1 Introduction
Edge Dominating Set (EDS) is a classical graph problem, equivalent to Minimum Dominating Set on line graphs. Despite the problem’s prominence, EDS has until recently received very little attention in the context of directed graphs. In this paper we investigate the complexity of a family of natural generalizations of this classical problem to digraphs, building upon recent work [23].
One of the reasons that EDS has not so far been well studied in digraphs is that there are several natural ways in which the undirected version can be generalized. For example, seeing as EDS is exactly Dominating Set in line graphs, one could define Directed EDS as (Directed) Dominating Set in line digraphs [24]. In this formulation, an arc dominates all arcs ; however does not dominate . Another natural way to define the problem would be to consider Dominating Set on the underlying graph of the line digraph, so as to maximize the symmetry of the problem, while still taking into account the directions of arcs. In this formulation, dominates arcs coming out of and arcs coming into , but not other arcs incident on .
A unifying framework for studying such formulations was recently given in [23], which defined dEDS for any two nonnegative integers . In this setting, an arc dominates every other arc which lies in a directed path of length at most that begins at , or lies in a directed path of length at most that ends at . In other words, dominates arcs in the forward direction up to distance , and in the backward direction up to distance . The interest in defining the problem in such a general manner is that it allows us to capture at the same time Directed Dominating Set on line digraphs (dEDS), Dominating Set on the underlying graph of the line digraph (dEDS), as well as versions corresponding to Dominating Set in the line digraph. We thus obtain a family of optimization problems on digraphs, with varying degrees of symmetry, all of which crucially depend on the directions of arcs in the input digraph.
Param.  FPT / Whard  Kernel  Approximability  

k  [23] [Thm.4]  vertices [Thm.11]  3apprx [Thm.5]  
[23] [Thm.3]  vertices [Thm.9]  8apprx [Thm.7]  
W[2]hard [23]    no approx [23]  
tw  any  W[1]hard [Thm.15]     
tw+p+q  any  FPT [Thm.16]  unknown   
Our contribution:
In this paper we advance the state of the art on the complexity of Directed Edge Dominating Set on two fronts.^{1}^{1}1We note that in the remainder we always assume that , as in the case where we can reverse the direction of all arcs and solve dEDS.
First, we study the complexity and approximability of the problem in general. The problem is NPhard for all values of (except ), even for planar boundeddegree DAGs [23], so it makes sense to study its parameterized complexity and approximability. We show that its two most natural cases, dEDS and dEDS, admit FPT algorithms with running times and respectively, where is the size of the optimal solution. These algorithms significantly improve upon the FPT algorithms given in [23], which uses the fact that the treewidth (of the underlying graph of the input) is at most and runs dynamic programming over a treedecomposition of width at most , obtained by the algorithm of [6]
. The resulting runningtime estimate for the algorithm of
[23] is thus around . Though both of our algorithms rely on standard branching techniques, we make use of several nontrivial ideas to obtain reasonable bases in their running times. We also show that both of these problems admit polynomial kernels. These are the only cases of the problem which may admit such kernels, since the problem is Whard for all other values of [23].Furthermore, we give an approximation for dEDS and a approximation for dEDS. We recall that [23] showed an approximation for general values of , and a matching logarithmic lower bound for the case . Therefore our result completes the picture on the approximability of the problem by showing that the only two currently unclassified cases belong in APX.
Finally, we consider the problem’s complexity parameterized by the treewidth of the underlying graph and show that, even though the problem is FPT when all of are parameters, it is in fact W[1]hard if parameterized only by tw; in fact, more strongly, we show that the problem is W[1]hard when parameterized by pathwidth and the size of the optimal (see Table 1).
Our second contribution in this paper is an analysis of the complexity of the problem on tournaments, which are one of the most wellstudied classes of digraphs (see Table 2). One of the reasons for focusing on this class is that the complexity of Dominating Set has a peculiar status on tournaments, as it is solvable in quasipolynomial time, W[2]hard, but neither in P nor NPcomplete (under standard assumptions). Here, we provide a complete classification of the problem which paints an even more surprising picture. We show that dEDS goes from being in P for ; to being APXhard and unsolvable in under the (randomized) ETH for ; to being equivalent to Dominating Set on tournaments, hence NPintermediate, quasipolynomialtime solvable, and W[2]hard, when one of and equals ; and finally to being polynomialtime solvable again if and neither nor equals 2. We find these results surprising, because few problems demonstrate such erratic complexity behavior when manipulating their parameters and because, even though in many cases the problem does seem to behave like Dominating Set, the fact that dEDS becomes significantly harder shows that the problem has interesting complexity aspects of its own. The most technical part of this classification is the reduction that establishes the hardness of dEDS, which makes use of several randomized
tournament constructions, that we show satisfy certain desirable properties with high probability; as a result our reduction itself is randomized.
Related Work:
On undirected graphs Edge Dominating Set, also known as Maximum Minimal Matching, is NPcomplete even on bipartite, planar, bounded degree graphs as well as other special cases [37, 25]. It can be approximated within a factor of 2 [20] (or better in some special cases [9, 32, 2]), but not a factor better than [10] unless P=NP. The problem has been the subject of intense study in the parameterized and exact algorithms community [35], producing a series of improved FPT algorithms [18, 4, 19, 33]; the current best is given in [26]. A kernel with vertices and edges is also known [22].
For dEDS, [23] shows the problem to be NPcomplete on planar DAGs, in P on trees, and W[2]hard and inapproximable on DAGs if . The same paper gives FPT algorithms for . Their algorithm performs DP on a treedecomposition of width in , and uses the fact that , and the algorithm of [6] to obtain a decomposition of width .
Dominating Set is known not to admit an approximation [13, 30], and to be W[2]hard and unsolvable in time under the ETH [14, 11]. The problem is significantly easier on tournaments, as the optimal is always at most , hence there is a trivial (quasipolynomial)time algorithm. It remains, however, W[2]hard [15]. The problem thus finds itself in an intermediate space between P and NP, as it cannot have a polynomialtime algorithm unless FPT=W[2], and it cannot be NPcomplete under the ETH (as it admits a quasipolynomial time algorithm). The generalization of Dominating Set where vertices dominate their neighborhood has also been wellstudied in general [8, 12, 16, 27, 29]. This problem is much easier on tournaments for , as the size of the solution is always a constant [5].
2 Definitions and Preliminaries
Graphs and domination:
We use standard graphtheoretic notation. If is a graph, a subset of vertices and a subset of edges, then denotes the subgraph of induced by , while denotes the subgraph of that includes and all its endpoints. We let denote the disjoint set union of and . For a vertex , the set of neighbors of in is denoted by , or simply , and will be written as . We define and .
Depending on the context, we use for to denote either an undirected edge connecting two vertices , or an arc (a directed edge) with tail and head . An incoming (resp. outgoing) arc for vertex is an arc whose head (resp. tail) is . In a directed graph , the set of outneighbors (resp.inneighbors) of a vertex is defined as (resp. ) and denoted as (resp. ). Similarly as for undirected graphs, and respectively stand for the sets and . For a subdigraph of and subsets , we let denote the set of arcs in whose tails are in and heads are in .
We use (resp. ) to denote the set (resp. the set ). If is a singleton consisting of a vertex , we write (resp. ) instead of (resp. ). The indegree (respectively outdegree ) of a vertex is defined as (resp. ), and we write to denote . We omit if it is clear from the context. If is for some vertex or arc set of , then we write in place of .
A source (resp. sink) is a vertex that has no incoming (resp. outgoing) arcs. A vertex is said to incover every incoming arc and outcover every outgoing arc for some . Here, for a path , the length of the path is defined as the number of arcs, that is, .
A directed graph is strongly connected if there is a path in each direction between each pair of vertices. A strongly connected component of a directed graph is a maximal strongly connected subgraph. The collection of strongly connected components forms a partition of the set of vertices of , while it also has a topological ordering, i.e. a linear ordering of its components such that for every arc , comes before in the ordering. If each strongly connected component of is contracted to a single vertex, the resulting graph is a directed acyclic graph (DAG).
For integers , an arc is said to dominate itself, and all arcs that are on a directed path of length at most to or on a directed path of length at most from . The central problem in this paper is Directed Edge Dominating Set (dEDS): given a directed graph , a positive integer and two nonnegative integers , we are asked to determine whether an arc subset of size at most exists, such that every arc is dominated by . Such a is called a edge dominating set of .
The Dominating Set problem is defined as follows: given an undirected graph , we are asked to find a subset of vertices , such that every vertex not in has at least one neighbor in : . For a directed graph , every vertex not in is required to have at least one incoming arc from at least one vertex of : .
We also use the Multicolored Clique problem, which is defined as follows: given a graph , with partitioned into independent sets , , we are asked to find a subset , such that forms a clique with . The problem Multicolored Clique is wellknown to be W[1]complete [17].
Complexity background:
We assume that the reader is familiar with the basic definitions of parameterized complexity, such as the classes FPT and W[1], as well as the Exponential Time Hypothesis (ETH, see [11]). For a problem , we let denote the value of its optimal solution. We also make use of standard graph width measures, such as vertex cover number vc, treewidth tw and pathwidth pw [11].
Treewidth and pathwidth:
A tree decomposition of a graph is a pair with a tree and a family of subsets of (called bags), one for each node of , with the following properties:

;

for all edges , there exists an with ;

for all , if is on the path from to in , then .
The width of a tree decomposition is . The treewidth of a graph is the minimum width over all tree decompositions of , denoted by .
Moreover, for rooted , let denote the terminal subgraph defined by node , i.e. the induced subgraph of on all vertices in bag and its descendants in . Also let denote the neighborhood of vertex in and denote the distance between vertices and in , while (absence of subscript) is the distance in .
In addition, a tree decomposition can be converted to a nice tree decomposition of the same width (in time and with nodes): the tree here is rooted and binary, while nodes can be of four types:

Leaf nodes are leaves of and have ;

Introduce nodes have one child with for some vertex and are said to introduce ;

Forget nodes have one child with for some vertex and are said to forget ;

Join nodes have two children denoted by and , with .
Nice tree decompositions were introduced by Kloks in [28] and using them does not in general give any additional algorithmic possibilities, yet algorithm design becomes considerably easier.
Replacing “tree” by “path” in the above, we get the definition of pathwidth pw. We recall the following wellknown relation:
Lemma 1.
For any graph we have .
Tournaments:
A tournament is a directed graph in which every pair of distinct vertices is connected by a single arc. Given a tournament , we denote by the tournament obtained from by reversing the direction of every arc. Every tournament has a king (sometimes also called a king), i.e. a vertex from which every other vertex can be reached by a path of length at most 2. One such king is the vertex of maximum outdegree (see e.g. [5]). It is folklore that any tournament contains a Hamiltonian path, i.e. a directed path that uses every vertex. The Dominating Set problem can be solved by brute force in time on tournaments, by the following lemma:
Lemma 2 ([11]).
Every tournament on vertices has a dominating set of size .
3 Tractability
3.1 FPT algorithms
In this section we present FPT branching algorithms for dEDS and dEDS. Both algorithms operate along similar lines, taking into consideration the particular ways available for domination of each arc.
Theorem 3.
The dEDS problem parameterized by solution size can be solved in time .
Proof.
We present an algorithm that works in two phases. In the first phase we perform a branching procedure which aims to locate vertices with positive outdegree or indegree in the solution. The general approach of this procedure is standard (as long as there is an uncovered arc, we consider all ways in which it may be covered), and uses the fact that at most vertices have positive in or outdegree in the solution. However, in order to speed up the algorithm, we use a more sophisticated branching procedure which picks an endpoint of the current arc and completely guesses its behavior in the solution. This ensures that this vertex will never be branched on again in the future. Once all arcs of the graph are covered, we perform a second phase, which runs in polynomial time, and by using a maximum matching algorithm finds the best solution corresponding to the current branch.
Let us now describe the branching phase of our algorithm. We construct three sets of vertices . The meaning of these sets is that when we place a vertex in or we guess that has (i) positive outdegree and zero indegree in the optimal solution; (ii) positive indegree and zero outdegree in the optimal solution; (iii) positive indegree and positive outdegree in the optimal solution, respectively. Initially all three sets are empty. When the algorithm places a vertex in one of these sets we say that the vertex has been marked.
Our algorithm now proceeds as follows: given a graph and three disjoint sets , we do the following:

If , reject.

While there exists an arc with both endpoints unmarked, do the following and return the best solution:

Call the algorithm with and the other sets unchanged.

Call the algorithm with and the other sets unchanged.

Call the algorithm with and the other sets unchanged.

Call the algorithm with and the other sets unchanged.

Call the algorithm with , , and unchanged.

It is not hard to see that Step 1 is correct as is a lower bound on the sum of the degrees of all vertices in the optimal solution and therefore cannot surpass .
Branching Step 2 is also correct: in order to cover , the optimal solution must either take an arc coming out of (2a,2b), or an arc coming into (2c,2d), or, if none of the previous cases apply, it must take the arc itself (2e).
Once we have applied the above procedure exhaustively, all arcs of the graph have at least one marked endpoint. We say that an arc with , or with is covered. We now check if the graph contains an uncovered arc with exactly one marked endpoint. We then branch by considering all possibilities for its other endpoint. More precisely, if and is unmarked, we branch into three cases, where is placed in , or , or (and similarly if is the marked endpoint). This branching step is also correct, since the degree specification for the currently marked endpoint does not dominate the arc , hence any feasible solution must take an arc incident on the other endpoint.
Once the above procedure is also applied exhaustively we have a graph where all arcs either have both endpoints marked, or have one endpoint marked but in a way that if we respect the degree specifications the arc is guaranteed to be covered. What remains is to find the best solution that agrees with the specifications of the sets .
We first add to our solution all arcs , i.e. all arcs such that and , since there is no other way to dominate these arcs. We then define a bipartite graph . That is, contains all vertices in along with a copy of on one side, all vertices of and a copy of on the other side and all arcs in with tails in and heads in . We now compute a minimum edge cover of this graph, that is, a minimum set of edges that touches every vertex. This can be done in polynomial time by finding a maximum matching and then adding an arbitrary incident edge for each unmatched vertex. It is not hard to see that a minimum edge cover of this graph corresponds exactly to the smallest edge dominating set that satisfies the specifications of the sets .
To see that the running time of our algorithm is , observe that there are two branching steps: either we have an arc with both endpoints unmarked; or we have an arc with exactly one unmarked endpoint. In both cases we measure the decrease of the quantity . The first case produces two instances with (2a,2c), and three instances with . We therefore have a recurrence satisfying , which gives . For the second case, we have three branches, all of which decrease and we therefore also have in this case. Taking into account that, initially we get a running time of at most . ∎
Theorem 4.
The dEDS problem parameterized by solution size can be solved in time .
Proof.
We give a branching algorithm that marks vertices of . During the branching process we construct three disjoint sets: contains vertices that will have indegree in the optimal solution; contains vertices that have positive indegree in the optimal solution and for which the algorithm has already identified at least one selected incoming arc; and contains vertices that have positive indegree in the optimal solution for which we have not yet identified an incoming arc. The algorithm will additionally mark some arcs as “forced”, meaning that these arcs have been identified as part of the solution.
Initially, the algorithm sets . These sets will remain disjoint during the branching. We denote and .
Before performing any branching steps we exhaustively apply the following rules:

If , we reject. This is correct since no solution can have more than vertices with positive indegree.

If there exists an arc with , we reject. Such an arc cannot be covered without violating the constraint that the indegrees of remain .

If there exists a source , we set . This is correct since a source will obviously have indegree in the optimal solution.

If there exists an arc with and , we set and . This is correct since the only way to cover is to take it. We mark all arcs with tail as forced.

If there exists an arc with and , we set . This is correct, since we cannot cover by selecting it (this would give positive indegree).

If there exists an arc with and which is not marked as forced, then we set . We explain the correctness of this rule below.
The above rules take polynomial time and can only increase . We observe that contains no sources (Rule 3). To see that Rule 6 is correct, suppose that there is a solution in which the indegree of is , therefore the arc is taken. However, since , we have already marked another arc that will be taken, so the indegree of will end up being at least . Since is not a source (Rule 3), we replace with an arbitrary incoming arc to . This is still a valid solution.
The first branching step is the following: suppose that there exists an arc with . In one branch we set , and in the other branch we set and and mark as forced. This branching is correct as any feasible solution will either take an arc incoming to to cover , or, if not, will take itself. In both branches the size of increases by one.
Suppose now that we have applied all the above rules exhaustively, and that we cannot apply the above branching step. This means that is a vertex cover. If there is a vertex that has two inneighbors we branch as follows: we either set ; or we set , , and and mark the arc as forced. This is correct, since a solution will either take an incoming arc to , or the arc . The first branch clearly increases . The key observation is that also increases in the second branch, as Rule 6 will immediately apply, and place in .
Suppose now that none of the above applies. Because of Rule 6 there are no arcs from to . Because the second branching Rule does not apply, and because of Rule 4, each vertex only has inneighbors in and at most one inneighbor in . For each that has an inneighbor we select in the solution; for every other we select an arbitrary incoming arc in the solution; for each we select the incoming arcs that the branching algorithm has identified. We claim that this is a valid solution. Because of Rule 4 all arcs coming out of are covered, because of Rule 2 no arcs are induced by , and because of Rule 5 all arcs going into have a tail with positive indegree in the solution. We have selected in the solution every arc from to , and there are no arcs induced by , otherwise we would have applied the first branching rule. All arcs from to are marked as forced and we have selected them in the solution. Finally, all arcs with tail in are covered.
Because of the correctness of the branching rules, if there is a solution, one of the branching choices will produce it. All rules can be applied in polynomial time, or produce two branches with larger values of . Since this value never goes above , we obtain an algorithm. ∎
3.2 Approximation algorithms
We present here constantfactor approximation algorithms for dEDS, and dEDS. Both algorithms appropriately utilize a maximal matching.
Theorem 5.
There are polynomialtime approximation algorithms for dEDS and dEDS.
Proof.
We present an approximation algorithm for dEDS. The algorithm for dEDS is obtained by reversing the orientation of each arc and applying the algorithm for dEDS.
Let be an input directed graph. We partition into so that and are the sets of sources and sinks respectively, and . A edge dominating set is constructed as follows.

Add the arc set to .

For each vertex of , choose precisely one arc from and add it to . In other words, as long as there exists a vertex for which we have not yet selected any of its incoming arcs and which has an outgoing arc to a sink, we select arbitrarily an arc coming into .

Let be the subdigraph of whose arc set consists of arcs not dominated by thus far constructed. Let be a maximal matching in (the underlying graph of) and be the set of vertices touched by . Let be the tails of the arcs in and let be the set of umatched vertices which are not sinks in , that is and . To , we add all arcs of , an arbitrary incoming arc of for every , and an arbitrary incoming arc of for every .
The above construction can be carried out in polynomial time. Furthermore, in all steps where we add an arbitrary arc to a vertex , we have , therefore such an arc exists. Let us first observe that the constructed solution is feasible. Let , and be the set of arcs added to at step 1, 2 and 3 respectively. contains all arcs incident on , so all these arcs are covered. For each arc with we have selected an arc going into into , so is covered. Finally, for each arc with we consider the following cases: If and is the head of an arc of , then is covered since we selected all arcs of ; If and is a tail of an arc in then contains an arc going into , so is covered; If then , so we have selected an arc going into . In all cases is covered.
Let us now argue about the approximation ratio. Fix an optimal solution . First, note for we must have , because the only arc that can dominate an arc of is itself. Let .
Consider the set . We claim that for each the set contains either at least one arc of or all arcs with tail and head in . Let be a set of arcs constructed by selecting for each a distinct element of , or if no such element exists all the arcs with . We have because all vertices of have an outneighbor in . Let .
We will now argue that . We first observe that any (optimal) solution must contain at least one arc of for every . In order to justify step 3, the following claim provides a key observation.
Claim 6.
It holds that . Furthermore is an independent set in the underlying graph of .
Proof.
If there is an arc from to then , which implies that all arcs coming out of are dominated by . This means that is a sink in , which is a contradiction. If there is an arc from to then there is an arc going into that belongs to , which again makes a sink in , contradiction. Therefore, .
Suppose that is not an independent set in and let be an arc with . However, is maximal and are unmatched, which implies that the arc does not appear in . This means that either , which makes a sink in , or an arc going into belongs in , which makes a sink in . In both cases we have a contradiction. ∎
Let us now use the above claim to show that . First, observe that , as all vertices of are sinks in . Furthermore, all arcs of have their heads in , hence none of them have their heads in . Similarly, no arc of has its head in , because this would make its head a sink in . Therefore, all arcs with tail in that exist in are dominated by . We now observe that since is an independent set, no arc of can dominate two arcs with tails in . Therefore, .
We now have
In order to dominate the entire arc set , one needs to take at least arcs, because is a matching. Therefore, we have and we deduce
∎
Theorem 7.
There is a polynomialtime approximation algorithm for dEDS.
Proof.
Let be an input directed graph. We partition into so that and are the sets of sources and sinks respectively, and .
We construct an edge dominating set as follows.

Add the arc set to .

For each vertex of , choose precisely one arc from and add it to .

For each vertex of , choose precisely one arc from and add it to .

Let be the subdigraph of whose arc set consists of those arcs not dominated by thus far constructed. Let be a maximal matching in (the underlying graph of) . Let and be respectively the tails and heads of the arcs in . To , we add all arcs of , an arc of for every , and also an arc of for every .
Clearly, the algorithm runs in polynomial time. In particular, for any vertex considered in Steps 24, both and are nonempty and choosing an arc from a designated set is always possible. We show that is indeed an edge dominating set. Suppose that an arc is not dominated by . As the first, second and third step of the construction ensures that any arc incident with is dominated, we know that is contained in the subdigraph constructed at step 4. For and being a maximal matching, one of the vertices must be incident with . Without loss of generality, we assume is incident with (and the other cases are symmetric). If , then clearly the arc whose tail coincides with would dominate , a contradiction. If , then the outgoing arc of added to at step 4 would dominate , again reaching a contradiction. Therefore, the constructed set is a solution to dEDS.
To prove the claimed approximation ratio, we first note that is contained in any (optimal) solution because any arc of can be dominated only by itself. Note that these arcs do not dominate any other arcs of . Further, we have because in order to dominate any arc of the form with and , one must take at least one arc from . Since the collection of sets are disjoint over all , the inequality holds. Likewise, it holds that . In order to dominate the entire arc set , one needs to take at least arcs. This is because an arc can dominate at most two arcs of . That is, we have Therefore, it is . ∎
3.3 Polynomial kernels
We give polynomial kernels for dEDS and dEDS. We first introduce a relation between the vertex cover number and the size of a minimum edge dominating set, shown in [23] and then proceed to show a quadraticvertex/cubicedge kernel for dEDS.
Lemma 8 ([23]).
Given a directed graph , let be the undirected underlying graph of , be the vertex cover number of , and be a minimum edge dominating set in . Then .
Theorem 9.
There exists an vertex/edge kernel for dEDS.
Proof.
Given a directed graph , we denote the underlying undirected graph of by . Let be a minimum edge dominating set and be the size of a minimum vertex cover in . First, we find a maximal matching in . If , we conclude this is a noinstance by Lemma 8 and the wellknown fact that [21]. Otherwise, let be the set of endpoints of edges in . Then is a vertex cover of size at most for the underlying undirected graph of and is an independent set.
We next explain the reduction step. For each , we arbitrarily mark the first tail vertices of incoming arcs of with “in” (or all, if the indegree of is ) and also arbitrarily the first head vertices of outgoing arcs of with “out” (or all, if the outdegree of is ). After this marking, if there exists a vertex without marks “in”, “out”, we can delete it.
We next show the correctness of the above. First, we can observe that if some has more than incoming arcs, then any feasible solution of size at most must select an arc with tail . Similarly, if has more than outgoing arcs, any feasible solution of size at most must select an arc with head . Consider now an unmarked vertex and suppose that it is the tail of an arc with (the case where is the head is symmetric). The vertex has other incoming arcs, besides , otherwise would have been marked. Therefore, in any solution of size at most in the graph where has been deleted we must select an arc coming out of . This arc dominates . Therefore, any feasible solution of the new graph remains feasible in the original graph. For the other direction, suppose a solution for the graph selects the arc . We consider the same solution without in the graph where is deleted. If this is already feasible, we are done. If not, any nondominated arc must have as its tail (every other arc dominated by has been deleted). All these arcs can be dominated by adding to the solution an arc going into .
After exhaustively applying the above rule every vertex of the independent set will be marked. We mark at most vertices of the independent set for each of the at most vertices of , so we have a total of at most vertices. Moreover, there exist at most arcs between the sets of the vertex cover and the independent set. Therefore, the number of arcs in the reduced graph is at most . ∎
Next, we note that the size of a minimum edge dominating set is equal to, or greater than the size of a minimum edge dominating set. Thus, we have where is a edge dominating set. We give a more strict relation, however, between vc and the size of a minimum edge dominating set that is then used to obtain Theorem 11.
Lemma 10.
Given a directed graph , let be the undirected underlying graph of , be the vertex cover number of , and be a minimum edge dominating set in . Then .
Proof.
For an arc , the head vertex covers all arcs (i.e. edges) dominated by in . Since dominates all edges in , the set of head vertices of is a vertex cover in . Thus, . ∎
Theorem 11.
There exists an vertex/edge kernel for dEDS.
Proof.
Our first reduction rule states that if there exists an arc where is a source () and is a sink () then we delete this arc and set . This rule is correct because the only arc that dominates is the arc itself, and does not dominate any other arc. In the remainder we assume that this rule has been applied exhaustively.
We then find a maximal matching in the underlying undirected graph. If , then by Lemma 10 we conclude that we can reject. Otherwise, the set of vertices incident on , denoted by is a vertex cover of size at most and is an independent set.
Now, suppose that there exist vertices in with positive outdegree. This means that there exist arcs with distinct tails in , and heads in . No arc of the graph dominates two of these arcs (since is independent), therefore any feasible solution has size at least and we can reject.
We can therefore assume that the number of nonsinks in is at most . We will now bound the number of sinks. Let be the set of sinks, that is, contains all vertices for which . We edit the graph as follows: delete all vertices of ; add a new vertex which is initially not connected to any vertex; and then for each vertex such that there is an arc with in we add the arc . We claim that this is an equivalent instance.
Before arguing correctness, observe that the new instance has at most vertices: has at most vertices, has at most nonsinks, and all sinks of have been replaced by . This graph clearly has edges.
Let be the original graph and the graph obtained after replacing all sinks in the independent set with the new vertex . Consider an optimal solution in . If the solution contains an edge where is a sink, then we know that is not a source (otherwise we would have simplified the instance by deleting ). We edit the solution by replacing with an arbitrary arc incoming to . Repeating this gives a solution which does not include any arc whose head is a sink of , but for each such arc contains an arc going into . This is therefore a valid solution of , as it dominates all arcs going into . For the converse direction we similarly edit a solution to by replacing any arc with an arbitrary arc going into (again, we can safely assume that such an arc exists). The result is a valid solution for with the same size. ∎
4 Treewidth
In this section we characterize the complexity of dEDS parameterized by the treewidth of the underlying graph of the input. Our main result is that, even though the problem is FPT when parameterized by , it becomes W[1]hard if parameterized only by tw (in fact, also by pw), even if we add the size of the optimal solution as a second parameter. The algorithm is based on standard dynamic programming techniques, while for hardness we reduce from the wellknown W[1]complete Multicolored Clique problem [17].
4.1 Hardness for Treewidth
Construction:
Before we proceed, let us define a more general version of dEDS which will be useful in our reduction. Suppose that in addition to a digraph we are also given as input a subset of “optional” arcs. In Partial dEDS we are asked to select a minimum set of arcs that dominate all arcs of , that is, it is not mandatory to dominate the optional arcs. We will describe a reduction from Multicolored Clique to a special instance of Partial dEDS, and then show how to reduce this to the original problem without significantly modifying the treewidth or the size of the optimal.
Given an instance of Multicolored Clique, with and , where we assume without loss of generality that is even, we will construct an instance of Partial dEDS. We set . We begin by adding to all vertices of and connecting each set into a directed cycle of length . Concretely, we add the arcs for all and where addition is performed modulo .
Intuitively, the idea up to this point is that selecting the vertex in the clique is represented in the new instance by selecting the arc of the cycle induced by whose head is . In order to make it easier to prove that the optimal solution will be forced to select one arc from each directed cycle we add to our instance the following: for each we construct a directed cycle of length and identify one of its vertices with . We call these cycles the “guard” cycles.
Finally, we need to add some gadgets to ensure that the arcs selected really represent a clique. For each pair of vertices of , which are not connected by an edge in we do the following: we construct two new vertices and an arc connecting them; if we construct a directed path of length from to
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