CSE331_Homework_2 - 1 a What is the total number of comparisons and the total number of swaps(comparing out-oforder pairs of adjacent elements in

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1. a. What is the total number of comparisons and the total number of swaps (comparing out-of-order pairs) of adjacent elements in the following list for both bubble and insertion sort. Bubble Sort Compares = 0 Swaps = 0 L: 8, 5, 7, 2, 3 Compares = 1 Swaps = 1 L: 5, 8, 7, 2, 3 Compares = 2 Swaps = 2 L: 5, 7, 8, 2, 3 Compares = 3 Swaps = 3 L: 5, 7, 2, 8, 3 Compares = 4 Swaps = 4 L: 5, 7, 2, 3, 8 Compares = 5 Swaps = 4 L: 5, 7, 2, 3, 8 Compares = 6 Swaps = 5 L: 5, 2, 7, 3, 8 Compares = 7 Swaps = 6 L: 5, 2, 3, 7, 8 Compares = 8 Swaps = 7 L: 2, 5, 3, 7, 8 Compares = 9 Swaps = 8 L: 2, 3, 5, 7, 8 Compares = 10 Swaps = 8 L: 2, 3, 5, 7, 8 Final Answer: Compares = 10 Swaps = 8 Insertion Sort Compares = 0 Swaps = 0 L: 8, 5, 7, 2, 3 Compares = 1 Swaps = 1 L: 5, 8, 7, 2, 3 Compares = 2 Swaps = 2 L: 5, 7, 8, 2, 3 Compares = 3 Swaps = 2 L: 5, 7, 8, 2, 3 Compares = 4 Swaps = 3 L: 5, 7, 2, 8, 3 Compares = 5 Swaps = 4 L: 5, 2, 7, 8, 3 Compares = 6 Swaps = 5 L: 2, 5, 7, 8, 3 Compares = 7 Swaps = 6 L: 2, 5, 7, 3, 8 Compares = 8 Swaps = 7 L: 2, 5, 3, 7, 8 Compares = 9 Swaps = 8 L: 2, 3, 5, 7, 8 Compares = 10 Swaps = 8 L: 2, 3, 5, 7, 8 Final Answer: Compares = 10 Swaps = 8 b. Why does the number of swaps give a lower bound on computation time for these two sorts? Bubble Sort give a lower bound on computation time compared to the number of swaps because the bigger numbers are at the front and it takes a lot more swaps to move them to the end because the way Bubble Sort swaps elements. Since every element is at most N- 1 away from its final position, the biggest number of operations to move the element to its sorted spot is n - 1 = O(n) operations. For Insertion Sort a lower bound computation is given based on the number of swaps because on average and in worse case it takes O(n^2). So more times than not the number of swaps will be a lower bound computation because contrary to bubble sort which takes less time than the previous pass, insertion may take more time on each pass than the previous. c. What is the number of inversions in list L? Why is the number of swaps of adjacent elements equal to the number of inversions for these sorts? There are 8 inversions in list L. The number of inversions is always the number of swaps because swapping two adjacent elements that are out of place removes one inversion.
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d. Why is the total number of inversions in the first two lists below equal to the number in the third list below? L: 8, 5, 7, 2, 3 Lreverse: 3, 2, 7, 5, 8 L3descending: 8, 7, 5, 3, 2
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This note was uploaded on 03/31/2008 for the course CSE 331 taught by Professor M.mccullen during the Spring '08 term at Michigan State University.

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CSE331_Homework_2 - 1 a What is the total number of comparisons and the total number of swaps(comparing out-oforder pairs of adjacent elements in

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