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

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Finally, it checks that the candidate star does not know any group member (while loop at Step 10). Each of three while loops executes at most n iterations, hence a complexity in O(n). 2: Algorithm identifying a star in a group. 3. First of all, we remark that the worst case is reached when the group of persons contains a star. To identify a star, it is necessary that each © 2014 by Taylor & Francis Group, LLC 16 Chapter 1. Introduction to complexity person, but the star, is involved in a question that identifies him or her as not being a star.

4: Naive algorithm to compute the minimum and the maximum of a set of n values. 2. 4, we considered one new value at each step. Here we will consider two new values at each step. We first compare them, and then we compare the largest one to the current maximum and compare the smallest one to the current minimum. 5 presents such an algorithm. If n is even, there are n2 pairs and thus n2 comparisons of pair elements. Then, there are n2 1 additional comparisons to compute the maximum and as many for the minimum.

Therefore, n n Pn,k z k Gn+1 (z) = z Pn,k z k +n = (z + n)Gn (z). k=1 k=1 Since G1 (z) = P1,1 z = z, we obtain Gn (z) = z(z + 1) (z + n 1). 3, to this function G1 (z). We remark that n n kPn,k z k−1 , Gn (1) = Gn (z) = n kPn,k , and Gn (1) = k=1 k=1 Therefore, avgA (n) = Gn (1) = [ln(G(z))] (1) Gn (1) n−1 = ln(z + i) (1) i=0 =1+ © 2014 by Taylor & Francis Group, LLC 1 + 2 Pn,k = n! k=1 + 1 = Hn n 20 Chapter 1. Introduction to complexity where Hn is the n-th partial sum of the diverging harmonic series (or the n-th harmonic number).