Haj\'os and Ore constructions for digraphs

The chromatic number $\overrightarrow{\chi}(D)$ of a digraph $D$ is the minimum number of colors needed to color the vertices of $D$ such that each color class induces an acyclic subdigraph of $D$. A digraph $D$ is $k$-critical if $\overrightarrow{\chi}(D) = k$ but $\overrightarrow{\chi}(D')<k$ for all proper subdigraphs $D'$ of $D$. We examine methods for creating infinite families of critical digraphs, the Dirac join and the directed and bidirected Haj\'os join. We prove that a digraph $D$ has chromatic number at least $k$ if and only if it contains a subdigraph that can be obtained from bidirected complete graphs on $k$ vertices by (directed) Haj\'os joins and identifying non-adjacent vertices. Building upon that, we show that a digraph $D$ has chromatic number at least $k$ if and only if it can be constructed from bidirected $K_k$'s by using directed and bidirected Haj\'os joins and identifying non-adjacent vertices (so called Ore joins), thereby transferring a well-known result of Urquhart to digraphs. Finally, we prove a Gallai-type theorem that characterizes the structure of the low vertex subdigraph of a critical digraph, that is, the subdigraph, which is induced by the vertices that have in-degree $k-1$ and out-degree $k-1$ in $D$.


Introduction
Recall that the chromatic number χ(G) of a graph G is the minimum number of colors needed to color the vertices of G so that each color class induces an edgeless subgraph of The topic of critical graphs has received much attention within the last century.Critical graphs were first introduced by G. A. Dirac in his doctoral thesis; famous mathematicians like G Hajós, T. Gallai and others continued developing the theory of critical graphs in the 1960's.

D such that D has no separating vertex.
A directed path is a non-empty digraph P with V (P ) = {v 1 , v 2 , . . ., v p } and A(P ) = {v 1 v 2 , v 2 v 3 , . . ., v p−1 v p } where the v i are all distinct.Furthermore, a directed cycle of length p ≥ 2 is a non-empty digraph C with V (C) = {v 1 , v 2 , . . ., v p } and A(C) = {v 1 v 2 , v 2 v 3 , . . ., v p−1 v p , v p v 1 } where the v i are all distinct.A directed cycle of length 2 is called a digon.A bidirected graph is a digraph that can be obtained from an undirected (simple) graph G by replacing each edge by two opposite arcs, we denote it by D(G).A bidirected complete graph is also called a complete digraph.By The Hajós join is a well-known tool for undirected graphs that can be used to create infinite families of k-critical graphs, see e. g. [8].For digraphs, an equivalent construction was defined by Hoshino and Kawarabayashi in [13].Let D 1 and D 2 be two disjoint digraphs and select an arc u 1 v 1 and an arc v 2 u 2 .Let D be the digraph obtained from the union D 1 ∪ D 2 by deleting the arcs u 1 v 1 as well as v 2 u 2 , identifying the vertices v 1 and v 2 to a new vertex v, and adding the arc u 1 u 2 .We say that D is the (directed) Hajós join of D 1 and D 2 and write D = (D 1 , v 1 , u 1 ) (D 2 , v 2 , u 2 ) or, briefly, D = D 1 D 2 .For the proof of the next theorem, recall that a k-coloring of a digraph D is a coloring of D, in which at most k colors are used.Statement (c) of the following theorem has already been mentioned in [13,Prop. 2]  Proof: ) and let v denote the vertex that is obtained from identifying v 1 and v 2 .For the proof of (a) let In order to prove (b), let We claim that ϕ is a k-coloring of D. For otherwise, D would contain a monochromatic directed cycle C with {u 1 , u 2 , v} ⊆ V (C) and u 1 u 2 ∈ A(C).But then, (C ∩ D 1 ) + u 1 v 1 is a monochromatic directed cycle in D 1 , which is impossible.
For the proof of (c) it suffices to show that assign the same color to v 1 and v 2 and taking the union of those colorings clearly leads to a (k − 1)coloring of D. Let a ∈ A(D) \ {u 1 u 2 }.By symmetry, we may assume that a ∈ A(D 1 ).Then, there is a . By taking the union of those colorings we obtain a (k − 1)-coloring of D and so D is k-critical, as claimed.
To prove statement (d) first assume that  We define the class of Hajós-k-constructible digraphs as the smallest family of digraphs that contains the bidirected complete graph of order k and is closed ander Hajós joins and identifying independent vertices.The next result was proved for undirected graphs by Hajós [8].
Theorem 3 Let k ≥ 3 be an integer.A digraph has chromatic number at least k if and only if it contains a Hajós-k-constructible subdigraph.
For the proof of the above theorem we need a result on perfect digraphs.Recall that the clique number ω(D) of a digraph D is the size of the largest bidirected complete subdigraph of D. A perfect digraph is a digraph D satisfying that for each induced subdigraph H of D it holds → χ(H) = ω(H).Recall that an odd hole is an (undirected) cycle of odd length at least 5 and an odd antihole is the complement of an odd hole.Moreover, a filled odd hole/antihole is a digraph D so that S(D) is an odd hole/antihole, where S(D) is the symmetric part of D, that is, the graph with vertex set V (D) and edge set E(S(D)) = {uv | uv ∈ A(D) and vu ∈ A(D)}.
Andres and Hochstttler [1, Corollary 5] proved the following result on perfect digraphs.Theorem 4 (Andres and Hochstttler) A digraph D is perfect if and only if it contains none of the following as an induced subdigraph: a filled odd hole, a filled odd antihole, or a directed cycle of length at least 3 as induced subdigraph.
This theorem is a really nice and powerful tool in many ways.If D = D(G) is a bidirected graph, then the theorem is equivalent to the Strong Perfect Graph Theorem (SPGT) by Chudnovsky, Robertson, Seymour, and Thomas [4], and hence, the SPGT follows from Andres and Hochstttler's result.Nevertheless, their proof heavily relies on the SPGT.
We distinguish between two cases and show that both of them lead to a contradiction.Case 1: ∼ is transitive.Then, in particular, the relation of being identical or non-adjacent in D is an equivalence relation.This implies that D is a semicomplete multipartite digraph with parts I 1 , I 2 , . . ., I , i.e., I j is an independent set in D for j ∈ {1, 2, . . ., } and for u ∈ I i , v ∈ I j with i = j, A(D) contains at least one of uv and vu.Suppose that there are vertices u ∈ I i , v ∈ I j such that u and v induce a digon in D. Then we claim that each pair of vertices (u , v ) ∈ I i × I j induces a digon in D. Otherwise, (by symmetry) there are vertices u, u ∈ I i and a vertex v ∈ I j such that {uv, vu, u v} ⊆ A(D) but vu ∈ A(D).But then, v ∼ u , u ∼ u and v ∼ u and hence, ∼ is not transitive, a contradiction.This proves the claim that all arcs between I i and I j are bidirected.Moreover, by using a similar argumentation, it is easy to see that if there are vertices u ∈ I i , v ∈ I j with uv ∈ A(D) and vu ∈ A(D), then A D (I j , I i ) = ∅.Now we claim that D is perfect.By Theorem 4 we only need to prove that D does neither contain a filled odd hole, nor a filled odd antihole, nor an induced directed cycle of length at least 3 as an induced subdigraph.
First assume that D contains a filled odd hole C as an induced subdigraph.Let v 1 , v 2 , . . ., v r , v 1 be a cyclic ordering of the vertices of the filled odd hole.By symmetry, we may assume that However, since C is a filled odd hole, this gives us v r ∼ v 3 and so v r ∼ v 3 , v 3 ∼ v 1 , but v 1 ∼ v r , a contradiction.Thus, D cannot contain a filled odd hole as an induced subdigraph.
Next assume that D contains a filled odd antihole C as an induced subdigraph.Let again v 1 , v 2 , . . ., v r , v 1 be a cyclic ordering of the vertices.By symmetry, we may assume that v 1 ∼ v 2 .Then, v 2 v 3 ∈ A(D) as otherwise ∼ would not be transitive.Continuing this argument, we obtain that v i ∼ v i+1 for i odd and Thus, D contains no filled antiholes as induced subdigraphs.
Finally, assume that D contains an directed cycle C of length at least 3 as an induced subdigraph.Again, let v 1 , v 2 , . . ., v r , v 1 be a cyclic ordering of the vertices of C.Then, As a consequence, D is perfect by Theorem 4 and so D contains a bidirected complete graph of order at least k as a subdigraph and, therefore, a Hajós-k-constructible sudigraph, which is impossible.Case 2: ∼ is not transitive.Then there are vertices u, v, w ∈ V (D) such that uv ∈ A(D), vw ∈ A(D), but uw ∈ A(D).By the maximality of D, there exist Hajós-k-constructible subdigraphs D uv ⊆ D + uv and D vw ⊆ D + vw.Let D be the graph obtained from the union (D uv − uv) ∪ (D vw − vw) by adding the arc uw.Then, D is a subdigraph of D that can be obtained from disjoint copies of D uv and D uw as follows.First we apply the Hajós join by removing the copies of the arcs uv and vw, identifying the two copies of v, and adding the arc uw.Afterwards, for each vertex x that belongs to both D uv and D vw , we identify the two copies of x.Hence, D is a Hajós-k-constructible subdigraph of D, a contradiction.This completes the proof.
In the last two decades Hajós' theorem (Theorem 3) became very popular among graph theorists.Hajós like theorems were established for the list chromatic number by Gravier [7] and Král [17], for the circular chromatic number by Zhu [30], for the signed chromatic number by Kang [16], for the chromatic number of edge weighted graphs by Mohar [19], for graph homomorphisms by Nešetril [22], and for Grassmann homomorphism (a homomorphism concept that provides a common generalization of graph colorings, hypergraph colorings and nowhere-zero flows) by Jensen [14].

The Ore construction
Regarding undirected graphs, Urquhart [29] proved that each graph with chromatic number at least k does not only contain a Hajós-k-constructible subgraph but itself is Hajósk-constructible. Recall that a Hajós join of two undirected disjoint graphs G 1 and G 2 is done by deleting two edges e = uv ∈ E(G 1 ) and e = u v ∈ E(G 2 ), identifying the vertices v and v , and adding the edge uu .The aim of this section is to point out that the same result does not hold for digraphs and to prove that, however, a slight modification of the Hajós join does the trick.The proof of the next theorem is straightforward and left to the reader.
Theorem 5 Let k ≥ 3 be an integer and let D be a Hajós-k-constructible digraph.Then, D is strongly connected.
As a consequence of the above theorem, every digraph with chromatic number at least k that is not strongly connected is not Hajós-k-constructible and so Urquhart's Theorem cannot be directly transferred to digraphs.Nevertheless, it turns out that we get an Urquhart-type theorem by further allowing the following join.Let D 1 and D 2 be two digraphs and let . Now let D be the digraph obtained from the union D 1 ∪ D 2 by deleting both arcs between u 1 and v 1 as well as both arcs between u 2 and v 2 , identifying the vertices v 1 and v 2 to a new vertex v, and adding both arcs u 1 u 2 and u 2 u 1 .We say that D is the bidirected Hajós join of D 1 and D 2 and write Note that for the proof of statement (b), we use the fact that k ≥ 3 and so we can choose ϕ 1 and ϕ 2 such that ϕ 1 (v 1 ) = ϕ 2 (v 2 ) and ϕ 1 (u 1 ) = ϕ 2 (u 2 ).For k = 2, the statement is not true: for example, ).The same trick works for statement (c).
For the proof of his Theorem, Urquhart even used a more restricted class of constructible (undirected) graphs than the class of Hajós-k-constructible graphs, which originally was introduced by Ore [25, Chapter 11].Transferred to digraphs, we get the following.Let D 1 and D 2 be two vertex-disjoint digraphs, let u 1 v 1 be an arc of D 1 , and let v 2 u 2 be an arc of D 2 .Furthermore, let ι : by identifying w with ι(w) for each w ∈ A 1 .Then, D is a directed Ore join of D 1 and D 2 and we write ) by identifying w with ι(w) for each w ∈ A 1 is a bidirected Ore join of D 1 and D 2 and we write We define the class of Ore-k-constructible digraphs as the smallest family of digraphs that contains ↔ K k and is closed under (directed and bidirected) Ore joins.The proof of Theorem 3 immediately implies the following theorem (see [25] for the undirected analogue).In particular, here we do not need any bidirected Ore joins.
Theorem 7 Let k ≥ 3 be an integer.A digraph has chromatic number at least k if and only if it contains an Ore-k-constructible subdigraph.
Urquhart proved the following, thereby answering a conjecture by Hanson, Robinson and Toft [9] (the conjecture was also proposed by Jensen and Toft in their book on graph coloring problems [15]).Recall that the Ore join of two undirected graphs is done via an undirected Hajós join and identification afterwards.
Theorem 8 Let k ≥ 3 be an integer.For a graph G the following conditions are equivalent: Note that if G is the Hajós join of two graphs G 1 and G 2 , then D(G) is the bidirected Hajós join of D(G 1 ) and D(G 2 ).Furthermore, → χ(D(G)) = χ(G) and so the above theorem immediately implies the following.
Observation 9 Each bidirected graph with chromatic number at least k ≥ 3 is Ore-kconstructible.
Now we have all the tools that we need in order to prove our Urquhart-type theorem.Thus, it suffices to show that each digraph with chromatic number ≥ k is Ore-kconstructible.We will do this via a sequence of claims.In the following, we will denote by Proof : It is clear that To this end, let D 1 (respectively D 2 ) be the bidirected graph obtained by identifying a vertex of 2a).Let ι be the bijection with ι(v i ) = v i for all i ∈ {1, 2, . . ., k} and let D 2 = (D 1 , u, v 1 ) o ι (D 2 , u 2 , u 1 ) (see Figure 2 (f) (g) Figure 2: The construction of Claim 2.
By using a similar construction starting with the graph that results from ↔ K k by adding a vertex v and joining it to two vertices of ↔ K k by arcs in both directions we obtain the following.For the exact construction see Figure 3.

Claim 3 The digraphs
From now on, we may argue similar to the original proof of Urquhart.The next claim can easily be deduced from Claim 1.  Proof : We distuingish between two cases.Case 1: One end-vertex of the arc a belongs to X.Then, let a = uv with u ∈ X and v ∈ V \ X (the case a = vu can be done analogously).Furthermore, let D be a copy and a high vertexof D, otherwise.Furthermore, let D L denote the digraph that is induced by the set of low vertices of D; we will call it the low vertex subdigraph of D. For undirected graphs, Gallai [5] proved that the low vertex subgraph has a specific structure.The next theorem transfers his result to digraphs.For the proof of Theorem 11 we will use a theorem of Harutyunyan and Mohar [12] concerning list-colorings of digraphs.Given a digraph D, a list-assignment L is a function that assigns each vertex v ∈ V (D) a set (list) of colors.An L-coloring ϕ of D is a coloring of D such that ϕ(v) ∈ L(v) for all v ∈ V (D).
Theorem 12 ([12]) Let D be a connected digraph, and let L be a list-assignment such that . Suppose that D is not L-colorable.Then, D is Eulerian and for every block B of D one of the following cases occurs: The next proposition states some important facts that will be needed for the proof of Theorem 11.Proof: Suppose (by symmetry) that there is a color α ∈ C such that α does not appear in N + (v).Then, coloring v with α cannot create a monochromatic cycle in D (as v has no out-neighbor with color α) and, thus, D would be (k − 1)-colorable, a contradiction.
For the proof of (b), assume (by symmetry) that uv ∈ A(D).Then it follows from (a) that after uncoloring u, v has no in-neighbor with color ϕ(u) and so coloring v with color ϕ(u) cannot create a monochromatic cycle.
In the following, we will call the procedure that is described in Proposition 13 (b) shifting the color from u to v and briefly write u → v. Now let D be a k-critical digraph, let C be a (not necessarily directed) cycle in D L and let v ∈ V (C).Moreover, let ϕ be a (k − 1) coloring of D − v and let u and w be the vertices such that u, v and w are consecutive in C.Then, beginning with u → v, we can shift each vertex of C, one after another, clockwise and obtain a new (k − 1)-coloring of D − v (see Figure 4).Similar, beginning with w → v, we can shift each vertex of C counter-clockwise and obtain a third (k − 1)-coloring of D − v.The main idea for this goes back to Gallai [5]; we will use this observation frequently in the following.B (v) = 0).Thus, it follows from Proposition 13(a) that ϕ(w) = ϕ(u), say ϕ(w) = β.Moreover, we obtain that the vertices of C (except from v) are colored alternately with β and α.Otherwise, there are two consecutive vertices x, x on C such that {ϕ(x), ϕ(x )} = {α, β}.Then we can shift the colors around the vertices of C such that u gets color ϕ(x) and w gets color ϕ(x ) and obtain a (k − 1)-coloring ϕ of D − v with {ϕ (u), ϕ (w)} = {α, β}, which contradicts Proposition 13(a) as C is induced and so no neighbors of v besides u and w take part in the shifting.
As a consequence, C has odd length.Now let v = v 1 , w = v 2 , v 3 , . . ., u = v r , v 1 be a cyclic ordering of the vertices of C. We claim that v 3 v 2 ∈ A(D).Assume, to the contrary, v 3 v 2 ∈ A(D).Then, we can shift w → v and obtain a coloring ϕ of D − w with ϕ (v) = β and ϕ (v 3 ) = α.In particular, v 3 is the only in-neighbor of w that has color α with respect to ϕ .On the other hand, beginning from ϕ with u → v, we can shift every vertex besides v clockwise around C (the last shift is w → v 3 ) and get a (k − 1)-coloring ϕ * of G − w with ϕ * (v) = α and ϕ * (v 3 ) = β.As vw ∈ A(D) and as C is induced, it follows that w has no in-neighbor that has color α with respect to ϕ * , a contradiction.Hence, v 3 v 2 ∈ A(D) and so v 2 v 3 ∈ A(D).By repeating this argumentation, we obtain that v i+1 v i ∈ A(D) but v i v i+1 ∈ A(D) for i ≥ 2 even and that v i v i+1 ∈ A(D) but v i+1 v i ∈ A(D) for i ≥ 3 odd.In particular, this leads to v r v ∈ A(D), a contradiction.This proves the claim.
. Moreover, B is not L-colorable, as the union of any L-coloring of B with ϕ would clearly lead to a (k − 1)-coloring of D. Hence, we can apply Theorem 12 and so B is a directed cycle, or an odd bidirected cycle, or a bidrected complete graph, as claimed.
In the undirected case, Gallai [5] showed that the only blocks of the low vertex graph are complete graphs or odd cycles.Although for digraphs the directed cycles arise naturally, it may surprise that there can also be blocks that consist of just one arc.That this indeed may happen is illustrated in Figure 5, where we show the Hajós join of two Gallai used the characterization of the low vertex subgraph of critical graphs he obtained in [5] to establish a lower bound for the number of edges of critical graphs.We can apply the same approach to obtain a similar bound for the number of arcs in critical digraphs, see also [27].
Since every vertex of U has total degree 2k in D (i.e., d On the other hand, since every vertex in W has total degree at least 2k + 1, we obtain that 2a( Adding the first inequality to the second inequality multiplied with (2k Thus the proof is complete.

Open Questions
Since the field of critical digraphs is still wide open, a lot of questions immediately come to mind.It follows from Theorem 3 that each k-critical digraph is Hajós-k-constructible.However, the proof of Theorem 3 is not constructive at all and the authors feel quite embarrassed in admitting that they could not even manage to construct a bidirected cycle of length five from ↔ K 4 's using Hajós joins and identification of non-adjacent vertices.Thus, we want to pose the following question.
Question 1 How can a bidirected C 5 be constructed from ↔ K 4 's by only using Hajós joins and identifying non-adjacent vertices?
Building upon this question, it is of particular interest to study the connection of the Hajós construction to computational complexity.In the undirected case, Mansfield and Welsh [18] stated the problem of determining the complexity of the Hajós construction.They noted that if for any k ≥ 3 there would exist a polynomial P such that every graph of order n with chromatic number k contains a Hajós-k-constructible subgraph that can be obtained by at most P (n) uses of the Hajós-join and identification of non-adjacent vertices, then NP = coNP.Hence, it is very likely that the Hajós construction is not polynomially bounded but not much progress has been made on this problem yet.Pitassi and Urquhart [26] found a linkage to another important open problem in logic; they proved that a restricted version of the Hajós construction is polynomially bounded if and only if extended Frege systems are polynomially bounded.
Question 2 For k ≥ 3, is there a polynomial P such that every digraph of order n contains a Hajós-k-constructible subdigraph that can be obtained from ↔ K k 's by at most P (n) uses of the Hajós-join and identification of non-adjacent vertices?
A beautiful theorem of Gallai [6] states that any k-critical graph with order at most 2k − 2 and k ≥ 2 is the Dirac join of two disjoint non-empty critical graphs (for the Dirac join of two undirected graphs just add all possible edges between the two graphs G 1 and G 2 ).Within the last decades, various different proofs of this theorem have been published (see e.g.[21] and [28]).Clearly, a graph G is the Dirac join of two disjoint non-empty graphs if and only if G is disconnected and so most of the proofs use matching theory for the complement graph G.However, it is yet unclear how to do this for digraphs.In coloring theory of digraphs, it is often of particular interest how digon-free digraphs behave.For example, it was shown by Harutyunyan in his PhD thesis [10] that almost all tournaments of order n have chromatic number at least 1 2 ( n log n+1 ).As a consequence, if n is large enough then for some k ≥ 1 2 ( n log n+1 ) there are k-critical digon-free digraphs on at most n vertices.This leads to our final question.In fact, Brooks' theorem for digraphs [20] implies that N (k) ≥ 2k for k ≥ 3.Moreover, some small values are already known: the directed triangle shows that N (2) = 3, and Neumann-Lara [24] proved that N (3) = 7, N (4) = 11, and 17 ≤ N (5) ≤ 19; he conjectured that N (5) = 17.

↔K 2 )
k we denote the bidirected complete graph on k vertices.It is easy to see that → χ(D(G)) = χ(G) and D(G) is critical with respect to → χ if and only if G is critical with respect to χ. 3 Construction of critical digraphs Let D 1 and D 2 be two disjoint digraphs.Let D be the digraph obtained from the union D 1 ∪ D 2 by adding all possible arcs in both directions between D 1 and D 2 , i.e., V (D) = V (D 1 ) ∪ V (D 2 ) and A(D) = A(D 1 ) ∪ A(D 2 ) ∪ {uv, vu | u ∈ V (D 1 ) and v ∈ V (D 2 )}.We say that D is the Dirac join of G 1 and G 2 and denote it by D = D 1 D 2 .The proof of the next theorem is straightforward and therefore left to the reader.Theorem 1 (Dirac Construction) Let D = D 1 D 2 be the Dirac join of two disjoint non-empty digraphs D 1 and D 2 .Then, and D is critical if and only if both D 1 and D 2 are critical.

Figure 1 : 3 Theorem 2 (
Figure 1: The Hajós join of two directed cycles of length 3 and the union of ϕ 1 and ϕ 2 is a (k − 1)-coloring of D. Hence, → χ(D 1 ) ≥ k and, by symmetry, we obtain → χ(D 2 ) ≥ k.In order to complete the proof we need to show that → χ(D i − a) < k for i ∈ {1, 2} and for a ∈ A(D i ).By symmetry, it suffices to show this for D 1 in D − a with respect to ϕ, a contradiction.This completes the proof.Another common operation for graphs and digraphs is the identification of independent sets.Let D be a digraph and let I be a non-empty independent set of D. Then, we can create a new digraph H from D − I by adding a new vertex v = v I and adding all arcs from v to N + H (I) = v∈I N + H (v) and all arcs from N − H (I) = v∈I N − H (v) to v. We say that H is obtained from D by identifying I with v, or briefly by identifying independent vertices and write H = D/(I → v) (briefly H = D/I).It is obvious that any k-coloring of D/I can be extended to a k-coloring of D by coloring each vertex of I with the color of v I .Thus,

Proof of Theorem 3 :
Let k ≥ 3 be an integer.Clearly, every Hajós-k-constructible digraph has chromatic number at least k (by Theorem 2 and since → χ(D/I) ≥ → χ(D) for each independent set I of a digraph D).This proves the "if"-implication.The proof of the "only if"-implication is by reductio ad absurdum.Let D be a maximal counter-example in the sense that D does not contain a Hajós-k-constructible subdigraph but adding a new arc a ∈ A(D) to D implies the existence of a Hajós-k-constructible subdigraph D a of D + a with a ∈ A(D a ).As D is not Hajós

1 ↔D 2
the bidirected Hajós join is the exact analogue of the undirected Hajos join.By a slight modification of the proof of Theorem 2(a)-(c) one can easily show that the following holds.Theorem 6 (Bidirected Hajós Construction) Let D = D result from the bidirected Hajós join of two disjoint non-empty digraphs D 1 and D 2 .Then, the following statements hold:

Theorem 10
Let k ≥ 3 be an integer.A digraph has chromatic number at least k if and only if it is Ore-k-constructible.Proof: It immediately follows from Theorem 2(a) and Theorem 6(a) that each Ore-kconstructible digraph has chromatic number at least k.

Claim 2
the digraph that results from ↔ K k by adding a vertex v and the arc uv (respectively vu) for some vertex u of the ↔ K k .Moreover, let ↔ K k + a be the digraph that results from ↔ K k by adding two new vertices u, v and the arc a = uv.Finally, O k denotes the class of Ore-k-constructible digraphs and O * k denotes the class of Ore-k-constructible digraphs containing a ↔ K k .It follows from Observation 9 that Claim 1 The digraph obtained from ↔ K k by adding an isolated vertex belongs to O * k .The digraph ↔ K k + a belongs to O * k .
(a) and (b)).This Ore-join leads to the digraph D 2 = D 2 − u 2 u 1 (see Figure 2(c)).By v * i we denote the vertex that results from identifying v i with ι(v i ) = v i .(a) The digraphs D 2 (on the left) and D 1 (on the right).Now we take a new copy of D 1 , define ι to be the bijection with ι (v * i ) = v i+1 for all i ∈ {1, 2, . . ., k} (where v k+1 = v 1 ), and set D = (D , u 1 , v * ) ↔ o ι (D 1 , u, v 1 ) (see Figure 2(d)(e)).Still, let v * i denote the vertex that results from identifying v * i with ι (v * i ).Finally, we take another copy of G 1 , set ι (v * i ) = v i+1 for i ∈ {1, 2, . . ., k} (where v k+1 = v 1 ) and perform the Ore join (D 2 , u 2 , v * ) ↔ o ι (G 1 , v 1 , u) (see Figure 2(f)(g)).This gives us the digraph K k + u 1 u 2 as required.(b) The graph we obtain after the directed Hajs join.We then identify the vertices of the same colour.(c) This leads to the graph D 2 .(d) (e) The graph D 2 .
(a) We start by performing a directed Hajós join between the two depicted graphs.(b) Afterwards, we identify the vertices of the same color.(c) End of the first step.(d) Now we perform a bidirected Hajós join.(e) Again we identify vertices of the same color.(f) This gives us the digraph K k + → v .

vCase 2 :
and let u be the vertex adjacent to v in D .Finally, let w, z ∈ X \ {u} and let w , z ∈ D \ {u , v }.By Claim 3, D ∈ O k .Now let ι be a bijection from (X \ {u}) ∪ {v} to (X \ {u }) ∪ {v } with ι(v) = v , ι(w) = z , and ι(z) = w .Then, (D, u, w) ↔ o ι (D , u , w ) ∈ O k is a copy of D + a, and we are done.No end-vertex of a belongs to X.Then, let a = uv, and D be a copy of ↔ K k + u v .By Claim 2, D belongs to O k .Now let x, y, z be three vertices from X and let {x , y , z } ⊆ D \ {u, v}.Finally, let ι be a bijection from X \ {x} ∪ {u, v} to D \ {x } with ι(u) = u , ι(v) = v , ι(y) = z , and ι(z) = y .Then, (D, x, y) ↔ o ι (D , x , y ) ∈ O k is a copy of D + a and the proof of the claim is complete.It follows from Claims 4 and 5 that each digraph containing a ↔ K k belongs to O * k .The remaining part of the proof is by reductio ad absurdum.Let D be a maximal counterexample in the sense that → χ(D) ≥ k, D is not Ore-k-constructible, and D has maximum number of edges with respect to this property.Then, D does not contain a ↔ K k and if a ∈ A(D), D + a belongs to O k .Now we argue as in the proof of Theorem 3.For two vertices u, v ∈ V (D), let u ∼ v denote the relation that uv ∈ A(D).If ∼ is transitive we again conclude that D contains a ↔ K k , a contradiction.Hence, ∼ is not transitive and so there are vertices u, v, w ∈ V (D) with uv ∈ A(D), vw ∈ A(D), but uw ∈ A(D).Then, both digraphs D + uv as well as D + vw belong to O k and D is the Ore join of two disjoint copies of these two digraphs.Thus, D belongs to O k , a contradiction.5 A Gallai-type theorem for critical digraphs Let D be a k-critical digraph.If v ∈ V (D), then D − v admits a (k − 1)-coloring and, since → χ(D) = k, v must have an out-and an in-neighbor in each color class of such a coloring.Hence, we have k − 1 ≤ min{d + D (v), d − D (v)} for every vertex v ∈ V (D), which gives us a natural way to classify the vertices of D. We say that a vertex

Theorem 11
Let D L be the low vertex subdigraph of a k-critical digraph D.Then, each block B of D L satisfies one of the following statements.
(a) B consists of just one single arc.(b)B is a directed cycle of length ≥ 2.(c) B is a bidirected cycle of odd length.
(d) B is a complete bidirected graph.

Proposition 13
Let D L be the low vertex subdigraph of a k-critical digraph D.Moreover, given a vertex v ∈ V (D L ), let ϕ be a (k − 1)-coloring of D − v with color set C. Then the following statements hold: (a) Each color from C appears exactly once in N + (v) and in N − (v).(b) If u ∈ V (D L ) is adjacent to v, then uncoloring u and coloring v with the color of u leads to a (k − 1)-coloring of D − u.

Proof of Theorem 11 :K 1
Let D L be the low vertex subdigraph of a k-critical digraph D and let B be an arbitrary block of D L .If |B| = 1, then B = ↔ and we are done.If |B| = 2, then either B consists of just one arc or B is a bidirected complete graph and so there is nothing to show.Thus, we may assume |B| ≥ 3. w

Figure 4 :Claim 1
Figure 4: The black vertex denotes the clockwise shifting around a cycle.

Question 3
Let k ≥ 3 be an integer.Is there a k-critical digraph D on at most 2k − 2 vertices that is not the Dirac join of two proper digraphs D 1 and D 2 ?

Question 4
For fixed k ≥ 3, what is the minimum integer N (k) such that there is a k-critical digon-free digraph on N (k) vertices?As k − 1 ≤ min{d + D (v), d − D (v)} for all vertices v of a k-critical digraph D, we trivially have N (k) ≥ 2k − 1.