2010
DOI: 10.1142/s0218195910003220
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Maximum Area Independent Sets in Disk Intersection Graphs

Abstract: Maximum Independent Set (MIS) and its relative Maximum Weight Independent Set (MWIS) are well-known problems in combinatorial optimization; they are NP-hard even in the geometric setting of unit disk graphs. In this paper, we study the Maximum Area Independent Set (MAIS) problem, a natural restricted version of MWIS in disk intersection graphs where the weight equals the disk area. We obtain: (i) Quantitative bounds on the maximum total area of an independent set relative to the union area; (ii) Practical cons… Show more

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Cited by 11 publications
(15 citation statements)
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“…The maximum area independent set (MAIS) problem for rectangles (squares, or disks, etc. ), is that of selecting a maximum area packing from a given set [3]; see classic papers such as [6,32,33,34,35] and also more recent ones [10,11,21] for quantitative bounds and constant approximations. These optimization problems are NP-hard, and there is a rich literature on approximation algorithms.…”
Section: Introductionmentioning
confidence: 99%
“…The maximum area independent set (MAIS) problem for rectangles (squares, or disks, etc. ), is that of selecting a maximum area packing from a given set [3]; see classic papers such as [6,32,33,34,35] and also more recent ones [10,11,21] for quantitative bounds and constant approximations. These optimization problems are NP-hard, and there is a rich literature on approximation algorithms.…”
Section: Introductionmentioning
confidence: 99%
“…Conjecture 3 (Bereg, Dumitrescu, and Jiang [3]). For any set S of (not necessary congruent) closed disks in the plane, there exists a subset I of pairwise-disjoint disks such that |I|/|F| ≥ Note that a disk D is centrally symmetric; for any finite family F of congruent disks (i.e., translates of a disk) in the plane, ν(F) ≥ Note.…”
Section: Resultsmentioning
confidence: 99%
“…4, we obtain an improved lower bound for the special case that the centrally symmetric convex body S is a disk. The previous best lower bound for disks was F (S) > 1/8.4898 [2]. Theorem 3 For a disk S, F (S) ≥ 1/λ disk , where λ disk = 8.3539 .…”
Section: Theoremmentioning
confidence: 98%
“…For the one-dimensional case, it is known that f (S) = F (S) = 1/2 for an interval S [2,15]. The aforementioned results of Zalgaller [22] and Ajtai [1], respectively, give lower and upper bounds of 1/8.6 ≤ F (S) ≤ 1/4 − 1/1728 for a square S. In Sect.…”
Section: Introductionmentioning
confidence: 95%