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Unformatted text preview: {1,… , 3t} such that for i=1,… , t? 2 4 b a b j < < ∑ ∈ = i S j j b a 3Partition is strongly NPhard. tb a t j j = ∑ = 3 1 Proof of NPHardness ■ A scheduling problem is NPhard in the ordinary sense if ➔ partition (or a similar problem) can be reduced to this problem with a polynomial time algorithm and ➔ there is an algorithm with pseudo polynomial time complexity that solves the scheduling problem. ■ A scheduling problem is strongly NPhard if ➔ 3partition (or a similar problem) can be reduced to this problem with a polynomial time algorithm....
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 Spring '10
 JohnFowler
 Computational complexity theory, polynomial time algorithm, CEI University Dortmund, positive integers a1, University Dortmund

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