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Unformatted text preview: Solutions 4 Question 1 Find the error in the “proof”. In the fifth sentence of the proof (“Because every ...”) it is argued that as (1) SAT ∈ TIME[ n k ], and (2) every language in NP is reducible in polynomial time to SAT, we can conclude that NP ⊆ TIME[ n k ], that is, that every language L in NP is in TIME[ n k ] (in other words, there is an O ( n k ) algorithm to decide L ). Whilst (1) is correct (if we accept the assumption that P = NP) and (2) is also true, they do not support this conclusion. (1) tells us that there is an O ( n k ) algorithm for SAT. (2) tells us that for any L ∈ NP , for some j , there is an O ( n j ) algorithm to reduce L to SAT. So to decide L , we can combine the reduction to SAT with the algorithm for SAT; the total running time is O ( n j + n k ). If j > k , then all we know is that L ∈ TIME[ n j ]; a weaker conclusion than L ∈ TIME[ n k ]. Question 2 Are there any problems in P that are complete with respect to linear time reductions? No, there cannot be a language that is Pcomplete with respect to linear reductions. We use proof by contradiction....
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 Fall '10
 Veneris
 Algorithms, Data Structures, Computational complexity theory, NJ, vertex cover, Clique, 2k, half clique

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