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soln7

# soln7 - CS 151 Complexity Theory Spring 2011 Solution Set 7...

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7-2 (b) As in part (a), we describe R separately for strings x of each length. Consider strings x of length m and assume | y | = | x | c . Set k = m 3 c and n = k 2 , and let E : { 0 , 1 } n × { 0 , 1 } t { 0 , 1 } m c be a ( k, ϵ ) extractor with ϵ < 1 / 6 and t = O (log n ). Define the language b R to be those triples ( x, ˆ y, ( z w ) w ∈{ 0 , 1 } t ) for which ( x, E y, w ) , z w ) R for more than half of the w ∈ { 0 , 1 } t . Since R is in P and t = O (log n ), b R is also in P . We now claim that If x L , then we claim |{ ˆ y |∀ ( z w ) w ∈{ 0 , 1 } t ( x, ˆ y, ( z w ) w ∈{ 0 , 1 } t ) ̸∈ b R }| ≤ 2 n 1 / 2 . Call a ˆ y in the above set “bad.” For ˆ y to be bad, it must be that | Pr y [ z ( x, y, z ) R ] Pr w [ z ( x, E y, w ) , z ) R ] | > 1 / 6 , (since the left probability is at least 2/3, and the right one must by less than 1/2 for bad ˆ y ). Thus there must be fewer than 2 k = 2 n 1 / 2 bad ˆ y
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