By Benoit A., Robert Y., Vivien F.
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Extra resources for A guide to algorithm design paradigms, methods, and complexity analysis
Therefore, we need at least d n2 e comparisons to get rid of all novices. Furthermore, at the end we must have n 2 average elements. A comparison creates at most one average element. Hence, we need at least n 2 additional comparisons to reach the desired number of average elements. Finally, we remark that no comparison implying a novice leads to an increase in the number of average elements. , the one to get rid of all novices and the one to create the required number of average elements) to obtain a global lower bound.
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. Therefore, we proceed otherwise. We start, as previously, by creating two pairs of two elements (a ! 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.
In our example, Ef3n g = 3, 9, 27, 81, . . = f3n+1 g. 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.