    # Hardness and approximation for the geodetic set problem in some graph classes

In this paper, we study the computational complexity of finding the geodetic number of graphs. A set of vertices S of a graph G is a geodetic set if any vertex of G lies in some shortest path between some pair of vertices from S. The Minimum Geodetic Set (MGS) problem is to find a geodetic set with minimum cardinality. In this paper, we prove that solving the MGS problem is NP-hard on planar graphs with a maximum degree six and line graphs. We also show that unless P=NP, there is no polynomial time algorithm to solve the MGS problem with sublogarithmic approximation factor (in terms of the number of vertices) even on graphs with diameter 2. On the positive side, we give an O(√(n)log n)-approximation algorithm for the MGS problem on general graphs of order n. We also give a 3-approximation algorithm for the MGS problem on the family of solid grid graphs which is a subclass of planar graphs.

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## 1 Introduction and results

Suppose there is a city-road network (i.e. a graph) and a bus company wants to open bus terminals in some of the cities. The buses will go from one bus terminal to another (i.e. from one city to another) following the shortest route in the network. Finding the minimum number of bus terminals required so that any city belongs to some shortest route between some pair of bus terminals is equivalent to finding the geodetic number of the corresponding graph. Formally, an undirected simple graph has vertex set and edge set . For two vertices , let denote the set of all vertices in that lie in some shortest path between and . A set of vertices is a geodetic set if . The geodetic number, denoted as , is the minimum integer such that has a geodetic set of cardinality . Given a graph , the Minimum Geodetic Set (MGS) problem is to compute a geodetic set of with minimum cardinality. In this paper, we shall study the computational complexity of the MGS problem in various graph classes.

The notion of geodetic sets and geodetic number was introduced by Harary et al. 

. The notion of geodetic number is closely related to convexity and convex hulls in graphs, which have applications in game theory, facility location, information retrieval, distributed computing and communication networks

[2, 19, 15, 22, 10]. In 2002, Atici  proved that finding the geodetic number of arbitrary graphs is NP-hard. Later, Dourado et al. [9, 8] strengthened the above result to bipartite graphs, chordal graphs and chordal bipartite graphs. Recently, Bueno et al.  proved that the MGS problem remains NP-hard even for subcubic graphs. On the positive side, polynomial time algorithms to solve the MGS problem are known for cographs , split graphs , ptolemaic graphs , outer planar graphs  and proper interval graphs . In this paper, we prove the following theorem.

The MGS problem is NP-hard for planar graphs of maximum degree .

Then we focus on line graphs. Given a graph , the line graph of , denoted by is a graph such that each vertex of represents an edge of and two vertices of are adjacent if and only if their corresponding edges share a common endpoint in . A graph is a line graph if for some . Some optimisation problems which are difficult to solve in general graphs admit polynomial time algorithms when the input is a line graph [14, 17]. We prove the following theorem.

The MGS problem is NP-hard for line graphs.

From a result of Dourado et al. , it follows that solving the MGS problem is NP-hard even for graphs with diameter at most . On the other hand, solving the MGS problem on graphs with diameter is trivial (since those are exactly complete graphs). In this paper, we prove that unless P=NP, there is no polynomial time algorithm with sublogarithmic approximation factor for the MGS problem even on graphs with diameter at most . A universal vertex of a graph is adjacent to all other vertices of the graph. We shall prove the following stronger theorem.

Unless P=NP, there is no polynomial time -approximation algorithm for the MGS problem even on graphs that have a universal vertex, where is the number of vertices in the input graph.

On the positive side, we show that a reduction to the Minimum Rainbow Subgraph of Multigraph problem (defined in Section 3.1) gives the first sublinear approximation algorithm for the MGS problem on general graphs.

Given a graph, there is a polynomial-time -approximation algorithm for the MGS problem where is the number of vertices.

Then we focus on solid grid graphs, an interesting subclass of planar graphs. A grid embedding of a graph is a collection of points with integer coordinates such that each point in the collection represents a vertex of the graph and two points are at a distance one if and only if the vertices they represent are adjacent in the graph. A graph is a grid graph if it has a grid embedding. A graph is a solid grid graph if it has a grid embedding such that all interior faces have unit area. Approximation algorithms for optimisation problems like Longest path, Longest Cycle, Node-Disjoint Path etc. on grid graphs and solid grid graphs have been studied [4, 20, 6, 25, 23, 27]. In this paper, we prove the following theorem.

Given a solid grid graph, there is an time -approximation algorithm for the MGS problem, even if the grid embedding is not given as part of the input. Here is the number of vertices in the input graph.

Note that recognising solid grid graphs is NP-complete .

Organisation of the paper: In Section 2, we prove the hardness results for planar graphs, line graphs and graphs with diameter . In Section 3, we present our approximation algorithms. Finally we draw our conclusions in Section 4.

## 2 Hardness results

In Section 2.1, we prove that the MGS problem is NP-hard for planar graphs with maximum degree (Theorem 1). Then in Section 2.2 we prove that the MGS problem is NP-hard for line graphs (Theorem 1). In Section 2.3 we prove the inapproximability result (Theorem 1).

### 2.1 NP-hardness on planar graphs

Given a graph , a subset is a dominating set of if any vertex in has a neighbour in . The problem Minimum Dominating set (MDS) consists in computing a dominating set of an input graph with minimum cardinality. To prove Theorem 1, we shall reduce the NP-complete MDS problem on subcubic planar graphs  to the MGS problem on planar graphs with maximum degree .

We will show that has a dominating set of size if and only if has a geodetic set of size .

Assume first that has a dominating set of size . We construct a geodetic set of of size as follows. For each vertex in , we add the three vertices () of to . If is in , we also add vertex to .

Let us show that is indeed a geodetic set. First, we observe that, in any vertex gadget that is part of , the unique shortest path between two distinct vertices , has length  and goes through vertices , and (where ). Thus, it only remains to show that vertices and () belong to some shortest path of vertices of . Assume that is a vertex of in . The shortest paths between and have length  and one of them goes through vertex . Thus, all vertices of belong to some shortest path between vertices of . Now, consider a vertex of adjacent to and let be the vertex of that is farthest from . The shortest paths between and have length ; one of them goes through vertices and ; two others go through the two other vertices and . Thus, is a geodetic set.

For the converse, assume we have a geodetic set of of size . We will show that has a dominating set of size . First of all, observe that all the vertices of type are necessarily in , since they have degree . As observed earlier, the shortest paths between those vertices already go through all vertices of type and . However, no other vertex lies on a shortest path between two such vertices: these shortest paths always go through the boundary -cycle of the vertex-gadgets. Let be the set of the remaining vertices of . These vertices are there to cover the vertices of type and . We construct a subset of as follows: contains those vertices of whose vertex-gadget contains a vertex of . We claim that is a dominating set of . Suppose by contradiction that there is a vertex of such that neither nor any of (with adjacent to in ) contains any vertex of . Here also, the shortest paths between vertices of always go through the boundary -cycle of and thus, they never include vertex , a contradiction. Thus, is a dominating set of size , and we are done.

### 2.2 NP-hardness on line graphs

In this section, we prove that the MGS problem remains NP-hard on line graphs. For a graph and edges , define if shares a vertex and if shares a vertex with an edge with . A path between two edges is defined in the usual way.

###### Observation A.

A path between two edges of a graph corresponds to a path between the vertices and in .

Given a graph , a set is a line geodetic set of if every edge belongs to some shortest path between some pair of edges . Observation A implies the following.

###### Observation B.

A graph has a line geodetic set of cardinality if and only if has a geodetic set of size .

We shall show (in Lemma 2.2) that finding a line geodetic set of a graph with minimum cardinality is NP-hard. Then Observation B shall imply that solving the MGS problem on line graphs is NP-hard. For the above purpose we need the following definition. Given a graph , a set is a good edge set if for any edge , there are two edges such that (i) lies in some shortest path between and , and (ii) is or . Figure 2: (a) A triangle-free graph G and (b) the graph HG.

Computing a good edge set of a triangle-free graph with minimum cardinality is NP-hard.

###### Proof.

We shall reduce the NP-complete Edge Dominating set problem on triangle-free graphs  to the problem of computing a good edge set of a graph with minimum cardinality on triangle-free graphs. Given a graph , a set is an edge dominating set of if any edge shares a vertex with some edge in . The Edge Dominating Set problem is to compute an edge dominating set of with minimum cardinality.

Let be a triangle-free graph. For each vertex , take a new edge . Construct a graph whose vertex set is the union of and the set and . Notice that is a triangle-free graph and we shall show that has an edge dominating set of cardinality if and only if has a good edge set of cardinality where .

Let be an edge dominating set of . For each , let be written as . Notice that the set forms a good edge set of and has cardinality . Let be a good edge set of of size at most . Notice that for each , must contain the edge . Hence, the cardinality of the set is at most . Moreover, for each , there is an edge which is at distance from . As is a good edge set of , any edge in shares a vertex with some edge of . Hence is an edge dominating set of of cardinality at most . ∎

For a triangle-free graph , let be the graph with and where . See Figure 2(a) and Figure 2(b) for an example. We prove the following proposition.

For a triangle-free graph , there is a line geodetic set of with minimum cardinality such that .

###### Proof.

For a set , an edge covers an edge , if there is another edge such that lies in the shortest path between and . Notice that the edges lie in any line geodetic set of and all edges in are covered by and . First we prove the following claims.

Let be a line geodetic set of and . If does not cover any edge of , then is a line geodetic set of .

The proof of the above claim follows from the fact that all edges in are covered by and .

Let be a line geodetic set of and . There is another edge such that is a line geodetic set of .

To prove the claim above, first we define the ecentricity of an edge to be the maximum shortest path distance between and any other edge in . Notice that the ecentricity of any edge in is two and the ecentricity of any edge of in is at most three. Now remove all edges from which do not cover any edge of . By Claim 1, the resulting set, say , is a line geodetic set of . Let be an edge and let be the set of edges covered by . Since the ecentricity of is two, there must exist in such that has a common endpoint with both and for each . Therefore the distance between and is two for each . As is triangle-free, for any . Choose any edge . Observe that the distance between and is two when . Therefore, for each , the edge lies in the shortest path between and . Therefore, is a line geodetic set of .

Given any line geodetic set of , we can use the arguments used in Claim 1 and Claim 2 repeatedly on to construct a line geodetic set of such that and . Thus we have the proof. ∎

Computing a line geodetic set of a graph with minimum cardinality is NP-hard.

###### Proof.

We shall reduce the NP-complete problem of computing a good edge set of a triangle-free graph with minimum cardinality (Lemma 2.2). Let be a triangle-free graph. Construct the graph as stated above (just before Lemma 2.2). The set is also defined as before. We shall show that a triangle-free graph has a good edge set of cardinality if and only if has a line geodetic set of cardinality .

Let be a good edge set of . Notice that, for each edge , there are two edges such that belongs to a shortest path between and in . Also any edge of belongs to a shortest path between the edges and in . Hence is a line geodetic set of with cardinality .

Let be a line geodetic set of of size . Notice that and let . Due to Lemma 2.2, we can assume that does not contain any edge of . Let be an edge in and let such that lies in some shortest path between and in . Since the distance between and is at most three in , it follows that is a good edge set of with cardinality . ∎

### 2.3 Inapproximability on graphs with diameter 2

Given a graph , a set is a 2-dominating set of if any vertex has at least two neighbours in . The 2-MDS problem is to compute a -dominating set of graphs with minimum cardinality. We shall use the following result.

[[5, 7]] Unless , there is no polynomial time -approximation algorithm for the 2-MDS problem on triangle-free graphs.

We observe the following.

Let be a triangle-free graph and be the graph obtained by adding an universal vertex to . A set of vertices of is a geodetic set if and only if is a -dominating set of .

###### Proof.

Let be a geodetic set of . Observe that for any vertex there must exist vertices such that and . Hence, is a -dominating set of . Conversely, let be any -dominating set of . For any two vertex there exist such that . Since is triangle-free, and are non-adjacent. Hence, and is a geodetic set of . ∎

The proof of Theorem 1 follows due to Lemma 2.3 and Theorem 2.3.

## 3 Approximation algorithms

In Section 3.1 and Section 3.2 we present approximation algorithms for the MGS problem on general graphs and solid grid graphs, respectively.

### 3.1 General graphs

We will reduce the Minimum Geodetic Set problem to the Minimum Rainbow Subgraph of Multigraph (MRSM) problem. A subgraph of an edge colored multigraph is colorful if contains at least one edge of each color. Given an edge colored multigraph , the MRSM problem is to find a colorful subgraph of of minimum cardinality. The following is a consequence of a result due to Tirodkar and Vishwanathan .

[] Given an edge colored multigraph , there is a polynomial time -approximation algorithm to solve the MRSM problem where .

We note that Tirodkar and Vishwanathan  proved the above theorem for simple graphs only, but the proof works for multigraphs as well.

Given a graph form an edge colored multigraph as follows. The vertex set of is the same as . For each subset such that lies in some shortest path between and , add an edge in between and having the color . Observe that, has a geodetic set of cardinality if and only if has a colorful subgraph with vertices. The proof of Theorem 1 follows from Theorem 3.1.

### 3.2 Solid grid graphs

In this section, we shall give a linear time -approximation algorithm for the MGS problem on solid grid graphs. From now on shall denote a solid grid graph and is a grid embedding of where every interior face has unit area. Figure 3: (a) The black and gray vertices are the vertices of the corner paths. The gray vertices indicate the corner vertices. (b) The gray vertices are vertices of the red path. Vertices in the shaded box form a rectangular block. (c) Example of a solid grid graph whose number of corner vertices is exactly three times the geodetic number.

Let be a solid grid graph. A path of is a corner path if (i) no vertex of is a cut vertex, (ii) both end-vertices of have degree , and (iii) all vertices except the end-vertices of have degree . See Figure 3(a) for an example. Observe that for a corner path , either the -coordinates of all vertices of are the same or the -coordinates of all vertices of are the same. Moreover, all vertices of a corner path lie in the outer face of . The next observation follows from the definition of corner path and the fact that is a solid grid graph.

###### Observation C.

Let be a corner path of . Consider the set . Then induces a path in . Moreover, if the -coordinates (resp. the -coordinates) of all the vertices of are the same, then the -coordinates (resp. the -coordinates) of all vertices in are the same.

We shall use Observation C to prove a lower bound on the geodetic number of in terms of the number of corner paths of .

Any geodetic set of contains at least one vertex from each corner path.

###### Proof.

Without loss of generality, we assume the -coordinates of all vertices of are the same. By Observation C, the set induces a path and the -coordinates of all vertices in are the same. Now consider any two vertices and with a path between and that contains one of the end-vertices, say , of . Observe that can be expressed as such that and . Then there is a path where for , is the vertex in which is adjacent to in . Observe that the length of is strictly less than that of . Therefore whenever . Hence any geodetic set of contains at least one vertex from . ∎

Any geodetic set of contains all vertices of degree . Inspired by the above fact and Lemma 3.2, we define the term corner vertex as follows. A vertex of is a corner vertex if has degree or is an end-vertex of some corner path. See Figure 3(a) for an example. Observe that two corner paths may have at most one corner vertex in common. Moreover, a corner vertex cannot be in three corner paths. Therefore it follows that the cardinality of the set of corner vertices is at most .

Note that there are solid grid graphs whose number of corner vertices is exactly three times the geodetic number. See Figure 3(c) for one such example.

Now we prove that the set of all corner vertices of is indeed a geodetic set of . We shall use the following proposition of Ekim and Erey .

[] Let be a graph and its biconnected components. Let be the set of cut vertices of . If is a minimum set such that is a minimum geodetic set of then is a minimum godetic set of .

The next observation follows from Theorem C.

###### Observation D.

Let be the set of corner vertices of and be the set of cut vertices of . Let be the set of biconnected components of . The set is a geodetic set of if and only if is a geodetic set of for all .

From now on, is the set of corner vertices of and are the biconnected components of . Due to Theorem C and Observation D, it is enough to show that for each , the set is a geodetic set of . First, we introduce some more notations and definitions below.

Let be a biconnected component of . Recall that each vertex of is a pair of integers and each edge is a line segment with unit length. An edge is an interior edge if all interior points of lie in an interior face of . For a vertex , let denote the maximal path such that all edges of are interior edges and each vertex in has the same -coordinate as . Similarly, let denote the maximal path such that all edges of are interior edges and each vertex in has the same -coordinate as . A path of is a red path if (i) there exists a such that and (ii) at least one end-vertex of is a cut-vertex or a vertex of degree . A vertex of is red if lies on some red path. See Figure 3(b) for an example.

A subgraph of is a rectangular block if satisfies the following properties.

1. For any two vertices of , we have that any pair with and is a vertex of .

2. Let be the maximum and minimum -coordinates of the vertices in . The -coordinate of any red vertex of must be equal to or . Similarly, let be the maximum and minimum -coordinates of the vertices in . The -coordinate of any red vertex of must be equal to or .

Observe that can be decomposed into rectangular blocks such that each non-red vertex belongs to exactly one rectangular block. See Figure 3(b) for an example. Let be a decomposition of into rectangular blocks. Recall that is the set of corner vertices of and is the set of cut vertices of . We have the following lemma.

For each , there are two vertices such that .

###### Proof.

Let be an arbitrary rectangular block. A vertex of is a northern vertex if the -coordinate of is maximum among all vertices of . Analogously, western vertices, eastern vertices and southern vertices are defined. A vertex of is a boundary vertex if it is either northern, western, southern or an eastern vertex of . Let be the vertex of which is both a northern vertex and a western vertex. Similarly, denotes the vertex which is both northern vertex and eastern vertex, denotes the vertex of which is both southern and western vertex and denotes the vertex of which is both southern and eastern vertex.

First we prove the lemma assuming that all boundary vertices of are red vertices. Let (resp. ) denote the vertex with minimum -coordinate such that (resp. ) contains (resp. ). Similarly, let (resp. ) denote the vertex with maximum -coordinate such that (resp. ) contains (resp. ). Let (resp. ) denote the vertex with minimum -coordinate such that (resp. ) contains (resp. ). Let (resp. ) denote the vertex with maximum -coordinate such that (resp. ) contains (resp. ). Observe that the vertices lie on the exterior face of the embedding.

For two vertices , let denote the path between that can be obtained by traversing the exterior face of the embedding in the counter-clockwise direction starting from . Observe that, if both and (resp. and ) contain a corner or cut vertex each, say , then (resp. ) and therefore . Now consider the case when at least one of the paths in does not contain any corner vertex or cut vertex and when at least one of the paths in does not contain any corner vertex or cut vertex. Due to symmetry of rotation and reflection on grids, without loss of generality we can assume that both and have no corner vertex or cut vertex. Observe that in this case there must be a corner vertex in whose -coordinate is the same as that of and therefore of . If contains a corner vertex , then ) and therefore . Otherwise, there must be a corner vertex in whose -coordinate is the same as that of and therefore of . Hence we have and therefore in this case also.

Now we consider the case when there are some non-red boundary vertices of . Let be a non-red vertex of . Without loss of generality, we can assume that is a western vertex of . Now we redefine the vertices as follows. Let , and (resp. ) be the vertex with minimum -coordinate such that there is a path from to (resp. from to ) containing vertices with the same -coordinate as that of (resp. ). Similarly, let (resp. ) be the vertex with maximum -coordinate such that there is a path from to (resp. from to ) containing vertices with the same -coordinate as that of (resp. ). Finally, let (resp. ) be the vertex with maximum -coordinate such that there is a path from to (resp. from to ) containing vertices with the same -coordinate as that of (resp. ). Using similar arguments on the paths with as before, we can show that there exists corner vertices such that . Thus we have the proof. ∎

By Observation D and Lemma 3.2, is a geodetic set of .

Time complexity: If the grid embedding of is given as part of the input, then the set of corner vertices can be computed in time by simply traversing the exterior face of the embedding. Otherwise, the set of corner vertices can be computed in time as follows (we shall only describe the procedure to find corner vertices of degree two as the other case is trivial). Let be a biconnected component of , be a vertex of having degree and be its neighbours. If both and have degree , then is not a corner vertex. Moreover, if at least one of and have degree then is a corner vertex. Otherwise, apply the following procedure. Assume has degree and denote as for technical reasons. Set . As is a biconnected solid grid graph, and must have exactly one common neighbour which is different from . Denote this vertex as . Let be the neighbour of different from both and . If or is a cut vertex in then terminate. If then is a corner vertex. Otherwise, set and repeat the above steps. Observe that, when the above procedure terminates either we know that is a corner vertex or there is no corner path that contains both and . Now swapping roles of and in the above procedure, we can decide if is a corner vertex. We can find all the corner vertices of by applying the above procedure to all vertices of degree of . Similarly by applying the above procedure to all the biconnected components of , we can find all corner vertices. Notice that, the total running time of the algorithm remains linear in the number of vertices of .

This completes the proof of Theorem 1.

## 4 Conclusion

In this paper, we studied the computational complexity of the MGS problem in various graph classes. We proved that the MGS problem remains NP-hard on planar graphs and line graphs. This motivates the following question.

###### Question 1.

Are there constant factor approximation algorithms for the MGS problem on planar graphs and line graphs?

We gave an -approximation algorithm for the MGS problem on general graphs and proved that unless P=NP, there is no polynomial time -approximation algorithm for the MGS problem even on graphs with diameter . The following is a natural question in this direction.

###### Question 2.

Is there a -approximation algorithm for the MGS problem on general graphs ?

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