On the complexity of Broadcast Domination and Multipacking in digraphs

We study the complexity of the two dual covering and packing distance-based problems Broadcast Domination and Multipacking in digraphs. A dominating broadcast of a digraph $D$ is a function $f:V(D)\to\mathbb{N}$ such that for each vertex $v$ of $D$, there exists a vertex $t$ with $f(t)>0$ having a directed path to $v$ of length at most $f(t)$. The cost of $f$ is the sum of $f(v)$ over all vertices $v$. A multipacking is a set $S$ of vertices of $D$ such that for each vertex $v$ of $D$ and for every integer $d$, there are at most $d$ vertices from $S$ within directed distance at most $d$ from $v$. The maximum size of a multipacking of $D$ is a lower bound to the minimum cost of a dominating broadcast of $D$. Let Broadcast Domination denote the problem of deciding whether a given digraph $D$ has a dominating broadcast of cost at most $k$, and Multipacking the problem of deciding whether $D$ has a multipacking of size at least $k$. It is known that Broadcast Domination is polynomial-time solvable for the class of all undirected graphs (that is, symmetric digraphs), while polynomial-time algorithms for Multipacking are known only for a few classes of undirected graphs. We prove that Broadcast Domination and Multipacking are both NP-complete for digraphs, even for planar layered acyclic digraphs of small maximum degree. Moreover, when parameterized by the solution cost/solution size, we show that the problems are W-hard. We also show that Broadcast Domination is FPT on acyclic digraphs, and that it does not admit a polynomial kernel for such inputs, unless the polynomial hierarchy collapses to its third level. In addition, we show that both problems are FPT when parameterized by the solution cost/solution size together with the maximum out-degree, and as well, by the vertex cover number. Finally, we give for both problems polynomial-time algorithms for some subclasses of acyclic digraphs.


Introduction
We study the complexity of the two dual problems Broadcast Domination and Multipacking in digraphs. These concepts were previously studied only for undirected graphs (which can be seen as symmetric digraphs, where for each arc (u, v), the symmetric arc (v, u) exists). Unlike most standard packing and covering problems, which are of local nature, these two problems have more global features since the covering and packing properties are based on arbitrary distances. This difference makes them algorithmically very interesting.
Broadcast domination. Broadcast domination is a concept modeling a natural covering problem in telecommunication networks: imagine we want to cover a network with transmitters placed on some nodes, so that each node can be reached by at least one transmitter. Already in his book in 1968 [26], with undirected graphs, we show that Broadcast Domination is NP-complete, even for planar layered acyclic digraphs (defined later) of maximum degree 4 and maximum finite distance 2. This holds for Multipacking, even for planar layered acyclic digraphs of maximum degree 3 and maximum finite distance 2, or acyclic digraphs with a single source and maximum degree 5. Moreover, when parameterized by the solution cost/solution size, we prove that Broadcast Domination is W [2]-hard (even for digraphs of maximum finite distance 2 or bipartite digraphs of maximum finite distance 6 without directed 2-cycles) and Multipacking is W[1]-hard (even for digraphs of maximum finite distance 3). On the positive side, we show that Broadcast Domination is FPT on acyclic digraphs (DAGs for short) but does not admit a polynomial kernel for layered DAGs of maximum finite distance 2, unless the polynomial hierarchy collapses to its third level. Moreover, we show that both Broadcast Domination and Multipacking are polynomial-time solvable for layered DAGs with a single source. We also show that both problems are FPT when parameterized by the solution cost/solution size together with the maximum out-degree, and as well, by the vertex cover number. Moreover it follows from a powerful meta-theorem of [21] that Broadcast Domination is FPT when parameterized by solution cost, on inputs whose underlying graphs belong to a nowhere dense class.
The resulting complexity landscape is represented in Fig. 1. We start with some definitions in Section 2. We prove our results for Broadcast Domination in Section 3. The results for Multipacking are presented in Section 4. We conclude in Section 5.

Preliminaries
Directed graphs. We mainly consider digraphs, usually denoted D = (V, A) 1 , where V is the set of vertices and A the set of arcs. For an arc uv ∈ A, we say that v is an out-neighbor of u, and u an inneighbor of v. Given a subset of vertices V ⊆ V , we define the digraph induced by V as D = (V , A ) where A = {uv ∈ A : u ∈ V and v ∈ V }. We denote such an induced subdigraph by D [V ]. A directed path from a vertex p 1 to p l is a sequence {p 1 , . . . , p l } such that p i ∈ V and p i p i+1 ∈ A for every 1 i < l. When p 1 = p l , it is a directed cycle. A digraph is acyclic whenever it does not contain any directed cycle as an induced subgraph. An acyclic digraph is called a DAG for short. The (open) out-neighborhood of a vertex v ∈ V is the set N + (v) = {u ∈ V : vu ∈ A}, and its closed out-neighborhood is N + [v] = N + (v) ∪ {v}. We define similarly the open and closed in-neighborhoods of v and denote them by N − (v) and For the sake of readability, we always mean out-neighborhood when speaking of the neighborhood of a vertex. A DAG D = (V, A) is layered when its vertex set can be partitioned into {V 0 , . . . , V t } such that N − (V 0 ) = ∅ and N + (V t ) = ∅ (vertices of V 0 and V t are respectively called sources and sinks), and uv ∈ A implies that u ∈ V i and v ∈ V i+1 , 0 i < t. A single-sourced layered DAG is a layered DAG with only one source, that is, satisfying |V 0 | = 1. A digraph is bipartite or planar if its underlying undirected graph has the corresponding property. Every layered digraph is bipartite. Given two vertices u and v, we denote by d(u, v) the length of a shortest directed path from u to v. For a vertex v ∈ V and an integer d, we define the ball of radius d centered The eccentricity of a vertex v in a digraph D is the largest (finite) distance between v and any vertex of D, denoted ecc(v) := max u∈V d (v, u). A digraph is strongly connected if for any two vertices u, v, there is a directed path from u to v, and weakly connected if its underlying undirected graph is connected. We will assume that all digraphs considered here are weakly connected (if not, each component can be considered independently). The diameter is the maximum directed distance max u,v∈V d(u, v) between any two vertices u and v of G. If the digraph is not strongly connected, then the diameter is infinite. The maximum finite distance of a digraph D is the largest finite directed distance between any two vertices of G, denoted mfd(D) :

Parameterized complexity.
A parameterized problem is a decision problem together with a parameter, that is, an integer k depending on the instance. A problem is fixed-parameter tractable (FPT for short) if it can be solved in time f (k) · |I| c for an instance I of size |I| with parameter k, where f is a computable function and c is a constant. Given a parameterized problem P , a kernel is a function which associates to each instance of P an equivalent instance of P whose size is bounded by a function h of the parameter. When h is a polynomial, the kernel is said to be polynomial. An FPT-reduction between two parameterized problems P and Q is a function mapping an instance (I, k) of P to an instance (f (I), g(k)) of Q, where f and g are computable in FPT time with respect to parameter k, and where I is a YES-instance of P if and only if f (I) is a YES-instance of Q. When moreover f can be computed in polynomial time and g is polynomial in k, we say that the reduction is a polynomial time and parameter transformation [4]. Both reductions can be used to derive conditional lower bounds: if a parameterized problem P does not admit an FPT algorithm (resp. a polynomial kernel) and there exists an FPTreduction (resp. a polynomial time and parameter transformation) from P to a parameterized problem Q, then Q is unlikely to admit an FPT algorithm (resp. a polynomial kernel). Both implications rely on certain standard complexity hypotheses; we refer the reader to the book [10] for details.

Theorem 1.
Broadcast Domination is NP-complete, even for planar layered DAGs of maximum degree 4 and maximum finite distance 2.
Proof. We will reduce from Exact Cover by 3-Sets, defined as follows.
Exact Cover by 3-Sets • Input: A set X of 3k elements (for some k ∈ N), and a set T = {t 1 , . . . , t n } of triples from X.
• Question: Does there exist a subset S of k triples from T such that each element of X appears in (exactly) one triple in S? Exact Cover by 3-Sets is NP-hard even when the incidence bipartite graph of the input is planar and each element appears in at most three triples [14]. We will reduce any such instance (X, T ) of Exact Cover by 3-Sets to an instance (D = (V , A ), k ) of Broadcast Domination.
We create V by taking two copies T 1 , T 2 of T and one copy of X. More precisely, we let T i = {t i j : 1 i n for i ∈ {1, 2}. We now add an arc from a vertex t 1 i ∈ T 1 to its corresponding vertex t 2 i in T 2 , and from a vertex t 2 i ∈ T 2 to all elements of X that are contained in t i in (X, T ). See also Figure 2. Formally: The construction can be done in polynomial time, and there is no cycle in D: arcs go either from T 1 to T 2 or from T 2 to X. Hence D is a layered DAG with three layers and thus, maximum finite distance 2.
In fact D is obtained from the bipartite incidence graph of (X, T ) (which is planar and of maximum degree 3) reproduced on the vertices of T 2 ∪ X, by adding pendant vertices (those from T 1 ) to those of T 2 , orienting the arcs as required. Thus, the maximum degree of D is 4 and D is planar. ⇒ Given a solution S of (X, T ), set f (t 1 i ) = 2 for all t i ∈ S, f (t 1 i ) = 1 for each of the n − k remaining vertices of T 1 and f (v) = 0 for all vertices of T 2 and X. For every vertex t 2 i ∈ T 2 , we have d(t 1 i , t 2 i ) = 1. Similarly, for every vertex x ∈ X, d(t 1 i , x) 2 holds for the vertex t 1 i such that t i is in S and contains x in (X, T ). Since every vertex t 1 i of T 1 satisfies f (t 1 i ) 1, it is covered by itself, and it follows that f is a dominating broadcast of size n + k. ⇐ Let us now consider the case where we are given a dominating broadcast for D of cost n + k. Note that since the maximum finite distance is 2, we can assume f : V → {0, 1, 2}. Remark that the vertices of T 1 are n sources. Therefore, any broadcast needs to set f (t 1 i ) 1 for each t 1 i ∈ T 1 , and this covers all vertices of T 1 and T 2 . It remains to cover vertices of X with a cost of k, which can be done by setting f (t 1 i ) = 2 for some vertices of T 1 and f (t 2 j ) = 1 for some vertices of T 2 . Notice that it is never useful to set f (x) = 1 for some vertex x ∈ X: setting an additional cost of 1 to any f (t 2 i ) such that t 2 i ∈ A is always better. Hence, the corresponding set of triples is a valid cover of (X, T ). (And it is an exact cover because there are 3k elements covered by k triples.) We next give two parameterized reductions for Broadcast Domination. Proof. We provide two reductions from the W[2]-hard Multicolored Dominating Set problem [8], defined as follows.

Multicolored Dominating Set
• Input: A graph G = (V, E) with V partitioned into k sets {V 1 , . . . , V k }, for an integer k ∈ N.
• Output: Does there exist A dominating set S of G such that |S ∩ V i | = 1 for every 1 i k.?
We first provide a reduction that gives digraphs with directed 2-cycles.

Construction 1.
We build an instance (D = (V , A ), k ) of Broadcast Domination as follows. To obtain the vertex set V , we duplicate V into two sets V 1 and V 2 . Following the partition of V into k sets, we let We then add every possible arc within V 1 i (1 i k), and an arc from a vertex v in V 1 to each vertex of V 2 corresponding to a vertex from the closed neighborhood of v in G. Altogether, V = V 1 ∪ V 2 . Finally, we set k = k. See Figure 3 for an illustration. Clearly mfd(D) = 2.  Proof. ⇒ Let S ⊆ V be a multicolored dominating set of size k of G. We claim that setting f (v) = 1 for every vertex v of V 1 such that the corresponding vertex v of G is in S, yields a dominating broadcast of cost k. To see this, notice that each vertex v ∈ V 1 i (1 i k) with cost 1 covers V 1 i . Now, since these vertices of cost 1 form a dominating set in G, they cover the vertices of V 2 corresponding to their closed neighborhood in G, and hence f is a dominating broadcast. ⇐ Assume now that D has a dominating broadcast f of cost k. Notice first that any set V 1 i (1 i k) must contain a vertex v such that f (v) 1. Since f has cost k, this means that for every vertex w ∈ V 2 , f (w) = 0. It follows that one needs to cover the vertices of V 2 using k vertices in V 1 , which can be done only if there is a multicolored dominating set of size k in G.
We now give a similar but more involved construction, which gives bipartite instances of maximum finite distance 6 and no directed 2-cycles.

Construction 2.
We build an instance (D = (V , A ), k ) of Broadcast Domination as follows. To obtain the vertex set V , we multiplicate V into four sets V 0 , V 1 , V 2 and V 3 and we will have a set M of subdivided vertices. The set V 0 ∪ V 1 will induce an oriented complete bipartite graph, while V 2 ∪ V 3 will induce a matching. Following the partition of V into k sets, for 0 i 3, we let We assume that |V i | 2, since otherwise one must take the only vertex in V i . For each 1 i k we then add the following arcs: • for every pair v, w of distinct vertices of V i , we add an arc from v 0 to w 1 ; • for every v ∈ V i , we add an arc from v 1 to v 0 ; • for every v ∈ V i , we add an arc from v 2 to v 3 . Figure 4: Sketch of the built digraph D in the second reduction of the proof of Theorem 3.
Moreover, for every edge vw in G, we add an arc from v 1 to w 2 , and we subdivide it once. The set of all subdivision vertices is called M . Finally, we set k = 3k. It is clear that mfd(D ) = 6 (shortest paths of length 6 exist from vertices of V 0 to vertices of V 3 , but no longer shortest paths exist). The digraph has clearly no directed 2-cycles, and is bipartite with

Claim 5. The graph G has a multicolored dominating set of size k if and only if the digraph D has a dominating broadcast of cost 3k.
Proof. ⇒ Let S ⊆ V be a multicolored dominating set of size k of G. We claim that setting f (v 1 ) = 3 for every vertex v 1 of V 1 such that v ∈ S yields a dominating broadcast of cost 3k. To see this, notice first that each such vertex belonging to V 1 i , 1 i k, covers the whole set V 0 i ∪ V 1 i and all the vertices of M with an in-neighbor in V 1 i . Now, each vertex v 1 with v ∈ S covers (at distance 3) each vertex w 2 and w 3 of V 2 ∪ V 3 such that w is in the closed neighborhood of v in G. Since S is dominating, f is thus a dominating broadcast. ⇐ Assume now that D has a dominating broadcast f of size 3k. First, we claim that for every i with 1 i k, we need a total cost of 3 for the vertices in V 0 Clearly, we cannot cover the vertices of V 0 i ∪ V 1 i with two vertices broadcasting at cost 1. Thus, we can assume that there is is a total cost of exactly 3 on the vertices of V 0 Thus, the proof is complete.

Complexity and algorithms for (layered) DAGs
We now address the special cases of (layered) DAGs. Note that Dominating Set remains W[2]-hard on DAGs by a reduction from [30,Theorem 6.11.2]. In contrast, we now give an FPT algorithm for Broadcast Domination on DAGs that counterbalances the W[2]-hardness result.

Theorem 6. Broadcast Domination parameterized by solution cost k can be solved in FPT time
The proof relies on the following proposition, which is reminiscent of a stronger statement of Dunbar et al. [13] for undirected graphs (stating that there always exists an optimal dominating broadcast where each vertex is covered exactly once, which is false for digraphs).

Proposition 7.
For any digraph D = (V, A), there exists an optimal dominating broadcast such that every broadcast dominator is covered by itself only.
Proof. Let f be an optimal dominating broadcast of D, and assume there exists two vertices u, v ∈ V such that f (v) 1 and f (u) d (u, v). In this case, v is covered by both u and itself. Notice that We can now prove Theorem 6.
Proof of Theorem 6. Let D = (V, A) be a DAG. We consider the set V 0 of sources of D. Observe that for every s ∈ V 0 , f (s) 1 must hold. In particular, this means that |V 0 | k (otherwise we return NO). We provide a branching algorithm based on this simple observation and on Proposition 7. We start with an initial broadcast f consisting of setting f (s) = 1 for every vertex s in V 0 . At each step of the branching algorithm, we let be the set of currently covered vertices, and we consider the Notice that D f is acyclic and hence contains a source u. Since every vertex of N f \ V f is covered, we may assume by Proposition 7 that in the sought optimal solution, u is only covered by itself or by a vertex in V f . This means that one needs to branch on at most k + 1 distinct cases: either setting f (u) = 1, or increasing the cost of one of its at most k broadcasting ancestors in V f . At every branching, the parameter k decreases by 1, which ultimately gives an O * (2 k log k )-time algorithm and completes the proof of Theorem 6.
We will now complement the previous result by a negative one, which can be proved using a reduction similar to the one in Theorem 1 but from Hitting Set, defined as follows.
Hitting Set • Input: A universe U of elements, a collection F of subsets of U , an integer k ∈ N.
• Question: Does there exist a hitting set S of size k, that is, a set of k elements from U such that each set of F contains an element of S?
Hitting Set is unlikely to have a polynomial kernel when parameterized by k + |U |, unless the polynomial hierarchy collapses to its third level [12, Theorem 5.1]. Theorem 8. Broadcast Domination parameterized by solution cost k does not admit a polynomial kernel even on layered DAGs of maximum finite distance 2, unless the polynomial hierarchy collapses to its third level.
Proof. We provide a reduction from HITTING SET. It is shown in [12,Theorem 5.1] that if Hitting Set admits a polynomial kernel when parameterized by |U | + k (a variant called Small Universe Hitting Set), then the polynomial hierarchy collapses to its third level.
We do the same reduction as the one from Exact Cover by 3-Sets from Theorem 1, except that the set T of triples is replaced by U and the set X of elements is replaced by F. We again obtain a DAG with three layers and maximum finite distance 2. The solution cost for the instance of Broadcast Domination is set to |U | + k, and the proof of validity of the reduction is the same.
Since this is clearly a polynomial time and parameter transformation, the result follows.
We now show that Broadcast Domination can be solved in polynomial time on special kinds of DAGs.

Theorem 9. Broadcast Domination is linear-time solvable on single-sourced layered DAGs.
Proof. Let D = (V, A) be a single-sourced layered DAG with layers {V 0 , . . . , V t }. For the sake of readability, sets V i such that |V i | = 1 are denoted {s i }, for 0 i t.
Our algorithm relies on the following structural properties of some optimal dominating broadcasts for single-sourced layered DAGs.

Claim 10.
There always exists an optimal dominating broadcast f of D such that: Proof. Let f be an optimal dominating broadcast of D having the properties of Proposition 7.
If j does not exist, then we are done. We hence assume j is well-defined. By the choice of i, we know that This means that v 2 j must be covered by s i , which in turn covers v 1 j , which is impossible by the choice of i (and the definition of f ). We thus have = 0 and increasing f (s i ) by 1 leads to a dominating broadcast of smaller cost, a contradiction.
We thus assume j < t. We claim that the dominating broadcast f i defined by setting: is optimal. Notice first that c fi (V ) c f (V ). Now, every vertex covered by both v 1 j and v 2 j is covered by s i : indeed, since s i corresponds to a layer with a single vertex, it has a directed path of length d(s i , v j−1 ) + max{f (v 1 j ), f (v 2 j )} + 1 to every vertex covered by both v 1 j and v 2 j , which are thus still covered.
Property (ii). Suppose that f satisfies Property (i). Assume there exist two vertices s i and s j with 0 i < i+1 < j t such that f (s i ) d(s i , s j ). In other words, vertex s i covers vertex s j . Consider that i is chosen to be minimum with this property. Notice that since f fulfills the properties of Proposition 7, we have f (s j ) = 0. We distinguish two cases: Notice that every vertex covered by s i in f is still covered in f i : indeed, s i covers all vertices up to V j−1 , and vertices in higher layers are now covered by s j , which covers itself. By construction, we have: the last inequality holding since f (s j ) = 0. This leads to a contradiction since f is an optimal dominating broadcast. Thus this case does not happen.
• We may hence assume that f (s i ) = d(s i , s j ). Since f fulfills the properties of Proposition 7 and Property (i), V j+1 has to dominate itself, and thus s j+1 must exist, unless j = t. Consider the dominating broadcast f i obtained from f by setting f i (s i ) = d(s i , s j ) − 1, f i (s j ) = 1 + f (s j+1 ) and f i (s j+1 ) = 0. If j = t we consider that f (s j+1 ) = 0. Notice that every vertex covered by s j+1 in f is covered by s j in f i . We have: the last equality holding since f (s j ) = 0.
We have thus obtained a dominating broadcast f i of the same cost as f , still satisfying Property (i) and Proposition 7, but where every vertex s l with l i satisfies (ii). If f i still does not satisfy (ii), we reiterate this process (each time, with increasing value of i) until (ii) is satisfied for all vertices. This concludes the proof of Claim 10.
We thus deduce a simple top-down procedure to compute an optimal dominating broadcast f . We initiate our solution by setting i = 0. While there remain uncovered vertices, we let f (s i ) = j − i − 1 for the smallest value j such that s j exists and j i + 2. In other words, s i will cover all vertices below it, until the closest vertex of the set t j=0 s j that is not a neighbour of s i . We then carry on by setting i = j. By Claim 10, this process leads to the construction of an optimal dominating broadcast.

Algorithms for structural parameters and structured classes
We now give some algorithms for structural parameters and classes. Proof. To solve Broadcast Domination by brute-force, we may try all the subsets of size k, and for each subset, try all possible k k broadcast functions. But we can assume that k d, since a single vertex with cost d covers all the digraph, which completes the proof.
We next consider jointly two parameters. Recall that by Theorems 1 and 3, such a result probably does not hold for each of them individually. Proof. Let (D = (V, A), k) be an instance of Broadcast Domination such that D has maximum outdegree d. Consider a dominating broadcast f of cost k. A vertex v with f (v) = i > 0 covers all vertices of its ball of radius i, which has size at most i j=0 (d − 1) j + 1 id i + 1. Thus, if the input has more than n = k(k + 1)d k vertices, we can reject. Otherwise, a simple brute-force algorithm over all possible 2 n possible subsets and, given a subset, all k k possible broadcasts, is FPT. The result follows.
Next, we consider the vertex cover number of input digraphs, that is, the smallest size of a set of vertices that intersects all arcs (or, in other words, the vertex cover number of the underlying undirected graph). For a given class, any broadcasting vertex out of the class either covers all vertices in the class, or none. Similarly, a vertex broadcasting at radius r inside the class covers the same set of vertices outside the class as any other vertex from the class would. Hence, we may assume that at most one selected vertex b i per class C i broadcasts with f (b i ) > 1. We can assume that the other vertices in the class either all satisfy f (v) = 0 or all f (v) = 1 (the latter may happen if they all need to cover themselves, for example if they are all sources). Moreover, mfd(D) 2c + 1 since every shortest path is either contained in S or has to alternate between a vertex of S and one of V \ S, but cannot have repeated vertices.
Hence, for each equivalence class C i , we have 2 × (2c + 1) choices: 2c + 1 for the value of f (b i ), and two possibilities for the other vertices of C i . Similarly, for each vertex of S, we have 2c + 1 possible broadcast values. In total, this gives (t + c) O(c) = 2 c O(c) different possible dominating broadcasts, and each of them can be checked in polynomial time.
We next see how to apply the following powerful theorem from [21], to show that Broadcast Domination is FPT for any class of digraphs whose underlying graph is nowhere dense. We will not give a definition of nowhere dense graph classes, and refer to the book [29] instead. Such classes include planar graphs, graphs excluding a fixed (topological) minor, graphs of bounded degree, graph classes of bounded expansion, etc.

Theorem 14 ([21]
). Let C be a nowhere dense graph class. There exists such that, given as inputs a graph G ∈ C and a first-order logic graph property ϕ, the problem of deciding whether G satisfies ϕ can be solved in time f (|ϕ|)|G| 1+ , that is, it is FPT when parameterized by the length of ϕ.

Corollary 15. For every fixed nowhere dense graph class C, Broadcast Domination parameterized by the solution cost of the input digraph is FPT for inputs whose underlying graphs are in C.
Proof. We want to show that for fixed parameter value k of the solution cost, Broadcast Domination can be expressed in first-order logic by a formula whose length is bounded by a function of k, and apply Theorem 14.
To do so, we extend the classic approach for defining k-Dominating Set in first-order logic (see e.g. [29,Chapter 18.4]).
We will use the property dp(x, y, i), stating that there is a directed path from x to y of length at most i. This can be expressed in first-order logic for fixed i. To this end, we state that either x = y, or there is an arc from x to y, or there is a directed path of length 2 from x to y (i.e. there exists a vertex z, x has an arc to z, and z, an arc to y), . . . or there exists a directed path of length i from x to y.
Let V 1 , . . . , V k denote the sets of broadcast dominators of a potential dominating broadcast f , where V i contains the vertices broadcasting at radius i. The union V f = k i=1 V i has size at most k, and since k is considered to be fixed, we can "guess" the size of each set V i . To this end, we let v 1 i , . . . , v k i be the potential vertices of V i . For a given partition Π of V f into sets V 1 , . . . , V k , we can express the fact that a given vertex x is dominated by f as the formula dom Π (x, v 1 1 , . . . , v k k ), which is composed of the conjunction of all formulae of type dp(v j i , x, i), where in Π, 1 j |V i |. Now, given the set Π 1 , . . . , Π t of all partitions of V f into sets V 1 , . . . , V k (note that t k k ), the first-order formula for Broadcast Domination is given as We remark that Corollary 15 does not imply Theorem 12, indeed there are digraph classes of bounded maximum out-degree whose underlying graphs do not form a nowhere dense class of graphs. For example, every d-degenerate graph can be oriented so as to have maximum out-degree at most d. Indeed, a graph is d-degenerate if its vertices can be ordered v 1 , . . . , v n such that for 2 i n, v i has at most d neighbors among v 1 , . . . , v i−1 . Thus, orienting every edge v i v j with i < j from v j to v i produces a digraph of maximum out-degree at most d. However, for every d, the class of d-degenerate graphs is not nowhere dense [21,29].

Complexity of Multipacking
We will need the following results to prove our results for Multipacking. The first one was proved for undirected graphs in [22]. Proof. Let D = (V, A) and let S ⊆ V be a multipacking of D of size at least k. Assume there exists a source s ∈ V that does not belong to S. We say that a vertex v ∈ V is full w.r.t. S whenever there exists an integer p > 0 such that |B + p (v) ∩ S| = p. Assume first that s is not full w.r.t. S: in that case, one can safely add s to the multipacking S and obtain a new solution of size at least k. Hence, we now consider the case where s is full. Notice that if s is full at distance 1 (i.e. B + 1 (s) ∩ S = 1), then the set (S \ {u}) ∪ {s} is a multipacking of size at least k (recall that s is a source), and thus we are done.
We hence assume that this is not the case. Proof. Let S ⊆ V be a multipacking of D of maximum size. By definition of a multipacking, considering each ball centered at the source s, the following holds for every 1 i t: We will prove the result inductively, by locally modifying S in a top-down manner until it has the desired property. Let j 2 be the smallest index such that |S ∩ V j | 2, and i < j be the largest index such that |S ∩ V i | = 0. Notice that i is well-defined due to (1). Moreover, let s 1 j and s 2 j be two vertices of S ∩ V j . Case 1. We assume first that i = j − 1. Let u 1 i and u 2 i be vertices of V i such that u 1 i s 1 j and u 2 i s 2 j belong to A (note that in a layered DAG every non-source vertex has a predecessor in the previous layer). Since S is a multipacking, we have u 1 i = u 2 i and neither u 1 i nor u 2 i is adjacent to both s 1 j and s 2 j . Moreover, a vertex s i−1 in S ∩ V i−1 cannot be adjacent to both u 1 i and u 2 i , since otherwise we would have B + 2 (s i−1 ) ∩ S > 2. Moreover by minimality of the index j, there is at most one vertex of S in V i−1 . Assuming w.l.o.g. that u 1 i has no predecessor in S, the set (S \ {s 1 j }) ∪ {u 1 i } is a multipacking having the same size than S.

Case 2.
We now consider the case where i < j −1. First, we will prove that there is a vertex We claim that s i−1 is not adjacent to every vertex of V i . Assume for a contradiction that this is the case. This means that s i−1 is within distance j − (i − 1) of every vertex contained in ∪ j l=i V l . By the choice of indices i and j we know that ∪ j l=i V l contains at least j − (i − 1) vertices from S, which in turn implies that B + j−(i−1) (s i−1 ) ∩ S = j − (i − 1) + 1, contradicting (1). Thus, there is a vertex v i in V i that has no in-neighbor in S. Now, we know by choices of i and j that |S ∩ V p | = 1 for i < p < j. Hence the set (S \ {s i+1 }) ∪ {v i }, where {s i+1 } = S ∩ V i+1 , is a multipacking of D having the same size than S. By iterating the above argument, we end up with i = j − 1, in which case we can apply the argument from Case 1. Overall, after each iteration of Case 1, j strictly increases. The procedure terminates when the value of j reaches t.

Theorem 19.
Multipacking is NP-complete, even for planar layered DAGs of maximum degree 3 and maximum finite distance 3.
Proof. We provide a reduction from the NP-complete Independent Set problem [19], which remains NP-complete on planar cubic graphs [20].

Independent Set
• Input: A graph G = (V, E), an integer k ∈ N.
• Question: Does there exist an independent set of G of size at most k?
The construction of the instance (D = (V , A ), k ) of Multipacking is done by setting This reduction can also be seen as follows: given any instance of Independent Set, we subdivide each edge uv by adding a new vertex w with wu, wv ∈ A and a pending source seeing w. Doing so, most properties of the given instance (such as planarity and maximum degree) are preserved. One can see that the graph G has an independent set of size k if and only if the digraph D has a multipacking of size k = m + k.
⇒ Let S be an independent set of G of size k, and let S = E 1 ∪ S. First, S is of size m + k. Then, for any e 1 ∈ E 1 , |N + [e 1 ] ∩ S | = 1 and |B + 2 (e 1 ) ∩ S | 2 hold since S is an independent set. By similar arguments, |N + [e 2 ] ∩ S | 1 holds for any e 2 ∈ E 2 , and thus no vertex of E 2 can have two out-neighbors in S. All other vertices of D are sinks (i.e. with empty out-neighborhood), so the multipacking property is trivially satisfied for them. Thus S is a multipacking of D of size m + k. ⇐ Let S be a multipacking of maximum size in D, such that |S| m + k. Each vertex of E 1 is a source of D, so by Lemma 17 we can assume that E 1 ⊆ S and then E 2 ∩ S = ∅. So S \ E 1 ⊆ V , and its size is at least k. Assume S contains two vertices u, v of V that are adjacent in G, then N + [e 2 i ] ∩ S 2 with e 2 i = uv, which contradicts the fact that S is a multipacking of D. Thus S \ E 1 is an independent set of G of size at least k. Proof. We provide a reduction from Independent Set problem [19], which remains NP-complete for cubic graphs [20]. We define the function f : V → E such that for v ∈ V , f (v) = e i if and only if e i is the first edge in which v appears (recall that E = {e 1 , . . . , e m }). We create the digraph D = (V , A) as follows (see Figure 5): Proof. ⇒ Let S be an independent set of size k of G. We set S = {s} ∪ {v i , y i : 1 i m} ∪ S. We need to show that S is a multipacking of D. Notice first that S contains exactly 2m+k +1 vertices. The vertices s and p satisfy the multipacking property since there is at most one vertex of S at distance exactly i from both these vertices, for any i (and there is no vertex of S at distance 1 from s and none at distance 0 from p). Each vertex of V and each vertex y i trivially satisfies the multipacking property since they are sinks. For 1 i m, notice that x i cannot have two out-neighbors in S since S is an independent set. Hence, x i and v i satisfy the multipacking property, since for the latter We can see that S ∩ {u i , v i } is either empty (which is sufficient to conclude since there remains only two distinct nonempty layers of D in W i ), or In both cases, we get that |S ∩ W i | 2. One can also easily see that both s and p cannot be together in S . Thus, the maximum size of a multipacking of D is 2m + 1.
Thus |S ∩ (V \ V )| 2m + 1, and |S ∩ V | k. We also know that for a, b ∈ S ∩ V , ab / ∈ E, otherwise there would exist an edge e i = ab and thus N + [x i ] ∩ S would be of size at least 2. So we can conclude that S ∩ V is an independent set of G of size at least k. This completes the proof.

Theorem 22. Multipacking parameterized by solution size k is W[1]-hard, even on digraphs of maximum finite distance 3.
Proof. We provide an FPT-reduction from Multicolored Independent Set, which is W[1]-hard when parameterized by k [10].
• Question: Does there exist an independent set S of G s.t. |S ∩ V i | = 1 for 1 i k?
Construction. We construct an instance (D = (V , A ), k ) of Multipacking as follows. We consider the bipartite incidence graph of G, that is we add V ∪ E to V . To construct A , we add an arc from a vertex e ∈ E to a vertex v ∈ V if and only if e contains v. We next group vertices of E into k 2 sets E i,j , 1 i < j k according to the colors of their corresponding endpoints, and add every possible arc within each set E i,j . We next duplicate the vertices of each set V i into a set V i such that there is an arc from each vertex v i ∈ V i to its corresponding copy v i in V i . Finally, we add k vertices {s 1 , . . . , s k } such that there is an arc from s i to every vertex of V i . Notice in particular that the maximum finite distance is 3.
See Figure 6 for an illustration.

Claim 23.
The graph G has a multicolored independent set of size k if and only if the digraph D has a multipacking of size k = 2k + k 2 . Proof. ⇒ Let S = {u 1 , . . . , u k } be an independent set of G of size k such that u i ∈ V i for every 1 i k. Let S ⊆ V be a set that contains exactly one arbitrary vertex e * i,j for every set E i,j (1 i < j n), together with each vertex of V i corresponding to each vertex u i of S. Finally, add {s 1 , . . . , s k } to S . We claim that S is the sought multipacking of D. To see this, notice first that |S | = 2k+ k 2 by construction. Moreover, every vertex contains at most one vertex from S in its closed out-neighborhood. We now prove that every vertex e i,j ∈ E i,j contains at most two vertices from S in B + 2 (e i,j ). Assume for a contradiction this is not the case; then, apart from e * i,j , there are two other vertices a and b in B + 2 (e i,j ). We have that a ∈ V i and b ∈ V j . By construction, this means that ab is an edge of G, contradicting the fact that S is an independent set. Finally, since every vertex s i (1 i k) has vertices from only one set V i in its distance 2 neighborhood, and since S is a multicolored set, the result follows. The only vertices for which checking their distance 3 neighborhood is needed are vertices from E i,j for every 1 i < j n. One can notice that for any )| = 2 and |S ∩ E i,j | = 1 by construction. ⇐ Assume that D has a multipacking S ⊆ V of size k = 2k + k 2 . By Lemma 17, we can assume that S contains {s 1 , . . . , s k }. In particular, this means that S ∩ V i = ∅ for every 1 i k. Moreover, at distance 2, we have |S ∩ V i | 1 for 1 i k since otherwise there would be three vertices from S in B + 2 (s i ), for some vertex s i . Moreover, for 1 i < j n, |E i,j ∩ S | 1 since E i,j is a bi-directed clique.
Thus, by the size of S , the only possibilities are to pick exactly one vertex in each set V i and one vertex e i,j in each set E i,j . This can be done only if there exists a multicolored independent set of size k in G: otherwise one would have to select two vertices a ∈ V i and b ∈ V j , i = j such that ab ∈ E, which in turn would imply that the vertex from E i,j corresponding to the edge ab has three vertices in its distance-2 neighborhood (namely e i,j , a and b). Thus, the proof is complete.

Algorithms
Next, we present a linear-time algorithm.

Theorem 24.
Multipacking can be solved in linear time on single-sourced layered DAGs.
Proof. Let D = (V, A) be a single-sourced layered DAG. By Lemma 18, in every single-sourced layered DAG there is a multipacking of maximum size that is a maximum-size set of vertices with at most one vertex per layer such that two chosen vertices of consecutive layers are not adjacent. We thus give a polynomial-time bottom-up procedure to find such a set of vertices. At each step of the procedure, a layer V i is partitioned into a set of active vertices and a set of universal ones, denoted respectively A i and U i . Our goal will be to select exactly one vertex in each set of active vertices. We initiate the algorithm by setting A t = V t and U t = ∅. Now, for every i with 0 i < t, we set In other words, U i contains the vertices of layer V i that are adjacent to all active vertices of V i+1 . During the procedure, if some layer V i satisfies A i = ∅, we let A i−1 = V i−1 and repeat this process until V 0 is reached.
To construct a multipacking of maximum size, we start from V 0 , and for each 0 i t we pick a vertex s i in each non-empty set A i of active vertices. Every time a vertex s i is picked, we remove its closed neighborhood from D. Notice that by construction, every time a vertex s i is picked, there exists a vertex s i+1 ∈ A i+1 such that s i s i+1 does not belong to A (otherwise s i would belong to U i ). To prove the optimality of our algorithm, let 0 i < t be such that A i = ∅, and j > i be the smallest integer greater than i such that V j = A j . Such a j exists since A t = V t . Claim 25. Let S be a multipacking with at most one vertex per layer. Then S satisfies: Proof. Let S be an optimal multipacking with at most one vertex per layer. Assume by contradiction that S ∩ j k=i V k > j − i + 1, and call s k the vertex in V k ∩ S for every i k j. We know that s i ∈ U i , and since every vertex in A i+1 is an out-neighbor of s i , then s i+1 ∈ U i+1 . By induction, for every i k j, we have s k ∈ U k , but U j = ∅ by choice of j, leading to a contradiction.
Notice that Claim 25 gives one less vertex than what Lemma 18 implies, and that it is the value reached by our algorithm, since for i k j the only layer with U k = V k is V i . Since the sets of active and universal vertices can be constructed by standard graph searching, the whole algorithm takes O(|V | + |A|) time.
We now give algorithms for structural parameters. We next give a simple algorithm for digraphs of bounded diameter. Proof. Let (D = (V, A), k) be an instance of Multipacking such that D has maximum out-degree d. By Lemma 16, if D has a shortest directed path of length 3k − 3, we can accept the input (this can be checked in polynomial time). Thus, we can assume that the length of any shortest path is at most 3k − 2. If a vertex u has a directed path to a vertex v, we say that u absorbs v, and a set S of vertices is absorbing if every vertex in D is absorbed by some vertex of S. If D has a set of k vertices, no two of which are absorbed by some common vertex (e.g. a set of k sources), we can accept, since this set forms a valid solution. Note that this property is satisfied by any minimum-size absorbing set S: indeed, if some vertex w absorbs two vertices u, v of S, we may replace them by w and obtain a smaller absorbing set, a contradiction.
We claim that we can find a minimum-size absorbing set in FPT time. Indeed, we can reduce this problem to Hitting Set (defined for the proof of Theorem 8) as follows. We let U = V (D), and F contains a set F v for every vertex v, where F v comprises every vertex which absorbs v (including v itself). Because D has out-degree at most d and the length of any shortest path is at most 3k − 2, every vertex of U is contained in at most d U =  [25], which proves the above claim. As mentioned before, if the obtained minimum-size absorbing set of D has size at least k, since it forms a valid multipacking, we can accept. Otherwise, D can be covered by k − 1 balls of radius at most 3k − 2. Each such ball has at most Next, we consider the vertex cover number, already considered for Theorem 13. Proof. Let (D = (V, A), k) be the input of Multipacking and let S be a vertex cover of D of size c. As for Theorem 13, we partition the set V \ S (which contains no arcs) into equivalence classes C 1 , . . . , C t according to their in-and out-neighborhoods in S. There are t 2 2c such classes.
By Lemma 17, we can assume that all sources belong to an optimal solution. Consider any class C i . Its vertices are either all sources, or none of them are. If they are not sources, they all have a common in-neighbor, and thus at most one vertex of C i can belong to a multipacking. It is not important which one is selected, since all vertices in C i are twins. We may thus simply try all possibilities of selecting at most one vertex per class C i , and all possibilities of selecting vertices of S. Thus, there are 2 t+c = 2 2 O(c) potential multipackings of D containing all sources. Each of them can be checked in polynomial time. This is an FPT algorithm.

Conclusion
We have studied Broadcast Domination and Multipacking on various subclasses of digraphs, with a focus on DAGs. It turns out that they behave very differently than for undirected graphs. We feel that Multipacking is slightly more challenging.
Indeed, we managed to solve some questions for Broadcast Domination, that we leave open for Multipacking. For example, it would be interesting to see whether Multipacking is FPT for DAGs, and whether it remains W[1]-hard for digraphs without directed 2-cycles. Also, Broadcast Domination is FPT for nowhere dense graphs (as it can be expressed in first-order logic), however, it is not clear whether this holds for Multipacking. It is also unknown whether Multipacking is NP-hard on undirected graphs, as asked in [31,32].
On the other hand, we showed that Multipacking is NP-complete for single-sourced DAGs, but we do not know whether the same holds for Broadcast Domination.
We note that in most of our hardness reductions, the maximum finite distance is very small (which helps us to control the problems at hand), but the actual diameter is infinite (as our digraphs are not strongly connected). It seems a challenging question to derive hardness results for strongly connected digraphs, which can be seen as an intermediate class between the two extremes that are undirected graphs, and DAGs.
We have also shown that both problems are FPT when parameterized by the vertex cover number. What about smaller parameters such as tree-width or DAG-width?
Finally, can our FPT algorithms for both problems parameterized by the solution cost/solution size and maximum out-degree be strengthened to a polynomial kernel?