Write down a recurrence relation for this version of QuickSort, and solve it asymp-totically. Show your work. Assume that the time it takes to find the pivot (eitherbest or worst, depending on the level of recursion) is Θ(n) for lists of lengthn.

As in merge sort, the time for a given recursive call on an n-element subarray is Θ(n). In merge sort, that was the time for merging, but in quicksort it's the time for partitioning.

Explanation:

Worst-case running time:

When quicksort always has the most unbalanced partitions possible, then the original call takes cncnc, n time for some constant ccc, the recursive call on n-1n−1n, minus, 1 elements takes c(n-1)c(n−1)c, left parenthesis, n, minus, 1, right parenthesis time, the recursive call on n-2n−2n, minus, 2 elements takes c(n-2)c(n−2)c, left parenthesis, n, minus, 2, right parenthesis time, and so on. cn+c(n−1)+c(n−2)+⋯+2c=c(n+(n−1)+(n−2)+⋯+2)

=c((n+1)(n/2)−1)

The last line is because 1 + 2 + 3 + n is the arithmetic series

Best-case running time:

Quicksort's best case occurs when the partitions are as evenly balanced as possible: their sizes either are equal or are within 1 of each other. The former case occurs if the subarray has an odd number of elements and the pivot is right in the middle after partitioning, and each partition has (n-1)/ 2 elements. The latter case occurs if the subarray has an even number n of elements and one partition has n/2 elements with the other having n/2-1.In either of these cases, each partition has at most n/2 elements.

Using big-Θ notation, we get the same result as for merge sort: Θ(nlog2n)

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the answer is d.

by: nicholas dooley