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

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**Example text**

Each branch of the tree corresponds to an execution of the algorithm, and the number of comparisons is the height of the branch. The number of comparisons in the worst case is then obtained as the height of the tree. (a) Prove that any decision tree that computes the maximum of n integers has at least 2n−1 leaves. (b) Prove that any binary tree of height h and with f leaves is such that 2h f . (c) Let A be a decision tree solving the problem. Give a lower bound on its number of leaves. Conclude with a lower bound on the number of comparisons in the worst case.

It is suboptimal for an algorithm to ask a question involving a person who is known not to be a star when there are still at least two star candidates. Indeed, in the worst case, this question will not provide any new knowledge. This is obviously the case if we already know that the two persons are not stars. Otherwise, the candidate involved will not be invalidated by the answer to the question but will eventually turn out not to be a star. In other words, in an optimal algorithm, each of the first n 1 questions should involve two persons who can still be stars, when taking into account the answers to all questions asked so far.

For instance, with Strassen’s algorithm, if we consider the number of additions to be executed in a matrix product, we have a = 7, b = 2, α = 2, and n is performed by c = 18 4 . Indeed, the product of two matrices of size n first computing 7 products of matrices of size n/2 n/2 and reconstructing the solution through 18 additions of matrices of size n/2 n/2; therefore, 2 R(n) = 18(n/2)2 = 18 4 n . In this case, the initial cost is S(1) = 0. Let us assume that there exists k 2 N such that n = bk ; thus, k = logb (n) and ak = nlogb (a) .