lec02a - Complexity Analysis Resource needs Insertion Sort...

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Complexity Analysis Resource needs
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Insertion Sort for (int i = 1; i < a.length; i++) {// insert a[i] into a[0:i-1] int t = a[i]; int j; for (j = i - 1; j >= 0 && t < a[j]; j--) a[j + 1] = a[j]; a[j + 1] = t; }
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Complexity Space/Memory Time Count a particular operation Count number of steps Asymptotic complexity
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Comparison Count for (int i = 1; i < a.length; i++) {// insert a[i] into a[0:i-1] int t = a[i]; int j; for (j = i - 1; j >= 0 && t < a[j] ; j--) a[j + 1] = a[j]; a[j + 1] = t; }
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Comparison Count Pick an instance characteristic … n, n = a.length for insertion sort Determine count as a function of this instance characteristic.
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Comparison Count for (j = i - 1; j >= 0 && t < a[j] ; j--) a[j + 1] = a[j]; How many comparisons are made?
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Comparison Count for (j = i - 1; j >= 0 && t < a[j] ; j--) a[j + 1] = a[j]; number of compares depends on a[]s and t as well as on i
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Comparison Count Worst-case count = maximum count Best-case count = minimum count Average count
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Worst-Case Comparison Count for (j = i - 1; j >= 0 && t < a[j] ; j--) a[j + 1] = a[j]; a = [1, 2, 3, 4] and t = 0 => 4 compares a = [1,2,3,…,i] and t = 0 => i compares
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Worst-Case Comparison Count
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This note was uploaded on 01/18/2012 for the course COP 3530 taught by Professor Davis during the Fall '08 term at University of Florida.

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lec02a - Complexity Analysis Resource needs Insertion Sort...

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