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The labor that F (n) estimates has two components. First there are the pairwise comparisons involved in choosing a splitting element and rearranging the array about the chosen splitting value. Second there are the comparisons that are done in the two recursive calls that follow the creation of a splitter. As we have seen, the number of comparisons involved in splitting the array is n − 1. Hence it remains to estimate the number of comparisons in the recursive calls. For this purpose, suppose we have rearranged the array about the splitting element, and that it has turned out that the splitting entry now occupies the ith position in the array.

1: Conditions (a), (b), (c) To see this, observe first that (a), (b), (c) are surely true at the beginning, when j = lef t + 1. Next, if for some j they are true, then the execution of lines 7, 8 guarantee that they will be true for the next value of j. Now look at (a), (b), (c) when j = right. It tells us that just prior to the execution of line 9 the condition of the array will be (a) x[lef t] = T and (b) x[r] < T for all lef t < r ≤ i and (c) x[r] ≥ T for all i < r ≤ right. When line 9 executes, the array will be in the correctly split condition.

Xi−1 ) in place; quicksortprelim(the subarray xi+1 , . . {quicksortprelim} * C. A. R. Hoare, Comp. , 5 (1962), 10-15. 33 This preliminary version won’t run, though. It looks like a recursive routine. It seems to call itself twice in order to get its job done. But it doesn’t. It calls something that’s just slightly different from itself in order to get its job done, and that won’t work. Observe the exact purpose of Quicksort, as described above. We are given an array of length n, and we want to sort it, all of it.

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