最大和与偶数 SAT

Tamio-Vesa Nakajima, Stanislav Živný
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引用次数: 0

摘要

一个(多)字面集合称为一个子句,如果没有一个字面的值为假,那么这个赋值就是强满足子句。寻找一个能最大限度地增加强满足子句数量的赋值是 NP 难的。我们提出了一种简单的算法,即在给定的条款集合中,如果有一个赋值能够强满足其中 $\rho$ 分数的条款,那么在这个赋值中,至少有 $\rho$ 分数的条款是弱满足的,也就是说,有偶数个字面的值是假的。特别地,这意味着一种高效的算法,可以在图中存在一个有向值$\rho$的切分的情况下,找到一个值为$\rho$的无向切分。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Maximum And- vs. Even-SAT
A (multi)set of literals, called a clause, is strongly satisfied by an assignment if no literal evaluates to false. Finding an assignment that maximises the number of strongly satisfied clauses is NP-hard. We present a simple algorithm that finds, given a set of clauses that admits an assignment that strongly satisfies a $\rho$-fraction of the clauses, an assignment in which at least a $\rho$-fraction of the clauses is weakly satisfied, in the sense that an even number of literals evaluates to false. In particular, this implies an efficient algorithm for finding an undirected cut of value $\rho$ in a graph given that a directed cut of value $\rho$ in the graph is promised to exist.
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