In this lab we will implement the Quicksort algorithm. As a reference, refer to the readings in §8.3 from the schedule.
Let us first review the main steps in Quicksort:
We will start this lab by first implementing the partition operation. For this part of the lab, we will fill in the following definition of the partition operation:
private static int partition(int arr, int lo, int hi, int pivotIndex)
The partition operation takes a sub-array (like merge sort, denoted by a
lo and a
hi bound) and the index of an element,
pivotIndex in that sub-array as input. Let’s walk through the general steps involved in this operation.
Let’s assume the array we want to partition has the shape below. Following the partition operation, we want the modified array to have the following properties (1) all the elements less than the pivot are to the left of the pivot, and (2) all the elements greater than or equal to the pivot are to the right.
[ ][ ][ ] pivot [< pivot][ ][>= pivot] pivot
To perform this operation in linear time, we will employ a two-fingered approach. Let’s visualize this operation with a sample array from the reading.
[3, 9, 2, 8, 6, 4, 1, 7, 5]
To start, you may choose the midpoint of the array to be the pivot. Let’s say we choose the element 6 as our pivot. Because we are sorting the array in-place, we want to first swap the pivot 6 with the last element of the array (or the sub-array, if we’re using it within Quicksort). Now we can consider partitioning every element but the last of the array according to the pivot.
[3, 9, 2, 8, 5, 4, 1, 7, 6]
Our implementation will use
hi to keep track of the sub-array limits. Then, we can initialize our pointers to keep track of the left and right indices relative to the sub-array limits.
[3, 9, 2, 8, 5, 4, 1, 7, 6] ^ ^
Now we will repeatedly compare elements on the left- and right-hand sides of the array and place them in the correct position relative to the pivot. In our array, the first swap occurs when we find an element on the left that is greater than the pivot, and on the right that is lesser than the pivot.
[3, 9, 2, 8, 5, 4, 1, 7, 6] ^ ^ [3, 1, 2, 8, 5, 4, 9, 7, 6] ^ ^
We will maintain the following loop invariant in our program design:
[ <pivot ][ ][ ][ >=pivot ][ ] ^ ^ pivot
What would be an appropriate condition to terminate the loop body?
When you’re done, you should swap the pivot into the location it belongs (in this version, at the left of the >= elements). You will thenreturn the index of the pivot.
And now to test our code! Test your code with common-case as well as edge-case scenarios. Does your code return the correct partitioning if the pivot is set to the smallest or the largest element in the list?
Now that we have implemented the partition operation, all that remains is to recursively apply the partition operation on our input array! Write an implementation of Quicksort that has the following definition:
private static void quickSort(int arr)
Note that we will also need to write a helper quickSort implementation which would take in the array bounds to recursively sort through the left and right sub-arrays.
private static void quickSort(int arr, int lb, int ub)
Test your code once again across both common-case and edge-case scenarios to make sure your sorting algorithm works.
Finally, we consider our choice of pivot. We have seen from the readings, that a poor choice of pivot can affect the run time of our Quicksort implementation and that a good choice of pivot would be the median of the list. A computationally efficient implementation is to use Median of three to select our pivot using the first, the middle, and the last elements of the array. Consider the following questions before proceeding to implement Median of three:
Write an implementation of median-of-three that has the following signature:
private static int medianOfThree(int arr, int left, int right)
For the few of you who have extra time.
Let’s say we wanted to select the kth smallest element in an unsorted list. How would we go about finding this element? Given a set of n distinct numbers and the number k, 1 <= k <= n, the selection problem computes the kth order statistic, or kth smallest number in the set of numbers.
We will now proceed to implement Quickselect which has a slight variation on Quicksort. The intuition in Quickselect is that we prune the sub-arrays where we know that the
kth value will not be found.
1. Pick a pivot at random from the input array.
2. Partition the array into sub-arrays as before and count the number of elements, L, in the left sub-array.
[< pivot][ ][> pivot] —— L— pivot
3. Selection: In this operation, we want to use the results of the partition, to select the sub-array where we will find the kth smallest value. a. If L is equal to k-1 then the pivot is the kth smallest value! b. If L is greater than k-1 then recursively call select on the left sub-array. c. If L is less than k-1 then recursively call select on the right sub-array.
select operation of Quickselect:
private static int select(int arr, int k)
This lab is copied nearly verbatim from a similar lab by Anya Vostinar and Peter-Michael Osera.