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An optimal parallel algorithm for the maximal element problem (abstract)

机译:最大元素问题(抽象)的最佳并行算法

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摘要

Given a set (p1, p2, …, pn) of n points in a plane, a point pi dominates another point pj iff X (pi) ≥ X(pj) and Y (pi) ≥ Y(pj) where X (p) and Y (p) denote the coordinates of a point p. Our problem is to find the points which are dominated by no other point. These points are called maximal elements. In this abstract we suggest an &Ogr; (log n) parallel algorithm with &Ogr; (n) processors in the EREW parallel model which is weaker than CREW. (Note that most parallel algorithms in the literature use the power of CREW PRAM model.) The main idea is to reduce the problem to the parallel prefix problem [2]. Our algorithm runs as follows: First, from P = {pl, …, pn}, find a subset P' = {p1 …, pm }, where each point pi is dominated by no other point having the same X -coordinate aspi. This can be done in &Ogr; (log n) time with &Ogr; (n) processors in the EREW model applying the parallel sort algorithm by [1]. Next sort P' according to the X -coordinate of each point and store each Y -coordinate in an array A in the increasing order of the corresponding X -coordinate. Assume that the Y -coordinate of pi is stored in A [i ] for i = 1, 2, …, m. Then for each A [i ], find an index j such that ij and A [j] ≥ A [k ] for each ikm. If i = j, then pi is one ofthe maximal elements. This can be done in &Ogr; (log n) time with &Ogr; (n) processors on the EREW parallel model by applying a parallel prefix algorithm |2|.

机译:

给定平面中 n 个点的集合( p1,p2,...,pn ),一个点 pi占主导地位另一个点< ITALIC> pj 且X( pi )≥X( pj )和Y( pi )≥Y( pj ),其中 X p )和 Y p )表示点的坐标< ITALIC> p 。我们的问题是要找到由其他点主导的点。这些点称为最大元素。在此摘要中,我们建议使用&Ogr; (log n)&Ogr;的并行算法(n)EREW并行模型中的处理器,比CREW弱。 (请注意,文献中大多数并行算法都使用CREW PRAM模型的功能。)主要思想是将问题简化为并行前缀问题[2]。我们的算法运行如下:首先,从 P = { pl,…,pn },找到子集 P '= { p1…,pm },其中每个点 pi 都没有其他点具有与 pi 相同的X坐标。这可以在&Ogr;中完成。 (log n)时间与&Ogr; (n)EREW模型中的处理器使用[1]的并行排序算法。接下来,根据每个点的X坐标对 P '进行排序,并将每个Y坐标以对应X坐标的升序存储在数组A中。假设 pi 的Y坐标存储在A [ i ]中,其中 i = 1,2,…, m < / ITALIC>。然后对于每个A [ i ],找到一个索引 j ,使 i j 和A [ j ]≥每个 i k m 的[ k ]。如果 i = j ,则 pi 是最大元素之一。这可以在&Ogr;中完成。 (log n)时间与&Ogr; (n)通过应用并行前缀算法| 2 |在EREW并行模型上处理处理器。

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