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Unformatted text preview: Notes on Complexity Theory Last updated: November, 2011 Lecture 19 Jonathan Katz 1 IP = PSPACE A small modification of the previous protocol gives an interactive proof for any language in PSPACE , and hence PSPACE ⊆ IP . Before showing this, however, we quickly argue that IP ⊆ PSPACE . To see this, fix some proof system ( P , V ) for a language L (actually, we really only care about the verifier algorithm V ). We claim that L ∈ PSPACE . Given an input x ∈ { , 1 } n , we compute exactly (using polynomial space) the maximum probability with which a prover can make V accept. (Although the prover is allowed to be allpowerful, we will see that the optimal strategy can be computed in PSPACE and so it suffices to consider PSPACE provers in general.) Imagine a tree where each node at level i (with the root at level 0) corresponds to some sequence of i messages exchanged between the prover and verifier. This tree has polynomial depth (since V can only run for polynomially many rounds), and each node has at most 2 n c children (for some constant c ), since messages in the protocol have polynomial length. We recursively assign values to each node of this tree in the following way: a leaf node is assigned 0 if the verifier rejects, and 1 if the verifier accepts. The value of an internal node where the prover sends the next message is the maximum over the values of that node’s children. The value of an internal node where the verifier sends the next message is the (weighted) average over the values of that node’s children. The value of the root determines the maximum probability with which a prover can make the verifier accept on the given input x , and this value can be computed in polynomial space. If this value is greater than 2 / 3 then x ∈ L ; if it is less than 1 / 3 then x 6∈ L . 1.1 PSPACE ⊆ IP We now turn to the more interesting direction, namely showing that PSPACE ⊆ IP . We will now work with the PSPACEcomplete language TQBF , which (recall) consists of true quantified boolean formulas of the form: ∀ x 1 ∃ x 2 ··· Q n x n φ ( x 1 ,...,x n ) , where φ is a 3CNF formula. We begin by arithmetizing φ as we did in the case of # P ; recall, if φ has m clauses this results in a degree3 m polynomial Φ such that, for x 1 ,...,x n ∈ { , 1 } , we have Φ( x 1 ,...,x n ) = 1 if φ ( x 1 ,...,x n ) is true, and Φ( x 1 ,...,x n ) = 0 if φ ( x 1 ,...,x n ) is false....
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This note was uploaded on 01/13/2012 for the course CMSC 652 taught by Professor Staff during the Fall '08 term at Maryland.
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