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

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CS 151 Complexity Theory Spring 2011 Solution Set 7 Posted: May 26 Chris Umans Obviously, if you have not yet turned in Problem Set 7, you shouldn’t consult these solutions. 1. (a) We describe R separately for strings x of each length. Consider strings x of length m and assume | z | = | 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, y, ˆ z ) for which ( x, y, E 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 there exists y for which |{ ˆ z : ( x, y, ˆ z ) ̸∈ b R }| ≤ 2 n 1 / 2 . To prove this, take y to be the y for which Pr z [( x, y, z ) R ] 2 / 3 (guaranteed by the definition), and call a ˆ z in the above set “bad.” For ˆ z to be bad, it must be that | Pr z [( x, y, z ) R ] Pr w [( x, y, E z, w )) R ] | > 1 / 6 , (since the left probability is at least 2/3, and the right one must be less than 1/2 for bad ˆ z ). Thus there must be fewer than 2 k = 2 n 1 / 2 bad ˆ z (because the set of bad ˆ z comprise a source with minentropy k on which the extractor fails). If x ̸∈ L , then for all y |{ ˆ z : ( x, y, ˆ z ) b R }| ≤ 2 n 1 / 2 . To prove this, fix a y and call a ˆ z in the above set “bad.” For ˆ z to be bad, it must be that | Pr z [( x, y, z ) R ] Pr w [( x, y, E z, w )) R ] | > 1 / 6 , (since the left probability is at most 1/3, and the right one must be at least 1/2 for bad ˆ z ). Thus there must be fewer than 2 k = 2 n 1 / 2 bad ˆ z for the same reason as above. Now we can define R . The idea is to split ˆ z into two equal-length halves: ˆ z = (ˆ z 1 , ˆ z 2 ). Then we define R to be those ( x, y = ( y, ˆ z 1 ) , z = ˆ z 2 ) for which ( x, y, ˆ z ) b R . Let’s check that this satisfies the requirements. If x L , then there exists a y and a ˆ z 1 for which for all ˆ z 2 , ( x, y, ˆ z ) b R (if not, then there would be at least 2 n/ 2 > 2 n 1 / 2 distinct ˆ z for which ( x, y, ˆ z ) ̸∈ b R , contradicting out analysis above). And, if x ̸∈ L , then we claim that for all y and all ˆ z 1 , Pr ˆ z 2 [( x, y, ˆ z ) b R ] < 1 / 3. If not, then for some y there would be at least (2 / 3)2 n/ 2 > 2 n 1 / 2 distinct ˆ z for which ( x, y, ˆ z ) b R , contradicting out analysis above. 7-1
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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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