HW5_sol - Fall Semester 2008 CS300 Algorithms Solution #5...

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Fall Semester 2008 CS300 Algorithms Solution #5 1. (a) 9 comparisons needed for MAX heap. (b) Every internal node performs FixHeap operation to construct a heap. Because it’s already sorted in the decreasing order, there is no key movement. When a node has two children nodes, you need two comparisons, and similarly when a node has one child, you need just one comparison. If n is even, the number of internal nodes is n/ 2 and the last internal node has one child node, so the number of comparison is 2 × ( n/ 2 - 1) + 1 = n - 1. If n is odd, the number of internal nodes is ( n - 1) / 2 and all internal nodes have two children nodes, so the number of comparisons is 2 × ( n - 1) / 2 = n - 1. (c) The best case. In the case of decreasing order, you do not need move a key during FixHeap operations, which minimizes the number of comparisons. 2. (a) Let us denote the list by L . Then the worst case is when L is sorted in the decreasing order like L = 111 ··· 1100 ··· 000. Let us assume that
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HW5_sol - Fall Semester 2008 CS300 Algorithms Solution #5...

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