近似最小的超体积贡献者：一般来说是NP困难的，但在实践中很快
Approximating the least hypervolume contributor: NP-hard in general, but fast in practice
摘要：
超体积指标是一种越来越流行的集合度量，用于比较两个帕累托集合的质量。大多数基于超体积指示器的优化算法的基本成分是关于帕累托集的单个解的超体积贡献的计算。我们证明了超体积贡献的精确计算是P-困难的，而它的近似是NP-困难的。最低缴款的计算也是如此。我们还证明了决定一个解是否具有最小超体积贡献是NP-困难的。即使决定解的贡献是否至多是最小贡献的$(1+\eps)$倍也是NP难的。这意味着既不可能有效地找到贡献最小的解（除非$P = NP$），也不可能对其进行近似（除非$NP = BPP$）。然而，在本文的第二部分，我们提出了一个快速的近似算法，这个问题。我们证明，对于任意给定的$\eps,δ>0$计算的解决方案的贡献最多$(1+\eps)$倍的最小贡献的概率至少$(1-δ)$。虽然它不能在所有实例中以多项式时间运行，但它在各种基准数据集上的执行速度非常快。该算法解决了非常大的问题实例，这些问题实例对于精确算法来说是棘手的（例如，100个维度的10000个解决方案）在几秒钟内。
英文摘要：
The hypervolume indicator is an increasingly popular set measure to compare the quality of two Pareto sets. The basic ingredient of most hypervolume indicator based optimization algorithms is the calculation of the hypervolume contribution of single solutions regarding a Pareto set. We show that exact calculation of the hypervolume contribution is #P-hard while its approximation is NP-hard. The same holds for the calculation of the minimal contribution. We also prove that it is NP-hard to decide whether a solution has the least hypervolume contribution. Even deciding whether the contribution of a solution is at most $(1+\eps)$ times the minimal contribution is NP-hard. This implies that it is neither possible to efficiently find the least contributing solution (unless $P = NP$) nor to approximate it (unless $NP = BPP$). Nevertheless, in the second part of the paper we present a fast approximation algorithm for this problem. We prove that for arbitrarily given $\eps,δ>0$ it calculates a solution with contribution at most $(1+\eps)$ times the minimal contribution with probability at least $(1-δ)$. Though it cannot run in polynomial time for all instances, it performs extremely fast on various benchmark datasets. The algorithm solves very large problem instances which are intractable for exact algorithms (e.g., 10000 solutions in 100 dimensions) within a few seconds.
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