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Show Source |    | About   «  13.10. Implementing Mergesort   ::   Contents   ::   13.12. Heapsort  »

13.11. Quicksort

13.11.1. Introduction

While Mergesort uses the most obvious form of divide and conquer (split the list in half then sort the halves), this is not the only way that we can break down the sorting problem. We saw that doing the merge step for Mergesort when using an array implementation is not so easy. So perhaps a different divide and conquer strategy might turn out to be more efficient?

Quicksort is aptly named because, when properly implemented, it is the fastest known general-purpose in-memory sorting algorithm in the average case. It does not require the extra array needed by Mergesort, so it is space efficient as well. Quicksort is widely used, and is typically the algorithm implemented in a library sort routine such as the UNIX qsort function. Interestingly, Quicksort is hampered by exceedingly poor worst-case performance, thus making it inappropriate for certain applications.

Before we get to Quicksort, consider for a moment the practicality of using a Binary Search Tree for sorting. You could insert all of the values to be sorted into the BST one by one, then traverse the completed tree using an inorder traversal. The output would form a sorted list. This approach has a number of drawbacks, including the extra space required by BST pointers and the amount of time required to insert nodes into the tree. However, this method introduces some interesting ideas. First, the root of the BST (i.e., the first node inserted) splits the list into two sublists: The left subtree contains those values in the list less than the root value while the right subtree contains those values in the list greater than or equal to the root value. Thus, the BST implicitly implements a "divide and conquer" approach to sorting the left and right subtrees. Quicksort implements this same concept in a much more efficient way.

Quicksort first selects a value called the pivot. (This is conceptually like the root node's value in the BST.) Assume that the input array contains \(k\) records with key values less than the pivot. The records are then rearranged in such a way that the \(k\) values less than the pivot are placed in the first, or leftmost, \(k\) positions in the array, and the values greater than or equal to the pivot are placed in the last, or rightmost, \(n-k\) positions. This is called a partition of the array. The values placed in a given partition need not (and typically will not) be sorted with respect to each other. All that is required is that all values end up in the correct partition. The pivot value itself is placed in position \(k\). Quicksort then proceeds to sort the resulting subarrays now on either side of the pivot, one of size \(k\) and the other of size \(n-k-1\). How are these values sorted? Because Quicksort is such a good algorithm, using Quicksort on the subarrays would be appropriate.

Unlike some of the sorts that we have seen earlier in this chapter, Quicksort might not seem very "natural" in that it is not an approach that a person is likely to use to sort real objects. But it should not be too surprising that a really efficient sort for huge numbers of abstract objects on a computer would be rather different from our experiences with sorting a relatively few physical objects.

Here is an implementation for Quicksort. Parameters i and j define the left and right indices, respectively, for the subarray being sorted. The initial call to quicksort would be quicksort(array, 0, n-1).

static void quicksort(Comparable[] A, int i, int j) { // Quicksort
  int pivotindex = findpivot(A, i, j);  // Pick a pivot
  Swap.swap(A, pivotindex, j);               // Stick pivot at end
  // k will be the first position in the right subarray
  int k = partition(A, i, j-1, A[j]);
  Swap.swap(A, k, j);                        // Put pivot in place
  if ((k-i) > 1) quicksort(A, i, k-1);  // Sort left partition
  if ((j-k) > 1) quicksort(A, k+1, j);  // Sort right partition
}
void quicksort(Comparable* A[], int i, int j) {
  int pivotindex = findpivot(i, j);
  swap(A, pivotindex, j); // Stick pivot at end
  // k will be the first position in the right subarray
  int k = partition(A, i, j-1,A[j]);
  swap(A, k, j);                       // Put pivot in place
  if ((k-i) > 1) quicksort(A, i, k-1); // Sort left partition
  if ((j-k) > 1) quicksort(A, k+1, j); // Sort right partition
}

Function partition will move records to the appropriate partition and then return k, the first position in the right partition. Note that the pivot value is initially placed at the end of the array (position j). Thus, partition must not affect the value of array position j. After partitioning, the pivot value is placed in position k, which is its correct position in the final, sorted array. By doing so, we guarantee that at least one value (the pivot) will not be processed in the recursive calls to qsort. Even if a bad pivot is selected, yielding a completely empty partition to one side of the pivot, the larger partition will contain at most \(n-1\) records.

Selecting a pivot can be done in many ways. The simplest is to use the first key. However, if the input is sorted or reverse sorted, this will produce a poor partitioning with all values to one side of the pivot. It is better to pick a value at random, thereby reducing the chance of a bad input order affecting the sort. Unfortunately, using a random number generator is relatively expensive, and we can do nearly as well by selecting the middle position in the array. Here is a simple findpivot function.

static int findpivot(Comparable[] A, int i, int j)
  { return (i+j)/2; }
int findpivot(int i, int j)
  { return (i+j)/2; }

13.11.2. Partition

We now turn to function partition. If we knew in advance how many keys are less than the pivot, partition could simply copy records with key values less than the pivot to the low end of the array, and records with larger keys to the high end. Because we do not know in advance how many keys are less than the pivot, we use a clever algorithm that moves indices inwards from the ends of the subarray, swapping values as necessary until the two indices meet. Here is an implementation for the partition step.

static int partition(Comparable[] A, int left, int right, Comparable pivot) {
  while (left <= right) { // Move bounds inward until they meet
    while (A[left].compareTo(pivot) < 0) left++;
    while ((right >= left) && (A[right].compareTo(pivot) >= 0)) right--;
    if (right > left) Swap.swap(A, left, right); // Swap out-of-place values
  }
  return left;            // Return first position in right partition
}
int partition(Comparable* A[], int left, int right, Comparable* pivot) {
  while (left <= right) { // Move bounds inward until they meet
    while (*A[left] < *pivot) left++;
    while ((right >= left)  && (*A[right] >= *pivot)) right--;
    if (right > left) swap(A, left, right);  // Swap out-of-place values
   }
  return left;           // Return first position in right partition
}

Note the check that right >= left in the second inner while loop. This ensures that right does not run off the low end of the partition in the case where the pivot is the least value in that partition. Function partition returns the first index of the right partition (the place where left ends at) so that the subarray bound for the recursive calls to qsort can be determined.

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And here is a visualization illustrating the running time analysis of the partition function

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13.11.3. Putting It Together

Here is a visualization for the entire Quicksort algorithm. This visualization shows you how the logical decomposition caused by the partitioning process works. In the visualization, the separate sub-partitions are separated out to match the recursion tree. In reality, there is only a single array involved (as you will see in the proficiency exercise that follows the visualization).

Here is a complete proficiency exercise to see how well you understand Quicksort.

13.11.4. Quicksort Analysis

This visualization explains the worst-case running time of Quick Sort

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This is terrible, no better than Bubble Sort. When will this worst case occur? Only when each pivot yields a bad partitioning of the array. If the pivot values are selected at random, then this is extremely unlikely to happen. When selecting the middle position of the current subarray, it is still unlikely to happen. It does not take many good partitionings for Quicksort to work fairly well.

This visualization explains the best-case running time of Quick Sort

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Quicksort's average-case behavior falls somewhere between the extremes of worst and best case. Average-case analysis considers the cost for all possible arrangements of input, summing the costs and dividing by the number of cases. We make one reasonable simplifying assumption: At each partition step, the pivot is equally likely to end in any position in the (sorted) array. In other words, the pivot is equally likely to break an array into partitions of sizes 0 and \(n-1\), or 1 and \(n-2\), and so on.

Given this assumption, the average-case cost is computed from the following equation:

\[{\bf T}(n) = cn + \frac{1}{n}\sum_{k=0}^{n-1}[{\bf T}(k) + {\bf T}(n - 1 - k)], \quad {\bf T}(0) = {\bf T}(1) = c.\]

This visualization will help you to understand how this recurrence relation was formed.

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This is an unusual situation that the average case cost and the worst case cost have asymptotically different growth rates. Consider what "average case" actually means. We compute an average cost for inputs of size \(n\) by summing up for every possible input of size \(n\) the product of the running time cost of that input times the probability that that input will occur. To simplify things, we assumed that every permutation is equally likely to occur. Thus, finding the average means summing up the cost for every permutation and dividing by the number of permuations (which is \(n!\)). We know that some of these \(n!\) inputs cost \(O(n^2)\). But the sum of all the permutation costs has to be \((n!)(O(n \log n))\). Given the extremely high cost of the worst inputs, there must be very few of them. In fact, there cannot be a constant fraction of the inputs with cost \(O(n^2)\). If even, say, 1% of the inputs have cost \(O(n^2)\), this would lead to an average cost of \(O(n^2)\). Thus, as \(n\) grows, the fraction of inputs with high cost must be going toward a limit of zero. We can conclude that Quicksort will run fast if we can avoid those very few bad input permutations. This is why picking a good pivot is so important.

The running time for Quicksort can be improved (by a constant factor), and much study has gone into optimizing this algorithm. Since Quicksort's worst case behavior arises when the pivot does a poor job of splitting the array into equal size subarrays, improving findpivot seems like a good place to start. If we are willing to do more work searching for a better pivot, the effects of a bad pivot can be decreased or even eliminated. Hopefully this will save more time than was added by the additional work needed to find the pivot. One widely-used choice is to use the "median of three" algorithm, which uses as a pivot the middle of three randomly selected values. Using a random number generator to choose the positions is relatively expensive, so a common compromise is to look at the first, middle, and last positions of the current subarray. However, our simple findpivot function that takes the middle value as its pivot has the virtue of making it highly unlikely to get a bad input by chance, and it is quite cheap to implement. This is in sharp contrast to selecting the first or last record as the pivot, which would yield bad performance for many permutations that are nearly sorted or nearly reverse sorted.

A significant improvement can be gained by recognizing that Quicksort is relatively slow when \(n\) is small. This might not seem to be relevant if most of the time we sort large arrays, nor should it matter how long Quicksort takes in the rare instance when a small array is sorted because it will be fast anyway. But you should notice that Quicksort itself sorts many, many small arrays! This happens as a natural by-product of the divide and conquer approach.

A simple improvement might then be to replace Quicksort with a faster sort for small numbers, say Insertion Sort or Selection Sort. However, there is an even better---and still simpler---optimization. When Quicksort partitions are below a certain size, do nothing! The values within that partition will be out of order. However, we do know that all values in the array to the left of the partition are smaller than all values in the partition. All values in the array to the right of the partition are greater than all values in the partition. Thus, even if Quicksort only gets the values to "nearly" the right locations, the array will be close to sorted. This is an ideal situation in which to take advantage of the best-case performance of Insertion Sort. The final step is a single call to Insertion Sort to process the entire array, putting the records into final sorted order. Empirical testing shows that the subarrays should be left unordered whenever they get down to nine or fewer records.

The last speedup to be considered reduces the cost of making recursive calls. Quicksort is inherently recursive, because each Quicksort operation must sort two sublists. Thus, there is no simple way to turn Quicksort into an iterative algorithm. However, Quicksort can be implemented using a stack to imitate recursion, as the amount of information that must be stored is small. We need not store copies of a subarray, only the subarray bounds. Furthermore, the stack depth can be kept small if care is taken on the order in which Quicksort's recursive calls are executed. We can also place the code for findpivot and partition inline to eliminate the remaining function calls. Note however that by not processing sublists of size nine or less as suggested above, about three quarters of the function calls will already have been eliminated. Thus, eliminating the remaining function calls will yield only a modest speedup.

Todo

type:Exercise

Consider the Quicksort implementation for this module, where the pivot is selected as the middle value of the partition. Give a permutation for the values 0 through 7 that will cause Quicksort to have its worst-case behavior.

There are a number of possible correct answers. To assess the answer, will need to run Quicksort over student's partition, and verify that at each step it will generate new partitions of size 6, 5, 4, 3, 2, then 1.

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