Solution to midterm02 session A

Solution to midterm02 session A - Spring 2010 CSE310...

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Unformatted text preview: Spring 2010 CSE310 Midterm Examination 02A (in class) Instructions: • There are five questions in this paper. Please use the space provided (below the questions) to write the answers. • Budget your time to answer various questions and avoid spending too much time on a particular question. • This is a closed book examination. You may not consult your books/notes. NAME ASUID Problems Score P1 P2 P3 P4 P5 Total P1. (20 points) (10 pts) This problem is related to order statistics. What is the minimum number of element-wise comparisons needed to find the smallest element and the largest element in an un-ordered sequence of n elements? Do not use asymptotic notation in your answer. Solution: Let f ( n ) denote the number of comparisons needed to find both the smallest and the largest elements in an array of n elements. First consider the case where n is even. Let n = 2 k . For k = 1, we only need one comparison. Therefore f (2) = 1. Suppose we have computed the largest and the smallest elements among the first 2( k − 1) elements, using f (2 k − 2) comparisons. For the next two elements, we need 3 comparisons. Therefore f (2 k ) = f (2 k − 2) + 3. This leads to f ( 2k ) = 3k − 2 for k = 1 , 2 , . . . . Now consider the case where n is odd. Let n = 2 k + 1. For k = 0, no comparison is needed, i.e., f (1) = 0. For k = 1 , 2 , . . . , f (2 k + 1) = f (2 k ) + 2, because the last element needs to be compared with both the candidate for largest and the candidate for smallest. Therefore f ( 2k + 1 ) = 3k for k = 0 , 1 , 2 , . . . . Combining both cases, we have f ( n ) = ⌈ 3 2 n ⌉ − 2 . Grading Keys: 5 pts for correct answer; 4 pts for the answer of the form 3 n 2 + const ; 3 pts for the answer 3 n 2 ; 1 pts for the answer 2 n − 3. Use the sequence 8, 4, 7, 3, 6, 2, 5, 1 to illustrate the process for finding the smallest element and the largest element. Show all the element-wise comparisons made for this example, in the correct order. Solution: The sequence of element-wise comparisons are: (8, 4), (7, 3), (8, 7), (3, 4), (6, 2), (8, 6), (3, 2), (5, 1), (8, 5), (1, 2)....
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This note was uploaded on 02/27/2011 for the course CSE 310 taught by Professor Davulcu,h during the Spring '08 term at ASU.

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Solution to midterm02 session A - Spring 2010 CSE310...

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