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Solving Medium-Density Subset Sum Problems in Expected Polynomial Time: An Enumeration Approach

机译:解决期望多项式时间内的中等密度子集和问题:一种枚举方法

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The subset sum problem (SSP) can be briefly stated as: givena target integer E and a set A containing n positive integer a_j, find a subset of A summing to E. The density d of an SSP instance is defined by the ratio of n to m, where m is the logarithm of the largest integer within A. Based on the structural and statistical properties of subset sums, we present an improved enumeration scheme for SSP, and implement it as a complete and exact algorithm (EnumPlus). The algorithm always equivalently reduces an instance to be low-density, and then solve it by enumeration. Through this approach, we show the possibility to design a sole algorithm that can efficiently solve arbitrary density instance in a uniform way. Furthermore, our algorithm has considerable performance advantage over previous algorithms. Firstly, it extends the density scope, in which SSP can be solved in expected polynomial time. Specifically, It solves SSP in expected O(n log n) time when density d ≥ c·n_(1/2)/log n, while the previously best density scope is d ≥ c·n/(log n)~2. In addition, the overall expected time and space requirement in the average case are proven to be O(n~5 log n) and O(n~5) respectively. Secondly, in the worst case, it slightly improves the previously best time complexity of exact algorithms for SSP. The worst-case time complexity of our algorithm is proved to be O(n·2~(n/2) - c·2(n/2) + n), while the previously best result is O(n·2/~(n/2)).
机译:子集和问题(SSP)可以简单地表示为:给定目标整数E和包含n个正整数a_j的集合A,找到求和为A的A的子集。SSP实例的密度d由n的比值定义到m,其中m是A内最大整数的对数。基于子集和的结构和统计属性,我们提出了一种改进的SSP枚举方案,并将其实现为完整而精确的算法(EnumPlus)。该算法总是将实例等效地降低为低密度,然后通过枚举来解决。通过这种方法,我们展示了设计唯一算法的可能性,该算法可以有效地以统一的方式求解任意密度的实例。此外,与以前的算法相比,我们的算法具有相当大的性能优势。首先,它扩展了密度范围,可以在期望的多项式时间内求解SSP。具体来说,当密度d≥c·n_(1/2)/ log n时,它可以在预期的O(n log n)时间内求解SSP,而以前最好的密度范围是d≥c·n /(log n)〜2。此外,平均情况下的总预期时间和空间需求被证明分别为O(n〜5 log n)和O(n〜5)。其次,在最坏的情况下,它会稍微提高SSP精确算法以前最佳的时间复杂度。我们的算法在最坏情况下的时间复杂度被证明为O(n·2〜(n / 2)-c·2(n / 2)+ n),而先前的最佳结果是O(n·2 /〜 (n / 2))。

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