hw_4_soln - CS 473: Algorithms, Fall 2010 HW 4 (due...

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Unformatted text preview: CS 473: Algorithms, Fall 2010 HW 4 (due Tuesday, September 28th) This homework contains four problems. Read the instructions for submitting homework on the course webpage . In particular, make sure that you write the solutions for the problems on separate sheets of paper; the sheets for each problem should be stapled together. Write your name and netid on each sheet. Collaboration Policy: For this home work, Problems 1-3 can be worked in groups of up to 3 students each. Problem 0 should be answered in Compass as part of the assessment HW4-Online and should be done individually. 1. (30 pts) You are given an array A of n distinct integers, and an integer k such that 1 k n . The square distance between a pair of integers x,y is defined as the quantity ( x- y ) 2 . Your goal is to design an O ( n ) time algorithm to find k elements in A with the smallest square distance to the median (i.e. the element of rank b n/ 2 c in A ). For instance, if A = [9 , 5 ,- 3 , 1 ,- 2] and k = 2, then the median element is 1, and the 2 elements in A with the smallest square distance to the median are { 1 ,- 2 } . If k = 3, then you can output either { 1 ,- 2 ,- 3 } or { 1 ,- 2 , 5 } . Hint: Use the linear time Selection algorithm. Solution : In the lecture notes we are given an O ( n ) time algorithm SELECT ( A,j ), which finds the j th smallest element in the array A . So by calling SELECT ( A, b n/ 2 c ) we get the median of A , call it m . We then can create an array B [1 ...n ], where B [ i ] = ( A [ i ]- m ) 2 . Filling in B can be done by making a single pass through A , doing O (1) work at each step, and hence takes O ( n ) time. We now wish to find the smallest k elements in B , since these are the elements with the smallest squared distance to the median. We now call SELECT ( B,k ), to find the k th smallest element in B . We can then do a linear time pivot operation to determine the indices of all the values less than the k th smallest element in B (i.e. one round of Quicksort, where we store the indices instead of the values). We can then retrieve the corresponding elements from A since the indices in B match the indices of the corresponding values from A ....
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hw_4_soln - CS 473: Algorithms, Fall 2010 HW 4 (due...

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