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Unformatted text preview: 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 : { , 1 } n × { , 1 } t → { , 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 (ˆ z,w )) ∈ R for more than half of the w ∈ { , 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 (ˆ 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 (ˆ 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 equallength 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. 71 72 (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 : { , 1 } n ×{ , 1 } t → { , 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 ∈{ , 1 } t ) for which ( x,E (ˆ y,w ) ,z w ) ∈ R for more than half of the w ∈ { , 1 } t . Since R is in P and t = O (log n ), b R is also in P . We now claim that....
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 Fall '09

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