Average case lower bounds on the construction and searching of partial orders

Harry G. Mairson
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引用次数: 5

Abstract

It is very well known in computer science that partially ordered files are easier to search. In the worst case, for example, a totally unordered file requires no preprocessing, but ¿(n) time to search, while a totally ordered file requires ¿(n log n) preprocessing time to sort, but can be searched in O(log n) time. Behind the casual observation, then, lurks the notion of a computational tradeoff between sorting and searching. We analyze this tradeoff in the average case, using the decision tree model. Let P be a preprocessing algorithm that produces partial orders given a set U of n elements, and let S be a searching algorithm for these partial orders. Assuming any of the n! permutations of the elements of U are equally likely, and that we search for any y isin; U with equal probability (in unsuccessful search, all "gaps" are considered equally likely), the average costs P(n) of preprocessing and S(n) of searching may be computed. We demonstrate a tradeoff of the form P(n) + n log S(n) = ¿(n log n), for both successful and unsuccessful search. The bound is tight up to a constant factor. In proving this tradeoff, we show a lower bound on the average case of searching a partial order. Let A be a partial order on n elements consistent with Π permutations. We show S(n) = ¿(Π3/n/n2) for successful search of A, and S(n) = ¿(Π2/n/n) for unsuccessful search. These lower bounds show, for example, that heaps require linear time to search on the average.
构造和搜索偏序的平均下界
在计算机科学中,部分排序的文件更容易搜索,这是众所周知的。在最坏的情况下,例如,一个完全无序的文件不需要预处理,但需要¿(n)时间来搜索,而一个完全有序的文件需要¿(n log n)预处理时间来排序,但可以在O(log n)时间内搜索。因此,在偶然观察的背后,隐藏着排序和搜索之间的计算权衡的概念。我们使用决策树模型在平均情况下分析这种权衡。设P是给定一个包含n个元素的集合U产生偏阶的预处理算法,设S是对这些偏阶的搜索算法。假设n中的任意一个!U中元素的排列是等可能的,我们搜索任意的y元素;U(在搜索不成功的情况下,所有“间隙”都被认为是等可能的),则可以计算出预处理的平均成本P(n)和搜索的平均成本S(n)。对于成功和不成功的搜索,我们展示了P(n) + n log S(n) =¿(n log n)的折衷形式。边界紧到一个常数因子。为了证明这种权衡,我们给出了搜索偏序的平均情况的下界。设A是符合Π排列的n个元素的偏序。对于A的成功搜索,我们显示S(n) =¿(Π3/n/n2),对于不成功的搜索,我们显示S(n) =¿(Π2/n/n)。例如,这些下界表明,堆平均需要线性时间来搜索。
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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