Subdivisions in Digraphs of Large Out-Degree or Large Dichromatic Number

Abstract

In 1985, Mader conjectured the existence of a function f such that every digraph with minimum out-degree at least f (k) contains a subdivision of the transitive tournament of order k. This conjecture is still completely open, as the existence of f (5) remains unknown. In this paper, we show that if D is an oriented path, or an in-arborescence (i.e., a tree with all edges oriented towards the root) or the union of two directed paths from x to y and a directed path from y to x, then every digraph with minimum out-degree large enough contains a subdivision of D. Additionally, we study Mader's conjecture considering another graph parameter. The dichromatic number of a digraph D is the smallest integer k such that D can be partitioned into k acyclic subdigraphs. We show that any digraph with dichromatic number greater than 4 m (n − 1) contains every digraph with n vertices and m arcs as a subdivision.Conjecture 2 (Mader [20]). There exists a least integer mader δ + (T T k ) such that every digraph D with δ + (D) ≥ mader δ + (T T k ) contains a subdivision of T T k .Mader proved that mader δ + (T T 4 ) = 3, but even the existence of mader δ + (T T 5 ) is still open. This conjecture implies directly that transitive tournaments (and thus all acyclic digraphs) are δ 0maderian. Conjecture 3. There exists a least integer maderIn fact, Conjecture 3 is equivalent to Conjecture 2 because if transitive tournaments are δ 0 -maderian, then mader δ + (T T k ) ≤ mader δ 0 (T T 2k ) for all k. Indeed, let D be a digraph with minimum out-degree mader δ 0 (T T 2k ), and let D ′ be the digraph obtained from disjoint copies of D and its converse (the digraph obtained by reversing all arcs) D by adding all arcs from D to D. Clearly, δ 0 (D ′ ) ≥ mader δ 0 (T T 2k ). Therefore D ′ contains a subdivision of T T 2k . Hence, either D or D (and so D) contains a subdivision of T T k .Both conjectures are equivalent, but the above reasoning does not prove that a δ 0 -maderian digraph is also δ + -maderian. The case of oriented trees (i.e. orientations of undirected trees) is typical. Using a simple greedy procedure, one can easily find every oriented tree of order k in every digraph with minimum in-and out-degree k (so mader δ 0 (T ) = |T | − 1 for any oriented tree T ). On the other hand, it is still open whether oriented trees are δ + -maderian and a natural important step towards Conjecture 2 would be to prove the following weaker one.Conjecture 4. Every oriented tree is δ + -maderian.We give evidences to this conjecture. First, in Subsection 2.1, we prove that every oriented path (i.e. orientation of an undirected path) P is δ + -maderian and that mader δ + (P ) = |V (P )| − 1. Next, in Subsection 2.2, we consider arborescences. An out-arborescence (resp. in-arborescence) is an oriented tree in which all arcs are directed away from (resp. towards) a vertex called the root. Trivially, the simple greedy procedure shows that mader δ + (T ) = |T | − 1 for every out-arborescence. In contrast, the fact that in-arborecences are δ + -maderian is no...

Bispindles in Strongly Connected Digraphs with Large Chromatic Number

“…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.…”

“…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.…”

“…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 (C 3 , 3 , v] is a subdipath of P 2,3 and so has length less than k. Therefore C 2 [t 1,2 , s 2,3 ] = Q 2 [t 1,2 , s 2,3 ] has length at least k because Q 2 has length at least 3k. It follows that the union of C 2 [s 2,3 , t 1,2 ], C 2 [t 1,2 , s 2,3 ] and R 3 [t 1,2 , s 2,3 ] is a subdivision of B(k, 1; k), a contradiction.…”

“…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 .…”

Extension of Gyárfás-Sumner Conjecture to Digraphs

“…A surge of recent results on the subject suggests that this concept is accepted as the natural extension of the chromatic number of graphs to digraphs by most authors. For example see: [3,7,15,20,22,25,26,30]. Note that the problem of deciding if a digraph has dichromatic number at most 2 is NP-complete (see [8]), even when restricted to tournaments (see [9]), which is in contrast to the undirected case, where 2-colorability is polynomial time to decide.…”