Finding N bits using O(N/logN) sums

S. Corlat, Veaceslav Guzun, Victor Vorona
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Abstract

The problem we are trying to solve sounds as follows: You are given N bits.Find the value of each bit. We will show a technique which enables finding the values of N bits using O(N/logN) subsequence-sum queries. The algorithm consists of two phases: Constructing the queries for each layer and using the queries for a particular layer to getthe value of every bit. We described the following technique in this blog [1], which wasinspired by this article [2]. It should be noted that this number of queries is indeed theoptimal one for finding all N bits of a binary array, since each subsequence-sum queriesoffers us at most log(2)N bits of information.
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用O(N/logN)和求N位
我们要解决的问题如下:给你N个比特。求每个位的值。我们将展示一种技术,它可以使用O(N/logN)子序列和查询来查找N位的值。该算法由两个阶段组成:为每个层构造查询,并使用特定层的查询来获取每个位的值。受本文[2]的启发,我们在本博客[1]中描述了以下技术。应该注意的是,这个查询数量确实是查找二进制数组中所有N位的最佳查询数量,因为每个子序列和查询最多为我们提供log(2)N位的信息。
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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