hwsol4.3 - CSci 5403, Spring 2010 Homework 4, Problem 3...

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CSci 5403, Spring 2010 Homework 4, Problem 3 Fedor Korsakov Even Better Amplification? In class we proved that for every ± = Ω( n - c ), BPP ± = BPP , where L BPP ± if there exists a polynomial-time TM M such th at Pr r [ M ( x,r ) = L ( x )] 1 2 + ± . Define the class PP = BPP 0 , i.e. the class of languages L such that there is a polytime TM M such that x L Pr r [ M ( x,r ) accepts] 1 2 and x 6∈ L Pr r [ M ( x,r ) accepts] < 1 2 . (a) Prove that NP PP . (In contrast, we have good evidence that BPP NP ) (b) Define PP ± to be the class of languages L such that for some polytime TM M , x L Pr r [ M ( x,r ) accepts] > ± . Prove that PP ± = PP . Use this to prove that PP 6 = NP unless PH = NP . Hint: Consider the implications for coPP . Solution. (a) We can prove that NP PP by demonstrating that SAT PP . Consider a boolean CNF formula C with n variables x i ,i [1 ,n ]. C SAT Pr x ( C ( x ) = 1) > 0 Pr x ( C ( x ) = 1) 2 - n . Introduce n + 1 new variables y i ,i = 1 , ··· ,n + 1. Define a new formula C 0 = y n +1 ( y 1 y 2 ∨ ··· ∨ y n ). Then C 0 ( y ) = 1 iff y n +1 = 1 and at least one of y 1 , ··· ,y n is 1. Thus, Pr y ( C 0 ( y ) = 1) = 1 2 (1 - 2 - n ) = 1 2 - 2 - n - 1 . Denote δ = Pr y
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This note was uploaded on 10/21/2011 for the course CSCI 5403 taught by Professor Sturtivant,c during the Spring '08 term at Minnesota.

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hwsol4.3 - CSci 5403, Spring 2010 Homework 4, Problem 3...

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