1 Introduction
Given a family of geometric objects in the plane, the intersection graph of has the objects in as its vertices and two vertices are adjacent in the graph if and only if . If is a set of curves in the plane, where a curve is a subset of homeomorphic to the unit interval , then the intersection graph of is a string graph. Many important graph classes like planar graphs and chordal graphs are subclasses of string graphs [24, 29] and so it is natural to study classic optimization problems such as Independent Set and Dominating Set on string graphs.
Asinowski et al. [2] introduced the class of Vertex intersection graphs of Paths on a Grid (VPG graphs for short). A graph is a VPG graph if one can associate a path on a grid with each vertex such that two vertices are adjacent if and only if the corresponding paths intersect on at least one gridpoint. It is not difficult to see that the class of VPG graphs coincides with that of string graphs [2]. If every path in the VPG representation has at most bends i.e., degrees turns at a gridpoint, the graph is a VPG graph and a segment of a path is a vertical or horizontal line segment in the polygonal curve constituting the path. We remark that VPG graphs are also known as DIR (see, e.g., [32]).
Golumbic et al. [23] introduced the class of Edge intersection graphs of Paths on a Grid (EPG graphs for short) as those graphs for which there exists a collection of paths on a grid in onetoone correspondence with their vertex set, two vertices being adjacent if and only if the corresponding paths intersect on at least one gridedge. It turns out that every graph is EPG [23] and EPG graphs have been defined similarly to VPG graphs. Notice that EPG graphs are the wellknown interval graphs.
Independent Set is known to be complete on VPG graphs even for [32]. Therefore, there has been a focus on providing approximation algorithms for restricted subclasses of string graphs. Fox and Pach [19] gave, for every , a approximation algorithm for string graphs i.e., string graphs in which every two curves intersect each other at most times. Lahiri et al. [33] provided a approximation algorithm for VPG graphs and a approximation algorithm for equilateral VPG graphs (a VPG graph is equilateral if, for each path, its horizontal and vertical segment have the same length), where denotes the ratio between the maximum and minimum length of segments of paths. Finally, they showed that Independent Set on equilateral VPG graphs where each horizontal and vertical segment have length is complete. Improving on [33], Biedl and Derka [6] provided a approximation algorithm for the weighted version of Independent Set on VPG graphs. The idea is to partition a VPG graph into outerstring graphs and then solve the problem optimally on each of them by [26]. In the case of VPG, this result was further improved by Bose et al. [7], who provided a approximation algorithm for the weighted version of Independent Set. They also showed that Independent Set can be solved in time for graphs admitting a grounded string representation (i.e. a string representation in which one endpoint of each string is attached to a grounding line and all strings lie on one side of the line), where the strings are monotone (not necessarily strict) polygonal paths, the length of each string is bounded by a constant and all the bends and endpoints are on integral coordinates. Mehrabi [36] considered the weighted version of Independent Set on VPG graphs for which the longest segment among all segments of paths in the graph has length , for some (not required to be a constant), and provided a approximation algorithm. Notice that, to the best of our knowledge, it is not known whether there exists a constantfactor approximation algorithm for Independent Set even on VPG graphs.
Concerning EPG graphs, Epstein et al. [18] showed that Independent Set is complete on EPG graphs and provided a approximation algorithm. Bougeret et al. [8] showed that the problem admits no PTAS on EPG graphs, unless , even if each path has its vertical segment or its horizontal segment of length at most and that it remains hard on EPG graphs even if all the paths have their horizontal segment and vertical segment of length at most . On the other hand, they provided a PTAS for EPG graphs such that each path has its horizontal segment of length at most , for some fixed constant . This was done by adapting the wellknown Baker’s shifting technique [3].
Let us now review Dominating Set. Observe first that it is hard on string VPG graphs. Indeed, every circle graph (i.e. intersection graph of chords in a circle) is a string VPG graph [2] and Dominating Set is hard on circle graphs [16]. Mehrabi [35] considered the subclass of string VPG graphs in which no endpoints of a path belong to any other path and provided an approximation algorithm. Bandyapadhyay et al. [4] considered intersection graphs of Lframes, where an Lframe is a path on a grid with exactly one bend, and provided a approximation algorithm in the case the bend of each path belongs to a diagonal line with slope . They also showed that the problem is hard if each Lframe intersects a diagonal line and that the same holds if instead all the frames intersect a vertical line. Chakraborty et al. [11] provided an approximation algorithm on intersection graphs of Lframes intersecting a common vertical line. They also showed that there is an approximation algorithm on unit VPG graphs^{1}^{1}1A VPG graph is unit if each path consists only of segments with unit length. Notice that every VPG graph is a unit VPG graph for some finite . and, on the negative side, that the problem is hard on unit VPG graphs. Similarly to Independent Set, it is not known whether there exists a constantfactor approximation algorithm for Dominating Set on VPG graphs.
Concerning EPG graphs, Bandyapadhyay et al. [4] showed that Dominating Set for EPG graphs is hard to approximate within a factor of even if all the paths intersect a vertical line.
1.1 Our results
As we have just mentioned, three natural constraints on VPG graphs have been considered in the search for efficient algorithms for Independent Set and Dominating Set: bound the number of bends on each path, bound the number of intersections between any two paths and bound the lengths of segments of paths. Unfortunately, combining these constraints is not enough to guarantee polynomialtime solvability: In Section 3, we show that Independent Set and Dominating Set remain complete when restricted to string VPG graphs such that each horizontal path has length at most , even if the representation is part of the input.
But what about approximation algorithms? Perhaps surprisingly, it turns out that the problems admit PTASes when those constraints are in place. More precisely, in Section 5, we provide PTASes for Independent Set and Dominating Set when restricted to VPG graphs admitting a representation such that:

each path in has a polynomial (in ) number of bends;

each gridedge in belongs to at most paths in ;

the horizontal part of each path in has length at most .
Here the horizontal part of a path is the interval corresponding to the projection of the path onto the horizontal axis.
Clearly, for fixed , VPG graphs satisfy condition 1. The class of VPG graphs satisfying 2 is rich as well: it contains string VPG graphs, for any fixed , VPG graphs with maximum degree at most and VPG graphs with maximum clique size at most .
Notice that recognizing string graphs (and so VPG graphs) is complete [28, 40]. Similarly, for each fixed , recognizing VPG graphs is complete [13, 30]. Therefore, in our PTASes, we assume that a representation of a VPG graph is always given as part of the input. The reason behind condition 1 is to avoid the following pathological behavior: there exist string graphs on vertices requiring paths with bends in any representation (this follows from [31]).
Our result on Dominating Set is best possible in the sense that, if we remove one of conditions 2 and 3, the problem does not admit a PTAS unless (Remark 15). The situation is more subtle for Independent Set, as it has been asked several times whether the problem is hard on VPG graphs (see, e.g., [6, 34]) and this remains open.
As observed above, the study of these two problems on VPG graphs has focused mostly on VPG graphs with and, to the best of our knowledge, ours are the first PTASes on a nontrivial subclass of VPG graphs. The PTAS for Dominating Set shows that the constantfactor approximation algorithm on unit VPG graphs in [11] can in fact be improved if input graphs satisfy also condition 2. Our PTASes are obtained by adapting Baker’s shifting technique [3] to the string graph setting and showing boundedness of an appropriate width parameter. Baker’s technique has already been applied to Independent Set in EPG graphs [8] and our main contribution is to pair it with powerful mimwidth arguments. Mimwidth is a graph parameter, introduced by Vatshelle [42], measuring how easy it is to decompose a graph along vertex cuts inducing a bipartite graph with small maximum induced matching size. Combining results in [5, 10], it is known that domination problems (a class of graph problems including Independent Set and Dominating Set introduced by Telle and Proskurowski [41]) can be solved in time, assuming a branch decomposition of mimwidth is provided as part of the input. Our key observation is that if a VPG graph admits a VPG representation with a bounded number of columns and such that each gridedge belongs to a bounded number of paths, then the graph has bounded mimwidth (Theorem 7). This result is of independent interest and best possible, in the sense that both conditions on the VPG graph are needed to guarantee boundedness of mimwidth (Remark 9). We then use Baker’s shifting technique by solving the problems optimally on each slice with bounded number of columns (Corollary 10).
2 Preliminaries
In this paper we consider only finite simple graphs. Given a graph , we usually denote its vertex set by and its edge set by . If is a subgraph of and contains all the edges of with both endpoints in , then is an induced subgraph of and we write .
Neighborhoods and degrees. For a vertex , the closed neighborhood is the set of vertices adjacent to in together with . The degree of a vertex is the number of edges incident to in . A vertex is a vertex of degree and a vertex is a vertex of degree at least . We refer to a vertex as a cubic vertex and to a vertex as an isolated vertex. The maximum degree of is the quantity and is subcubic if .
Graph operations. Given a graph and , the operation of deleting the set of vertices from results in the graph . A subdivision of an edge is the operation replacing with a path of length .
Graph classes and special graphs. A graph is free if it does not contain induced subgraphs isomorphic to graphs in a set . A complete graph is a graph whose vertices are pairwise adjacent and the complete graph on vertices is denoted by . A triangle is the graph . A graph is bipartite if its vertex set admits a partition into two classes such that every edge has its endpoints in different classes. A split graph is a graph whose vertices can be partitioned into a clique and an independent set. A caterpillar is a tree whose nonleaf vertices form a path.
Graph properties and parameters. A set of vertices or edges of a graph is minimum with respect to the property if it has minimum size among all subsets having property . The term maximum is defined analogously. An independent set of a graph is a set of pairwise nonadjacent vertices. The size of a maximum independent set of is denoted by . A clique of a graph is a set of pairwise adjacent vertices. A dominating set of is a subset such that each vertex in is adjacent to a vertex in . The size of a minimum dominating set of is denoted by . A matching of a graph is a set of pairwise nonincident edges. An induced matching in a graph is a matching such that no two vertices belonging to different edges in are adjacent in the graph.
VPG, CPG and EPG graphs. Given a rectangular grid , its horizontal lines are referred to as rows and its vertical lines as columns. The gridpoint lying on row and column is denoted by .
A graph is VPG if there exists a collection of paths on a grid such that is in onetoone correspondence with and two vertices are adjacent in if and only if the corresponding paths intersect. A graph is CPG if there exists a collection of interiorly disjoint paths on a grid such that is in onetoone correspondence with and two vertices are adjacent in if and only if the corresponding paths touch. A graph is EPG if there exists a collection of paths on a grid such that is in onetoone correspondence with and two vertices are adjacent in if and only if the corresponding paths share a gridedge.
A VPG graph is a VPG graph if there exists a collection as above such that every path in has at most bends i.e., degree turns at a gridpoint. For a VPG graph , the pair is a VPG representation of and, more specifically, a VPG representation if every path in has at most bends. The path in corresponding to the vertex is denoted by . For a path , we denote by the set of endpoints of . The length of a path is the number of gridedges used and an interior point of is a point belonging to and different from its endpoints. A bendpoint of is a gridpoint corresponding to a bend of and a segment of is either a vertical or horizontal line segment in the polygonal curve constituting . Analogous definitions hold for CPG and EPG graphs.
We now need to describe how a VPG or CPGrepresentation of a graph is encoded. For the grid , we only keep track of the gridstep , that is, the length of a gridedge. For each path , we have three sequences of points in . The first sequence consists of the endpoints and of and all the bendpoints of in their order of appearance while traversing from to . In other words, for any , is a segment of . The second and third sequences contain, for any , the points and (if any) corresponding to the first intersection of with and to the last intersection of with , respectively, i.e there is no intersection with on the portion of from to and on the portion of from to . The intersection points in are ordered according to their appearance while traversing from to , whereas the intersection points in are ordered according to their appearance while traversing from to . The first intersection point of is the smallest intersection point in according to the above order and the last intersection point of is the smallest intersection point in according to the above order. Notice that if each path in has a number of bends bounded by a polynomial in , then the size of this data structure is polynomial in . Moreover, knowing for each , we can compute in time polynomial in the second and third sequences of each path. Given , we can also easily determine the horizontal part of the path as follows. Let and let . Then is the segment . The following easy observation will be used in the proof of Theorem 13.
Lemma 1
Let be a VPG graph with representation and such that each path in has a number of bends polynomial in . Let be an induced subgraph of . Then we can compute in time a VPG representation of .
Proof
We obtain a VPG representation of from as follows. Denote by the set of paths whose corresponding vertices belong to . Then, a path is described by the same sequence as in together with the sequence of first intersection points and the sequence of last intersection points .
The refinement of a grid having gridstep is the operation adding a new column (resp. row) between any pair of consecutive columns (resp. rows) in and setting the gridstep to . Notice that this operation does not change the sequences above.
3 Hardness results
In this section, we show that Independent Set and Dominating Set remain complete when restricted to VPG graphs admitting a representation such that each gridedge belongs to at most path and each horizontal path has length at most . In fact, our results hold for a subclass of VPG graphs, namely that of CPG graphs defined in Section 2.
Similarly to VPG graphs, it was recently shown that recognizing CPG graphs is complete, for each fixed [12, 17]. Nonetheless, as it will become evident from the proofs, both hardness results (Theorems 5 and 4) hold even if the CPG representation is given as part of the input.
Before turning to the proofs we need the following technical result.
Lemma 2
For any subcubic trianglefree CPG graph on vertices and with bend CPG representation , we can update in time polynomial in so that the following hold:

contains columns;

A path strictly contains one endpoint of another path if and only if the vertex corresponding to is a vertex;

Each horizontal path corresponding to a vertex has length at least ;

For each horizontal path corresponding to a vertex , denoting by and its left and right endpoint, respectively, and by the contactpoint contained in its interior, the segments and have length at least ;

Each path corresponding to a vertex has length .
Proof
Let be the input bend CPG representation. Without loss of generality, the gridstep is . We begin by preprocessing the grid so that it contains columns. Let be the sequence of coordinates defined as follows:

and are the smallest and largest coordinates of paths in , respectively, that is, there exist and in such that and and, for any with , we have and ;

for any , if and only if and there exists a path such that for some ;

for any , .
For a fixed , let be a row of such that is nonempty. Let be the sequence of coordinates of endpoints of paths in in increasing order, that is, if and only if there exists a path such that with , and for any , . We then let . Observe that .
Suppose now there exists such that and consider the smallest such index . The idea is that, since there is no vertical path between columns and , we can “shrink” the slice keeping fixed so that and repeat. More precisely, for each row of such that , we proceed as follows (note that there are at most such rows). Consider the sequence defined above. If , we replace the occurrence of in with , otherwise we replace the at most two occurrences of in with . Now suppose that there exists such that and consider the smallest such index . By minimality, for any . We then replace the at most two occurences of in with . Repeating this process, we obtain that for any . Finally, if , we replace the occurrence of in with . Suppose now that there exists such that (and so ) and consider the smallest such index . For any , we replace in with . Moreover, if , we replace in with , otherwise we replace in with . Repeating this process, we obtain that for any , and so . For any path with endpoints and such that , we then replace in with . Finally, we update to , and either for any , we have , in which case uses at most columns, or there exists such that , in which case we repeat the procedure above until we obtain a representation satisfying (a).
Consider now . Since has no bend, . Moreover, paths pairwise touch at most once and so , for any . Suppose first is such that its first intersection point and last intersection point coincide, that is, corresponds to a vertex. If , we replace with in . Suppose finally that is such that and are distinct, that is, corresponds to a vertex. If , we replace with in . If , we replace with in . Applying this procedure to any path , the updated representation satisfies (b) and clearly still satisfies (a).
Consider now, in the updated representation, the set of horizontal paths whose corresponding vertex has degree at least . Clearly, the length of any path corresponding to a vertex is at least . Similarly, if corresponds to a vertex and denotes the contactpoint in , we have that the paths and on the grid have both length at least . Therefore, by refining the grid times, that is, setting the gridstep to , any path corresponding to a vertex has length at least and any path corresponding to a vertex is such that and have length at least . The updated representation satisfies (c) and (d), and still satisfies (a) and (b).
Consider now a path in the updated representation such that , that is, corresponds to a vertex. Suppose first that . If is a vertical path (i.e., ), we set if , and if . Otherwise, is a horizontal path (i.e., ) and we set if , and if . We proceed similarly in the case that . By repeating this procedure for any such path , the updated representation satisfies (e). It still satisfies (a), (b), (c) and (d). Clearly, any of the above operations can be done in polynomial time.
Remark 3
Observe that Lemma 2 remains true if we replace in (c) and (d) with .
Theorem 4
Independent Set is complete when restricted to CPG graphs admitting a CPG representation where each horizontal path has length at most 2.
Proof
We reduce from Independent Set restricted to trianglefree subcubic CPG graphs which was shown to be hard even if a CPG representation is part of the input [12]. Given a trianglefree subcubic CPG graph with CPG representation , we construct a graph as follows. First, we update in polynomial time so that it satisfies (a) to (e) in Lemma 2. We then introduce two operations which will be applied to each vertex corresponding to a horizontal path . They depend on whether or .
Suppose first that and denote by the length of (hence, ). Let and be the quotient and remainder, respectively, of the division of by 2. The splitting of is the operation replacing with horizontal paths (from left to right) of length 1 and one horizontal path (at the right extremity) of length .
Suppose now that and let , and be the three grid segments obtained by dividing as follows. is the segment strictly containing the contactpoint and with length , is the remaining part of to the left of (hence with length ) and is the remaining part of to the right of (hence with length ) (see Figure 1).
For , let and be the quotient and remainder, respectively, of the division of by 2. The splitting of is the operation replacing:

with horizontal paths (from left to right) of length 1 and one horizontal path (at the right extremity) of length ;

with a horizontal path of length 2;

with horizontal paths (from left to right) of length 1 and one horizontal path (at the right extremity) of length .
Notice that the splitting of a vertex removes and replaces it with new vertices^{2}^{2}2We remark that this operation can be obtained by several applications of the vertex stretching introduced in [1] and is in fact equivalent to edge subdivisions. (if , then ). The graph is then obtained from by splitting every vertex whose corresponding path is horizontal. It is easy to see that this operation can be performed in polynomial time, that has vertices (by Lemma 2) and that it admits a CPG representation where each horizontal path has length at most . To complete the proof, it is then enough to show the following:
Claim 1
Let be the graph obtained by splitting a vertex whose corresponding path in is horizontal. We have that .
Denote by the set of vertices introduced by the splitting of , where for any .
Given a maximum independent set of , we construct an independent set of as follows. If , then
otherwise,
In both cases, is easily seen to be independent and so .
Conversely, let be a maximum independent set of . We construct an independent set of as follows. First we add to every vertex in . Then we decide whether to add or not according to the following cases.
Suppose first that . If and both belong to , then we add to (note that, by maximality, contains vertices of ). Otherwise, one of and does not belong to , in which case we do not add to (again, by maximality, contains vertices of ). In both cases the constructed is easily seen to be independent and .
Suppose now that . If , and belong to , then we add to (note that, by maximality, contains vertices of ). Otherwise, one of , and does not belong to and we do not add to (note that in this case contains at most vertices of ). In both cases the constructed is easily seen to be independent and .
Therefore, , concluding the proof of creftypecap 1.
In the following hardness proof, we reduce from Dominating Set restricted to subcubic planar bipartite graphs. This problem is easily seen to be hard by recalling that a subdivision of an edge of a graph increases the domination number by exactly one [27] and that Dominating Set restricted to subcubic planar graphs is hard [21].
Theorem 5
Dominating Set is complete when restricted to CPG graphs admitting a CPG representation where each horizontal path has length at most 2.
Proof
We reduce from Dominating Set restricted to subcubic planar bipartite graphs which is hard by the paragraph above. Given a subcubic planar bipartite graph , we construct, similarly to Theorem 4, a graph as follows. First, since is planar and bipartite, we obtain in linear time a CPG representation of [15]. Since is trianglefree and subcubic, we then update in polynomial time so that it satisfies (a) to (e) in Lemma 2 where the ’s in the statements are replaced by ’s (see Remark 3). We now introduce two operations which will be applied to each vertex corresponding to a horizontal path . They depend on whether or .
Suppose first that and denote by the length of (hence, ). Let and be the quotient and remainder, respectively, of the division of by 3. The splitting of is the operation replacing with horizontal paths (from left to right) of length 1 and two horizontal paths (touching ) and (at the right extremity) of lengths and , respectively.
Suppose now that and let , and be the three grid segments obtained by dividing as follows. is the segment strictly containing the contactpoint and with length , is the remaining part of to the left of (hence with length ) and is the remaining part of to the right of (hence with length ). For , let and be the quotient and remainder, respectively, of the division of by 3. The splitting of is the operation replacing:

with horizontal paths (from left to right) of length 1 and two horizontal paths (touching ) and (at the right extremity) of lengths and , respectively;

with a horizontal path of length 2;

with horizontal paths (from left to right) of length 1 and two horizontal paths (touching ) and (at the right extremity) of lengths and , respectively.
Notice that the splitting of a vertex removes and replaces it with new vertices (if , then ). The graph is then obtained from by splitting every vertex whose corresponding path is horizontal. It is easy to see that this operation can be performed in polynomial time, that has vertices (by Lemma 2) and that admits a CPG representation where each horizontal path has length at most 2. To complete the proof, it is then enough to show the following:
Claim 2
Let be the graph obtained by splitting a vertex whose corresponding path in is horizontal. We have that .
Denote by the set of vertices introduced by the splitting of , where for any .
Given a minimum dominating set of , we construct a dominating set of as follows. If , then
Otherwise, there exists which belongs to and we distinguish cases depending on which vertex is adjacent to in . If is adjacent to , then
If is adjacent to , then
Otherwise, is adjacent to and
In all cases, is easily seen to be dominating and .
Conversely, let be a minimum dominating set of . We construct a dominating set of as follows. First we put in every vertex in . Then we decide whether to add or not according to the following cases.
Suppose first that . Since the vertices with have pairwise disjoint neighborhoods, we have that . We now claim that if , then none of and belongs to and one of them is dominated by some vertex in . Indeed,
and the unions are over pairwise disjoint neighborhoods. Therefore, if , then
Comments
There are no comments yet.