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+\epsilon)-approximation . Furthermore, we obtain a PTAS in case all the input L-frames are anchored at the diagonalfrom 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 resultsis an NP-hardness proof of the third version which answers a question posed by Mehrabi (WAOA 2017).
We consider the Minimum Dominating Set ( MDS ) problem on the intersection graphs of geometric objects. Even for simple and widely-used geometric objects such as rectangles, no sub-logarithmic approximation is known for the problem and (perhaps surprisingly) the problem is NP-hard even when all the rectangles are "anchored" at a diagonal line with slope -1 (Pandit, CCCG 2017). In this paper, we first show that for any \epsilon>0, there exists a (2+\epsilon)-approximation algorithm for the MDS problem on "diagonal-anchored" rectangles, providing the first O(1)-approximation for the problem on a non-trivial subclass of rectangles. It is not hard to see that the MDS problem on "diagonal-anchored" rectangles is the same as the MDS problem on "diagonal-anchored" L-frames : the union of a vertical and a horizontal line segment that share an endpoint. As such, we also obtain a (2+\epsilon)-approximation for the problem with "diagonal-anchored" L-frames. On the other hand, we show that the problem is APX-hard in case the input L-frames intersect the diagonal, or the horizontal segments of the L-frames intersect a vertical line . However, as we show, the problem is linear-time solvable in case the L-frames intersect a vertical as well as a horizontal line . Finally, we consider the MDS problem in the so-called "edge intersection model" and obtain a number of results, answering two questions posed by Mehrabi (WAOA 2017).
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Minimum
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DS
MDS
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Surprisingly,
Even
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widely used
widely-used geometric
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the problem
no sub-logarithmic approximation is known for the problem and (perhaps surprisingly) the problem
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diagonal
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-1. It is easy
-1 (Pandit, CCCG 2017). In this paper, we first show that for any \epsilon>0, there exists a (2+\epsilon)-approximation algorithm for the MDS problem on "diagonal-anchored" rectangles, providing the first O(1)-approximation for the problem on a non-trivial subclass of rectangles. It is not hard
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for the anchored rectangles , the problem reduces to one with even simpler objects:
the MDS problem on "diagonal-anchored" rectangles is the same as the MDS problem on "diagonal-anchored"
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. An L-frame is
:
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line
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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
an endpoint. As such, we also obtain a
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. Furthermore, we obtain a PTAS in case all
for the problem with "diagonal-anchored" L-frames. On the other hand, we show that the problem is APX-hard in case
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are anchored at the diagonalfrom one side. Next, we consider the version, where all input
intersect the diagonal, or the horizontal segments of the
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, and prove APX-hardness of this version. Moreover, we prove NP-hardness of this version even
. However, as we show, the problem is linear-time solvable
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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
L-frames intersect a vertical as well as
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, and show that this version is linear-time solvable. We also consider these versions of the
. Finally, we consider the MDS
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, and obtain several interesting results. One of the resultsis an NP-hardness proof of the third version which answers a question
and obtain a number of results, answering two questions