# Oriented Trees in $O(k\sqrt{k})$-Chromatic Digraphs, a Subquadratic Bound for Burr's Conjecture

Stéphane Bessy, Daniel Gonçalves, Amadeus Reinald

► To cite this version:

Stéphane Bessy, Daniel Gonçalves, Amadeus Reinald. Oriented Trees in $O(k\sqrt{k})$-Chromatic Digraphs, a Subquadratic Bound for Burr's Conjecture. WG 2024 - 50th International Workshop on Graph-Theoretic Concepts in Computer Science, Jun 2024, Gozd Martuljek, Slovenia. pp.79-91, 10.1007/978-3-031-75409-8_6. lirmm-04938265

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https://hal-lirmm.ccsd.cnrs.fr/lirmm-04938265v1

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# Oriented trees in $O(k\sqrt{k})$-chromatic digraphs, a subquadratic bound for Burr's conjecture *

Stéphane Bessy$^a$, Daniel Gonçalves$^a$, and Amadeus Reinald$^a$

$^a$LIRMM, Univ Montpellier, CNRS, Montpellier, France.

## Abstract

In 1980, Burr conjectured that every directed graph with chromatic number $2k-2$ contains any oriented tree of order $k$ as a subdigraph. Burr showed that chromatic number $(k-1)^2$ suffices, which was improved in 2013 to $\frac{k^2}{2} - \frac{k}{2} + 1$ by Addario-Berry et al.

We give the first subquadratic bound for Burr's conjecture, by showing that every directed graph with chromatic number $8\sqrt{\frac{2}{15}}k\sqrt{k} + O(k)$ contains any oriented tree of order $k$. Moreover, we provide improved bounds of $\sqrt[4]{3}k\sqrt{k} + O(k)$ for arborescences, and $(b-1)(k-3) + 3$ for paths on $b$ blocks, with $b \ge 2$.

## 1 Introduction

In essence, many of the questions in structural graph theory ask for necessary conditions for a (di)graph to contain some other given (di)graph. A natural way to specify this question is to ask, when imposing that a (di)graph $G$ has chromatic number $c$, which (di)graphs are necessarily contained in $G$. A well-known result by Erdős [9] states that there exist graphs with arbitrarily large girth and chromatic number. Therefore, all (di)graphs one can hope to be necessarily contained in (di)graphs of chromatic number $c$ are (directed) trees. This question is easily settled when the graph is undirected, as all $k$-chromatic graphs contain all trees of size $k$, which is tight by considering the clique $K_k$. When asking for *induced* subtrees instead, and with the additional restriction of forbidding a fixed clique, this is the notorious Gyárfás–Sumner conjecture [11] [18], which is still wide open.

In this paper, we investigate the directed variant of the question, asking which directed trees are contained as (non-induced) subdigraphs in digraphs with sufficiently large chromatic number. We refer the reader to [14] for lecture

<footnote>*This paper was partially supported by the ANR DIGRAPHS (ANR-19-CE48-0013)</footnote>
notes on the topic. A digraph is *c*-universal for some value *c* > 0 if every digraph with chromatic number at least *c* contains it as a subdigraph. Since the chromatic number of a digraph is constant when replacing digons with arcs, if *H* is *c*-universal in the class of oriented graphs, it is also *c*-universal. Moreover, considering oriented graphs with arbitrarily large girth and chromatic number yields that *H* must be an oriented tree. We are thus interested in upper bounds on the universality of oriented trees, for which Burr posed the following conjecture in 1980.

**Conjecture 1 (Burr [4])** Every oriented tree of order *k* is (*2k* − 2)-universal.

Observe that this is the best one can ask for, as the out-star on $k$ vertices $S_k^+$ is not contained in the regular tournament of order $2k-3$. In the same paper, Burr showed all oriented trees are $(k-1)^2$-universal. This stood as the only bound for the problem for decades, until Addario-Berry, Havet, Linhares-Sales, Reed and Thomassé gave the following improvement in 2013.

**Theorem 2 (Addario-Berry et al. [1])** Every oriented tree of order $k$ is ($\frac{k^2}{2} - \frac{k}{2} + 1$)-universal.

Until now, this has remained the best bound for the general case. Improvements have nevertheless been obtained when asking for particular oriented trees, or by further restricting the directed graphs on which one imposes the chromatic number requirement. We provide the first subquadratic bound for Burr’s conjecture, by showing the following (with $8\sqrt{2/15} \approx 2.92$).

**Theorem 3** Every oriented tree of order $k$ is $(8\sqrt{\frac{2}{15}}k\sqrt{k} + \frac{11}{3}k + \sqrt{\frac{5}{6}}\sqrt{k} + 1)$-universal.

Moreover, we give improved functions for the case of arborescences, showing in Theorem 12 that they are $(\sqrt{4/3} \cdot k\sqrt{k} + k/2)$-universal (where $\sqrt{4/3} \approx 1.15$).

Regarding results on specific oriented trees, a folklore observation yields the (2k − 2)-universality of out-stars and in-stars. Indeed, a digraph forbidding $S_k^+$ (or $S_k^-$) is necessarily (2k − 4)-degenerate, yielding chromatic number at most 2k − 3. One of the most important results on the matter is the Gallai-Roy-Hasse-Vitaver theorem [10] [12] [17] [19], dating 1962, and stating that directed paths of order *k* are *k*-universal. A natural extension is the case of *b*-blocks paths, which are oriented paths that can be arc-partitioned into *b* maximal directed paths. In 2007, Addario-Berry et al. proved the following, improving on a result by El Sahili [7].

**Theorem 4 (Addario-Berry et al. [2])** Every oriented path of order *k* with 2 blocks is *k*-universal.

A linear bound has also been achieved for 3-block paths by El Joubbeh [5], and for 4-block paths by El Joubbeh and Ghazal [6]. In the same paper, they improve the general bound for *b*-block paths when *b* is small compared to *n*. We provide the first general linear bounds, with *b* fixed, for paths with *b* blocks.
**Theorem 5** For $b \ge 2$, every oriented path of order $k$ with $b$ blocks is (($b-1)(k-3)+3$)-universal.

Along the way, this settles the conjecture for 3-block paths, giving us a bound of $2k-3$. Nevertheless, reflecting the idea that oriented paths appear to give some slack on Burr's bound, Havet [13] conjectured that every oriented tree of order $k$ with $\ell$-leaves is $k+\ell+1$-universal. Another notable example where the bound is known to be linear is when looking for antidirected trees [1].

When considering the question over restricted classes of graphs, only a few cases are known. For instance, acyclic digraphs [1], or more generally bikernel-perfect digraphs, satisfy the conjecture. This is the building block for the $k^2/2 + O(k)$ bound of [1], and will also be key in our proof of Theorem 3. Perhaps the most fruitful restriction to date is Sumner's conjecture [18], predating Burr's conjecture, and stating that every tournament of order $2k-2$ contains every oriented tree of order $k$. A linear bound of $3k-3$ was achieved by El Sahili [8], and the conjecture has been proved for sufficiently large $k$ by Kühn and Osthus [15]. Another interesting restriction is to consider digraphs with sufficiently large chromatic number compared to their order, as shown by Naia [16].

**Roadmap of our strategy** A natural technique when trying to establish bounds for the universality of trees is to proceed by induction, usually starting from a star or a path. A step of the induction goes as follows: given an oriented tree $T'$ which we know to be $c'$-universal, we consider a tree $T$ obtained by "gluing" simple subtrees to $T'$. Here, gluing or appending a tree $T''$ to the tree $T'$ means considering the disjoint union of $T'$ and $T''$ and identifying one vertex of $T''$ to one vertex of $T'$. We then derive a bound for the universality of $T$ in terms of $|T'|$, $c'$, and the subtree being glued. It was shown in [1] that augmenting $T'$ with $\ell$ out-neighbours appended to arbitrary vertices yields a tree $T$ that is ($c' + 2|T| - 4$)-universal. This already yields a quadratic bound for the conjecture, where the worst cases are given by orientations of paths, and more generally trees with few leaves.

The cornerstone of our contributions is the establishment of such a gluing lemma appending oriented paths, and an improved version for directed paths. We leverage this in the following win-win strategy when deriving universality bounds for some $T$ by induction. If $T$ has many leaves (more than $O(\sqrt{k})$), we apply the gluing lemma of [1] to the tree obtained by removing the leaves of $T$. Otherwise, we observe $T$ is decomposable into few oriented paths, which we can glue one by one a limited number of times through Lemma 14.

**Further directions** Our ability to control the growth of our universality bound for trees as $O(k\sqrt{k})$ stems from our construction of trees, and corresponding bounds, by gluing both paths and leaves. In the border cases where a tree can be built by adding many leaves at each step, or by gluing few paths, better bounds can be obtained (see Proposition 8 in [1] for leaves). Nevertheless, each gluing operation to some tree $T$ increases the bound by at least $|T|$. Thus, if a tree falls between these two categories, such as a star of degree $\sqrt{k}$
subdivided $\sqrt{k}$ times, our current techniques do not allow us to improve on the
$O(k\sqrt{k})$ bound. Adapting one of the gluing lemmas to get rid of this $O(|T|)$
cost may allow us to skew our win-win strategy towards that type of gluing,
and yield a better bound. Another way to improve our strategy is to manage to
glue multiple paths in parallel, or even more general trees. Note that the cost of
gluing a path depends quadratically on its size, so we are still far from a linear
bound for paths. Nevertheless, even an improvement in that direction would
not directly yield bounds better than $O(k\sqrt{k})$ for the general case, as witnessed
by the case of arborescences.

**Structure of the paper** In Section 2, we introduce various technical tools, and the gluing lemma for leaves. In Section 3, we prove our gluing lemma for directed paths, deriving a $(b-1)(k-3)+3$ bound for $b$-block paths in Theorem 5, and a $\sqrt{4/3 \cdot k\sqrt{k}}+O(k)$ bound for arborescences in Theorem 12. In Section 4, we prove our gluing lemma for oriented paths and derive Theorem 3, showing that oriented trees of order $k$ are $8\sqrt{2/15 \cdot k\sqrt{k}}+O(k)$-universal. The proofs for arborescences and general oriented trees follow the same approach, with only the technical lemmas differing in nature.

## 2 Preliminaries

Throughout the paper, all graphs and digraphs we consider are simple. In
general, we follow standard terminology, which can be found in Bang-Jensen
and Gutin [3]. The *order* $|D|$ of a (di)graph $D$ is the size of its vertex set.
A *digon* in a digraph is an oriented cycle of length 2. Digraphs may contain
digons, whereas *oriented graphs* do not. The union $D \cup H$ of two (di)graphs
is the (di)graph on vertex set $V(D) \cup V(H)$, with edge (arc) set the union of
the edges (arcs) of $D$ and $H$. When clear from the context, we may say tree
or path to refer to an oriented tree or oriented path. A (proper) *colouring* of
a (di)graph $D$ is an assignment of colours to $V$ such that any pair of adjacent
vertices is coloured differently. Then, the *chromatic number* of $D$, denoted $\chi(D)$,
is the minimal number of colours required in such a colouring. A *directed acyclic
graph*, *DAG* for short, is a directed graph containing no directed cycles. Given
a digraph $D$, an *in-kernel* (respectively *out-kernel*) of $D$ is an independent $K$
set that in-dominates (respectively out-dominates) $D$. That is, every vertex
in $V(D) - K$ admits an in-neighbour (respectively out-neighbour) in $K$. It is
well-known that DAGs both an in-kernel and an out-kernel [20].

## 2.1 Universality in Directed Acyclic Graphs

The following lemma is an easy extension of a result from [1] stating that oriented
trees on $k$ vertices are $k$-universal in the class of acyclic digraphs. For the sake
of completeness, we provide its proof below.
**Lemma 6** Let $T$ be an oriented tree of order $k$, and $D$ be a directed acyclic graph such that $\chi(D) \ge k$, then $D$ contains $T$. Moreover, rooting $T$ in any vertex $r$, $D$ admits a partition $(X, K)$ of its vertex set with $\chi(D[K]) \le k-1$ and such that for every vertex $x$ of $X$, there exists a copy of $T$ in $D$ where $r$ is identified to $x$ and all other vertices of $T$ lie in $K$.

**Proof.** We prove the stronger second statement by induction on $k$. If $k=1$, then we choose $X=V(D)$ and $K=\emptyset$. Now, consider $k>1$ and a tree $T$ on $k$ vertices with root $r$. As $k>1$, there exists a leaf $l$ in $T$ different from $r$. Denote $T\setminus\{l\}$ by $T'$ and let $l'$ be the (only) neighbour of $l$ in $T$. Without loss of generality, we can assume that $l'$ is an in-neighbour of $l$. Let $K_0$ be an in-kernel of $D$, which exists has $D$ is acyclic. Since $K_0$ is independent, $\chi(D\setminus K_0) \ge k-1$, for otherwise colouring $K_0$ with a single colour would yield $\chi(D) \le k-1$. Then, the induction hypothesis provides a partition $(X', K')$ of $V(D\setminus K_0)$ for the tree $T'$. Now, let us show that $(X=X', K=K'\cup K_0)$ is the desired partition of $D$ for $T$.

First, note that $\chi(D[K]) \le \chi(D[K']) + \chi(D[K_0]) \le (k-2)+1 = k-1$. Moreover, for any vertex $x$ in $X$, there exists a copy of $T'$ in $D\setminus K_0$ where $r$ is identified to $x$ and the other vertices of $T'$ lie in $K'$. In particular $l'$ is in $K'$ and as $K_0$ is an in-kernel of $D$, $l'$ has an out-neighbour in $K_0$ to which we can identify $l$. $\square$

## 2.2 Gluing leaves

We introduce the gluing lemma of [1], the first example of deriving a universality bound for a tree constructed by some other tree for which we have a bound. Let $T$ be an oriented tree. Let $u$ be a leaf of $T$ and consider $u'$ its unique neighbour in $T$. If the arc between $u$ and $u'$ is oriented towards $u$, we say that $u$ is an *out-leaf* of $T$, otherwise, we say that $u$ is an *in-leaf* of $T$. We let $\text{Out}(T)$ (resp. $\text{In}(T)$) denote the set of all the out-leaves (resp. in-leaves) of $T$.

**Lemma 7 (Addario-Berry et al. [1])** For $k \ge 3$, let $T$ be an oriented tree of order $k$ other than $S_k^+$ or $S_k^-$ . If $T - \text{Out}(T)$, respectively $T - \text{In}(T)$ is $c$-universal, then $T$ is $(c+2k-4)$ universal.

The next corollary follows by performing two gluing operations in a row, one for $\text{Out}(T)$ and one for $\text{In}(T)$.

**Corollary 8** For $k \ge 3$, let $T$ be an oriented tree of order $k$ other than $S_k^+$ or $S_k^-$ . Let $L(T)$ denote the set of all leaves of $T$. If $T - L(T)$ is $c$-universal, then $T$ is $(c+4k-9)$-universal.

**Proof.** We may assume without loss of generality that $T$ has at least one out-leaf, dealing with the case where $T$ has at least in-leaf symmetrically. Our goal is to apply two gluing operations in a row, once for the out-leaves, then once for the in-leaves. Before being able to do so, we must take care of the case where $T - \text{Out}(T)$ is an in-star or an out-star. Then, since $T$ has an out-leaf,
$|T - \text{Out}(T)| \le k-1$, and recall Burr's conjecture holds for $T - \text{Out}(T)$, hence this tree is $2(k-1) - 2 = 2k - 4$-universal. In turn, Lemma 7 yields that $T$ is $4k-8$-universal. Since $T - L(T)$ consists of a single vertex, which is 1-universal, $T$ satisfies the lemma.

We may now assume $T' = T - \text{Out}(T)$ is neither an out-star nor an in-star. Then, we let $T''$ denote the tree $T' - \text{In}(T')$. Observe $T''$ is a subgraph of $T - L(T)$, so any digraph containing a copy of $T - L(T)$ contains a copy of $T''$. Therefore, $T''$ is $c$-universal, and since $|T'| \le k-1$, Lemma 7 yields that $T'$ is $(c+2k-6)$-universal. Applying the same lemma again on $T$, with $|T| = k$, we conclude that $T$ is $(c+4k-10)$-universal, concluding the proof. □

## 2.3 Decomposing trees into paths

The following decomposition of a tree into paths is rather classical, we prove it for the sake of completeness. Here, we state the lemma in terms of undirected trees. We will later apply it to oriented trees, and when doing so, we are implicitly decomposing the underlying tree into paths, and assume the paths given by the decomposition preserve their initial orientation. Let $T$ be an undirected tree, rooted in some vertex $r$. A path $P = v_1, \dots, v_\ell$ of $T$ is *descending* if $v_i$ is the parent of $v_{i+1}$ in $T$ for $i=1, \dots, \ell-1$. Then, we consider $v_1$ as the *beginning* of $P$.

**Lemma 9** Let $T$ be an undirected rooted tree with $p$ leaves. Then, there exist $p$ descending paths $P_1, \dots, P_p$ of $T$ such that for every $i \in [1, p-1]$ the path $P_{i+1}$ and $T_i = \cup_{j=1}^i P_j$ intersect in the vertex beginning $P_{i+1}$, and $T = T_p = \cup_{j=1}^p P_j$.

**Proof.** Let us denote the leaves of $T$ as $(l_1, \dots, l_p)$. First, we define $P_1$ as the path from the root $r$ of $T$ to $l_1$, which contains no other leaf. Then, assume we have defined descending paths $P_1, \dots, P_i$ for $i < p$ satisfying the conditions of the lemma. Assume also that $T_i = \cup_{j=1}^i P_j$ contains exactly leaves $(l_j)_{j \le i}$, and in particular $l_{i+1} \notin T_i$. We consider the descending path $P'_{i+1}$ from $r$ to $l_{i+1}$, noting $r \in T_i$, $l_{i+1} \notin T_i$, and $T_i$ is a subtree of $T$. This allows us to define $P_{i+1}$ as the subpath of $P'_{i+1}$ beginning at the last vertex in $T_i$, and ending in $l_{i+1}$. Then, $T_i$ and $P_{i+1}$ only intersect in the vertex beginning $P_{i+1}$, and $T_{i+1} = T_i \cup P_{i+1}$ contains exactly $(l_j)_{j \le i+1}$. When this process ends, we have defined $p$ paths $P_1, \dots, P_p$, such that $\bar{T}_p = \cup_{j=1}^p P_j = T$. □

# 3 Arborescences and b-block paths

In this section, we tackle the universality of arborescences and paths with $b$ blocks. Consider a tree $T'$ for which we have a universality bound $c'$. Our goal is to obtain a universality bound for the tree $T$ obtained by gluing a directed path of length $\ell$ to some vertex in $T'$. It turns out that gluing directed paths as such is considerably cheaper, in terms of the increase on the bound, than gluing arbitrary orientations of paths. This gives us improved bounds for arborescences
and b-blocks paths, and more generally for any oriented tree that decomposes
into few directed paths.

**3.1 Gluing a directed path**

We first show a technical result, which provides us with all the structure required
to prove the gluing lemma for directed paths.

**Lemma 10** For every integer $\ell \ge 0$, and every digraph $D$, there exists a partition of $V(D)$ into sets $X, Y, Z$ such that:

(X) Every vertex $x \in X$ admits both an in-neighbour $y^-$ and an out-neighbour $y^+$ in $Y$,

(Y) $D[Y]$ is a DAG, and every sink of $D[Y]$ is the beginning of a directed path of length $\ell$ whose remaining vertices are all in $Z$.

(Z) $\chi(D[Z]) \le \ell$.

Symmetrically, there also exists a partition $(X, Y, Z)$ for $D$ satisfying $(X)$, $(Z)$ and the following in place of $(Y)$:

$(\overleftarrow{Y})$ $D[Y]$ is a DAG, and every source of $D[Y]$ is the endpoint of a directed path of length $\ell$ whose other vertices are all in $Z$.

**Proof.** Throughout this proof, we substitute $D$ with an oriented subdi-
graph obtained by replacing each digon of $D$ with an arc arbitrarily, noting this
preserves the chromatic number. The resulting partition will also satisfy the
conditions in the initial digraph. Note that to obtain the second partition, it
suffices to reverse all the arcs of $D$ and consider the first partition on the result-
ing digraph. Therefore, we show how to obtain the first partition, proceeding
by induction on $\ell$.

For $\ell = 0$, let $Z = \emptyset$, let $Y$ be the vertex set of a maximal induced acyclic subdigraph of $D$, and let $X = V(D) \setminus Y$. Properties (Y) and (Z) hold trivially. For property (X), note that by maximality of $Y$, for any vertex $x \in X$ the graph $D[Y \cup \{x\}]$ contains a cycle passing through $x$. The neighbours $y^-, y^+$ of $x$ on the cycle belong to $Y$, which yields property (X).

Assume by induction that the lemma holds for a fixed $\ell \ge 0$, and let us
denote $X', Y', Z'$ the corresponding subsets of $V(D)$. We will build $X, Y, Z$
satisfying the lemma for $\ell + 1$, and refer the reader to Figure 1 for a sketch
of the construction. Informally, in trying to satisfy (Y), we transfer sinks $S'$
of $Y'$ to $Z'$, defining $Z$. Then, to satisfy (X), we transfer vertices from $X'$ to
the remaining vertices of $Y'$, defining $Y$, after which $X$ consists of the vertices
left in $X'$. We then argue that the sinks $S$ of $Y$ admit an out-neighbour in $S'$,
allowing us to extend the directed path given by induction by an arc.

Let $S' \subseteq Y'$ be the set of sinks in $D[Y']$, shown in darker red, note that $S'$ is a stable set in $D[Y']$, and thus also in $D$. Letting $Z = Z' \cup S'$, we then have $\chi(D[Z]) \le \chi(D[Z']) + 1 \le \ell + 1$, satisfying (Z).
<figure><img src="image_2.png" /><figcaption>Figure 1: The partition of $V(D)$ into sets $X', Y', Z'$. The next step of the induction yields partition $X, Y, Z$, shown in grey, green, and red respectively. DAGs $D[Y']$ and $D[Y]$ are layered such that all arcs go from left to right. The sinks $S'$ of $D[Y']$ begin a directed path of length $\ell$ continuing in $Z'$, in dash-dotted. Then, the sinks $S$ of $D[Y]$ begin a directed path of length $\ell + 1$ continuing in $Z$.</figcaption></figure>

We now turn to defining $Y$ and $X$. Consider the subdigraph $D[Y' \setminus S']$, depicted in lighter green, which is acyclic since $D[Y']$ is. We define $Y$ as a vertex set containing $Y' \setminus S'$, and which induces a maximal directed acyclic graph in $D[V \setminus Z]$. Then, we let $X = X' \setminus Y$. The set $Y$ can be obtained by starting with $Y = Y'$ and iteratively adding vertices of $X'$ (shown in darker green), while maintaining that $D[Y]$ is acyclic. After having considered all the vertices of $X'$, we are guaranteed that for every $x \in X$, there is a cycle going through $x$ in $D[Y \cup \{x\}]$. The neighbours $y^-, y^+$ of $x$ on the cycle belong to $Y$, which yields property (X).

We now turn to property (Y), letting $S \subseteq Y$ be the set of sinks in $D[Y]$. Recall $Y = (Y' \setminus S') \cup (Y \cap X')$, and let us first show that $S \subseteq Y' \setminus S'$, meaning none of the sinks of $D[Y]$ belong to $X'$. Indeed, consider any $x' \in Y \cap X'$, and note the vertex $y'^+ \in Y'$ given by property (X') must belong to $Y' \setminus S'$. This is because of our assumption that $D$ is oriented, so $y'^+ \neq y'^-$, and the $y'^+y'^-$-directed path then ensures that $y'^+$ is not a sink of $D[Y']$. Hence, $x'$ has an out-neighbour $y'^+$ in $Y' \setminus S' \subseteq Y$, meaning it is not a sink of $D[Y]$.

We have just shown that sinks $S$ of $D[Y]$ are sinks of $D[Y' \setminus S']$. Since these vertices are not sinks in $D[Y']$, they must have an out-neighbour in $S'$. By induction, all such out-neighbours are the beginning of a directed path of length $\ell$ whose remaining vertices are in $Z'$. Therefore, every vertex in $S$ is the beginning of some directed path of length $\ell + 1$, whose remaining vertices are all in $Z = Z' \cup S'$, ensuring (Y). $\square$

We are now ready to prove the gluing lemma for directed paths.

**Lemma 11** Let $T'$ be an oriented tree of order $k' \ge 1$, and let $T$ be the oriented tree obtained by appending a directed path of length $\ell$ from any of its endpoints to some vertex of $T'$. If $T'$ is $c'$-universal, then $T$ is $(c' + k' + 2\ell - 3)$-universal.

**Proof.** Let $T'$ be any oriented tree of order $k'$, and consider a digraph $D$ with
chromatic number at least $c' + k' + 2\ell - 3$. We will first show how to append a directed path of length $\ell$ to $T'$ by identifying the source of the path to some vertex of $T'$. Let us apply Lemma 10 to $D$ with $\ell-2$ in place of $\ell$. and let We let $X, Y, Z$ be the corresponding partition of $V(D)$, satisfying properties (X), (Y) and (Z). If $D[Y]$ has chromatic number at least $k' + \ell = |T|$, then by Lemma 6, it contains the tree $T$.

Otherwise, $D[Y]$ has chromatic number at most $k' + \ell - 1$. Since $D[Z]$ has chromatic number at most $\ell - 2$, we have that $D[X]$ has chromatic number at least $c' + k' + 2\ell - 3 - (k' + \ell - 1) - (\ell - 2) = c'$. By hypothesis $D[X]$ must contain a copy of $T'$, and we denote by $x$ the vertex on this copy corresponding to the vertex of $T'$ to which we wish to identify the source of the directed path. Then, property (X) ensures that $x$ has an out-neighbour $y^+ \in Y$. If $y^+$ is not a sink, we may follow a directed path from $y^+$ towards a sink $s$ of $D[Y]$, if it is we let $y^+ = s$. Then, property (Y) yields a path $P$ of length $\ell - 2$ starting in $s$ and wholly contained in $Z$. Now, $(x, y^+, \dots, s)$ along with the path $P$ yield a directed path of length at least $\ell$ starting in $x$. Considering $T'$ in $X$ along with the path obtained from $x$ yields the desired copy of $T$ in $D$. The case is symmetrical when appending a directed path from its sink, using the second kind of partition and property $(\bar{Y})$ in Lemma 10. This achieves to show the lemma. $\square$

## 3.2 Growing arborescences and $b$-block paths

With the gluing lemma for directed paths in hand, the proof for $b$-block paths is now straightforward. We start with the bound for 2-block paths given by Theorem 4, and glue the remaining $b-2$ paths one by one using Lemma 11.

**Theorem 5** For $b \ge 2$, every oriented path of order $k$ with $b$ blocks is $((b-1)(k-3)+3)$-universal.

**Proof.** We proceed by induction on $b$. The case $b=2$ holds by Theorem 4. We thus assume that $P$ has $b \ge 3$ blocks, and let $P'$ be a subpath of $P$ obtained by deleting one of the end-blocks of $P$, that is, a block containing an end-vertex of $P$. We let $\ell$ be the length of this end-block, and note that $P'$ has $k-\ell$ vertices.

By induction hypothesis, every graph with chromatic number at least $(b-2)(k-\ell-3)+3$ contains a copy of $P'$. Now, $P$ is obtained from $P'$ by appending a directed path of length $\ell$. We apply Lemma 11, which yields that $P$ is $((b-2)(k-\ell-3)+3+(k-\ell)+2\ell-3)$-universal. Then, we have $(b-2)(k-\ell-3)+3+(k-\ell)+2\ell-3 \le (b-1)(k-\ell-3)+2\ell+3$. Since $2\ell+3 \le (b-1)\ell+3$ as $b \ge 3$, we finally obtain $(b-1)(k-\ell-3)+2\ell+3 \le (b-1)(k-3)+3$. This achieves to show that paths of order $k$ with $b$-block are $((b-1)(k-3)+3)$ universal. $\square$

We now move on to the case of arborescences. The proof is by induction, and follows a win-win argument that will also be used in the general case. At a given step, if there are more than $O(\sqrt{k})$ leaves, we use the gluing lemma for leaves, otherwise we leverage our gluing lemma for directed paths.
**Theorem 12** Every k-vertex arborescence is $(\sqrt{\frac{4}{3}}k\sqrt{k} + \frac{k}{2})$-universal.

**Proof.** Let us show the result for out-arborescences, the case of in-arborescences is symmetrical. We set $g(k) = \sqrt{4/3} \cdot k\sqrt{k} + k/2$. First, notice that the statement holds trivially for $k=1$ and $k=2$, as we have $g(1) \ge 1$ and $g(2) \ge 2$. Then, if $T$ is an out-star $S_k^+$, recall it is $2k-2$-universal, and since $2k-2 \le g(k)$, it is $g(k)$-universal.

Let us proceed by induction on $k$, the order of $T$. By the above, we can assume that $k \ge 3$ and take any out-arborescences of size $k$ distinct from $S_k^+$. We distinguish two cases to show that $T$ is $f(k)$-universal, according to the number $p$ of (out-)leaves in $T$. We first settle the case $p \ge \sqrt{4k/3}$, then deal with the case $p < \sqrt{4k/3}$.

Assume first that $T$ contains at least $\sqrt{4k/3}$ (out-)leaves. Let $T'$ be the tree obtained from $T$ by removing its leaves, that is $T' = T - \text{Out}(T)$, and note $|T'| \le k - \sqrt{4k/3}$. By induction, we know that $T'$ is $g(k - \sqrt{4k/3})$-universal. Then, recall $T \ne S_k^+$, so Lemma 7 yields that $T$ is $g(k - \sqrt{4k/3}) + 2k - 4$-universal. To show that $T$ is $g(k)$-universal, it suffices to verify $g(k) \ge g(k - \sqrt{4k/3}) + 2k - 4$, that is:

$$ \sqrt{\frac{4}{3}} \left( k - \sqrt{\frac{4k}{3}} \right)^{\frac{3}{2}} + \frac{k - \sqrt{4k/3}}{2} \le \sqrt{\frac{4}{3}} k \sqrt{k} + \frac{k}{2} \quad \forall k \ge 2 \quad (1) $$

which we numerically verify for all $k \ge 3$.

From now on, we may therefore assume $T$ contains $p < \sqrt{4k/3}$ leaves. Using Lemma 9 on the tree underlying $T$ rooted in an arbitrary vertex, we obtain $p$ descending paths $P_1, \dots, P_p$. We consider these paths with their initial orientation, and as $T$ is an out-arborescence, each $P_i$ is a directed path. We let $T_i = \bigcup_{j=1}^i P_j$, such that $T = T_p$, and we know that $T_i$ and $P_{i+1}$ only intersect at the beginning of $P_{i+1}$.

Now, we show by induction on $i$ that $T_i$ is $(i \cdot |T_i|)$-universal. For $i=1$, as $T_1$ is a directed path, it is $|T_1|$-universal by the Gallai-Roy-Hasse-Vitaver Theorem. Assume the induction hypothesis holds for some $i < p$. Consider $T_{i+1}$, which is obtained from $T_i$ by appending the path $P_{i+1}$ rooted at their intersecting vertex. Then, Lemma 11 yields that $T_{i+1}$ is $(i \cdot |T_i| + |T_i| + 2l_{i+1} - 3)$-universal, where $l_{i+1}$ is the length of $P_{i+1}$. As $|T_{i+1}| = |T_i| + l_{i+1}$, we obtain that $i \cdot |T_i| + |T_i| + 2l_{i+1} - 3 \le (i-1) \cdot |T_i| + 2(|T_i| + l_{i+1}) \le (i-1) \cdot |T_{i+1}| + 2(|T_{i+1}|) = (i+1) \cdot |T_{i+1}|$. Thus we conclude that $T_{i+1}$ is $(i+1) \cdot |T_{i+1}|$-universal, proving step $i+1$ of the induction. Finally we obtain that $T = T_p$ is $pk$-universal, and $pk < \sqrt{4k/3} \cdot k$. Since $g(k) > \sqrt{4/3} \cdot k\sqrt{k}$, this achieves to show that $T$ is $g(k)$-universal, concluding the proof. $\square$

# 4 Oriented trees

In this section, we obtain universality bounds for general oriented trees. As in the case of arborescences, we consider an oriented tree $T'$, for which we have
a bound $c'$. Then, we derive a universality bound for the tree $T$ obtained by appending an oriented path $Q$ to $T'$. We adapt the techniques of the previous section from directed paths to general oriented paths.

## 4.1 Gluing an oriented path

Given an oriented path $Q$, we obtain a *rooted oriented path* from $Q$ by choosing one of the extremities of $Q$ as the *root* of $Q$. Now, given a rooted oriented path $Q$, with root $r$, a digraph $D$ and a vertex $x$ of $D$, we say that $D$ contains a copy of $Q$ *starting at* $x$ if $D$ contains a copy of $Q$ where $r$ is identified to $x$.

The next two lemmas are the respective counterparts of Lemma 10 and Lemma 11. In both of those, the obtained bounds now depend quadratically on the length of the path being appended, where this dependency was linear in the last section.

**Lemma 13** Let $\ell \ge 0$ be an integer, and $Q$ a rooted oriented path of length $\ell$. Then, for every digraph $D$, there exists a partition of $V(D)$ into sets $X, Y, Z$ such that:

(X) Every vertex $x \in X$ admits both an in-neighbour $y^-$ and an out-neighbour $y^+$ in $Y$,

(Y) $Y$ induces a directed acyclic graph, and for every vertex $y \in Y$, there exists a copy of $Q$ starting at $y$ and contained in $D[Y \cup Z]$,

(Z) $\chi(D[Z]) \le \frac{\ell(\ell+1)}{2}$.

**Proof.** We prove the result by induction on $\ell$. For $\ell = 0$, let $Z = \emptyset$, let $Y$ be the vertex set of a maximal acyclic subdigraph of $D$, and $X = V(D) \setminus Y$. Properties (Y) and (Z) hold trivially, while property (X) follows from the maximality of $D[Y]$ as an acyclic subdigraph of $D$.

Assume by induction that the lemma holds for any rooted oriented path of length $\ell \ge 0$. To show it holds for paths of length $\ell + 1$, let us consider any oriented path $Q$ of length $\ell + 1$ rooted in $r$. Let $r'$ be the unique neighbour of $r$ in $Q$ and let $Q'$ be the oriented path $Q - r$ rooted in $r'$. Consider the partition $X', Y', Z' \subseteq V(D)$ given by induction to satisfy the properties of the lemma for $Q'$. We will build partition $X, Y, Z$ satisfying the properties for $Q$, and refer the reader to Figure 2 for a sketch. Informally, in trying to satisfy (Y), we transfer the vertices of a suitably chosen set $K' \subseteq Y'$, from $Y'$ to $Z'$, defining $Z$. Then, to satisfy (X), we transfer vertices from $X'$ to the remaining vertices of $Y'$, defining $Y$, after which $X$ consists of the vertices left in $X'$. We show how to build copies of $Q$ starting in $Y$, either thanks to $K'$, or by appending a neighbour to some copy of $Q'$ obtained by induction.

Let us start by defining the set $K' \subseteq Y'$, such that $Z = Z' \cup K'$. Let us start by defining $Z$, which is built from $Z'$ by adding a set of vertices $K'$, which we now define. If $\chi(D[Y']) \le \ell + 1$, we simply define $K' = Y'$. Otherwise, $D[Y']$ is an acyclic digraph such that $\chi(D[Y']) \ge \ell + 2$. We may now apply Lemma 6 to
$D[Y']$, with $Q$ as the oriented tree $T$ rooted in $r$ (note that $|Q| = \ell + 2$). This gives us a subset $K' \subseteq Y'$ (shown in darker red), with $\chi(D[K']) \le \ell + 1$, and such that every $y \in Y' \setminus K$ is the beginning of some $Q$ whose remaining vertices belong to $K'$. We let $Z = Z' \cup K'$. We have shown that $\chi(D[K']) \le \ell + 1$ in both cases above, and by induction we have $\chi(D[Z']) \le \frac{\ell(\ell+1)}{2}$. Colouring $D[K']$ and $D[Z']$ with a different set of colours gives $\chi(D[Z]) \le \frac{\ell(\ell+1)}{2} + (\ell+1) = \frac{(\ell+1)(\ell+2)}{2}$, satisfying (Z).

We now turn to defining $Y$ and $X$. Consider the (possibly empty) subdigraph $D[Y' \setminus K']$, depicted in lighter green, which is acyclic since $D[Y']$ is. We define $Y$ as a vertex set containing $Y' \setminus K'$, and which induces a maximal directed acyclic graph in $D[V \setminus Z]$. Then, we let $X = X' \setminus Y$. The set $Y$ can be obtained by starting with $Y = Y'$ and iteratively adding vertices of $X'$ (shown in darker green), while maintaining that $D[Y]$ is acyclic. After having considered all the vertices of $X'$, we are guaranteed that for every $x \in X$, there is a cycle going through $x$ in $D[Y \cup \{x\}]$. The neighbours $y^-, y^+$ of $x$ on the cycle belong to $Y$, which yields property (X).

What is left to show is that property (Y) holds. Recall $Y = (Y' \setminus K') \cup (Y \cap X')$, to build $Q$ starting from any vertex of $Y$, we use two strategies by considering either $y \in (Y' \setminus K')$ or $x' \in (Y \cap X')$. In the first case, $Y' \setminus K' \ne \emptyset$, and we already argued that $\chi(D[Y']) \ge \ell + 2$. Then, recall $y$ must be the beginning of a subpath $Q$ whose remaining vertices lie in $K'$. Since $y \in Y$ and $K' \subseteq Z$, $Q$ is contained in $D[Y \cup Z]$, ensuring (Y) here. In the second case, we have $x' \in X'$ in particular, so $x'$ admits both an in-neighbour $y'^-$ and an out-neighbour $y'^+$ in $Y'$ by induction. Now, $Q$ may be obtained from $Q'$ by appending either an out-neighbour, or respectively an in-neighbour, to its root $r'$. By induction, we may then consider a copy of $Q'$ starting at either $y'^-$, or $y'^+$ respectively. See two examples for $y'^-$ in Figure 2. This copy is contained in $D[Y' \cup Z']$, and in particular it cannot contain $x'$ since $(Y \cap X') \cap ((Y' \setminus K') \cup Z') = \emptyset$. Therefore, $(x', y'^-)$, or $(x', y'^+)$ respectively, along with the copy of $Q'$ yields a copy of $Q$ starting in $x'$, achieving to show (Y). $\square$

We now move on to show the gluing lemma for oriented paths.

**Lemma 14** Let $T'$ be an oriented tree of order $k' \ge 1$, and let $T$ be the oriented tree obtained by appending a rooted oriented path of length $\ell$ to any vertex of $T'$. If $T'$ is $c$-universal, then $T$ is $(c' + k' + \frac{\ell(\ell+1)}{2} - 1)$-universal.

**Proof.** Let us consider a digraph $D$ with chromatic number at least $c' + k' + \ell(\ell+1)/2 - 1$, and $Q$ the rooted oriented path of length $\ell$ we want to append to $T$. We denote by $r$ the root of $Q$ and by $r'$ its unique neighbour in $Q$. Let $Q'$ be the oriented path of length $\ell - 1$ obtained by deleting $r$ from $Q$ and rooted in $r'$. We start by applying Lemma 13 to $D$ with the rooted path $Q'$. We let $X, Y, Z$ be the corresponding partition of $V(D)$, satisfying properties (X), (Y) and (Z). If $D[Y]$ has chromatic number at least $k' + \ell = |T|$, since it is a directed acyclic graph, it contains the tree $T$ by Lemma 6.

Otherwise, $\chi(D[Y]) \le k' + \ell - 1$, and we also know $\chi(D[Z]) \le (\ell - 1)\ell/2$ by property (Z). Therefore, $D[X]$ has chromatic number at least $(c' + k' + \ell(\ell +
<figure><img src="image_3.png" /><figcaption>Figure 2: The partition of $V(D)$ into sets $X', Y', Z'$. The next step of the induction yields partition $X, Y, Z$, shown in grey, green, and red respectively. Vertices of $Y'$ begin a copy of $Q'$, in dash-dotted. Then, vertices of $Y$ begin a copy of $Q$, either in $K'$ (see $y$), or by appending an arc to some $Q'$ (see $x'$).</figcaption></figure>

$1)/2 - 1) - (k' + l - 1) - (l - 1)l/2 = c'$. By hypothesis, $D[X]$ must contain a copy of $T'$, which we identify with $T'$ implicitly, and we denote $x$ the vertex on this copy where $Q$ has to start to form $T$. Then, property (X) ensures us that $x$ has both an in-neighbour $y^-$ and an out-neighbour $y^+$ in $Y$. Both $y^-$ and $y^+$ are the beginning of a rooted copy of $Q'$ contained in $Y \cup Z$. Therefore, according to the orientation of the arc between $r$ and $r'$ in $Q$, we obtain a rooted copy of $Q$ starting in $x$ with either $(y^-, x)$ or $(y, x^+)$, with all remaining vertices in $Y \cup Z$. Except from $x$, this copy of $Q$ is disjoint from the copy of $T'$ obtained in $x$, yielding a copy of $T$ in $D$. $\square$

## 4.2 Growing oriented trees

We are now ready to prove our main result, bounding the universality of oriented trees. As for the case of arborescences, we proceed by induction, and use the same win-win argument. If there are more than $O(\sqrt{k})$ leaves, we glue leaves through Corollary 8, otherwise we glue oriented paths through Lemma 14.

**Theorem 3** Every oriented tree of order $k$ is $(8\sqrt{\frac{2}{15}}k\sqrt{k} + \frac{11}{3}k + \sqrt{\frac{5}{6}}\sqrt{k} + 1)$-universal.

**Proof.** We set $f(k) = 8\sqrt{2/15} \cdot k\sqrt{k} + 11k/3 + \sqrt{5/6} \cdot \sqrt{k} + 1$. First, notice that the statement holds trivially for $k=1$ and $k=2$, as we have $f(1) \ge 1$ and $f(2) \ge 2$. On the other hand, if $T$ is an out-star $S_k^+$ or an in-star $S_k^-$, recall it is $(2k-2)$-universal, and since $2k-2 \le f(k)$, it is $f(k)$-universal.

Let us now proceed by induction on $k$, the order of $T$. By the above, we can assume $k \ge 3$, and take any oriented tree $T$ of size $k$ distinct from $S_k^+$ and $S_k^-$. We show $T$ is $f(k)$-universal by using two strategies according to the number $p$ of leaves in $T$. We first settle the case $p \ge \sqrt{5k/6}$, then deal with the case $p < \sqrt{5k/6}$.
Assume first that $T$ contains at least $\sqrt{5k/6}$ leaves. Consider the subtree $T'$ of $T$ obtained by removing all of its leaves, and note $|T'| \le k - \sqrt{5k/6}$. By induction, we know that $T'$ is $f(k - \sqrt{5k/6})$-universal, then Corollary 8 yields that $T$ is $(f(k - 5k/6) + 4k - 9)$-universal. To show that $T$ is $f(k)$-universal, it suffices to verify $f(k) \ge f(k - \sqrt{5k/6}) + 4k - 9$, that is:

$$ \frac{8\sqrt{30}}{15}k^{3/2} + \frac{11}{3}k + \sqrt{\frac{5}{6}}\sqrt{k} + 1 \\ \ge \frac{8\sqrt{30}}{15}(k - \sqrt{\frac{5}{6}}k)^{3/2} + \frac{11}{3}(k - \sqrt{\frac{5}{6}}k) + \sqrt{\frac{5}{6}}\sqrt{k - \sqrt{\frac{5}{6}}k} + 4k - 8 $$

which we numerically verify for all $k \ge 3$.

From now on, we may therefore assume that $T$ contains $p < \sqrt{5k/6}$ leaves. Our goal is to split $T$ into oriented paths, which we use to build $T$ by using Lemma 14. Let us first apply Lemma 9 on the tree underlying $T$, rooted in an arbitrary vertex $r$. This yields descending paths $Q_1, \dots, Q_p$, which we root in their endpoint closest to $r$. We consider these rooted paths with their initial orientation, such that $T = T_p = \cup_{j=1}^p Q_j$. Note then that $T$ can be obtained recursively by starting from $T_1 = Q_1$, and building $T_{i+1}$ by identifying the root of $Q_{i+1}$ to some vertex of $T_i$.

At this point, the number of paths is bounded by $p$, but in order to apply Lemma 14 successfully, we also need to control their length. To do so, we cut paths of length exceeding $\ell = \lceil \sqrt{6k/5} \rceil \le \sqrt{6k/5} + 1$ into smaller ones as follows. For $j \in [1, p]$, we split $Q_j$ into a minimal number of consecutive subpaths $(Q_{j,h})_h$, such that all have length exactly $\ell$ except for possibly the last one, which may have length strictly less than $\ell$. We denote $(P_i)_{i \in [1, m]}$ the $m$ resulting paths, and order them according to $(Q_j)_j$ they stem from, then with respect to their distance to the root of $Q_j$. Observe that $T$ can still be constructed recursively using $(P_i)_i$, starting with $T_1 = P_1$, and constructing $T_{i+1}$ by appending $P_{i+1}$ to $T_i$. Then, we have $T = T_m = \cup_{i=1}^m P_i$.

To bound the number $m$ of newly created paths $(P_i)_i$, we count those of length less than $\ell$ separately from those with length exactly $\ell$. By definition, at most one $P_i$ of length strictly less than $\ell$ is created per $Q_j$, so there are at most $p < \sqrt{5k/6}$ of them. Let us now bound the number of paths of length exactly $\ell = \lceil \sqrt{6k/5} \rceil$. Note $T$ consists of exactly $k-1$ arcs, and since the paths $(P_i)_i$ are arc-disjoint, there must be at most $\lfloor (k-1)/(\sqrt{6k/5}) \rfloor \le \sqrt{5k/6}$ of them. Finally, $(P_i)_i$ consists of $m \le 2\sqrt{5k/6}$ paths, each of length at most $\ell \le \sqrt{6k/5} + 1$.

Recall that $T = T_m$. We are now ready to show that $T$ is $f(k)$-universal, which we do by recursively computing bounds $(c_i)_i$ such that $T_i$ is $c_i$-universal. Assume $T_i$, of order $k_i$, is $c_i$-universal. Recall $T_{i+1}$ is obtained from $T_i$ by appending $P_{i+1}$, which has length at most $\ell$. We apply Lemma 14, which yields that $T_{i+1}$ is $c_{i+1}$-universal for $c_{i+1} = c_i + k_i + \ell(\ell+1)/2 - 1$. Now, since $T_i$ is obtained from $P_1$, of order at most $\ell+1$, by appending $(i-1)$ paths of length
at most $\ell$, we have $k_i \le i\ell + 1$. This yields:

$$c_{i+1} \le c_i + i\ell + 1 + \frac{\ell(\ell+1)}{2} - 1 = c_i + i\ell + \frac{\ell(\ell+1)}{2}$$

$$c_{i+1} \le c_1 + \ell \left( \sum_{j=1}^{i} j \right) + \frac{i\ell(\ell+1)}{2} \quad \text{by an immediate induction}$$

$$c_{i+1} \le c_1 + \frac{1}{2} \ell i (\ell + i + 2)$$

Now, since $|T_1| \le \ell + 1$ we have $c_1 \le \ell(\ell+1)/2 + 1$ by Theorem 2. We set $i = m-1$, and with the inequalities $\ell \le \sqrt{6k/5} + 1$ and $m \le 2\sqrt{5k/6}$ we obtain:

$$c_m \le \frac{\ell(\ell+1)}{2} + 1 + \frac{1}{2}\ell(m-1)(\ell+m+1)$$

$$c_m \le \frac{\ell m}{2}(\ell + m) + 1$$

$$c_m \le \frac{1}{2} \left( \sqrt{\frac{6}{5}}k + 1 \right) \left( 2\sqrt{\frac{5}{6}}k \right) \left( \sqrt{\frac{6}{5}}k + 1 + 2\sqrt{\frac{5}{6}}k \right) + 1$$

$$c_m \le 8\sqrt{\frac{2}{15}}k^{3/2} + \frac{11}{3}k + \sqrt{\frac{5}{6}}\sqrt{k} + 1$$

This is exactly saying $c_m \le f(k)$, which achieves to prove that $T$ is $f(k)$-universal and concludes the proof. $\square$

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