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Unformatted text preview: CS 151 Complexity Theory Spring 2011 Final Solutions Posted: June 3 Chris Umans 1. (a) The procedure that traverses a fanin 2 depth O (log i n ) circuit and outputs a formula runs in L i – this can be done by a recursive depthfirst traversal, which only requires 1 bit of information (“left” or ”right”) at each level of recursion. The procedure for FVAL (Lecture 2) runs in logspace, so on a formula of size 2 O (log i n ) , it runs in O (log i n ) space. Using spaceeﬃcient composition of the logspace procedure that generates the circuit together with these two procedures we obtain a procedure to evaluate an NC i circuit on a given input in only O (log i n ) space, as required. (b) The configuration graph for an NL i machine on input x of length n has at most 2 O (log i n ) nodes. The input x is accepted if and only if there is a path from the start node s to the accept node t in this graph. We can construct the incidence matrix A of this graph (with ones on the diagonal), and we observe that A ∗ = A 2 m , for m = O (log i n ) has a one in position s,t if and only if there is a path of length at most 2 m from s to t (here we are using Boolean matrix multiplication). We can square matrix A with a O (log  A  ) = O (log i n ) depth circuit. We repeat this squaring m times, to compute A ∗ . The repeated squaring entails m sequential copies of the squaring circuit, which has depth O (log i n ). The total depth is O (log 2 i n ). (c) Suppose we show NL i ⊆ NC 2 i for some i > 1. Then we have L i ⊆ NL i ⊆ NC 2 i ⊆ P . However, we know by the Space Hierarchy Theorem that L is strictly contained in L i for i > 1. Thus we would have proved L ̸ = P . In fact, we would have proved something stronger: that NC 1 ̸ = NC 2 , since an equality would collapse all of the hierarchy to NC 1 , including NC 2 i (and then we would have NC 1 = L = L i = NL i = NC 2 i , contradicting the Space Hierarchy Theorem). 2. (a) Fix an x ∈ { , 1 } n and a y ∈ { , 1 } k . Imagine that we have already chosen M . In order to have h M,b ( x ) = y , we must have Mx + b = y or equivalently y − Mx = b . This happens with probability exactly 2 − k since b is chosen uniformly from { , 1 } k . For the second part, we know that x 1 ̸ = x 2 . Thus there must be a position i in which they differ. WLOG, assume ( x 1 ) i = 1 and ( x 2 ) i = 0. Imagine that we have already chosen all of M except for the ith column, and denote by M ′ the matrix M with 0s in the ith column. Let us denote by a ∈ { , 1 } k our choice of the ith column of M . Note that h M,b ( x 1 ) = M ′ x 1 + a + b and h M,b ( x 2 ) = M ′ x 2 + b . Thus we are interested in the probability that a + b = y 1 − M ′ x 1 and b = y 2 − M ′ x 2 . Since each of a and b are chosen uniformly and independently from { , 1 } k , this happens with probability 2 − 2 k . More precisely, there is a 2 − k chance of choosing b equal to the fixed vector y 2 − M ′ x 2 , and then given the choice of...
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 Fall '09

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