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PairwiseInd1

# PairwiseInd1 - Pairwiee Independence and Derandemizatien...

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Unformatted text preview: Pairwiee Independence and Derandemizatien Michael Lubyr 3.: Avi Wigdersen Feundetiuns and Trends in Thenre‘tical Computer Science EDGE 0 Introduction to Pairwise Independence o Generating Pairwise Independent Bits 0 Introduction to l Derandomization o Brute Force Derandomization o Derandomization via Pairwise Independence Pairwise Independent Events 0 Consider the set of events A1, An 0 The events A1, ...,A,, are mutually independent if Pr[A1 0 A2... 0 An] : Pr[A1]Pr[A2]...Pr[A,,] o The events A; and A1- are pairwise independent if PrlAi ﬂ Aj] : Pr[A,-]Pr[AJ-] for all I,j Example of Pairwise Independent Events 0 Consider the set 0 2 {abc, acb, aaa, bac, bca, bbb, cab, cba, ccc} and probability of any of them being picked is 5 e Let event Ak denote that the kth letter is a. Clearly Pr[Ak] : § 0 Pr[A1 0 A2] = Pr[A2 0 A3] = MM; 0 A1] = % o Pr[A1 0 A2 0 A3] = :15 75 Pr[A1]Pr[A2]Pr[A3] 0 Thus, the events pairwise independent but not mutually independent Pairwise Independence is weaker than Mutual Independence Generating Pairwise Independent Bits 0 We know how to generate mutually independent bits (Coin Flips) 0 Can we use mutually independent bits to generate pairwise independent bits? Construction of n pairwise independent random bits with uniform distribution from a small sample space 0 For given n, let t = Iog(n+ 1) and let I = {1,..., t}. 0 Let random bits b1, ..., b, be obtained by tosses of a fair coin. 0 Let J1, ..., J,, be pairwise distinct nonempty subsets of I. o For‘i = 1, ..., n let r,~ : €919.13}- . Observe that there are 2‘ - 1 pairwise distinct nonempty subsets of I Generating Pairwise Independent Bits Lemma: Verification of the construction o The random bits r1, ..., I), obtained by the construction above are pairwise independent and each bit is uniformly distributed. 0 Fix any nonempty set J g {1, ..., t} and let r = @jejbj o For any ﬁxed index 5 6 J we have, r = ejerj = (\$jeA{s}l),-)\bs o If the bj's have already been determined for all j e J. then depending on the value of b5, the value of r will either agree with or will differ from @jéerj 0 But b, attains its two possible values with probability; -— each and the b are mutually independent hence r will be uniformly2 distributed In {0 1} Generating Pairwise Independent Bits Proof(continued) 0 Consider any two set S, T(S ;£ T) g [t]. We have: R5 = Rsm‘ 69 R\$\T RT = RSnT 6 RT\\$ o R5\T, RSnT and RT\5 are independent as they depend on disjoint subsets of the bi's 0 At least two of these subsets are nonempty (because 5 gé T). This implies that (R5, R7) each take value in {0, 1} with probability %. o The random bits generated are pairwise independent bits Brute Force Derandomization o Brute Force Derandomization refers to simulating a randomized algorithm for every possible value of its random source. 0 Consider a randomized algorithm that runs for poly(n) steps on all inputs of size n. o The algorithm might uses r random bits o Once the random bits are known, each execution can be seen as a deterministic algorithm 0 There are 2’ possible combination of random bits 0 Algorithm: Execute randomized algorithm for each possible combination 0 Time Complexity of deterministic algorithm 2'poly(n) steps 0 Done? No! Brute Force Derandomization c We want efﬁcient deterministic algorithm 0 2’poly(n) must be 0(n‘) for some constant c o :> r : 0(Iogn) 0 Thus, when r = 0(Iogn), then even brute force derandomization works well 0 First transform a randomized algorithm into one that uses only few random bits, then apply brute force derandomization ...
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