Question

Suppose we modify the deterministic version of the quick-sort algorithm so that, instead of selecting the last element in an \(n\) -element sequence as the pivot, we choose the element at index \(\lfloor n / 2\rfloor\). What is the running time of this version of quick-sort on a sequence that is already sorted?

Short answer

The running time is \( O(n \log n) \).

Step-by-step solution

Understand the Problem

We need to determine the running time of a modified quick-sort algorithm that chooses the pivot element at index \( \lfloor n / 2 \rfloor \) on a sequence that is already sorted.

Base Case Analysis

If the sequence has one or zero elements, it is already sorted, so the running time is \( O(1) \).

Recursive Step Analysis

For the modified quick-sort, the middle element is chosen as the pivot. In an already sorted sequence, this splits the sequence into two nearly equal halves.

Analyze Running Time for Balanced Splits

If the pivot always splits the sequence into two equal halves, the modified quick-sort will have the same running time as merge-sort, which is \( O(n \log n) \).

Conclusion

Since choosing the middle element ensures balanced splits, the running time of this version of quick-sort on an already sorted sequence is \( O(n \log n) \).

Key concepts

Running Time Analysis

The running time of any algorithm is a crucial factor to consider. For quick-sort, we generally look at the best, average, and worst-case scenarios. When we modify quick-sort to pick the middle element \( \lfloor n / 2 \rfloor \) as the pivot, we must analyze how it affects the running time, especially for an already sorted sequence.
In the base case, where the sequence has one or zero elements, the running time is \( O(1) \). This is because there's nothing to sort.
For larger sequences, if we're lucky enough to always split the sequence into two equal halves (or almost equal in size), quick-sort behaves similarly to merge-sort, having a running time of \ O(n \log n) \. This is considered optimal for comparison-based sorting algorithms.
On the other hand, in the worst-case scenario where splits are highly unbalanced, quick-sort degenerates to a running time of \( O(n^2)\). However, by choosing the middle element as the pivot, we greatly reduce the chance of hitting the worst case, ensuring more balanced splits and an average running time much closer to \( O(n \log n )\).

Pivot Selection

The pivot in quick-sort is the element used to partition the array into subarrays which are then recursively sorted.
In this modified quick-sort version, selecting the pivot element at index \( \lfloor n / 2 \rfloor \) (the middle element) ensures that even for an already sorted sequence, the partitions are reasonably balanced.
When we choose consistent and balanced pivots:

• It leads to roughly equal-sized partitions in each recursion.
• This makes the sorting process more efficient.

In contrast, choosing the last element as a pivot in a sorted array would always create highly unbalanced partitions, leading to poor performance and a worst-case running time.
Thus, by choosing the middle element as the pivot, the algorithm maximizes the chances of balanced splits, contributing significantly to achieving the running time of \( O(n \log n)\) for sorted sequences.

Divide and Conquer

Quick-sort is a classic example of the divide-and-conquer strategy. Here's how it works:

• Divide: The array is divided into two smaller sub-arrays using the pivot.
• Conquer: Each sub-array is recursively sorted.
• Combine: Because the sub-arrays are sorted in place, there's no need for a separate combine step (as in merge-sort).

This approach ensures that the problem-solving process is broken down into more manageable sub-problems.
By selecting the middle element as the pivot, the divide-step ensures more balanced sub-problems. The more balanced these sub-problems are, the more efficient the sorting process becomes.
Every division results in approximately \( n / 2 \) elements on either side of the pivot, keeping the recursion tree fairly shallow and ensuring a running time of \( O(n \log n)\) for the overall process.
Hence, the divide-and-conquer strategy, combined with smart pivot selection, results in efficient sorting and is particularly effective in minimizing the running time for sorted sequences.