1985
DOI: 10.1109/tc.1985.6312202
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The power of parallel prefix

Abstract: The prefix computation problem is to compute all η initial products «ι ° · · · ° α" ι = 1, · · ·, n, of a set of η elements, where ° is an associative operation. We present an 0(((log n)/\og(2n/p)) · (n/p)) time deterministic parallel algorithm using ρ < η processors to solve the prefix computation problem, when the order of the elements is specified by a linked list. For ρ ^ 0(n l~€ ) (c > 0 any constant), this algorithm achieves linear speedup. Such optimal speedup was previously achieved only by probabilist… Show more

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Cited by 206 publications
(63 citation statements)
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“…Hence, we assume that the intervals in F have been labelled in such a way that b i < b j if and only if i < j. Such a labelling is easily obtained from L in O(log n) time using n/ log n processors by parallel prefixing [14,16]. Let s (respectively, t) be the smallest left endpoint (respectively, largest right endpoint) among those in L. Then, s = a l for some l ∈ {1, 2, .…”
Section: Preliminariesmentioning
confidence: 99%
“…Hence, we assume that the intervals in F have been labelled in such a way that b i < b j if and only if i < j. Such a labelling is easily obtained from L in O(log n) time using n/ log n processors by parallel prefixing [14,16]. Let s (respectively, t) be the smallest left endpoint (respectively, largest right endpoint) among those in L. Then, s = a l for some l ∈ {1, 2, .…”
Section: Preliminariesmentioning
confidence: 99%
“…, x k of numbers stored in an array, compute all partial sums r 1 x i . It can be solved with k/log(k) processors (CREW), essentially by covering the array with a balanced tree structure and calculating partial sums of subintervals covered by nodes of the tree [19]. The runtime of log(k) achieved is more or less optimal [23].…”
Section: Is An Array Of Processors Some Of Them Marked In a Crcw Mamentioning
confidence: 99%
“…We call the leaves thus sampled the marked leaves of T . 19 The marked leaves form a subsequence such that between any two leaves in the sequence (in inorder) there are O(k) nodes of T . The span of an internal node v is its set of marked leaf descendants: this is easily calculated from the indexes i and j of the leftmost and rightmost entries in A covering the descendants of v.…”
Section: 10mentioning
confidence: 99%
“…Steps (2) and (3) are performed by using [5] and parallel prefix [16,17]. Steps (4) and (5) are easily handled by using parallel prefix [16,17].…”
Section: For Every Vertexmentioning
confidence: 99%