Choice Is Hard

被引:16
作者
Arkin, Esther M. [1 ]
Banik, Aritra [2 ]
Carmi, Paz [2 ]
Citovsky, Gui [1 ]
Katz, Matthew J. [2 ]
Mitchell, Joseph S. B. [1 ]
Simakov, Marina [2 ]
机构
[1] SUNY Stony Brook, Dept Appl Math & Stat, Stony Brook, NY 11794 USA
[2] Ben Gurion Univ Negev, Dept Comp Sci, Beer Sheva, Israel
来源
ALGORITHMS AND COMPUTATION, ISAAC 2015 | 2015年 / 9472卷
关键词
SATISFIABILITY;
D O I
10.1007/978-3-662-48971-0_28
中图分类号
TP301 [理论、方法];
学科分类号
081202 ;
摘要
Let P = {C-1, C-2,..., C-n} be a set of color classes, where each color class C-i consists of a pair of objects. We focus on two problems in which the objects are points on the line. In the first problem (rainbow minmax gap), given P, one needs to select exactly one point from each color class, such that the maximum distance between a pair of consecutive selected points is minimized. This problem was studied by Consuegra and Narasimhan, who left the question of its complexity unresolved. We prove that it is NP-hard. For our proof we obtain the following auxiliary result. A 3-SAT formula is an LSAT formula if each clause (viewed as a set of literals) intersects at most one other clause, and, moreover, if two clauses intersect, then they have exactly one literal in common. We prove that the problem of deciding whether an LSAT formula is satisfiable or not is NP-complete. We present two additional applications of the LSAT result, namely, to rainbow piercing and rainbow covering. In the second problem (covering color classes with intervals), given P, one needs to find a minimum-cardinality set I of intervals, such that exactly one point from each color class is covered by an interval in I. Motivated by a problem in storage systems, this problem has received significant attention. Here, we settle the complexity question by proving that it is NP-hard.
引用
收藏
页码:318 / 328
页数:11
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