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The complexity in O(log(n)) is a hint: One should use a binary search. Indeed, if we have k dlog(n)e, we know the result for the floors whose indices range from i to j by dropping a box from the m-th floor where and then by iterating with floors i to m 1 if the box broke, m = i+j 2 and by iterating with floors m to j otherwise. The principle of the binary search guarantees that we will obtain the desired result (when i = j) and in at most dlog(n)e steps, and, thus, after having broken at most dlog(n)e boxes.

1. Compute M (n) and A(n) for the usual algorithm to multiply two npolynomials. 2. We assume that n is even, n = 2 m. We can then write P = P1 + X m P2 and Q = Q1 +X m Q2 . What is the degree of the polynomials P1 , P2 , Q1 , and Q2 ? 3. Let R1 = P1 Q1 , R2 = P2 Q2 , and R3 = (P1 + P2 ) (Q1 + Q2 ). Can you express R = P Q as a function of R1 , R2 , and R3 ? What is the degree of these three new polynomials? Compute M (n) and A(n), assuming that we use the classical multiplication algorithm to compute R1 , R2 , and R3 .

2 presents the comparison tree (to be read bottom up). In this tree, the dotted lines mark the trajectory of the maximum value. n 1 comparisons are performed to determine the maximum. The second maximum is one of the values that lost their comparison against the maximum. 2. There are k such values, and, therefore, k 1 = log(n) 1 comparisons are needed to determine the second maximum. Hence, the overall complexity is of n + log(n) 2 comparisons. © 2014 by Taylor & Francis Group, LLC 24 Chapter 1.

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