# On the minimum number of arcs in 4-dicritical oriented graphs

Frédéric Havet, Lucas Picasarri-Arrieta, Clément Rambaud

► To cite this version:

Frédéric Havet, Lucas Picasarri-Arrieta, Clément Rambaud. On the minimum number of arcs in 4-dicritical oriented graphs. Journal of Graph Theory, 2024, 107 (4), pp.778-809. 10.1002/jgt.23159. hal-04664504

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# On the minimum number of arcs in 4-dicritical oriented graphs *

Frédéric Havet¹, Lucas Picasarri-Arrieta¹, and Clément Rambaud¹,²

¹Université Côte d'Azur, CNRS, Inria, I3S, Sophia Antipolis, France

²DIENS, École Normale Supérieure, CNRS, PSL University, Paris, France

`{frederic.havet, lucas.picasarri-arrieta, clement.rambaud}@inria.fr`

## Abstract

The dichromatic number $\tilde{\chi}(D)$ of a digraph $D$ is the minimum number of colours needed to colour the vertices of a digraph such that each colour class induces an acyclic subdigraph. A digraph $D$ is $k$-dicritical if $\tilde{\chi}(D) = k$ and each proper subdigraph $H$ of $D$ satisfies $\tilde{\chi}(H) < k$.

For integers $k$ and $n$, we define $d_k(n)$ (respectively $o_k(n)$) as the minimum number of arcs possible in a $k$-dicritical digraph (respectively oriented graph). Kostochka and Stiebitz have shown [10] that $d_4(n) \ge \frac{10}{3}n - \frac{4}{3}$. They also conjectured that there is a constant $c$ such that $o_k(n) \ge cd_k(n)$ for $k \ge 3$ and $n$ large enough. This conjecture is known to be true for $k = 3$ (Aboulker et al. [1]).

In this work, we prove that every 4-dicritical oriented graph on $n$ vertices has at least $(\frac{10}{3} + \frac{1}{51})n - 1$ arcs, showing the conjecture for $k = 4$. We also characterise exactly the 4-dicritical digraphs on $n$ vertices with exactly $\frac{10}{3}n - \frac{4}{3}$ arcs.

# 1 Introduction

Let $G$ be a graph. We denote by $V(G)$ its vertex set and by $E(G)$ its edge set; we set $n(G) = |V(G)|$ and $m(G) = |E(G)|$. A $k$-colouring of $G$ is a function $\varphi : V(G) \to [k]$. It is proper if for every edge $uv \in E(G)$, $\varphi(u) \neq \varphi(v)$. The smallest integer $k$ such that $G$ has a proper $k$-colouring is the chromatic number, and is denoted by $\chi(G)$. Since $\chi$ is non-decreasing with respect to the subgraph relation, it is natural to consider the minimal graphs (for this relation) which are not $(k-1)$-colourable. Following this idea, Dirac defined $k$-critical graphs as the graphs $G$ with $\chi(G) = k$ and $\chi(H) < k$ for every proper subgraph

<footnote>*A 12-page extended abstract of this paper has been published in the proceedings of WG 2023 [8].</footnote>

<footnote>Research supported by research grant DIGRAPHS ANR-19-CE48-0013 and by the French government, through the EUR DS4H Investments in the Future project managed by the National Research Agency (ANR) with the reference number ANR-17-EURE-0004.</footnote>
H of G. A first property of k-critical graph is that their minimum degree is at least $k-1$. Indeed, if a vertex $v$ has degree at most $k-2$, then a $(k-1)$-colouring of $G-v$ can be easily extended to G, contradicting the fact that $\chi(G) = k$. As a consequence, the number of edges in a k-critical graph is at least $\frac{k-1}{2}n$. This bound is tight for complete graphs and odd cycles, but Dirac [3] proved an inequality of the form $m \ge \frac{k-1+\varepsilon_k}{2}n - c_k$ for every n-vertex k-critical graph with $m$ edges, for some $c_k$ and $\varepsilon_k > 0$. This shows that, for $n$ sufficiently large, the average degree of a k-critical graph is at least $k-1+\varepsilon_k$. This initiated the quest after the best lower bound on the number of edges in n-vertex k-critical graphs. This problem was almost completely solved by Kostochka and Yancey in 2014 [11].

**Theorem 1** (Kostochka and Yancey [11]).

*Every k-critical graph on n vertices has at least $\frac{1}{2}(k - \frac{2}{k-1})n - \frac{k(k-3)}{2(k-1)}$ edges. For every k, this bound is tight for infinitely many values of n.*

Kostochka and Yancey [12] also characterised k-critical graphs for which this inequality is an equality, and all of them contain a copy of $K_{k-2}$, the complete graph on $k-2$ vertices. This motivated the following conjecture of Postle [13].

**Conjecture 2** (Postle [13]). For every integer $k \ge 4$, there exists $\varepsilon_k > 0$ such that every k-critical $K_{k-2}$-free graph $G$ on $n$ vertices has at least $\frac{1}{2}\left(k - \frac{2}{k-1} + \varepsilon_k\right)n - \frac{k(k-3)}{2(k-1)}$ edges.

For $k=4$, the conjecture trivially holds as there is no $K_2$-free 4-critical graph. Moreover, this conjecture has been confirmed for $k=5$ by Postle [13], for $k=6$ by Gao and Postle [5], and for $k \ge 33$ by Gould, Larsen, and Postle [6].

Let $D$ be a digraph. We denote by $V(D)$ its vertex set and by $A(D)$ its arc set; we set $n(D) = |V(D)|$ and $m(D) = |A(D)|$. A **k-colouring** of $D$ is a function $\varphi : V(D) \to [k]$. It is a **k-dicolouring** if every directed cycle $C$ in $D$ is not monochromatic for $\varphi$ (that is $|\varphi(V(C))| > 1$). Equivalently, it is a $k$-dicolouring if every colour class induces an acyclic subdigraph. The smallest integer $k$ such that $D$ has a $k$-dicolouring is the **dichromatic number** of $D$ and is denoted by $\vec{\chi}(D)$.

A **digon** in $D$ is a pair of opposite arcs between two vertices. Such a pair of arcs $\{uv, vu\}$ is denoted by $[u, v]$. We say that $D$ is a **bidirected graph** if every pair of adjacent vertices forms a digon. In this case, $D$ can be viewed as obtained from an undirected graph $G$ by replacing each edge $\{u, v\}$ of $G$ by the digon $[u, v]$. We say that $D$ is a bidirected $G$, and we denote it by $\overleftrightarrow{G}$. Observe that $\chi(G) = \vec{\chi}(\overleftrightarrow{G})$. Thus every statement on proper colouring of undirected graphs can be seen as a statement on dicolouring of bidirected graphs.

Exactly as in the undirected case, one can define **k-dicritical** digraphs to be digraphs $D$ with $\vec{\chi}(D) = k$ and $\vec{\chi}(H) < k$ for every proper subdigraph $H$ of $D$. It is easy to check that if $G$ is a $k$-critical graph, then $\overleftrightarrow{G}$ is $k$-dicritical. Kostochka and Stiebitz [10] conjectured that the $k$-dicritical digraphs with the minimum number of arcs are bidirected graphs. Thus they conjectured the following generalisation of Theorem 1 to digraphs.

**Conjecture 3** (Kostochka and Stiebitz [10]). Let $k \ge 2$. Every $k$-dicritical digraph on $n$ vertices has at least $(k - \frac{2}{k-1})n - \frac{k(k-3)}{k-1}$ arcs. Moreover, equality holds only if $D$ is bidirected.
In the case $k=2$, this conjecture is easy and weak as it states that a 2-dicritical digraph on $n$ vertices has at least two arcs, while, for all $n \ge 2$, the unique 2-dicritical digraph of order $n$ is the directed $n$-cycle which has $n$ arcs. The case $k=3$ of the conjecture has been confirmed by Kostochka and Stiebitz [10]. Using a Brooks-type result for digraphs due to Harutyunyan and Mohar [7], they proved the following: if $D$ is a 3-dicritical digraph of order $n \ge 3$, then $m(D) \ge 2n$ and equality holds if and only if $n$ is odd and $D$ is a bidirected odd cycle. The conjecture has also been proved for $k=4$ by Kostochka and Stiebitz [10]. However, the conjecture is open for every $k \ge 5$. Recently, this problem has been investigated by Aboulker and Vermande [2] who proved the weaker bound $(k - \frac{1}{2} - \frac{2}{k-1})n - \frac{k(k-3)}{k-1}$ for the number of arcs in an $n$-vertex $k$-dicritical digraph.

For integers $k$ and $n$, let $d_k(n)$ denote the minimum number of arcs in a $k$-dicritical digraph of order $n$. By the above observations, $d_2(n) = n$ for all $n \ge 2$, and $d_3(n) \ge 2n$ for all possible $n$, and equality holds if and only if $n$ is odd and $n \ge 3$. Moreover, if $n$ is even then $d_3(n) = 2n + 1$ (see [1]).

Kostochka and Stiebitz [9] showed that if a $k$-critical graph $G$ is triangle-free (that is has no cycle of length 3), then $m(G)/n(G) \ge k - o(k)$ as $k \to +\infty$. Informally, this means that the minimum average degree of a $k$-critical triangle-free graph is (asymptotically) twice the minimum average degree of a $k$-critical graph. Similarly to this undirected case, it is expected that the minimum number of arcs in a $k$-dicritical digraph of order $n$ is larger than $d_k(n)$ if we impose this digraph to have no short directed cycles, and in particular if the digraph is an **oriented graph**, that is a digraph with no digon. Let $o_k(n)$ denote the minimum number of arcs in a $k$-dicritical oriented graph of order $n$ (with the convention $o_k(n) = +\infty$ if there is no $k$-dicritical oriented graph of order $n$). Clearly $o_k(n) \ge d_k(n)$.

**Conjecture 4 (Kostochka and Stiebitz [10]).** For any $k \ge 3$, there is a constant $\alpha_k > 0$ such that $o_k(n) > (1 + \alpha_k)d_k(n)$ for $n$ sufficiently large.

For $k=3$, this conjecture has been recently confirmed by Aboulker, Bellitto, Havet, and Rambaud [1] who proved that $o_3(n) \ge (2 + \frac{1}{3})n + \frac{2}{3}$.

In view of Conjecture 2, Conjecture 4 can be generalized to $\overleftarrow{K}_{k-2}$-free digraphs.

**Conjecture 5.** For any $k \ge 4$, there is a constant $\beta_k > 0$ such that every $k$-dicritical $\overleftarrow{K}_{k-2}$-free digraph $D$ on $n$ vertices has at least $(1 + \beta_k)d_k(n)$ arcs.

Together with Conjecture 3, this conjecture would imply the following generalisation of Conjecture 2.

**Conjecture 6.** For every integer $k \ge 4$, there exists $\varepsilon_k > 0$ such that every $k$-dicritical $\overleftarrow{K}_{k-2}$-free digraph $D$ on $n$ vertices has at least $(k - \frac{2}{k-1} + \varepsilon_k)n - \frac{k(k-3)}{k-1}$ arcs.

A $\overleftarrow{K}_2$-free digraph is an oriented graph, and there are infinitely many 4-dicritical oriented graphs. Thus, while Conjecture 2 holds vacuously for $k=4$, this is not the case for Conjecture 6. In this paper, we prove that Conjectures 4, 5, and 6 hold for $k=4$.
**Theorem 7.** If $\vec{G}$ is a 4-dicritical oriented graph, then

$$m(\vec{G}) \geq \left(\frac{10}{3} + \frac{1}{51}\right) n(\vec{G}) - 1.$$

To prove Theorem 7, we use an approach similar to the proof of the case $k = 5$ of Conjecture 2 by Postle [13]. This proof is based on the potential method, which was first popularised by Kostochka and Yancey [11] when they proved Theorem 1. The idea is to prove a more general result on every 4-dicritical digraphs that takes into account the digons.

With a slight abuse, we call **digon** a subdigraph isomorphic to $\vec{K}_2$, the bidirected complete graph on two vertices. We also call **bidirected triangle** a subdigraph isomorphic to $\vec{K}_3$, the bidirected complete graph on three vertices. A **packing** of digons and bidirected triangles is a set of vertex-disjoint digons and bidirected triangles. To take into account the digons, we define a parameter $T(D)$ as follows.

$$T(D) = \max\{d + 2t \mid \text{there exists a packing of } d \text{ digons and } t \text{ bidirected triangles}\}$$

Clearly, $T(D) = 0$ if and only if $D$ is an oriented graph.

Let $\varepsilon, \delta$ be fixed non-negative real numbers. We define the **potential** (with respect to $\varepsilon$ and $\delta$) of a digraph $D$ to be

$$\rho(D) = \left(\frac{10}{3} + \varepsilon\right) n(D) - m(D) - \delta T(D).$$

Thus Theorem 7 can be rephrased as follows.

**Theorem 7.** Set $\varepsilon = \frac{1}{51}$ and $\delta = 6\varepsilon = \frac{2}{17}$. If $\vec{G}$ is a 4-dicritical oriented graph, then $\rho(\vec{G}) \leq 1$.

In fact, we prove a more general statement which holds for every 4-dicritical digraph (with or without digons), except for some exceptions called the **4-Ore digraphs**. Those digraphs, which are formally defined in Section 2, are the bidirected graphs whose underlying graph is one of the 4-critical graphs reaching equality in Theorem 1. In particular, every 4-Ore digraph $D$ has $\frac{10}{3}n(D) - \frac{4}{3}$ arcs. Moreover, the statement holds for all non-negative constants $\varepsilon$ and $\delta$ satisfying the following inequalities:

* $\delta \geq 6\varepsilon$;

* $3\delta - \varepsilon \leq \frac{1}{3}$;

**Theorem 8.** Let $\varepsilon, \delta \geq 0$ be constants satisfying the aforementioned inequalities. If $D$ is a 4-dicritical digraph with $n$ vertices, then

(i) $\rho(D) \leq \frac{4}{3} + \varepsilon n - \delta \frac{2(n-1)}{3}$ if $D$ is 4-Ore, and

(ii) $\rho(D) \leq 1$ otherwise.
In order to provide some intuition to the reader, let us briefly describe the main ideas of our proof. We will consider a minimum counterexample $D$ to Theorem 8, and show that every subdigraph of $D$ must have large potential. To do so, we need to construct some smaller 4-dicritical digraphs to leverage the minimality of $D$. These smaller 4-dicritical digraphs will be constructed by identifying some vertices of $D$. This is why, in the definition of the potential, we consider $T(D)$ instead of the number of digons: when identifying a set of vertices, the number of digons may be arbitrary larger in the resulting digraph, but $T(D)$ increases at most by 1. Using the fact that every subdigraph of $D$ has large potential, we will prove that some subdigraphs are forbidden in $D$. Using this, we get the final contradiction by a discharging argument.

In addition to Theorem 7, Theorem 8 has also the following consequence when we take $\epsilon = \delta = 0$.

**Corollary 9.** If $D$ is a 4-dicritical digraph, then $m(D) \ge \frac{10}{3}n(D) - \frac{4}{3}$. Moreover, equality holds if and only if $D$ is 4-Ore, otherwise $m(D) \ge \frac{10}{3}n(D) - 1$.

This is a slight improvement on a result of Kostochka and Stiebitz [10] who proved the inequality $m(D) \ge \frac{10}{3}n(D) - \frac{4}{3}$ without characterising the equality case.

Another interesting consequence of our result is the following bound on the number of vertices in a 4-dicritical oriented graph embedded on a fixed surface. Since a graph on $n$ vertices embedded on a surface of Euler characteristic $c$ has at most $3n - 3c$ edges, we immediately deduce the following from Theorem 7.

**Corollary 10.** If $\vec{G}$ is a 4-dicritical oriented graph embedded on a surface of Euler characteristic $c$, then $n(\vec{G}) \le \frac{17}{6}(1 - 3c)$.

The previous best upper bound was $n(\vec{G}) \le 4 - 9c$ [10].

In Section 2 we prove some first preliminary results on 4-Ore digraphs, before proving Theorem 8 in Section 3. In Section 4, we show that $o_k(n) \le (2k - \frac{7}{2})n$ for every fixed $k$ and infinitely many values of $n$. The proof is strongly based on the proof of [1, Theorem 4.4], which shows $o_k(n) \le (2k - 3)n$ for every fixed $k, n$ (with $n$ large enough). For $k = 4$, the construction implies in particular that there is a 4-dicritical oriented graph with 76 vertices and 330 arcs, and there are infinitely many 4-dicritical oriented graphs with $m/n \le 9/2$.

## 2 The 4-Ore digraphs and their properties

We start with a few notations. We denote by $[[x_1, \dots, x_n]]$ the bidirected path with vertex set $\{x_1, \dots, x_n\}$ in this order. If $x_1 = x_n$, $[[x_1, \dots, x_n]]$ denotes the bidirected cycle of order $n$ with cyclic order $x_1, \dots, x_n$. If $D$ is a digraph, for any $X \subseteq V(D)$, $D-X$ is the subdigraph induced by $V(D) \setminus X$. We abbreviate $D - \{x\}$ into $D-x$. Moreover, for any $F \subseteq V(D) \times V(D)$, $D \setminus F$ is the subdigraph $(V(D), A(D) \setminus F)$ and $D \cup F$ is the digraph $(V(D), A(D) \cup F)$

Let $D_1, D_2$ be two bidirected graphs, $[x, y] \subseteq A(D_1)$, and $z \in V(D_2)$. An **Ore-composition** $D$ of $D_1$ and $D_2$ with **replaced digon** $[x, y]$ and **split vertex** $z$ is a digraph obtained by removing $[x, y]$ of $D_1$ and $z$ of $D_2$, and adding the set of arcs $\{xz_1 | zz_1 \in A(D_2) \text{ and } z_1 \in Z_1\}$,
<figure><img src="image_2.png" /><figcaption>Figure 1: An example of a 4-Ore digraph obtained by an Ore-composition of two smaller 4-Ore digraphs, with replaced digon $[x, y]$ and split vertex $z$.</figcaption></figure>

$\{z_1x \mid z_1z \in A(D_2) \text{ and } z_1 \in Z_1\}, \{yz_2 \mid zz_2 \in A(D_2) \text{ and } z_2 \in Z_2\}, \{z_2y \mid z_2z \in A(D_2) \text{ and } z_2 \in Z_2\}$, where $(Z_1, Z_2)$ is a partition of $N_{D_2}(z)$ into non-empty sets. We call $D_1$ the **digon side** and $D_2$ the **split side** of the Ore-composition. The class of the **4-Ore digraphs** is the smallest class containing $\overleftrightarrow{K_4}$ which is stable under Ore-composition. See Figure 1 for an example of a 4-Ore digraph. Observe that all the 4-Ore-digraphs are bidirected.

**Proposition 11** (Dirac [4], see also [12]). 4-Ore digraphs are 4-dicritical.

*Proof.* One can easily show that a bidirected digraph is 4-dicritical if and only if its undirected underlying graph is 4-critical. Then the result follows from the undirected analogous proved by [4]. □

**Lemma 12.** Let $D$ be a 4-dicritical bidirected digraph and $v \in V(D)$. Let $(N_1^+, N_2^+)$ and $(N_1^-, N_2^-)$ be two partitions of $N(v)$. Consider $D'$ the digraph with vertex set $V(D) \setminus \{v\} \cup \{v_1, v_2\}$ with $N^+(v_i) = N_i^+, N^-(v_i) = N_i^-$ for $i=1,2$ and $D'\langle V(D) \setminus \{v\} \rangle = D-v$. Then $D'$ has a 3-dicolouring with $v_1$ and $v_2$ coloured the same except if $N_1^+ = N_1^-$ (that is $D'$ is bidirected).

*Proof.* Suppose that $D'$ is not bidirected. Consider a vertex $u \in N_D(v)$ such that $v_1u, uv_2 \in A(D')$ or $v_2u, uv_1 \in A(D')$. Without loss of generality, suppose $v_1u, uv_2 \in A(D')$. As $D$ is 4-dicritical, $D \setminus [u, v]$ has a proper 3-dicolouring $\varphi$. We set $\varphi(v_1) = \varphi(v_2) = \varphi(v)$ and claim that $\varphi$ is a 3-dicolouring of $D'$. To show that, observe that $\varphi$ is a proper 3-colouring of the underlying undirected graph of $D' \setminus \{v_1u, uv_2\}$, and so $\varphi$ is a 3-dicolouring of $D'$ as wanted. □

**Lemma 13.** Let $D$ be a digraph. If $v$ is a vertex of $D$, then $T(D-v) \ge T(D)-1$.

*Proof.* Let $M$ be a packing of $d$ digons and $t$ bidirected triangles in $H$ such that $d+2t = T(D)$. If $v$ belongs to a digon $[u, v]$ in $M$, then $M \setminus \{[u, v]\}$ witnesses the fact that $T(D-v) \ge T(D)-1$. If $v$ belongs to a bidirected triangle $[[u, v, w, u]]$, then $M \setminus \{[[u, v, w, u]] \cup [u, w]\}$ witnesses the fact that $T(D-v) \ge T(D)-2+1$. Otherwise $T(D-v) \ge T(D)$. □
**Lemma 14.** If $D_1, D_2$ are two digraphs, and $D$ is an Ore-composition of $D_1$ and $D_2$, then $T(D) \ge T(D_1) + T(D_2) - 2$. Moreover, if $D_1$ or $D_2$ is isomorphic to $\overleftrightarrow{K}_4$, then $T(D) \ge T(D_1) + T(D_2) - 1$.

*Proof.* Let $D$ be the Ore-composition of $D_1$ (the digon side with replaced digon $[x, y]$) and $D_2$ (the split side with split vertex $z$). One can easily see that $T(D) \ge T(D_1-x) + T(D-z) \ge T(D_1)+T(D_2)-2$ by Lemma 13. Moreover, if $D_1$ (resp. $D_2$) is a copy of $\overleftrightarrow{K}_4$, then $T(D_1-x) = 2 = T(D_1)$ (resp. $T(D_2-z) = 2 = T(D_2)$) and therefore $T(D) \ge T(D_1) + T(D_2) - 1$. $\square$

**Lemma 15.** If $D$ is 4-Ore, then $T(D) \ge \frac{2}{3}(n(D) - 1)$.

*Proof.* If $D$ is $\overleftrightarrow{K}_4$, then the result is clear. Suppose now that $D$ is an Ore-composition of $D_1$ and $D_2$. Then $n(D) = n(D_1) + n(D_2) - 1$ and, by Lemma 14, $T(D) \ge T(D_1) + T(D_2) - 2$. By induction, $T(D_1) \ge \frac{2}{3}(n(D_1) - 1)$ and $T(D_2) \ge \frac{2}{3}(n(D_2) - 1)$, and so $T(D) \ge \frac{2}{3}(n(D_1) + n(D_2) - 1 - 1) = \frac{2}{3}(n(D) - 1)$. $\square$

Let $D$ be a digraph. A **diamond** in $D$ is a subdigraph isomorphic to $\overleftrightarrow{K}_4$ minus a digon $[u, v]$, with vertices different from $u$ and $v$ having degree 6 in $D$. An **emerald** in $D$ is a subdigraph isomorphic to $\overrightarrow{K_3}$ whose vertices have degree 6 in $D$.

Let $R$ be an induced subdigraph of $D$ with $n(R) < n(D)$. The **boundary** of $R$ in $D$, denoted by $\partial_D(R)$, or simply $\partial(R)$ when $D$ is clear from the context, is the set of vertices of $R$ having a neighbour in $V(D) \setminus R$. We say that $R$ is **Ore-collapsible** if the boundary of $R$ contains exactly two vertices $u$ and $v$ and $R \cup [u, v]$ is 4-Ore.

**Lemma 16.** If $D$ is 4-Ore and $v \in V(D)$, then there exists either an Ore-collapsible subdigraph of $D$ disjoint from $v$ or an emerald of $D$ disjoint from $v$.

*Proof.* If $D$ is a copy of $\overrightarrow{K_4}$, then $D-v$ is an emerald. Otherwise, $D$ is the Ore-composition of two 4-Ore digraphs: $D_1$ the digon side with replaced digon $[x, y]$, and $D_2$ the split side with split vertex $z$. If $v \in V(D_2-z)$, then $D_1$ is an Ore-collapsible subdigraph with boundary $\{x, y\}$. Otherwise $v \in V(D_1)$ and we apply induction on $D_2$ to find an emerald or an Ore-collapsible subdigraph in $D_2$ disjoint from $z$. $\square$

**Lemma 17.** If $D \ne \overleftrightarrow{K_4}$ is 4-Ore and $T$ is a copy of $\overrightarrow{K_3}$ in $D$, then there exists either an Ore-collapsible subdigraph of $D$ disjoint from $T$ or an emerald of $D$ disjoint from $T$.

*Proof.* As $D$ is not $\overrightarrow{K_4}$, it is an Ore-composition of two 4-Ore digraphs: $D_1$ the digon side with replaced digon $[x, y]$, and $D_2$ the split side with split vertex $z$. As $x$ and $y$ are not adjacent, we have either $T \subseteq D_1$, $T \subseteq D_2 - z$, or $T$ contains a vertex $w \in \{x, y\}$ and two vertices in $V(D_2-z)$.

If $T \subseteq D_1$, then by Lemma 16, in $D_2$ there exists either an Ore-collapsible subdigraph $O$ or an emerald $E$ disjoint from $z$. In the former case $O$ is an Ore-collapsible subdigraph of $D$ disjoint from $T$, and in the latter one $E$ is an emerald in $D$ disjoint from $T$.
If $T \subseteq D_2 - z$, then $D_1 \setminus \{x, y\}$ is an Ore-collapsible subdigraph disjoint from $T$.

Assume now that $T$ contains a vertex $w \in \{x, y\}$ and two vertices in $V(D_2 - z)$. Without loss of generality, we may assume that $y \notin T$. Let $z_1$ and $z_2$ be the two vertices of $T$ disjoint from $w$. Then $\{z, z_1, z_2\}$ induces a bidirected triangle $T'$ in $D_2$. If $D_2 \neq \overleftrightarrow{K}_4$, then by induction in $D_2$, there exists either an Ore-collapsible subdigraph $O$ or an emerald $E$ disjoint from $T'$. In the former case $O$ is an Ore-collapsible subdigraph of $D$ disjoint from $T$, and in the latter one $E$ is an emerald in $D$ disjoint from $T$.

Henceforth we may assume that $D_2 = \overleftrightarrow{K}_4$. This implies that $y$ has exactly one neighbour in $D_2 - z$ and so its degree is the same in $D_1$ and $D$. By Lemma 16, in $D_1$ there exists either an Ore-collapsible subdigraph $O$ or an emerald $E$ disjoint from $x$. In the former case $O$ is an Ore-collapsible subdigraph of $D$ disjoint from $T$, and in the latter one $E$ is an emerald in $D$ disjoint from $T$ even if $y \in V(E)$ because $y$ has the same degree in $D_1$ and $D$. $\square$

**Lemma 18.** If $R$ is an Ore-collapsible induced subdigraph of a 4-Ore digraph $D$, then there exists a diamond or an emerald of $D$ whose vertices lie in $V(R)$.

*Proof.* Let $D$ be a digraph. Let $R$ be a minimal counterexample to this lemma, and let $\partial(R) = \{u, v\}$ and $H = D\langle R \rangle \cup [u, v]$. If $H = \overleftrightarrow{K}_4$, then $R$ is a diamond in $D$. Suppose now that $H$ is the Ore-composition of two 4-Ore digraphs $H_1$ (the digon side with replaced digon $[x, y]$) and $H_2$ (the split side with split vertex $z$). If $\{u, v\} \not\subseteq V(H_2)$, then by Lemma 16 there exists an Ore-collapsible subdigraph in $H_2$ disjoint from $z$. As it is smaller than $H$, it contains an emerald or a diamond as desired, a contradiction.

Now assume that $\{u, v\} \subset V(H_2)$, then $H_1$ is an Ore-collapsible subdigraph of $D$ smaller than $H$, and by induction, $H_1$ contains a diamond or an emerald in $D$. $\square$

**Lemma 19.** If $D$ is a 4-Ore digraph and $v$ is a vertex in $D$, then $D$ contains a diamond or an emerald disjoint from $v$.

*Proof.* Follows from Lemmas 16 and 18. $\square$

**Lemma 20.** If $D$ is a 4-Ore digraph and $T$ is a bidirected triangle in $D$, then either $D = \overleftrightarrow{K}_4$ or $D$ contains a diamond or an emerald disjoint from $T$.

*Proof.* Follows from Lemmas 17 and 18. $\square$

The following theorem was formulated for undirected graphs, but by replacing every edge by a digon, it can be restated as follows:

**Theorem 21** (Kostochka and Yancey [12], Theorem 6). Let $D$ be a 4-dicritical bidirected digraph.

If $\frac{10}{3}n(D) - m(D) > 1$, then $D$ is 4-Ore and $\frac{10}{3}n(D) - m(D) = \frac{4}{3}$.

**Lemma 22.** If $D$ is a 4-Ore digraph with $n$ vertices, then $\rho(D) \leq \frac{4}{3} + \varepsilon n - \delta \frac{2(n-1)}{3}$.

*Proof.* Follows from Theorem 21 and Lemma 15. $\square$
**Lemma 23** (Kostochka and Yancey [12], Claim 16). Let $D$ be a 4-Ore digraph. If $R \subseteq D$ and $0 < n(R) < n(D)$, then $\frac{10}{3}n(R) - m(R) \ge \frac{10}{3}$.

**Lemma 24.** Let $D$ be a 4-Ore digraph obtained from a copy $J$ of $\overleftrightarrow{K}_4$ by successive Ore-compositions with 4-Ore digraphs, vertices and digons in $J$ being always on the digon side. Let $[u, v]$ be a digon in $D\langle V(J)\rangle$. For every 3-dicolouring $\varphi$ of $D \setminus [u, v]$, vertices in $V(J)$ receive distinct colours except $u$ and $v$.

*Proof.* We proceed by induction on $n(D)$, the result holding trivially when $D$ is $\overleftrightarrow{K}_4$. Now assume that $D$ is the Ore-composition of $D_1$, the digon side containing $J$, and $D_2$, with $D_1$ and $D_2$ being 4-Ore digraphs. Let $[x, y] \subseteq A(D_1)$ be the replaced digon in this Ore-composition, and let $z \in V(D_2)$ be the split vertex. Let $\varphi$ be a 3-dicolouring of $D \setminus [u, v]$. Then $\varphi$ induces a 3-dicolouring of $D\langle V(D_2 - z) \cup \{x, y\} \rangle$. Necessarily $\varphi(x) \ne \varphi(y)$, for otherwise $\varphi_2$ defined by $\varphi_2(w) = \varphi(w)$ if $w \in V(D_2 - z)$ and $\varphi_2(z) = \varphi(x)$ is a 3-dicolouring of $D_2$, contradicting the fact that 4-Ore digraphs have dichromatic number 4 by Lemma 11. Hence $\varphi$ induces a 3-dicolouring of $D_1 \setminus [u, v]$. So, by the induction hypothesis, vertices in $V(J)$ have distinct colours in $\varphi$, except $u$ and $v$. $\square$

**Lemma 25.** Let $D$ be a 4-Ore digraph obtained from a copy $J$ of $\overleftrightarrow{K}_4$ by successive Ore-compositions with 4-Ore digraphs, vertices and digons in $J$ being always on the digon side. Let $v$ be a vertex in $V(J)$. For every 3-dicolouring $\varphi$ of $D - v$, vertices in $J$ receive distinct colours.

*Proof.* We proceed by induction on $n(D)$, the result holding trivially when $D$ is $\overleftrightarrow{K}_4$. Now assume that $D$ is the Ore-composition of $D_1$, the digon side containing $J$, and $D_2$, with $D_1$ and $D_2$ being 4-Ore digraphs. Let $[x, y] \subseteq A(D_1)$ be the replaced digon in this Ore-composition, and let $z \in V(D_2)$ be the split vertex. Let $\varphi$ be a 3-dicolouring of $D - v$. If $v \in \{x, y\}$, then $\varphi$ is a 3-dicolouring of $D_1 - v$ and the result follows by induction. Now assume $v \notin \{x, y\}$. Then $\varphi$ induces a 3-dicolouring of $D\langle V(D_2 - z) \cup \{x, y\} \rangle$. Necessarily $\varphi(x) \ne \varphi(y)$, for otherwise $\varphi_2$ defined by $\varphi_2(w) = \varphi(w)$ if $w \in V(D_2 - z)$ and $\varphi_2(z) = \varphi(x)$ is a 3-dicolouring of $D_2$, contradicting the fact that 4-Ore digraphs have dichromatic number 4 by Lemma 11. Hence $\varphi$ induces a 3-dicolouring of $D_1 - v$. So, by the induction hypothesis, vertices in $V(J)$ have distinct colours in $\varphi$. $\square$

### 3 Proof of Theorem 8

Let $D$ be a 4-dicritical digraph, $R$ be an induced subdigraph of $D$ with $4 \le n(R) < n(D)$ and $\varphi$ a 3-dicolouring of $R$. The $\varphi$-identification of $R$ in $D$, denoted by $D_\varphi(R)$ is the digraph obtained from $D$ by identifying for each $i \in [3]$ the vertices coloured $i$ in $V(R)$ to a vertex $x_i$, adding the digons $[x_i, x_j]$ for all $1 \le i < j \le 3$. Observe that $D_\varphi(R)$ is not 3-dicolourable. Indeed, assume for a contradiction that $D_\varphi(R)$ has a 3-dicolouring $\varphi'$. Since $\{x_1, x_2, x_3\}$ induces a $\overleftrightarrow{K}_3$, we may assume without loss of generality that $\varphi'(x_i) = i$ for $i \in [3]$. Consider
the 3-colouring $\varphi''$ of $D$ defined by $\varphi''(v) = \varphi'(v)$ if $v \notin R$ and $\varphi''(v) = \varphi(v)$ if $v \in R$. One easily checks that $\varphi''$ is a 3-dicolouring of $D$, a contradiction to the fact that $\vec{\chi}(D) = 4$.

Now let $W$ be a 4-dicritical subdigraph of $D_\varphi(R)$ and $X = \{x_1, x_2, x_3\}$. Then we say that $R' = D\langle(V(W) \setminus X) \cup R\rangle$ is the **dicritical extension** of $R$ with **extender** $W$. We call $X_W = X \cap V(W)$ the **core** of the extension. Note that $X_W$ is not empty, because $W$ is not a subdigraph of $D$. Thus $1 \le |X_W| \le 3$. See Figure 2 for an example of a $\varphi$-identification and a dicritical extension.

<figure><img src="image_3.png" /><figcaption>Figure 2: A 4-dicritical digraph $D$ together with an induced subdigraph $R$ of $D$ and $\varphi$ a 3-dicolouring of $R$, the $\varphi$-identification $D_\varphi(R)$ of $R$ in $D$ and the dicritical extension $R'$ of $R$ with extender $W$ and core $X_W$. For clarity, the digons are represented by undirected edges.</figcaption></figure>

Let $D$ be a counterexample to Theorem 8 with minimum number of vertices. By Lemma 22, $D$ is not 4-Ore. Thus $\rho(D) > 1$.

**Claim 1.** If $\tilde{D}$ is a 4-dicritical digraph with $n(\tilde{D}) < n(D)$, then $\rho(\tilde{D}) \le \frac{4}{3} + 4\varepsilon - 2\delta$.

*Proof of claim.* If $\tilde{D}$ is not 4-Ore, then $\rho(\tilde{D}) \le 1$ by minimality of $D$. Thus $\rho(\tilde{D}) \le \frac{4}{3} + 4\varepsilon - 2\delta$ because $4\varepsilon - 2\delta \ge \frac{-1}{3}$. Otherwise, by Lemma 22, $\rho(\tilde{D}) \le \frac{4}{3} + \varepsilon n(\tilde{D}) - \delta \frac{2(n(\tilde{D})-1)}{3} \le \frac{4}{3} + 4\varepsilon - 2\delta$ because $\delta \ge \frac{3}{2}\varepsilon$ and $n(\tilde{D}) \ge 4$. $\diamond$

**Claim 2.** Let $R$ be a subdigraph of $D$ with $4 \le n(R) < n(D)$. If $R'$ is a dicritical extension of $R$ with extender $W$ and core $X_W$, then

$$ \rho(R') \le \rho(W) + \rho(R) - \left( \rho(\overleftarrow{K_{|X_W|}}) + \delta \cdot T(\overleftarrow{K_{|X_W|}}) \right) + \delta \cdot (T(W) - T(W - X_W)) $$

and in particular

$$ \rho(R') \le \rho(W) + \rho(R) - \frac{10}{3} - \varepsilon + \delta. $$

*Proof of claim.* We have

$$ \bullet n(R') = n(W) - |X_W| + n(R), $$
• $m(R') \ge m(W) + m(R) - m(\overleftarrow{K}_{|X_W|})$,

• $T(R') \ge T(W - X_W) + T(R)$

and by summing these inequalities, we get the first result.

Now observe that $T(W) - T(W - X_W) \le |X_W|$ by Lemma 13, and that the maximum of $- (\rho(\overleftarrow{K}_{|X_W|}) + \delta T(\overleftarrow{K}_{|X_W|})) + \delta |X_W|$ is reached when $|X_W| = 1$, in which case it is equal to $-\frac{10}{3} - \varepsilon + \delta$. The second inequality follows. $\diamond$

**Claim 3.** If $R$ is a subdigraph of $D$ with $4 \le n(R) < n(D)$, then $\rho(R) \ge \rho(D) + 2 - 3\varepsilon + \delta > 3 - 3\varepsilon + \delta$.

*Proof of claim.* We proceed by induction on $n - n(R)$. Let $R'$ be a dicritical extension of $R$ with extender $W$ and core $X_W$. By Claim 2, we have

$$\rho(R') \le \rho(W) + \rho(R) - \frac{10}{3} - \varepsilon + \delta.$$

Either $V(R') = V(D)$ and so $\rho(R') \ge \rho(D)$ or $V(R')$ is a proper subset of $V(D)$ and, since $R$ is a proper subdigraph of $R'$, by induction $\rho(R') \ge \rho(D) + 2 - 3\varepsilon + \delta \ge \rho(D)$. In both cases, $\rho(R') \ge \rho(D)$. Now $W$ is smaller than $D$ so $\rho(W) \le \frac{4}{3} + 4\varepsilon - 2\delta$ by Claim 1. Thus

$$\rho(D) \le \rho(R') \le \frac{4}{3} + 4\varepsilon - 2\delta + \rho(R) - \frac{10}{3} - \varepsilon + \delta.$$

This gives $\rho(R) \ge \rho(D) + 2 - 3\varepsilon + \delta > 3 - 3\varepsilon + \delta$, because $\rho(D) > 1$. $\diamond$

As a consequence of Claim 3, any subdigraph (proper or not) of size at least 4 has potential at least $\rho(D)$.

We say that an induced subdigraph $R$ of $D$ is **collapsible** if, for every 3-dicolouring $\varphi$ of $R$, its dicritical extension $R'$ (with extender $W$ and core $X_W$) is $D$, has core of size 1 (i.e. $|X_W| = 1$), and the border $\partial_D(R)$ of $R$ is monochromatic in $\varphi$.

**Claim 4.** Let $R$ be an induced subdigraph of $D$ and $\varphi$ a 3-dicolouring of $R$ such that $\partial(R)$ is not monochromatic in $\varphi$. If $D$ is a dicritical extension of $R$ dicoloured by $\varphi$ with extender $W$ and core $X_W$ with $|X_W| = 1$, then

$$\rho(R) \ge \rho(D) + 3 - 3\varepsilon + \delta.$$

*Proof of claim.* Assume $D$ is a dicritical extension of $R$ dicoloured by $\varphi$ with extender $W$ and core $X_W$ with $|X_W| = 1$. Observe that each of the following inequalities holds:

• $n(D) = n(W) - |X_W| + n(R) = n(W) + n(R) - 1$,

• $m(D) \ge m(W) + m(R) - m(\overleftarrow{K}_{|X_W|}) + 1 = m(W) + m(R) + 1$ because $\partial_D(R)$ is not monochromatic in $\varphi$, and
• $T(D) \ge T(W - X_W) + T(R) \ge T(W) + T(R) - 1$ by Lemma 13.

By Claim 1, we have

$$ \rho(D) \le \rho(W) + \rho(R) - \left(\frac{10}{3} + \varepsilon\right) - 1 + \delta \le \left(\frac{4}{3} + 4\varepsilon - 2\delta\right) + \rho(R) - \frac{13}{3} - \varepsilon + \delta $$

$$ \text{and so } \rho(R) \ge \rho(D) + 3 - 3\varepsilon + \delta. \quad \diamond $$

**Claim 5.** If $R$ is a subdigraph of $D$ with $4 \le n(R) < n(D)$ and $R$ is not collapsible, then $\rho(R) \ge \rho(D) + \frac{8}{3} - \varepsilon - \delta > \frac{11}{3} - \varepsilon - \delta$.

*Proof of claim.* Let $R'$ be a dicritical extension of $R$ dicoloured by $\varphi$ with extender $W$ and core $X_W$.

(i) If $R'$ is not $D$, then it has a dicritical extension $R''$ with extender $W'$. By (the consequence of) Claim 3, we have $\rho(D) \le \rho(R'')$; by Claim 2 (applied twice), $\rho(R'') \le \rho(R) + \rho(W') + \rho(W) + 2(-\frac{10}{3} - \varepsilon + \delta)$; both $W$ and $W'$ are smaller than $D$, so, by Claim 1, $\rho(W), \rho(W') \le \frac{4}{3} + 4\varepsilon - 2\delta$. Those three inequalities imply

$$ \rho(D) \le \rho(R'') \le \rho(R) + 2\left(\frac{4}{3} + 4\varepsilon - 2\delta\right) + 2\left(-\frac{10}{3} - \varepsilon + \delta\right) = \rho(R) - 4 + 6\varepsilon - 2\delta $$

$$ \text{and so } \rho(R) \ge \rho(D) + 4 - 6\varepsilon + 2\delta \ge \rho(D) + \frac{8}{3} - \varepsilon - \delta. $$

(ii) If $R' = D$ and $|X_W| = 2$, then $\rho(\overleftarrow{K_{|X_W|}}) + \delta T(\overleftarrow{K_{|X_W|}}) = \frac{14}{3} + 2\varepsilon$, and, by Lemma 13, $T(W) - T(W - X_W) \le |X_W| = 2$. Thus, by Claim 2,

$$ \rho(D) \le \rho(W) + \rho(R) - \frac{14}{3} - 2\varepsilon + 2\delta $$

Now, since $W$ is smaller than $D$, $\rho(W) \le \frac{4}{3} + 4\varepsilon - 2\delta$ by Claim 1. Thus

$$ \rho(D) \le \rho(R) + \frac{4}{3} + 4\varepsilon - 2\delta - \frac{14}{3} - 2\varepsilon + 2\delta = \rho(R) - \frac{10}{3} + 2\varepsilon $$

$$ \text{and so } \rho(R) \ge \rho(D) + \frac{10}{3} - 2\varepsilon \ge \rho(D) + \frac{8}{3} - \varepsilon - \delta. $$

(iii) If $R' = D$ and $|X_W| = 3$, then $\rho(\overleftarrow{K_{|X_W|}}) + \delta T(\overleftarrow{K_{|X_W|}}) = 4 + 3\varepsilon$, and, by Lemma 13, $T(W) - T(W - X_W) \le |X_W| = 3$. Thus, by Claim 2,

$$ \rho(D) \le \rho(W) + \rho(R) - 4 - 3\varepsilon + 3\delta. $$

Now, since $W$ is smaller than $D$, $\rho(W) \le \frac{4}{3} + 4\varepsilon - 2\delta$ by Claim 1. Thus

$$ \rho(D) \le \rho(R) + \frac{4}{3} + 4\varepsilon - 2\delta - 4 - 3\varepsilon + 3\delta = \rho(R) - \frac{8}{3} + \varepsilon + \delta $$
and so $\rho(R) \ge \rho(D) + \frac{8}{3} - \varepsilon - \delta$.

(iv) If $R' = D$, $|X_W| = 1$ and $\partial(R)$ is not monochromatic in $\varphi$, then, by Claim 4, we have $\rho(R) \ge \rho(D) + 3 - 3\varepsilon + \delta \ge \rho(D) + \frac{8}{3} - \varepsilon - \delta$.

If $R$ is not collapsible, then, by definition, it has a dicritical extension $R'$ satisfying the hypothesis of one of the cases (i)–(iv). In any case, $\rho(R) \ge \rho(D) + \frac{8}{3} - \varepsilon - \delta$. $\diamond$

Recall that a $k$-cutset in a graph $G$ is a set $S$ of $k$ vertices such that $G - S$ is not connected. A graph is $k$-connected if it has more than $k$ vertices and has no $(k-1)$-cutset. A $k$-cutset in a digraph is a $k$-cutset in its underlying graph, and a digraph is $k$-connected if its underlying graph is $k$-connected.

**Claim 6.** *D is 2-connected.*

*Proof of claim.* Suppose for contradiction that $\{x\}$ is a 1-cutset in $D$. Let $(A_0, B_0)$ be a partition of $V(D-x)$ into non-empty sets such that there is no edge between $A_0$ and $B_0$, and set $A = A_0 \cup \{x\}$ and $B = B_0 \cup \{x\}$.

Since $D$ is 4-dicritical, there exist a 3-dicolouring $\varphi_A$ of $D\langle A\rangle$ and a 3-dicolouring $\varphi_B$ of $D\langle B\rangle$. Free to swap the colours, we may assume $\varphi_A(x) = \varphi_B(x)$. Let $\varphi$ be defined by $\varphi(v) = \varphi_A(v)$ if $v \in A$ and $\varphi(v) = \varphi_B(v)$ if $v \in B$. Since $\vec{\chi}(D) = 4$, $D$, coloured with $\varphi$, must contain a monochromatic directed cycle. Such a directed cycle must be contained in $D\langle A\rangle$ or $D\langle B\rangle$, a contradiction. $\diamond$

**Claim 7.** *D is 3-connected. In particular, D contains no diamond.*

*Proof of claim.* Suppose for contradiction that $\{x, y\}$ is a 2-cutset of $D$. Let $(A_0, B_0)$ be a partition of $V(D) \setminus \{x, y\}$ into non-empty sets such that there is no edge between $A_0$ and $B_0$, and set $A = A_0 \cup \{x, y\}$ and $B = B_0 \cup \{x, y\}$.

Assume for a contradiction that there exists a 3-dicolouring $\varphi_A$ of $D\langle A\rangle$ and a 3-dicolouring $\varphi_B$ of $D\langle B\rangle$ such that $\varphi_A(x) \ne \varphi_A(y)$ and $\varphi_B(x) \ne \varphi_B(y)$. Free to swap the colours, we may assume $\varphi_A(x) = \varphi_B(x)$ and $\varphi_A(y) = \varphi_B(y)$. Let $\varphi$ be defined by $\varphi(v) = \varphi_A(v)$ if $v \in A$ and $\varphi(v) = \varphi_B(v)$ if $v \in B$. Every directed cycle either is in $D\langle A\rangle$, or is in $D\langle B\rangle$ or contains both $x$ and $y$. Therefore it cannot be monochromatic with $\varphi$ because $\varphi_A$ and $\varphi_B$ are 3-dicolourings of $D\langle A\rangle$ and $D\langle B\rangle$ respectively, and $\varphi(x) \ne \varphi(y)$. Thus $\varphi$ is a 3-dicolouring of $D$, a contradiction. Henceforth either $D\langle A\rangle$ or $D\langle B\rangle$ has no 3-dicolouring $\varphi$ such that $\varphi(x) \ne \varphi(y)$. Suppose without loss of generality that it is $D\langle A\rangle$.

Let $D_A = D\langle A\rangle \cup [x, y]$. $D_A$ is not 3-dicolourable because in every 3-dicolouring of $D\langle A\rangle$, $x$ and $y$ are coloured the same. Let $D_B$ be the digraph obtained from $D\langle B\rangle$ by identifying $x$ and $y$ into a vertex $z$. Assume for a contradiction that $D_B$ has a 3-dicolouring $\psi_B$. Set $\psi(x) = \psi(y) = \psi_B(z)$, and $\psi(u) = \psi_B(u)$ for every $u \in B \setminus \{x, y\}$. Then consider a 3-dicolouring $\psi_A$ of $D\langle A\rangle$ such that $\psi_A(x) = \psi(x) = \psi_A(y) = \psi(y)$ (such a colouring exists because $A$ is a proper subdigraph of $D$) and we set $\psi(u) = \psi_A(u)$ for every $u \in V(A) \setminus \{x, y\}$. As $D$ is not 3-dicolourable, it contains a monochromatic directed cycle $C$ (with respect to $\psi$). The cycle $C$ is not included in $D\langle A\rangle$ nor in $D_B$. As a consequence, there is a monochromatic
directed path from {$x, y$} to {$x, y$} in $B$, and so there is a monochromatic directed cycle in $D_B$ for $\psi_B$, a contradiction. Therefore $D_B$ is not 3-dicolourable

Now $D_A$ has a 4-dicritical subdigraph $W_A$ which necessarily contains {$x, y$}, and $D_B$ has a 4-dicritical subdigraph $W_B$ which necessarily contains $z$. As $W_A$ and $W_B$ are 4-dicritical digraphs smaller than $D$, we have $\rho(W_A), \rho(W_B) \le \frac{4}{3} + 4\varepsilon - 2\delta$ by Claim 1. Let $H$ be the subdigraph of $D$ induced by $V(W_A) \cup V(W_B - z)$.

Note that $n(H) = n(W_A) + n(W_B) - 1$ and $m(H) \ge m(W_A) + m(W_B) - 2$. Moreover $T(H) \ge T(W_A - x) + T(W_B - z) \ge T(W_A) + T(W_B) - 2$, by Lemma 13. Hence we have

$$ 
\begin{align*}
\rho(H) &\le \rho(W_A) + \rho(W_B) - \left(\frac{10}{3} + \varepsilon\right) + (m(W_A) + m(W_B) - m(H)) + 2\delta \\
&\le \rho(W_A) + \rho(W_B) - \frac{10}{3} - \varepsilon + 2 + 2\delta \\
&= \rho(W_A) + \rho(W_B) - \frac{4}{3} - \varepsilon + 2\delta \tag{1} \\
&\le 2\left(\frac{4}{3} + 4\varepsilon - 2\delta\right) - \frac{4}{3} - \varepsilon + 2\delta \\
&= \frac{4}{3} + 7\varepsilon - 2\delta
\end{align*}
$$

By Claim 3, if $n(H) < n(D)$ then $\rho(H) > 3 - 3\varepsilon + \delta$. As $10\varepsilon - 3\delta \le \frac{5}{3}$, we deduce that $H = D$. In the above chain of inequalities, we upper bounded $m(W_A) + m(W_B) - m(H)$ by 2, doing the same computation with $2 + m(W_A) + m(W_B) - m(H) - 2$ instead of 2, we get $1 < \rho(D) = \rho(H) \le \frac{4}{3} + 7\varepsilon - 2\delta + (m(W_A) + m(W_B) - m(H) - 2)$ and so $m(H) = m(W_A) + m(W_B) - 2$ because $2\delta - 7\varepsilon \le \frac{2}{3}$. In particular, there is no arc between $x$ and $y$ in $D$. Moreover, no arc was suppressed when identifying $x$ and $y$ into $z$ to obtain $D_B$, so $x$ and $y$ have no common out-neighbour (resp. in-neighbour) in $B_0$.

We first show that either $W_A$ or $W_B$ is not 4-Ore. Assume for contradiction that both $W_A$ and $W_B$ are 4-Ore. If $H = D$ is not bidirected, then by Lemma 12, $D\langle B\rangle$ admits a 3-dicolouring $\varphi_B$ such that $\varphi_B(x) = \varphi_B(y)$. Now let $\varphi_A$ be a 3-dicolouring of $D\langle A\rangle$. We have $\varphi_A(x) = \varphi_A(y)$. Free to exchange colours, we may assume $\varphi_A(x) = \varphi_A(y) = \varphi_B(x) = \varphi_B(y)$. Hence we can define the 3-colouring $\varphi$ of $D$ by $\varphi(v) = \varphi_A(v)$ if $v \in A$, and $\varphi(v) = \varphi_B(v)$ if $v \in B$. Observe that, since $A$ is bidirected, all neighbours of $x$ and $y$ in $D\langle A\rangle$ have a colour distinct from $\varphi(x)$. Therefore there is no monochromatic directed cycle in $D$ coloured by $\varphi$. Thus $\varphi$ is a 3-dicolouring of $D$, a contradiction. Therefore, $H = D$ is bidirected, and so $H$ is an Ore-composition of $W_A$ and $W_B$ (because $D$ is 2-connected by Claim 6), and so $D$ is 4-Ore, a contradiction. Henceforth, we may assume that either $W_A$ or $W_B$ is not 4-Ore.

If none of $W_A$ and $W_B$ are a 4-Ore, then by minimality of $D$, $\rho(W_A) \le 1$ and $\rho(W_B) \le 1$. Together with Equation (1), this yields

$$ \rho(H) \le \frac{2}{3} - \varepsilon + 2\delta \le 1 $$

because $2\delta - \varepsilon \le \frac{1}{3}$, a contradiction.
If none of $W_A$ and $W_B$ is $\overleftrightarrow{K_4}$, then $\rho(W_A) + \rho(W_B) \le 1 + (\frac{4}{3} + 7\varepsilon - 4\delta)$ (recall that if a digraph is 4-Ore but not $\overleftrightarrow{K_4}$, then it has potential at most $\frac{4}{3} + 7\varepsilon - 4\delta$ by Lemma 22). Thus, with Equation (1), we get

$$ \rho(H) \le 1 + \left(\frac{4}{3} + 7\varepsilon - 4\delta\right) - \frac{4}{3} - \varepsilon + 2\delta = 1 + 6\varepsilon - 2\delta \le 1 $$

because $\delta \ge 3\varepsilon$.

Finally, if exactly one of $W_A$ or $W_B$ is isomorphic to $\overleftrightarrow{K_4}$, then either $T(W_A - x) = T(W_A) = 2$ (if $W_A = \overleftrightarrow{K_4}$) or $T(W_B - z) = T(W_B) = 2$ (if $W_B = \overleftrightarrow{K_4}$). Therefore $T(H) \ge T(W_A - x) + T(W_B - z) \ge T(W_A) + T(W_B) - 1$ by Lemma 13, and so

$$ \rho(H) \le \rho(W_A) + \rho(W_B) - \left(\frac{10}{3} + \varepsilon\right) + 2 + \delta. $$

Now the non 4-Ore digraph among $W_A$, $W_B$ has potential at most 1 and the other has potential $\rho(\overleftrightarrow{K_4}) = \frac{4}{3} + 4\varepsilon - 2\delta$. Thus

$$ \rho(H) \le 1 + \left(\frac{4}{3} + 4\varepsilon - 2\delta\right) - \left(\frac{10}{3} + \varepsilon\right) + 2 + \delta = 1 + 3\varepsilon - \delta \le 1 $$

because $\delta \ge 3\varepsilon$.

In all three cases, $\rho(D) = \rho(H) \le 1$, which is a contradiction. Hence $D$ is 3-connected. $\diamond$

**Claim 8.** If $R$ is a collapsible subdigraph of $D$, $u, v$ are in the boundary of $R$ and $D\langle R\rangle \cup [u, v]$ is 4-Ore, then there exists $R' \subseteq R$ such that

(i) either $R'$ is an Ore-collapsible subdigraph of $D$, or

(ii) $R'$ is an induced subdigraph of $R$, $n(R') < n(R)$, and there exist $u', v'$ in $\partial_D(R')$ such that $R' \cup [u', v']$ is 4-Ore.

*Proof of claim.* If $\partial(R) = \{u, v\}$, then $R$ is Ore-collapsible and we are done. Suppose now that there exists $w \in \partial(R)$ distinct from $u$ and $v$. Let $H = D\langle R\rangle \cup [u, v]$. Observe that $H \ne \overleftrightarrow{K_4}$ as $u, v$ and $w$ receive the same colour in any 3-dicolouring of $D\langle R\rangle$ because $R$ is collapsible. Hence $H$ is the Ore-composition of two 4-Ore digraphs $H_1$ (the digon side with replaced digon $[x, y]$) and $H_2$ (the split side with split vertex $z$).

If $u$ or $v$ is in $V(H_2)$, then $R' = D\langle V(H_1)\rangle$ with $u' = x, v' = y$ satisfies (ii). Now we assume that $u, v \in V(H_1) \setminus V(H_2)$. By repeating this argument successively on $H_1$, and then on the digon-side of $H_1$, etc, either we find a subdigraph $R'$ satisfying (ii) or $u$ and $v$ are in a copy $J$ of $\overleftrightarrow{K_4}$ such that $H$ is obtained by Ore-compositions between $J$ and some 4-Ore digraphs with $J$ being always in the digon side.

Observe that $w \notin V(J)$ because in any 3-dicolouring of $H \setminus [u, v]$, vertices in $J$ receive different colours by Lemma 24, except $u$ and $v$. Hence at one step in the succession of
Ore-compositions, $w$ was in the split-side $S$ when a digon $e$ in $J$ has been replaced. However $e \neq [u, v]$, so either $u$ or $v$ is not in $e$. Suppose without loss of generality that $e$ is not incident to $v$.

We claim that $H' = R - v \cup [u, w]$ is not 3-dicolourable. Otherwise, let $\varphi$ be a 3-dicolouring of $H'$. Then $\varphi$ is a 3-dicolouring of $H - v$ with $H$ 4-Ore, so vertices in $J - v$ must receive pairwise different colours by Lemma 25. Let $\varphi'$ be a 3-dicolouring of $R$. Without loss of generality, we may assume that $\varphi(x) = \varphi'(x)$ for every $x \in V(J - v)$. If $y \in S$, let $\varphi''(y) = \varphi(y)$, and let $\varphi''(y) = \varphi'(y)$ if $y \notin S$. Then $\varphi''$ is a 3-dicolouring of $R$ but with $\varphi(u) \neq \varphi(w)$, contradicting the fact that $R$ is collapsible. This shows that $H' = R - v \cup [u, w]$ is not 3-dicolourable.

Hence $R - v \cup [u, w]$ contains a 4-dicritical digraph $K$. By Lemma 23, $R' = D\langle V(K)\rangle$, as a subdigraph of $H$ which is a 4-Ore, satisfies $\frac{10}{3}n(R') - m(R') \geq \frac{10}{3}$. This implies that $\frac{10}{3}n(K) - m(K) \geq \frac{4}{3}$. Note also that $K$ is bidirected because $R - v$ is bidirected. Thus, by Theorem 21, $K$ is 4-Ore. Hence $R'$ with $u, w$ satisfies (ii). $\diamond$

**Claim 9.** If $R$ is a subdigraph of $D$ with $n(R) < n(D)$ and $u, v \in V(R)$, then $R \cup [u, v]$ is 3-dicolourable. As a consequence, there is no collapsible subdigraph in $D$.

*Proof of claim.* Assume for a contradiction that the statement is false. Consider a smallest induced subdigraph $R$ for which the statement does not hold. Then $K = R \cup [u, v]$ is 4-vertex-dicritical, that is for every vertex $v \in V(K)$, $\chi(K - v) < 4 = \chi(K)$. Note that 4-vertex-dicritical digraphs smaller than $D$ satisfy the outcome of Theorem 8 since adding arcs does not increase the potential. Note that $\rho(R) \leq \rho(K) + 2 + \delta$.

If $R$ is not collapsible, then, by Claim 5, $\rho(R) \geq \rho(D) + \frac{8}{3} - \varepsilon - \delta > \frac{11}{3} - \varepsilon - \delta$. But we also have $\rho(R) \leq \rho(K) + 2 + \delta \leq \frac{10}{3} + 4\varepsilon - \delta$ by Claim 1, which is a contradiction because $5\varepsilon \leq \frac{1}{3}$. Hence $R$ is collapsible.

Let $\varphi$ be a 3-dicolouring of $R$. Observe that $\varphi(u) = \varphi(v)$ for otherwise $R \cup [u, v]$ would be 3-dicolourable. Let $R'$ be the dicritical extension of $R$ with extender $W$ and core $X_W$. We have $R' = D$ and $|X_W| = 1$. Since $R$ is collapsible, for every two vertices $u'$, $v'$ on the boundary of $R$, $R \cup [u', v']$ is not 3-dicolourable. Hence, free to consider $u'$, $v'$ instead of $u, v$, we can suppose that $u$ and $v$ are on the boundary of $R$. If $K$ is 4-Ore, then, by Claim 8 and by minimality of $R$, we have that $R$ is Ore-collapsible, and so has boundary of size 2. This contradicts the fact that $D$ is 3-connected. Hence $K$ is not 4-Ore.

By Claim 2, we have

$$1 < \rho(D) = \rho(R') \leq \rho(W) + \rho(R) - \frac{10}{3} - \varepsilon + \delta \\ \leq \rho(W) + (\rho(K) + 2 + \delta) - \frac{10}{3} - \varepsilon + \delta$$

and as $\rho(K) \leq 1$ (because it is not 4-Ore and by minimality of $D$) we get

$$1 < 1 + \rho(W) - \left(\frac{4}{3} + \varepsilon - 2\delta\right)$$
that is $\rho(W) > \frac{4}{3} + \varepsilon - 2\delta$. But as $W$ is smaller than $D$, it satisfies Theorem 8. Thus, since $\varepsilon - 2\delta \ge -\frac{1}{3}$, $W$ must be 4-Ore. Moreover, $W$ must be isomorphic to $\overleftrightarrow{K_4}$, for otherwise $\rho(W)$ would be at most $\frac{4}{3} + 7\varepsilon - 4\delta$, and $\frac{4}{3} + 7\varepsilon - 4\delta \ge \rho(W) > \frac{4}{3} + \varepsilon - 2\delta$ would contradict $\delta \ge 3\varepsilon$. Hence $\rho(W) = \rho(\overleftrightarrow{K_4}) = \frac{4}{3} + 4\varepsilon - 2\delta$ and $T(W - X_W) = 2 = T(W)$. Thus, by Claim 2 and because $\delta \ge 3\varepsilon$, we have

$$1 < \rho(D) \le \rho(W) + \rho(K) + 2 + \delta - \frac{10}{3} - \varepsilon \le \rho(K) + 3\varepsilon - \delta \le \rho(K) \le 1,$$

a contradiction.

This implies that $D$ does not contain any collapsible subdigraph. Indeed, assume for a contradiction that $D$ contains a collapsible subdigraph $R$, and let $u, v$ be two vertices in its boundary. Then there exists a 3-dicolouring $\varphi$ of $R \cup [u, v]$, for which $\partial(R)$ is not monochromatic, a contradiction. $\diamond$

**Claim 10.** If $R$ is a subdigraph of $D$ with $n(R) < n(D)$ and $u, v, u', v' \in R$, then $R \cup \{uv, u'v'\}$ is 3-dicolourable. In particular, $D$ contains no copy of $\overleftrightarrow{K_4}$ minus two arcs.

*Proof of claim.* Assume for a contradiction that the statement is false. Consider a smallest subdigraph $R$ for which the statement does not hold. Then $K = R \cup \{uv, u'v'\}$ is 4-dicritical and smaller than $D$, so $\rho(K) \le \frac{4}{3} + 4\varepsilon - 2\delta$ by Claim 1. By Claim 9, $R$ is not collapsible, so, by Claim 5, we have $\rho(R) \ge \rho(D) + \frac{8}{3} - \varepsilon - \delta > \frac{11}{3} - \varepsilon - \delta$. But $\rho(R) \le \rho(K) + 2 + 2\delta \le \frac{10}{3} + 4\varepsilon$, which is a contradiction as $5\varepsilon + \delta \le \frac{1}{3}$. $\diamond$

For any $v \in V(D)$, we denote by $n(v)$ its number of neighbours, that is $n(v) = |N^+(u) \cup N^-(v)|$, and by $d(v)$ its number of incident arcs, that is $d(v) = d^+(v) + d^-(v)$.

**Claim 11.** Vertices of degree 6 in $D$ have either three or six neighbours.

*Proof of claim.* Let $x$ be a vertex of degree 6.

If $n(x) = 4$, then let $a, b, c, d$ be its neighbours such that $N^+(x) = \{a, b, c\}$ and $N^-(x) = \{a, b, d\}$. Consider $D' = D - x \cup dc$. By Claim 10, $D'$ has a 3-dicolouring $\varphi$. If $|\varphi(N^-(x))| < 3$, then choosing $\varphi(x)$ in $\{1, 2, 3\} \setminus \varphi(N^-(x))$, we obtain a 3-dicolouring of $D$, a contradiction. Hence $\varphi(N^-(x)) = \{1, 2, 3\}$. We set $\varphi(x) = \varphi(d)$. As $D$ is not 3-dicolourable, $D$ contains a monochromatic directed cycle $C$. This cycle $C$ must contain the arc $dx$, and an out-neighbour $z$ of $x$. Since $\varphi(a), \varphi(b)$ and $\varphi(d)$ are all distinct, necessarily $z = c$. But then $C - x \cup dc$ is a monochromatic directed cycle in $D'$, a contradiction.

Similarly, if $n(v) = 5$, let $N^+(x) = \{a, b, c\}$ and $N^-(x) = \{a, d, e\}$, and consider $D' = D - x \cup db \cup dc$. By Claim 10, $D'$ has a 3-dicolouring $\varphi$. If $|\varphi(N^-(x))| < 3$, then choosing $\varphi(x)$ in $\{1, 2, 3\} \setminus \varphi(N^-(x))$, we obtain a 3-dicolouring of $D$, a contradiction. Hence $\varphi(N^-(x)) = \{1, 2, 3\}$. We set $\varphi(x) = \varphi(d)$. As $D$ is not 3-dicolourable, there is a monochromatic directed cycle $C$, which must contain the arc $dx$ and an out-neighbour $z$ of $x$. Note that $z$ must be $b$ or $c$ because $\varphi(a) \ne \varphi(d)$. Then $C - x \cup dz$ is a monochromatic directed cycle in $D'$, a contradiction. $\diamond$
**Claim 12.** There is no bidirected triangle containing two vertices of degree 6. In particular, D contains no emerald.

*Proof of claim.* Suppose that $D\langle\{x, y, z\}\rangle = \overleftrightarrow{K}_3$ and $d(x) = d(y) = 6$. By Claim 11, $x$ and $y$ have exactly three neighbours, and $N[x] \neq N[y]$ because $D$ contains no copy of $\overleftrightarrow{K}_4$ minus two arcs by Claim 10. Let $u$ (resp. $v$) be the unique neighbour of $x$ distinct from $y$ and $z$ (resp. $x$ and $z$). Consider $D' = D - \{x, y\} \cup [u, v]$. By Claim 9, $D'$ has a 3-dicolouring $\varphi$. Without loss of generality, suppose that $\varphi(u) = 1$ and $\varphi(v) = 2$. If $\varphi(z) = 1$ (resp. $\varphi(z) = 2$, $\varphi(z) = 3$), we set $\varphi(x) = 2$ and $\varphi(y) = 3$ (resp. $\varphi(x) = 3$ and $\varphi(y) = 1$, $\varphi(x) = 2$ and $\varphi(y) = 1$). In each case, this yields a 3-dicolouring of $D$, a contradiction. $\diamond$

So now we know that $D$ contains no emerald, and no diamond by Claim 7.

**Claim 13.** If $R$ is an induced subdigraph of $D$ with $4 \le n(R) < n(D)$, then $\rho(R) \ge \rho(D) + 3 + 3\varepsilon - 3\delta$, except if $D-R$ contains a single vertex which has degree 6 in $D$.

*Proof of claim.* Let $R$ be an induced subdigraph of $D$ with $4 \le n(R) < n(D)$. By Claim 9, $R$ is not collapsible. Let $\varphi$ be a 3-dicolouring of $R$, $R'$ be a dicritical extension of $R$ with extender $W$ and core $X_W$ (with respect to $\varphi$). By (the consequence of) Claim 3, we know that $\rho(R') \ge \rho(D)$.

Assume first that $R' \neq D$. Then, by Claims 3 and 2,

$$ \rho(D) + 2 - 3\varepsilon + \delta \le \rho(R') \le \rho(W) + \rho(R) - \frac{10}{3} - \varepsilon + \delta. $$

Since $\rho(W) \le \frac{4}{3} + 4\varepsilon - 2\delta$ by Claim 1, we have $\rho(R) \ge \rho(D) + 4 - 6\varepsilon + 2\delta \ge \rho(D) + 3 + 3\varepsilon - 3\delta$, because $1 \ge 9\varepsilon - 5\delta$. In the following we suppose that $R' = D$. We distinguish three cases depending on the cardinality of $|X_W|$.

* Assume first that $|X_W| = 2$. Then, by Claim 2 and Lemma 13,

$$ \rho(D) \le \rho(R') \le \rho(W) + \rho(R) - \frac{20}{3} - 2\varepsilon + 2 + 2\delta $$

and, as $\rho(W) \le \frac{4}{3} + 4\varepsilon - 2\delta$ by Claim 1, we have $\rho(R) \ge \rho(D) + \frac{10}{3} - 2\varepsilon \ge \rho(D) + 3 + 3\varepsilon - 3\delta$ because $5\varepsilon - 3\delta \le \frac{1}{3}$.

* Assume now that $|X_W| = 3$. If there is a vertex $v \in V(D-R)$ with two out-neighbours (resp. two in-neighbours) in $V(R)$ with the same colour for $\varphi$, then

$$ -n(R') = n(W) - |X_W| + n(R), $$

$$ -m(R') \ge m(W) + m(R) - m(\overleftrightarrow{K}_{|X_W|}) + 1 \text{ because } v \text{ has two in- or out-neighbour in } V(R) \text{ with the same colour for } \varphi, $$

$$ -T(R') \ge T(W - X_W) + T(R). $$
It follows that

$$ \rho(D) \le \rho(R') \le \rho(W) + \rho(R) - (10 + 3\epsilon - 6) + 3\delta - 1 $$

and so $\rho(R) \ge \rho(D) - \frac{4}{3} - 4\epsilon + 2\delta + 5 + 3\epsilon - 3\delta \ge \rho(D) + \frac{11}{3} - \epsilon - \delta \ge \rho(D) + 3 + 3\epsilon - 3\delta$ because $4\epsilon - 2\delta \le \frac{2}{3}$. Now we assume that there is no vertex with two out-neighbours (resp. two in-neighbours) in $R$ with the same colour for $\varphi$. In other words, the in-degrees and out-degrees of vertices in $D - R$ are the same in $D$ and in $W$.

If $W$ is not 4-Ore, then by Claim 2

$$ \rho(D) \le \rho(R') \le \rho(W) + \rho(R) - (10 + 3\epsilon - 6) + 3\delta $$

and, as $\rho(W) \le 1$, we have $\rho(R) \ge \rho(D) + 3 + 3\epsilon - 3\delta$.

Now suppose $W$ is 4-Ore. If $W \ne \overleftrightarrow{K}_4$, then, by Lemma 20, $W$ contains a diamond or an emerald disjoint from $X$, and this gives a diamond or an emerald in $D$ because the degrees of vertices in $D - R$ are the same in $D$ and in $W$, which is a contradiction.

Now suppose that $W = \overleftrightarrow{K}_4$. Then $D - R$ has a single vertex of degree 6 in $D$.

* Assume finally that $|X_W| = 1$. Since $R$ is not collapsible by Claim 9, $\varphi$ may have been chosen so that $\partial(R)$ is not monochromatic in $\varphi$. Then, by Claim 4, $\rho(R) \ge \rho(D) + 3 - 3\epsilon + \delta \ge \rho(D) + 3 + 3\epsilon - 3\delta$, because $6\epsilon - 4\delta \le 0$.

◇

In $D$, we say that a vertex $v$ is a **simple in-neighbour** (resp. **simple out-neighbour**) if $v$ is a in-neighbour (resp. out-neighbour) of $u$ and $[u, v]$ is not a digon in $D$. If $v$ is a simple in-neighbour or simple out-neighbour of $u$, we simply say that $v$ is a **simple neighbour** of $u$.

**Claim 14.** Vertices of degree 7 have seven neighbours. In other words, every vertex of degree 7 has only simple neighbours.

*Proof of claim.* Let $x$ be a vertex of degree 7. We suppose, without loss of generality, that $d^-(x) = 3$ and $d^+(x) = 4$.

If $n(x) = 4$, then $x$ has a unique simple out-neighbour $a$. As $D$ is 4-dicritical, $D \setminus xa$ has a 3-dicolouring $\varphi$. But then every directed cycle is either in $D \setminus xa$ or it contains $xa$ and thus an in-neighbour $t$ of $x$. In the first case, it is not monochromatic because $\varphi$ is a 3-dicolouring of $D \setminus xa$, and in the second case, it is not monochromatic because $[t, x]$ is a digon and so $\varphi(t) \ne \varphi(x)$. Hence $\varphi$ is a 3-dicolouring of $D$, a contradiction.

If $n(x) = 5$, let $N^-(x) = \{a, b, c\}$ and $N^+(x) = \{a, b, d, e\}$. By Claim 10, $D' = D - x \cup \{cd, ce\}$ has a 3-dicolouring $\varphi$. If $|\varphi(N^-(x))| < 3$, then choosing $\varphi(x)$ in $\{1, 2, 3\} \setminus \varphi(N^-(x))$ gives a 3-dicolouring of $D$, a contradiction. If $|\varphi(N^-(x))| = 3$, then we set $\varphi(x) = \varphi(c)$. Suppose for a contradiction that there is a monochromatic directed cycle $C$ in $D$ (with $\varphi$).
Necessarily $C$ contains $x$ (since $\varphi$ is a 3-dicolouring of $D-x$) and so it must contain $c$ and one vertex $y$ in $\{d,e\}$ because $\varphi(a), \varphi(b)$, and $\varphi(c)$ are all distinct. Then $C-x \cup cy$ is a monochromatic directed cycle in $D'$, a contradiction. Therefore $\varphi$ is a 3-dicolouring of $D$, a contradiction.

If $n(x) = 6$, let $N^-(x) = \{a, b, c\}$ and $N^+(x) = \{a, d, e, f\}$. Consider $D' = D - x \cup \{bd, be, bf\}$.

We first show that $D'$ is not 3-dicolourable. Assume for a contradiction that there is a 3-dicolouring $\varphi$ of $D'$. If $|\varphi(N^-(x))| < 3$, then choosing $\varphi(x)$ in $\{1, 2, 3\} \setminus \varphi(N^-(x))$ gives a 3-dicolouring of $D$, a contradiction. Hence $|\varphi(N^-(x))| = 3$. We set $\varphi(x) = \varphi(b)$. Since $D$ is not 3-dicolourable, there exists a monochromatic directed cycle $C$ in $D$ (with $\varphi$). Necessarily $C$ contains $x$ (since $\varphi$ is a 3-dicolouring of $D-x$) and so it must contain $b$ and one vertex $y$ in $\{d, e, f\}$ because $\varphi(a), \varphi(b)$, and $\varphi(c)$ are all distinct. Then $C - x \cup by$ is a monochromatic directed cycle in $D'$, a contradiction. This gives a 3-dicolouring of $D$, a contradiction.

Henceforth $D'$ is not 3-dicolourable, and so it contains a 4-dicritical digraph $\tilde{D}$, smaller than $D$. If $\tilde{D}$ does not contain the three arcs $bd, be, bf$, then it can be obtained from a proper induced subdigraph of $D$ by adding at most two arcs, and so it is 3-dicolourable by Claim 10, a contradiction.

Hence $\{b, d, e, f\} \subseteq V(\tilde{D})$. Now consider $U = D \langle V(\tilde{D}) \cup \{x\} \rangle$.

• Assume first that $a \notin V(U)$ or $c \notin V(U)$. Then we have

• $n(U) = n(\tilde{D}) + 1$,

• $m(U) \ge m(\tilde{D}) + 1$ and

• $T(U) \ge T(\tilde{D} - b) \ge T(\tilde{D}) - 1$ by Lemma 13.

Hence

$$ 
\begin{aligned}
\rho(U) &\le \rho(\tilde{D}) + \frac{10}{3} + \varepsilon - 1 + \delta \\
&\le \frac{4}{3} + 4\varepsilon - 2\delta + \frac{10}{3} + \varepsilon - 1 + \delta \quad \text{by Claim 1,} \\
&= 1 + \frac{8}{3} + 5\varepsilon - \delta \\
&< \rho(D) + \frac{8}{3} + 5\varepsilon - \delta \\
&\le \rho(D) + 3 + 3\varepsilon - 3\delta \quad \text{because } \frac{1}{3} \ge 2\delta + 2\varepsilon.
\end{aligned}
$$

Hence by Claim 13, $D - U$ has a single vertex of degree 6 (in $D$), which must be either $a$ or $c$. Then we have

• $n(D) = n(\tilde{D}) + 2$,

• $m(D) \ge m(\tilde{D}) - 3 + 11$ and

• $T(D) \ge T(\tilde{D} - b) \ge T(\tilde{D}) - 1$.
Thus

$$
\begin{align*}
\rho(D) &\le \rho(\tilde{D}) + 2\left(\frac{10}{3} + \varepsilon\right) - 8 + \delta \\
&\le \left(\frac{4}{3} + 4\varepsilon - 2\delta\right) - \frac{4}{3} + 2\varepsilon + \delta \quad \text{by Claim 1,} \\
&\le 1 \qquad \text{because } 6\varepsilon - \delta \le 1.
\end{align*}
$$

This is a contradiction.

• Assume now that $a, c \in V(U)$, then we have

$$
\bullet n(U) = n(\tilde{D}) + 1,
$$

$$
\bullet m(U) \geq m(\tilde{D}) + 4 \text{ and}
$$

$$
\bullet T(U) \geq T(\tilde{D} - b) \geq T(\tilde{D}) - 1 \text{ by Lemma 13.}
$$

Thus

$$
\begin{align*}
\rho(U) &\le \rho(\tilde{D}) + \frac{10}{3} + \varepsilon - 4 + \delta \\
&\le \left(\frac{4}{3} + 4\varepsilon - 2\delta\right) + \frac{10}{3} + \varepsilon - 4 + \delta \quad \text{by Claim 1,} \\
&\le 1 \qquad \text{because } 5\varepsilon - \delta \le \frac{1}{3}.
\end{align*}
$$

Together with the consequence of Claim 3, we get that $\rho(D) \le \rho(U) \le 1$, a contradiction.
$\diamond$

The $8^+$-valency of a vertex $v$, denoted by $\nu(v)$, is the number of arcs incident to $v$ and
a vertex of degree at least 8.

Let $D_6$ be the subdigraph of $D$ induced by the vertices of degree 6 incident to digons.
Let us describe the connected components of $D_6$ and their neighbourhoods. Remember that
vertices of degree 7 are incident to no digon by Claim 14, and so they do not have neighbours
in $V(D_6)$. If $v$ is a vertex in $D_6$, we define its **neighbourhood valency** to be the sum of
the $8^+$-valency of its neighbours of degree at least 8. We denote the neighbourhood valency
of $v$ by $\nu_N(v)$.

**Claim 15.** If $[x, y]$ is a digon and both $x$ and $y$ have degree 6, then either

(i) the two neighbours of $y$ distinct from $x$ have degree at least 8, or

(ii) the two neighbours of $x$ distinct from $y$ have degree at least 8 and $\nu_N(x) \ge 4$.

Proof of claim. Let [x, y] be a digon in D with d(x) = d(y) = 6. By Claim 11 n(x) = n(y) = 3. Let u and v be the two neighbours of x different from y. By Claim 14, u and v have degree 6 or at least 8.
If $u$ and $v$ are linked by a digon, then by Claim 12, $u$ and $v$ do not have degree 6, so they have degree 8. Moreover $\nu(u) \ge 2$ and $\nu(v) \ge 2$. Thus $\nu_N(x) = \nu(u) + \nu(v) \ge 4$ and (ii) holds. Henceforth, we may assume that $u$ and $v$ are not linked by a digon.

Let $D'$ the digraph obtained by removing $x$ and $y$ and identifying $u$ and $v$ into a single vertex $u \star v$. We claim that $D'$ is not 3-dicolourable. To see that, suppose for contradiction that there exists a 3-dicolouring $\varphi$ of $D'$. Then set $\varphi(u) = \varphi(v) = \varphi(u \star v)$, choose $\varphi(y)$ in $\{1, 2, 3\} \setminus \varphi(N(y) \setminus \{x\})$, and finally choose $\varphi(x)$ in $\{1, 2, 3\} \setminus \{\varphi(u \star v), \varphi(y)\}$. One can easily see that $\varphi$ is now a 3-dicolouring of $D$, a contradiction. This proves that $D'$ is not 3-dicolourable and so it contains a 4-dicritical digraph $\tilde{D}$, which must contain $u \star v$ because every subdigraph of $D$ is 3-dicolourable. Let $R$ be the subdigraph of $D$ induced by $(V(\tilde{D}) \setminus \{u \star v\}) \cup \{u, v, x\}$. We have

* $n(R) = n(\tilde{D}) + 2$,

* $m(R) \ge m(\tilde{D}) + 4$ and

* $T(R) \ge T(\tilde{D} - u \star v) + 1 \ge T(\tilde{D})$ because $[x, u]$ is a digon, and by Lemma 13.

If $\tilde{D}$ is not 4-Ore, then $\rho(\tilde{D}) \le 1$ by minimality of $D$, and so

$$ 
\begin{aligned}
\rho(R) &\le \rho(\tilde{D}) + 2\left(\frac{10}{3} + \varepsilon\right) - 4 \\
&\le 1 + \frac{8}{3} + 2\varepsilon \\
&< \rho(D) + 3 + 3\varepsilon - 3\delta \quad \text{because } \varepsilon - 3\delta \ge -\frac{1}{3}.
\end{aligned}
$$

Similarly, if $\tilde{D}$ is 4-Ore but not $\overleftrightarrow{K}_4$, then

$$ 
\begin{aligned}
\rho(R) &\le \rho(\tilde{D}) + 2\left(\frac{10}{3} + \varepsilon\right) - 4 \\
&\le \left(\frac{4}{3} + 7\varepsilon - 4\delta\right) + \frac{8}{3} + 2\varepsilon \quad \text{by Lemma 22,} \\
&= 1 + 3 + 9\varepsilon - 4\delta \\
&< \rho(D) + 3 + 9\varepsilon - 4\delta \\
&\le \rho(D) + 3 + 3\varepsilon - 3\delta \quad \text{because } \delta \ge 6\varepsilon.
\end{aligned}
$$

In both cases (that is when $\tilde{D}$ is not $\overleftrightarrow{K}_4$), by Claim 13, $D-R$ is a single vertex of degree 6, namely $y$. Then every neighbour $w$ of $y$ different from $x$ has degree at least 6 in $\tilde{D}$ (because $\tilde{D}$ is 3-dicritical) and so has degree at least 8 in $D$ and (i) holds.

Assume now that $\tilde{D}$ is a copy of $\overleftrightarrow{K}_4$. Let us denote by $a, b, c$ the vertices of $\tilde{D}$ different from $u \star v$. Suppose for a contradiction that $u$ has degree 6. Then $u$ has exactly three
<figure><img src="image_4.png" /><figcaption>Figure 3: The possible connected components of $D_6$.</figcaption></figure>

neighbours by Claim 11. If $|N(u) \cap \{a, b, c\}| = 2$, then $D\langle\{u, a, b, c\}\rangle$ is a copy of $\overleftrightarrow{K_4}$ minus a digon, contradicting Claim 9. If $|N(u) \cap \{a, b, c\}| \le 1$, then $v$ must be adjacent to at least two vertices of $\{a, b, c\}$ with a digon, and so $D\langle\{v, a, b, c\}\rangle$ contains a copy of $\overleftrightarrow{K_4}$ minus a digon, contradicting Claim 9. Hence $u$ has degree at least 8, and by symmetry so does $v$. Moreover $D\langle\{a, b, c\}\rangle$ is a bidirected triangle, and so by Claim 12, at least two of these vertices have degree at least 8 (remember that vertices of degree 7 are in no digon by Claim 14). Hence at least four arcs between $\{u, v\}$ and $\{a, b, c\}$ are incident to two vertices of degree at least 8. In other word, $\nu_N(x) = \nu(u) + \nu(v) \ge 4$, so (ii) holds. $\diamond$

**Claim 16.** Let $C$ be a connected component of $D_6$. Then $C$ is one of the following (see Figure 3):

(i) a single vertex, or

(ii) a bidirected path on two vertices, or

(iii) a bidirected path on three vertices, whose extremities have neighbourhood valency at least 4, or

(iv) a star on four vertices, whose non-central vertices have neighbourhood valency at least 4.

*Proof of claim.* First observe that $C$ does not contain a bidirected path $[[x, y, z, w]]$ on four vertices, because otherwise, by Claim 15 applied on $[y, z]$, either $y$ or $z$ has two neighbours of degree at least 8, a contradiction. Observe also that $C$ contains no bidirected triangle by Claim 12.

Moreover, if $[[x, y, z]]$ is a bidirected path in $C$ on three vertices, then by Claim 15 applied both on $[y, z]$ and $[z, y]$, $x$ and $z$ have both neighbourhood valency at least 4. The statement of the claim follows. $\diamond$

An arc $xy$ is said to be **out-chelou** if

(i) $yx \notin A(D)$,

(ii) $d^+(x) = 3$,
<figure><img src="image_5.png" /><figcaption>Figure 4: An example of an out-chelou arc xy.</figcaption></figure>

(iii) $d^{-}(y) = 3$, and

(iv) there exists $z \in N^{-}(y) \setminus N^{+}(y)$ distinct from $x$.

Symmetrically, we say that an arc $xy$ is **in-chelou** if $yx$ is out-chelou in the digraph obtained from $D$ by reversing every arc. See Figure 4 for an example of an out-chelou arc.

**Claim 17.** *There is no out-chelou arc and no in-chelou arc in D.*

*Proof of claim.* By directional duality, it suffices to prove that $D$ has no out-chelou arcs.

Let $xy$ be an out-chelou arc with $z \in N^{-}(y) \setminus (N^{+}(y) \cup \{x\})$. Consider $D' = D - \{x, y\} \cup \{zz' | z' \in N^{+}(y) \setminus N^{-}(y)\}$. We claim that $D'$ is not 3-dicolourable. To see that, suppose for contradiction that there is a 3-dicolouring $\varphi$ of $D'$. As $d^{+}(x) = 3$, we can choose $\varphi(x)$ in $\{1, 2, 3\} \setminus \varphi(N^{+}(x) \setminus \{y\})$ to obtain a 3-dicolouring of $D-y$. If $|\varphi(N^{-}(y))| < 3$, then choosing $\varphi(y)$ in $\{1, 2, 3\} \setminus \varphi(N^{-}(y))$ gives a 3-dicolouring of $D$, a contradiction. Hence $|\varphi(N^{-}(y)| = 3$. Set $\varphi(x) = \varphi(z)$. Suppose there is a monochromatic directed cycle $C$ in $D$. It must contain $y$ and thus $z$, its unique in-neighbour with its colour. Let $z'$ be the out-neighbour of $y$ in $C$. It must be in $N^{+}(y) \setminus N^{-}(y)$, so $zz'$ is an arc in $D'$. Thus $C - y \cup zz'$ is a monochromatic directed cycle in $D'$, a contradiction. Therefore $\varphi$ is a 3-dicolouring of $D$, a contradiction. Hence $D'$ is not 3-dicolourable.

Consequently, $D'$ contains a 4-dicritical digraph $\tilde{D}$, which is smaller than $D$ and contains $z$, for otherwise $\tilde{D}$ would be a subdigraph of $D$. Consider $U = D \langle V(\tilde{D}) \cup \{y\} \rangle$. We have

$$ \bullet \ n(U) = n(\tilde{D}) + 1, $$

$$ \bullet \ m(U) \geq m(\tilde{D}) + 1 \text{ and} $$

$$ \bullet \ T(U) \geq T(\tilde{D} - z) \geq T(\tilde{D}) - 1 \text{ by Lemma 13.} $$

First if $\tilde{D}$ is not 4-Ore, then by minimality of $D$ we have $\rho(\tilde{D}) \leq 1$, so

$$ \rho(U) \leq \rho(\tilde{D}) + \frac{10}{3} + \varepsilon - 1 + \delta \leq \frac{10}{3} + \varepsilon + \delta \leq \frac{11}{3} - \varepsilon - \delta $$

$$ \text{because } 2\varepsilon + 2\delta \leq \frac{1}{3}. $$

Next if $\tilde{D}$ is 4-Ore, but not isomorphic to $\tilde{K}_4$, then $\rho(\tilde{D}) \leq \frac{4}{3} + 7\varepsilon - 4\delta$ by Lemma 22, and

$$ \rho(U) \leq \rho(\tilde{D}) + \frac{10}{3} + \varepsilon - 1 + \delta \leq \frac{11}{3} + 8\varepsilon - 3\delta \leq \frac{11}{3} - \varepsilon - \delta $$
because $9\epsilon - 2\delta \le 0$.

Finally if $\tilde{D}$ is isomorphic to $\overleftrightarrow{K}_4$, then we have $T(U) \ge T(\tilde{D} - z) \ge T(\tilde{D})$ and $\rho(\tilde{D}) = \frac{4}{3} + 4\epsilon - 2\delta$. So the same computation yields

$$ \rho(U) \le \rho(\tilde{D}) + \frac{10}{3} + \epsilon - 1 \le \frac{11}{3} + 5\epsilon - 2\delta \le \frac{11}{3} - \epsilon - \delta $$

because $6\epsilon - \delta \le 0$.

In all cases, we have $\rho(U) \le \frac{11}{3} - \epsilon - \delta$. This contradicts Claim 5 because $U$ is not collapsible by Claim 9. $\diamond$

We now use the discharging method. For every vertex $v$, let $\sigma(v) = \frac{\delta}{|C|}$ if $v$ has degree 6 and is in a component $C$ of $D_6$ of size at least 2, and $\sigma(v) = 0$ otherwise. Clearly $T(D)$ is at least the number of connected components of size at least 2 of $D_6$ so $\sum_{v \in V(D)} \sigma(v) \le \delta T(D)$. We define the **initial charge** of $v$ to be $w(v) = \frac{10}{3} + \epsilon - \frac{d(v)}{2} - \sigma(v)$. We have

$$ \rho(D) \le \sum_{v \in V(D)} w(v). $$

We now redistribute this total charge according to the following rules:

(R1) A vertex of degree 6 incident to no digon sends $\frac{1}{12} - \frac{\epsilon}{8}$ to each of its neighbours.

(R2) A vertex of degree 6 incident to digons sends $\frac{2}{d(v)-\nu(v)}(-\frac{10}{3} + \frac{d(v)}{2} - \epsilon)$ to each neighbour $v$ of degree at least 8 (so $\frac{1}{d(v)-\nu(v)}(-\frac{10}{3} + \frac{d(v)}{2} - \epsilon)$ via each arc of the digon).

(R3) A vertex of degree 7 with $d^-(v) = 3$ (resp. $d^+(v) = 3$) sends $\frac{1}{12} - \frac{\epsilon}{8}$ to each of its in-neighbours (resp. out-neighbours).

For every vertex $v$, let $w^*(v)$ be the final charge of $v$.

**Claim 18.** If $v$ has degree at least 8, then $w^*(v) \le 0$.

*Proof of claim.* Let $v$ be a vertex of degree at least 8. If $v$ is not adjacent to a vertex of degree at most 7, then $w^*(v) = w(v) = \frac{10}{3} + \epsilon - \frac{d(v)}{2} \le 0$ (because $\epsilon \le \frac{2}{3}$). Otherwise, $d(v) - \nu(v) \ge 1$ and

$$ \frac{1}{d(v) - \nu(v)} \left( -\frac{10}{3} + \frac{d(v)}{2} - \epsilon \right) \ge \frac{1}{d(v)} \left( -\frac{10}{3} + \frac{d(v)}{2} - \epsilon \right) \\ \ge \frac{1}{12} - \frac{\epsilon}{8}. $$

Thus $v$ receives at most $\frac{1}{d(v)-\nu(v)}(-\frac{10}{3} + \frac{d(v)}{2} - \epsilon)$ per arc incident with a vertex of degree 6 or 7. Since there are $d(v) - \nu(v)$ such arcs, $w^*(v) \le w(v) - \frac{10}{3} - \epsilon + \frac{d(v)}{2} = 0$. $\diamond$
**Claim 19.** If $v$ has degree 7, then $w^*(v) \le 0$.

*Proof of claim.* By Claim 14, $v$ has seven neighbours. Without loss of generality, let us suppose that $d^-(v) = 3$ and $d^+(v) = 4$. By Claim 17, the in-neighbours of $v$ can not have out-degree 3. In particular, they do not have degree 6, and if they have degree 7, they do not send anything to $v$ by Rule (R3). Hence $v$ receives at most four times the charge $\frac{1}{12} - \frac{\epsilon}{8}$ by (R1) or (R3), and it sends three times this charge by (R3). Hence

$$ w^*(v) \le w(v) + \frac{1}{12} - \frac{\epsilon}{8} \\ = -\frac{1}{12} + \frac{7}{8}\epsilon $$

and the result comes because $\epsilon \le \frac{2}{21}$.

◇

**Claim 20.** If $v$ is a vertex of degree 6 incident to no digon, then $w^*(v) \le 0$.

*Proof of claim.* The vertex $v$ sends $\frac{1}{12} - \frac{\epsilon}{8}$ to each of its neighbours, and it receives no charge as all its in-neighbours (resp. out-neighbours) have out-degree (resp. in-degree) at least 4, by Claim 17. As a consequence,

$$ w^*(v) = w(v) - 6 \left( \frac{1}{12} - \frac{\epsilon}{8} \right) = -\frac{1}{6} + \frac{7\epsilon}{4} $$

and the result comes because $\epsilon \le \frac{2}{21}$.

◇

**Claim 21.** Let $v$ be a vertex in $D_6$ having at least two neighbours of degree at least 8. Then $w^*(v) \le 0$. Moreover, if $v$ is not an isolated vertex in $D_6$ and $\nu_N(v) \ge 4$, then $w^*(v) \le -\frac{1}{9} + \frac{5}{3}\epsilon - \frac{\delta}{4}$.

*Proof of claim.* Observe that $v$ receives no charge and sends the following charge to each of its neighbour $u$ with degree at least 8:

$$ \begin{aligned} \frac{2}{d(u) - \nu(u)} \left( -\frac{10}{3} - \epsilon + \frac{d(u)}{2} \right) &\ge \frac{2}{d(u)} \left( -\frac{10}{3} - \epsilon + \frac{d(u)}{2} \right) \\ &= 1 - \frac{2}{d(u)} \left( \frac{10}{3} + \epsilon \right) \\ &\ge \frac{2}{8} \left( -\frac{10}{3} - \epsilon + 4 \right) \\ &= \frac{1}{6} - \frac{\epsilon}{4}. \end{aligned} $$

Assume first that $v$ is isolated in $D_6$. By Claim 14, its three neighbours do not have degree 7, and so have degree at least 8. Thus $v$ sends three times at least $\frac{1}{6} - \frac{\epsilon}{4}$, and so

$$ w^*(v) \le w(v) - 3 \left( \frac{1}{6} - \frac{\epsilon}{4} \right) = -\frac{1}{6} + \frac{7}{4}\epsilon $$
and the result comes because $\varepsilon \le \frac{2}{21}$.

Assume now that $v$ is in a connected component $C$ of $D_6$ of size at least 2. By Claim 16, $\sigma(v) \ge \frac{\delta}{4}$, so $w(v) \le \frac{1}{3} + \varepsilon - \frac{\delta}{4}$. Moreover, it sends twice at least $\frac{1}{6} - \frac{\varepsilon}{4}$. Hence

$$w^*(v) \le \left(\frac{1}{3} + \varepsilon - \frac{\delta}{4}\right) - 2\left(\frac{1}{6} - \frac{\varepsilon}{4}\right) = \frac{3}{2}\varepsilon - \frac{\delta}{4}$$

and the result comes because $\delta \ge 6\varepsilon$. This shows the first part of the statement.

We will now prove the second part of the statement. Assume that $v$ is not an isolated vertex in $D_6$ and $\nu_N(v) \ge 4$. Let $u_1$ and $u_2$ be the two neighbours of $v$ with degree at least 8. For every $i \in \{1, 2\}$ we have

$$\frac{2}{d(u_i) - \nu(u_i)} \left( -\frac{10}{3} - \varepsilon + \frac{d(u_i)}{2} \right) = 1 - \frac{1}{d(u_i) - \nu(u_i)} \left( \frac{20}{3} + 2\varepsilon - \nu(u_i) \right)$$

**Case 1:** $\nu(u_i) \ge 7$ for some $i \in \{1, 2\}$. Without loss of generality suppose $i = 1$. Then we have

$$1 - \frac{1}{d(u_1) - \nu(u_1)} \left( \frac{20}{3} + 2\varepsilon - \nu(u_1) \right) \ge 1$$

because $\nu(u_1) \ge 7 \ge \frac{20}{3} + 2\varepsilon$ as $\varepsilon \le \frac{1}{6}$. Then the total charge sent by $v$ is at least 1, and thus

$$w^*(v) \le w(v) - 1 \le \left(\frac{1}{3} + \varepsilon - \frac{\delta}{4}\right) - 1 = -\frac{2}{3} + \varepsilon - \frac{\delta}{4}$$

Thus, we have $w^*(v) \le -\frac{1}{9} + \frac{5}{3}\varepsilon - \frac{\delta}{4}$ because $\varepsilon, \delta \ge 0$.

**Case 2:** $\nu(u_1), \nu(u_2) \le 6$. Let $f: [0, 6] \to \mathbb{R}$ be the function defined by

$$f(x) = \frac{2}{8-x} \left( -\frac{10}{3} - \varepsilon + \frac{8}{2} \right) = 1 - \frac{1}{8-x} \left( \frac{20}{3} - 2\varepsilon - x \right)$$

for every $x \in [0, 6]$. Observe that $f$ is non-decreasing and convex on $[0, 6]$ because $-\frac{10}{3} - \varepsilon + \frac{8}{2} \ge 0$. For $i = 1, 2$, we have

$$\frac{2}{d(u_i) - \nu(u_i)} \left( -\frac{10}{3} - \varepsilon + \frac{d(u_i)}{2} \right) \ge f(\nu(u_i))$$

because the function $d \mapsto 1 - \frac{1}{d-\nu(u_i)} \left(\frac{20}{3} + 2\varepsilon - \nu(u_i)\right)$ is non-decreasing on $[8, +\infty[$ as $\nu(u_i) \le 6 \le \frac{20}{3} + 2\varepsilon$. Hence the charge sent by $v$ to $u_i$ is at least $f(\nu(u_i))$. By hypothesis we have $\nu_N(v) = \nu(u_1) + \nu(u_2) \ge 4$. It follows that the total charge sent by $v$ is at least

$$\begin{aligned} f(\nu(u_1)) + f(\nu(u_2)) &\ge 2f\left(\frac{\nu(u_1) + \nu(u_2)}{2}\right) \quad \text{by convexity of } f \\ &\ge 2f(2) \quad \text{because } f \text{ is non-decreasing} \\ &= \frac{4}{9} - \frac{2}{3}\varepsilon. \end{aligned}$$
Hence

$$w^*(v) \le w(v) - \left(\frac{4}{9} - \frac{2}{3}\varepsilon\right) \le \left(\frac{1}{3} + \varepsilon - \frac{\delta}{4}\right) - \frac{4}{9} + \frac{2}{3}\varepsilon = -\frac{1}{9} + \frac{5}{3}\varepsilon - \frac{\delta}{4}.$$

showing the second part of the statement.

◇

**Claim 22.** If C is a connected component of $D_6$, then $\sum_{v \in V(C)} w^*(v) \le 0$.

*Proof of claim.* If C has a unique vertex v, then, by Claim 21, we have $w^*(v) \le 0$ as wanted.

If C has two vertices x and y, then, again by Claim 21, $w^*(x), w^*(y) \le 0$, and so $w^*(x) + w^*(y) \le 0$.

If C is a bidirected path $[x, y, z]$, then, by Claim 16, x and z have both neighbourhood valency at least 4 and so by Claim 21 $w^*(x), w^*(z) \le -\frac{1}{9} - \frac{\varepsilon}{6}$. Moreover, y sends at least $\frac{2}{8}(-\frac{10}{3} + 4 - \varepsilon) = \frac{1}{6} - \frac{\varepsilon}{4}$ to its neighbour out of C. Hence

$$w^*(y) \le w(y) - \left(\frac{1}{6} - \frac{\varepsilon}{4}\right) \le \frac{1}{3} + \varepsilon - \frac{\delta}{3} - \frac{1}{6} + \frac{\varepsilon}{4} = \frac{1}{6} + \frac{5}{4}\varepsilon - \frac{\delta}{3}.$$

Altogether, we get that

$$w^*(x) + w^*(y) + w^*(z) \le \frac{1}{6} + \frac{5}{4}\varepsilon - \frac{\delta}{3} + 2\left(-\frac{1}{9} - \frac{\varepsilon}{6}\right) = -\frac{1}{18} + \frac{11}{12}\varepsilon - \frac{\delta}{3} \le 0$$

because $\delta \ge 6\varepsilon$.

Finally, if C is a bidirected star with centre x and three other vertices y, z, w, then $w^*(x) \le w(x) = \frac{1}{3} + \varepsilon - \frac{\delta}{4}$. Moreover, each of y, z, w has neighbourhood valency at least 4 by Claim 16 and so has final charge at most $-\frac{1}{9} + \frac{5}{3}\varepsilon - \frac{\delta}{4}$ by Claim 21. Hence

$$w^*(x) + w^*(y) + w^*(z) + w^*(w) \le \frac{1}{3} + \varepsilon - \frac{\delta}{4} + 3\left(-\frac{1}{9} + \frac{5}{3}\varepsilon - \frac{\delta}{4}\right) \le 6\varepsilon - \delta \le 0$$

because $\delta \ge 6\varepsilon$.

◇

As a consequence of these last claims, we have $\rho(D) \le \sum_{v \in V(D)} w(v) = \sum_{v \in V(D)} w^*(v) \le 0 \le 1$, a contradiction. This proves Theorem 8. □

# 4 An upper bound on $o_k(n)$

In this section, we show that, for every fixed k, there are infinitely many values of n such that $o_k(n) \le (2k - \frac{7}{2})n$. The proof is strongly based on the proof of [1, Theorem 4.4], which shows $o_k(n) \le (2k - 3)n$ for every k, n (with n large enough). For k = 4, the construction implies in particular that there is a 4-dicritical oriented graph with 76 vertices and 330 arcs, and there are infinitely many 4-dicritical oriented graphs with $m/n \le 9/2$.
**Proposition 26.** Let $k \ge 3$ be an integer. For infinitely many values of $n \in \mathbb{N}$, there exists a $k$-dicritical oriented graph $\vec{G}_k$ on $n$ vertices with at most $(2k - \frac{7}{2})n$ arcs.

*Proof.* Let us fix $n_0 \in \mathbb{N}$. We will show, by induction on $k$, that there exists a $k$-dicritical oriented graph $\vec{G}_k$ on $n$ vertices with at most $(2k - \frac{7}{2})n$ arcs, such that $n \ge n_0$.

When $k=3$, the result is known ([1, Corollary 4.3]). We briefly describe the construction for completeness. Start from any orientation of an odd cycle on $2n_0+1$ vertices. Then for each arc $xy$ in this orientation, add a directed triangle $\vec{C}_3$ and every arc from $y$ to $V(\vec{C}_3)$ and every arc from $V(\vec{C}_3)$ to $x$ (see Figure 5). This gadget forces $x$ and $y$ to have different colours in every 2-dicolouring. Since we started from an orientation of an odd cycle, the result is a 3-dicritical oriented graph on $4(2n_0+1)$ vertices and $10(2n_0+1)$ arcs.

<figure><img src="image_6.png" /><figcaption>Figure 5: A 3-dicritical oriented graph with $\frac{5}{2}n$ arcs.</figcaption></figure>

Let us fix $k \ge 4$ and assume that there exists such a $(k-1)$-dicritical oriented graph $\vec{G}_{k-1}$ on $n_{k-1} \ge n_0$ vertices with $m_{k-1} \le (2(k-1) - \frac{7}{2})n_{k-1}$ arcs. We start from any tournament $T$ on $k$ vertices. Then we add, for each arc $xy$ of $T$, a copy $\vec{G}_{k-1}^{xy}$ of $\vec{G}_{k-1}$, all arcs from $y$ to $\vec{G}_{k-1}^{xy}$ and all arcs from $\vec{G}_{k-1}^{xy}$ to $x$. Figure 6 illustrates a possible construction of $\vec{G}_4$, where $T$ is the transitive tournament on 4 vertices.

<figure><img src="image_7.png" /><figcaption>Figure 6: A 4-dicritical oriented graph with at most $\frac{9}{2}n$ arcs.</figcaption></figure>
Let $\vec{G}_k$ be the resulting oriented graph. By construction, $n_k = |V(\vec{G}_k)|$ and $m_k = |A(\vec{G}_k)|$ satisfy:

$$ 
\begin{align*}
n_k &= k + \binom{k}{2} n_{k-1} \\
m_k &= \binom{k}{2} + \binom{k}{2} \times 2 \times n_{k-1} + \binom{k}{2} \times m_{k-1} \\
&\le \binom{k}{2} + \binom{k}{2} \left(2 + 2(k-1) - \frac{7}{2}\right) n_{k-1} \\
&= \binom{k}{2} + \binom{k}{2} \left(2k - \frac{7}{2}\right) n_{k-1} \\
&= \binom{k}{2} + \left(2k - \frac{7}{2}\right) (n_k - k) \\
&\le \left(2k - \frac{7}{2}\right) n_k
\end{align*}
$$

where in the last inequality we used $k \left(2k - \frac{7}{2}\right) \ge \binom{k}{2}$, which holds when $k \ge 2$. We will now prove that $\vec{G}_k$ is indeed $k$-dicritical.

We first prove that $\vec{\chi}(\vec{G}_k) = k$. Assume that there exists a $(k-1)$-dicolouring $\alpha$ of $\vec{G}_k$. Then there exist $x, y \in V(T)$ such that $\alpha(x) = \alpha(y)$. Since $\vec{\chi}(\vec{G}_{k-1}) = k-1$, there exists $z \in V(\vec{G}_{k-1}^{xy})$ such that $\alpha(z) = \alpha(x)$. But then $(x, y, z, x)$ is a monochromatic directed triangle in $\alpha$: a contradiction.

Let us now prove that $\vec{\chi}(\vec{G}_k \setminus \{uv\}) \le k-1$ for every arc $uv \in A(\vec{G}_k)$. This implies immediately that $\vec{\chi}(\vec{G}_k = k$ and shows the result.

Consider first an arc $uv$ in $A(T)$. We colour each copy $\vec{G}_{k-1}^{xy}$ of $\vec{G}_{k-1}$ with a $(k-1)$-dicolouring of $\vec{G}_{k-1}$. We then choose a distinct colour for every vertex in $T$, except $u$ and $v$ which receive the same colour. This results in a $(k-1)$-dicolouring of $\vec{G}_k \setminus \{uv\}$.

Consider now an arc $uv$ of $\vec{G}_{k-1}^{xy}$ for some $xy \in A(T)$. Because $\vec{G}_{k-1}$ is $(k-1)$-dicritical, there exists a $(k-2)$-dicolouring $\xi$ of $\vec{G}_{k-1}^{xy} \setminus \{uv\}$. Hence we colour $\vec{G}_{k-1}^{xy} \setminus \{uv\}$ with $\xi$, every other copy of $\vec{G}_{k-1}$ a $(k-1)$-dicolouring of $\vec{G}_{k-1}$, and we choose a distinct colour for every vertex in $T$, except $x$ and $y$ which both receive colour $k-1$. This results in a $(k-1)$-dicolouring of $\vec{G}_k \setminus \{uv\}$.

Consider finally an arc $uv$ arc from $u \in V(T)$ to $v \in V(\vec{G}_{k-1}^{uy})$ (the case of $u \in V(\vec{G}_{k-1}^{xv})$ and $v \in V(T)$ being symmetric). Because $\vec{G}_{k-1}$ is dicritical, there exists a $(k-1)$-dicolouring $\gamma$ of $\vec{G}_{k-1}^{uy}$ in which $v$ is the only vertex coloured $k-1$. Hence, we colour $\vec{G}_{k-1}^{uy}$ with $\gamma$, every other copy of $\vec{G}_{k-1}$ with a $(k-1)$-dicolouring of $\vec{G}_{k-1}$, and we choose a distinct colour for every vertex in $T$, except $u$ and $y$ which both receive colour $k-1$. This results in a $(k-1)$-dicolouring of $\vec{G}_k \setminus \{uv\}$. $\square$
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