构造和搜索偏序的平均下界

Harry G. Mairson
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引用次数: 5

摘要

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