sol3 - Massachusetts Institute of Technology...

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Massachusetts Institute of Technology Handout 8 6.854J/18.415J: Advanced Algorithms Wednesday, October 7, 2009 David Karger Problem Set 3 Solutions Wednesday, October 7, 2009 Problem 1. (a) Consider the ( k + 1) st item inserted. Since only k buckets (at worst) are occupied, the probability that both candidate locations are occupied is only ( k/n 1 / 5 ) 2 . Thus, the expected number of times an item is actually inserted into an already-occupied bucket is at most n 1 s k =0 ( k/n 1 . 5 ) 2 = ( n 1)( n )(2 n 1) 6 n 3 1 / 3 Now let’s consider pairwise collisions. Item k collides with item j < k only if (i) one of the candidate locations of item k is the location of item j (this has probability at most 2 /n 1 . 5 ) and (ii) the other candidate location for item k contains at least one element (probability k/n 1 . 5 ). Thus, the probability k collides with j is at most k/n 3 . Summing over the k possible values of j < k , we Fnd the expected number of collisions for item k is at most k 2 /n 3 . Summing over all k , we get the same result as above: O (1) expected collisions. (b) Start with a 2-universal family of hash functions mapping n items to 2 n 1 . 5 lo- cations. Consider any particular set of n items. Consider choosing a random function from the hash family. The probability that item k collides with item j is 1 / 2 n 1 . 5 by pairwise independence, implying by the union bound that the probability k collides with any item is at most 1 / 2 n . Now suppose that we allocate two arrays of size 2 n 1 . 5 and choose a random 2- universal hash function from the family independently for each array. If an item has no collision in either array, then it will be placed in an empty bucket by the bash function. We need merely analyze the probability that this happens for every item (this would make the bash function perfect). The probability that item k has a collision in both arrays is at most (1 / 2 n ) 2 = 1 / 4 n . It follows that the expected number of items colliding with some other item is at most 1 / 4. This implies in turn that with probability 3/4, every item is placed in an empty bucket by the (perfect) bash function. This in turn implies that some pair of 2-universal hash functions deFnes a perfect bash for our set of n items. Since every set of items gets a perfect bash from this scheme, it follows that the family of pairs of 2-universal functions above is a perfect bash family. Since the 2-universal family has size polynomial in the universe, so does the family of pairs of 2-universal functions.
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2 Handout 8: Problem Set 3 Solutions (c) When we sample a hash function from the above 2-universal family, we get a 3 / 4 probability of having no collisions. It follows that if we make 2 or more attempts, we can expect to Fnd a collision-free hash function.
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sol3 - Massachusetts Institute of Technology...

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