On the average-case complexity of selecting k-th best

A. Yao, F. Yao
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引用次数: 7

Abstract

Let Vk (n) be the minimum average number of pairwise comparisons needed to find the k-th largest of n numbers (k≥2), assuming that all n! orderings are equally likely. D. W. Matula proved that, for some absolute constant c, Vk(n)- n ≤ ck log log n as n → ∞. In the present paper, we show that there exists an absolute constant c′ ≫ 0 such that Vk(n) - n ≥ c′k log log n as n → ∞, proving a conjecture by Matula.
选择k个最优的平均复杂度
让Vk (n)成为pairwise最低平均当家》《k-th comparisons需要努力去发现n的最大数字(k≥2),assuming那所有n !订单是一样的。D . W . Matula proved that c为一些绝对康斯坦,Vk (n) - n≤美国ck对数log n n→∞。《现在,这篇文章我们秀那绝对有exists an康斯坦c′≫0如此那Vk (n)美国c - n≥对数log k′n n→∞,证实a conjecture Matula偏。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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