A parallel time/hardware tradeoff T ter dot H = O(2 sup n/2 ) for knapsack problem

Journal Article · Fri Feb 01 04:00:00 EST 1991 · IEEE Transactions on Computers (Institute of Electrical and Electronics Engineers); (United States)

DOI:https://doi.org/10.1109/12.73593· OSTI ID:6097938

Ferreira, A G ^[1]

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1. Lab. l'Informatique du Parallelisme-IMAG, Ecole Normale Superieure de Lyon, 69364 Lyon Cedex 07 (FR)

In this paper, the authors propose a parallel algorithm that solves a knapsack problem of size {ital n} in time {ital T} = {ital O}({ital n} * (2{sup {ital n}/2}){sup {epsilon}}) when {ital P} = {ital O}(2{sup {ital n}/2}){sup (1{minus}{epsilon}}), 0 {le} {epsilon} {le} 1, processors are available. The algorithm needs {ital S} = {ital O}(2{sup {ital n}/2}) memory space in a shared memory. Let {ital H} (for hardware) be the number of processors plus the number of memory cells used by a parallel algorithm. The parallel algorithm that the authors describe here takes a linear time proportional to ({ital n}/2) to find a solution when {ital P} = {ital O}(2{sup {ital n}/2}), leading to a tradeoff {ital T} {center dot} {ital H} = {ital O}(2{sup {ital n}/2}).

• OSTI ID: 6097938
• Journal Information: IEEE Transactions on Computers (Institute of Electrical and Electronics Engineers); (United States), Journal Name: IEEE Transactions on Computers (Institute of Electrical and Electronics Engineers); (United States) Vol. 40:2; ISSN ITCOB; ISSN 0018-9340
• Country of Publication: United States
• Language: English

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