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Unformatted text preview: Notes on Complexity Theory Last updated: October, 2011 Lecture 18 Jonathan Katz 1 The Power of IP We have seen a (surprising!) interactive proof for graph nonisomorphism. This begs the question: how powerful is IP ? 1.1 co NP ⊆ IP As a “warmup” we show that co NP ⊆ IP . We have seen last time that co NP is unlikely to have a constantround interactive proof system (since this would imply 1 that the polynomial hierarchy collapses). For this reason it was conjectured at one point that IP was not “too much more powerful” than NP . Here, however, we show this intuition wrong: any language in co NP has a proof system using a linear number of rounds. We begin by arithmetizing a 3CNF formula φ to obtain a polynomial expression that evaluates to 0 iff φ has no satisfying assignments. (This powerful technique, by which a “combinatorial” statement about satisfiability of a formula is mapped to an algebraic statement about a polynomial, will come up again later in the course.) We then show how to give an interactive proof demonstrating that the expression indeed evaluates to 0. To arithmetize φ , the prover and verifier proceed as follows: identify 0 with “false” and positive integers with “true.” The literal x i becomes the variable x i , and the literal ¯ x i becomes (1 x i ). We replace “ ∧ ” by multiplication, and “ ∨ ” by addition. Let Φ denote the polynomial that results from this arithmetization; note that this is an nvariate polynomial in the variables x 1 ,...,x n , whose total degree is at most the number of clauses in φ . Now consider what happens when the { x i } are assigned boolean values: all literals take the value 1 if they evaluate to “true,” and 0 if they evaluate to “false.” Any clause (which is a disjunction of literals) takes a positive value iff at least one of its literals is true; thus, a clause takes a positive value iff it evaluates to “true.” Finally, note that Φ itself (which is a conjunction of clauses) takes on a positive value iff all of its constituent clauses are positive. We can summarize this as: Φ( x 1 ,...,x n ) > 0 if φ ( x 1 ,...,x n ) = true , and Φ( x 1 ,...,x n ) = 0 if φ ( x 1 ,...,x n ) = false . Summing over all possible (boolean) settings to the variables, we see that φ ∈ SAT ⇔ X x 1 ∈{ , 1 } ··· X x n ∈{ , 1 } Φ( x 1 ,...,x n ) = 0 ....
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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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