Quicksort and Equal Elements: An Analysis

When sorting an array of N items that are all equal using the Quicksort algorithm, the number of comparisons made can vary widely depending on the implementation of Quicksort, particularly the choice of pivot. This article will explore how Quicksort behaves in such scenarios and discuss the implications for comparison counts and time complexities.

Worst-Case Scenario: Unbalanced Partitions

In the worst-case scenario, if the pivot is always chosen as the first element or any fixed position, leading to unbalanced partitions, Quicksort performs poorly. Here's a detailed look at why this happens.

The first element is chosen as the pivot. Since all elements are equal, every element will be compared to the pivot, but none will be placed in a different partition. This results in a recursive call on the remaining N - 1 elements. The number of comparisons for each level of recursion can be summarized as follows: At the first level: N - 1 comparisons are made. At the second level: N - 2 comparisons are made. This continues until the last level, which makes 0 comparisons. The total number of comparisons CN can be calculated as:

CN  (N-1) * (N-2) * (N-3) * ... * 1 * 0  frac{N(N-1)}{2}

Thus, the total number of comparisons made by Quicksort in this scenario is:

CN  frac{N(N-1)}{2}

This demonstrates that Quicksort can perform as many as O(N^2) comparisons in the case of sorting an array where all elements are equal, which is the worst-case time complexity for this algorithm.

Best-Case Scenario: Middle Value Pivot

Assuming the middle value is used for the pivot, Quicksort behaves differently:

In the Lomuto partition scheme, all equal elements are considered a worst-case scenario. However, in the Hoare partition scheme, the same scenario is a best-case scenario because it leads to a perfect 1/2 split, yielding an efficient NlogN comparison count.

Trivial Implementation: Perfect Midpoint Split

The most trivial implementation of Quicksort assumes that:

Pivot is picked. Starting from the beginning, elements are compared until the midpoint is reached. No swaps are performed, and the pivot element is placed in the middle. Recursion proceeds with a perfect 1/2 split.

This results in a comparison count of NlogN. This approach assumes an even distribution of elements, which is ideal in the case of a perfectly balanced partition.

However, we must be aware that when all elements are equal, we have to compare each element to the pivot, making the overall complexity O(N^2). Therefore, the recurrence for this scenario is:

Tn  Tn-1   cn

which gives us On^2.