Advanced Computational Complexity-Lecture 19 Slides-Computer Science, Slides of Advanced Computational Complexity

Download Advanced Computational Complexity-Lecture 19 Slides-Computer Science and more Slides Advanced Computational Complexity in PDF only on Docsity! Lecture 19: More on Randomized Complexity Matthew Johnson matthew.johnson2@dur.ac.uk Advanced Computational Complexity More on Randomized Complexity Revision: Polynomial Hierarchy Definition Σi is the class of all languages reducible to deciding the satisfiability of a formula of type ∃x1∀x2∃x3 · · ·R(x1, x2, x3, . . .) where there are i alternating quantifiers Advanced Computational Complexity More on Randomized Complexity A useful result We will use the following (without proof): Lemma If L is in BPP then there is a polynomial probabilistic Turing Machine M such that x ∈ L ⇒ Prob[M(x) = 1] ≥ 1− 1 3m x /∈ L ⇒ Prob[M(x) = 1] < 1 3m where m is number of random bits used in the computation M(x). Advanced Computational Complexity More on Randomized Complexity BPP Theorem BPP ⊆ Σ2. Proof: We must show that, for any L ∈ BPP there is a problem in Σ2 that is equivalent to deciding L. We use the following: x ∈ L if and only if ∃y1, . . . , ym∀z (M(x , y1 ⊕ z) ∨ · · · ∨M(x , y1 ⊕ z)) where — each yi and z are strings of zeros and ones of length m; — M(x , r) denotes the output of M on instance x with guess input r ; — ⊕ denotes addition modulo 2. Advanced Computational Complexity More on Randomized Complexity Proof continued This says that given an instance x ∈ L, we can find a set of m strings y1, . . . , ym, such that given any string z, M will output 1 at least once if run m times with each yi ⊕ z used once as the guess input (and that if x /∈ L we cannot find these strings). We will not show how to find these m strings, but we will show that they must exist. We will show that given a random choice of y1, . . . , ym the probability that it does not have the required property is less than 1 (if the strings did not exist the probability would be exactly 1). Advanced Computational Complexity More on Randomized Complexity Proof continued The probability that a random choice of y1, . . . , ym is “bad” is the probability that there exists a string z such that M outputs 0 for every guess input yi ⊕ z, 1 ≤ i ≤ m. Let us fix a choice of y1, . . . , ym. For a particular z and yi , the probability that M outputs 0 with guess input yi ⊕ z is 1/3m (from the Lemma). For a particular z, the probability that M outputs 0 for every guess input yi ⊕ z is 1(3m)m . The probability that there is not at least one choice of z such that M outputs 0 for every guess input yi ⊕ z is 2m 1(3m)m . This is less than 1. Advanced Computational Complexity More on Randomized Complexity Proof continued The probability that a random choice of y1, . . . , ym is “bad” is the probability that there exists a string z such that M outputs 0 for every guess input yi ⊕ z, 1 ≤ i ≤ m. Let us fix a choice of y1, . . . , ym. For a particular z and yi , the probability that M outputs 0 with guess input yi ⊕ z is 1/3m (from the Lemma). For a particular z, the probability that M outputs 0 for every guess input yi ⊕ z is 1(3m)m . The probability that there is not at least one choice of z such that M outputs 0 for every guess input yi ⊕ z is 2m 1(3m)m . This is less than 1. Advanced Computational Complexity More on Randomized Complexity Proof continued The probability that a random choice of y1, . . . , ym is “bad” is the probability that there exists a string z such that M outputs 0 for every guess input yi ⊕ z, 1 ≤ i ≤ m. Let us fix a choice of y1, . . . , ym. For a particular z and yi , the probability that M outputs 0 with guess input yi ⊕ z is 1/3m (from the Lemma). For a particular z, the probability that M outputs 0 for every guess input yi ⊕ z is 1(3m)m . The probability that there is not at least one choice of z such that M outputs 0 for every guess input yi ⊕ z is 2m 1(3m)m . This is less than 1. Advanced Computational Complexity More on Randomized Complexity Proof continued Now we must prove that if x /∈ L, then we cannot find such a y1, . . . , ym. That is, we must show that for every possible choice, there exists a string z such that M returns 0 on every guess input yi ⊕ z. We prove that a randomly chosen z does not have this property with probability less than 1. So some z does have this property. Advanced Computational Complexity More on Randomized Complexity Proof continued Let us fix a choice of y1, . . . , ym and of z. The probability that M outputs 1 with guess input yi ⊕ z is 1/3m (from the Lemma). The probability that M outputs 1 on at least one guess input yi ⊕ z is at most m(1/3m) = 1/3. So there must exist some z such that M outputs 0 for every guess input yi ⊕ z. Advanced Computational Complexity More on Randomized Complexity Proof continued Let us fix a choice of y1, . . . , ym and of z. The probability that M outputs 1 with guess input yi ⊕ z is 1/3m (from the Lemma). The probability that M outputs 1 on at least one guess input yi ⊕ z is at most m(1/3m) = 1/3. So there must exist some z such that M outputs 0 for every guess input yi ⊕ z. Advanced Computational Complexity More on Randomized Complexity