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Published byÏἸάϊρος Κόρακας Modified over 5 years ago
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Adding numbers n data items, p processors ts = O(n)
tp = O(n/p) if data on each proc => S=ts/tp=O(p) tp = O(n + n/p) if data needs broadcasting => S=ts/tp=o(1)
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Sequential Recursion
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Parallel Recursion tcomm = O(n/2 +n/4 + ..+ n/p) = O(n) S=o(1)
tcomp = O(n/2 +n/ n/p) = O(n)
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tcomm = O(1 +1 + ..+ 1) = O(log p)
S=O(n / log p) tcomp = O( ) = O(log p)
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Parallel Bucket Sort
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Sequential m buckets , n numbers ts = O(n + m((n/m) log (n/m))) = O(n log(n/m))
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m buckets , n numbers, p=m processors tp = O(n + (n/p) log (n/p))
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tp = O(n/p + (n/p) log (n/p)) = O( (n/p) log (n/p)) => S=O(p)
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Det. Sample Sort
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Det. Sample Sort sort locally and create p-sample
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Det. Sample Sort send all p-samples to processor 1
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Det. Sample Sort proc.1: sort all received samples and compute global p-sample
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Det. Sample Sort broadcast global p-sample
bucket locally according to global p-sample send bucket i to proc.i resort locally
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Det. Sample Sort Lemma: Each proc. receives at most 2 n/p data items
global sample global sample
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Det. Sample Sort Post-Processing: “Array Balancing” 2 Rounds:
n/p n/p n/p n/p n/p n/p n/p n/p 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 Post-Processing: “Array Balancing” 2 Rounds: Each proc. sends rec. data size to all other proc. Move data to right location via one h-relation
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Det. Sample Sort 5 MPI_AlltoAllv for n/p > p2
O(n/p log n) local comp. Goodrich (FOCS'98): O(1) rounds for n/p > pe
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Performance: Det. Sample Sort
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Numerical Integration
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static assignment of processors
to segments of [a,b] area = d (f(p)+f(q))/2
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Problem: precision depends on curve’s shape
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Adaptive Quadrature Terminate when C is sufficiently small
Problem: different parts of the curve need different resolution
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segment 1 segment 3 segment 4 segment 2 segment 5
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Gravitational N-Body Problem
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ts = O(n2) per time step
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for each time step: for each object: traverse tree to determine its forces Problem: traversals have different lengths
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object 1 object 3 object 5 object 2 object 4
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