选择k个最优的平均复杂度

19th Annual Symposium on Foundations of Computer Science (sfcs 1978) Pub Date : 1978-10-16 DOI:10.1109/SFCS.1978.29

A. Yao, F. Yao

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引用次数: 7

摘要

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。

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

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.

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19th Annual Symposium on Foundations of Computer Science (sfcs 1978)

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