A guide to algorithm design paradigms, methods, and by Benoit A., Robert Y., Vivien F.

By Benoit A., Robert Y., Vivien F.

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Solutions to exercises 17 the binary search using k 1 boxes in order to narrow as much as possible the search interval around the desired floor. We then use the last box to scan the remaining interval floor by floor, from the lowest to the highest. After throwing k 1 boxes, if the target floor has not been identified, there are at most n/2k−1 floors in the search interval, hence a worst-case complexity of O(k + n/2k−1 ). 3. When k = 2 we do not want to have to test each floor one after the other, thereby ending up with a linear complexity.

What is the complexity of this algorithm? In the general case, how can we adapt the algorithm for any value of n? 3. Prove the optimality of this algorithm by providing a lower bound on the number of comparisons. The idea is to use decision trees. The decision tree of an algorithm is a tree that represents all the possible executions of the algorithm, on every possible input of size n. The internal nodes correspond to tests. In our case, the test is a comparison; if the answer is “yes” we move to the left child, otherwise to the right child, hence having a binary tree.

B) and (c ! d) with one comparison for each pair; then we compare the two largest elements with an additional comparison; finally, we insert c with a binary search in the sorted list containing a and b (we already know that c d), which requires two more comparisons. Hence, we sort four numbers in 2 1 + 1 + 2 = 5 comparisons, which is optimal. 2. Sorting 5 numbers. Sorting four numbers and then inserting the fifth one with a binary search would cost: 5 + 3 = 8 comparisons, which would be suboptimal.

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