soln4 - CS 151 Complexity Theory Spring 2011 Solution Set 4...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 non-empty 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 pairwise-independent, 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 ] 4-1 4-2 ≤ 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 second-to-last line invokes Chebyshev’s Inequality.where the second-to-last line invokes Chebyshev’s Inequality....
View Full Document

This document was uploaded on 01/05/2012.

Page1 / 4

soln4 - CS 151 Complexity Theory Spring 2011 Solution Set 4...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online