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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

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