lec7 - CS151 Complexity Theory Lecture 7 April 19, 2011...

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CS151 Complexity Theory Lecture 7 April 19, 2011
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April 19, 2011 2 Outline 3 examples of the power of randomness communication complexity polynomial identity testing complexity of finding unique solutions randomized complexity classes
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April 19, 2011 3 1. Communication complexity Goal : compute f(x, y) while communicating as few bits as possible between Alice and Bob count number of bits exchanged (computation free) at each step: one party sends bits that are a function of held input and received bits so far two parties: Alice and Bob function f:{0,1} n x {0,1} n {0,1} Alice holds x {0,1} n ; Bob holds y {0,1} n
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April 19, 2011 4 Communication complexity simple function (equality): EQ(x, y) = 1 iff x = y simple protocol: Alice sends x to Bob (n bits) Bob sends EQ(x, y) to Alice (1 bit) total: n + 1 bits (works for any predicate f)
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April 19, 2011 5 Communication complexity Can we do better? deterministic protocol? probabilistic protocol ? at each step: one party sends bits that are a function of held input and received bits so far and the result of some coin tosses required to output f(x, y) with high probability over all coin tosses
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April 19, 2011 6 Communication complexity Theorem : no deterministic protocol can compute EQ(x, y) while exchanging fewer than n+1 bits. Proof: “input matrix”: X = {0,1} n Y =  {0,1} n f(x,y)
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April 19, 2011 7 Communication complexity assume 1 bit sent at a time, alternating (same proof works in general setting) A sends 1 bit depending only on x: X = {0,1} n Y =  {0,1} n inputs  x c a us ing  A  to  s e nd 1 inputs  x c a us ing  A  to  s e nd 0
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April 19, 2011 8 Communication complexity B sends 1 bit depending only on y and received bit: X = {0,1} n Y =  {0,1} n inputs  y c a us ing  B  to  s e nd 1 inputs  y c a us ing  B  to  s e nd 0
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April 19, 2011 9 Communication complexity at end of protocol involving k bits of communication , matrix is partitioned into at most 2 k combinatorial rectangles bits sent in protocol are the same for every input (x, y) in given rectangle conclude: f(x,y) must be constant on each rectangle
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April 19, 2011 10 Communication complexity any partition into combinatorial rectangles with constant f(x,y) must have 2 n + 1 rectangles protocol that exchanges ≤ n bits can only create 2 n rectangles, so must exchange at least n+1 bits. X = {0,1} n Y =  {0,1} n 1 1 1 1 0 0 Ma trix fo r EQ :
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April 19, 2011 11 Communication complexity protocol for EQ employing randomness? Alice picks random prime p in {1. ..4n 2 }, sends: p (x mod p) Bob sends: (y mod p) players output 1 if and only if: (x mod p) = (y mod p)
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April 19, 2011 12 Communication complexity O(log n) bits exchanged if x = y, always correct if x ≠ y, incorrect if and only if: p divides |x – y| # primes in range is ≥ 2n # primes dividing |x – y| is ≤ n probability incorrect ≤ 1/2 Randomness gives an exponential advantage!!
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lec7 - CS151 Complexity Theory Lecture 7 April 19, 2011...

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