lecture12

# lecture12 - COMMUNICATION COMPLEXITY CSci 5403 COMPLEXITY...

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1 COMPLEXITY THEORY CSci 5403 LECTURE XII: CAUTION MAY BE UNPREDICTABLE x 2 {0,1} n y 2 {0,1} n COMMUNICATION COMPLEXITY X = Y ? How many bits do Alice and Bob need to send? Deterministic worst case: |X| = n Pick random prime r 2 {p 1 ,…,p 2n } r, X mod r X=Y mod r O(log n) bits x=y: Then Pr r [x=y mod r] = 1. x y: imagine x=y mod r i , for {r 1 ,…r m } ½ {p 1 ,…p 2n } Two cases for x vs y: Claim. m < n. Proof. Suppose not. Then Π i r i > 2 n , and by Chinese Remainder Theorem x=y. Thus Pr r [x=y mod r] ½ . We can amplify the certainty by iterating k times: If we find r such that x y mod r, then x y Else, say x=y. Pr r1…rk [x y | Alice says x=y] · 2 -k CHECKING LINEAR ALGEBRA Alice claims that A £ B = C mod p. Can we check? Best deterministic algorithms: Θ (n 2.38 ) time Pick random r 2 {0,1} n . Check A(Br) = Cr mod p. Claim. Pr r [A(Br)=Cr mod p | AB C mod p] · ½ . Proof. Let AB C. Then D = AB-C 0. Fix r 1 …r n-1 and wlog let D i,n 0. Then: Pr[ Σ j<n D i,j r j = r n D i,n mod p] · ½ . Time for check: O(n 2 ).

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2 RP Definition. A 2 RP if there is a polynomial time NTM such that: 1. M accepts A 2. If x 2 A, at least ½ the paths accept Alternatively: x 2 A ) At least half of the paths of M(x) accept x A ) None of the paths for M(x) accept
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lecture12 - COMMUNICATION COMPLEXITY CSci 5403 COMPLEXITY...

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