A key aspect of the Quick Sort algorithm is how the pivot element is chosen. In my earlier post on the Python code for Quick Sort, my implementation takes the first element of the unsorted array as the pivot element.

However with some mathematical analysis it can be seen that such an implementation is *O(n*^{2}) in complexity while if a pivot is randomly chosen, the Quick Sort algorithm is *O(nlog*_{2}n).

To witness this in action, one can measure the work done by the algorithm comparing two cases, one with a randomized pivot choice – and one with a fixed pivot choice, say the first element of the array (or the last element of the array).

**Implementation**

A decent proxy for the amount of work done by the algorithm would be the number of pivot comparisons. These comparisons needn’t be computed one-by-one, rather when there is a recursive call on a subarray of length m, you should simply add m−1 to your running total of comparisons.

**3 Cases**

To put things in perspective, let’s look at 3 cases. (This is basically straight out of a homework assignment from Tim Roughgarden’s course on the Design and Analysis of Algorithms).

**Case I** with the pivot being the *first element*.

**Case II** with the pivot being the *last element.*

**Case III** using the *“median-of-three”* pivot rule. The primary motivation behind this rule is to do a little bit of extra work to get much better performance on input arrays that are nearly sorted or reverse sorted.

**Median-of-Three Pivot Rule**

Consider the first, middle, and final elements of the given array. (If the array has odd length it should be clear what the “middle” element is; for an array with even length 2k, use the k^{th} element as the “middle” element. So for the array 4 5 6 7, the “middle” element is the second one —- 5 and not 6! Identify which of these three elements is the median (i.e., the one whose value is in between the other two), and use this as your pivot.

**Python Code**

This file contains all of the integers between 1 and 10,000 (inclusive, with no repeats) in unsorted order. The integer in the i^{th} row of the file gives you the i^{th} entry of an input array. I downloaded this file and named it *QuickSort_List.txt*

You can run the code below and see for yourself that the number of comparisons for Case III are 138,382 compared to 162,085 and 164,123 for Case I and Case II respectively. You can play around with the code in an IPython / Jupyter notebook here.

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