Approximating Dominating Set on Intersection Graphs of L-frames

We consider the Dominating Set (DS) problem on the intersection graphs of geometric objects. Surprisingly, for simple and widely used objects such as rectangles, the problem is NP-hard even when all the rectangles are "anchored" at a line with slope -1. It is easy to see that for the anchored rectangles, the problem reduces to one with even simpler objects: L-frames. An L-frame is the union of a vertical and a horizontal segment that share one endpoint (corner of the L-frame). In light of the above discussion, we consider DS on the intersection graphs of L-frames. In this paper, we consider three restricted versions of the problem. First, we consider the version in which the corners of all input L-frames are anchored at a line with slope -1, and obtain a polynomial-time (2+ϵ)-approximation. Furthermore, we obtain a PTAS in case all the input L-frames are anchored at the diagonal from one side. Next, we consider the version, where all input L-frames intersect a vertical line, and prove APX-hardness of this version. Moreover, we prove NP-hardness of this version even in case the horizontal and vertical segments of each L-frame have the same length. Finally, we consider the version, where every L-frame intersects a vertical and a horizontal line, and show that this version is linear-time solvable. We also consider these versions of the problem in the so-called "edge intersection model", and obtain several interesting results. One of the results is an NP-hardness proof of the third version which answers a question posed by Mehrabi (WAOA 2017).

Authors

• 8 publications
• 21 publications
• 18 publications
• 3 publications
• Approximating Dominating Set on Intersection Graphs of Rectangles and L-frames

We consider the Minimum Dominating Set (MDS) problem on the intersection...
03/16/2018 ∙ by Sayan Bandyapadhyay, et al. ∙ 0

• Constrained Orthogonal Segment Stabbing

Let S and D each be a set of orthogonal line segments in the plane. A li...
04/30/2019 ∙ by Sayan Bandyapadhyay, et al. ∙ 0

• Robust Approach to Restricted Items Selection Problem

We consider the robust version of items selection problem, in which the ...
07/22/2019 ∙ by Maciej Drwal, et al. ∙ 0

• Stick Graphs with Length Constraints

Stick graphs are intersection graphs of horizontal and vertical line seg...
07/11/2019 ∙ by Steven Chaplick, et al. ∙ 0

• Packing Rotating Segments

We show that the following variant of labeling rotating maps is NP-hard,...
07/30/2019 ∙ by Ali Gholami Rudi, et al. ∙ 0

• The Dominating Set Problem in Geometric Intersection Graphs

We study the parameterized complexity of dominating sets in geometric in...
09/15/2017 ∙ by Mark de Berg, et al. ∙ 0

• Recognition and Drawing of Stick Graphs

A Stick graph is an intersection graph of axis-aligned segments such tha...
08/29/2018 ∙ by Felice De Luca, et al. ∙ 0

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1 Introduction

Minimum Dominating Set () is an -hard problem in graph theory and discrete optimization. Given a graph , the objective of the  problem is to compute a minimum-size subset such that every vertex not in is adjacent to at least one vertex in . For general graphs, it is known that a greedy algorithm for  achieves an -factor approximation and within a constant factor this is the best one can hope for, unless = [27]. As such, the problem has been extensively studied on many subclasses of graphs, one of which is the intersection graphs of geometric objects in the plane [11, 22, 16, 15, 23, 26]. Here, each vertex of the graph is in one-to-one correspondence with a geometric object in the plane and two vertices are adjacent if and only if the corresponding objects have a non-empty intersection.

In this paper, we consider the approximability and hardness of the  problem on the intersection graphs of geometric objects. The  problem is known to admit es on disk graphs [16] and the intersection graphs of non-piercing111Two connected objects and are called non-piercing if both and are connected. objects [18]. On the other hand, it is -hard to obtain a -approximation in polynomial time for sufficiently complicated shapes, e.g. rectilinear polygons [14, 15]. However, even for simple shapes such as axis-parallel rectangles no sub-logarithmic approximation is known. The only approximation for rectangles we are aware of is due to Erlebach et al. [15] who gave an -approximation on rectangles with aspect-ratio at most . In fact, the problem is -hard [15] on rectangles, and (perhaps surprisingly) the problem is shown to be -hard even on diagonal-anchored rectangles [26]; that is, the intersection of every rectangle and a diagonal line with slope -1 is exactly one corner of the rectangle. See Figure 2(a) for an example. However, to the best of our knowledge no sub-logarithmic approximation is known even in this case. We note that optimization problems on “diagonal-intersecting” geometric objects have been studied before through the lenses of approximation algorithms; e.g. maximum independent set [8, 4, 21] and minimum hitting set [8, 7, 24].

Our results.

In this paper, we first give a -approximation algorithm for the  problem on diagonal-anchored rectangles, providing the first -approximation for the problem on a non-trivial subclass of rectangles. For any , there exists a -approximation algorithm for the  problem on diagonal-anchored rectangles.

To prove Theorem 1, we first divide the problem into two subproblems and then give a for the subproblems using the local search technique [5, 25]. Each such subproblem involves diagonal-anchored rectangles that lie on only one side of the diagonal. The key to obtain our is in showing a planar drawing of a bipartite graph that is required for the analysis of the local search algorithm. We note that, even in these simpler cases the problem remains sufficiently challenging due to the geometry of the rectangles, and the existing schemes are not useful to obtain a near-optimal solution. For example, the local search analyses for non-piercing objects in [18] do not hold here, as the diagonal-anchored rectangles can still “pierce” each other.

It is not hard to see that the  problem on “diagonal-anchored” rectangles is the same as the  problem on “diagonal-anchored” -frames [4]. An -frame is the union of a vertical and a horizontal line segment that share an endpoint (corner). Indeed, each rectangle in the input can be replaced by a diagonal-anchored -frame without altering the underlying intersection graph (see Figure 1 for an illustration). Thus, a dominating set for the instance with the -frames is also a dominating set for the original instance with rectangles, and vice versa. Hence, we also obtain a -approximation for the problem with diagonal-anchored -frames. For the  problem on general -frames, the only approximation we are aware of is due to Mehrabi [23] who gave an -approximation algorithm when every two -frames intersect in at most one point. Asinowski et al. [1] proved that every circle graph is an intersection graph of -frames. Since  is -hard on circle graphs [9], the problem is also -hard on -frames.

Note that, by definition, we can have four different types of -frames depending on the two endpoints that define the corner. Considering the problem on -frames we extend the -hardness result in the general case to two constrained cases. First, we show that the problem does not admit a -approximation for any on “diagonal-intersecting” -frames (see Figure 2(b) for an example). The  problem is -hard on -frames when every -frame intersects a diagonal line.

As the construction in proving Theorem 1 shows, the theorem holds even if the input consists of only one type of -frames and all intersection points of the -frames lie on only one side of the diagonal. This is in contrast to the diagonal-anchored -frames case where we obtain a . We also show that one cannot hope for a -approximation for any even when all the -frames intersect a vertical line; see Figure 2(c) for an example. We refer to these -frames as vertical-intersecting -frames. The  problem is -hard on vertical-intersecting -frames even if all the -frames intersect the vertical line from one side. Moreover, the problem is -hard even if for each -frame, the horizontal and vertical segments have the same length.

Moreover, we show that the -hardness of Theorem 1 is almost tight in the sense that the problem admits a polynomial-time algorithm on -frames, where each -frame intersects a vertical line and a horizontal line. See Figure 2(d) for an example. Note that, all the -frames in the input are of the same type. The  problem is linear-time solvable on -frames that intersect a vertical line and a horizontal line.

To prove Theorem 1, we show that this class of graphs are the same as permutation graphs for which given the permutation of the vertices, the  problem can be solved in linear time [6]. As given a set of -frames that intersect a vertical line and a horizontal line, the corresponding permutation can be computed in linear time, the theorem follows.

While the standard notion of intersection dates back to 1970s, Golumbic et al. [17] introduced the notion of edge intersection of -frames. In this model, two -frames are considered adjacent if and only if they overlap in strictly more than a single point in the plane. More formally, the -frames corresponding to the vertices of the graph are drawn on a grid and two vertices are adjacent in the graph if and only if their corresponding -frames share at least one grid edge; see Figure 2(e) for an example. To distinguish between the two models we will explicitly refer the edge intersection model whenever discussing a result on this model. Otherwise, we always mean the standard intersection model.

For the edge intersection model, there is a 4-approximation algorithm for  on -frames [3, 19]. Moreover, the problem was recently shown to be -hard by Mehrabi [23], where two “types” of -frames are needed for the construction. He left open whether the problem remains -hard when the input consists of only one type of -frames or when the -frames intersect a vertical line. We answer both questions affirmatively. In the edge intersection model, the  problem on -frames of a single type is hard to approximate within a factor of 1.1377 even if all the -frames intersect a vertical line from one side.

Furthermore, we show that even intersecting two lines does not help: the  problem is -hard on -frames in the edge intersection model even if every -frame intersects a vertical line and a horizontal line. Observe that this is in contrast to the existence of the linear-time algorithm of Theorem 1 under the standard intersection model.

Organization.

In Section 2, we give some definitions and revisit some necessary background. We prove Theorem 1 and 1 in Section 3. The proofs of Theorem 1 and 1 are given in Section 4. Finally, we show the results for the edge intersection model in Section 5 and conclude the paper in Section 6. Throughout this paper, the proofs of lemmas and theorems marked with are given in the full version of the paper due to space constraints.

2 Preliminaries

We denote the - and -coordinates of a point by and , respectively. For two points and , we denote the Euclidean distance between and by . Given a graph , we denote the -frame corresponding to a vertex by ; we use and interchangeably. We denote the corner of an -frame by .

Local search.

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 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 re-iterate. 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 solution returned by the algorithm and an optimal solution, respectively. We can assume that ; otherwise, we can remove the common elements of and and analyze the algorithm with the new sets, which guarantees that the approximation factor of the original instance is upper bounded by that of the new sets. The following result establishes the connection between local search technique and obtaining a . [[5, 25]] Consider the solutions and for a minimization problem, and suppose that there exists a planar bipartite graph that satisfies the local exchange property, which is as follows: 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.

3 Diagonal-intersecting Rectangles

In this section, we prove Theorem 1 and 1. To prove Theorem 1, we first give a for the problem when the rectangles are anchored at the diagonal from only one side and will then prove the theorem by applying the twice.

Recall the class of intersection graphs of diagonal-anchored rectangles is same as that of diagonal-anchored -frames [4]. As such, to simplify the presentation of the result, we prove Theorem 1 for -frames.

3.1 Ptas

Suppose that we are given a set of -frames each of which is anchored at the diagonal from above; let be the corresponding graph. Consider any two -frames and that intersect each other. We say that and are coincident if =. is said to intersect from left (resp. from below) if (resp. ). Notice that intersects from left if and only if intersects from below.

Consider the  problem on and run the local search algorithm with for some constant . Let be the solution returned by the local search algorithm and denote an optimal solution. Consider the bipartite graph in which the edge set is defined as follows. For any vertex , consider the set of all -frames in that intersect and let , where and , be a pair for which (i.e., the Euclidean distance between their corners) is minimum over all such pairs. Then, . We call a witness vertex for the pair . See Figure 3 for an example of how an edge of is determined.

In the following, we show that is planar (Lemma 3.13.1) and will then prove that satisfies the local exchange property (Lemma 6). Then, by Theorem 2, our local search algorithm yields a for the problem on . To distinguish between the edges of and those of , we refer to the edges of as arcs. Let and such that . We say that is a top arc if there is a witness for , such that both and intersect from left; choose one such arbitrarily and denote by . Otherwise, if there is a witness for , such that both and intersect from below, is a down arc; choose one such arbitrarily and denote by . Otherwise, for any witness of , and intersect from different sides; we call a mixed arc, and choose one such arbitrarily and denote by .

Drawing of H.

To draw , we map to on the diagonal for all . To draw the arcs of , the idea is to draw each arc either above or below the diagonal depending on whether it is a top arc or a down arc, respectively. A mixed arc is drawn in such a way that some part of the arc is drawn above and some part is drawn below the diagonal (and, hence crosses the diagonal). We next give the details of each case.

Let and such that and assume w.l.o.g. that . Moreover, let

 O:=(x(cor(bi))+x(cor(rj))2,y(% cor(bi))+y(cor(rj))2),

and consider the circle centred at with radius . If the arc is a top arc (resp., down arc), then we draw it on the half circle of that lies above the diagonal (resp., that lies below the diagonal ) starting from and ending at ; see Figure 4(a)—(b) for an illustration. A mixed arc is drawn in a slightly different way. For a mixed arc , assume w.l.o.g. that intersects from left while intersects from below. Notice that . To draw , let

 O1:=(x(cor(bi))+x(cor(w(bi,rj)))2,y(cor(bi))+y(cor(w(bi,rj)))2),
 O2:=(x(cor(w(bi,rj)))+x(cor(rj))2,y(cor(w(bi,rj)))+y(cor(rj))2),

and consider the following two circles: the circle that is centred at and has the radius , and the circle that is centred at with radius . See Figure 4(c). We draw the first part of arc on the half circle of that lies above the diagonal starting from and ending at and then the second part on the half circle of that lies below the diagonal starting from and ending at . See Figure 5 for an example.

Planarity of H.

Clearly, no top arc crosses a down arc (except perhaps at their endpoints). To show the planarity of , we show that — no top arc crosses another top arc, no down arc crosses another down arc, and no mixed arc crosses a top, a down or another mixed arc. No two top arcs in cross each other.

Proof.

Suppose for a contradiction that two top arcs cross each other, and w.l.o.g. assume that . Since is a top arc, we must have ; for a similar reason, we must have . We now consider two cases. (i) If , then must intersect , which is a contradiction because in that case we should have added or to corresponding to instead of . (ii) If , then must intersect — this is also a contradiction because in that case we should have added or to corresponding to instead of . ∎

No two down arcs in cross each other.

Proof.

Suppose for a contradiction that two down arcs cross each other, and w.l.o.g. assume that . Since is a down arc, we must have ; for a similar reason, we must have . We now consider two cases. (i) If , then must intersect , which is a contradiction because in that case we should have added either or to corresponding to the witness instead of the arc . (ii) If , then must intersect — this is a contradiction because in that case we should have added either or to corresponding to the witness , instead of the arc . ∎

It remains to show that no mixed arc crosses a top arc, a down arc or another mixed arc. No mixed arc crosses a top arc in .

Proof.

Suppose for a contradiction that a mixed arc crosses a top arc and assume w.l.o.g. that and . If , then we must have — otherwise, the two arcs would not cross. First, we know that . If , then intersects before intersecting , which is a contradiction because in that case we should have added either the top arc or the top arc to corresponding to instead of ; see Figure 6(a). If , then intersects before intersecting , which is again a contradiction as we should have added either the top arc or the mixed arc to corresponding to instead of . See Figure 6(b).

Now, suppose that . Then, we must have as the two arcs would not intersect otherwise. Since is a top arc, we must have and so would intersect before intersecting ; see Figure 6(c). This is a contradiction because we should have added either the top arc or the mixed arc to corresponding to instead of . ∎

No mixed arc crosses a down arc in .

Proof.

Suppose for a contradiction that a mixed arc crosses a down arc and assume w.l.o.g. that and . First, suppose that . Then, we must have — otherwise, the two arcs would not cross. Since is a down arc, we know that . If , then must intersect , which is a contradiction because we should have added either or to corresponding to the witness , instead of the arc ; see Figure 7(a). If , then must intersect , which is a contradiction as in that case we should have added or to corresponding to the witness , instead of the arc ; see Figure 7(b).

Now, suppose that . Then, we must have that as otherwise the two arcs would not cross; see Figure 7(c). Since is a down arc, we must have and so must intersect — this is a contradiction because in that case we should have added or to corresponding to the witness , instead of the arc . ∎

No two mixed arcs cross each other in .

Proof.

Let and be two mixed arcs in . First, and do not cross each other on the diagonal because crossing on the diagonal means that and are coincident and so we should have had only one arc among in for both and . As such, we assume in the following that and are not coincident.

If and cross each other, then they can have at most two crossings, and in case of two, one is above and the other is below the diagonal. We show that no such types of crossings are possible. First, suppose that and cross each other above the diagonal, and assume w.l.o.g. that . See Figure 8(a). Then, it is easy to see that would intersect before intersecting and so we should have added either the top arc or the mixed arc to corresponding to instead of (see Figure 8(a)). Now, suppose that and cross each other below the diagonal and assume w.l.o.g. that . See Figure 8(b). Then must intersect before intersecting and so we should have added either the down arc or the mixed arc to corresponding to the witness instead of the arc (see Figure 8(b)). This concludes the proof.

By Lemma 3.13.1, it follows that is a planar graph. The following lemma along with the planarity of gives a for the problem on . Graph satisfies the local exchange property.

Proof.

To prove the lemma, it is sufficient to show that for any vertex , there are and such that both and intersect and . Take any vertex and let be the set of all -frames that intersect . Notice that and because each of and is a feasible solution to the  problem on . Now, consider and for which is minimum over all such pairs . We know by definition of that , which proves the lemma. ∎

3.2 Proof of Theorem 1

We are now ready to prove Theorem 1. Here, the rectangles are anchored at the diagonal from both sides; let denote the corresponding graph.

Proof of Theorem 1.

Let be a partition of the vertices of such that the -frames corresponding to vertices in (resp. ) are anchored at from above (resp. below). By abusing notation, we refer to these two sets of -frames also as and . For any , set . We apply the of Section 3.1 with parameter to the -frames in and independently, and let and be the solutions returned by the algorithm, respectively. We return as the final solution. Let , and denote an optimal solution for the  problem on the -frames in (i.e., graph ), and , respectively.

Consider the solution . Let be a minimum size set of -frames that dominates all the -frames in . If there is an -frame that is in , then can only dominate those -frames in that are anchored at the diagonal at the same point as , in which case we can replace by one of those -frames from . As such, there exists a set of size at most that dominates the -frames of . It follows that . Similarly, one can show that . Now, by using the result from Section 3.1, we have and . Then,

 |S|≤|SX|+|SY|≤(1+ϵ′)|OPTX|+(1+ϵ′)|OPTY|≤2(1+ϵ′)|OPT|=(2+ϵ)|OPT|,

which completes the proof of the theorem. ∎

3.3 Proof of Theorem 1

Recall that Theorem 1 claims -hardness of the problem on -frames in the diagonal-intersecting case. We show a gap-preserving reduction from the  problem on circle graphs, which is known to be -hard [9]. Recall that a graph is called a circle graph, if it is the intersection graph of chords of a circle. Take any circle graph with vertices, and consider a geometric representation of . By a closer look at the -hardness proof [9], we can assume that no two chords share an endpoint; that is, there are exactly distinct points on the circle determining the endpoints of the chords. Cut the circle at an arbitrary point and consider the ordering of the endpoints of chords visited in counter-clockwise along the circle starting at . Now, let be the diagonal line . Consider each endpoint (where ) in the order given by . Then, we map each point to the point on . Let be a chord of . Then, the -frame corresponding to is the unique -frame that lies below and connects the two points mapped on corresponding to and ; see Figure 9 for an example. Let be the graph corresponding to the constructed -frames. Clearly, has slope -1 and the construction can be done in polynomial time. Moreover, since the endpoints of chords of the input representation are pairwise distinct, the points mapped to the line are in general position; that is, no two such points have the same - or the same -coordinates. The following lemma completes the proof of Theorem 1.

has a dominating set of size if and only if has a dominating set of size .

Proof.

First, we show that there is a one-to-one correspondence between the vertices of and those of such that if and only if . The one-to-one correspondence between the vertices of these two graphs is clear from the construction. Now, suppose that . This means that the endpoints of vertices and appear in as either or , where and with (resp., and with ) correspond to the endpoints of (resp., ). By the mapping used in the construction, this ordering is preserved on (when going from to along ) and so the -frames and intersect each other below ; that is, . Conversely, if , then their endpoints appear in as either or , where and with (resp., and with ) correspond to the endpoints of (resp., ). Consequently, this ordering is preserved on by the mapping (when going from to along ) and so and do not intersect each other inside ; that is, . It is now straightforward to see that has a dominating set of size if and only if has a dominating set of size , which is clearly a gap-preserving reduction. ∎

Remark.

Using a similar reduction in the other direction one can show that, for every intersection graph of a set of diagonal-intersecting -frames that intersect each other only below the diagonal, one can find a circle graph such that has a dominating set of size if and only if has a dominating set of size . As one can obtain a -approximation for  on circle graphs [10],  can be approximated within a factor of in this special case of diagonal-intersecting -frames. Note that this is the version which we have just proved to be -hard.

4 Vertical-intersecting L-frames

In this section, we prove Theorem 1 and 1. To prove Theorem 1, we first show that the  problem is -hard even when each -frame intersects a vertical line. The proof is essentially the same as that of Theorem 1, but here we replace the diagonal with a quarter of a circle and then extend the horizontal segment of each -frame until it hits a vertical line that is placed far to the right.

Next we show the gap-preserving reduction from the  problem on circle graphs, which is known to be -hard [9]. Recall that a graph is called a circle graph, if it is the intersection graph of chords of a circle. Take any circle graph with vertices, and consider a geometric representation of . Recall that we can assume no two chords share an endpoint; that is, there are exactly distinct points on the circle determining the endpoints of the chords. Cut the circle at an arbitrary point and consider the ordering of the endpoints of chords visited in counter-clockwise along the circle starting at . Now, let be a circle with radius one that is centred at the origin of the Cartesian coordinate system. Consider each endpoint (where ) in the order given by , and let . Then, we map to the point on the top right quadrant of circle . Let be a chord of . Then, the -frame corresponding to is the unique -frame that lies inside and connects the two points mapped on corresponding to and ; see Figure 10(b) for an illustration. Finally, we complete the construction by extending the horizontal segment of every -frame to the right until it intersects the vertical line . See Figure 10(c). Clearly, the construction can be done in polynomial time, and since the endpoints of chords of the input representation are pairwise distinct, the points mapped to the circle are in general position; that is, no two such points have the same - or the same -coordinates. Now, one can easily prove a lemma similar to Lemma 3.3 for this case as well, and so the -hardness of the problem follows.

To complete the proof of Theorem 1, we show that the  problem is -hard on -frames even if the horizontal and vertical segments of every -frame have the same length. The reduction is from a variant of the 3SAT problem. For any 3SAT instance, one can define a bipartite graph on the clauses and the variables which we refer to as the incidence graph. For any clause , if contains a literal corresponding to a variable , the edge is added to the graph. A drawing of a planar incidence graph is called planar rectilinear if it has the following properties. Each variable vertex is drawn as a horizontal segment on the -axis and the clause vertices are drawn as horizontal segments above and below the -axis. For each edge , the horizontal segment corresponding to has a vertical connection to the horizontal segment corresponding to . Moreover, this vertical connection does not intersect any other horizontal segments. See the left figure of Figure 11 for an example. An instance of the 3SAT problem is called monotone if, for every clause in the instance, the literals of the clause are either all positive (called a positive clause) or all negative (called a negative clause). A planar rectilinear drawing of an incidence graph is called monotone if it corresponds to a monotone 3SAT instance such that all the positive (resp., negative) clauses are drawn above (resp., below) the -axis. In the Planar Monotone Rectilinear 3SAT problem the incidence graph of any instance has a planar rectilinear and monotone drawing. The problem is known to be -hard even when the drawing is given [12]. Given an instance of the Planar Monotone Rectilinear 3SAT, we construct a set of -frames such that the horizontal and vertical segments of each -frame have the same length and they all intersect a vertical line.

Let be the number of variables. Due to the hierarchical structure of the clauses lying above and below the -axis, respectively in the given drawing, one can modify the drawing in a way such that for each clause, the length of the horizontal segment corresponding to it is the same as the length of the vertical segments corresponding to it (see the left figure of Figure 11). Let be the new drawing. We construct the instance of  from in the following way. For each clause , removing all the vertical connections except the leftmost one creates an -frame. We add this -frame (denoted by ) to our instance. For each variable , we add two -frames: one (denoted by ) above and the other (denoted by ) below the -axis. The vertical segments of these two -frames are the maximum length vertical connections corresponding to lying above and below -axis, respectively. For the variable -frames, their horizontal segments lie on the right of their vertical segment. Moreover, for any variable -frame above (resp., below) the -axis, the corner is the topmost (resp., bottommost) point of the vertical segment. The two -frames corresponding to each variable are shifted accordingly so that they intersect at a point on the -axis (see Figure 11). Also, for each variable , we add an auxiliary -frame that intersects only the two -frames corresponding to it as shown in Figure 11. Note that in the constructed instance, one endpoint of each -frame lies on the -axis. Now it could be the case that, for some and , an intersects though does not contain any literal of . To get rid of these intersections, we extend each vertically and then horizontally (to maintain the symmetry) so that it does not intersect any such that does not contain . Note that, as we do not modify the ’s, if contains the literal (resp. ), (resp. ) still intersects . Finally, we rotate the created instance clockwise by 90 with respect to the origin. Hence all the -frames in the constructed instance now intersect the vertical line or the -axis, and thus together they form an instance of the problem with vertical-intersecting -frames. Lemma 4 below completes the proof of the second part of Theorem 1. We first need the following observation whose proof is immediate from the construction.

Observation .

For all , (resp., ) intersects if and only if the clause contains the literal (resp., ).

The instance has a dominating set of size if and only if the 3SAT instance is satisfiable.

Proof.

Suppose has a dominating set of size . We construct a satisfiable assignment in the following way. Consider any variable . As intersects only and , one of these three -frames must be in the dominating set. The size of being , only one of the above mentioned three -frames can be in and no clause -frame can be in . If is in , we set to be true. Otherwise, we set to be false. Now consider any clause . Let be a variable such that a -frame corresponding to is in that intersects . Note that there exists such an . Now if is the chosen -frame that dominates , then is in and by Observation 4 must be a positive clause. It follows that, we have set to be true and thus is satisfied. Similarly, one can show that if is the chosen -frame that dominates , then also is satisfied.

Now, suppose that we are given a satisfiable assignment of the 3SAT instance. We construct a dominating set in the following way. For any variable , if is set to be true, then we select . If is set to be false, then we select . We claim that the selected -frames form a dominating set. It is easy to see that the selected -frames dominate the variable and auxiliary -frames. Now, consider any clause -frame . There must be a variable that satisfies . If is positive, the literal must be set to true and we select . By Observation 4, intersects , and thus is being dominated. If is negative, one can similarly show that is also being dominated. ∎

We now prove Theorem 1. To this end, we show that the intersection graph of -frames that inersect a vertical and a horizontal line is a permutation graph and so  is linear-time solvable on such a graph [6]. Geometrically, a graph is a permutation graph if there are two embeddings of its vertices on two parallel lines such that, when connecting a vertex from the first line to the same vertex on the other line using a line segment, the edge set of the graph is realized exactly by the intersections of those segments. That is, two vertices appear in different order on the parallel lines (and, so their line segments intersect) whenever they are adjacent in the graph. By taking the two orderings in which the endpoints of the -frames in the input graph intersect the two lines, we can construct the geometric representation of a permutation graph having the same edges. Hence, Theorem 1 follows.

5 Edge Intersection Model

In this section, we show our results for the edge intersection model [17].

Proof of Theorem 2.

We show a reduction from the Vertex Cover problem. In Vertex Cover, we are given a graph with vertices and the goal is to find a subset such that for any edge , . Let be the size of any minimum size vertex cover of . As noted in [2], from the work of Dinur and Safra [13] the following theorem can be derived. [Dinur and Safra [13]] Let and . For any constant , given any unweighted graph with bounded degrees, it is -hard to distinguish between “Yes”: VC(G) , and “No”: VC(G) .

Given the vertex cover graph , we construct an instance of . In the instance , all the -frames lie on the left of the line and intersect the line, and thus each has the shape. The construction is as follows. For each vertex , we have an -frame that intersects at the point as shown in Figure 12. Note that these -frames do not intersect each other. For each edge with , we have an -frame that intersects at , and thus intersects (i.e., has a common horizontal grid edge) with . It also intersects (has a common vertical grid edge) with (see Figure 12). Also, for each vertex , we have two auxiliary -frames and that intersect at , and shares a vertical grid edge with . Note that the only -frame intersects is . Also, all the -frames intersect the vertical line . We first need the following observations.

Observation .

For , the only -frames that intersects are and , where with .

Observation .

The only -frames that get dominated by the -frames in are and for .

Observation .

Any dominating set contains at least one of and for each . Also for any dominating set of size , there is a dominating set of size at most such that contains for all and does not contain any for .

has a vertex cover of size if and only if has a dominating set of size .

Proof.

Suppose has a vertex cover of size . We construct a dominating set of size in the following way. At first add all the -frames to . Then for each in the vertex cover, add to . Clearly . By Observation 5, dominates any and for . Now consider any . Note that we have added at least one of and to , and hence dominates as well.

Now, suppose we have a dominating set of size . By Observation 5, we can assume that contains for all and does not contain any . Thus contains -frames corresponding to the vertices and the edges. Now suppose contains an edge -frame . By Observation 5 and 5, the only -frames that cannot get dominated by by the removal of from are , where with . Thus if we replace by in , still remains a dominating set. Thus, we can assume w.l.o.g. that does not contain any . Now we construct a subset by selecting vertices corresponding to the vertex -frames contained in . Clearly . We claim that is a vertex cover. Consider any edge . Then by Observation 5, at least one of and must be in . It follows that at least one of and is present in which finishes the proof. ∎

By Theorem 5 and Lemma 5, we have the following lemma. Let