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Unformatted text preview: Notes on Complexity Theory Last updated: October, 2011 Lecture 17 Jonathan Katz 1 Graph Non-Isomorphism is in AM The proof system we showed earlier for graph non-isomorphism relied on the fact that the verifiers coins are kept hidden from the prover. Is this inherent? Somewhat surprisingly, we now show a public-coin proof for graph non-isomorphism. Before doing so, we take a brief detour to discuss pairwise-independent hash functions (which are useful in many other contexts as well). 1.1 Pairwise-Independent Hash Functions Fix some domain D and range R . Let H = { h k } k K be a family of functions, where each k K defines a function h k : D R . We say that H is 1 pairwise independent family if for all distinct x,x D and all (not necessarily distinct) y,y R we have Pr k K h k ( x ) = y V h k ( x ) = y / = 1 / | R | 2 . Put differently, let D = { x 1 ,...,x } and consider the random variables Y i = h K ( x i ) (where K is uniform). If H is pairwise independent then each Y i is uniformly distributed, and moreover the random variables Y 1 ,...,Y are pairwise independent; i.e., for any i 6 = j the random variables Y i and Y j are independent. We show a simple construction of a pairwise-independent family for D = R = F , where F is any finite field. Setting F = GF (2 n ), and viewing strings of length n as field elements, we obtain a construction with D = R = { , 1 } n . By truncating the output, we obtain a construction with D = { , 1 } n and R = { , 1 } for any n . By padding the input with 0s, we obtain a construction for any n . Fix D = R = F and let H = { h a,b } a,b F where h a,b ( x ) = ax + b . We claim that H is pairwise independent. Indeed, fix any distinct x,x F and any y,y F , and consider the probability (over choice of a,b ) that y = ax + b y = ax +...
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lecture17 - Notes on Complexity Theory Last updated:...

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