By Benoit A., Robert Y., Vivien F.
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Additional resources for A guide to algorithm design paradigms, methods, and complexity analysis
B) and (c ! d) with one comparison for each pair. Then we compare the two largest elements with an additional comparison. After three comparisons, we obtain the same configuration as previously: a b c © 2014 by Taylor & Francis Group, LLC d 30 Chapter 1. Introduction to complexity Then, we insert the fifth element, e, in the chain a ! b ! d with two comparisons. Finally, we are left only with inserting c in the sorted chain made of the three elements a, b, and e, which costs two additional comparisons (we already know that c d).
We, however, will solve this problem in a simple way. 4. 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.
More generally, Efsn g = fsn+1 g. We then define the following operations on sequences: cfsn g = fcsn g, (E1 + E2 )fsn g = E1 fsn g + E2 fsn g, (E1 E2 )fsn g = E1 (E2 fsn g). For instance, (E 3)fsn g = fsn+1 3sn g, and (2 + E 2 )fsn g = f2sn + sn+2 g. We are looking for annihilators of the sequences. That is, we are looking for operators P (E) such that P (E)fsn g = f0g. For our example, (E 3)f3n g = f3n+1 3 3n g = f0g. We provide a few more examples, where Qk (n) is a polynomial in n of degree k: fcn sequence fcg fQk (n)g fcn g Qk (n)g annihilator E 1 (E 1)k+1 E c (E c)k+1 The first three lines are special cases of the fourth line; therefore, we need to prove only the last relation.