A dominating set in a digraph is a set of vertices of such every vertex not in has an in-neighbour in . The domination number of is the smallest size of a dominating set of . The area of domination is one of the main topics in graph theory: see the two classic books [10, 11] on the subject. While there are hundreds of papers on domination in undirected graphs,444A set of vertices of an undirected graph is a dominating set if it is a dominating set of the digraph obtained from by replacing each edge by two symmetric arcs. domination in digraphs is less studied; we refer to the papers [8, 14, 15] for some examples. One particular variation of domination is the concept of location-domination, introduced by Slater for undirected graphs in  (see also [17, 18]). For a set of vertices of a digraph , two vertices and of are located by if there is a vertex of that is an in-neighbour of exactly one vertex among and . The set is a locating set of if it locates all the pairs of (but does not necessarily dominate the graph). Equivalently, every vertex not in has a distinct set of in-neighbours in . The location number of a graph is the size of a smallest locating set and is denoted by . A set of vertices of a digraph is locating-dominating if it is both locating and dominating. The location-domination number of a digraph is the smallest size of a locating-dominating set of . Note that , since at most one vertex is not dominated in a locating set.
Our goal is to investigate bounds on the location-domination number of digraphs. Such bounds are absent from the literature; in fact, the only paper on location-domination in digraphs we are aware of is , which deals with the computational complexity of the problem. Such bounds on digraphs have been studied for the closely related concept of identifying codes in .
We now introduce our terminology. We will assume that all the considered digraphs are loopless and have no multiple arcs. A digraph contains a set of vertices and a set
of arcs, that are ordered pairs of vertices. If there is an arc fromto , we say that is an in-neighbour of , and that is an out-neighbour of . The open in-neighbourhood and the open out-neighbourhood of a vertex , denoted by and , respectively, are the set of in-neighbours of and the set of out-neighbours of , respectively. Further, the closed in-neigbourhood of is and the closed out-neigbourhood of is . A source is a vertex with no in-neighbours, and a sink is a vertex with no out-neighbours. Two vertices are called twins if they have the same open in-neighbourhood or the same closed in-neighbourhood. If two vertices and satisfy , they are called quasi-twins. See Figure 1 for an illustration. A directed path is a sequence of vertices where each vertex has an arc to the next vertex in the sequence. A directed cycle is a directed path where the first and the last vertex are the same. An acyclic digraph is a digraph with no directed cyle. A tournament is a digraph in which there is a unique arc between any pair of vertices. A tournament is transitive if it is acyclic. A digraph is called strongly connected if there exists a directed path in both directions between every pair of vertices.
Perhaps the most classic result in domination of undirected graphs is the theorem of Ore  which states that any undirected graph without isolated vertices has a dominating set of size at most half the order. Such a theorem does not hold for the location-domination number of undirected graphs, for example complete graphs and stars of order have location-domination number , see . Nevertheless, Garijo, González and Márquez have conjectured in  that in the absence of twins, the upper bound of Ore’s theorem also holds for the location-domination number of undirected graphs; they also proved that an upper bound of roughly two thirds the order holds in this context (see [4, 5, 6] for further developments on this matter).
Therefore, it is natural to ask whether similar bounds exist in the case of locating-dominating sets of digraphs. However, Ore’s theorem does not hold for digraphs. Indeed, not only every isolated vertex but also every source of a digraph belongs to any dominating set of . For example, a star of order with sources has domination number (it is easy to see that this is the only digraph of order with no isolated vertices and domination number , see ). Lee showed (as part of a more general result) in  that every source-free digraph of order has domination number at most . Moreover, the digraph of order consisting of vertex-disjoint directed triangles is source-free and has domination number , so this bound is almost tight. Better bounds have been obtained for specific classes: every tournament of order has domination number at most  (the order is tight for random tournaments), and every strongly connected digraph of order has domination number at most  (this is tight for directed cycles).
What happens for the location-domination number? A first question of interest is to determine which graphs have largest possible location-domination number. We address this question in Section 3, where we describe the class of graphs of order with location-domination number . Of course it includes stars with sources (that have domination number ), bidirected complete graphs and bidirected stars (that correspond to undirected graphs with location-domination number ), but as we will see, there are many more examples.
In Section 4, we devise a general technique to obtain small locating-dominating sets in twin-free digraphs. This technique is a refinement of those used in [6, 9, 12]. We use this technique in Section 5 to show that every source-free and twin-free digraph of order has a locating-dominating set of size at most . This bound is improved to if moreover the digraph has no quasi-twins. By adding one to these bounds, they also hold even in the presence of sources. We will show that there exist strongly connected twin-free and quasi-twin-free digraphs of order with location-domination number .
We then show in Section 6 that any tournament of order has a locating-dominating set of size at most (this is tight for transitive tournaments and other examples).
In Section 7, we show that the same bound holds for twin-free acyclic digraphs (this is also tight, for directed paths).
We start with some useful propositions.
The following propositon generalizes a well-known fact known for locating-dominating sets of undirected graphs .
Let be a digraph with a set of pairwise twins (open or closed) or quasi-twins. There are at least vertices of in any locating-dominating set of .
By contradiction, let be a locating-dominating set that does not contain two mutual twins or quasi-twins and . The vertices and have the same in-neighbourhood in , and also in . This is a contradiction. ∎
However, we note that, unlike twins, there cannot exist a set of three pairwise quasi-twins.
Let be three vertices. If are quasi-twins and are quasi-twins, then cannot be quasi-twins.
Suppose without loss of generality that we have the arc from to . There are two cases. If we have the arc from to , then we must have both arcs from to and from to , so and cannot be quasi-twins. If we have the arc from to , then we also have the arc from to because and are quasi-twins. But then is an in-neighbour only of and not , so and cannot be quasi-twins. ∎
We now present a family of twin-free digraphs of order that have location-domination number almost . The graph is drawn in Figure 2.
Let be the strongly connected twin-free and quasi-twin-free digraph of order obtained from vertex-disjoint directed triangles by adding a new vertex that is an out-neighbour all vertices of each triangle, a vertex that is an in-neighbour of all vertices of each triangle, and an arc from to . Then, we have .
To see that , consider the locating-domnating set shown in Figure 2: take , one vertex of some directed triangle, and two vertices of all the other directed triangles.
To see that , consider a locating-dominating set of . First, each of the original directed triangle contains a vertex of , because otherwise the three vertices in this triangle are not located. Second, there is at most one directed triangle that contains only one vertex of , because otherwise in both triangles there is a vertex only dominated by , and they are not located. Finally, if some triangle contains only one vertex of , then because otherwise some vertex of the triangle is not dominated. So, in total there are two vertices of for each directed triangle. Since there are original directed triangles, the proof is finished. ∎
3 Characterizing digraphs with location-domination number
In this section, we characterize those digraphs with maximum possible location-domination and location numbers.
For every digraph of order , any set of size is a locating set, thus . This is not true for locating-dominating sets: consider a digraph with no arcs. However, this is the only such example: if has some arc, then it has a locating-dominating set of size at most (consider the set obtained from by removing the endpoint of one arbitrary arc).
We say that a vertex is universal if it is an in-neighbour of all other vertices. We first characterize those digraphs of order with (of course, unless it has no arc, such a digraph also satisfies ).
Let be a digraph of order . We have if and only if every vertex is either universal, or a sink.
Recall that always holds. If some vertex has an out-neighbour and there exists another vertex that is not an out-neighbour of , then is a locating set of of size . Thus, if , then every vertex in is either universal or a sink.
On the other hand, suppose that every vertex of is either universal or is a sink. Let be the set of universal vertices, and , the set of sinks. Every vertex is dominated by the set . Thus, if two distinct vertices are not in a locating set , they are both dominated exactly by the vertices in , a contradiction. Thus we have . ∎
We call a digraph a directed star if it has a special vertex that belongs to all the arcs, and there are no isolated vertices (see Figure 3). In other words, the underlying undirected graph of is a star.
Directed stars form another family of connected digraphs with large location-domination number.
For any directed star of order , we have .
Let be the center of (the vertex belonging to all arcs). Since , there is at least one arc in , and so . Let be a locating-dominating set of . Clearly, every source of belongs to . If two neighbours of do not belong to , then they are not located. Thus, at most one neighbour of is missing from . In the case where exactly one such vertex is not in , in order to be dominated, must belong to . This shows that . ∎
The only connected undirected graphs of order with location-domination number are stars and complete graphs  (seen as digraphs, they correspond to bidirected stars and bidirected complete graphs). As we already saw, there are more digraph examples. We now characterize all of them.
For a connected digraph of order , we have if and only if at least one of the following conditions holds:
is a directed star;
can be partitoned into three (possibly empty) sets , and , such that and are independent sets, is a bidirected clique, and the remaining arcs in are all the possible arcs from to and those from to .
Assume first that (then is a directed star) or . If two vertices of are not in some locating-dominating set of , then these two are either not located or one of them is not dominated, a contradiction. Thus, necessarily and we may assume in the remainder that .
By Proposition 5, (b) implies that . Next, assume that (c) holds. Then, is obtained from a digraph where all vertices are either universal (those in ) or sinks (those in ), by adding the set . Since all vertices in are sources, they all belong to any locating-dominating set of . Moreover, every vertex in is dominated by all vertices in , that is, we need a locating set of in every locating-dominating set of . By Proposition 4, such a set has size exactly , thus .
We must now prove the converse: assume that (and ). We must prove that (b) or (c) holds.
Case 1. Suppose first that contains some sources, and let be the set of these sources. If there is a vertex in and two vertices and of that are located by , then is a locating-dominating set of of size , a contradiction. Thus all the vertices of have the same neighbourhood and contains all the arcs between and .
Consider now the subdigraph of induced by . If it has a locating set of size (assume that the two vertices and are those not in ), then the set would be a locating-dominating set of , a contradiction. Thus, we have , and by Proposition 4 the vertices of can be partitioned into sinks of (set ) and universal vertices of (set ). We let . If is empty, we are done, because then satisfies the condition (c), with , and . Thus, we assume that is nonempty. If contains an arc from vertex to vertex such that is nonempty, then we could construct a locating-dominating set of of size from by removing and any vertex of . Thus, must be an independent set, and if there is an arc from to , then . But since contains no sources, there is necessarily an arc from to , and so . But then, is a directed star and (b) holds, so we are done.
Case 2. Now, we assume that has no sources. If every vertex is either universal or a sink, satisfies (c) (with ) and we are done. Thus, we may assume that there exists a vertex with an out-neighbour , and a third vertex that is not an out-neighbour of . By assumption, is not a locating-dominating set; but since locates and , it is a locating set, and it certainly dominates . Thus, it does not dominate , that is, is the only in-neighbour of (recall that has an in-neighbour since has no sources). Let . If had an out-neighbour in , then would be locating-dominating, a contradiction. Similarly, if had an in-neighbour in , then would be locating-dominating, a contradiction. If there is an arc from a vertex to a vertex inside , then is a locating-dominating set of , a contradiction. Thus is an independent set. Now, if has no neighbour in , is a directed star and we are done. Thus, must have an out-neighbour in . But now, is locating-dominating, a contradiction. This completes the proof.∎
4 A general method to obtain locating-dominating sets of twin-free digraphs
Note that all graphs described in Section 3 of order have twins. What happens for twin-free digraphs?
In this section, we propose a general method to obtain locating-dominating sets of twin-free digraphs, based on special dominating sets. This method was used in  for the case of undirected graphs (a similar argument was also used in ). It was adapted to digraphs in  for quasi-twin-free digraphs, and here we extend it to all twin-free digraphs.
First we start with some definitions.
Let be a set of vertices of a digraph . The -partition of is the partition of where two vertices are in the same part if and only if they have the same set of in-neighbours in .
Given a vertex , an -external private neighbour of is a vertex outside that is an out-neighbour of but of no other vertex of in .
Suppose that is a twin-free digraph of order . Let be a dominating set of such that the -partition of contains at least parts (with ). Then, . Moreover, if is also quasi-twin-free, then .
Let be the -partition of , where are the parts of size and are the parts of size at least .
We assume that is maximal with the property that has at least parts (this is ensured by adding vertices to while this property holds).
Now, we let . We have the following property of .
Two vertices in are located by , unless they form a pair of quasi-twins.
Proof of claim. Clearly, if two vertices are in different parts of , they are located by some vertices in . Thus, by contradiction, let and be two vertices of belonging to some part of that are not quasi-twins but are not located by . Since is twin-free, there is a vertex in that can locate and : without loss of generality is an in-neighbour of but not . By our assumption . Now, consider and the corresponding -partition of (with and defined like before). Since , we have (because in , has been split into two parts). So (because ). This contradicts the choice of , which we assumed to be maximal with this property. ()
Since is a dominating set, Claim 8.A shows that in the absence of quasi-twins, is locating-dominating. Next, we address the case of quasi-twins. We first prove the following fact.
Any two pairs of quasi-twins in are disjoint.
Proof of claim. Note that two quasi-twins and in must belong to the same part of , since they have the same in-neighbours in . Let , , , be four vertices in such that and are two distinct pairs of quasi-twins (with and being in-neighbours of and , respectively). Assume by contradiction that the two pairs are not disjoint. Then, all the vertices in the two pairs belong to the same part of . If , then and must be twins, a contradiction. Similarly, if then and are twins. Thus, we may assume without loss of generality that . Then, we proceed similarly as in Claim 8.A: the set still satisfies the property that , contradicting the maximality of . This proves the claim. ()
Consider now the set that is without one (arbitrary) vertex from each part of . It is clear that is locating-dominating: all vertices of are located and dominated by .
We now assume that has no quasi-twins: then and are two locating-dominating sets of . If , we are done. So, assume that . We have , so . Recall that . Therefore,
If has some quasi-twins, we use the locating-dominating sets and . Again, if , we are done. So, assume that . Then, . We obtain:
and the proof is finished. ∎
5 A general bound for twin-free digraphs with few sources
We now apply the method of Section 4 for source-free digraphs.
Any source-free digraph has a minimum-size dominating set with at least distinct parts in the -partition of .
Let be a minimum-size dominating set. We choose with a maximum number of parts in the partition of .
Let be the set of vertices of that have an -external private neighbour. Next, we choose as a minimum-size set of vertices of that dominates all the vertices of (that is, dominates those vertices that are not in and are not an external private neighbour of any vertex in ). Finally, we let .
Note that is at most the number of parts of containing vertices with at least two neighbours in , since a vertex of suffices to dominate the vertices of each such part. Therefore, the number of parts in is at least . If , we are done. We may thus assume in the remainder that , that is, .
No vertex of has an in-neighbour in , for otherwise would be a dominating set, contradicting the minimality of . Since is source-free, has an in-neighbour in . Let be an arbitrary in-neighbour of , and let . The function is necessarily injective: if we had for two distinct vertices and of , then the set would be a smaller dominating set than . Thus .
Now, we let . Clearly, is a dominating set of of size . Moreover, every vertex of is an -external private neighbour of . Thus, the partition of has at least parts, which is more than . This contradicts the choice of and concludes the proof. ∎
We remark that the bound of Proposition 9 is tight by considering any digraph consisting only of vertex-disjoint directed triangles.
For any source-free and twin-free digraph of order , we have . If moreover is quasi-twin-free, then .
One can extend this result to twin-free digraphs with sources (note that there can only be one source, otherwise they would be open twins).
Let be a twin-free digraph of order . Then . If moreover is quasi-twin-free, then .
Let be the unique source of . Let be the collection of all subsets for . The set has order and thus there is a non-empty subset of that is not in , and thus is not the open in-neighbourhood of some vertex in . Let be the graph obtained from in which we add all the arcs between and . The graph is now source-free and twin-free. By Corollary 11, there is a locating-dominating set of of size at most . Then is a locating-dominating set of . Indeed, all the vertices are dominated and two vertices not in are still located by . Thus .
For the second part, we do the same reasonning but with containing all the subsets and , for . There are at most such sets and thus again there is a nonempty set such that adding the arcs between and does not create twins or quasi-twins in . The end of the proof is the same as in the first part. ∎
It is clear that there are no twins in tournaments. The method of Section 4 was applied to tournaments in  (in a weaker form): the bound was proved for every tournament by showing that has a dominating set with at least parts in the -partition of .
In fact, using a different technique, we prove a much stronger bound, which is also tight. We first prove this bound in transitive tournaments (for which it is actually the exact value).
For a transitive tournament of order , we have and .
Let where has each with as its out-neighbour. To see that , consider the dominating set containing all ’s with odd. Two vertices and () not in are located by . Now, let be a locating-dominating set of . We need in , otherwise it is not dominated. Then, every two vertices , are quasi-twins, thus by Proposition 1, one of them needs to be in . Consider the sets with even and : they are disjoint, and each contains one vertex of . There are such sets, so we have
For locating sets, if is odd, consider the set that contains all ’s with even. Then is a locating set of size . Since , this set is optimal. If is even, at least one vertex in any set with odd must be in the locating set. This gives at least vertices and thus . ∎
We now extend the upper bound to any tournament.
For any tournament of order , we have and .
We prove the result by induction. By Proposition 12, this is true for any transitive tournament, and in particular if . Let and assume that this is true for any tournament of order .
Let be a tournamement of order that is not transitive. We first find a locating set of size .
Let be any vertex. Let and be the -partition of . Let and be the sizes of and , respectively. Let and be two optimal locating sets of the tournaments induced by and , respectively. By induction, and have size at most and . Consider the set . It is a locating set of since any pair , with and is located by . Its size is at most which is equal to if or is odd. In this case we are done, so we can assume that both and are even and thus is odd. Since we chose an arbitrary vertex , one can also assume that all the vertices have even out-degree and in-degree (if not, we are done).
Consider now two arbitrary vertices and with an arc from to . Let , , , be the -partition of ( contains the vertices that have but not in their in-neighbourhood and the other notations follow the same logic). As before, if one takes a minimum locating set in each part of the partition and add and , one obtains a locating set of . If there are three odd-sized sets among the four sets, one obtains, using the induction hypothesis, a locating set of size at most . Note that if has odd size, then so does since has even out-degree and all its out-neighbours are in or . Since the total number of vertices is odd, there is another odd-sized set among and (which must actually be ). In total, there are three odd-sized sets among the sets of the partition and thus the locating set has size at most . Thus, one can assume that for any pair of vertices of , the number of common out-neighbours is even.
We now consider three vertices , and that induce a directed triangle (this exists since is not transitive). Again, we consider the -partition of , we consider a minimum locating set of each part and we add and and to obtain a locating set. If there are four odd-sized sets in the partition, the obtained locating set has, by induction, size at most . In fact, this is always the case: if (the vertices that have , and as in-neighbours) is odd-sized, then is also odd-sized since is exactly the set of common out-neighbours of and , which is even-sized. In the same way, and are odd-sized and we are done. If is even-sized, then using the same argument, , , are also even-sized. But then, is the out-neighbourhood of and has even size. Thus is odd-sized, and similarly, and are also odd-sized. Since the order of is odd-sized, must also be odd-sized and we are done.
We now prove in a similar way that . If is odd, this is clear since by the previous result. Thus we assume that is even. We first take an arbitrary vertex and consider the -partition ,. If one takes a locating-dominating set for , a locating set for and adds , one obtains a locating-dominating set. Since is even, exactly one set among and has odd size. If is even, by induction, this gives a locating-dominating set of size at most and we are done. Thus, one can assume that all the vertices have odd in-degree and even out-degree.
Consider now two arbitrary vertices and with an arc from to in and the associated -partition. Taking a locating-dominating set for , locating sets for the three other parts and and gives a locating-dominating set. If two sets of the partition that are not are odd-sized, this gives a locating-dominating set of size , and we are done. So, we can assume that among and , exactly one set is odd. Indeed, has even out-degree and its out-neighbourhood is . Thus, if is odd-sized, we are done. Hence, we can assume that is even-sized, which implies that is also even-sized. In particular, we can now suppose that all pairs of vertices have a common out-neighbourhood that has even size.
Finally, consider an oriented triangle and the associated -partition. As before, by taking , , , a locating-dominating set of and locating sets in the seven other parts, we obtain a locating-dominating set. Since is even, there is an odd number of odd-sized sets in the partition. If there are three odd-sized sets that are not , then the total locating-dominating set has size at most (indeed, if is also odd-sized, then there will be a fourth odd-sized set that is not ). As before, if is odd-sized, then , , are also odd-sized and we are done. If it is even-sized, since , and have even out-degree, we have , and that are odd-sized and we are also done.
Thus, there is always a locating-dominating set of of size . ∎
Transitive tournaments are not the only tight example. For an integer , let be the tournament of order obtained from a collection of vertex-disjoint directed triangles, with arcs going from all vertices of to all vertices of whenever .
The tournament of order satisfies .
Let be an optimal locating-dominating set of . There must be at least one vertex of in any , otherwise the vertices of are not located. Furtheremore, there must be two vertices of in to dominate the three vertices of .
Assume that there are two consecutive triangles, and , each with only one vertex in . Let and be the vertices of that are in and . Then, the out-neighbour of in and the in-neighbor of in are not located, a contradiction.
Thus, there must be at least three vertices of in any two consecutive triangles, and at least two vertices of in the first triangle, which gives a total of at least vertices in . ∎
7 Twin-free acyclic digraphs
Twins generalize sources (since two sources are twins) but in a twin-free digraph we may have up to one source. This allows us to consider twin-free acyclic digraphs (they have exactly one source) in this section.
We will need the following theorem of Bondy, rephrasesd for our context.
Theorem 15 (Bondy ).
Let , be two disjoint sets of vertices in a digraph such that the vertices of have distinct in-neighbourhoods in . Then, there is a subset of size at most such that the vertices of have distinct in-neighbourhoods in .
Now we can prove the following. The proof technique is similar, but more complicated, as the one used to prove the same bound for the domination number of twin-free digraphs in .
If is a twin-free acyclic digraph of order , then .
Let be the unique source. We partition the vertex set step by step. For , the set contains all sources in . Since is acyclic and has only one source all the vertices are in some . Let be the last non-empty . See Figure 4 for a picture.
The following claims are a direct consequence of the construction.
Let (). Then has an in-neighbour in .
There is no arc inside any set .
There is no arc from to for .
Now we construct a locating-dominating set of . For , we construct step by step sets such that dominates and locates the vertices of , where .
Let and assume that the sets for are constructed and let (we start for with .)
Let be the -partition of . In particular, vertices in this partition are are twins with respect to . For each , let be any vertex of . By Bondy’s theorem (Theorem 15), there is a set of at most vertices in that locates and dominates .
Another time by applying Bondy’s theorem, for any there is a set of vertices in that locates . Note that the vertices of must belong to . Let . Then dominates and that locates .
Now we prove that is a locating-dominating set. The set is a dominating set because for any out of (), is dominated by . It is also locating because if there are two vertices in , they are located by and if they are from different and for , is located by its in-neighbour in .
For the size of , because at each step of the construction, we have . Thus there are at most as many vertices in as in ( only contains vertices that are in with , thus is exactly all the vertices in the sets ). So, if (that is, contains ), we have . But if does not contain , we have in fact that there are at most as many vertices in as in . Thus, and
as wished. ∎
The bound of Theorem 16 is best possible by considering directed paths.
We conclude the paper with Table 1, summarizing the known upper bounds on the domination and location-domination numbers for certain classes of digraphs.
|class of digraphs|
|twin-free and||[Cor 11] ( [Prop 3])|
|twin-free, source-free and||[Cor 11] ( [Prop 3])|
|acyclic twin-free||[Thm 16]||[Thm 16]|
|tournaments||||[Thm 13, Prop 12]|
|strongly connected||||( [Prop 3])|
The main question arising from our results is whether every twin-free digraph of order admits a locating-dominating set of size . Also, it would be interesting to determine which are the tournaments and the twin-free acyclic digraphs of order with location-domination number exactly .