Majority Colorings of Sparse Digraphs

A majority coloring of a directed graph is a vertex-coloring in which every vertex has the same color as at most half of its out-neighbors. Kreutzer, Oum, Seymour, van der Zypen and Wood proved that every digraph has a majority 4-coloring and conjectured that every digraph admits a majority 3-coloring. We verify this conjecture for digraphs with chromatic number at most 6 or dichromatic number at most 3. We obtain analogous results for list coloring: We show that every digraph with list chromatic number at most 6 or list dichromatic number at most 3 is majority 3-choosable. We deduce that digraphs with maximum out-degree at most 4 or maximum degree at most 7 are majority 3-choosable. On the way to these results we investigate digraphs admitting a majority 2-coloring. We show that every digraph without odd directed cycles is majority 2-choosable. We answer an open question posed by Kreutzer et al. negatively, by showing that deciding whether a given digraph is majority 2-colorable is NP-complete. Finally we deal with a fractional relaxation of majority coloring proposed by Kreutzer et al. and show that every digraph has a fractional majority 3.9602-coloring. We show that every digraph with minimum out-degree $\Omega\left((1/\varepsilon)^2\ln(1/\varepsilon)\right)$ has a fractional majority $(2+\varepsilon)$-coloring.

The relevant property of acyclic digraphs is that there is an ordering of its vertices, in which every vertex is preceded by its complete out-neighborhood. Then coloring vertices along this ordering with two colors such that for each vertex we use the color that appears least frequently in the (already colored) out-neighborhood will produce a majority 2-coloring.
It is easy to construct digraphs which do require 3 colors for a majority coloring. The canonical examples are the odd directed cycles C 2k+1 , k ≥ 1, which are not majority 2-colorable since for digraphs with maximum out-degree one majority-coloring and proper graph coloring of the underlying graph are equivalent. However, no example of a digraph is known that requires the use of four colors. Kreutzer et al. conjectured that there are none.

Conjecture 1 ([KOS + 17]). Every digraph is majority 3-colorable.
Kreutzer et al. [KOS + 17] also provide ample evidence for their conjecture by establishing that it holds for "most" digraphs. They show, using the Lovász Local Lemma, that the uniform random 3-coloring is a majority 3-coloring with non-zero probability if certain local density conditions hold, namely if , In [KOS + 17] it is also mentioned at the end that a more careful analysis of the Local Lemma approach works for r-regular digraphs provided r ≥ 144. Subsequently Girão, Kittipassorn, and Popielarz [GaKP17] studied tournaments in particular, and showed, also using the probabilistic method, that if the minimum out-degree is at least 55, then the tournament is majority 3colorable.
These are all the results we are aware of about Conjecture 1. All the proofs use some probabilistic idea and require some lower bound on the minimum out-degree. However, digraphs with small minimum out-degree seem to be outside the realm of any such probabilistic methods and it looks like they constitute a main difficulty of the problem. This is also illustrated by the fact that it was not even known whether planar digraphs are majority 3-colorable.
In this paper our main motivation is to complement the existing results for locally dense digraphs and provide approaches to this intriguing conjecture on the opposite end of the spectrum.

Majority 3-Colorability
Since a proper coloring is also a majority coloring, Conjecture 1 is immediately true for digraphs with chromatic number at most three. For four-chromatic digraphs this is already not obvious. Our first result resolves the conjecture for digraphs with low chromatic number, including planar digraphs.
Theorem 1. Let D be a digraph such that χ(D) ≤ 6. Then D is majority 3-colorable.
Another coloring concept for digraphs which greatly grew in importance in the last two decades is the dichromatic number. This parameter of a digraph D, denoted by χ(D), was introduced already in 1982 by Victor Neumann-Lara [NL82]. 1 It is defined as the smallest k allowing a partition X 1 , . . . , X k of V (D) such that D[X i ] is acyclic for all i = 1, . . . , k.
In the introduction above we mentioned how to give a majority 2-coloring of acyclic digraphs, i.e. digraphs with dichromatic number one. In our second main result we prove Conjecture 1 for digraphs with dichromatic number at most three.
Theorem 2. Let D be a digraph such that χ(D) ≤ 3. Then D is majority 3-colorable.
The results of [KOS + 17] and [GaKP17] cited in the introduction indicate that the case of r-regular digraphs for constant r is probably the most important benchmark in the study of Conjecture 1. Recall in particular that the Local Lemma approach works for r-regular digraphs provided r ≥ 144. Note however the crucial non-monotonicity in the problem: even though we do not know whether Conjecture 1 is true for r = 143, it does hold (quite easily) for r = 1 and 2. Indeed, a 1-regular digraph is the disjoint union of directed cycles, and hence we can 3-color it properly to obtain a majority-coloring. Then Conjecture 1 also follows for 2-regular digraphs. Even more generally, the validity of the conjecture for any odd regularity r − 1 implies it for the next even regularity r. This is the consequence of the fact that for even r any r-regular digraph D contains a 1-regular spanning subgraph F and any 3-majority coloring of the (r − 1)-regular digraph D − F is also a majority coloring of D. Most generally, if a digraph D is obtained from a digraph D by adding an edge (u, v) whose tail has odd out-degree d + D (u) then a majority coloring of D is also a majority coloring of D .
From our next main result it will follow that majority 3-colorings also exist in the cases when r = 3 or 4. Since our proof of Conjecture 1 for 3-regular digraphs relies crucially on a natural list coloring extension of majority coloring and also implies a stronger statement, we first introduce this stronger concept in the next subsection and then present our results there.

Majority 3-Choosability
The notion of majority choosability of digraphs was already proposed in [KOS + 17]. We call a digraph k-majority-choosable, if for any assignment of lists of size at least k to the vertices, we can choose colors from the respective lists such that the arising coloring is a majority coloring.
It was already noted in [KOS + 17] that all their results about dense digraphs using probabilistic methods, including in particular the one about r-regular digraphs for r ≥ 144, remain valid for majority 3-choosability instead of majority 3-colorability. Here we obtain results at the other end of the spectrum, involving digraphs with bounded maximum (out-)degrees. This implies Conjecture 1 for these cases, and in particular also for 3-regular digraphs.
Next we derive choosability analogues of our first two theorems. The analogue of Theorem 1 connects the choosability of the underlying graph to majority choosability.
Theorem 4. Let D be a digraph whose underlying undirected graph is 6-choosable. Then D is majority 3-choosable. In particular any digraph with a 5-degenerate underlying graph is majority 3-choosable.
The list dichromatic number χ (D) of a digraph D was introduced by Bensmail, Harutyunyan, and Le [BHL18]. It is defined as the minimum integer k ≥ 1 such that for any assignment of lists of size at least k to the vertices, we can choose colors without producing monochromatic directed cycles. We have the following analogue of Theorem 2 involving this parameter.

Fractional Majority Colorings
The concept of fractional majority coloring emerges as an LP-relaxation of the problem of majority coloring, much in the same way as the usual fractional colorings of graphs. This notion was already introduced in [KOS + 17]. The definition is somewhat technical and we postpone it to Section 4. To appreciate our results here, it is sufficient to keep in mind that the minimum total weight of a fractional majority coloring is at most the majority chromatic number.
Kreutzer et al [KOS + 17] ask what is the smallest constant K such that every digraph admits a fractional majority coloring with total weight at most K. This is yet another direction to approach Conjecture 1 from. Proving that there is a fractional majority coloring with total weight 3 for every digraph would certainly be an easier task. Here we take the first step in this direction and show that the upper bound of 4, which follows from the fact that every digraph is majority 4-colorable, can be slightly improved.
Theorem 6. Every digraph D admits a fractional majority coloring with total weight at most 3.9602.
Our proof is the combination of an intricate probabilistic coloring with some deterministic alteration.
In the second theorem of the section we show that digraphs with sufficiently large minimum out-degree have fractional majority colorings with total weight arbitrarily close to 2. The results in [KOS + 17] obtained using the Local Lemma instead only give an upper bound of 3 under stronger assumptions (an upper bound on the maximum degree). This result further highlights that the main difficulty of deciding Conjecture 1 might lie with digraphs of low out-degrees.

Majority 2-Colorability
We prepare the investigation of majority 3-colorable digraphs in Section 3 with an analysis of majority 2-colorings in Section 2. An open question posed in [KOS + 17] asked whether there is a characterisation of digraphs that have a majority 2-coloring (or a polynomial time algorithm to recognise such digraphs). This was answered (most likely) in the negative by Bang-Jensen, Bessy, Havet, and Yeo [BJBHY18] who showed that deciding whether a 3-out-regular digraph is majority 2-colorable is NP-complete. With our hope for an efficient characterization of majority 2-colorability being shattered, any simple sufficient condition comes in handy.
For a condition, it is natural to exclude odd directed cycles, as they are canonical examples of graphs with no majority 2-coloring. It turns out that excluding them already implies 2choosability.

Theorem 8. If D is a digraph without odd directed cycles, then D is majority 2-choosable.
Organization of the paper. In Section 2 we obtain Theorem 8 as a consequence of a more general result (Theorem 10). This result is crucial for the proofs of Theorems 1, 2, 3, 4, 5, which are presented in Section 3. In Section 4 we treat fractional majority colorings and prove Theorems 6 and 7. We conclude with final remarks and some open problems in Section 5.

Digraphs without Odd Directed Cycles
We have seen that acyclic digraphs as well as bipartite digraphs are majority 2-colorable. We have also seen that odd directed cycles are canonical examples of digraphs having no majority 2-coloring. It is therefore natural to try unifying these results and ask whether every digraph without an odd directed cycle is majority 2-colorable. In this section, we answer this question positively. We start with a simple observation: Proof. Sufficiency of this condition is obvious, as a directed cycle is always contained in a single strong component. For the reverse direction, it suffices to observe that if D is strongly connected and all directed cycles have even length, then D is bipartite. However, this statement can be easily verified by considering an ear decomposition of D.
Proposition 1. Let D be a digraph which contains no odd directed cycles. Then D is majority 2-colorable. Moreover, such a coloring can be chosen to extend any given pre-coloring of the sinks of D with colors 1, 2.
Proof. We prove the statement by induction on the number s ≥ 1 of strong components of D. Suppose first that s = 1, i.e. D is strongly connected. Then by Lemma 9 D is bipartite and therefore majority 2-colorable. As D is either a single vertex or contains no sinks, the claim follows. Now let s ≥ 2 and suppose that the statement holds true for all digraphs with at most s − 1 strong components. We now distinguish two cases: Either, D is an independent set of s vertices, and therefore, the claim holds trivially true. If there exists at least one arc in D, there has to be a strong component of D containing no sinks such that there are no arcs entering it. Let X be the vertex set of this component. We now claim that there exists a subset U ⊆ X and a 1, 2-coloring c U of D − U which extends c, such that In order to find such a set, we apply the following procedure: We keep track of a pair (W, c W ), consisting of a subset W ⊆ X and a vertex-coloring c W : V (D)\W → {1, 2} extending c. As an invariant we will keep the first of the two above properties, i.e. we assert that every vertex x ∈ V (D) \ W has at least d + (x) 2 out-neighbors with a different color according to c W .
We initialize W := X, c W := c. It is clear that this assignment satifies the invariant (remember that c is a majority coloring of D − X, and that there are no edges entering X).

As long as a vertex
.
It is easily verified that the coloring c W also fulfills the invariant, since by definition x 0 has at least max{d(c U , 1, Theorem 8 is now obtained from Theorem 10 as a direct consequence.

Majority 3-Colorings of Sparse Digraphs
As a consequence of Theorem 10, we obtain our main result: Theorem 11. Let D be a digraph. Suppose there is a partition {X 1 , X 2 , X 3 } of the vertex set such that for every i ∈ {1, 2, 3}, D[X i ] contains no odd directed cycles. Then D is majority 3-colorable.
Proof. We assign lists of size two to the vertices of D, namely, we assign the list {2, 3} to all vertices in X 1 , the list {1, 3} to all vertices in X 2 , and the list {1, 2} to all vertices in X 3 . Because D[X i ], i = 1, 2, 3 contains no odd directed cycle, we can apply Theorem 10 to conclude that there exists a majority-coloring of D which uses only colors 1, 2 and 3. This proves the claim.
From this we now directly derive Theorem 1 and 2.
Proof of Theorem 1. If χ(D) ≤ 6, then D admits a partition Y 1 , . . . , Y 6 into independent sets. Using the partition {Y 1 ∪ Y 2 , Y 3 ∪ Y 4 , Y 5 ∪ Y 6 } of the vertex set to apply Theorem 11 now shows that D is indeed majority 3-colorable.
Proof of Theorem 2. If χ(D) ≤ 3, then there exists a partition {X 1 , X 2 , X 3 } of the vertex set such that D[X i ] contains no directed cycles, for i = 1, 2, 3. The claim now follows by Theorem 11.
The fact that Theorem 10 deals with an assignment of lists can be further exploited to show analogues of Theorem 11, Theorems 1 and 2 for list colorings.
For this purpose we need the following notion: Call a digraph D OD-3-choosable if for any assignment of color lists L(x), x ∈ V (D) of size 3 to the vertices, there exists a choice function c (i.e. c(x) ∈ L(x) for all x ∈ V (D)) such that no odd directed cycle in (D, c) is monochromatic.

Theorem 12. Let D be a digraph. If D is OD-3-choosable, then D is majority 3-choosable.
Proof. Let L(v) for all v ∈ V (D) be a given color list of size three. We have to show that there is a majority-coloring c of D such that c(v) ∈ L(v) for all v ∈ V (D). For every v ∈ V (D), we let L * (v) := {{C 1 , C 2 }|C 1 = C 2 ∈ L(v)} contain all three unordered color-pairs in L(v). Since D is OD-3-choosable, there exists a choice function c * on V (D) such that c * (v) ∈ L * (v) for each vertex v ∈ V (D) is a subset of L(v) of size two and such that there exists no odd directed cycle in D which is monochromatic with respect to c * . If we now consider c * (v), v ∈ V (D) as an assignment of lists of size two to the vertices of D, we can apply Theorem 10 to conclude that there is a majority-coloring c of D such that c(v) ∈ c * (v) ⊆ L(v) for every vertex v ∈ V (D). As L(·) was arbitrary, we conclude that D is majority 3-choosable.
We are now ready to prove Theorem 4 and Theorem 5.
Proof of Theorem 4. We show that D is OD-3-choosable, the claim then follows by Theorem 12. Let L(v) for each vertex v ∈ V (D) be an assigned list of three colors. For each color C used in one of the lists, let C be a distinct copy of this color. We now consider the assignment L 6 (·) of lists of size 6 to the vertices of D, where for each vertex v ∈ V (D), L 6 (v) := {C 1 , C 1 , C 2 , C 2 , C 3 , C 3 } if C 1 , C 2 , C 3 denote the colors contained in L(v). Because the underlying graph of D is 6-choosable, there is a proper coloring c 6 of D such that c 6 (v) ∈ L 6 (v) for all v ∈ V (D). Now consider the coloring c of D obtained from c 6 by identifying each copy C of an original color C with C again. We then have c(v) ∈ L(v) for every v ∈ V (D). Since c 6 was a proper coloring of the undirected underlying graph of D, each color class with respect to c induces a bipartite subdigraph of D, and hence there are no monochromatic odd directed cycles in (D, c). Hence, D is OD-3-choosable.
Proof of Theorem 5. This follows directly since any digraph with χ (D) ≤ 3 is clearly OD-3choosable.
The rest of this section is devoted to proving Theorem 3. We prepare it with the following Lemma, whose proof makes use of Theorems 4 and 5.
Proof. Suppose the claim was false and consider a counterexample D minimizing |V (D)|+|E(D)|. We have |V (D)| ≥ 4, D is connected and every proper subdigraph of D must be OD-3-choosable.
We first consider the case that there is We claim that c is a coloring of D without monochromatic odd directed cycles. In fact, such a cycle would have to pass v, however no edge entering v is monochromatic. Therefore D is OD-3-choosable, a contradiction.
Hence we know for every . Consequently, the underlying simple graph U (D) has maximum degree ∆(U (D)) ≤ 6. If U (D) is 6-choosable, then it follows as in the proof of Theorem 4 that D is OD-3-choosable, a contradiction.
Finally, since we obtained that D is OD-3-choosable in each case, the initial assumption was wrong, which concludes the proof by contradiction.
Proof. For a proof by contradiction, suppose the claim was false and consider a counterexample D minimizing the number of edges.
Consider first the case that there is a v ∈ V (D) with d + (v) = 4. Let e be an edge leaving v and put D := D −e. By the minimality of D, D is majority 3-choosable. We now claim that any majority-coloring of D also defines a majority-coloring of D. Clearly, such a coloring satisfies the condition for a majority-coloring at any vertex distinct from v. Since v has out-degree 3 in D , it has at most one out-neighbor in D of the same color. Thus there are at most two out-neighbors of v in D which share its color, and so the majority condition is fulfilled at v. We conclude that also D must be majority 3-choosable, which gives the desired contradiction. Now for the second case, assume that no vertex has out-degree 4. This means that for every x ∈ V (D), we either have d + (x) ≤ 3 or d + (x) ≥ 5 and therefore d − (x) ≤ 2. We can therefore apply Lemma 13 to D, which shows that D is OD-3-choosable. From Theorem 12 we get that D is majority 3-choosable. This again is a contradiction to D being a counterexample to the claim.
Therefore the initial assumption was wrong, and this concludes the proof.
If ∆(U (D)) ≤ 6, then by the list coloring version of Brook's Theorem either U (D) is 6-choosable, and then the claim follows from Theorem 4, or U (D) = K 7 . Now let L(v 1 ), . . . , L(v 7 ) be lists of size three assigned to the vertices {v 1 , . . . , v 7 } of D. We first consider the case that all lists are equal, i.e., show that D is majority 3-colorable.
If there exists a vertex v ∈ V (D) which is contained in at most 3 digons, then there are vertices u 1 = u 2 ∈ V (D) \ {v} such that u 1 , u 2 , v do not form a directed triangle. Therefore, any partition {X 1 , X 2 , X 3 } of V (D) where X 1 = {v, u 1 , u 2 } and |X 2 | = |X 3 | shows, by Theorem 11, that D is majority 3-colorable. Otherwise, every vertex in D is contained in at least 4 digons and thus has out-degree at least 4. Now any 3-coloring of D with color classes of sizes 2, 2, 3 defines a majority-coloring of D.
Now suppose that not all lists are equal. In this case we can choose for each vertex v i a sublist L 2 (v i ) ⊆ L(v i ) of size two such that no three vertices are assigned the same sublist (minimize the number of edges whose ends are assigned the same sublist). By Theorem 10 we obtain a majority-coloring Hence, D is majority 3-choosable in each case, which concludes the proof.

Fractional Majority Colorings
Another concept introduced in [KOS + 17] is that of a fractional majority coloring. Given a subset S ⊆ V (D), a vertex v is popular in S if v ∈ S and more than half of its out-neighbors are in S. A subset S ⊆ V (D) is stable if it contains no popular elements. Let S(D) be the set of all stable sets of D, and S(D, v) the set of all stable sets containing v. A fractional majority coloring is a function that assigns a weight w T ≥ 0 to every set T ∈ S(D), satisfying T ∈S(D,v) w T ≥ 1 for every v ∈ V (D). The total weight of a fractional majority coloring is simply T ∈S(D) w T . Kreutzer et al. asked for the minimum constant K such that every digraph admits a fractional majority coloring with total weight at most K.
We will show two results related to this question, namely Theorem 6 and Theorem 7. The proof of these two theorems will be based on the dual of the linear program defined by the restrictions on a fractional majority coloring: Observation 1. For a digraph D, the minimum possible total weight of a fractional majority coloring is also the maximum total weight v∈V (D) w v in a non-negative weight assignment of V (D) in which every stable set T satisfies v∈T w v ≤ 1.
The main idea of the proof of both theorems is that, given any choice of weights on V (D), we can construct a stable set in which the weight is at least a given fraction of the total weight, using the probabilistic method.
Lemma 15. Let D be a digraph and let 0 < p < 1. Suppose that one can take a random subset X ⊆ V (D) with the property that, for every v ∈ V (D), the probability that v is in X but not popular in X is at least p. Then D admits a fractional majority coloring with total weight at most 1 p . Proof. Suppose that D is a counterexample to our statement, and we will reach a contradiction. By Observation 1, we can assign weights to V (D) so that the total weight is w > 1 p , and every stable set in D has a sum of weights at most one. Let Y be the set of popular vertices in X. By linearity of expectation, the expected total weight of X \ Y is at least pw > 1. Take an instance of X \ Y with weight greater than 1. Every vertex in X \ Y has at least half of its out-neighbors outside of X, which implies that it is not popular in X \ Y . Hence X \ Y is stable in D and has total weight greater than 1, producing a contradiction.
The proof of Theorem 7 is a straightforward application of this lemma: Proof of Theorem 7. Let N be a large enough positive integer. Let D be a digraph with δ + (D) ≥ N . Set p = 1 2 − ln N N . Let X be a random subset of V (D) in which every element is included independently with probability p. By Hoeffding's inequality, for any vertex v the probability that at least half of its out-neighbors are in X is at most Setting q = N −2 , from Lemma 15 we find a fractional majority coloring of total weight at For N large enough, we have 1 p−q < 2 + ε.
For Theorem 6, we need to be more careful. Consider again the set X containing each vertex independently with probability p, where p is slightly lower than 1 2 . If the out-degree of v is not 1, one can show that the probability that v is popular in X is upper-bounded by a constant, strictly smaller than p − 1 4 . However, if v has out-degree 1, the probability that v is popular in X is p 2 > p − 1 4 . For this reason, the vertices with out-degree 1 deserve extra consideration. Observe that, in the graph induced by the vertices of out-degree 1, all cycles are directed, pairwise disjoint and act as sinks. Consequently, removing one vertex from each directed cycle produces an acyclic graph, where the vertices can be given an ordering in which every edge goes from a larger vertex to a smaller one.
Proof of Theorem 6. Set p 1 = 0.4594 and p 2 = 0.4503. Assign independently to each vertex v a random indicating variable X v , which takes the value 1 with probability p 1 if d + (v) = 1 and with probability p 2 otherwise. Now construct the random subset X as follows: • Add to X all vertices v with d + (v) = 1 and X v = 1.
• For every cycle C formed by vertices with d + (v) = 1 and X v = 1, select a vertex v ∈ C uniformly at random and set X v = 0.
• Take an ordering of the vertices v with d + (v) = 1 and X v = 1, in which if we have an edge (v, w) then v comes after w (this is possible because these vertices form an acyclic digraph). Following this order, add v to X if its out-neighbor is not in X.
We will show that, for every vertex v, the probability that v is in X but not popular in X is at least 1 4 + ε, for a fixed value of ε > 0. Suppose first that d + (v) = 1. If we draw the vertices with out-degree 1 in red and those with other out-degrees in blue, then the successive out-neighborhoods of v must have one of these forms: We label the cases as Case 1 through Case 4, left to right and top to bottom in Figure 1. We denote v = v 0 , and v i+1 as the out-neighbor of v i , if it is unique. We go through each case: • If v is in Case 1, then whenever X v = 1 and X v1 = 0 we have v ∈ X. This happens with probability p 1 (1 − p 2 ). Applying Lemma 15, there is a fractional majority coloring of D with total weight at most The methods used in this paper are unlikely to resolve Conjecture 1 for the open cases of 5and 6-regular digraphs. One possible approach could be via an extension to hypergraphs: Given a 5-regular digraph D, consider the hypergraph H(D) with vertex set V (D) and whose edges are {v} ∪ N + (v), v ∈ V (D). This hypergraph is 6-regular and 6-uniform. If we could now find a vertex-3-coloring of H(D) such that no hyperedge contains four vertices of the same color, this coloring would certainly be a majority coloring of D. We would therefore be interested in deciding the following question. Problem 1. Let H be a 6-regular 6-uniform hypergraph. Is there a 3-coloring of V (H) such that no hyperedge contains four vertices of the same color?
The setting of k-regular k-uniform hypergraphs could be fruitful, as it is known that these hypergraphs have property B for all k ≥ 4 (as noted in [Vis03]). We want to conclude with a small selection of open questions.
• Is every 5-regular digraph 1 3 -majority 5-colorable? We can show that it is possible to color with 5 colors such that in each connected component, at most one vertex violates the majority condition.