Bispindles in strongly connected digraphs with large chromatic number

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Introduction
Throughout this paper, the chromatic number of a digraph D, denoted by χ(D), is the chromatic number of its underlying graph.In a digraph D, a directed path, or dipath is an oriented path where all the arcs are oriented in the same direction, from the initial vertex towards the terminal vertex.
A classical result due to Gallai, Hasse, Roy and Vitaver is the following.
This raises the following question.
Question 2. Which digraphs are subdigraphs of all digraphs with large chromatic number ?
A famous theorem by Erdős [9] states that there exist graphs with arbitrarily high girth and arbitrarily large chromatic number.This means that if H is a digraph containing a cycle, there exist digraphs with arbitrarily high chromatic number with no subdigraph isomorphic to H. Thus the only possible candidates to generalise Theorem 1 are the oriented trees that are orientations of trees.Burr [6] proved that every (k − 1) 2 -chromatic digraph contains every oriented tree of order k and made the following conjecture.
The best known upper bound, due to Addario-Berry et al. [2], is in (k/2) 2 .However, for paths with two blocks (blocks are maximal directed subpaths), the best possible upper bound is known.
Theorem 4 (Addario-Berry et al. [1]).Let P be an oriented path with two blocks on n > 3 vertices, then every digraph with chromatic number (at least) n contains P .
The following celebrated theorem of Bondy shows that the story does not stop here.
Theorem 5 (Bondy [4]).Every strong digraph of chromatic number at least k contains a directed cycle of length at least k.
The strong connectivity assumption is indeed necessary, as transitive tournaments contain no directed cycle but can have arbitrarily high chromatic number.
Observe that a directed cycle of length at least k can be seen as a subdivision of C k , the directed cycle of length k.Recall that a subdivision of a digraph F is a digraph that can be obtained from F by replacing each arc uv by a directed path from u to v. Cohen et al. [8] conjecture that Bondy's theorem can be extended to all oriented cycles.Conjecture 6 (Cohen et al. [8]).For every oriented cycle C, there exists a constant f (C) such that every strong digraph with chromatic number at least f (C) contains a subdivision of C.
The strong connectivity assumption is also necessary in Conjecture 6 as shown by Cohen et al. [8].This follows from the following result.On the other hand, Cohen et al. [8] proved conjecture for cycles with two blocks and the antidirected cycle of length 4.More precisely, denoting by C(k, ℓ) the cycle on two blocks, one of length k and the other of length ℓ, Cohen et al. [8] proved the following result.
A p-spindle is the union of k internally disjoint (x, y)-dipaths for some vertices x and y.Vertex a is said to be the tail of the spindle and b its head.A (p + q)-bispindle is the internally disjoint union of a p-spindle with tail x and head y and a q-spindle with tail y and head x.In other words, it is the union of k 1 (x, y)-dipaths and k 2 (y, x)-dipaths, all of these dipaths being pairwise internally disjoint.Note that 2-spindles are the cycles with two blocks and the (1 + 1)-bispindles are the directed cycles.In this paper, we generalize this and study the existence of spindles and bispindles in strong digraphs with large chromatic number.First, we give a construction of digraphs with arbitrarily large chromatic number that contains no 3-spindle and no (2 + 2)-bispindle.Therefore, the most we can expect in all strongly connected digraphs with large chromatic number are (2 + 1)-bispindle.Let B(k 1 , k 2 ; k 3 ) denote the digraph formed by three internally disjoint paths between two vertices x, y, two (x, y)-directed paths, one of size at least k 1 , the other of size at least k 2 , and one (y, x)-directed path of size at least k 3 .We conjecture the following.Conjecture 9.There is a function g : N 3 → N such that every strong digraph with chromatic number at least g(k 1 , k 2 , k 3 ) contains a subdivision of B(k 1 , k 2 ; k 3 ).
As an evidence, we prove this conjecture for k 2 = 1 and arbitrary k 1 and k 3 .In Section 4, we first investigate the case k 2 = k 3 = 1.We first prove in Proposition 22 that very strong digraph D with χ(D) > 3 contains a subdivision of B(2, 1; 1).We then prove the following.In Section 5, using the same approach but in a more complicated way, we prove our main result: We prove the above theorem for a huge constant γ k .It can easily be lowered.However, we made no attempt to it here for two reasons: firstly, we would like to keep the proof as simple as possible; secondly using our method, there is no hope to get an optimal or near optimal value for γ k .
Similar questions with χ replaced by another graph parameter can be studied.We refer the reader to [3] and [8] for more exhaustive discussions on such questions.Let us just give one result proved by Aboulker et al. [3] which can be seen as an analogue to Conjecture 9.

Definitions and preliminaries
We follow standard terminology as used in [5].We denote by [k] the set of integers {1, . . ., k}.
Let F be a digraph.A digraph D is said to be F -subdivision-free, if it contains no subdivision of F .The union of two digraphs D 1 and D 2 is the digraph A digraph is connected (resp.2-connected) if its underlying graph is connected (resp.2-connected.The connected components of a digraph are the connected components of its underlying graph.A digraph D is strongly connected or strong if for any two vertices x, y there is directed path from x to y.The strong components of a digraph are its maximal strong subdigraphs. Let G be a graph or a digraph.A proper k-colouring of G is a mapping φ : V (G) → [k] such that φ(u) = φ(v) whenever u is adjacent to v. G is k-colourable if it admits a proper k-colouring.The chromatic number of G, denoted by χ(G), is the least integer k such that G is k-colourable.
A (directed) graph G is k-degenerate if every subgraph H of G has a vertex of degree at most k.The following proposition is well-known.
Theorem 14 (Brooks).Let G be a connected graph.Then χ(G) ≤ ∆(G) unless G is a complete graph or an odd cycle.
The following easy lemma is well-known.
For each i, let φ i be a proper colouring of D i using colours in [k 1 ] and let φ ′ be a proper colouring of . Let x and y be adjacent vertices of D. If they belong to the same subdigraph D i , then φ i (x) = φ i (y) and so φ(x) = φ(y).If they do not belong to the same component, then the vertices corresponding to these vertices in D C are adjacent and so φ(x) = φ(y).Thus φ is a proper colouring of D using k 1 • k 2 colours.
Proof.Let T be a strong tournament of order 2k − 1.By Camion's Theorem, it has a hamiltonian directed cycle C = (v 1 , v 2 , . . ., v 2k−1 , v 1 ).If there exists an arc v i v j with j − i ≥ k (indices are modulo 2k − 1), then the union of C[v i , v j ], (v i , v j ) and C[v j , v i ] is a B(k, 1; 1)-subdivision.Henceforth, we may assume that Let F be a subdigraph of a digraph D. A directed ear of F in D is a directed path in D whose ends lie in F but whose internal vertices do not.The following lemma is well known.
Lemma 18 (Proposition 5.11 in [5]).Let F be a nontrivial proper 2-connected strong subdigraph of a 2connected strong digraph D. Then F has a directed ear in D.
We will need the following lemmas: , and let l be a positive integer.If p ≥ l k , then there exists a set L of l indices such that for any i, j ∈ L with i < j the following holds : u i = u j and u t > u i , for all i < t < j.
Proof.By induction on k, the result holding trivially when k = 1.Assume now that k > 1.Let L 1 be the elements of the sequence with value 1.If L 1 has at least l elements, we are done.If not, then there is a subsequence σ , then there exists a subsequence of m consecutive integers such that the last one is the largest.Proof.By induction on k, the result holding trivially when k = 1.Let i be the smallest integer such that u t ≤ k − 1 for all t ≥ i.If i > m, then u i−1 = k, and the subsequence of the i − 1 first elements of σ is the desired sequence.If i ≤ m, apply the induction on σ ′ = (u t ) i≤t≤p which is a sequence of more than (k − 1)(m − 1) integers in [k − 1], to get the result.
Proof.Let D k,4 an acyclic digraph with chromatic number greater than k and with every cycle having at least four blocks obtained by Theorem 7. Let S = {s 1 , . . ., s l } be the set of vertices of D k,4 with out-degree 0 and T = {t 1 , . . ., t m } the set of vertices with in-degree 0.
Consider the digraph D obtained from D k,4 as follows.Add a dipath P = (x 1 , x 2 , . . ., x l , z, y 1 , y 2 , . . ., y m ) and the arc s i x i for all i ∈ [l] and y j t j for all j ∈ [m].It is easy to see that D is strong.Moreover, in D, every directed cycle uses the arc x l z.Therefore D does not contain a (2 + 2)-bispindle, which has two arc-disjoint directed cycles.
Suppose now that D has a 3-spindle with tail u and head v, and let Q 1 , Q 2 , Q 3 be its three (u, v)-dipaths.Observe that u and v are not vertices of P , because all vertices of this dipath have either in-degree 2 or out-degree 2. In D each cycle with two blocks between vertices outside P must use the arc x l z.The union of Q 1 and Q 2 form a cycle on two blocks, which means one of the two paths, say Q 1 , contains x l z.But Q 2 and Q 3 also form a cycle on two blocks, but they cannot contain x l z, a contradiction.We will prove that every B(k, 1; 1)-subdivision-free strong digraph D has bounded chromatic number in the following way: Take C a maximal nice collection of cycles.We will prove that every component S of C induces a digraph D[S] on D of bounded chromatic number.Then we will prove that, since it contains no long directed cycle and it is strong, D C has bounded chromatic number, which, by Lemma 16, allows us to conclude.
We will need the following lemma: Lemma 23.Let C be a nice collection of cycle in a B(k, 1; 1)-subdivision-free digraph D and let C, C ′ be two cycles of the same component S of C.There is no dipath P from C to C ′ whose arcs are not in A( S).
Proof.By the contrapositive.We suppose that there exists such a dipath P and show that there is a subdivision of B(k, 1; 1) in D.
By definition of S, there exists a dipath Q from C to C ′ in S. By choosing C and C ′ such that Q is as small as possible, then s(Q) = t(P ) and t(Q) = s(P ) (note that s(Q) and t(Q) can be the same vertex).
Since C has length at least 2k − 2, either C[t(Q), s(P )] has length at least k − 1 or C[s(P ), t(Q)] has length at least k.
] is a subdivision of B(k, 1; 1) between s(Q) and t(P ).
] is a subdivision of B(k, 1; 1) between s(P ) and t(Q).Proof.First note that since D is strong, then so is D C .Suppose χ(D C ) ≥ 2k − 2. By Bondy's Theorem, there exists a directed cycle C = x 1 . . .x l of length at least 2k − 2 in D C .We derive a cycle C ′ in D the following way: Suppose the vertex x i corresponds to a component S i of C: the arc x i−1 x i corresponds in D to an arc whose head is a vertex p i of S i , and the arc x i x i+1 corresponds to an arc whose tail is a vertex l i of S i .Let P i be a dipath from p i to l i in D[S i ].Note that P i intersects each cycle of S i on a, possibly empty, subpath of P i .Then C ′ is the cycle obtained from C by replacing the vertices x i by the path P i .
C ′ is a cycle of D of length at least 2k − 2 because it is no shorter than C. Let C 1 be a cycle of C. By construction of C ′ and D C , C ′ and C 1 can intersect only along a subpath of one P i .Suppose this dipath is more than just one vertex.Let x and y the initial and terminal vertex of this path.Then the union of So C ′ is a cycle of length at least 2k − 2, intersecting each cycle of C on at most one vertex, and which does not belong to C, for otherwise would be reduced to one vertex in D C .This contradicts the fact that C is maximal.
So we can finally prove Theorem 10.
Proof of Theorem 10.Let C be a maximal nice collection of cycle in D. Lemmas 24, 25 and 16 give the result.

B(k, 1; k)
In this section, we present a proof of Theorem 11.
We prove the result by the contrapositive.We consider a digraph D that contains no subdivision of B(k, 1; k).We shall prove that χ(D) Our proof heavily uses the notion of k-suitable collection of directed cycles, which can be seen as a generalization of the notion of nice collection of cycles used to prove Theorem 10.
A collection C of directed cycles is k-suitable if all cycles of C have length at least 8k, and any two distinct cycles C i , C j ∈ C intersect on a subpath P i,j of order at most k.We denote by s i,j (resp.t i,j ) the initial (resp.terminal) vertex of P i,j .the notion of nice collection seen before.
The proof of Theorem 11 uses the same idea as Theorem 10: take a maximal k-suitable collection of directed cycles C; show that the digraph D C obtained by contracting the components of C has bounded chromatic number, and that each component also has bounded chromatic; conclude using Lemma 16.However, because the intersection of cycles in this collection are more complicated and because there might be arcs between cycles of the same component, bounding the chromatic number of the components is way more challenging.The next subsection is devoted to this.

k-suitable collections of directed cycles
. Then exactly one of the following holds: Proof.Observe first that since C 2 has length at least 8k and P 1,2 has length at most k − 1, the sum of lengths of C 2 [t 1,2 , v] and C[v, s 1,2 ] is at least 7k + 1.Similarly, the sum of lengths of C 2 [t 1,3 , v] and C[v, s 1,3 ] is at least 7k + 1.In particular, if (i) holds, then (ii) does not hold and vice-versa.
Suppose for a contradiction that both (i) and (ii) do not hold.By symmetry and the above inequalities, we may assume that both C 2 [t 1,2 , v] and C 3 [v, s 1,3 ] have length more than 3k.But v / ∈ V (C 1 ), so v / ∈ V (P 1,3 ).Thus C 3 [v, t 1,3 ] has also length at least 3k.
If there is a vertex in ] would have length less than 2k (since it would be contained in P 2,3 ∪ P 1,3 and each of those paths has length less than k), a contradiction.Hence Let C be a k-suitable collection of directed cycles.For every set of vertices or digraph S, we denote by C ∩ S the set of cycles of C that intersect S.
Let C 1 ∈ C. For each C j ∈ C ∩ C 1 , let Q j be the subdipath of C j containing all the vertices that are at distance at most 3k from P 1,j in the cycle underlying C j .Then the dipath C j [s(Q j ), s 1,j ] and Observe that Lemma 26 implies directly the following.Proof.Suppose it is not the case.Then there are two distinct cycles C 2 , C 3 of C ∩ A that intersect with C 1 .
Observe that there is a sequence of distinct cycles . Free to consider the first C * j = C 2 in this sequence such that V (C * j ) ⊆ A in place of C 3 , we may assume that all C * j , 2 ≤ j ≤ q − 1, have all their vertices in A. In particular, there exist a Lemma 29.Let C be a k-suitable collection of directed cycles in a B(k, 1; k)-subdivision-free digraph.For any cycle C 1 ∈ C, the digraph I + (C 1 ) has no directed cycle.
Proof.Suppose for a contradiction that I + (C 1 ) contains a directed cycle C ′ .Clearly, it must contain arcs from at least two Q + j .Assume that C ′ contains several vertices of Q + j .Necessarily, there must be two vertices x, y of x] is also a directed cycle in I + (C 1 ).Free to consider this cycle, we may assume that C ′ ∩ Q + j is a directed path.Doing so, for all j, we may assume that C ′ ∩ Q + j is a directed path for every C j ∈ C ∩ C 1 .Without loss of generality, we may assume that there are cycles C 2 , . . ., C p such that • for all 2 ≤ j ≤ p, C ′ ∩ Q + j is a directed path P + j with initial vertex a j and terminal vertex b j ; • the a j and the b j appear according to the following order around with possibly a j+1 = b j for some 1 ≤ j ≤ p where a p+1 = a 2 .
For 2 ≤ j ≤ p, set B j = C j [b j , a j ].Note that B j has length at least 4k, because Q + 2 has length less than 3k.Consider the closed directed walk W contains a directed cycle C W . Wihtout loss of generality, we may assume that this cycle is of the form for some vertex v ∈ B 2 ∩ B q .(The case when W is a directed cycle corresponds to p + 1 and B 2 = B p+1 .) Note that necessarily, q ≥ 4, for ] has length at least k.Indeed, if q = p + 1, then it follows from the fact that B 2 has length as least 4k; if 5 ≤ q ≤ p, then it comes form the fact that Set Lemma 30.Let C be a k-suitable collection of directed cycles in a B(k, 1; k)-subdivision-free digraph.Let φ be a partial colouring of a cycle C 1 ∈ C such that only a path of length at most 7k is coloured and this path is rainbow-coloured.Then φ can be extended into a colouring of I(C 1 ) using α k colours, such that every subpath of length at most 7k of C 1 is rainbow-coloured and Q j is rainbow-coloured, for every Proof.We can easily extend φ to C 1 using 14k colours (including the at most 7k already used colours) so that every subpath of C 1 of length 7k is rainbow-coloured.
We shall now prove that there exists a colouring φ The union of the three colourings φ, φ + , and φ − is clearly the desired colouring of I(C 1 ).(Observe that a vertex of I(C 1 ) is coloured only once because C 1 , I + (C 1 ) and I − (C 1 ) are disjoint by Corollary 27.) It remains to prove the existence of φ + and φ − .By symmetry, it suffices to prove the existence of φ + .To do so, we consider an auxiliary digraph D + 1 .For each C j ∈ C ∩ C 1 , let T + j be the transitive tournament whose hamiltonian dipath is The arcs of the A(T + j ) \ A(Q + j ) are called fake arcs.Clearly, φ + exists if and only if D + 1 admits a proper (6k 2 ) 3k -colouring.Henceforth it remains to prove the following claim.
For 1 ≤ i ≤ p, let m i = min Dis(v i ).Lemma 19 applied to (m i ) 1≤i≤p yields a set L of 6k 2 indices of such that for any i < j ∈ L, m i = m j and m k > m i , for all i < k < j.Let . By definition M j ≤ 3k.Applying Lemma 20 to (M j ) 1≤j≤6k 2 , we get a sequence of size 2k M j0+1 . . .M j0+2k such that M j0+2k is the greatest.For sake of simplicity, we set ℓ i = j 0 + i for 1 ≤ i ≤ 2k.Let f the smallest index not smaller than ℓ 2k for which Let j 1 be an index such that C j1 [t 1,j1 , v ℓ1 ] has length m and set Note that any internal vertex x of P 1 or P 2 has an integer in Dis(x) which is smaller than m and every internal vertex y of P 3 has an integer in Dis(y) which is greater than M ℓ 2k , or does not belong to I + (C 1 ).Hence, P 1 , P 2 and P 3 are disjoint from We distinguish between the intersection of P 1 , P 2 and P 3 : -Assume first that P 1 and P 2 are disjoint.If s(P 1 ) is in C 1 [t(P 3 ), s(P 2 )], then the union of ), s(P 2 )], and P 2 is a subdivision of B(k, 1; k), a contradiction.
-Assume now P 1 and P 2 intersect.Let u be the last vertex along P 2 on which they intersect.The union of ), s(P 1 )] ⊙ P 1 [s(P 1 ), u], and P 2 [u, v ℓ k ] is a subdivision of B(k, 1; k), a contradiction.
• Assume P 3 intersect P 1 ∩ P 2 .Let v be the first vertex along P 3 in P 1 ∩ P 2 and let u be the last vertex of P 1 ∩ P 2 along P 2 .The union of is a subdivision of B(k, 1; k), a contradiction.
• Assume now that P 3 intersects P 1 ∪ P 2 but not P 1 ∩ P 2 .Let v be the first vertex along P 3 in P 1 ∪ P 2 .
-If v ∈ P 2 , let u be the last vertex on P 2 ∩ P 3 along P 3 .Observe that P 3 [v, u] is also a subpath of P 2 and therefore contains no vertex of P 1 .Furthermore, there is a dipath -If v ∈ P 1 , let u be the last vertex on P 1 ∩ P 3 along P 3 .Observe that P 3 [v, u] is also a subpath of P 1 and therefore contains no vertex of P 2 .Furthermore, there is a dipath

♦
Claim 30.1 shows the existence of φ + and completes the proof of Lemma 30.
Lemma 31.Let C be a k-suitable collection of directed cycles in a B(k, 1; k)-subdivision-free digraph.There exists a proper colouring φ of C with α k colours, such that, each subpath of length 7k of each cycle of C is rainbow-coloured.
Proof.We prove by induction on the number of cycles in C the following stronger statement: if there exists a partial colouring φ such that one of the cycle C 1 has a path of length less than 7k which is rainbow-coloured, then we can extend this colouring to all D[C] using less than α k colours such that, on each cycle, every subpath of length 7k is rainbow-coloured.Consider a rainbow-colouring of a subpath of length less than 7k of a cycle C 1 ∈ C. By Lemma 30, we can extend this colouring to a colouring φ 1 of I(C 1 ) at most α k colours.Note that the non-coloured vertices of C are in one of the connected components of C − I(C 1 ).Let A be a connected component of C − I(C 1 ).The coloured (by φ 1 ) vertices of C ∩ A are those of (C ∩ A) − A. Hence, by Lemma 28, they all belong to some cycle C j and so to the diptah Q j which has length at most 7k.Hence, by the induction hypothesis, we can extend φ 1 to A. Doing this for each component, we extend φ 1 to the whole C.
Lemma 32.Let C be a k-suitable collection of directed cycles in a B(k, 1; k)-subdivision-free digraph D. For every component S of C, we have χ(D[S]) ≤ β k .
Proof.We define a sort of Breadth-First-Search for S. Let C 0 be a cycle of S and set L 0 = {C 0 }.For every cycle C s of S ∩ C 0 , we put C s in level L 1 and say that C 0 is the father of C s .We build the levels L i inductively until all cycles of S are put in a level : L i+1 consists of every cycle C l not in j≤i L j such that there exists a cycle in L i intersecting C l .For every C l ∈ L i+1 , we choose one of the cycles L i intersecting it to be its father.Henceforth every cycle in L i+1 has a unique father even though it might intersect many cycles of L For a vertex x of S, we say that x belongs to level L i if i is the smallest integer such that there exists a cycle in L i containing x. Observe that the vertices of each cycle C l of S belong to consecutive levels, that is there exists i such that V (C l ) ⊆ L i ∪ L i+1 .
To bound the chromatic number of D[S], we partition its arc set of in (A 0 , A 1 , A 2 ), where Subproof.Let φ 1 be the colouring that assigns to all vertices of level L i the colour i modulo k, it is easy to see that φ 1 is a proper colouring of D 1 .♦ Let C l be a cycle of L i , i ≥ 1 and C l ′ its father.Let p + l and r + l be the vertices such that C l [t l,l ′ , p + l ] and l the set of vertices of C l ]t l,l ′ , r l [, and finally let R ′ l be the set of vertices belonging to L i in C l \ {R + l ∪ R − l }.Claim 32.2.Let x be a vertex in L i with i ≥ 1.Let C l and C m be two cycles of L i containing x. Then either x ∈ P + l and x ∈ P + m , or x ∈ P − l and x ∈ P − m .
Subproof.Suppose for a contradiction that x ∈ P + l and x ∈ P + m .Let C l ′ and C m ′ be the fathers of C l and C m respectively (they can be the same cycle).By definition of the L j 's, there exists a dipath P from t l,l ′ to s m,m ′ only going through C l ′ , C s ′ and their ancestors.In particular P is disjoint from C l − C l ′ and C s − C s ′ .Observe that C l [s l,l ′ , t l,m ] has length at most 3k because it is contained in the union of P l,l ′ , P l,m , and C l [t l.l ′ , x] which has length at most k because x ∈ P + l .Hence C l [t l,m , s l,l ′ ] has length at least k.
Let x and y be two adjacent vertices of D 2 [X + ∪ X ′ ].Let L i be the level of x and L j be the level of y.Without loss of generality, we may assume that j ≥ i + k.Let C x be the cycle of L i such that x ∈ C x and C y the cycle of L j such that y ∈ C y .By considering ancestors of C x and C y , there is a shortest sequence of cycles C 1 . . .C p such that C 1 = C x and C p = C y and for all l ∈ [p − 1], either C l is the father of C l+1 or C l+1 is the father of C l .In particular C p−1 is the father of C p .Since y ∈ X + ∪ X ′ , then C[y, t p−1,p ] has length at least k.
Assume that xy is an arc.In p−1 l=1 C l , there is a dipath P from t p−1,p to x.This path has length at least k − 1 because it must go through all levels L i ′ , i ≤ i ′ ≤ j − 1 because the vertices of any cycle of S are in two consecutive levels.Hence the union of P ⊙ (x, y), C p [t p−1,p , y], and C p [y, t p−1,p ] is a subdivision of B(k, 1; k), a contradiction.Hence yx is an arc.
Suppose that C x is not an ancestor of C y .In particular, C 2 is the father of C 1 and there exists a path P from t 1,2 to y in p−1 l=2 C l of length at least k − 1 and internally disjoint from C 1 .Hence the union of P ⊙ yx, C 1 [x, t 1,2 ] and C 1 [t 1,2 , x] is a subdivsion of B(k, 1; k).Hence C x is an ancestor of C y .
In particular, C l is the father of C l+1 for all l ∈ [p − 1].Let P be the dipath from t 1,2 to y p l=2 C l .It has length at least k − 1 because it must go through all levels L i , 1 ≤ i ≤ p − 1. C 1 [x, t 1,2 ] has length less than k , for otherwise the union of P ⊙ yx, C 1 [x, t 1,2 ] and C 1 [t 1,2 , x] would be a subdivision of B(k, 1; k).
To summarize, the only arcs of D 2 [X + ∪ X ′ ] are arcs yx such that C x is an ancestor of C y and C 1 [x, t 1,2 ] has length less than k with C 1 . . .C p is the sequence of cycles such that C 1 = C x to C p = C y and C l is the father of C l+1 for all l ∈ [p − 1].In particular, D 2 [X + ∪ X ′ ] is acyclic.
Let y be a vertex of D 2 [X + ∪ X ′ ].Let L p be the level of y and let C 0 , . . ., C p be the sequence of cycles such that C l−1 is the father of C l for all l ∈ [p].For 0 ≤ l ≤ p − 1, let Q l be the subdipath of C l of length k − 1 terminating at t l,l+1 .By the above property, the out-neighbbours of y are in p−1 l=0 Q l .Suppose for a contradiction that y has out-degree at least 2k 2 + 1.Then there are 2k + 1 distinct indices l 1 < • < l 2k+1 such that for all i ∈ [2k + 1], C li contains an out-neighbour X i of y.Let P be the shortest dipath from x 1 to y in p l=l1 C l .This dipath intersect all cycles C l l 1 ≤ l ≤ p.Let z be first vertex of P along C l k+1 [x k+1 , t l k+1 ,l k+2 ].Vertex z belongs to either L l k+1 −1 or L l k+1 .Thus P [x 1 , z] and P [z, y] have length at least k − 1 and k respectively since P goes through all levels from L l1 to L p .Hence the union of (y, x 1 ) ⊙ P [x 1 , z], (y, x k+1 ⊙ C l k+1 [x k+1 , z], and P [z, y] is a subdivision of B(k, 1; k), a contradiction.Therefore D 2 [X + ∪ X ′ ] has maximum out-degree at most 2k 2 .

Theorem 7 .
For any positive integer b and k, there exists an acyclic digraph D k,b such that any cycle in D k,b has at least b blocks and χ(D k,b ) > k.
we denote by D the union of the digraphs, i.e.V ( D) = D∈D V (D) and A( D) = D∈D A(D).Let P be a path.We denote by s(P ) its initial vertex and by t(P ) its terminal vertex.If D is a directed path or a directed cycle, then we denote by D[a, b] the subdipath of D with initial vertex a and terminal vertex b.We denote by D[a, b[ the dipath D[a, b] − b, by C]a, b] the dipath D[a, b] − a, and by D]a, b[ the dipath D]a, b[−{a, b}.If P and Q are two directed paths such that V

4 B(k, 1 ; 1 )
Proposition 22.Let D be a strong digraph.If χ(D) ≥ 4, then D contains a B(2, 1; 1)-subdivision.Proof.Assume χ(D) ≥ 4. Since every digraph contains a 2-connected strong subdigraph with the same chromatic number, we may assume that D is 2-connected.Let C be a shortest directed cycle in D. It must be induced, so χ(D[C]) = χ(C) ≤ 3. Now by Lemma 18, C has a directed ear P in D. Necessarily, P has length at least 2 since C is induced.Thus the union of P and C is a B(2, 1; 1)-subdivision.The bound 4 in Proposition 22 is best possible because a directed odd cycle has chromatic number 3 and contains no B(2, 1; 1)-subdivision.In the remaining of this section, we present a proof of Theorem 10.Let C be a collection of directed cycles.It is nice if all cycles of C have length at least 2k − 2, and any two distinct cycles C i , C j ∈ C intersect on at most one vertex.A component of C is a connected component in the adjacency graph of C, where vertices correspond to cycles in C and two vertices are adjacent if the corresponding cycles intersect.Note that if S is a component of C, then S is both a connected component and a strong component of C. Call D C the digraph obtained from D by contracting each component of C into one vertex.For sake of simplicity, we denote by D[S] the digraph D[ S].Observe that this digraph contains S but has more arcs.

Lemma 24 .
Let k ≥ 3 be an integer, and let C be a nice collection of cycles in a B(k, 1; 1)-subdivision-free digraph D and S a component of C. Then χ(D[S]) ≤ 2k − 2. Proof.By induction on the number of cycles in S. Let C be a cycle of S.There is no chord between x and y in C such that C[x, y] has length at least k, for otherwise there would be a B(k, 1; 1)-subdivision.Hence D[C] has maximum degree at most 2k − 2.Moreover, by Proposition 17, D[C] is not a tournament of order 2k − 1.Thus, by Brooks' Theorem (14), χ(D[C]) ≤ 2k − 2. Let c be a proper colouring of C with 2k − 2 colours.Let S 1 , S 2 , . . ., S r be the components of S \ C. Since S is the union the S l , l ∈ [r], and {C}, each S l has less cycles than S. By the induction hypothesis, there exists a proper colouring c l using 2k − 2 colours for each D[S i ].Now, we claim that each D[S l ] intersects C in exactly one vertex.It is easy to see that C must intersect at least one cycle of each S l .Now suppose there exist two vertices of C, x and y in D[S l ].By definition of a nice collection, they cannot belong to the same cycle of S l , so there exist two cycles C i and C j of S l such that x ∈ C i and y ∈ C j .Now C[x, y] is a dipath form C i to C j whose arcs are not in A( S l ).This contradicts Lemma 23.Consequently, free to permute the colours of the c l , we may assume that each vertex of C receives the same colour in c and in the c l .In addition, by Lemma 23, there is no arc between different D[S l ] nor between D[S l ] and C. Hence the union of the c l and c is a proper colouring of D[S] using 2k − 2 colours.Lemma 25.Let C be a maximal nice collection of cycle in a B(k, 1; 1)-subdivision-free strong digraph D. Then χ(D C ) ≤ 2k − 3

Corollary 27 .
Let C be a k-suitable collection of directed cycles and let C 1 ∈ C.

1 .
(i) I + (C 1 ) and I − (C 1 ) are vertex-disjoint digraphs.(ii) I − (C 1 ) ∩ C j = Q − j and I + (C 1 ) ∩ C j = Q + j , for all C j ∈ C ∩ C Lemma 28.Let C be a k-suitable collection of directed cycles in a B(k, 1; k)-subdivision-free digraph D. Let C 1 be a cycle of C and let A be a connected component of C − I(C 1 ).All vertices of (C ∩ A) − A belongs to a unique cycle C A of C.
then it follows from Lemma 26 applied to C 3 , C 2 , C 4 in the role of C 1 , C 2 , C 3 respectively.In both case, C W [b 2 , a 4 ] has length at least k.Furthermore, C W [a 4 , b 2 ] has length at least k because it contains B 3 .Therefore the union of C W [b 2 , a 4 ], C W [a 4 , b 2 ] and C ′ [b 2 , a 4 ] = C 3 [b 3 , a 4 ] is a subdivision of of B(k, 1; k), a contradiction.Let φ be a colouring of G.A subset of vertices or a subgraph S of G is rainbow-coloured by φ if all vertices of S have distinct colours.
the set of arcs of D[S] which ends belong to the same level, and • A 1 is the set of arcs of D[S] which ends belong to different levels i and j with |i − j| < k. • A 2 is the set of arcs of D[S] which ends belong to different levels i and j with |i − j| ≥ k.For i ∈ [3], let D i be the spanning subdigraph of D[S] with arc set A i .We shall now we bound the chromatic numbers of D 0 , D 1 and D 2 .Claim 32.1.χ(D 1 ) ≤ k.