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Unformatted text preview: CS5314 Randomized Algorithms Homework 2 (Suggested Solution) 1. A fixed point of a permutation : [1 ,n ] [1 ,n ] is a value for which ( x ) = x . Find the variance in the number of fixed points of a permutation chosen uniformly at random from all permutations. Hint: Let X i be an indicator such that X i = 1 if ( i ) = i . Then, n i =1 X i is the number of fixed points. You cannot use linearity to find Var [ n i =1 X i ], but you can calculate it directly. Ans. For each i , we have E [ X i ] = Pr( X i = 1) = 1 /n and E [ X 2 i ] = Pr( X 2 i = 1) = 1 /n. For any i,j with i 6 = j , we have E [ X i X j ] = Pr( X i X j = 1) = Pr( X i = 1 X j = 1) = Pr( X i = 1)Pr( X j = 1 | X i = 1) = 1 n ( n- 1) . Let X be the number of fixed points. So, X = n i =1 X i , and E [ X ] = n i =1 E [ X i ] = 1. Then, we have Var[ X ] = E [ X 2 ]- ( E [ X ]) 2 = E n X i =1 X i ! 2 - 1 = n X i =1 E [ X 2 i ] + X i 6 = j E [ X i X j ]- 1 = n 1 n + n ( n- 1) 1 n ( n- 1)- 1 = 1 . 2. Generalize the median-finding algorithm to find the k th largest item in a set of n items for any given value of k . Prove that your resulting algorithm is correct, and bound its running time. Ans. We first consider the case where k n/ 2. For this case, we scan the n items and obtain the minimum value, say, t . Then, we add n- 2 k + 1 items to the set of n items, each item having a value equal to t . It is easy to check that the k th largest item among the original n items will be the median of the new set of 2 n- 2 k + 1 items. Therefore, we can apply the median-finding algorithm for the new set to obtain the desired k th largest item of the original set. The running time is O ( n ). For the case where k > n/ 2, we proceed by adding items with maximum value instead. The running time for this case is also O ( n ). 3. The weak law of large numbers state that, if X 1 ,X 2 ,X 3 ,... are independent and identically distributed random variables with finite mean and finite standard deviation , then for any constant > 0 we have lim n Pr fl fl fl fl X 1 + X 2 + + X n n- fl fl fl fl > = 0 ....
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- Spring '08