle $uvw$ of color 1, a contradiction. Thus, $\chi(A^{\prime})\geq k+1$, completing the proof of the claim. $\Diamond$

We now can finish the proof of the theorem. Since the class of $t$-local digraphs with independence number at most $\alpha$ is tamed, by applying tamed property for $k=t+f_{\alpha-1}(t)+1$, we have that there exist $(K,\ell)$ such that every $t$-local digraph $D$ with $\alpha(D)\leq\alpha$ and $\chi(D)\geq K$ contains a set $A$ of $\ell$ vertices and $\chi(A)\geq t+f_{\alpha-1}(t)+1$. We claim that $A$ is a dominating set. If not, then there is a vertex $v$ such that $A\subseteq N^{o}(v)\cup N^{+}(v)$. Then $\chi(A)\leq\chi(N^{o}(v))+\chi(N^{+}(v))\leq t+f_{\alpha-1}(t)$, a contradiction. Hence, $A$ is a dominating set of $D$. Thus, $\chi(D)=\chi(\cup_{x\in A}(N^{+}(x)\cup\{x\}))\leq(t+1)|A|=\ell+\ell t$. Consequently, $t$-local digraphs have chromatic number at most $f(t):=\max(K,\ell+\ell t)$. ∎

By reversing the directions of all arcs, we have the following theorem.

Theorem 2.5**.**

For every $t$ and $\alpha$, there is $c$ such that every directed graph $D$ with $\alpha(D)\leq\alpha$ and $\chi(N^{-}(v))\leq t$ for every $v\in V(D)$ has chromatic number at most $c$.

3 Proof of Theorem 1.4

Theorem 1.4 states that if $H_{1}$ and $H_{2}$ are superheroes, then so is $H_{1}\Rightarrow H_{2}$. 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: $(C_{3}\Rightarrow C_{3})$-free tournaments have bounded chromatic number. Given a $(C_{3}\Rightarrow C_{3})$-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 part of this partition has chromatic number at least 2, and so contains a copy of $C_{3}$. Let $Y_{Q_{1},Q_{2}}\subseteq V(D)\setminus{Q}$ 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 $C_{3}\Rightarrow C_{3}$. Note that $V(T)\setminus Q$ is covered by only $2^{|Q|}$ such sets $Y_{Q_{1},Q_{2}}$, and hence $\chi(T)$ 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 $N^{-}(u)\cap N^{+}(v)$ 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 $\chi(T)$ 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 $r,s\geq 1$,

• •

An arc $e=uv$ of $D$ is $r$-thick if $N^{-}(u)\cap N^{+}(v)$ contains an $r$-mountain. An $r$-mountain in $N^{-}(u)\cap N^{+}(v)$ is a certificate of $r$-thickness of $e$.

• •

An $(r,s)$-clique of $D$ is a set $S\subseteq V(D)$ such that $|S|=s$, and for every distinct vertices $u,v\in S$, 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 $u,v\in S$, then the tournament induced on $S\cup(\bigcup_{u,v\in S}C_{u,v})$ 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.

Lemma 3.1** ([2], Lemma 3.3).**

Every $r$-mountain has chromatic number at least $r$, and has at most $(r!)^{2}$ vertices.

Fix two superheroes $H_{1}$ and $H_{2}$.

Lemma 3.2**.**

Fix $\alpha\geq 2$ and $r,s\geq 1$, suppose that there are $b_{0},b_{1},b_{2}$ such that

• •

Every $(H_{1}\Rightarrow H_{2})$-free digraph $D$ with $\alpha(D)\leq\alpha-1$ has $\chi(D)\leq b_{0}$.

• •

Every $(H_{1}\Rightarrow H_{2})$-free digraph $D$ with $\alpha(D)\leq\alpha$ and containing no $(r,s)$-clique has $\chi(D)\leq b_{1}$.

• •

Every $(H_{1}\Rightarrow H_{2})$-free digraph $D$ with $\alpha(D)\leq\alpha$ and containing no $r$-mountain has $\chi(D)\leq b_{2}$.

Then there is $b_{3}$ such that every $(H_{1}\Rightarrow H_{2})$-free digraph $D$ with $\alpha(D)\leq\alpha$ and containing no $(r,s+1)$-clique has $\chi(D)\leq b_{3}$.

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.

First, note that 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 $\alpha(D)\leq\alpha$ has $\chi(D)\leq b_{4}$. We first identify all $r$-thick arcs of $D$. Fix an arbitrary vertex $v$. Then $V(D)\setminus{\{v\}}$ can be partitioned into four sets: $N^{*}$ the set of neighbors of $v$ that are connected to $v$ by an $r$-thick arc; $N^{-}=N^{-}(v)\setminus{N^{*}}$; $N^{+}=N^{+}(v)\setminus{N^{*}}$; and $N^{o}(v)$.

Note that $N^{o}(v)$ is $(H_{1}\Rightarrow H_{2})$-free and $\alpha(N^{o}(v))\leq\alpha-1$, and so by the first hypothesis, $\chi(N^{o}(v))\leq b_{0}$. 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, $\chi(N^{*})\leq b_{1}$.

Claim 3.3**.**

There is $b_{5}$ such that either $\chi(N^{-})\leq b_{5}$ or $\chi(N^{+})\leq b_{5}$.

Proof..

Suppose that $\chi(N^{-})>b_{4}$, then $N^{-}$ contains a copy of $H_{1}$, say $\hat{H}_{1}$. Note that

[TABLE]

• •

If $\chi(M^{+}(\hat{H}_{1})\cap N^{+})>b_{4}$, then $M^{+}(\hat{H}_{1})\cap N^{+}$ contains a copy of $H_{2}$, say $\hat{H}_{2}$, and we have $\hat{H}_{1}\Rightarrow\hat{H}_{2}$ forming a copy of $H_{1}\Rightarrow H_{2}$, a contradiction. Hence, $\chi(M^{+}(\hat{H}_{1})\cap N^{+})\leq b_{4}$

• •

For each $u\in\hat{H}_{1}$, we have $\alpha(N^{o}(u)\cap N^{+})\leq\alpha-1$, so $\chi(N^{o}(u)\cap N^{+})\leq b_{0}$.

• •

For each $u\in\hat{H}_{1}$, if $\chi(N^{-}(u)\cap N^{+})\geq b_{2}$, then $N^{-}(u)\cap N^{+}$ contains a $r$-mountain. This means that $uv$ is an $r$-thick arc, contradicting $u\notin N^{*}$. Hence $\chi(N^{-}(u)\cap N^{+})\leq b_{2}$, for each $u\in\hat{H}_{1}$.

Thus we have,

[TABLE]

Set $b_{5}:=b_{4}+|H_{1}|(b_{0}+b_{2})$. We have just shown that if $\chi(N^{-})>b_{4}$, then $\chi(N^{+})\leq b_{5}$. Hence either $\chi(N^{-})\leq b_{4}\leq b_{5}$ or $\chi(N^{+})\leq b_{5}$. This proves the claim. $\Diamond$

Note that $N^{+}(v)\subseteq N^{+}\cup N^{*}$, and so $\chi(N^{+}(v))\leq\chi(N^{+})+\chi(N^{*})$. Similarly, $\chi(N^{-}(v))\leq\chi(N^{-})+\chi(N^{*})$. Hence for every $v\in V(D)$, either $\chi(N^{+}(v))\leq b_{5}+b_{1}$ or $\chi(N^{-}(v))\leq b_{5}+b_{1}$. Let $R$ be the set of all vertices $v\in V(D)$ with $\chi(N^{+}(v))\leq b_{5}+b_{1}$ and $B$ be the set of all vertices $v\in V(D)$ with $\chi(N^{-}(v))\leq b_{5}+b_{1}$. Note that $R\cup B=V(D)$.

Observe that $R$ is a digraph with $\alpha(R)\leq\alpha$, and $\chi(N^{+}_{R}(v))\leq\chi(N^{+}_{D}(v))\leq b_{5}+b_{1}$ for every $v\in R$. Then applying Theorem 2.3 to $R$ with $t=b_{5}+b_{1}$, there is $b_{6}$ such that $\chi(R)\leq b_{6}$. Similarly, by Theorem 2.5, there is $b_{7}$ such that $\chi(B)\leq b_{7}$. Hence $\chi(D)\leq\chi(R)+\chi(B)\leq b_{6}+b_{7}$. Setting $b_{3}:=b_{6}+b_{7}$ completes the proof of Lemma 3.2. ∎

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 that do not contain an $r$-mountain have bounded chromatic number.

Lemma 3.4**.**

Let $\alpha\geq 2$, and suppose that every $(H_{1}\Rightarrow H_{2})$-free digraph $D$ with $\alpha(D)\leq\alpha-1$ has $\chi(D)\leq b_{0}$ for some $b_{0}$. Then for every $r$, there exists $g_{\alpha}(r)$ such that every $(H_{1}\Rightarrow H_{2})$-free digraph $D$ with $\alpha(D)\leq\alpha$ and not containing an $r$-mountain has $\chi(D)\leq g_{\alpha}(r)$.

Proof.

We proceed by induction on $r$. If $D$ contains no $1$-mountain, then $D$ has no vertices, and we can set $g_{{\alpha}}(r):=0$. Now suppose by induction that $g_{{\alpha}}({r})$ exists. We will show that $g_{{\alpha}}({r}+1)$ exists. First, we claim the following.

1. (A)

For every $s$, there exists function $g^{\prime}_{{\alpha},{r}}(s)$ such that if $D$ contains no $({r},s)$-clique, then $\chi(D)\leq g^{\prime}_{{\alpha},{r}}(s)$.

We prove (A) by induction on $s$. For $s=1$, if $D$ contains no $({r},1)$-clique, then $D$ has no vertex, so $g^{\prime}_{{\alpha},{r}}(s)=0$. Suppose, by induction, that $g^{\prime}_{{\alpha},{r}}(s)$ exists. Let $D$ be a digraph not containing a $({r},s+1)$-clique. Applying Lemma 3.2 with $b_{1}=g^{\prime}_{{\alpha},{r}}(s)$ and $b_{2}=g_{{\alpha}}({r})$, we deduce that $g^{\prime}_{{\alpha},{r}}(s+1)$ exists. This proves (A).

If $D$ contains no $({r}+1)$-mountain, then $D$ contains no $({r},{r}+1)$-clique, implying $\chi(D)\leq g^{\prime}_{{\alpha},{r}}({r}+1)$. Set $g_{{\alpha}}({r+1}):=g^{\prime}_{{\alpha},{r}}({r}+1)$. This completes the proof. ∎

To prove Theorem 1.4, it suffices to prove the following lemma.

Lemma 3.5**.**

For every integer $\alpha\geq 1$, there exists $f(\alpha)$ such that every $(H_{1}\Rightarrow H_{2})$-free digraph $D$ with $\alpha(D)\leq\alpha$ has $\chi(D)\leq f(\alpha)$.

Proof.

We proceed by induction on $\alpha$. Since $H_{1}\Rightarrow H_{2}$ is a hero (see [2], Theorem 3.2), Lemma 3.5 is true for $\alpha=1$. Suppose that Lemma 3.5 is true for $\alpha-1$ and let $c_{0}=f(\alpha-1)$. 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 $\chi(D)\leq c_{1}-1$. Let $D$ be a $(H_{1}\Rightarrow H_{2})$-free digraph with vertex set $V$ and $\alpha(D)\leq\alpha$. 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 $\chi(D)\leq c_{2}$. Thus, it remains to consider the case that $D$ contains a $(c_{0}+2c_{1})$-mountain.

Claim 3.6**.**

There is $c_{3}$ such that if $D$ contains a $(c_{0}+2c_{1})$-mountain, then $\chi(D)\leq c_{3}$.

Proof..

Let $Q$ be a $(c_{0}+2c_{1})$-mountain of $D$. Then by Lemma 3.1, $|Q|\leq((c_{0}+2c_{1})!)^{2}$ and $\chi(Q)\geq c_{0}+2c_{1}$. For every partition $Q$ into three sets $Q_{0},Q_{1},Q_{2}$, let $Y_{Q_{0},Q_{1},Q_{2}}$ be the set of vertices $v\in V\setminus{Q}$ such that $Q_{0}\subseteq N^{o}(v),Q_{1}\subseteq N^{+}(v)$, and $Q_{2}\subseteq N^{-}(v)$. Note that for every vertex $v\in V\setminus{Q}$, 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, $V\setminus{Q}$ can be written as the union of all possible $Y_{Q_{0},Q_{1},Q_{2}}$. There are $3^{|Q|}$ sets $Y_{Q_{0},Q_{1},Q_{2}}$.

1. (B)

$\chi(Y_{Q_{0},Q_{1},Q_{2}})\leq c_{1}$ for every partition $(Q_{0},Q_{1},Q_{2})$ of $Q$.

Indeed, if $Y_{Q_{0},Q_{1},Q_{2}}=\emptyset$, then (B) clearly holds. Otherwise, $Q_{0}\subseteq N^{o}(v)$ for any $v\in Y_{Q_{0},Q_{1},Q_{2}}$. Note that the digraph $Q_{0}$ is $(H_{1}\Rightarrow H_{2})$-free and $\alpha(Q_{0})\leq\alpha-1$, and so by induction hypothesis, $\chi(Q_{0})\leq c_{0}.$ This gives $\chi(Q_{1}\cup Q_{2})\geq\chi(Q)-\chi(Q_{0})\geq 2c_{1}$, implying that either $\chi(Q_{1})\geq c_{1}$ or $\chi(Q_{2})\geq c_{1}$. If $\chi(Q_{1})\geq c_{1}$, then $Q_{1}$ contains a copy of $H_{2}$, say $\hat{H}_{2}$. If $\chi(Y_{Q_{0},Q_{1},Q_{2}})\geq c_{1}$, then $Y_{Q_{0},Q_{1},Q_{2}}$ contains a copy of $H_{1}$, say $\hat{H}_{1}$. Then $\hat{H}_{1}\Rightarrow\hat{H}_{2}$ forms a copy of $H_{1}\Rightarrow H_{2}$, a contradiction. Hence, $\chi(Y_{Q_{0},Q_{1},Q_{2}})\leq c_{1}$. A similar argument establishes the case $\chi(Q_{2})\geq c_{1}$. This proves (B).

Hence

[TABLE]

Set $c_{3}:=((c_{0}+2c_{1})!)^{2}+3^{((c_{0}+2b_{1})!)^{2}}c_{1}$. This completes proof of the claim. $\Diamond$

Hence $\chi(D)\leq\max(c_{2},c_{3})$. Setting $f(\alpha):=\max(c_{2},c_{3})$ completes the proof of Lemma 3.5, thus proving Theorem 1.4. ∎

4 Proof of Theorem 1.5

It is proved in [14] that for each integer $k\geq 1$, every tournament with at least $2^{k-1}$ vertices contains a copy of $T_{k}$. Let ${\cal R}(a,b)$ 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 $b$).

Proposition 4.1**.**

For each integer $k\geq 1$, every $T_{k}$-free digraph $D$ with $\alpha(D)\leq\alpha$ has at most ${\cal R}(\alpha+1,2^{k-1})$ vertices.

Proof.

Suppose for a contradiction that there is a $T_{k}$-free digraph $D$ with at least ${\cal R}(\alpha+1,2^{k-1})$ vertices. Then the underlying graph of $D$ contains either an independent set of size $\alpha+1$ or a clique of size $2^{k-1}$. The former case is impossible since $\alpha(D)\leq\alpha$. 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 $\Delta(H,T_{k},T_{1})$ and $\Delta(H,T_{1},T_{k})$ for any integer $k\geq 1$. We will prove that if $H$ is a superhero, then so is $\Delta(H,T_{k},T_{1})$ for any $k\geq 1$. 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, $\Delta(H_{rev},T_{k},T_{1})_{rev}=\Delta(H,T_{1},T_{k})$ is also a superhero.

Theorem 4.2**.**

For every superhero $H$ and every pair of integers $k,\alpha\geq 1$, there is a number $f(H,k,\alpha)$ such that every $\Delta(H,T_{k},T_{1})$-free digraph $D$ with $\alpha(D)\leq\alpha$ has $\chi(D)\leq f(H,k,\alpha)$.

The idea of the proof of Theorem 4.2 is as follows. Fix a large number $c$ and call a subset $B$ of $V(D)$ such that $\chi(B)=c$ a bag. We aim to find a longest chain of disjoint bags $B_{1},\dots,B_{t}$ in $V(D)$, together with a partition $V(D)\setminus{\bigcup B_{i}}$ into sets we call zones $Z_{0},\dots,Z_{t}$ 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. Once proving this, we observe that all $B_{i}$ and $Z_{i}$ have bounded chromatic number and since $D$ has no backward arc, $D$ 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.

For an integer $c\geq 1$ and a $\Delta(H,T_{k},T_{1})$-free digraph $D$ with $\alpha(D)\leq\alpha$, a set $B\subseteq V(D)$ such that $\chi(B)=c$ is called a $c$-bag. A family of pairwise disjoint $c$-bags $B_{1},\dots,B_{t}$ is a $c$-bag-chain if for every $i$ and every $v\in B_{i}$, we have $\chi(N^{+}(v)\cap B_{i-1})\leq c_{1}$ and $\chi(N^{-}(v)\cap B_{i+1})\leq c_{1}$, where $c_{1}$ is a fixed number satisfying

• •

$c_{1}\geq\mathcal{R}(\alpha+1,2^{k-1})$, and

• •

$c_{1}\geq\chi(D)$ for every $H$-free digraph $D$ with $\alpha(D)\leq\alpha$.

Since $H$ is a superhero, such $c_{1}$ clearly exists. Note that every $T_{k}$-free digraph $D$ with $\alpha(D)\leq\alpha$ 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 $\alpha$. Since $\Delta(H,T_{k},T_{1})$ is a hero (see [2], Theorem 4.1), the theorem holds for $\alpha=1$. Suppose that Theorem 4.2 is true for $\alpha-1$ and let $c_{0}=f(H,k,\alpha-1)$. Let $D$ be a $\Delta(H,T_{k},T_{1})$-free digraph with vertex set $V$ and $\alpha(D)\leq\alpha$. An important observation is that for every $v\in V$, the digraph $N^{o}(v)$ is $\Delta(H,T_{k},T_{1})$-free and has independence number at most $\alpha-1$, and hence

[TABLE]

We also would like to recall some useful formulas. For every $v\in X\subseteq V$,

[TABLE]

and if $Y\subseteq V$ and $Y\cap X=\emptyset$, then (recalling that $M^{+}(X)$ is the set of vertices seen by all vertices of $X$)

[TABLE]

Let $h=|H|$ and set $c:=2(c_{0}+c_{1})(h+k)$. Let us assume that $B_{1},\dots,B_{t}$ is a $c$-bag-chain of $D$ with $t$ as large as possible. In the proof of this theorem, we drop prefix $c$- of $c$-bag and $c$-bag-chain for convenience. By definition of bag-chain, every bag has few backward arcs with bags preceding or succeeding it. In the following claim, we show that in fact every bag has few backward arcs with any other bag.

Claim 4.3**.**

For every $i$ and $v\in B_{i}$, and for every $r>0$,

1. (a)

$\chi(N^{+}(v)\cap B_{i-r})\leq c_{1}$, and 2. (b)

$\chi(N^{-}(v)\cap B_{i+r})\leq c_{1}$.

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 $\chi(N^{+}(v)\cap B_{i-r})>c_{1}$. Then $N^{+}(v)\cap B_{i-r}$ has a copy of $H$, say $\hat{H}$. Then by applying (4.3) we have

[TABLE]

and

[TABLE]

Thus (by using the fact that if $A=B\cup C=B^{\prime}\cup C^{\prime}$, then $A=(B\cap B^{\prime})\cup C\cup C^{\prime}$) we have

[TABLE]

For each $x\in\hat{H}$, by (4.1) we have $\chi(N^{o}(x)\cap B_{i-1})\leq c_{0}$, and by induction hypothesis of (b) applied to $x$ and $r-1$, we have $\chi(N^{-}(x)\cap B_{i-1})\leq c_{1}$. We also have $\chi(N^{o}(v)\cap B_{i-1})\leq c_{0}$ by (4.1) and $\chi(N^{+}(v)\cap B_{i-1})\leq c_{1}$ by definition of a bag-chain. Combining with (4.4) and (4.2) we have

[TABLE]

Then there exists a copy of $T_{k}$ in $M^{+}(\hat{H})\cap N^{-}(v)\cap B_{i-1}$, say $\hat{T_{k}}$. Note that by construction, we have all arcs from $\hat{H}$ to $\hat{T}_{k}$, from $\hat{T}_{k}$ to $v$ and from $v$ to $\hat{H}$. Then $\Delta(\hat{H},\hat{T_{k}},v)$ forms a copy of $\Delta(H,T_{k},T_{1})$, a contradiction.

The proof of (b) for $r$ is similar but not symmetric. In order to obtain a copy of $\Delta(H,T_{k},T_{1})$, 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. $\Diamond$

We next prove a stronger statement that every bag has few backward arcs with the union of all other bags preceding or succeeding it. Let $B_{i,j}=\bigcup_{s=i}^{j}B_{s}$ for any $1\leq i\leq j\leq t$. (If $i<1$ or $j>t$ or $j<i$, we set $B_{i,j}=\emptyset$.)

Claim 4.4**.**

For every $i$ and $v\in B_{i}$,

• •

$\chi(N^{+}(v)\cap B_{1,i-2})\leq c_{1}$, and

• •

$\chi(N^{-}(v)\cap B_{i+2,t}))\leq c_{1}$.

Proof..

We repeat the same argument as in the proof of Claim 4.3. Towards a contradiction, suppose that the first statement is false: there is $v$ in some $B_{i}$ such that $\chi(N^{+}(v)\cap B_{1,i-2})>c_{1}$. Then $N^{+}(v)\cap B_{1,i-2}$ has a copy of $H$, say $\hat{H}$. For each $x\in\hat{H}$, we have $\chi(N^{o}(x)\cap B_{i-1})\leq c_{0}$ by (4.1), and $\chi(N^{-}(x)\cap B_{i-1})\leq c_{1}$ by Claim 4.3. We also have $\chi(N^{o}(v)\cap B_{i-1})\leq c_{0}$ and $\chi(N^{+}(v)\cap B_{i-1})\leq c_{1}$. Thus by the same computation as in Claim 4.3, we obtain a copy of $T_{k}$ in $B_{i-1}$ and reach the contradiction. The proof of the second statement is similar. $\Diamond$

From Claim 4.4 we have the following immediate corollary.

Claim 4.5**.**

For every $i$ and $v\in B_{i}$,

• •

$\chi(N^{+}(v)\cap B_{1,i-1})\leq 2c_{1}$.

• •

$\chi(N^{-}(v)\cap B_{i+1,t})\leq 2c_{1}$.

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 $\chi(B_{i})$ is bounded for every $i$. We will re-use the arguments in this proof for subsequent claims.

Claim 4.6**.**

$\chi(B_{1,t})\leq 8c(c_{1}+1)$.

Proof..

An arc $uv$ with $u\in B_{j},v\in B_{i},$ and $j>i$ is called a backarc with span $j-i$. For every $i$ and every $v\in B_{i}$, if $|N^{-}(v)\cap B_{i+1,t}|<c_{1}$, let $F_{v}:=N^{-}(v)\cap B_{i+1,t}$. If $|N^{-}(v)\cap B_{i+1,t}|\geq c_{1}$, let $F_{v}\subseteq N^{-}(v)\cap B_{i+1,t}$ consist of $c_{1}$ vertices whose backarcs to $v$ have largest possible spans. Let us show that

1. (A)

For every backarc $uv$ with $u\notin F_{v}$, if $u\in B_{j}$ and $v\in B_{i}$ then $\chi(B_{i,j})\leq 4c$.

If $j=i+1$, then (A) clearly holds since $\chi(B_{i,i+1})\leq\chi(B_{i})+\chi(B_{i+1})\leq 2c$. Thus we may suppose that $j\geq i+2$. Since $u\in N^{-}(v)\cap B_{i+1,t}$ but $u\notin F_{v}$, it follows from the definition of $F_{v}$ that $|F_{v}|=c_{1}$. Hence $|F_{v}\cup\{u\}|=c_{1}+1$. Then there is a copy of $T_{k}$ in $F_{v}\cup\{u\}$, say $\hat{T}_{k}$. Note that $\hat{T}_{k}\subseteq B_{j,t}$.

We have a formula similar to (4.4):

[TABLE]

For each $x\in\hat{T}_{k}\cup\{v\}$, we have $\chi(N^{o}(x)\cap B_{i+1,j-1})\leq c_{0}$. Note also that from Claim 4.5, we have $\chi(N^{+}(x)\cap B_{i+1,j-1})\leq 2c_{1}$ for every $x\in\hat{T}_{k}$ and $\chi(N^{-}(v)\cap B_{i+1,j-1})\leq 2c_{1}$. Furthermore, if $\chi(M^{-}(\hat{T}_{k})\cap N^{+}(v)\cap B_{i+1,j-1})>c_{1}$, then $M^{-}(\hat{T}_{k})\cap N^{+}(v)\cap B_{i+1,j-1}$ contains a copy of $H$, say $\hat{H}$. Then $\Delta(\hat{H},\hat{T}_{k},v)$ forms a copy of $\Delta(H,T_{k},T_{1})$, a contradiction. Hence

[TABLE]

since $c=2(c_{0}+c_{1})(h+k)$. This gives $\chi(B_{i,j})\leq\chi(B_{i})+\chi(B_{i+1,j-1})+\chi(B_{j})\leq 4c$, which proves (A).

Now let $G$ be the undirected graph with vertex set $B_{1,t}$ and $uv\in E(G)$ if $u\in F_{v}$ or $v\in F_{u}$. Then $B_{i}$ is a stable set in $G$ for every $i$. We now color the vertices of $G$ by $c_{1}+1$ colors as follows. First, color all $B_{t}$ properly by color

1. Suppose that we have already colored $B_{i+1},\dots,B_{t}$. Every vertex $v$ in $B_{i}$ is incident (in $G$) with at most $c_{1}$ vertices in $B_{i+1,t}$ (those belonging 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 $c_{1}+1$ sets of colors, say $X_{1},\dots,X_{c_{1}+1}$, 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.

2. (B)

$\chi(X_{s})\leq 8c$ for every $1\leq s\leq c_{1}+1$.

To prove (B), we define a sequence of indices $i_{1},i_{2},\dots$ inductively as follows. Let $i_{1}=1$, and for every $r\geq 1$, let $i_{r+1}>i_{r}$ be the smallest index such that $\chi(B_{i_{r},i_{r+1}})>4c$. The sequence ends by $i_{\ell}$ with $\chi(B_{i_{\ell},t})\leq 4c$ (i.e., there is no $i_{\ell+1}$ satisfying the condition). Set $A_{r}:=B_{i_{r},i_{r+1}-1}$ for every $1\leq r\leq\ell-1$ and $A_{\ell}:=B_{i_{\ell},t}$. Then $B_{1,t}=\bigcup_{r=1}^{\ell}A_{r}$, and by definition of the sequence, $\chi(B_{i_{r},i_{r+1}-1})\leq 4c$ for every $1\leq r\leq\ell-1$. In other words, for every $1\leq r\leq\ell$,

[TABLE]

Suppose that there is a backarc $uv$ with $u\notin F_{v}$ and $u\in A_{r},v\in A_{r^{\prime}}$, where $r\geq r^{\prime}+2$. Suppose that $u\in B_{j}$ and $v\in B_{j^{\prime}}$. Then $j\geq i_{r}$ since $B_{j}\subseteq A_{r}=B_{i_{r},i_{r+1}-1}$, and $j^{\prime}<i_{r^{\prime}+1}$ since $B_{j^{\prime}}\subseteq A_{r^{\prime}}=B_{i_{r^{\prime}},i_{r^{\prime}+1}-1}$, and so $j^{\prime}<i_{r-1}$ since $r^{\prime}+1\leq r-1$. Also note that $\chi(B_{i_{r-1},i_{r}})>4c$ by definition of the sequence. Thus we have

[TABLE]

which contradicts (A) in that $\chi(B_{j^{\prime},j})\leq 4c$ if $u\notin F_{v}$. This shows that for any $r\geq r^{\prime}+2$, there is no backarc $uv$ with $u\in A_{r},v\in A_{r^{\prime}}$ and $u\notin F_{v}$.

Now fix an arbitrary $X_{s}$ and let $X_{s,r}=X_{s}\cap A_{r}$ for every $1\leq r\leq\ell$. Observe that if $uv$ is a backarc with $u,v\in X_{s}$, then $u\notin F_{v}$ (otherwise, $u\in F_{v}$, so $uv\in G$ 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 $X_{s,r^{\prime}}$ for any $r,r^{\prime}$ with $r\geq r^{\prime}+2$; in other words, $X_{s,r^{\prime}}\to X_{s,r}$ for any $r,r^{\prime}$ with $r\geq r^{\prime}+2$. Recall from (4.5) that $\chi(X_{s,r})\leq 4c$. Thus \chi\big{(}\bigcup_{r\geq 1}X_{s,2r-1}\big{)}\leq 4c and \chi\big{(}\bigcup_{r\geq 1}X_{s,2r}\big{)}\leq 4c. This gives \chi(X_{s})=\chi\big{(}\bigcup_{r\geq 1}X_{s,r}\big{)}\leq 8c, which proves (B).

Hence

[TABLE]

This completes the proof of Claim 4.6. $\Diamond$

We now turn our attention to the vertices not in the bags. We partition $V\setminus{B_{1,t}}$ into sets $Z_{0},\dots,Z_{t}$ called zones as follows. For every $x\in V\setminus{B_{1,t}}$, if $i$ is the largest index such that $\chi(N^{-}(x)\cap B_{i})>c_{1}$, then $x\in Z_{i}$; otherwise, $x\in Z_{0}$. We first show that every $Z_{i}$ has few backward arcs with any $B_{j}$ sufficiently far from it.

Claim 4.7**.**

For every $i$ and $v\in Z_{i}$,

• •

$\chi(N^{-}(v)\cap B_{i+r})\leq c_{1}$* for every $r\geq 1$, and*

• •

$\chi(N^{+}(v)\cap B_{i-r})\leq c_{1}$* for every $r\geq 2$.*

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 $\chi(N^{+}(v)\cap B_{i-r})>c_{1}$ for some $r\geq 2$. Then there is a copy of $H$ in $N^{+}(v)\cap B_{i-r}$, say $\hat{H}$. Since $\chi(N^{-}(v)\cap B_{i})>c_{1}$ by partition criterion, there is a copy of $T_{k}$ in $N^{-}(v)\cap B_{i}$, say $\hat{T}_{k}$. Then by applying (4.3) we have

[TABLE]

[TABLE]

and

[TABLE]

Let R=M^{+}(\hat{H})\cap M^{-}(\hat{T}_{k})\cap\big{(}N^{+}(v)\cup N^{-}(v)\big{)}\cap B_{i-1}. Then we have

[TABLE]

For each $x\in\hat{H}\cup\hat{T}_{k}\cup\{v\}$, we have $\chi(N^{o}(x)\cap B_{i-1})\leq c_{0}$. By Claim 4.3, we have $\chi(N^{-}(x)\cap B_{i-1})\leq c_{1}$ for each $x\in\hat{H}$ and $\chi(N^{+}(x)\cap B_{i-1})\leq c_{1}$ for each $x\in\hat{T}_{k}$. Then

[TABLE]

Let $R_{1}=R\cap N^{+}(v)$ and $R_{2}=R\cap N^{-}(v)$. Note that $R=R_{1}\cup R_{2}$, and so either $\chi(R_{1})>c_{1}$ or $\chi(R_{2})>c_{1}$. If $\chi(R_{1})>c_{1}$, there is a copy of $H$ in $R_{1}$, say $\hat{H}^{\prime}$. Then $\Delta(\hat{H}^{\prime},\hat{T}_{k},v)$ forms a copy of $\Delta(H,T_{k},T_{1})$, a contradiction. If $\chi(R_{2})>c_{1}$, there is a copy of $T_{k}$ in $R_{2}$, say $\hat{T}_{k}^{\prime}$. Then $\Delta(\hat{H},\hat{T}_{k}^{\prime},v)$ forms a copy of $\Delta(H,T_{k},T_{1})$, a contradiction again. This proves the claim. $\Diamond$

Claim 4.7 shows that every zone has few backward arcs with every bag sufficiently far from it. The next claim shows the converse (i.e., every bag has few backward arcs with every zone sufficiently far from it).

Claim 4.8**.**

For every $i$ and $v\in B_{i}$,

• •

$\chi(N^{+}(v)\cap Z_{i-r})\leq c_{1}$* for every $r\geq 2$, and*

• •

$\chi(N^{-}(v)\cap Z_{i+r})\leq c_{1}$* for every $r\geq 3$.*

Proof..

We repeat the argument in the proof of Claim 4.3. Suppose for a contradiction that the first statement is false: there is $v$ in some $B_{i}$ and $r\geq 2$ such that $\chi(N^{+}(v)\cap Z_{i-r})>c_{1}$. Then $N^{+}(v)\cap Z_{i-r}$ has a copy of $H$, say $\hat{H}$. For each $x\in\hat{H}$, we have $\chi(N^{o}(x)\cap B_{i-1})\leq c_{0}$ by (4.1), and $\chi(N^{-}(x)\cap B_{i-1})\leq c_{1}$ by the first inequality in Claim 4.7. We also have $\chi(N^{o}(v)\cap B_{i-1})\leq c_{0}$ and $\chi(N^{+}(v)\cap B_{i-1})\leq c_{1}$ by definition of a bag-chain. Thus by the same computation as in Claim 4.3, we obtain a copy of $T_{k}$ in $B_{i-1}$ and reach the contradiction. The proof of the second statement is similar. $\Diamond$

The next claim is a counterpart of Claim 4.5 for zones, and will be used to bound the chromatic number of the union of zones.

Claim 4.9**.**

For every $i$ and $v\in Z_{i}$,

• •

$\chi(N^{+}(v)\cap\bigcup_{s=0}^{i-3}Z_{s})\leq c_{1}$, and

• •

$\chi(N^{-}(v)\cap\bigcup_{s=i+3}^{t}Z_{s})\leq c_{1}$.

Proof..

Suppose for a contradiction that $\chi(N^{+}(v)\cap\bigcup_{s=0}^{i-3}Z_{s})>c_{1}$, then there is a copy of $H$ in $N^{+}(v)\cap\bigcup_{s=0}^{i-3}Z_{s}$, say $\hat{H}$. For each $x\in\hat{H}$, we have $\chi(N^{o}(x)\cap B_{i-2})\leq c_{0}$ by (4.1), and $\chi(N^{-}(x)\cap B_{i-2})\leq c_{1}$ by the first inequality in Claim 4.7. We also have $\chi(N^{o}(v)\cap B_{i-2})\leq c_{0}$ and $\chi(N^{+}(v)\cap B_{i-2})\leq c_{1}$ by the second inequality in Claim 4.7. Thus by the same computation as in Claim 4.3, we obtain that $\chi(M^{+}(\hat{H})\cap N^{-}(v)\cap B_{i-2})>c_{1}$ implying that there is a copy of $T_{k}$ in $M^{+}(\hat{H})\cap N^{-}(v)\cap B_{i-2}$, reaching a contradiction. The proof of the second statement is similar, where we use $B_{i+1}$ instead of $B_{i-2}$. $\Diamond$

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 $B_{1},\dots,B_{t}$ is of maximal length.

Claim 4.10**.**

No zone $Z_{i}$ contains a bag-chain of length 6.

Proof..

Suppose that some $Z_{i}$ contains a bag-chain of length 6, say $Y_{1},\dots,Y_{6}$. Note that we have

• •

$\chi(N^{+}(v)\cap B_{i-3})\leq c_{1}$ for every $v\in Y_{1}$ by Claim 4.7 and $\chi(N^{-}(v)\cap Y_{1})\leq c_{1}$ for every $v\in B_{i-3}$ by Claim 4.8; and

• •

$\chi(N^{-}(v)\cap B_{i+3})\leq c_{1}$ for every $v\in Y_{6}$ by Claim 4.7 and $\chi(N^{+}(v)\cap Y_{6})\leq c_{1}$ for every $v\in B_{i+3}$ by Claim 4.8.

Thus by definition of a bag-chain, $B_{1},\dots,B_{i-3},Y_{1},Y_{2},\dots,Y_{6},B_{i+3},\dots,B_{t}$ 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. $\Diamond$

Lemma 4.11**.**

There is $c^{\prime}$ such that if a $\Delta(H,T_{k},T_{1})$-free digraph $D^{\prime}$ with $\alpha(D^{\prime})\leq\alpha$ does not contain any $c$-bag-chain of length $6$, then $\chi(D^{\prime})\leq c^{\prime}$.

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. Fix $0\leq s\leq 2$, and for every $i\equiv s\mod 3$, $1\leq i\leq t$, let $Z^{s}_{\lfloor i/3\rfloor}:=Z_{i}$. By Claim 4.9, for every $j$ and for every $v\in Z^{s}_{j}$, we have

• •

$\chi(N^{+}(v)\cap\bigcup_{r<j}Z^{s}_{r})\leq c_{1}$, and

• •

$\chi(N^{-}(v)\cap\bigcup_{r>j}Z^{s}_{r})\leq c_{1}.$

By applying Lemma 4.11 on $D^{\prime}=Z_{i}$ and using Claim 4.10, it follows that $\chi(Z_{i})\leq c^{\prime}$ for every $i$. Thus, $\chi(Z^{s}_{j})\leq c^{\prime}$ for every $j$. We can repeat exactly the argument of Claim 4.6 (with $c$ replaced by $c^{\prime}$ and by using Claim 4.9 instead of Claim 4.5) to deduce

[TABLE]

for every $s=0,1,2$. (Remark: we may assume that $c^{\prime}>c$, which is necessary for the argument at the end of the proof of Property (A) that $3(c_{1}+c_{0})(k+1)\leq 2c<2c^{\prime}$.)

Hence

[TABLE]

From Claims 4.6 and the above inequality, we have

[TABLE]

which proves Theorem 4.2. ∎

4.1 Proof of Lemma 4.11

Lemma 4.11 asserts that having no bag-chain of length 6 is enough to force a digraph (with bounded independence number and $\Delta(H,T_{k},T_{1})$-free) to have bounded chromatic number. It is trivial by definition that having no bag-chain of length 1 forces bounded chromatic number. In the next lemma, we show that having no bag-chain of length 2 can force bounded chromatic number, which contains most of the difficulty of 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 $c=2(c_{0}+c_{1})(h+k)$.

Lemma 4.12**.**

For every $d\geq c$, there is $g(d)$ such that if a $\Delta(H,T_{k},T_{1})$-free digraph $D$ with $\alpha(D)\leq\alpha$ contains no $d$-bag-chain of length $2$, then $\chi(D)\leq g(d)$.

Proof.

Let $J$ be the tournament $H\Rightarrow T_{k}$. 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 $V$ satisfying the hypotheses of the lemma. A copy of $J$ in $D$ is called a ball. A ball $\hat{J}$ is colored red if $\chi(M^{+}(\hat{J}))\leq d$ and blue if $\chi(M^{-}(\hat{J}))\leq d$ (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 ball $\hat{J}$ is neither red nor blue. Then there are $B_{1}\subseteq M^{-}(\hat{J})$ and $B_{2}\subseteq M^{+}(\hat{J})$ such that $\chi(B_{1})=d$ and $\chi(B_{2})=d$. Suppose that $\chi(N^{+}(v)\cap B_{1})>c_{1}$ for some $v\in B_{2}$. Let $\hat{H}$ be a copy of $H$ in $N^{+}(v)\cap B_{1}$. Let $\hat{T}_{k}$ be a copy of $T_{k}$ in $\hat{J}$. Then $\Delta(\hat{H},\hat{T}_{k},v)$ forms a copy of $\Delta(H,T_{k},T_{1})$, a contradiction. Hence $\chi(N^{+}(v)\cap B_{1})\leq c_{1}$ for every $v\in B_{2}$. Similarly, $\chi(N^{-}(v)\cap B_{2})\leq c_{1}$ for every $v\in B_{1}$, for otherwise a copy of $T_{k}$ in $N^{-}(v)\cap B_{2}$ would yield a contradiction. Hence $B_{1},B_{2}$ is a $d$-bag-chain of length 2, a contradiction. $\Diamond$

For every vertex $v$ of $V$, we color $v$ as follows. If there are $c_{1}+1$ vertex-disjoint red balls $\hat{J}_{1},\dots,\hat{J}_{c_{1}+1}$ such that $\hat{J}_{i}$ has complete arcs to $\hat{J}_{j}$ for every $i<j$, and $v\in\hat{J}_{c_{1}+1}$, then we color $v$ red. If there are $c_{1}+1$ blue balls $\hat{J}_{1},\dots,\hat{J}_{c_{1}+1}$ such that $\hat{J}_{i}$ has complete arcs to $\hat{J}_{j}$ for every $i<j$, and $v\in\hat{J}_{1}$, 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.

Claim 4.14**.**

There is $d_{1}$ such that $\chi(U)\leq d_{1}$.

Proof..

Let $K$ be the tournament of $J\Rightarrow J\Rightarrow\dots\Rightarrow J$ ($2c_{1}+2$ 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 $D^{\prime}$ with $\alpha(D^{\prime})\leq\alpha$ has chromatic number at most $d_{1}$. Suppose that $U$ contains a copy of $K$, say $\hat{K}$. Since every ball is either red or blue, we can find $c_{1}+1$ vertex-disjoint monochromatic balls $\hat{J}_{1},\dots,\hat{J}_{c_{1}}$ in $\hat{K}$ such that $\hat{J}_{i}$ has complete arcs to $\hat{J}_{j}$ for every $i<j$. Then either vertices of $\hat{J}_{1}$ are blue or vertices of $\hat{J}_{c_{1}+1}$ are red, a contradiction with the fact that all vertices of $U$ are uncolored. Hence $U$ is $K$-free, and so $\chi(U)\leq d_{1}$. $\Diamond$

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 $v\in R$.

Claim 4.15**.**

There is $d_{2}$ such that $\chi(N^{+}(v))\leq d_{2}$ for every $v\in R$.

Proof..

Fix $v\in R$. Then there are vertex-disjoint red balls $\hat{J}_{1},\dots,\hat{J}_{c_{1}+1}$ where $v\in\hat{J}_{c_{1}+1}$ and $\hat{J}_{i}$ has complete arcs to $\hat{J}_{j}$ for every $i<j$. Let $L=\bigcup_{i=1}^{c_{1}}\hat{J}_{1}$. Note that $v$ is seen by all vertices of $L$. For every $u\in L$, we have $\chi(N^{+}(v)\cap N^{o}(u))\leq c_{0}$. Hence

[TABLE]

For every partition of $L$ into $L_{1},L_{2}$ ($L_{1}$ or $L_{2}$ may be empty), let $Y_{L_{1},L_{2}}=N^{+}(v)\cap M^{-}(L_{1})\cap M^{+}(L_{2})$. Observe that for every vertex $x\in N^{+}(v)\setminus{N^{o}(L)}$ (note that $x\notin L$ since $v\in\hat{J}_{c_{1}+1}$), 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 $x\in Y_{L_{1},L_{2}}$. Hence

[TABLE]

We now show that $\chi(Y_{L_{1},L_{2}})\leq d$ for every $Y_{L_{1},L_{2}}$.

• •

If, for some $1\leq i\leq c_{1}$, there is $\hat{J}_{i}\subseteq L_{2}$, recall that $\chi(M^{+}(\hat{J}_{i}))\leq d$ since $\hat{J}_{i}$ is red. Since $Y_{L_{1},L_{2}}\subseteq M^{+}(L_{2})\subseteq M^{+}(\hat{J}_{i})$, we have that $\chi(Y_{L_{1},L_{2}})\leq d$.

• •

Otherwise, if there is no $\hat{J}_{i}\subseteq L_{2}$, then $L_{1}$ contains at least one vertex of each $\hat{J}_{i}$, $1\leq i\leq c_{1}$, and so $|L_{1}|\geq c_{1}$. Hence $L_{1}$ has a copy of $T_{k}$, say $\hat{T}_{k}$. Note that all vertices of $\hat{T}_{k}$ see $v$ since $v\in\hat{J}_{c_{1}+1}$. If $\chi(Y_{L_{1},L_{2}})>c_{1}$, then $Y_{L_{1},L_{2}}$ contains a copy of $H$, say $\hat{H}$, then $\Delta(\hat{H},\hat{T}_{k},v)$ forms a copy of $\Delta(H,T_{k},T_{1})$, a contradiction. Hence $\chi(Y_{L_{1},L_{2}})\leq c_{1}\leq d$.

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

[TABLE]

Set $d_{2}:={(h+k)c_{1}}c_{0}+2^{{(h+k)c_{1}}}d.$ This completes the proof of the claim. $\Diamond$

Then $\chi(N^{+}(v)\cap R)\leq\chi(N^{+}(v))\leq d_{2}$ for every $v\in R$. Then by applying Theorem 2.3 for digraph $R$ with $t=d_{2}$, we have $\chi(R)\leq d_{3}$ 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 $\Delta(H,T_{k},T_{1})$.

Claim 4.16**.**

There is $d_{4}$ such that $\chi(N^{-}(v))\leq d_{4}$ for every $v\in B$.

Proof..

Fix $v\in B$. Then there are vertex-disjoint blues balls $\hat{J}_{1},\dots,\hat{J}_{c_{1}+1}$ where $v\in\hat{J}_{1}$ and $\hat{J}_{i}$ has complete arcs to $\hat{J}_{j}$ for every $i<j$. Let $L=\bigcup_{i=2}^{c_{1}+1}\hat{J}_{i}$. Note that $v$ sees all vertices of $L$. For every $u\in L$, we have

[TABLE]

For every partition of $L$ into $L_{1},L_{2}$ ($L_{1}$ or $L_{2}$ may be empty), let $Y_{L_{1},L_{2}}=N^{-}(v)\cap M^{-}(L_{1})\cap M^{+}(L_{2})$. Observe that for every vertex $x\in N^{-}(v)\setminus{N^{o}(L)}$ (note that $x\notin L$ since $v\in\hat{J}_{1}$), 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 $x\in Y_{L_{1},L_{2}}$. Hence

[TABLE]

We now show that $\chi(Y_{L_{1},L_{2}})\leq d$ for every $Y_{L_{1},L_{2}}$.

• •

If there is $\hat{J}_{i}\subseteq L_{2}$, then $\hat{J}_{i}$ contains a copy of $H$, say $\hat{H}$. Note that $v$ is in the first ball, so $v$ sees all vertices of $\hat{H}$. If $\chi(Y_{L_{1},L_{2}})>c_{1}$, then $Y_{L_{1},L_{2}}$ contains a copy of $T_{k}$, say $\hat{T}_{k}$, then $\Delta(\hat{H},\hat{T}_{k},v)$ forms a copy of $\Delta(H,T_{k},T_{1})$, a contradiction. Hence $\chi(Y_{L_{1},L_{2}})\leq c_{1}\leq d$.

• •

If $\hat{J}_{c_{1}+1}\subseteq L_{1}$, recall that $\chi(M^{-}(\hat{J}_{c_{1}+1}))\leq d$ since $\hat{J}_{c_{1}+1}$ is blue. Since $Y_{L_{1},L_{2}}\subseteq M^{-}(L_{1})\subseteq M^{-}(\hat{J}_{c_{1}+1})$, we have that $\chi(Y_{L_{1},L_{2}})\leq d$.

• •

Otherwise, we have two remarks: (1) $\hat{J}_{c_{1}+1}$ must have a vertex in $L_{2}$, say $z$. (2) Every $\hat{J}_{i},2\leq i\leq c_{1}$ must have a vertex in $L_{1}$. Then $|L_{1}\cup\{v\}|>c_{1}$, and so $L_{1}\cup\{v\}$ contains a copy of $T_{k}$, say $\hat{T}_{k}^{\prime}$, such that all vertices of $\hat{T}_{k}^{\prime}$ are in one of the $\hat{J}_{i}$, $1\leq i\leq c_{1}$ (note that $v\in\hat{J}_{1}$). Observe that $z$ is seen by all vertices of $\hat{T}_{k}^{\prime}$ since $z\in\hat{J}_{c_{1}+1}$. If $\chi(Y_{L_{1},L_{2}})>c_{1}$, then $Y_{L_{1},L_{2}}$ contains a copy of $H$, say $\hat{H}^{\prime}$, then $\Delta(\hat{H}^{\prime},\hat{T}_{k}^{\prime},z)$ forms a copy of $\Delta(H,T_{k},T_{1})$, a contradiction. Hence $\chi(Y_{L_{1},L_{2}})\leq c_{1}\leq d$.

Hence using a computation similar to the claim above, we have $\chi(N^{-}(v))\leq d_{4}$, where $d_{4}:={(h+k)c_{1}}c_{0}+2^{{(h+k)c_{1}}}d.$ This completes the proof of the claim. $\Diamond$

Then $\chi(N^{-}(v)\cap B)\leq\chi(N^{-}(v))\leq d_{4}$ for every $v\in B$. Then by applying Theorem 2.5 to the digraph $B$ with $t=d_{4}$, we have $\chi(B)\leq d_{5}$ for some $d_{5}$. Hence

[TABLE]

This completes the proof of Lemma 4.12. ∎

We are now ready to prove Lemma 4.11.

Claim 4.17**.**

If a $\Delta(H,T_{k},T_{1})$-free digraph $D$ with $\alpha(D)\leq\alpha$ contains no $c$-bag-chain of length $6$, then $\chi(D)\leq g(g(g(c)))$ where $g$ is the function in Lemma 4.12.

Proof..

Suppose for a contradiction that $\chi(D)\geq g(g(g(c)))$. 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 $\chi(X_{1})=\chi(X_{2})=g(g(c))$ and

• •

$\chi(N^{+}(v)\cap X_{1})\leq c_{1}$ for every $v\in X_{2}$, and

• •

$\chi(N^{-}(v)\cap X_{2})\leq c_{1}$ for every $v\in X_{1}$.

Apply Lemma 4.12 again to $X_{1}$ (respectively, $X_{2}$) with $d:=g(c)$, we obtain a $g(c)$-bag-chain of length two, say $Y_{1},Y_{2}$ in $X_{1}$ (respectively, $Y_{3},Y_{4}$ in $X_{2}$). Since $Y_{2}\subset X_{1}$ and $Y_{3}\subset X_{2}$, we have

• •

$\chi(N^{+}(v)\cap Y_{2})\leq c_{1}$ for every $v\in Y_{3}$, and

• •

$\chi(N^{-}(v)\cap Y_{3})\leq c_{1}$ for every $v\in Y_{2}$.

Hence by definition, $Y_{1},\dots,Y_{4}$ forms a $g(c)$-bag-chain of length 4. Note that $\chi(Y_{s})=g(c)$ for every $1\leq s\leq 4$. Repeating the argument we obtain a $c$-bag-chain of length 2 inside each $Y_{s}$, and hence obtain a $c$-bag-chain $B_{1},\dots,B_{8}$ of length 8 inside $D$. This contradicts the fact that $D$ has no $c$-bag-chain of length 6, and so completes the proof. $\Diamond$

Claim 4.17 proves Lemma 4.11, concluding the proof of Theorem 4.2.

5 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 $\alpha(D)\leq\alpha$ with at most $35^{\alpha-1}\alpha!$ 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 (implicit) value of $k$ may be different.

For each integer $\alpha\geq 1$, define $h(\alpha)$ to be the minimum number such that every $C_{3}$-free digraph $D$ with $\alpha(D)\leq\alpha$ has chromatic number at most $h(\alpha)$. Conjecture 1.6 asserts that $h(\alpha)\leq\alpha^{\ell}$ for some ${\ell}$. Clearly $h(1)=1$ since every $C_{3}$-free tournament is acyclic. However, $h(\alpha)$ is still unknown for all $\alpha\geq 2$. We believe that $h(2)=2$ or $3$ even though the best bound we have is around 25 (by tweaking the proof of Theorem 1.7). Theorem 1.7 gives an exponential upper bound for $h(\alpha)$ by the function $g(\alpha):=35^{\alpha-1}\alpha!$. Since our proof of Theorem 1.7 will use induction, we will assume that a $C_{3}$-free digraph with independence number $\alpha-1$ can be colored with at most $g(\alpha-1)$ colors and we will prove it for $\alpha$. Note that $g(1)=1$. Let us restate Theorem 1.7.

Theorem 5.1**.**

Let $D$ be a $C_{3}$-free digraph where $\alpha(D)\leq\alpha$. Then $\chi(D)\leq 35^{\alpha-1}\alpha!$ and this coloring can be found in time $poly(n)$.

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.

Proposition 5.2**.**

A digraph $D$ has an acyclic dominating set, and this set can be found in time $poly(n)$.

Proof.

We proceed by induction on $n$. The statement clearly holds for $n=1$. For $n>1$, pick an arbitrary vertex $v$. Then $V(D)\setminus{v}=N^{o}(v)\cup N^{-}(v)\cup N^{+}(v)$. Applying induction to the subgraph, $D[N^{o}(v)\cup N^{-}(v)]$, we obtain an acyclic dominating set $S^{\prime}$. Then $S:=S^{\prime}\cup\{v\}$ is a dominating set of $D$. Note that $S^{\prime}\subseteq N^{o}(v)\cup N^{-}(v)$, so $v$ does not see any vertex of $S^{\prime}$. Hence $S$ is an acyclic set since $S^{\prime}$ is an acyclic set. The running time for this procedure is $poly(n)$. ∎

Proposition 5.3**.**

Given a $C_{3}$-free digraph $D$, there is a set $Y\subseteq V(D)$ of size at most $\alpha(D)$ such that $V(D)=Y\cup N^{o}(Y)\cup N^{+}(Y)$, and this set can be found in time $poly(n)$.

Proof.

First apply Proposition 5.2 to obtain an acyclic dominating set $S$ of $D$. Then apply Proposition 2.1 to $D[S]$ to obtain a stable dominating set $Y$ of $D[S]$ of size at most $\alpha(D)$ in time $poly(n)$.

We now show that $Y$ is a set with the desired properties. Suppose for a contradiction that there is $v\notin Y\cup N^{o}(Y)\cup N^{+}(Y)$. Then $Y\subseteq N^{+}(v)$ and $v\notin S$ since $Y$ dominates all vertices of $S$. There is $u\in S$ seeing $v$ since $S$ is a dominating set of $D$. Note that $u\notin Y$; otherwise this contradicts $Y\subseteq N^{+}(v)$. There is $y\in Y$ 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 retains the useful properties of a bag as defined in Section 4 and so that it can be tested efficiently.

Definition 5.4**.**

For a digraph $D$, we say that $B\subseteq V(D)$ is a bag of $D$, if every three distinct vertices $\{x,y,z\}\in V(D)\setminus{B}$ have a common neighbor in $B$.

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 $V(D)\setminus B$. 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 $\alpha-1$. Let $t(\alpha-1,n)$ 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 $g(\alpha-1)$ colors. The following claim follows from this definition.

Claim 5.5**.**

Suppose $\sum_{i=1}^{\ell}n_{i}=n$, where $n_{i}\geq 1$ is an integer. Then $\sum_{i=1}^{\ell}t(\alpha,n_{i})\leq t(\alpha,n)$.

We now fix an arbitrary $C_{3}$-free digraph $D$ such that $\alpha(D)\leq\alpha$, and we omit the subscript $D$ from the relevant notation when the context is clear.

Claim 5.6**.**

If $S\subset V(D)$ is not a bag of $D$, then:

1. (a)

We can partition $S$ into three disjoint sets, $S_{1},S_{2}$ and $S_{3}$, such that $\alpha(D[S_{i}])\leq\alpha-1$ for $i\in\{1,2,3\}$. This procedure takes time $poly(n)$. 2. (b)

We can color $S$ with $3\cdot g(\alpha-1)$ colors in time $t(\alpha-1,|S|)$.

Proof..

If $S$ is not a bag of $D$, then we can, in time $poly(n)$, find a triple $\{x,y,z\}\in V(D)\setminus S$ such that every $v\in S$ is not incident to at least one of $x,y$ or $z$. Thus, each vertex $v\in S$ belongs to either $N^{o}(x)\cap S,N^{o}(y)\cap S$ or $N^{o}(z)\cap S$. Each of these sets has independence number at most $\alpha-1$. The total time for this procedure is $poly(n)$. The second assertion follows from (a) and from the definition of the function $t(\alpha,n)$. $\Diamond$

Definition 5.7**.**

A bag $B\subseteq V(D)$ is poor if for every vertex $v\in B$, either $N^{-}(v)\cap B$ or $N^{+}(v)\cap B$ is not a bag of $D$.

We can check in $poly(n)$ time (e.g. $O(n^{5})$) if a bag $B$ is poor by testing whether or not $N^{-}(v)\cap B$ and $N^{+}(v)\cap B$ are bags for every $v\in B$.

Claim 5.8**.**

If $B\subseteq V(D)$ is a poor bag, then we can color $B$ with $8\alpha\cdot g(\alpha-1)$ colors in time $poly(n)+t(\alpha-1,|B|)$.

Proof..

Since $B$ is poor, then for each $v\in B$, either $N^{-}(v)\cap B$ or $N^{+}(v)\cap B$ is not a bag of $D$. Then we can partition $B$ into two sets, $L$ and $R$, where $N^{-}(v)\cap B$ is not a bag for every $v\in L$, and $N^{+}(v)\cap B$ is not a bag for every $v\in R$.

Applying Proposition 5.3 to $R$, we can find a set $Y_{R}\subseteq R$ (respectively $Y_{L}\subseteq L$) such that $|Y_{R}|\leq\alpha$ and $R\subseteq Y_{R}\cup N^{+}(Y_{R})\cup N^{o}(Y_{R})$. And so:

[TABLE]

For $v\in Y_{R}$, note that $N^{+}(v)\cap R$ is not a bag, and so by (a) from Claim 5.6, we can partition $N^{+}(v)\cap R$ into three sets, each with independence number at most $\alpha-1$. Additionally, we have the set $N^{o}(v)\cap R$, which also has independence number at most $\alpha-1$. Overall, we can partition $R\setminus{Y_{R}}$ into $4\alpha$ sets, each with independence number $\alpha-1$. The same argument can be applied to $Y_{L}$ to obtain $8\alpha$ disjoint sets. Therefore, in time $poly(n)$, we can partition $B\setminus{Y_{R}\cup Y_{L}}$, and in time $t(\alpha-1,|B|)$ we can color $B\setminus{\{Y_{r}\cup Y_{L}\}}$ with $8\alpha\cdot g(\alpha-1)$ colors. We can then color each $v\in Y_{R}$ (respectively, $v\in Y_{L}$) with an arbitrary color used to color the set $N^{o}(v)\cap R$ (respectively, $N^{o}(v)\cap L$). $\Diamond$

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.

Definition 5.9**.**

A sequence of pairwise disjoint bags $B_{1},\dots,B_{t}$ forms a chain of bags if $B_{i}\to B_{i+1}$ for every $i\in[1,t).$

Recall that $B_{i}\to B_{i+1}$ means there is no arc from $B_{i+1}$ to $B_{i}$. Moreover, if each $B_{i}$ is a poor bag, then this sequence is a chain of poor bags. Given a chain of bags $C=\{B_{1},B_{2},\dots,B_{t}\}$ for $D$, we say that $v\in C$ if $v\in B_{i}$ for some $i\in[1,t]$. We can partition the vertices in $V(D)\setminus{C}$ into sets $Z_{0},\dots,Z_{t}$, which we call zones, as follows. For every $v\in V(D)\setminus{C}$, 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.

Claim 5.10**.**

Let $C=\{B_{1},\dots,B_{t}\}$ be a chain of bags, and let $Z_{0},Z_{1},\dots,Z_{t}$ be a partition of the vertices in $V(D)\setminus{C}$. For every $i$, the following properties hold:

1. (a)

$B_{i}\to B_{i+r}$* for every $r\geq 1$,* 2. (b)

$Z_{i}\to B_{i+r}$* for every $r\geq 1$,* 3. (c)

$B_{i}\to Z_{i+r}$* for every $r\geq 2$,* 4. (d)

$Z_{i}\to Z_{i+r}$* for every $r\geq 3$.*

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 $u\in B_{i+r}$ and $v\in B_{i}$. Since $B_{i+1}$ is a bag, there is $x\in B_{i+1}$ 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 $r\geq 1$ by the partitioning criterion of vertices into zones.

To prove property (c), suppose that there is an arc $zv$ with $z\in Z_{i+r}$ and $v\in B_{i}$ for some $r\geq 2$. Then there is $u\in B_{i+r}$ such that $uz$ is an arc by the partitioning criterion of vertices into zones. Since $B_{i+1}$ is a bag, there is $x\in B_{i+1}$ 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 $r\geq 2$.

To prove property (d), suppose that there is an arc $uv$ with $u\in Z_{i+r}$ and $v\in Z_{i}$ for some $r\geq 3$. Since $B_{i+1}$ is a bag, there is $x\in B_{i+1}$ such that $x$ is a common neighbor of both $u$ 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 $r\geq 3$. $\Diamond$

We now show how to find a chain of poor bags, which can be colored using Claim 5.8.

Find-Chain$(D,B)$

If there is $v\in B$ such that both $N^{+}(v)\cap B$ and $N^{-}(v)\cap B$ are bags of $D$, then:

Return (Find-Chain$(D,N^{-}(v)\cap B)$, $v$, Find-Chain$(D,N^{+}(v)\cap B)$ ).

Otherwise, return $B$.

The routine Find-Chain$(D,B)$ returns $B_{1},v_{1},B_{2},\dots,v_{t-1},B_{t}$ (i.e., a sequence of sets $B_{i}$ alternating with vertices $v_{i}$). We say that the sequence $B_{1},B_{2},\dots,B_{t}$ is the chain of poor bags output by the procedure Find-Chain$(D,B)$. Later on, we will use the vertices $v_{i}$ in the output sequence to facilitate the coloring of vertices outside the chain. Observe that if $B$ is a poor bag or if $B$ is not a bag, then Find-Chain$(D,B)$ returns a single set, namely $B$.

Claim 5.11**.**

If bag $B\subseteq V(D)$ is not poor, then Find-Chain$(D,B)$ returns a chain of poor bags $B_{1},\dots,B_{t}$ for some $t\geq 2$ in time $poly(n)$.

Proof..

Let $B_{1},\dots,B_{t}$ 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, $B_{i+1}\subseteq N^{+}(v_{i})$ and $B_{i}\subseteq N^{-}(v_{i})$. If there is an arc $xy$ with $x\in B_{i+1}$ and $y\in B_{i}$, then $v_{i}xy$ is a copy of $C_{3}$, a contradiction. Hence $B_{1},\dots,B_{t}$ is a chain of poor bags by definition. The procedure Find-Chain$(D,B)$ runs in time $poly(n)$. $\Diamond$

Claim 5.12**.**

If Find-Chain$(D,B)$ returns a chain of $t$ poor bags, then we can color $B$ with $8\alpha\cdot g(\alpha-1)+(t-1)\cdot g(\alpha-1)$ colors in time $poly(n)+t(\alpha-1,|B|)$.

Proof..

Suppose that Find-Chain$(D,B)$ returns $B_{1},v_{1},B_{2},\dots,v_{t-1},B_{t}$ and that $C=\{B_{1},\dots,B_{t}\}$ 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, $B_{i}\to B_{j}$ for every $i<j$, and so we can color the vertices in $C$ using $8\alpha\cdot g(\alpha-1)$ colors in time $poly(n)+t(\alpha-1,|C|)$ time.

Observe that each $v\in B\setminus{C}$ 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 $v\in V(D)$, the set $N^{o}(v)$ has independence number $\alpha-1$. Therefore, the vertices in $\bigcup_{i=1}^{t-1}N^{o}(v_{i})$ can be colored with $(t-1)\cdot g(\alpha-1)$ 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 $poly(n)+t(\alpha-1,|B\setminus{C}|)$. $\Diamond$

Corollary 5.13**.**

Either Find-Chain$(D,B)$ returns a chain of $t$ poor bags, or $B$ can be colored using $8\alpha\cdot g(\alpha-1)+(t-2)\cdot g(\alpha-1)$ colors.

We now have the tools to outline our main coloring algorithm.

Color-Digraph$(D)$

If $D$ is acyclic, color $D$ with one color and terminate.

Otherwise:

Run Find-Chain$(D,V(D))$ and let $C:=\{B_{1},\dots,B_{t}\}$ denote the chain of poor bags that is output.

Assign each vertex in $V(D)\setminus{C}$ to a zone $Z_{i}$ for $i\in[0,t]$.

While Find-Chain$(D,Z_{i})$ returns a chain of poor bags $B_{1}^{\prime},B_{2}^{\prime},\dots,B_{k}^{\prime}$ for $k\geq 3$:

(a)

Update chain: $C:=\{B_{1},\dots,B_{i-2},B_{1}^{\prime},B_{2}^{\prime},\dots,B_{k}^{\prime},B_{i+1},\dots,B_{t}\}.$

(b)

Re-assign each vertex in $V(D)\setminus{C}$ to a zone.

Color all vertices in the chain $C$ with $8\alpha\cdot g(\alpha-1)$ colors.

Color all vertices in the zones of $C$ with $3(8\alpha+1)\cdot g(\alpha-1)$ colors.

Claim 5.14**.**

Color-Digraph$(D)$* colors $D$ with at most $35\alpha\cdot g(\alpha-1)$ colors.*

Proof..

If $V(D)$ is a poor bag, then we can apply Claim 5.8. Hence we may suppose that $V(D)$ is not a poor bag.

Note that the updated chain $C$ resulting from Step 3 (a) is still a chain of poor bags, due to properties (b) and (c) from Claim 5.10. After Step 3 finishes, the chain $C$ is maximal in that the procedure Find-Chain$(D,Z_{i})$ will not find a chain of three poor bags in any zone $Z_{i}$. Using Corollary 5.13, we can therefore color each zone using $(8\alpha+1)\cdot g(\alpha-1)$ colors. Applying property (d) from Claim 5.10, we can use at most $(24\alpha+3)\cdot g(\alpha-1)$ colors to color all vertices in $V(D)\setminus{C}$. Each bag $B_{i}$ in the chain $C$ is a poor bag, so we can color $B_{i}$ with $8\alpha\cdot g(\alpha-1)$ colors by Claim 5.8. Moreover, since $B_{i}\to B_{j}$ for every $i<j$ (by property (a) from Claim 5.10), we need $8\alpha\cdot g(\alpha-1)$ colors to color the entire chain $C$. Thus, we can color $D$ with $(32\alpha+3)\cdot g(\alpha-1)\leq 35\alpha\cdot g(\alpha-1)$ colors. $\Diamond$

Claim 5.15**.**

The procedure Color-Digraph$(D)$ uses $g(\alpha)$ colors.

Proof..

We proceed by induction on $\alpha$. If $\alpha=1$, then we use one color, since every $C_{3}$-free tournament is acyclic. Suppose that the algorithms colors each $C_{3}$-free digraph $D^{\prime}$ where $\alpha(D^{\prime})\leq\alpha-1$ using at most $g(\alpha-1)$ colors. Then by Claim 5.14, the procedure Color-Digraph$(D)$ colors $D$ with at most

[TABLE]

colors. $\Diamond$

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 $V(D)\setminus{C}$ into zones. Once we have found this partitioning, we can color the vertices in $C$ in time $poly(n)+t(\alpha-1,|C|)$ using Claim 5.8, and we can color each zone $Z_{i}$ in time $poly(n)+(\alpha-1,|Z_{i}|)$ using Claim 5.12 and Corollary 5.13. So applying Claim 5.5, the total running time of Color-Digraph$(D)$ is at most $poly(n)+t(\alpha-1,n)$. This leads to the following recurrence relation:

[TABLE]

Since $\alpha\leq n$, we have that $t(\alpha,n)=poly(n)$.

We note that our algorithm is actually just partitioning $V(D)$ into disjoint subsets, where each subset has independence number at most $\alpha-1$. 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 $\alpha-1$. (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 $\alpha-1$. Similarly, in the proof of Claim 5.8, a poor bag is partitioned into $8\alpha$ disjoint sets, each with independence number at most $\alpha-1$.) Once the subgraphs on these induced subsets are colored (recursively) using $g(\alpha-1)$ 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 $t(\alpha-1,n)$ time. $\Diamond$

Theorem 5.1 follows from Claims 5.14, 5.15 and 5.16.

Finally, we remark that Theorem 1.7 yields a bound on the size of a maximum acyclic subgraph of a $C_{3}$-free digraph $D$ in terms of $\alpha$.

Theorem 5.17**.**

Let $D=(V,A)$ be a $C_{3}$-free digraph where $\alpha(D)\leq\alpha$. Then $D$ contains an acyclic subset of arcs, $A^{\prime}\subseteq A$, with cardinality $|A^{\prime}|\geq|A|\cdot(\frac{1}{2}+c_{\alpha})$, where $c_{\alpha}$ is a constant depending on $\alpha$.

Proof.

Since $D$ can be colored with $g(\alpha)$ colors, there is a color class containing at least $n/g(\alpha)$ vertices, and by Turan’s Theorem, its induced subgraph contains at least

[TABLE]

arcs. Thus, $D$ contains a maximum acyclic subgraph of size at least

[TABLE]

Let $c$ be an absolute constant. Then

[TABLE]

satisfies the theorem. ∎