Given a finite family $\mathcal{F}$ of graphs, we say that a graph $G$ is “$\mathcal{F}$-free” if $G$ does not contain any graph in $\mathcal{F}$ as a subgraph. We abbreviate $\mathcal{F}$-free to just “$F$-free” when $\mathcal{F}=\{F\}$.
A vertex-colored graph \(H\) is called “rainbow” if no two vertices of \(H\) have the same color. Given an integer \(s\) and a finite family of graphs \(\mathcal{F}\), let $\ell(s,\mathcal{F})$ denote the smallest integer such that any properly vertex-colored $\mathcal{F}$-free graph $G$ having $\chi(G)\geq \ell(s,\mathcal{F})$ contains an induced rainbow path on $s$ vertices. Scott and Seymour showed that \(\ell(s,K)\) exists for every complete graph $K$. A conjecture of N. R. Aravind states that $\ell(s,C_3)=s$. The upper bound on $\ell(s,C_3)$ that can be obtained using the methods of Scott and Seymour setting $K=C_3$ are, however, super-exponential. Gyárfás and Sárközy showed that \(\ell(s,\{C_3,C_4\}) = \mathcal{O}\big((2s)^{2s}\big)\). For $r\geq 2$, we show that \(\ell(s,K_{2,r}) \leq (r-1)(s-1)s/2+s\) and therefore, \(\ell(s,C_4) \leq \frac{s^2+s}{2}\). This significantly improves Gyárfás and Sárközy's bound and also covers a bigger class of graphs. We adapt our proof to achieve much stronger upper bounds for graphs of higher girth: we prove that \(\ell(s,\{ C_3,C_4,\ldots,C_{g-1}\})\leq s^{1+\frac{4}{g-4}}\), where $g\geq 5$. Moreover, in each case, our results imply the existence of at least \(s!/2\) distinct induced rainbow paths on \(s\) vertices. Along the way, we obtain some new results on an oriented variant of the Gyárfás-Sumner conjecture. For $r\geq 2$, let $\mathcal{B}_r$ denote the orientations of $K_{2,r}$ in which one vertex has out-degree $r$. We show that every $\mathcal{B}_r$-free oriented graph having chromatic number at least $(r-1) (s-1) (s-2) + 2s +1$ and every bikernel-perfect oriented graph with girth $g\geq 5$ having chromatic number at least $2s^{1+\frac{4}{g-4}}$ contains every oriented tree on at most $s$ vertices as an induced subgraph.

Keywords: Induced rainbow paths, Induced oriented trees, chromatic number, Gyárfás-Sumner conjecture, Gallai-Roy-Vitaver theorem.

§ INTRODUCTION

All the graphs considered in this paper are simple and finite. They are also considered to be undirected unless otherwise mentioned. Please refer [5] for standard graph theoretic terminology and notation that is not defined in this paper. Given a (directed or undirected) graph \(G=(V,E)\), its vertex set and edge set are denoted by \(V(G)\) and \(E(G)\), respectively. Given a directed graph $G$, we let $\hat{G}$ denote the underlying undirected graph of $G$: more precisely, $V(\hat{G})=V(G)$ and $E(\hat{G})=\{uv\colon (u,v)\in E(G)$ or $(v,u)\in E(G)\}$. If $G$ is an undirected graph, then we define $\hat{G}=G$. If \(\mathcal{F}\) is a finite set of (directed) graphs, a (directed) graph \(G\) is said to be \(\mathcal{F}\)-free if there is no subgraph in \(G\) which is isomorphic to a member of \(\mathcal{F}\). If \(\mathcal{F}\) is a singleton with the element \(F\), we shall simply write \(F\)-free instead of \(\{F\}\)-free.
If $\mathcal{F}$ is a finite family of undirected graphs, a directed graph $G$ is said to be $\mathcal{F}$-free if $\hat{G}$ is $\mathcal{F}$-free.

All the colorings considered in this paper are proper vertex colorings—i.e. they are assignments of colors to the vertices such that no two adjacent vertices get the same color. The sets of colors that we use shall always be subsets of natural numbers. The minimum number of colors required to properly color a graph \(G\) is called its chromatic number, denoted by \(\chi(G)\). A proper coloring which uses exactly \(\chi(G)\) colors is called an optimal coloring. Given a properly colored graph \(G\), a subgraph \(H\) of \(G\) is said to be rainbow if no two vertices of \(H\) have the same color. Additionally, if the subgraph \(H\) is induced, then \(H\) is said to be an induced rainbow subgraph of \(G\). A \(t\)-path is a path on \(t\) vertices. A cycle of length \(n\) is denoted by \(C_n\) (the length of a path or cycle is the number of edges in it). A complete graph on $n$ vertices is denoted by $K_n$.

If $G$ is a directed graph, then a path in $G$ simply refers to a path in $\hat{G}$ and a cycle in $G$ simply refers to a cycle in $\hat{G}$. The girth of a (directed or undirected) graph \(G\), denoted by \(g(G)\), is the minimum length of a cycle in $G$. A directed path in a directed graph $G$ is a sequence of pairwise distinct vertices $v_1,v_2,\ldots,v_t$ such that for $1\leq i<t$, $(v_i,v_{i+1})\in E(G)$.  A directed cycle in a directed graph $G$ is a sequence of vertices $v_1,v_2,\ldots,v_t,v_1$ such that $v_1,v_2,\ldots,v_t$ is a directed path in $G$ and $(v_t,v_1)\in E(G)$. An orientation of an undirected graph \(G\) is a directed graph obtained by assigning a direction to each edge in \(E(G)\). An acyclic orientation of a graph is one in which there are no directed cycles. A directed graph $G$ is said to be an oriented graph if it is an orientation of $\hat{G}$; in other words, it is a directed graph with no directed cycle of length 2.
An oriented tree is an orientation of a tree. Let \(T\) be an oriented tree with a vertex \(r \in V(T)\) designated as the root. If for each vertex \(v \in V(T)\), there is a directed path starting from \(r\) and ending at \(v\), then \(T\) is called an out-tree. On the other hand, if for each vertex \(v \in V(T)\), there is a directed path starting from \(v\) and ending at \(r\), then \(T\) is called an in-tree. A directed graph without any directed cycles is called an directed acyclic graph. There is a linear ordering of the vertices of every directed acyclic graph, called a topological ordering of the graph, such that for every edge $(u,v)$ in the graph, the vertex $u$ comes before the vertex $v$ in the ordering. A kernel of a directed graph $G$ is an independent set $S$ of $G$ having the additional property that for every vertex $u\in V(G)\setminus S$, there exists a vertex $v\in S$ such that $(u,v)\in E(G)$. Similarly, an antikernel of $G$ is an independent set $S\subseteq V(G)$ having the property that for every vertex $u\in V(G)\setminus S$, there exists a vertex $v\in S$ such that $(v,u)\in E(G)$ (note that some authors use the terms “kernel” and “antikernel” with their meanings interchanged). A directed graph is called bikernel-perfect if every induced subgraph of it has both a kernel and an antikernel. It is well known that directed acyclic graphs are all bikernel-perfect [14].

§ PRELIMINARIES

Erdős [6] proved that given any integer \(k\), there exist graphs with girth more than \(k\) and arbitrarily large chromatic number. This is equivalent to the following statement.

If $\mathcal{F}$ is a finite family of graphs, none of which is a forest, then for every integer $c$, there exists an $\mathcal{F}$-free graph $G$ having $\chi(G)\geq c$.

It is folklore that every graph $G$ contains every tree (and hence every forest) on $\chi(G)$ vertices as a subgraph. In fact, every graph having minimum degree $k-1$ contains every tree on $k$ vertices as a subgraph.
Now combining this with Theorem <ref>, we get the following.

Let $\mathcal{F}$ be a finite family of graphs. There exists a constant $c$ such that $\chi(G)<c$ for every $\mathcal{F}$-free graph $G$ if and only if $\mathcal{F}$ contains a forest.

§.§ The Gyárfás-Sumner Conjecture

The clique number $\omega(G)$ of a graph $G$ is the maximum size of a clique in $G$.
For any graph \(G\), a natural lower bound for its chromatic number \(\chi(G)\) is its clique number \(\omega(G)\). There are several constructions in the literature that show that there exist triangle-free graphs having arbitrarily large chromatic number (for example, see Mycielski's construction in [18]), which means that
\(\chi(G)\) can be arbitrarily large compared to \(\omega(G)\). Thus, it is interesting to study those classes of graphs for which \(\chi(G)\) can be bounded above by a function of \(\omega(G)\). Formally, a class of graphs is said to be \(\chi\)-bounded if there exists a function \(f\) such that for every graph \(G\) in the class, \(\chi(G) < f(\omega(G))\). We phrase this in slightly different language: for a family $\mathcal{G}$ of graphs and a complete graph $K$, let $f(\mathcal{G},K)$ be the smallest integer such that every $K$-free graph $G$ in $\mathcal{G}$ has $\chi(G)<f(\mathcal{G},K)$. Thus, a class of graphs $\mathcal{G}$ is $\chi$-bounded if and only if $f(\mathcal{G},K)$ exists for every complete graph $K$. Clearly, for a positive integer $t$, $f(\mathcal{G},K_t)$ exists only if $f(\mathcal{G},K_i)$ exists for each $i<t$. From the existence of constructions like that of Mycielski's, we know that the class of all graphs, in fact, even the class of triangle-free graphs, is not \(\chi\)-bounded—i.e., if we denote the class of all graphs by $\mathcal{G}$, then even $f(\mathcal{G},K_3)$ does not exist.

We denote by $\mathcal{G}_A$ the class of all graphs that do not contain the graph $A$ as an induced subgraph. For brevity, we abuse notation and let $f(A,K)$ denote $f(\mathcal{G}_A,K)$. Now we can think of $f(A,K)$ in the following way: $f(A,K)$ is the smallest integer such that every $K$-free graph $G$ having $\chi(G)\geq f(A,K)$ contains $A$ as an induced subgraph, or put in other words, every graph that has chromatic number at least $f(A,K)$ contains either $A$ as an induced subgraph or $K$ as a subgraph.
Since the class of all graphs is not $\chi$-bounded, it is natural to ask the following question: Does there exist a graph \(A\) such that $\mathcal{G}_A$ is \(\chi\)-bounded? In other words, is there a graph $A$ such that $f(A,K)$ exists for every complete graph $K$? Clearly, there are such graphs—for example, a $K_2$—and they are called \(\chi\)-bounding graphs. Even though $K_2$ is \(\chi\)-bounding, Mycielski's construction shows that \(K_3\) is not. In fact, it follows from Theorem <ref> that no \(\chi\)-bounding graph can have a cycle. Thus, the only possible \(\chi\)-bounding graphs are forests.
Gyárfás [8] and Sumner [17] independently conjectured that the converse must also be true, i.e. all forests are \(\chi\)-bounding. In fact, as noted in [9], since every forest is an induced subgraph of some tree, this is equivalent to saying that all trees are $\chi$-bounding.

For every tree $T$ and complete graph $K$, $f(T,K)$ exists. That is, there exists an integer $f(T,K)$ such that all \(K\)-free graphs \(G\) with \(\chi(G) \geq f(T,K)\) contain \(T\) as an induced subgraph.

Note that if the above conjecture is true, then for any positive integer $s$ and complete graph $K$, there exists a number $f(s,K)$ such that every $K$-free graph $G$ having $\chi(G)\geq f(s,K)$ contains every tree on $s$ vertices as an induced subgraph—in fact, $f(s,K)=\max\{f(T,K)\colon T$ is a tree on $s$ vertices$\}$. Despite considerable work, this conjecture is known to be true only for simple types of trees (see [16] for a complete list of trees for which the conjecture has been solved). It is not even known whether the conjecture holds for triangle-free graphs; i.e. whether or not $f(s,K_3)$ exists for every integer $s$.

Generalizing the above notation, for a finite family $\mathcal{F}$ of graphs and a graph $A$, we can define $f(A,\mathcal{F})$ to be the least integer such that every $\mathcal{F}$-free graph $G$ having $\chi(G)\geq f(A,\mathcal{F})$ contains $A$ as an induced subgraph. By Proposition <ref>, we have that $f(A,\mathcal{F})$ exists whenever $\mathcal{F}$ contains a forest, irrespective of the graph $A$. Since our aim is to study those graphs $A$ that appear as induced subgraphs in $\mathcal{F}$-free graphs when their chromatic number becomes sufficiently large, we shall limit our study to those families $\mathcal{F}$ that contain no forest. Then from Theorem <ref>, it follows that $f(A,\mathcal{F})$ exists only if $A$ is a forest. As before, it can be seen that $f(A,\mathcal{F})$ exists for all forests $A$ if and only if $f(T,\mathcal{F})$ exists for all trees $T$. Following our earlier notation, we define $f(s,\mathcal{F})=\max\{f(T,\mathcal{F})\colon T$ is a tree on $s$ vertices$\}$.

Some versions of the Gyárfás-Sumner conjecture obtained by relaxing the condition of $K$-freeness to $\mathcal{F}$-freeness, where $\mathcal{F}$ is some finite family of undirected graphs, are known to be true. In other words, $f(s,\mathcal{F})$ exists for some families $\mathcal{F}$ of graphs. For example, as the following theorem shows, even though it is unknown whether $f(s,C_3)$ exists (note that $C_3=K_3$), we know that $f(s,\{C_3,C_4\})$ exists.

Every $\{C_3,C_4\}$-free graph $G$ with $\chi(G)\geq s$ contains every tree on $s$ vertices as an induced subgraph. In other words, $f(s,\{C_3,C_4\})\leq s$.

The detailed survey of Scott and Seymour [16] lists many known results on the Gyárfas-Sumner conjecture, and discusses two well-studied variants of the conjecture: the “oriented” variant and the “rainbow” variant. We mainly concern ourselves with these two variants, which as we shall see, are closely related. Below, we give a brief introduction to each of these variants and present our contributions.

§.§ Our results

The Gallai-Roy-Vitaver theorem states that every properly colored graph $G$ contains a (not necessarily induced) rainbow $\chi(G)$-path. It is natural to ask if a stronger statement, along the lines of the Gyárfás-Sumner conjecture, holds: for every integer $s\geq 1$, does there exist an integer $h(s)$ such that every properly colored graph $G$ having $\chi(G)\geq h(s)$ contains every tree on $s$ vertices as a rainbow subgraph? This is not true as shown by Kierstead and Trotter [13], who constructed graphs with arbitrarily large chromatic number that can be properly colored in such a way that the neighbourhood of every vertex is colored using just two colors. Thus, there are properly colored graphs with arbitrarily large chromatic number that do not even contain a rainbow claw. Note that the graphs produced by the construction of Kierstead and Trotter are also triangle-free. N. R. Aravind (see [2]) proposed to modify the Gallai-Roy-Vitaver theorem in another way by asking the following question: does every properly colored triangle-free graph $G$ contain an induced rainbow $\chi(G)$-path? Note that such a statement cannot be true for general graphs, as for example, the complete graph does not contain even an induced 3-path in spite of having arbitrarily large chromatic number. On the other hand, it is known that every triangle-free graph $G$ contains an induced $\chi(G)$-path [9]. We state Aravind's conjecture below.

Any properly colored triangle-free graph \(G\) contains an induced rainbow path on \(\chi(G)\) vertices.

This conjecture can be generalized along the lines of the Gyárfás-Sumner conjecture as follows. Let $\mathcal{F}$ be a finite family of graphs. We define \(\ell(s,\mathcal{F})\) to be the smallest integer such that all properly colored \(\mathcal{F}\)-free graphs \(G\) with \(\chi(G) \geq \ell(s,\mathcal{F})\) contain an induced rainbow $s$-path. Clearly, $\ell(s,\emptyset)$ does not exist for $s\geq 3$ (since, as noted above, there exist graphs having arbitrarily large chromatic number and containing no induced 3-paths). Scott and Seymour [15] showed that $\ell(s,K)$ exists for every complete graph $K$, or in other words, that all rainbow paths are $\chi$-bounding (here we mean that the class of all properly colored graphs that do not contain an induced rainbow $s$-path, for some integer $s\geq 0$, is $\chi$-bounded). But the upper bounds provided by their construction are very fast growing functions. For example, their bound for \(\ell(s,C_3)\) is super-exponential in \(s\), whereas Conjecture <ref> states that \(\ell(s,C_3) = s\).

Note that Conjecture <ref> holds trivially for the case when \(g(G) > \chi(G)\) because the rainbow \(\chi(G)\)-path in $G$ (which exists by the Gallai-Roy-Vitaver theorem)

is also an induced path in this case. Babu et al. [2] showed Conjecture <ref> to be true for the case when \(g(G) = \chi(G)\). Therefore, as a special case, Conjecture <ref> holds true for 4-chromatic triangle-free graphs. Gyárfás and Sárközy [11] showed that $\ell(s,\{C_3,C_4\})$ exists. From their proof, it follows that \(\ell(s,\{C_3,C_4\}) = \mathcal{O}\big((2s)^{2s}\big)\).
In this paper, we improve this result significantly and show the following.

Let $G$ be a properly colored graph. In each of the following cases, $G$ contains at least $s!/2$ distinct
induced rainbow paths on $s$ vertices:

*  \(G\) is \(K_{2,r}\)-free (\(r \geq 2\)) and \(\chi(G)\geq (r-1)(s-1)s/2 + s\).

*  \(G\) has girth $g\geq 5$ and \(\chi(G) \geq s^{1+\frac{4}{g-4}}\).

As the theorem above implies that $\ell(s,C_4)\leq \frac {s^2 + s} {2} $, it improves the bound given by Gyárfás and Sárközy and also covers a bigger class of graphs. For graphs of higher girth, our results imply that \(\ell(s,\mathcal{C}_g) \leq s^{1+\frac{4}{g-4}}\) where \(\mathcal{C}_{g} = \{C_3,C_4,\dots,C_{g-1}\}\) and \(g \geq 5\). Note that the method of Gyárfás and Sárközy does not yield a bound which improves as the girth of the graph increases.

Although Babu et al. [2] proved that \(\ell(s,\mathcal{C}_s) =s\), it seems extremely difficult by their method to even show that \(\ell(s,\mathcal{C}_{s-1})\) is linear in $s$. However, our result above (Theorem <ref><ref>) shows that linear bounds on \(\ell(s,\mathcal{C}_g)\) can be achieved even when \(g = \Omega(\log s)\). Note that this is equivalent to saying that
there exists a function $h$ such that for every $\epsilon<1$, every properly colored graph $G$ having girth at least $h(\epsilon)\log\chi(G)$ contains an induced rainbow path on at least $\epsilon\cdot\chi(G)$ vertices (for example, we could let $h(\epsilon)=\frac{4}{\log(1/\epsilon)}$ ).

Another feature of our method, in comparison to the previous approaches, is that we are able to show the presence of at least \(s!/2\) distinct induced rainbow \(s\)-paths as opposed to a single rainbow $s$-path.

Let $G$ be an undirected graph, $\alpha$ be a proper vertex coloring of $G$, and $<$ a total order defined on the colors used in the coloring $\alpha$.
Assign a direction to each edge $uv$ of $G$ as follows: If $\alpha(u) <  \alpha(v)$, direct the edge from $v$ to $u$; otherwise,
direct it from $u$ to $v$. Let the resulting oriented graph be $D$. It is easy to verify that $D$ is a directed acyclic graph, since along any directed path, the colors of the vertices are in decreasing order. We refer to $D$ as the natural orientation of $G$ with respect to $(\alpha, <)$. Whenever the ordering $<$ is clear from the context, we may drop $<$ and just call it “the natural orientation of $G$ with respect to $\alpha$”. Note that any induced directed path of $D$ would correspond to an induced rainbow path in $G$ and any induced rainbow path in $G$ along which the colors decrease corresponds to an induced directed path in $D$.

Let $\mathcal{F}$ be a finite collection of directed or undirected graphs. Let $\overline{\ell}(s,\mathcal{F})$ denote the least integer such that every $\mathcal{F}$-free directed acyclic graph $G$ with $\chi(G)\geq\overline{\ell}(s,\mathcal{F})$ contains an induced directed $s$-path.

Note that a properly colored graph contains an induced rainbow path on $s$ vertices if its natural orientation $D$ (with respect to an arbitrarily chosen ordering of the colors) contains an induced directed path on $s$ vertices. It follows that $\ell(s,\hat{\mathcal{F}})\leq\overline{\ell}(s,\mathcal{F})$, where $\hat{\mathcal{F}}=\{\hat{G}\colon G\in\mathcal{F}\}$.
This means that in order to prove Theorem <ref><ref>, it suffices to show that $\overline{\ell}(s,K_{2,r}) \le (r-1)(s-1)s/2 + s$ and in order to prove Theorem <ref>(ii), it suffices to show that for $g\geq 5$, $\overline{\ell}(s,\{C_3,C_4,\ldots,C_{g-1}\}) \le s^{1+\frac{4}{g-4}}$. In fact, we prove a stronger statement: we show that if the chromatic number of a directed acyclic graph is at least the bounds mentioned above, then it contains every out-tree and every in-tree on $s$ vertices as an induced subgraph. Also, for the former case, we prove a slightly stronger statement, namely instead of $K_{2,r}$-free directed acylic graphs, we prove the bound for the more general class of $\mathcal{B}_r$-free directed acyclic graphs, where $\mathcal{B}_r$ denotes the family of orientations of $K_{2,r}$ in which one vertex has out-degree $r$ (note that we assume that $r\geq 2$).

Let $G$ be a directed acyclic graph. Then in each of the following cases, $G$ contains every out-tree and every in-tree on at most $s$ vertices as an induced subgraph:

* $G$ is $\mathcal{B}_r$-free ($r\geq 2$) and $\chi(G)\geq (r-1)(s-1)s/2+s$.

* $G$ has girth $g\geq 5$ and \(\chi(G) \geq s^{1+\frac{4}{g-4}}\).

Could it be possible that a theorem such as the one above could be made for arbitrary oriented trees instead of just out-trees and in-trees, and also without restricting ourselves to directed acyclic graphs? In fact, if we only want the oriented trees to occur as subgraphs that are not necessarily induced, then such a result exists due to Burr [3].

Every directed graph $G$ having $\chi(G)\geq (s-1)^2$ contains every oriented forest on $s$ vertices as a subgraph.

But what if we want the oriented tree to be an induced subgraph? As we shall see, it turns out that we can show the existence of every oriented tree on $s$ vertices as an induced subgraph in any $\mathcal{B}_r$-free oriented graph whose chromatic number is just a little over two times the bound in Theorem <ref>(i) (see Theorem <ref>), and also in any bikernel-perfect oriented graph having girth $g\geq 5$ whose chromatic number is roughly two times the bound in Theorem <ref>(ii) (see Theorem <ref>). A natural question now would be whether we can remove the restrictions we impose on the oriented graphs, i.e. is it true that for every integer $s$, every oriented graph of high enough chromatic number contains every oriented tree on $s$ vertices as an induced subgraph? As the discussion below shows, this is not true.

Given a finite collection of directed or undirected graphs ${\cal F}$ and an oriented graph $A$, let $\overline{f}(A,{\cal F})$ be the smallest integer such that every ${\cal F}$-free oriented graph $G$ having $\chi(G)\geq \overline{f}(A,{\cal F})$ contains $A$ as an induced subgraph. Since a graph contains $\hat{A}$ as an induced subgraph if some orientation of it contains $A$ as an induced subgraph, we have $f(\hat{A},\hat{\mathcal{F}})\leq\overline{f}(A,\mathcal{F})$. Note that if $\mathcal{F}$ contains an oriented forest, then by Theorem <ref>, we have that $\overline{f}(A,\mathcal{F})$ exists for every oriented graph $A$. So the families $\mathcal{F}$ that we consider do not contain any oriented forest. Then as noted before, $f(\hat{A},\hat{\mathcal{F}})$ exists only when $\hat{A}$ is a forest, which implies by the previous inequality that $\overline{f}(A,\mathcal{F})$ exists only if $A$ is an oriented forest. Thus, as before, we will focus on the case when $A$ is an oriented tree. As with undirected trees, we say that an oriented tree $T$ is $\chi$-bounding if $\overline{f}(T,K)$ exists for all complete graphs $K$.
Clearly, if $\overline{f}(T,K)$ exists for every oriented tree $T$ and any complete graph $K$, then the Gyárfás-Sumner conjecture would also hold. However, it is known that even \(\overline{f}(T,C_3)\) does not exist when $T$ corresponds to certain orientations of a 4-path: Kierstead and Trotter showed that \(\rightarrow\rightarrow\rightarrow\) is not \(\chi\)-bounding [13] and Gyárfás showed that \(\rightarrow\leftarrow\rightarrow\) is not \(\chi\)-bounding [10]. Therefore, $\overline{f}(T,K)$ can exist for every complete graph $K$ only for certain restricted kinds of oriented trees. Recently, Chudnovsky, Scott and Seymour showed that the family of oriented stars and the remaining orientations of a 4-path are all \(\chi\)-bounding [4].

Further questions in this direction could involve finding out whether specific orientations of larger trees are \(\chi\)-bounding. However, proceeding in this direction may be tedious. A more fruitful approach would be to alter  the constraint of triangle-freeness and ask whether $\overline{f}(T,C_4)$ exists for every oriented tree $T$. In other words, is it true that every oriented tree appears as an induced subgraph of any \(C_4\)-free oriented graph with a high enough chromatic number?

We show that this is indeed the case: in particular, we show that $\overline{f}(T,\mathcal{B}_r)$ exists for every oriented tree $T$ and $r\geq 2$. As every $C_4$-free oriented graph is also $\mathcal{B}_2$-free, this result is an oriented variant of Gyárfás, Szemeredi and Tuza's result that $f(T,\{C_3,C_4\})$ exists for every undirected tree $T$ (Theorem <ref>).
As before, for a positive integer $s$ and a finite collection of directed or undirected graphs ${\cal F}$, we let $\overline{f}(s,{\cal F})=\max\{\overline{f}(T,{\cal F})\colon T$ is an oriented tree on $s$ vertices$\}$ — i.e., it is the minimum integer such that every ${\cal F}$-free oriented graph having chromatic number at least $\overline{f}(s,{\cal F})$ contains every oriented tree on $s$ vertices as an induced subgraph. Clearly, $f(s,\hat{\cal F})\leq \overline{f}(s,{\cal F})$. We show that $\overline{f}(s,\mathcal{B}_r)\leq (r-1)(s-1)(s-2)+2s+1$, which is stated as the theorem below.

Let $G$ be a $\mathcal{B}_r$-free oriented graph ($r\geq 2$) with $\chi(G) \geq (r-1)(s-1)(s-2)+2s+1$. Then $G$ contains every oriented tree on $s$ vertices as an induced subgraph.

Thus, $\overline{f}(s,C_4)\leq s^2+s+1$.
We then give stronger bounds for graphs of higher girth, albeit only in a more restricted setting.

Let $G$ be a bikernel-perfect oriented graph having girth $g\geq 5$. Then $G$ contains every oriented tree on $s$ vertices as an induced subgraph if $\chi(G)\geq 2s^{1+\frac{4}{g-4}}$.

§.§ Notation

We define some common notation that we shall use in our proofs.
Let $G$ be an undirected or directed graph.
An undirected edge between the vertices $u$ and $v$ of $G$ shall be denoted as $uv$, whereas an edge directed from $u$ to $v$ shall be denoted as $(u,v)$. Two vertices are said to be adjacent in $G$ if there is a (directed or undirected) edge between them in $G$. The set of all vertices adjacent to a vertex $v$ in $G$ is called the neighbourhood of \(v\) in $G$ and is denoted by \(N_G(v)\) (the subscript $G$ is omitted when the graph under consideration is clear). In a directed graph $G$, if $(u,v)\in E(G)$, then $v$ is said to be an out-neighbour of $u$, and $u$ is said to be an in-neighbour of $v$. For a directed graph $G$ and a vertex $v\in V(G)$, the out-neighbourhood (resp. in-neighbourhood) of $v$, i.e. the set of out-neighbours (resp. in-neighbours) of $v$ in $G$, is denoted by $N^+_G(v)$ (resp. $N^-_G(v)$). (Again, the subscript is omitted when the graph under consideration is clear from the context.) Note that for every vertex $v\in V(G)$, $N(v)=N^+(v)\cup N^-(v)$.
Also out-degree  (resp. in-degree) of a vertex $v$ is defined as $|N^+(v)|$ (resp. $|N^-(v)|$) and is denoted as $d^+(v)$ (resp. $d^-(v)$), with sometimes a subscript denoting the graph under consideration.  For a graph or digraph $G$ and a vertex $v\in V(G)$, we denote by $N_G[v]$ the closed neighbourhood of $v$, i.e. $N_G[v]=N_G(v)\cup\{v\}$.
Given a coloring $\alpha$ of a (directed or undirected) graph $G$ and $S\subseteq V(G)$, we denote by $\alpha(S)$ the set $\{\alpha(v)\colon v\in S\}$.
For a graph (resp. digraph) and a set $S\subseteq V(G)$, we denote by $G[S]$ the subgraph (resp. subdigraph) of $G$ induced by the vertices $S$. We denote by $G-S$ the graph $G[V(G)\setminus S]$.

§ INDUCED RAINBOW PATHS

We first prove Lemma <ref>, which will be our main tool for later results. In order to prove the lemma, we need the following Turán-type results from extremal graph theory.

Given a finite family of graphs \(\mathcal{L}\), let \(\text{\textbf{ex}}(n,\mathcal{L})\) denote the maximum number of edges that an \(\mathcal{L}\)-free graph \(G\) on \(n\) vertices can have.

*  \(\textnormal{\textbf{ex}}(n,\{C_3,C_4,\dots,C_{2k}\}) < \frac{1}{2}n^{1+\frac{1}{k}} + \frac{1}{2}n.\)

*  \(\textnormal{\textbf{ex}}(n,\{C_3,C_4,\dots,C_{2k+1}\}) < \big(\frac{n}{2}\big)^{1+\frac{1}{k}} + \frac{1}{2}n.\)

Let $\alpha$ be a proper vertex coloring of an oriented graph $G$ and let $<$ be a total ordering of $\alpha(V(G))$. We say that $(\alpha,<)$ is an out-tree coloring of $G$ if for every $v\in V(G)$, $\alpha(N^+(v))\supseteq\{i\in\alpha(V(G))\colon i<\alpha(v)\}$.  When the ordering $<$ on the colors is clear from the context, we shorten “$(\alpha,<)$ is an out-tree coloring” to just “$\alpha$ is an out-tree coloring”.

Let $F$ be an oriented graph and let $G$ be a subgraph of $F$. Suppose that $G$ has an out-tree coloring using at least $k$ colors. In each of the following cases, every out-tree tree $T$ on $s$ vertices is present as an induced rainbow subgraph of $G$, which is also an induced subgraph of $F$:

*  \(F\) is $\mathcal{B}_r$-free (\(r \geq 2\)) and \(k\geq (r-1)(s-1)s/2 + s\).

*  $F$ has girth $g\geq 5$ and \(k \geq s^{1+\frac{4}{g-4}}\).

We assume that the given out-tree coloring $(\alpha,<)$ of $G$ uses natural numbers as colors, where the ordering $<$ between the colors is the usual ordering $<$ on natural numbers. For each vertex $v\in V(G)$ and $i\in\{1,2,\ldots,\alpha(v)-1\}$, we denote by $v^i$ an arbitrarily chosen vertex in $N_G^+(v)$ such that $\alpha(v^i)=i$ (such a vertex exists as $\alpha$ is an out-tree coloring of $G$).
Let \(T\) be any out-tree on \(s\) vertices and let \(w_1,w_2,\ldots,w_s\) be a topological ordering of \(T\). For each $i\in\{2,3,\ldots,s\}$, let $p(i)$ be the integer such that $w_{p(i)}$ is the parent of $w_i$ in $T$. Clearly $w_1$ corresponds to the root of $T$, and since the parent of every vertex appears before it in the ordering, we have $p(i)<i$.

Let $i\in\{1,2,\ldots,s\}$. Define $T_i=T[\{w_1,w_2,\ldots,w_i\}]$. Suppose that $T'_i$ is an induced tree in $G$ as well as $F$ that is isomorphic to $T_i$. Let $v_1,v_2,\ldots,v_i$ be the vertices of $T'_i$ corresponding to $w_1,w_2,\ldots,w_i$. Then clearly, for each $j\in\{2,3,\ldots,i\}$, $v_{p(j)}$ is the parent of $v_j$ in $T'_i$. We say that $T'_i$ is a “good” tree if $\alpha(v_1)=k$ and for $2\leq j\leq i$,

* $\alpha(v_j)<\alpha(v_{j-1})$, and

* for $\alpha(v_j)<t<\alpha(v_{j-1})$, the subgraph induced by $\{v_1,v_2,\ldots,v_{j-1},v_{p(j)}^t\}$ in $F$ is not isomorphic to $T_j$. This means that in $F$, the vertex $v_{p(j)}^t$ has a neighbor other than $v_{p(j)}$ in $\{v_1,v_2,\ldots,v_{j-1}\}$, i.e. $N_F(v_{p(j)}^t)\cap (\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\})\neq\emptyset$.  (Refer Figure <ref> for an example that shows how a good tree might look like.)

[above=7] at (5,5) $v_1$;
at (5,5) $k$;
[left=7] at (4,4) $v_2$;
at (4,4) $k\!\!-\!\!1$;
at (4.5,4) $k\!\!-\!\!2$;
at (5,4) $k\!\!-\!\!3$;
at (5.5,4) $k\!\!-\!\!4$;
[right=7] at (6,4) $v_3=v_{p(6)}=v_1^{k-5}$;
at (6,4) $k\!\!-\!\!5$;
at (3,3) $k\!\!-\!\!6$;
[below=7] at (3.5,3) $v_4$;
at (3.5,3) $k\!\!-\!\!7$;
at (4,3) $k\!\!-\!\!8$;
at (4.5,3) $k\!\!-\!\!9$;
[below=7] at (5,3) $v_5$;
at (5,3) $k\!\!-\!\!10$;
at (5.5,3) $k\!\!-\!\!11$;
[below=7] at (5.5,3) $v_{p(6)}^{k\!-\!11}$;
at (6,3) $k\!\!-\!\!12$;
[below=7] at (6,3) $v_{p(6)}^{k\!-\!12}$;
[below=7] at (6.5,3) $v_6$;
at (6.5,3) $k\!\!-\!\!13$;
[decorate,decoration=brace,amplitude=4pt] (4.65,2.8) – (3.85,2.8);
at (4.25,2.6) $X_5$;

A possible way the good tree $T'_6$ might look like. The edges of $T'_6$ are drawn in bold. For each vertex $u$, its color $\alpha(u)$ is written inside it. For each $j\in\{2,3,\ldots,6\}$ and $\alpha(v_j)<t<\alpha(v_{j-1})$, a gray arc shows the edge from $v_{p(j)}$ to $v_{p(j)}^t$. A red edge shows an adjacency in $F$ between $v_{p(j)}^t$ and some vertex in $\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\}$.

Notice that every good tree is an induced rainbow subgraph of $G$.
In order to prove the statement of the lemma, we shall prove the stronger statement that there is a good tree isomorphic to $T_s=T$.
Suppose for the sake of contradiction that there is no good tree isomorphic to $T_s$. Let $i$ be the largest integer in $\{1,2,\ldots,s-1\}$ such that there is a good tree $T'_i$ isomorphic to $T_i$ (recall that there is a vertex $y\in V(G)$ such that $\alpha(y)=k$; this means that the single vertex $y$ is a good tree $T'_1$ isomorphic to $T_1$, and so $i$ exists).
As before, let $v_1,v_2,\ldots,v_i$ be the vertices of $T'_i$ corresponding to the vertices $w_1,w_2,\ldots,w_i$ of $T_i$. For $2\leq j\leq i$, define $X_j=\left\{v_{p(j)}^{\alpha(v_j)+1},v_{p(j)}^{\alpha(v_j)+2},\ldots,v_{p(j)}^{\alpha(v_{j-1})-1}\right\}$ and $X'=\left\{v_{p(i+1)}^1,v_{p(i+1)}^2,\ldots,v_{p(i+1)}^{\alpha(v_i)-1}\right\}$. Further, define $X=X'\cup\bigcup_{j=2}^i X_j$. (It is not difficult to see that we always have $X_2=\emptyset$, although we do not use this fact.) Note that for each color in $\{1,2,\ldots,k\}\setminus\{\alpha(v_1),\alpha(v_2),\ldots,\alpha(v_i)\}$, the set $X$ contains exactly one vertex having that color. Therefore, we have

\begin{equation}\label{eq1}
k = i + |X|
\end{equation}

As $T'_i$ is a good tree, we know that for every vertex $x\in X_j$, where $2\leq j\leq i$, $N_F(x)\cap(\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\})\neq\emptyset$.
If for some integer $t\in\{1,2,\ldots,\alpha(v_i)-1\}$, we have $N_F(v_{p(i+1)}^t)\cap(\{v_1,v_2,\ldots,v_i\}\setminus\{v_{p(i+1)}\})=\emptyset$, then taking $t$ to be the largest such integer, we have that $G[\{v_1,v_2,\ldots,v_i\}\cup\{v_{p(i+1)}^t\}]$ is a good tree isomorphic to $T_{i+1}$, which contradicts our choice of $i$. We can therefore conclude that for every $x\in X'$, $N_F(x)\cap(\{v_1,v_2,\ldots,v_i\}\setminus\{v_{p(i+1)}\})\neq\emptyset$.

Proof of <ref>

Let $j\in\{2,3,\ldots,i\}$. Since \(F\) is \(\famK\)-free (\(r \geq 2\)), any vertex $z\in\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\}$ can have at most $r-1$ vertices in $X_j$ as neighbours in $F$, as otherwise there is a subgraph in $F$ that is isomorphic to a graph in $\famK$,  whose vertex set is $\{v_{p(j)},z\}$ together with $r$ vertices in $N_F(z)\cap X_j$.

As every vertex in $X_j$ has some neighbour in $\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\}$, we can conclude that $|X_j|\leq (r-1)(j-2)$. Similarly, no vertex in $\{v_1,v_2,\ldots,v_i\}\setminus\{v_{p(i+1)}\}$ can have $r$ neighbours in $X'$ (in $F$), implying that $|X'|\leq (r-1)(i-1)$. We thus get:

\[|X| \leq (r-1)(i-1)+\sum_{j=2}^i (r-1)(j-2) = (r-1)(i)(i-1)/2\]

By plugging this into Equation (<ref>) and by using the bound on $k$ given in the statement of the lemma, we get:

\[\frac {(r-1)(s)(s-1)} {2}  + s  \: \leq \: k \: \leq \: i + \frac{(r-1)(i)(i-1)}{2} \]

From this, we get \(i \geq s\), which is a contradiction to the assumption that $i<s$.

Proof of <ref>

First, let us show that $i\geq g$. Suppose that $i<g$. If there exists $x\in X_j$ for some $j\in\{2,3,\ldots,i\}$, then we know that there exists $z\in N_F(x)\cap\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\}$. But then the path in $T'_i$ between $z$ and $v_{p(j)}$ (this path contains only vertices in $\{v_1,v_2,\ldots,v_{j-1}\}$) together with $x$ forms a cycle of length at most $j\leq i<g$ in $F$. Since this contradicts the fact that $g$ is the girth of $F$, we can conclude that $X_j=\emptyset$ for $2\leq j\leq i$. Further, if there exists $x\in X'$, then since it has as neighbours the vertex $v_{p(i+1)}$ and also a vertex $z\in N_F(x)\cap\{v_1,v_2,\ldots,v_i\}\setminus\{v_{p(i+1)}\}$, we can conclude that the path in $T'_i$ between $v_{p(i+1)}$ and $z$ contains at least $g-2\geq i-1$ edges. Since $T'_i$ contains only $i$ vertices, this means that $\hat{T'_i}$ is the path $v_1,v_2,\ldots,v_i$ and that $N_F(x)\cap\{v_1,v_2,\ldots,v_i\}=\{v_1,v_i\}$. Further, if there are two vertices $x,y\in X'$, this means that $x,v_1,y,v_i,x$ is a cycle of length 4 in $F$, which is a contradiction to the fact that $g\geq 5$. This implies that $|X'|\leq 1$. Then $|X|\leq 1$, which by Equation (<ref>) means that $i\geq k-1$. From the statement of the lemma, we have $k\geq s+1$ as $k$ and $s$ are integers, implying that \(i\geq s\), contradicting the assumption that $i<s$. We can therefore assume that $i\geq g$.

We shall construct an auxiliary undirected graph \(H\) with \(V(H) = V(T'_i)\) and \(E(H) = E(\hat{T'_i}) \cup E'\), where \(E'\) is a set of edges defined as follows. For $j\in\{2,3,\ldots,i\}$ and vertex \(x \in X_j\), the edge \(v_{p(j)}y\) is added to \(E'\), where $y$ is an arbitrarily chosen vertex in $N_F(x)\cap (\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\})$. Note that if $v_{p(j)}$ and $y$ are adjacent in $F$, then $v_{p(j)},x,y$ form a triangle in $F$, contradicting the fact that $g(F)=g\geq 5$. Also, no other neighbour of $v_{p(j)}$ can be adjacent to $y$ in $F$, since otherwise there will be a cycle of length four in $F$, again contradicting the fact that $g(F)=g\geq 5$. These two observations together imply that we never add an edge of $E(\hat{F})$ to $E'$ and also that we never add the same edge twice to $E'$. Similarly, for each $x\in X'$, the edge $v_{p(i+1)}y$ is added to $E'$, where $y$ is an arbitrarily chosen vertex in $N_F(x)\cap (\{v_1,v_2,\ldots,v_i\}\setminus\{v_{p(i+1)}\})$. Again, arguing as before, we can see that we never add the same edge twice to $E'$ and that $E'$ and $E(\hat{F})$ are disjoint. We can therefore conclude that $H$ is a simple graph and that \(|E'| = |X'|+\sum_{j=2}^i |X_j|= |X|\).

Now, we claim that \(g(H) \geq g/2\). Given any cycle \(C\) of \(H\), each edge of \(C\) can be replaced by at most 2 edges of \(F\) (which in effect is like subdividing that edge) to obtain a closed walk (in fact, a cycle) in $F$ containing at most $2|E(C)|$ edges. This can be seen as follows. Let $C=u_0,u_1,\ldots,u_{l-1},u_0$ be a cycle in $H$ and let the edges of $E'$ that belong to $C$ be $u_{j_1}u_{j_1+1},u_{j_2}u_{j_2+2},\ldots,u_{j_q}u_{j_q+1}$ where $j_1<j_2<\cdots<j_q$ and subscripts are modulo $l$. Recalling the definition of $E'$, we know that for each $t\in\{1,2,\ldots,q\}$, there exists a vertex $x_t\in X_2\cup X_3\cup\cdots\cup X_i\cup X'$ such that $u_{j_t}x_t,u_{j_t+1}x_t\in E(F)$. Then $C'=u_0,u_1,\ldots,u_{j_1},x_t,u_{j_1+1},\ldots,u_{j_2},x_2,u_{j_2+1},\ldots,u_{j_q},x_q,u_{j_q+1},\ldots,u_{l-1},u_0$ is a closed walk in $F$ that contains at most $2l$ edges. Thus, if \(g(H) < g/2\), then there exists a cycle in \(F\) of length less than \(g\), which is a contradiction.

From Lemma [ext1]<ref><ref>, if \(g(H)\) is odd, we get (note that if $g(H)=3$, then the inequality below holds trivially):

\[|E(H)| \  < \  \frac{i^{1+2/(g(H)-1)} + i}{2} \  \leq \  \frac{i^{1+4/(g-2)} + i}{2} \ < \ \frac{i^{1+4/(g-4)} + i}{2}\]

From Lemma [ext2]<ref><ref>, if \(g(H)\) is even, we get:

\[|E(H)| \  < \  \bigg(\frac{i}{2}\bigg)^{1+2/(g(H)-2)} + \frac{i}{2} \  \leq \  \bigg(\frac{i}{2}\bigg)^{1+4/(g-4)} + \frac{i}{2}  \ < \ \frac{i^{1+4/(g-4)} + i}{2}\]

Combining both the cases, we get

\[|E(H)| \ < \ \frac{i^{1+4/(g-4)} + i}{2}\]

Since \(|E(H)| = |E'|+(i-1)\), we have:

\[|E'| < \frac{i^{1+4/(g-4)} - i + 2}{2}\]

Since \(|E'| = |X|\), by plugging the above inequality into Equation (<ref>),

we get:

\begin{eqnarray*}
k \: &<& \: i + \frac{i^{1+4/(g-4)} - i + 2}{2}=\frac{i(1 + 2/i+i^{4/(g-4)})}{2}\\
&\leq&\frac{i(e^{2/i}+i^{4/(g-4)})}{2}\qquad\mbox{(since }1+2/i\leq e^{2/i}\mbox{)}\\
&\leq&\frac{i(e^{2/(g-4)}+i^{4/(g-4)})}{2}\qquad\mbox{(since $i\geq g\geq g-4$)}\\
\end{eqnarray*}

By our assumption that $i\geq g$, we get that \(i \geq 2 > \sqrt{e}\). Using this in the above inequality, we get:

\[k < i^{1+4/(g-4)}\]

Now by using the bound on $k$ given in the statement of the lemma, we get:

$$ s^{1+4/(g-4)} \leq k < i^{1+4/(g-4)}$$

This implies that $i\geq s$, which is a contradiction.

The following corollary simply states the case when $F=G$ in Lemma <ref>.

Let $G$ be an oriented graph that has an out-tree coloring using at least $k$ colors. Then in each of the following cases, every out-tree on $s$ vertices is present as an induced rainbow subgraph of $G$:

*  \(G\) is $\mathcal{B}_r$-free (\(r \geq 2\)) and \(k\geq (r-1)(s-1)s/2 + s\).

*  $G$ has girth $g\geq 5$ and \(k \geq s^{1+\frac{4}{g-4}}\).

Recall that our intention is to prove Theorem <ref>. Given a graph $G$ with a proper vertex coloring $\beta$ and a total ordering $<$ on the set of
colors, a convenient orientation of $G$ that we can consider is the
natural orientation $D$ of $G$ with respect to $(\beta, <)$, as we had discussed in Section <ref>. If $(\beta,<)$ were an out-tree coloring of $D$, we could have immediately
inferred that there is an induced rainbow path on $s$ vertices in $G$ by applying Corollary <ref> on $D$, setting $T$ to be a directed path on $s$ vertices.
But note that it is possible that $(\beta,<)$ is not an out-tree coloring of $D$ for any choice of the total ordering $<$ of the colors of $\beta$.

To overcome this difficulty, we recolor $G$ using the following Greedy Refinement Algorithm  developed and used in [2]. The algorithm takes as input a graph with a proper coloring $\beta$
where some total order on  the set of colors is also given. The algorithm gives a new proper coloring $\alpha$ as output.

A graph \(G=(V,E)\) with a proper coloring \(\beta\) and an ordering $<$ of the colors used by $\beta$.
The graph \(G=(V,E)\) with a refined coloring \(\alpha\)
Let $\beta_1,\beta_2,\ldots,\beta_h$ be the colors used by $\beta$, where $\beta_1<\beta_2< \cdots<\beta_h$.

\(v \in V(G)\)
\(\alpha(v) \gets 0\)
\(i \gets 1\) \(h\)

\(v\in V(G)\) such that \(\beta(v)=\beta_i\) and \(\alpha(v)=0\)

\(\alpha(v)\leftarrow\min (\mathbb{N}^{>0} \setminus \{\alpha(u)\colon u\in N(v)\})\)

(i.e., \(\alpha(v)\) is set equal to the least positive integer that has not already been assigned to a neighbour of \(v\)).

Greedy Refinement Algorithm

For every vertex \(v\), there exists a subset \(S=\{u_1,u_2, \dots, u_{j-1}\}\) of \(N(v)\), where $j=\alpha(v)$, such that for \(1 \leq i \leq j-1\), \(\alpha(u_i)=i\) and \(\beta(u_i) < \beta(v)\).

Let \(v\) be any vertex with \(\alpha(v)=j\). From Algorithm <ref>, it is clear that at the point in time when \(v\) was colored with \(j\), there existed a subset \(S=\{u_1,u_2,\ldots, u_{j-1}\}\) of \(N(v)\) such that \(\alpha(u_i)=i\) for each $i\in\{1,2,\ldots,j-1\}$. Because each such \(u_i \in S\) should have been colored before \(v\), we have \(\beta(u_i) \leq\beta(v)\), and since $u_i\in N(v)$, we further have $\beta(u_i)<\beta(v)$. As Algorithm <ref> colors each vertex only once, the subset \(S\) of \(N(v)\) exists even when the algorithm terminates.

The following proposition is easy to see.

The coloring $\alpha$ generated by the Greedy Refinement Algorithm (Algorithm <ref>) always uses at most as many colors as that used by the input coloring $\beta$.

The coloring $\alpha$ generated by Algorithm <ref> will be referred to as $\refinement(\beta,<)$, the “refinement” of the proper coloring $\beta$ with respect to the total
order $<$ on the set of colors. Note that the colors used by $\alpha$ are from $\mathbb{N}^{>0}$ and unless otherwise specified, the color set of $\alpha$ is ordered by the usual order on natural numbers.
Note that $\alpha =\refinement(\beta,<)$ is an out-tree coloring of $D$, the natural orientation of $G$ with respect to $\alpha$. This follows from Proposition <ref> and the definition of
natural orientation (note that the vertices $u_1,u_2, \ldots, u_{\alpha(v)-1}$ that are guaranteed to exist for every vertex $v$ in Proposition <ref> are all out-neighbours of $v$ in $D$).

Given an undirected graph $G$ with a proper vertex coloring $\beta$ and an ordering of the colors of $\beta$, a tree  $T$ in $G$ with a chosen root is said to be a decreasing tree in $G$ if the colors on every path from root to leaf in $T$ are in decreasing order.

Let $G$ be an undirected graph with a proper coloring $\beta$ on whose colors a total order $<$ is defined. Then in each of the following cases, every rooted tree on $s$ vertices is present in $G$ as an induced decreasing tree:

* \(G\) is \(K_{2,r}\)-free (\(r \geq 2\)) and \(\chi(G)\geq (r-1)(s-1)s/2 + s\).

* $G$ has girth \(g \geq 5\) and \(\chi(G) \geq s^{1+4/(g-4)}\).

Let $T$ be a rooted tree on $s$ vertices. Assign directions to the edges of $T$ so that we get an out-tree $T'$ with the same root as $T$. Let $\alpha=\refinement(\beta,<)$ be a coloring that uses the colors $\{1,2,\ldots,k\}$. Clearly, $k\geq \chi(G)$. Let $D$ be the natural orientation of $G$ with respect to $\alpha$. Let $D'$ be the subgraph of $D$ that has vertex set $V(D')=V(D)$ and edge set $E(D')=\{(u,v)\in E(D)\colon\beta(u)>\beta(v)\}$. By Proposition <ref>, we know that $\alpha$ is an out-tree coloring of $D'$ as well. Since $V(D')=V(D)$, we know that there is at least one vertex $x$ in $D'$ having $\alpha(x)=k$.
It can be seen that if one of the conditions in the statement of the lemma is satisfied, then Lemma <ref> can be applied to $D$, $D'$ and $T'$ in order to conclude there is an induced subgraph of $D'$, which is also induced in $D$, that is isomorphic to $T'$. Since the colors in $\beta$ decrease along any directed path in $D'$, we have that the induced subgraph of $D$ that is isomorphic to $T'$ is a decreasing tree with respect to $(\beta,<)$.

Theorem <ref> follows from Lemma <ref> in a straightforward way.

Let $k$ be the number of colors used in the proper coloring of $G$. For a given ordering of the colors in $G$, we can apply Lemma <ref> on $G$ and a path on $s$ vertices (choosing any endpoint of the path as its root), and thereby conclude that there is a decreasing path on $s$ vertices that is induced in $G$.
Note that a particular decreasing path $P$ can be produced by the application of Lemma <ref> for at most $2k!/s!$ different orderings of the $k$ colors of the proper coloring of $G$. Thus, if we apply Lemma <ref> for all the $k!$ different orderings possible for the colors in $G$, we get at least $s!/2$ different decreasing paths in $G$.

Theorem [rainbowpathk2r]<ref><ref> shows that every properly colored $C_4$-free graph $G$ having $\chi(G)\geq\frac{s^2+s} {2}$ contains $s!/2$ induced rainbow paths on $s$ vertices. This improves the result of Gyárfás and Sarközy, who showed that there exists a function $h(s)=\Omega((2s)^{2s})$ such that every graph having girth at least 5 and chromatic number at least $h(s)$ contains an induced rainbow path on $s$ vertices, not only in the bound, but also in the fact that it guarantees the existence of several rainbow paths. Notice that Theorem [rainbowpathk2r]<ref><ref> applies to a wider class of graphs than the graphs with girth at least 5 and that Theorem [rainbowpathgirth]<ref><ref> gives better and better bounds as the girth increases.

If \(c\) is a proper coloring of a graph \(G\), let \(\mu(G,c)\) be the number of vertices in the largest induced rainbow path in \(G\) with respect to \(c\). Let \(\mu(G) := \min_c\mu(G,c)\) over all proper colorings \(c\) of \(G\).
In the context of induced rainbow paths, Theorem [rainbowpathgirth]<ref><ref> shows that linear bounds on $\ell(s,\mathcal{C}_g)$ can be achieved when \(g = \Omega(\log s)\). More precisely, if \(G\) is a graph with girth \(g \geq 4 + 4\log s\) and \(\chi(G) \geq 2s\), then \(\mu(G) \geq s\). Babu et al. [2] proved that \(\ell(s,\mathcal{C}_s) =s\), i.e. $\mu(G) = \chi(G)$ if $\chi(G) \ge g(G)$. They asked whether there exists some constant $c<1$ for which it can be shown that $\mu(G)=\chi(G)$ for every graph $G$ having girth at least $c\cdot\chi(G)$. Note that our result above implies that

there exists a function $h$ such that for every $\epsilon<1$, $\mu(G)\geq\epsilon\cdot\chi(G)$ for every graph $G$ having girth at least $h(\epsilon)\log\chi(G)$.

§.§ Induced rainbow trees in optimal colorings

As noted in Section <ref>, a counterexample given by Kierstead and Trotter <cit.> shows that Conjecture <ref> cannot be extended from induced rainbow paths to induced rainbow trees.   However, similar counterexamples are not known for \(C_4\)-free graphs. Therefore the following question remains to be answered.

Does there exist an integer \(h(s,C_4)\) corresponding to every positive integer $s$ such that all properly colored \(C_4\)-free graphs \(G\) with \(\chi(G) \geq h(s, C_4)\) contain every tree on \(s\) vertices as an induced rainbow subgraph?

Gyárfás, Szemeredi and Tuza [12] showed that given a graph $G$ with  $\chi(G)\geq s$ and any optimal coloring of it, every tree on $s$ vertices occurs as a rainbow subgraph (which may not be an induced subgraph) of $G$.

Let $G$ be an optimally colored graph. Then every tree on $\chi(G)$ vertices occurs as a rainbow subgraph of $G$.

Such a result cannot be obtained for non-optimal colorings, even for triangle-free graphs, due to Kierstead and Trotter's counterexample <cit.>, which is a triangle-free graph with a coloring such that there is no rainbow claw in the graph. Does every tree on $s$ vertices occur as an induced rainbow subgraph of any optimally colored $K$-free graph with high enough chromatic number (where $K$ is some complete graph)?
Clearly, an affirmative answer to this question would imply the Gyárfás-Sumner conjecture.
We now show that it can be inferred  from Theorem <ref> that an optimally colored $C_4$-free graph $G$ having $\chi(G)\geq (2s)^s$ contains every tree on $s$ vertices as an induced rainbow subgraph. In fact, the following more general statement is true.

Let $G$ be a $K_{2,r}$-free graph ($r\geq 2$) such that $\chi(G)\geq (rs)^s$. Then for every optimal coloring of $G$, every tree on $s$ vertices is present as an induced rainbow subgraph of $G$.

Consider an optimal coloring of $G$.
Let $T$ be a complete $(rs)$-ary tree having $s$ levels (a rooted tree in which all internal vertices have exactly $rs$ children and every leaf is at a distance of $s-1$ from the root). It follows from Theorem <ref> that $T$ occurs as a rainbow subgraph of $G$. It can be seen in the following way that any tree $H$ having at most $s$ vertices occurs as a subgraph of $T$ that is also an induced subgraph of $G$. We prove this by induction on $|V(H)|$. If $|V(H)|\leq 1$, then the statement is trivially true. Suppose that $|V(H)|\geq 2$. Let $u$ be a leaf of $H$ and $v$ its unique neighbour in $H$. Further, let $H'=H-u$. Let $T'$ be the subtree induced in $T$ by the internal vertices of $T$. Note that $T'$ contains as a subgraph a complete $(r(s-1))$-ary tree having $s-1$ levels. Therefore, since $|V(H')|\leq s-1$, by the inductive hypothesis, there is a subgraph $H''$ of $T'$ isomorphic to $H'$ that is also an induced subgraph of $G$. Let $v'$ be the vertex corresponding to $v$ in $V(H'')\subseteq V(T)$. Note that since $|N_{H''}(v')|\leq s-2$, we have $|N_T(v')\setminus N_{H''}(v')|\geq rs-(s-2)$. Therefore, we can choose distinct vertices $u_1,u_2,\ldots,u_{rs-s}\in N_T(v')\setminus N_{H''}(v')$. Note that since $H''$ is an induced subgraph of $G$, $\{u_1,u_2,\ldots,u_{rs-s}\}\cap V(H'')=\emptyset$. For every vertex $x\in V(H'')\setminus\{v'\}$, we have $|N_G(x)\cap\{u_1,u_2,\ldots,u_{rs-s}\}|\leq r-1$, as otherwise, $r$ neighbours of $x$ in $\{u_1,u_2,\ldots,u_{rs-s}\}$ together with $x$ and $v'$ will form a $K_{2,r}$ in $G$. Thus $\sum_{x\in V(H'')\setminus\{v'\}} |N_G(x)\cap\{u_1,u_2,\ldots,u_{rs-s}\}|\leq (r-1)(s-2)<rs-s$. Therefore, there exists $i\in\{1,2,\ldots,rs-s\}$ such that $N_G(u_i)\cap (V(H'')\setminus\{v'\})=\emptyset$. Now $G[V(H'')\cup\{u_i\}]$ is an induced subgraph of $G$ that is isomorphic to $H$ and is also a subgraph of $T$. Thus every tree on $s$ vertices occurs as a subgraph of $T$ that is also an induced subgraph of $G$. Since $T$ is rainbow, it follows that every tree on $s$ vertices occurs as an induced rainbow subgraph in $G$.

Next, we show that if we are ready to weaken the statement of Theorem <ref> so that we no longer want every optimal coloring of $G$ to have the property stated in the theorem, but only want to guarantee the existence of an optimal coloring of $G$ such that every tree on $s$ vertices occurs as an induced rainbow subgraph of $G$, then we can get much better bounds for $\chi(G)$. Notice that we could also ask whether any $K$-free graph (where $K$ is some complete graph) with a high enough chromatic number has an optimal coloring such that the graph contains every tree on $s$ vertices as an induced rainbow subgraph. Again, an affirmative answer to this question would imply the Gyárfás-Sumner conjecture.

Let $G$ be an undirected graph with a proper coloring $\beta$. Let $<$ be a total order on the colors of $\beta$ and $\alpha=\refinement(\beta,<)$. Then in each of the following cases, every tree on $s$ vertices is present as an induced rainbow subgraph of $G$ with respect to the coloring $\alpha$:

* \(G\) is \(K_{2,r}\)-free (\(r \geq 2\)) and \(\chi(G)\geq (r-1)(s-1)s/2 +s\).

* $G$ has girth $g\geq 5$ and \(\chi(G) \geq s^{1+4/(g-4)}\).

Let $T$ be any tree on $s$ vertices. Assign directions to the edges of $T$ so that we get an out-tree $T'$.
Consider the natural orientation $D$ of $G$ with respect to $\alpha$.
By Proposition <ref>, it follows that $\alpha$ is an out-tree coloring of $D$.
Let the colors used by $\alpha$ be $\{1,2,\ldots,k\}$. Clearly, $k\geq\chi(G)=\chi(D)$. Notice that if $G,s$ satisfy one of the conditions in the statement of the lemma, then $D,s$ satisfy the corresponding condition in Corollary <ref>.
Thus we can apply Corollary <ref> to conclude that there is an induced rainbow subgraph of $D$ isomorphic to $T'$ with respect to the coloring $\alpha$, which implies that $T$ is an induced rainbow subgraph of $G$ with respect to $\alpha$.

Let $G$ be a graph. In each of the following cases, there exists an optimal coloring of $G$ such that every tree $T$ on $s$ vertices occurs as an induced rainbow subgraph of $G$:

* \(G\) is \(K_{2,r}\)-free (\(r \geq 2\)) and \(\chi(G)\geq (r-1)(s-1)s/2 + s\).

* $G$ has girth \(g\geq 5\) and \(\chi(G) \geq s^{1+4/(g-4)}\).

Let $\beta$ be an arbitrary optimal coloring of $G$, and $<$ be any total order on the set of colors. Let $\alpha=\refinement(\beta, <)$. By Proposition <ref>, we have that $\alpha$ is an optimal coloring of $G$.
Then by Lemma <ref>, every tree on $s$ vertices is present as an induced rainbow subgraph with respect to the optimal coloring $\alpha$ in $G$.

§ INDUCED ORIENTED TREES

§.§ Out-trees and in-trees in directed acyclic graphs

Let $\beta: V(G) \rightarrow \mathbb{N}^{>0}$ be a proper vertex  coloring of $\hat G$,  such that $G$ is the natural orientation of $\hat G$ with respect to  $\beta$.
Such a coloring $\beta$ can be obtained as follows.
Let $G_0=G$. For each $i\geq 1$, define $C_i=\{u\in V(G_{i-1})\colon d^+_{G_{i-1}}(u)=0\}$ and $G_i=G_{i-1}-C_i$. Let $k$ be the number of vertices in the longest directed path in $G$. Clearly, $V(G_i)=\emptyset$ for every $i\geq k$. For each $u\in V(G)$, we define $\beta(u)$ to be the unique integer $i\in\{1,2,\ldots,k\}$ such that $u\in C_i$.

It well-known (and easy to verify)  that $\beta$ is a proper vertex coloring of $\hat G$ using $k$ colors, and  whenever there is
an edge between $u$ and $v$, with $\beta(u) >  \beta(v)$, it is oriented in $G$ from $u$ to $v$, i.e. $G$ is indeed the natural orientation of $\hat G$ with respect to $\beta$.

Now applying  Lemma <ref>, all decreasing (with respect to $\beta$) trees on at most $s$ vertices appear as induced subgraphs of $G$. It is easy to
see that these decreasing trees are indeed out-trees in $G$, since along any path  from the root to leaf, the colors are in  decreasing order and thus it will be a  directed path
from root to the leaf.

To show that all in-trees on $s$ vertices are contained as induced subgraphs in $G$, we consider the coloring $\beta'$ defined as $\beta'(u)=k-\beta(u)+1$ for every vertex $u\in V(G)$.
Now it is easy to see that the natural orientation of $\hat G$ with respect to $\beta'$ is $G'$ where $G'$ is obtained from $G$ by reversing the direction of each edge.  Thus
with respect to $\beta'$, $G'$ contains all out-trees on $s$ vertices, which are all in-trees in $G$.

§.§ Oriented trees in bikernel-perfect oriented graphs of girth at least 5

A proper coloring $\alpha$ of an oriented graph using colors from $\mathbb{N}^{>0}$ is said to be a  if for each vertex $u\in V(G)$, $\alpha(N^+(u))\supseteq\{i\in\{1,2,\ldots,\alpha(u)-1\}\colon i\equiv 0\mod 2\}$ and $\alpha(N^-(u))\supseteq\{i\in \{1,2,\ldots,\alpha(u)-1\}\colon i\equiv 1\mod 2\}$.

We now prove a lemma using techniques similar to the ones used in Lemma <ref>.

Let $G$ be an oriented graph having girth $g\geq 5$. Suppose that $G$ has a  using colors $\{1,2,\ldots,k\}$ in which the color $k$ has been given to some vertex. If $k\geq 2s^{1+4/(g-4)}$, then $G$ contains every oriented tree on $s$ vertices as an induced subgraph.

For every vertex $u\in V(G)$ and every even integer $i\in\{1,2,\ldots,\alpha(u)-1\}$, let $u^i$ denote an arbitrarily chosen vertex in $N^+(u)$ having color $i$. Similarly, for every vertex $u\in V(G)$ and every odd integer $i\in \{1,2,\ldots,\alpha(u)-1\}$, let $u^i$ denote an arbitrarily chosen vertex in $N^-(u)$ having color $i$.
Let $T$ be any oriented tree on $s$ vertices and let $w_1,w_2,\ldots,w_s$ be an ordering of $V(T)$ obtained by a depth-first traversal of $\hat{T}$ starting from a vertex $w_1$ that we fix as the root. For each $i\in\{2,3,\ldots,s\}$, let $p(i)$ be the integer such that $w_{p(i)}$ is the parent of $w_i$ in the rooted tree $\hat{T}$. As $w_1$ corresponds to the root of $\hat{T}$, the parent of every vertex appears before it in the ordering $w_1,w_2,\ldots,w_s$, and so we have $p(i)<i$.

For any $i\in\{1,2,\ldots,s\}$, we denote by $T_i$ the oriented tree $T[\{w_1,w_2,\ldots,w_i\}]$. Let $T'_i$ be an induced subgraph of $G$ that is isomorphic to $T_i$ and having vertices $v_1,v_2,\ldots,v_i$ corresponding to $w_1,w_2,\ldots,w_i$. Clearly, for $j\in\{2,3,\ldots,i\}$, $v_{p(j)}$ is the parent of $v_j$ in the rooted tree $\hat{T'_i}$ having $v_1$ as its root. We say that $T'_i$ is a “good” tree if $\alpha(v_1)=k$, and for $2\leq j\leq i$:

* $\alpha(v_{j-1})>\alpha(v_j)$, and

* for $\alpha(v_{j-1})>t>\alpha(v_j)$, $G[\{v_1,v_2,\ldots,v_{j-1},v_{p(j)}^t\}]$ is not isomorphic to $T_j$.

Let $i$ be the maximum integer such that there is a good tree $T'_i$ isomorphic to $T_i$ in $G$ (since there is a vertex having color $k$, we have that $i$ exists). If $i\geq s$, then we are done. So let us assume for the sake of contradiction that $i\leq s-1$. For $2\leq j\leq i$, let $U_j=\{t\colon\alpha(v_j)<t<\alpha(v_{j-1})\}$.

As $T'_i$ is a good tree in $G$, if $v_j\in N^+(v_{p(j)})$ (resp. $v_j\in N^-(v_{p(j)})$) then for every even (resp. odd) $t\in U_j$,

we have $N(v_{p(j)}^t)\cap(\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\})\neq\emptyset$.
Let $U'=\{t\colon 1\leq t<\alpha(v_i)\}$.

If $w_{i+1}\in N^+_T(w_{p(i+1)})$ (resp. $w_{i+1}\in N^-_T(w_{p(i+1)})$) and for some even (resp. odd) $t\in U'$, we have $N(v_{p(i+1)}^t)\cap(\{v_1,v_2,\ldots,v_i\}\setminus\{v_{p(i+1)}\})=\emptyset$, then taking $t$ to be the largest such integer in $U'$, we have that $G[\{v_1,v_2,\ldots,v_i\}\cup\{v_{p(i+1)}^t\}]$ is a good tree isomorphic to $T_{i+1}$, which contradicts our choice of $i$. We can therefore conclude that if $w_{i+1}\in N^+_T(w_{p(i+1)})$ (resp. $w_{i+1}\in N^-_T(w_{p(i+1)})$), then for every

even (resp. odd) $t\in U'$, we have $N(v_{p(i+1)}^t)\cap(\{v_1,v_2,\ldots,v_i\}\setminus\{v_{p(i+1)}\})\neq\emptyset$.
Let $U=U'\cup\bigcup_{j=2}^i U_i$. Clearly, $|U|+i=k$.

First, let us show that $i\geq g$. Assume for the sake of contradiction that $i<g$.  We claim that $|U_j|\leq 1$ for $2\leq j\leq i$. Suppose that there exists $j\in\{2,3,\ldots,i\}$ such that $|U_j|\geq 2$. Then if $v_j\in N^+(v_{p(j)})$ (resp. $v_j\in N^-(v_{p(j)})$), there is an even (resp. odd) $t\in U_j$ such that there exists $z\in N(v_{p(j)}^t)\cap (\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\})$. But then the path in $T'_i$ between $z$ and $v_{p(j)}$ together with $v_{p(j)}^t$ forms a cycle of length at most $j\leq i<g$ in $G$, which is a contradiction. So we can assume that $|U_j|\leq 1$ for $2\leq j\leq i$. Now suppose that $w_{i+1}\in N^+_T(w_{p(i+1)})$ (resp. $w_{i+1}\in N^-_T(w_{p(i+1)})$ and there is an even (resp. odd) $t\in U'$. Then there exists $z\in N(v_{p(i+1)}^t)\cap (\{v_1,v_2,\ldots,v_i\}\setminus\{v_{p(i+1)}\})$. Note that the path in $T'_i$ between $z$ and $v_{p(i+1)}$ together with $z$ forms a cycle of length at most $i+1$. Since we assumed that $i<g$, we also have that this cycle has length at least $i+1$, which implies that this cycle has length exactly $i+1$. This means that the path between $z$ and $v_{p(i+1)}$ contains all the vertices of $T'_i$, or to be more precise, $z=v_1,v_2,\ldots,v_i=v_{p(i+1)}$ is a path. Thus, if $w_{i+1}\in N^+_T(w_{p(i+1)})$ (resp. $w_{i+1}\in N^-_T(w_{p(i+1)})$ and there exist distinct even (resp. odd) $t,t'\in U'$, we have $v_1\in N(v_{p(i+1)}^t)\cap N(v_{p(i+1)}^{t'})$, which implies that there is a cycle $v_{p(i+1)},v_{p(i+1)}^t,v_1,v_{p(i+1)}^{t'},v_{p(i+1)}$ of length 4 in $G$, which is a contradiction. It follows that $|U'|\leq 3$. Then $|U|\leq 2+i$, which gives $i\geq k-2-i$, or in other words, $i\geq (k-2)/2$. From the statement of the lemma, we have $k\geq 2s+1$, as $k$ and $s$ are integers, implying that \(i\geq (2s-1)/2>s-1\), contradicting the assumption that $i\leq s-1$. We can therefore assume that $i\geq g$.

We shall construct an auxiliary undirected graph \(H\) with \(V(H) = V(T'_i)\) and \(E(H) = E(\hat{T'_i}) \cup E'\), where \(E'\) is a set of edges defined as follows. For $j\in\{2,3,\ldots,i\}$ such that $v_j\in N^+(v_{p(j)})$ (resp. $v_j\in N^-(v_{p(j)})$) and even (resp. odd) \(t \in U_j\), the edge \(v_{p(j)}x\) is added to \(E'\), where $x$ is an arbitrarily chosen vertex in $N(v_{p(j)}^t)\cap (\{v_1,v_2,\ldots,v_{j-1}\}\setminus\{v_{p(j)}\})$. Note that since $v_{p(j)}$ is a neighbor of $v_{p(j)}^t$ in $G$ and $g(G)=g\geq 5$, $v_{p(j)}x$ is not an edge of $\hat{G}$. Also, if any other neighbour of $v_{p(j)}$ was adjacent to $x$, there would be a cycle of length four in $G$, again contradicting the fact that $G$ has girth at least 5. Thus, no edge of $E(\hat{G})$ is present in $E'$ and no edge is added twice to $E'$. Similarly, if $w_{i+1}\in N^+(w_{p(i+1)})$ (resp. $w_{i+1}\in N^-(w_{p(i+1)})$), then for each even (resp. odd) $t\in U'$, the edge $v_{p(i+1)}x$ is added to $E'$, where $x$ is an arbitrarily chosen vertex in $N(v_{p(i+1)}^t)\cap (\{v_1,v_2,\ldots,v_i\}\setminus\{v_{p(i+1)}\})$. Thus, $H$ is a simple graph and \(|E'| \geq \left\lfloor\frac{|U'|}{2}\right\rfloor+\sum_{j=2}^i \left\lfloor\frac{|U_j|}{2}\right\rfloor\geq\frac{|U|-i}{2}=\frac{k}{2}-i\). This gives $k\leq 2|E'|+2i$.

Now, we claim that \(g(H) \geq g/2\). Given any cycle \(C\) of \(H\), we can replace each edge of \(C\) by at most 2 edges of \(G\) to obtain a closed walk in $G$ containing at most $2g(H)$ edges (the closed walk will actually be a cycle). Thus, if \(g(H) < g/2\), then there is a cycle in \(G\) of length less than \(g\), which is a contradiction.

From Lemma [ext1]<ref><ref>, if \(g(H)\) is odd, we get:

\[|E(H)| \  < \  \frac{i^{1+2/(g(H)-1)} + i}{2} \  \leq \  \frac{i^{1+4/(g-2)} + i}{2} \ < \ \frac{i^{1+4/(g-4)} + i}{2}\]

From Lemma [ext2]<ref><ref>, if \(g(H)\) is even, we get:

\[|E(H)| \  < \  \bigg(\frac{i}{2}\bigg)^{1+2/(g(H)-2)} + \frac{i}{2} \  \leq \  \bigg(\frac{i}{2}\bigg)^{1+4/(g-4)} + \frac{i}{2}  \ < \ \frac{i^{1+4/(g-4)} + i}{2}\]

Combining both the cases, we get

\[|E(H)| \ < \ \frac{i^{1+4/(g-4)} + i}{2}\]

Since \(|E(H)| = |E'|+(i-1)\)

\[|E'| < \frac{i^{1+4/(g-4)} - i + 2}{2}\]

Since \(k\leq 2|E'|+2i\), we get:

\begin{eqnarray*}
k \: &<& \: i + i^{1+4/(g-4)} + 2= i(1 + 2/i+i^{4/(g-4)})\\
&\leq& i(e^{2/i}+i^{4/(g-4)})\qquad\mbox{(since }1+2/i\leq e^{2/i}\mbox{)}\\
&\leq& i(e^{2/(g-4)}+i^{4/(g-4)})\qquad\mbox{(since $i\geq g\geq g-4$)}\\
\end{eqnarray*}

By our assumption that $i\geq g$, we get that \(i \geq 2 > \sqrt{e}\). Using this in the above inequality, we get:

\[k < 2i^{1+4/(g-4)}\]

Now by using the bound on $k$ given in the statement of the lemma, we get:

$$ 2s^{1+4/(g-4)} \leq k < 2i^{1+4/(g-4)}$$

This implies that $i\geq s$, which is a contradiction.

Let $G_0=G$. For each odd (resp. even) $i\geq 1$, let $C_i$ be an antikernel (resp. kernel) of $G_{i-1}$ and let $G_i=G_{i-1}-C_i$. Since $G$ is bikernel-perfect, there exists a smallest integer $k$ for which $V(G_k)=\emptyset$. Note that $\{C_1,C_2,\ldots,C_k\}$ is a partition of $V(G)$. For each vertex $u\in V(G)$, define $\beta(u)$ to be the unique integer $i$ in $\{1,2,\ldots,k\}$ such that $u\in C_i$. Since for $1\leq i<j\leq k$, every vertex in $C_j\subseteq V(G_i)$ has an in-neighbour (resp. out-neighbour) in $C_i$ when $i$ is odd (resp. even), we can conclude that $\beta$ is a parity coloring of $G$ using the colors in $\{1,2,\ldots,k\}$. Since $k\geq\chi(G)\geq 2s^{1+\frac{4}{g-4}}$, it follows from Lemma <ref> that $G$ contains every oriented tree on $s$ vertices as an induced subgraph.

§.§ Oriented trees in $\mathcal{B}_r$-free oriented graphs

In this section, we will show a much stronger result than what is required to prove Theorem <ref>(i), albeit for a slightly higher bound on the chromatic number. We show that if the chromatic number of an oriented $\mathcal{B}_r$-free graph is at least $(r-1)(s-1)(s-2)+2s+1$, then it contains every oriented tree on at most $s$ vertices as induced subgraph. This generalizes Theorem <ref>(i) from out-trees in directed acyclic graphs to arbitrarily oriented trees in general oriented graphs. Recall that if $A$ is an oriented graph and $\mathcal{F}$ a family of oriented graphs, then $\overline{f}(A,\mathcal{F})$ is the smallest positive integer
such that every $\mathcal{F}$-free oriented graph having chromatic number at least $\overline{f}(A,\mathcal{F})$ contains $A$ as an induced subgraph. Also recall that $\overline{f}(s,\mathcal{F})$ is the smallest positive integer such that every $\mathcal{F}$-free oriented graph having chromatic number at least $\overline{f}(s,\mathcal{F})$ contains every oriented tree on $s$ vertices as an induced subgraph. Thus, in this section, we prove that $\overline{f}(s,\mathcal{B}_r)\leq (r-1)(s-1)(s-2)+2s+1$.

For $s \ge 1$, let $q(s)$ be the smallest positive integer such that every $q(s)$-chromatic directed graph contains every oriented tree on $s$ vertices as a (not necessarily induced) subgraph. Burr [3] proved that $q(s) \le (s-1)^2$, and conjectured that $q(s) = 2s-2$. Addario-Berry et al. [1] obtained the following improved bound for $q(s)$.

$q(s) \le s^2/2 - s/2 + 1$.

We need an analogous version of this result, but for induced subgraphs. But as noted in Section <ref>, we cannot hope to have such a result even for triangle-free graphs, as $\overline{f}(T,C_3)$ does not exist even for the case when $T$ corresponds to certain orientations of a 4-path. We thus restrict ourselves to $\mathcal{B}_r$-free graphs.
In their paper, Addario-Berry et al. describe  several approaches to prove Theorem <ref>, with slight changes in the upper bounds for $q(s)$. We show that their most basic approach can be
adapted to get an induced version of the theorem for $\mathcal{B}_r$-free graphs; the other methods in [1] seem difficult to adapt for the induced version.

An undirected graph $G$ is said to be $k$-degenerate if every subgraph of $G$ has minimum degree at most $k$. It is a well-known and easy to see fact that the chromatic number of a $k$-degenerate graph is at most $k+1$. The following observation will be useful for us.

Let $k>2$. If $G$ is a graph whose every subgraph has average degree at most $k$, then either $G$ contains a clique on $k+1$ vertices or $\chi(G)\leq k$.

Suppose that $\chi(G)>k$. Then $G$ is not $(k-1)$-degenerate. From $G$, we repeatedly remove vertices of degree at most $k-1$ until we get a graph $G'$ whose minimum degree is $k$.
Notice that we will be done if we prove that $\chi(G')\leq k$ or $G'$ contains a clique on $k+1$ vertices.
From the fact that every subgraph of $G$ has average degree at most $k$, we know that the average degree of $G'$ is at most $k$. Since the minimum degree of $G'$ is $k$, it follows that the maximum degree of $G'$ is also $k$, or in other words, $G'$ is a $k$-regular graph.
This means that if $G'$ is a complete graph, then it is a complete graph on $k+1$ vertices, in which case we are done. As $k>2$, $G'$ is not an odd cycle, and therefore by Brooks' theorem, we have $\chi(G')\leq k$.

A vertex $v$ is said to be an out-leaf of an oriented tree $T$ if $d^-_T(v)=1$ and $d^+_T(v)=0$. Similarly, a vertex $v$ is said to be an in-leaf of an oriented tree $T$ if $d^+_T(v)=1$ and $d^-_T(v)=0$. A vertex is said to be a leaf of an oriented tree $T$ if it is either an out-leaf or an in-leaf of $T$.
It is easy to see that for $r\geq 2$, $\overline{f}(T,\mathcal{B}_r)=1$ when $|V(T)|=1$. The following lemma will help us in bounding $\overline{f}(T,\mathcal{B}_r)$ for arbitrary oriented trees $T$.

Let $T$ be an oriented tree on $s$ vertices, where $s\geq 2$, and let $L$ be either the set of all out-leaves of $T$ or the set of all in-leaves of $T$. Let $t=|L|$ and $T'=T-L$.
Then for $r\geq 2$, $\overline{f}(T,\mathcal{B}_r)\leq \overline{f}(T',\mathcal{B}_r) + 2(\val)$.

If $s=2$, then any oriented graph $D$ having $\chi(D)\geq 2$ contains an induced subgraph isomorphic to $T$, implying that $\overline{f}(T,\mathcal{B}_r)\leq 2$ (since in this case, we have $t=1$), and so the bound given in the statement of the lemma holds. Therefore, from here onward, we assume that $s\geq 3$ (which implies that $t\geq 1$).

We assume that $L$ is a set of out-leaves of $T$; the case when $L$ is a set of in-leaves is symmetric. Let $D$ be any oriented graph with $\chi(D) \ge \overline{f}(T',\mathcal{B}_r)+2(\val)$. Define $S= \{ x \in V(D) : d^+_D(x) \le \val\}$.

We claim that $\chi(D[S]) \le 2(\val)$.
Indeed for any $S'\subseteq S$, each vertex in the graph $D[S']$ has out degree at most $\val$ and therefore the number of edges in $D[S']$ is at most $(\val)|S'|$. It follows that the average degree of $\hat{D}[S']$ is at most $2(\val)$.
Thus every subgraph of $\hat{D}[S]$ has average degree at most $2(\val)$.

It is easy to see that $2(\val)>2$. Then by Lemma <ref>, we have that either $\chi(D[S])\leq 2(\val)$ or $D[S]$ contains a tournament on $2(\val)+1\geq 2r+1$ vertices. It follows that there exists a vertex having out-degree at least $r$ in this tournament, which implies that the tournament contains a subgraph isomorphic to a graph in $\mathcal{B}_r$. Since this is a contradiction, we can conclude that $\chi(D[S])\leq 2(\val)$.

From this, we can infer that $\chi(D') \ge  \overline{f}(T', \mathcal{B}_r)$, where $D' = D-S$. Then by the definition of $\overline{f}(T', \mathcal{B}_r)$, we have that $D'$ contains an induced subgraph that is isomorphic to $T'$.

Let $L=\{v_1,v_2,\ldots,v_t\}$.
For each $j\in\{1,2,\ldots,t\}$, we denote by $T_j$ the oriented tree $T[V(T')\cup\{v_1,v_2,\ldots,v_j\}]$. We let $T_0=T'$.
Let $i$ be the maximum integer such that there is an

induced subgraph of $D$ that is isomorphic to $T_i$ having the property that the vertices in it corresponding to the vertices in $V(T_i-L)=V(T')$ are all from $V(D')$.
Since there is an induced subgraph of $D'$ that is isomorphic to $T'=T_0$, we know that $i$ exists. We shall prove that $i=t$, which will complete the proof.

Suppose that $i<t$.
By the definition of $i$, we know that there is an induced subgraph $F$ of $D$ that is isomorphic to $T_i$ in which the vertices corresponding to $V(T')$ are all from $V(D')$.
Clearly, $|V(F)|<|V(T)|=s$.
Let $u$ be the only neighbor of $v_{i+1}$ in $T$ and let $u'$ be the vertex corresponding to $u$ in $F$. Clearly, $u\in V(T')$, and therefore $u'\in V(D')$.

This implies that $|N^+_D(u')|>\val$.
For each vertex $x\in V(F)\setminus\{u'\}$, define $W_x=N[x]\cap N^+(u')$.
For each $x\in V(F)\setminus N^+_{F}[u']$, $|W_x|\leq r-1$, as otherwise $r$ vertices from $W_x$ together with $x$ and $u'$ will form a subgraph of $D$ that is isomorphic to a graph in $\mathcal{B}_r$ (note that since $F$ is an induced subgraph of $D$, we have $x\notin W_x$ in this case).
Similarly, for each $x\in N^+_{F}(u')$, $|W_x|\leq r$, as otherwise $r$ vertices from $W_x\setminus\{x\}$ together with $x$ and $u'$ will form a subgraph of $D$ that is isomorphic to a graph in $\mathcal{B}_r$.
Let $Z\subseteq N^+_{F}(u')$ be the leaves of $F$ in $N^+_{F}(u')$. Since the vertices in $Z$ are all out-leaves of $F$, and $F$ has at most as many out-leaves as $T$, it follows that $|Z|\leq t$.
For every vertex $x\in N^+_{F}(u')\setminus Z$, let $y(x)$ be an arbitrarily chosen vertex in $N_{F}(x)\setminus\{u'\}$. Let $Y=\{y(x)\colon x\in N^+_{F}(u')\setminus Z\}$. Clearly, as $F$ is a tree that is induced in $D$, we have that for each $x\in N^+_{F}(u')\setminus Z$, $y(x)\notin N[u']$ and for any $x'\in N^+_{F}(u')\setminus Z$ that is distinct from $x$, we have $y(x)\neq y(x')$. Thus, we have $Y\cap N^+_{F}(u')=\emptyset$ and $|Y|=|N^+_{F}(u')\setminus Z|$. For each $x\in Y$, $|W_x\setminus\{y^{-1}(x)\}|\leq r-2$, since otherwise $r-1$ vertices from $W_x\setminus\{y^{-1}(x)\}$, $x$, $y^{-1}(x)$ and $u'$ will form a $\mathcal{B}_r$ in $D$.

The number of vertices in $N^+_D(u')$ that are either in $V(F)$ or are adjacent to a vertex in $V(F)$ other than $u'$ is at most

\begin{eqnarray*}
\left|\bigcup_{x\in V(F)\setminus\{u'\}} W_x\right|&\leq&\sum_{x\in N^+_{F}(u')} |W_x|+ \sum_{x\in Y} |W_x\setminus\{y^{-1}(x)\}|+\sum_{x\in V(F)\setminus (N^+_{F}[u']\cup Y)} |W_x|
\end{eqnarray*}

The second term comes from the fact that for every $x\in N^+_F(u')$, we have $x\in W_x$ and $x\in W_{y(x)}$. Substituting the upper bounds derived above, we get

\begin{eqnarray*}
\left|\bigcup_{x\in V(F)\setminus\{u'\}} W_x\right|&\leq&\sum_{x\in N^+_{F}(u')} r + \sum_{x\in Y} r-2 + \sum_{x\in V(F)\setminus (N^+_{F}[u']\cup Y)} r-1\\
&=& \sum_{x\in N^+_{F}(u')\setminus Z} r+ \sum_{x\in Z} r + \left(\sum_{x\in Y} r-1\right) - |Y|+ \sum_{x\in V(F)\setminus (N^+_{F}[u']\cup Y)} r-1\\
\end{eqnarray*}

Since $|Y|=|N^+_F(u')|\setminus Z$, we can take the term $-|Y|$ inside the first term of the right hand side, which gives us

\begin{eqnarray*}
\left|\bigcup_{x\in V(F)\setminus\{u'\}} W_x\right|&\leq& \sum_{x\in N^+_{F}(u')\setminus Z} r-1+ \sum_{x\in Z} r + \sum_{x\in Y} r-1 + \sum_{x\in V(F)\setminus (N^+_{F}[u']\cup Y)} r-1\\
&=&\sum_{x\in N^+_{F}(u')\setminus Z} r-1 + |Z| + \sum_{x\in Z} r-1 + \sum_{x\in Y} r-1 + \sum_{x\in V(F)\setminus (N^+_{F}[u']\cup Y)} r-1\\
&=&|Z| + \sum_{x\in V(F)\setminus\{u'\}} r-1\\
\end{eqnarray*}

Since $|N^+_D(u')|>(r-1)(s-2)+t$, there exists a vertex $v'$ in $N^+_D(u')$ that is neither a vertex of $F$ nor adjacent to any vertex of $F$ other than $u'$. It can be seen that $V(F)\cup\{v'\}$ induces a subgraph of $D$ that is isomorphic to $T_{i+1}$ in which the vertices corresponding to $V(T')$ are all from $D'$. This contradicts our choice of $i$.

As defined in [1], for an oriented tree $T$, the parameter $\st(T)$ is defined as follows. Let $L_1$ be the set of out-leaves of $T$ and $L_2$ the set of in-leaves of $T$. Then $\st(T)=0$ if $|V(T)|=1$, and otherwise $\st(T)=1+\min\{\st(T-L_1),\st(T-L_2)\}$.

For any oriented tree $T$ having $s$ vertices and $r\geq 2$,
$\overline{f}(T,\mathcal{B}_r) \le  (r-1)(2s-\st(T)-3)\st(T)+2s+1$.

We prove this by induction on $\st(T)$. If $\st(T)=0$, then $|V(T)|=1$, implying that $\overline{f}(T,\mathcal{B}_r)=1$, and therefore the upper bound given by the theorem holds.
So we shall assume that $\st(T)>0$. Let $T'$ be the tree obtained by removing all the out-leaves or all the in-leaves of $T$ so that $\st(T')=\st(T)-1$.

Let $t$ denote the number of out-leaves or in-leaves that were removed from $T$ to obtain $T'$. Then $|V(T')|=s-t$.

By the induction hypothesis, we have $\overline{f}(T',\mathcal{B}_r)\leq (r-1) (2|V(T')| - \st(T') -3) \st(T') + 2|V(T')| + 1 =(r-1)(2(s-t)-\st(T)-2)(\st(T)-1) + 2(s-t) + 1$. By Lemma <ref>, we now have

\begin{eqnarray*}
\overline{f}(T,\mathcal{B}_r)&\leq&\overline{f}(T',\mathcal{B}_r)+2((r-1)(s-2)+t)\\
&\leq&(r-1) (2s-2t-\st(T)-2)(\st(T)-1)+  2(s-t) +1 +2((r-1)(s-2)+t)\\
&=&(r-1)(2s-2t-\st(T)-2)\st(T)-(r-1)(2s-2t-\st(T))+2s+1-2t +2(r-1)(s-2) +2t\\
%&=&(r-1)(2s-1-\st(T))\st(T)+2s+1 +2(r-1)(s-2)\\
%&\leq&(r-1)\st(T)(2s-1-\st(T)) + 2s +1
\end{eqnarray*}

Note that the upper bound given in Theorem <ref> is maximized for a tree $T$ such that $\st(T)\in\{|V(T)|-1,|V(T)|-2\}$. We thus get the following result.

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g = h(e) log chi

if chi >= e.chi^h(e) log chi/(h(e) log chi -4), then I should have path on e.chi vertices
we want 2^log (e.chi) h(e) log (chi)/(h(e) log chi - 4) to be less than or equal to chi
this will be satisfied if 2^log (e.chi) h(e) log (chi)/(h(e) log chi - 4) <= chi
this will be satisfied if 2^log (e.chi) h(e) log (chi)/(h(e) log chi - 4 log(chi)) <= chi

f(T) = f(T_1) + 2((r-1)(s-2)+t_1)
= f(T_2) + 2((r-1)(s-t_1-2)+t_2) + 2((r-1)(s-2)+t_1)
= f(T_3) + 2((r-1)(s-t_1-t_2-2)+t_3) + 2((r-1)(s-t_1-2)+t_2) + 2((r-1)(s-2)+t_1)
= f(T_4) + 2((r-1)(s-t_1-t_2-t_3-2)+t_4) + 2((r-1)(s-t_1-t_2-2)+t_3) + 2((r-1)(s-t_1-2)+t_2) + 2((r-1)(s-2)+t_1)
.    k = (T)
= 1 + 2 (t_k+...+t_4+t_3+t_2+t_1) + 2(r-1)(ks - (k-1)t_1 - (k-2)t_2 - (k-3)t_3 - (k-4)t_4 -... - 2k)

Assuming t_1 = t_2 = ... = 1, we have k = s-1

= 1 + 2(s-1) + 2(r-1)(s(s-1) - (s-1)(s-2)/2 - 2(s-1))
= 1 + 2s - 2 + (r-1)(2s^2 - 2s - s^2 + 3s - 2 - 4s + 4)
= 1 + 2s - 2 + (r-1)(s^2 - 3s + 2)
= 2s - 1 + (r-1)(s-1)(s-2)

f(T) = f(T_1) + 2((r-1)(s-2)) + t_1
= 1 +