Linear Bound for Majority Colourings of Digraphs

Given $\eta \in [0, 1]$, a colouring $C$ of $V(G)$ is an $\eta$-majority colouring if at most $\eta d^+(v)$ out-neighbours of $v$ have colour $C(v)$, for any $v \in V(G)$. We show that every digraph $G$ equipped with an assignment of lists $L$, each of size at least $k$, has a $2/k$-majority $L$-colouring. For even $k$ this is best possible, while for odd $k$ the constant $2/k$ cannot be replaced by any number less than $2/(k+1)$. This generalizes a result of Anholcer, Bosek and Grytczuk, who proved the cases $k=3$ and $k=4$ and gave a weaker result for general $k$.


Introduction
Given a digraph G, we write V (G) and E(G) for the vertex and edge set of a digraph G, respectively. For v ∈ V (G), we denote by d + (v) the out-degree of v. Given η ∈ [0, 1], a (not necessarily proper) colouring C of V (G) is an η-majority colouring if at most ηd + (v) out-neighbours of v have colour C(v), for any v ∈ V (G). A 1/2-majority colouring is referred to simply as a majority colouring. This concept was introduced by van der Zypen [5], who asked whether every digraph has a majority colouring with a bounded number of colours. This question was answered by Kreutzer, Oum, Seymour, van der Zypen and Wood [4], who showed that 4 colours always suffice.
We consider the list-colouring version of this problem. For a set S, we denote by P(S) the power set of S. Given a digraph G and an assignment L : V (G) → P(N) of lists to vertices of G, an L-colouring C : V (G) → N of G is a colouring of V (G) such that C(v) ∈ L(v) for every v ∈ V (G). If G has an η-majority L-colouring for any such assignment L whose lists are all of size at least k, we say that G is η-majority k-choosable. Anholcer, Bosek and Grytczuk [1] showed that every graph G is 1/n-majority n 2 -choosable for every n ≥ 2. Theorem 1 improves on this result. Theorem 1. For any integer k ≥ 2, every digraph G is 2/k-majority kchoosable.
Theorem 1 was proved independently by Girão, Kittipassorn and Popielarz [2]. The case k = 2 is trivial. Previously, Anholcer, Bosek and Grytczuk [1] showed that Theorem 1 holds in the cases k = 3 and k = 4 and conjectured that 2/k can be replaced by 1/2 when k = 3. Theorem 1 is best possible when k is even, as shown by the example of a k/2-regular tournament on k + 1 vertices (that is, all vertices have both in-degree and out-degree equal to k/2). If we make all lists equal, then some vertex must have an outneighbour of the same colour, and this out-neighbour represents 2/k of its out-neighbourhood. When k is odd, a similar example shows that we cannot replace 2/k by any number less than 2/(k + 1).

Proof of Theorem 1
We denote by vw an edge from a vertex v of a digraph to another vertex w. The proof of Theorem 1 relies on Lemma 2, which follows.
Lemma 2. Let k ≥ 2 be an integer and let G be a digraph on a vertex set V = S ∪ T , such that G[S] is strongly connected, G[T ] is edgeless and there are no edges from T to S. Let C T be any colouring of T and let L : S → P(N) be an assignment of lists, each of size at least k, to vertices in S. Then there exists an extension C of C T to V with C(v) ∈ L(v) for each v ∈ S, such that no vertex v ∈ S has more than 2d + (v)/k out-neighbours with the same colour as v.
Proof. For any colouring C of V , we define the function f C : S → R by for each vertex v ∈ S; i.e., f C (v) is the proportion of out-neighbours of v which have the same colour as v under C. Given v ∈ S, we write d + Let A be the non-negative real S × S matrix with entries A vw = 1/d + S (v) if vw is an edge of G and A vw = 0 otherwise. We have Aj = j (where j is the vector of all 1's). On the other hand if Ay = cy for any vector y, then choose v ∈ S such that |y v | is maximal; now |cy v | = | w∈S A vw y w | ≤ w∈S A vw |y v | = |y v | and so |c| ≤ 1. Thus, the spectral radius of A is 1. By applying the Perron-Frobenius Theorem (see, e.g., [3, Theorem 8.8.1]) to A ⊤ , noting that G[S] is strongly connected, we obtain an eigenvector x of A ⊤ with positive entries and eigenvalue 1. We remark that by normalizing x we could obtain a stationary distribution of the uniform random walk on G[S].
Consider an extension C of We claim that C satisfies the requirements of the lemma. For brevity we write f for f C . It suffices to show that f (v) ≤ 2/k for every v ∈ S. Observe that .
Fix a vertex v ∈ S. We define g : L(v) → R by for i ∈ L(v). Observe that if v were recoloured with colour i, then (1) would change by g(i) − g(C(V )). By the minimality of C and the definition of g we have that g Since x v > 0, we have f (v) ≤ 2/k. It follows immediately that C satisfies the requirements of the lemma.
Proof of Theorem 1. We partition V (G) into strongly connected components S 1 , S 2 , . . . , S r , where there are no edges from S i to S j for any i < j. Let L : V (G) → P(N) be an assignment of lists, each of size at least k, to vertices of G. We write A i for j∈[i] S j (taking A 0 = ∅); let C 0 be the unique colouring of A 0 . For each i = 0, 1, 2, . . . , r − 1 in turn, we apply Lemma 2 with S = S i+1 , T = A i and G = G[A i+1 ] \ G[A i ] to obtain an extension of C i to an L-colouring C i+1 of A i+1 such that no v ∈ S i+1 (and hence no v ∈ A i+1 ) has more than 2d + (v)/k out-neighbours of the same colour. At the end of this process we obtain C r , which is the desired 2/k-majority L-colouring of V (G).