线性近似压实的下界

S. Chaudhuri
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引用次数: 6

摘要

lambda -近似压缩问题是:给定一个包含n个值的输入数组,每个值要么为0,要么为1,将每个值放入输出数组中,以便所有的1都位于数组的前(1+ lambda)k个位置,其中k是输入中1的数量。Lambda是一个精度参数。这个问题在并行计算中具有重要的基础意义,因为它适用于处理器分配和近似计数。当lambda为常数时,这个问题称为线性近似压缩(LAC)。在CRCW PRAM模型上,有一种算法可以在O((log logn) /sup 3/)时间内解决lambda =/sup 1///sub loglogn/的近似压缩问题,使用/sup n///sub (loglogn)3/处理器。这几乎是最好的了。具体来说,作者证明了LAC使用O(n)个处理器需要Omega (log log n)时间。它们还在lambda和处理时间之间进行了权衡。For in >
本文章由计算机程序翻译,如有差异,请以英文原文为准。
A lower bound for linear approximate compaction
The lambda -approximate compaction problem is: given an input array of n values, each either 0 or 1, place each value in the output array so that all the 1s are in the first (1+ lambda )k array locations, where k is the number of 1's in the input. lambda is an accuracy parameter. This problem is of fundamental importance in parallel computation because of its applications to processor allocation and approximate counting. When lambda is a constant, the problem is called linear approximate compaction (LAC). On the CRCW PRAM model, there is an algorithm that solves approximate compaction in O((log log n)/sup 3/) time for lambda =/sup 1///sub loglogn/, using /sup n///sub (loglogn)3/ processors. This is close to the best possible. Specifically, the authors, prove that LAC requires Omega (log log n) time using O(n) processors. They also give a tradeoff between lambda and the processing time. For in <1, and lambda =n/sup in /, the time required is Omega (log/sup 1///sub in /).<>
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