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Unformatted text preview: Solutions for CAS 702 Midterm 2007 Problem 1a Let k j and i j be the values of k and i respectively after state- ment i := i × i is executed j times. Then we observe that i = 2 , k = 1 and i j = 2 2 j , k j = i 2 j = 2 2 j +1 . The loop ends when k j = 2 2 j +1 ≥ n , or j ≥ lg lg n- 1. So j = Θ(lg lg n ). Problem 1b Consider a red-black tree with black-height k . If every node is black the total number of internal nodes is 2 k- 1. If only every other nodes is black we can construct a tree with 2 2 k- 1 nodes. Probelm 1c One of the counter example is making changes for 8 cents with coin denominations 1, 4 and 6. Problem 2a Loop invariant : At the start of each iteration of the for loop of lines 1-4, the subarray A [1 ..i- 1] consists of the i- 1 smallest values originally in A [1 ..n ], in sorted order, and A [ i..n ] consists of the n- i + 1 remaining values originally in A [1 ..n ]....
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- Fall '09
- Array, Loop invariant, 1 j, sorted order