k-SAT和约束满足问题的概率算法

U. Schöning
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引用次数: 470

摘要

我们提出了一个简单的概率算法来求解k-SAT,更一般地说,用于求解约束满足问题(CSP)。该算法遵循一个简单的局部搜索范式(S. Minton et al., 1992):随机猜测一个初始赋值,然后在那些不满足的子句(约束)的指导下,依次从该子句中选择一个随机文字并翻转相应的位,试图找到一个满意的赋值。如果在O(n)步之后没有找到满意的分配,则重新开始。我们的分析表明,对于任何有n个变量的可满足的k- cnf公式,这个过程平均只需要重复t次,就能找到一个令人满意的赋值,其中t在多项式因子(2(1-1/k))/sup n/内。这是目前已知的最快(也是最简单)的3-SAT算法。我们还考虑了具有n个变量,每个变量最多取d个值,约束为l阶的CSP的更一般情况,并分析了相应的(广义)算法m的复杂度。结果表明,任何CSP都可以以最多(d/spl middot/(1-1/l)+/spl epsiv/)/sup n/的复杂度求解。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
A probabilistic algorithm for k-SAT and constraint satisfaction problems
We present a simple probabilistic algorithm for solving k-SAT and more generally, for solving constraint satisfaction problems (CSP). The algorithm follows a simple local search paradigm (S. Minton et al., 1992): randomly guess an initial assignment and then, guided by those clauses (constraints) that are not satisfied, by successively choosing a random literal from such a clause and flipping the corresponding bit, try to find a satisfying assignment. If no satisfying assignment is found after O(n) steps, start over again. Our analysis shows that for any satisfiable k-CNF-formula with n variables this process has to be repeated only t times, on the average, to find a satisfying assignment, where t is within a polynomial factor of (2(1-1/k))/sup n/. This is the fastest (and also the simplest) algorithm for 3-SAT known up to date. We consider also the more general case of a CSP with n variables, each variable taking at most d values, and constraints of order l, and analyze the complexity of the corresponding (generalized) algorith m. It turns out that any CSP can be solved with complexity at most (d/spl middot/(1-1/l)+/spl epsiv/)/sup n/.
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