k-拟阵中近似最大独立集的一换二交换启发式分析
Analysis of the two-for-one swap heuristic for approximating the maximum independent set in a k-polymatroid
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中文总结 AI 辅助
本文分析k-拟阵中最大独立集的一换二重复交换启发式，给出简单证明，得到2/(k+1)-近似比，弱于已知的2/k近似算法。
中文摘要 AI 辅助
设 f:2^N --> \cZ^+ 为一个拟阵（一个整数值非递减子模集函数，满足 f(emptyset) = 0）。一个 k-拟阵满足对所有 e in N 有 f(e) <= k。我们称 N 的子集 S 是独立的，如果 f(S) 等于 S 中元素 e 的 f(e) 之和，且对所有 e in S 有 f(e) > 0。在 2-拟阵中寻找最大规模独立集已被研究，对于线性拟阵已知有多项式时间算法。对于 k >= 3，该问题是 NP 难的，已知一个近似比为接近 2/k 的近似算法，该算法通过尽可能长时间地用当前解中的一个“大”子集交换一个多一个元素的集合来获得。这里我们给出更具体的重复一换二交换启发式的简单分析，得到一个（较弱的）2/(k+1)-近似。
英文摘要
Let f:2^N --> \cZ^+ be a polymatroid (an integer-valued non-decreasing submodular set function with f(emptyset) = 0). A k-polymatroid satisfies that f(e) <= k for all e in N. We call a subset S of N independent if f(S) equals the sum of f(e) over the elements e of S and f(e) > 0 for all e in S. Finding a maximum-size independent set in a 2-polymatroid has been studied and polynomial-time algorithms are known for linear polymatroids. For k >= 3, the problem is NP-hard, and an approximation algorithm with ratio approaching 2/k is known and is obtained by swapping as long as possible a "large" subset from the current solution by a set with one more element. Here we give a simple analysis of the more particular two-for-one repeated swapping heuristic, obtaining a (weaker) 2/(k+1)-approximation.
发表机构
- Research Institute of the University of Bucharest（布加勒斯特大学研究院）
- Illinois Institute of Technology（伊利诺伊理工学院）
- University of Illinois（伊利诺伊大学）
机构由 AI 辅助整理，请以论文原文为准。
