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cse101_10_17_11

cse101_10_17_11 - Median finding 1 Median of a set of...

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Median finding 1. Median of a set of numbers a. Given n numbers, find their median Ex. 10 1 60 8 15 10 21 7 3 38 72 Simple algorithm: Sort the numbers Return the one in the middle O(n log n) 1, 3, 7, 8, 10, 10 , 15, 21, 38, 60, 72 If even number, hen anything in the middle of the 2 middle numbers can be the median b. We’ll give a fast algorithm for a more general problem Selection – given: a list of n numbersk S integer k, return: the kth smallest number in S Ex: k = 1, minimum element k(ceiling n/2) k = ceiling (0.9n) 2. A divide-and-conquer scheme for selection a. Ex. S = 10, 1, 60, 8, 15, 10, 21, 7, 3, 38, 72 k = 6 (median) b. Pick an element v from the list S. Split S into three lists. This takes linear time. S c : elements less than v S = :the elements equal to v S > : the elements greater than v Ex. V = 7. S < : 1 3 S = : 7 S > : 10, 60, 8, 15, 10, 21, 38, 72 Median must be 3 rd smallest element in S > since the size of the other lists are 2 and 1 and the median is the 6 th slot. c. Now recurse: Select(S,k) = Select(S c , k) if k < abs(S < ) V if abs(S < ) < k < abs(S c ) + abs(S = ) Select (S > , k – abs(S c ) – abs(S = )) if k > abs(S c ) + S = ) Ex. V = 21: S < : 10, 1, 8, 15, 10, 7, 3 Rest don’t’ matter because you know that it is the 6
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