Approximability of Covering Cells with Line Segments^{†}^{†}thanks: Paz Carmi is supported by Grant 2016116 from the United StatesIsrael Binational Science Foundation. Anil Maheshwari is supported in part by Natural Sciences and Engineering Research Council of Canada (NSERC). Saeed Mehrabi is supported by a CarletonFields postdoctoral fellowship.
Abstract
In COCOA 2015, Korman et al. studied the following geometric covering problem: given a set of line segments in the plane, find a minimum number of line segments such that every cell in the arrangement of the line segments is covered. Here, a line segment covers a cell if is incident to . The problem was shown to be hard, even if the line segments in are axisparallel, and it remains hard when the goal is cover the “rectangular” cells (i.e., cells that are defined by exactly four axisparallel line segments).
In this paper, we consider the approximability of the problem. We first give a for the problem when the line segments in are in any orientation, but we can only select the covering line segments from one orientation. Then, we show that when the goal is to cover the rectangular cells using line segments from both horizontal and vertical line segments, then the problem is hard. We also consider the parameterized complexity of the problem and prove that the problem is when parameterized by the size of an optimal solution. Our algorithm works when the line segments in have two orientations and the goal is to cover all cells, complementing that of Korman et al. [9] in which the goal is to cover the “rectangular” cells.
1 Introduction
Set Cover is a wellstudied problem in computer science. The input to the problem is a ground set of elements and a set of subsets of ; that is, such that for all . The objective is to find a minimumcardinality subset of whose union is . Set Cover is known to be hard [5] and even hard to approximate [8].
In this paper, we consider a geometric variant of the set cover problem that was first studied by Korman et al. [9]. A set of line segments in the plane is said to be nonoverlapping if any two line segments from the set intersect in at most one point. Given a set of nonoverlapping line segments in the plane, a cell in the arrangement of is a maximally connected region that is not intersected by any line segment in [9]. Then, the objective of the Line Segment Covering () problem is to select a minimum number of line segments such that every cell in the arrangement of the line segments is covered. Here, a cell is covered by a line segment if it is incident to the line segment (i.e., the line segment is in the set of line segments defining the boundary of the cell). We assume that at most two line segments may share a fixed point in the plane.
Related work.
Korman et al. [9] proved that when the line segments are only horizontal and vertical, the problem is hard and it remains hard when the goal is to cover the “rectangular” cells. By a closer look at their hardness proof, one can see that the problem is hard even if we are only allowed to select the line segments from one orientation (they only select vertical line segments when constructing a solution from a given truth assignment for the corresponding 3SAT problem). Moreover, the authors gave an time algorithm for covering the rectangular cells when parameterized by , the size of an optimal solution. However, the algorithm does not work when the goal is to cover all cells of the arrangement. The authors leave open studying the approximability of the problem.
The problem is closely related to a guarding problem studied by Bose et al. [3]. Given a set of lines in the plane, they studied the problems of guarding cells of the arrangement by selecting a minimum number of lines, or guarding the lines by selecting a minimum number of cells. Here, “guarding” has the same meaning as “covering” in the problem. However, their results do not extend to the problem, because (as also noted by Korman et al. [9]) they use some properties of lines that are not true for the case of line segments.
Our results.
In this paper, we prove the following results.

We give a for the problem when the line segments in can have any arbitrarily orientations, but we are allowed to select the covering line segments from only one orientation. Given the hardness of the problem [9], this settles the complexity of this variant of the problem.

When we allow selecting the covering line segments from more than one direction, we show that the problem is hard when the line segments in have two orientations and the goal is to cover the rectangular cells.

We give an algorithm for the problem when the line segments in have only two orientations and the goal is to cover all cells of the arrangement. This complements the algorithm of Korman et al. [9] as we do not restrict the covering only to rectangular cells.
Organization.
2 Preliminaries
In the following, we revisit some techniques and background that are used throughout this paper.
Local search.
Our for the problem is based on the local search technique, which was introduced independently by Mustafa and Ray [11], and Chan and HarPeled [4]. Consider an optimization problem in which the objective is to compute a feasible subset of a ground set whose cardinality is minimum over all such feasible subsets of . Moreover, it is assumed that computing some initial feasible solution and determining whether a subset is a feasible solution can be done in polynomial time. The local search algorithm for a minimization problem is as follows. Fix some fixed parameter , and let be some initial feasible solution for the problem. In each iteration, if there are and such that , and is a feasible solution, then set and reiterate. The algorithm returns and terminates when no such local improvement is possible.
Clearly, the local search algorithm runs in polynomial time. Let and be the solutions returned by the algorithm and an optimal solution, respectively. The following result establishes the connection between local search technique and obtaining a .
Theorem 2.1 ([4, 11]).
Consider the solutions and for a minimization problem, and suppose that there exists a planar bipartite graph that satisfies the local exchange property: for any subset , is a feasible solution, where denotes the set of neighbours of in . Then, the local search algorithm yields a for the problem.
The local search was used by Mustafa and Ray [11] to obtain a for geometric hitting set problem and by Chan and HarPeled [4] to obtain a for geometric independent set problem. Since then, the technique has been used to get a for several other geometric problems, such as geometric dominating set [2] and unique covering [1].
Fixedparameter tractability.
The theory of parameterized complexity was developed by Downey and Fellows [6]. Let be a finite alphabet. Then, a parameterized problem is a language in which the second component is called the parameter of the problem. A parameterized problem is said to be fixedparameter tractable or , if the question “” can be decided in time , where is an arbitrary function. We call an algorithm with such running time , an algorithm.
For the rest of this paper, we denote a set of line segments in the plane by (i.e., ) and the resulting arrangement by .
3
In this section, we show that the problem admits a when the line segments in are in any orientations, but we can select line segments from only one orientation to cover the cells. To this end, we run the local search algorithm with parameter for some , where is a constant. Let be the solution returned by the algorithm and let be an optimal solution. We can assume that . This is because if , then we can consider the sets and , and analyze the algorithm with and . Here, we mark the faces covered by a line segment in as “covered” so as we do not need to cover then in the new variant of the problem. This guarantees that the approximation factor of the original instance is upper bounded by that of the new instance with these two new sets and .
We now construct a planar bipartite graph that satisfies the local exchange property, hence proving that the problem admits a by Theorem 2.1. To this end, we first construct an auxiliary planar graph and then show how to obtain from by edge contraction. For each cell , let and be two line segments that cover ; we select a point and and connect them by a curve that lies in the interior of (except its endpoints and ). Notice that since both and are feasible solutions, we know that contains at least one line segment that covers and also contains at least one line segment that covers , for all . We add and to and to . We complete the definition of by connecting every pair of consecutive points in , for all , by an edge that is exactly the portion of that lies between the pair of points. See Figure 1(a) for an example. Clearly, is planar because the first set of edges are drawn in the interior of cells and each cell contains at most one edge. Moreover, the second set of edges are aligned with the line segments in . Since the line segments in are nonoverlapping and all have the same orientation, the second set of edges are also noncrossing. To obtain the graph , for each segment , we contract the edges of that are contained in such that we get a single point corresponding to ; see Figure 1(b). So, . Graph is planar since remains planar after this edge contraction. Moreover, is a bipartite graph as the edges of with both endpoints belonging to a line segment in or both endpoints belonging to a line segment in are collapsed into a single point (i.e., ).
Lemma 3.1.
Graph is planar and bipartite.
We next show that satisfies the exchange property.
Lemma 3.2.
Graph satisfies the local exchange property.
Proof.
It is sufficient to show that for every cell , there are vertices and such that both segments corresponding to these vertices cover and . Take any cell and let be the set of all line segments that cover . Notice that and because and are each a feasible solution. Then, by definition, there must be a and for which . This completes the proof of the lemma. ∎
Putting everything together, we have the main result of this section.
Theorem 3.1.
There exists a for the line segment covering problem when the line segments in can have any orientation and we are allowed to select the covering line segments from only one orientation.
4 Hardness
In this section, we show that the problem is hard when the line segments in have only two orientations and the goal is to cover the rectangular cells. To this end, we give an reduction from the Minimum Vertex Cover () problem on graphs with maximumdegree three to this variant of the problem. Our reduction is inspired by the construction of Mehrabi [10]. As a reminder, we first give a formal definition of reduction [12], which is one of the gappreserving reductions. Let and be two optimization problems with the cost functions and , respectively. We say that reduces to if there are two polynomialtime computable functions and such that the followings hold.

For any instance of , is an instance of .

If is a solution to , then is a solution to .

There exists a constant such that
where denotes the cost of an optimal solution for problem on its instance .

There exists a constant such that for every solution for ,
where denotes the absolute value of .
Lemma 4.1.
The minimum vertex cover problem on graphs with maximumdegree three is reducible to the problem, where is a set of horizontal and vertical line segments and the goal is to cover the rectangular cells of .
Proof.
Let be an instance of on graphs of maximumdegree three; let be the graph corresponding to and let be the size of the smallest vertex cover in . First, let be an arbitrary ordering of the vertices of , where . In the following, we give a polynomialtime computable function that takes as input and outputs an instance of the problem.
We first describe the vertex gadgets. For each vertex , , construct a horizontal line segment and a vertical line segment , and connect them as shown in Figure 2. We call the (blue) horizontal line segment used in the connection of and the horizontal connector of . Moreover, there are four (small, dashed) line segments used in the connection of and that we call the small connectors of . Notice that these five “connectors” along with and form exactly two rectangular cells. For each edge , where , we add two small line segments, one horizontal and one vertical, at the intersection point of and such that they intersect each other as well as each intersects one of and , hence forming a rectangular cell; see the two (red, dashed) line segments at the intersection of and in Figure 2 for an example. We call such a pair edge line segments and denote them by . Finally, for every rectangular cell whose four sides are all defined by the line segments corresponding to a 4subset of (i.e., the cell is not covered by a horizontal connector or edge line segments), we insert a vertical line segment into the cell so as to make it nonrectangular; see the vertical (red) line segment in Figure 2. This ensures that every rectangular cell is incident either to a horizontal connector or to edge line segments for some and . This gives the instance of the problem. Notice that is a polynomialtime computable function. In the following, we denote an optimal solution for the instance of a problem by . We now prove that all the four conditions of reduction hold.
First, let be a vertex cover of of size . Denote by the set of horizontal line segments induced by and define analogously. Moreover, let be the set of horizontal connectors whose corresponding vertex is not in . We show that is a feasible solution for covering all the rectangular cells of . Let be a rectangular cell. Then, must be incident either to a horizontal connector or to edge line segments for some and . First, if is incident to a horizontal connector , then either or and by the construction of and so is covered either way. Next, if is incident to edge line segments for some and , where w.l.o.g. , then either or because we know that or . So, is again covered in this case. Therefore, is a feasible solution.
Second, let be any feasible solution for . Notice that we can construct a feasible solution for such that and consists of only and for some , or a horizontal connector. This is because (i) any rectangular cell covered by a small connector is also covered by a horizontal connector, and (ii) any cell covered by a pair of edge line segments (for some and ) is also covered by and . For (ii), if exactly one of the line segments in is in , then we replace it with exactly one of or . Otherwise, if both line segments of are in , then we replace both of them with and . So, and is a feasible solution for . Now, let . To show that is a vertex cover for , consider any edge , where . Then, we know that there exists a rectangular cell at the intersection of and that must be covered by . Since none of the two edge line segments of are in , we conclude that at least one of and is in , which means that or . Hence, is a vertex cover.
Third, observe that and also . Given that has degree three, and so .
We now prove the last condition of reduction. First, define ; that is, the paths of a vertex , where both its horizontal and vertical line segments appear in . Also, define to be the remaining line segments corresponding to either or for some ; i.e., those of , where only one of its line segments appears in . Take any vertex . To cover the two rectangular cells incident to the horizontal connector of , we must have or ; this is true for all . Then, . Moreover, since is a vertex cover of . Therefore, . By this and our earlier inequality , we have . Now, suppose that for some . Then,
That is, . This concludes our reduction from on graphs of maximumdegree three to with and . ∎
Theorem 4.1.
The line segment covering problem is hard when the line segments in are either horizontal or vertical and the goal is to cover the rectangular cells of .
5
In this section, we show that the problem is fixedparameter tractable (parametrized by the size of an optimal solution) when the line segments in are either horizontal or vertical, and the goal is to cover all the cells in . This complements the result of Korman et al. [9], where the goal is to cover the rectangular cells. Throughout this section, let be the size of an optimal solution.
Our follows the framework of Korman et al. [9]. That is, we formulate the problem as a hitting set problem and argue that we only need to hit an number of sets; hence, obtaining a kernel of size for the problem. The of Korman et al. [9] is based on the fact that any three orthogonal line segments can cover at most two “rectangular” cells (i.e., at most two rectangular cells can be incident to all the three line segments). As an analogous result, we prove in Lemma 5.1 that the number of such cells can be at most six when the goal is to cover all cells, including nonrectangular ones. We will then apply this result to obtain the desired kernel.
Lemma 5.1.
Let be a set of axisparallel line segments in the plane. Then, for any three line segments , there are at most six cells in that can be covered by all three line segments and .
Proof.
Take any three line segments and in and let be the set of all cells in that are covered by all three line segments and . We need to show that . To this end, we construct a planar graph corresponding to and the cells in and will then argue that this graph must contain a subdivision of if . We next give the details. Let be a cell in . Consider a point in the interior of as well as a distinct point in , for all (notice that is on the boundary of ). These points together form the set of vertices of ; that is, . Now, for each , consider an ordering of the points on , , and connect every two consecutive points by an edge, which is exactly the portion of that lies between the two points. Moreover, for each cell , we connect to by a curve that lies strictly in the interior of (except at its endpoints) for all . Then, the edge set of consists of the set of all edges connecting the consecutive points as we as the curves , for all and . Clearly, is a planar graph. In the following, we consider several cases depending on whether the line segments and intersect each other; observe that there can be at most two intersection points between them.
Case 1.
There is no intersection point; that is, the line segments and are pairwise disjoint. In this case, we show that in fact . To this end, suppose for a contradiction that . Take any three cells and consider the subgraph of induced by . Now, consider the graph constructed from as follows. For each , , we place a new vertex close to and connect it to the three vertices for all such that the resulting graph remains planar. One can easily verify that this is doable since the three line segments are disjoint and so there are a few cases for where to place depending on which side of the three cells lie; see Figure 3. Observe that the resulting graph is a planar drawing of a subdivision of , which is not possible. So, .
Case 2.
There is exactly one intersection point; assume w.o.l.g. that is horizontal, and are vertical and intersects . Here, we show that . Again, suppose for a contradiction that . Then, considering the graph , there must be at least three vertices in that lie w.l.o.g. to the right of . Take any three such vertices and denote the corresponding cells by . We can now construct the graph analogous to the one in Case 1 with these three cells and so obtain a planar drawing of a subdivision of , which is a contradiction.
Case 3.
There are two intersection points. Here, we show that and we use a similar argument to those in the previous cases. Denote the endpoints of by and , and let and be the intersection points of with and ; assume w.l.o.g. that is the left endpoint of and that lies to the left of . If , then at least one of the line segments and must contain three vertices of ; assume w.l.o.g. that it is . Then, take any three such vertices on and consider the three cells and corresponding to these vertices. We can now construct the graph analogous to the one in Case 1 and so obtain a planar drawing of a subdivision of , which is a contradiction. As such, .
By the three cases described above, we conclude that . ∎
We note that the upper bound in Lemma 5.1 is tight as Figure 4 shows an example with three line segments that cover six cell. We now apply Lemma 5.1 to obtain our . We first formulate the problem as a hitting set problem as follows. The ground set is and for each cell in , there exists a set that contains the line segments that cover the cell. Let be the resulting set of subsets of . Then, the problem is equivalent to selecting a minimum number of elements from such that each set in is hit by at least one selected element.
We first reduce the set to a set as follows. For every pair of line segments , if they appear in more than sets , then we remove all such sets form and add the set to . Let be the resulting set.
Lemma 5.2.
A set with is a minimumsize cover of if and only if it is a minimumsize cover of .
Proof.
We prove the lemma by an argument similar to the one by Korman et al. [9]. The lemma clearly follows if . So, assume that and are two nonempty sets. Let with be a minimumsize cover for . First, is also a cover for because for every set there exists a pair of line segment and such that both and are in and we have . We now prove that is also a cover of minimum size for .
Suppose for a contradiction that there exists a cover for such that . Then, cannot be a cover because is a cover of minimum size for . Since covers , there must exist such that neither nor is in . But, we introduced the set into because there were more than sets containing both and . If neither nor is in , then every other line segment can cover at most six of such subsets by Lemma 5.1. Therefore, — a contradiction. By a similar argument, we can show that a minimumsize cover of is also a minimumsize cover for . This completes the proof of the lemma. ∎
Next, we reduce to a new set as follows. For each line segment , we count how many sets in contain . If appears in more than , then we remove all those sets and add the set to . Let denote the resulting set.
Lemma 5.3.
A set with is a minimumsize cover for if and only if it is a minimumsize cover for .
Proof.
The lemma follows if . So, assume that and are two nonempty sets. Let with be a minimumsize cover for . For any set , there exists a singleton set in whose member is in . This means that is also a cover for . We next show that is also a minimumsize cover for .
Suppose for a contradiction that there exists a cover for such that . Therefore, is not a cover of . Since covers , there must exist a set in that is not covered by . Notice that this set must be of size 1 from the construction of ; let be such a set, where . The reason we have the set in is that because there were more than sets in containing . If is not in , then all such sets of must be covered by other line segments. But, from the construction of , every pair of line segments can appear in at most sets. So, must be greater than , which is a contradiction. A similar argument can be used to show that a minimumsize cover for is also is minimumsize cover for . This completes the proof of the lemma. ∎
Consider the set . By Lemma 5.3, no line segment of appears in more than sets in . Therefore, if , then the problem does not have a cover of size at most . Since the construction of can be done in polynomial time, we have the following result.
Lemma 5.4.
For the problem on a set of axisparallel line segments, in polynomial time, we can either obtain a kernel of size or conclude that the problem does not have a cover of size at most , where is the size of an optimal cover.
Since having a kernel of size implies that the problem is [7], we have the main result of this section.
Theorem 5.1.
The line segment covering problem on a set of axisparallel line segments is with respect to the size of an optimal cover.
6 Conclusion
In this paper, we considered the problem of covering the cells in the arrangement of a set of line segments in the plane. We proved that the problem admits a when the covering line segments can be selected from only one orientation. We then showed that if we allow selecting the covering line segments from more than one orientation, then the problem is hard when we are interested in covering the rectangular cells. Finally, we gave an algorithm for the problem when the line segments have only two orientations, but the goal is to cover all the cells. Our hardness rules out the possibility of a for “covering rectangular faces” variant of the problem, but is there a 2approximation algorithm for the problem? For the more general variant, where the line segments are in any orientation, covering line segments can be selected from any orientation and the goal is to cover all the cells, can we obtain a approximation algorithm for some small constant ?
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