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Unformatted text preview: CS 151 Complexity Theory Spring 2011 Solution Set 4 Posted: April 28 Chris Umans If you are turning in this problem set late, obviously you shouldn’t consult these solutions. 1. Consider a language L ∈ ] ZPP decided by a machine M that runs in expected time n k for some constant k . By Markov’s inequality, we see that Pr y [# steps before M ( x,y ) halts ≥ 2  x  k ] ≤ 1 / 2 . We build a new machine M ′ that simulates M and after 2  x  k simulation steps, it outputs “fail”. The new machine certainly runs in polynomial time, with probability at most 1 / 2 it outputs “fail,” and otherwise it accepts or rejects x correctly. Thus L ∈ ZPP . In the other direction, consider a language L ∈ ZPP with associated probabilistic TM M that runs in time n k and outputs “fail” with probability 1 / 2 and otherwise it accepts or rejects its input correctly. We will simulate M repeatedly until it does not output “fail” (and then accept or reject the input based on M ’s answer). The expected number or repetitions is ∞ ∑ i =1 2 − i i = 1 / 2 (1 − (1 / 2)) 2 = 2 , and so the expected running time is 2 n k . Thus L ∈ ] ZPP . We conclude that ] ZPP = ZPP as required. 2. (a) Since S is nonempty it contains some j ∈ { 1 , 2 , 3 ,...,ℓ } . Imagine choosing v j last. We have u S = α iff v j = α − ∑ i ∈ ( S −{ j } ) v i . This happens with probability exactly 2 − k . Now consider S ̸ = T , and WLOG let j be an element of S that is not in T , and let j ′ be an element of T (which is nonempty). Imagine choosing all the v i except v j and v ′ j . There is a unique pair of values for v j and v ′ j that make u S = α and u T = β , and so Pr[ u S = α ∧ u T = β ] = 2 − 2 k as required. (b) As suggested, we define the indicator random variables X S for the event R u S + e i = C u S + e i and X = ∑ S ̸ = ∅ X S . Note that E[ X S ] ≥ 1 / 2 + ϵ and Var[ X S ] ≤ 1 / 4. Since the { U S } are pairwiseindependent, the { X S } are as well. Thus Var[ X ] ≤ (2 ℓ − 1) / 4 (since the variance of the sum is the sum of the variances for pairwise independent random variables), and E[ X ] ≥ (1 / 2 + ϵ )(2 ℓ − 1) (by linearity of expectation). Now we have: Pr [ { S ̸ = ∅ : C u S + R u S + e i = m i } ≤ 2 ℓ − 1 2 ] = Pr [ X ≤ 2 ℓ − 1 2 ] 41 42 ≤ Pr[  X − E[ X ]  ≥ ϵ (2 ℓ − 1)] ≤ Var[ X ] ϵ 2 (2 ℓ − 1) 2 ≤ 1 4 ϵ 2 (2 ℓ − 1) where the secondtolast line invokes Chebyshev’s Inequality.where the secondtolast line invokes Chebyshev’s Inequality....
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