By Julienne Walker
License: Public Domain

I'm confident that you already know what sorting is. Not only is it a concept that saturates computer science, it's also pretty intuitive. Basically you take things that are out of order, and put them into order. Examples of sorting are a phone book being ordered by name, or a coin machine that puts the same type of coins into those neat little packages so that the bank will give you paper money in exchange. :-) The idea behind sorting is simple. It makes life easier if things are in order, not the least important is finding things. It'd be a nightmare to try and find somebody's phone number if the phone book wasn't ordered by name.

Beyond the practical uses for sorting, the algorithms are fundamental to computer science, and fluency in the techniques used by sorting algorithms provides insight into the craft of programming in many unexpected areas. Sorting also gives one a certain measure of confidence in designing and analyzing algorithms because most of the techniques used appear in the context of sorting as well.

Finally, it has been said that computers have spent more time sorting data than anything else, so it makes sense to have a strong background in the concept and theory of sorting, as well as a firm understanding of the different sorting methods and their relative strengths and weaknesses. Since sorting efficiently is surprisingly complicated, programmers should try to avoid adding to the time wasted by using a poor algorithm for the job.

Many generic sorting libraries already exist and are ready to be used, such as C's qsort, or C++'s std::sort or std::partial_sort. However, these general algorithms are not always the best option. Yes, they should be first on your list of possible solutions because you're basically given a tested and debugged algorithm that's easy to use. That's far superior to writing your own algorithm, testing it, and debugging it before it can be used with any confidence.

However, there are two distinct problems with the standard library sorting algorithms. First, they're designed to work for as many cases as possible, and may not be practical or efficient for your needs. For example, if qsort is too slow, how to do you fix the problem? You can find another library, but what are the chances that it's any better than qsort? What if you need a sorting algorithm that's tailored to your data? The chances of finding a library that meets your needs are slim indeed unless your needs are very common.

Second, libraries are generally designed as black boxes, so users of the functions are have no idea what algorithm is actually being used. This isn't as much of a problem in the standard C++ library, where the sorting algorithms have well defined performance attributes that must be met, but many people are surprised to learn that qsort may actually be implemented as anything as long as it eventually sorts the array correctly. The name suggests the quicksort algorithm, but in reality it could be bubble sort, or worse, a very poor implementation of quicksort that works for many cases but runs very very slow in common worst cases.

As such, it's beneficial to understand what algorithms are available, what their advantages and disadvantages are, and how to implement your own algorithm that's custom tailored to your data if the need arises. And you can brag that you know half a dozen sorting algorithms and sneer at your friends who can barely write bubble sort off the top of their heads. Bragging rights are a pretty darned good reason to learn something in my opinion. ;-)


Sorting Considerations

To sort or not to sort. That's the question, isn't it? Sorting an unordered array or list is time consuming, even if you do it efficiently, so it makes sense to avoid sorting when you can. That usually amounts to using a better data structure such as a hash table (where sorting probably isn't necessary), or a binary search tree (where items are sorted by default). Sometimes you don't even need to sort! It could be just as fast, or faster, to just brute force your way through things. It really isn't noticeably slower to use a sequential search on an array of about 50 items than it is to sort the same array and then use a binary search on it. To use a crude analogy, it doesn't take a wrecking ball to destroy a sand castle.

Okay, let's pretend that we want to sort. What about stability? Stability, you say? What's stability? No, it's not a part of my psychological diagnosis that's immediately preceeded by “lack of”. Stability is what happens when two items compare as equal when sorting. There are two kinds of sorting algorithms concerning stability, stable and unstable (obviously!). The only difference is how they treat duplicate items.

In a stable algorithm, two items that compare as equal will retain their original relation to each other as in the original unsorted array or list. This makes absolutely no difference if the items being compared consist only of, say, integral values. When 2 compares equal to 2, it really doesn't matter which 2 is placed before the other in the final sorted sequence. However, consider the more common situation where items consist of more than one value, and only one of those values is used as the item for sorting:


When sorting on the first half of the item, a stable sort has to give a result that retains the original order of the second half of the item. In other words, the second values have to be in sorted ascending order after the sort is finished, or the algorithm isn't stable:


An unstable sort has no such restriction. Of the algorithms that we'll cover in this tutorial, the following are naturally stable, but the others can be made stable at the cost of using more space or taking more time. For example, selection sort is not usually stable, but can easily be made so by inserting the new item (shifting all items after its location to make a hole, like insertion sort) instead of simply swapping items:

What about performance? This is a tricky area because sorting algorithms are all across the board. The performance complexities you'll see primarily are O(N2), O(Nlog N), and O(N). It should be relatively obvious that the upper bound for any sorting algorithm is infinite as long as it manages to sort the items eventually. Naturally, this suggests a realistic upper bound of O(N!), because sorting can be thought of as picking a sorted permutation of the items. However, the canonical awful algorithm, called bogosort, is probably the worst possible sorting algorithm. Bogosort is equivalent to throwing a deck of cards in the air, then testing to see if they're sorted when you pick them up at random. Bogosort could potentially never end due to the its random nature.

Lower bounds aren't as obvious. The lowest possible bound for most sorting algorithms is Ω(Nlog N). This is because most sorting algorithms use item comparisons to determine the relative order of items. Any algorithm that sorts by comparison will have a minimum lower bound of Ω(Nlog N) because a comparison tree is used to select a permutation that's sorted. A comparison tree for the three numbers 1,2, and 3 can be easily constructed:

                             1 < 2

               1 < 3                       1 < 3

       2 < 3           3,1,2       2,1,3           2 < 3

  1,2,3     1,3,2                            2,3,1     3,2,1

Notice how every item is compared with every other item, and that each path results in a valid permutation of the three items. The height of the tree determines the lower bound of the sorting algorithm. Because there must be as many leaves as there are permutations for the algorithm to be correct, the smallest possible height of the comparison tree is logn!, which is equivalent to Ω(Nlog N).

So are we doomed to Ω(Nlog N) as the best possible performance for sorting algorithms? Well, yes, if you use item comparisons to determine relative order. Therefore, the most general of sorting algorithms will be limited by this lower bound. However, it's possible to meet the safe lower bound of O(N) for sorting. We'll look at some of these algorithms later when we look at the counting and radix sorts.

It's easy to see that O(N) is a safe lower bound on general sorting algorithms for the obvious reasons; you can't sort what you haven't seen. Well, technically, you could rely on the law of quantum indeterminacy whereby not observing any of the items, you place them in a state of superposition and it's both sorted and unsorted, and the sorting process would be O(1). But as of yet, nobody has suggested a practical use for such a sorting algorithm, or even whether it would work correctly in anything but a theoretical sense. :-)



Before we start, there are a few helper functions that I'll be using in much of the code. Rather than keep repeating myself, I'll supply it all here. The first is a swap function that takes pointers to two items and swaps the items:

1 void jsw_swap ( int *a, int *b ) 2 { 3 int c = *a; 4 *a = *b; 5 *b = c; 6 }

Next, it's useful to be able to quickly determine if an array is already sorted or not for testing, and we'll use it as an optimization feature later, so here is a function that returns 1 if the array is sorted and 0 if it isn't:

1 int is_sorted ( int a[], int n ) 2 { 3 int i; 4 5 for ( i = 0; i < n - 1; i++ ) { 6 if ( a[i] > a[i + 1] ) 7 return 0; 8 } 9 10 return 1; 11 }

Finally, some of the algorithms work with linked lists. To save you the trouble of figuring out what I mean, here's the node and list structure that I built the lists with:

1 struct jsw_node { 2 int data; 3 struct jsw_node *next; 4 }; 5 6 struct jsw_list { 7 struct jsw_node *head; 8 int size; 9 };


Selection Sort

One of the two most obvious ways to sort data is to find the largest (or smallest) item in the collection, and then move it to the back. This straight selection process is repeated for all other items in the list except for the last of them, then again for all items except for the last two, and so on until all items are ignored. When all items are ignored, they are all in their sorted position, and the algorithm is complete. An example of this simple process follows:




The code to do this is as simple as finding the largest item in the array. A few notes should be made though. If the largest item is the current end of the array, no swap needs to be done, so we use a conditional test to avoid a function call and the operations of swapping an item with itself. This algorithm looks at every item in the array to find the largest. It has to, because no other order has been imposed on the array to make selection easier, so it's slow:

1 void jsw_selection ( int a[], int n ) 2 { 3 while ( --n > 0 ) { 4 int i, max = n; 5 6 for ( i = 0; i < n; i++ ) { 7 if ( a[i] > a[max] ) 8 max = i; 9 } 10 11 if ( max != n ) 12 jsw_swap ( &a[n], &a[max] ); 13 } 14 }

There are a few important little factoids about straight selection sort that you need to remember. First, all of the items have to be present before the sort can begin. That means that selection sort isn't a viable solution for sorting items as they come though an input stream or a random number generator. The array has to be completely filled before you can sort it. Second, straight selection sort is slow because the process of selection is slow. Since straight selection is an O(N) operation, and the algorithm does it approximately N times, straight selection sort is a quadratic sorting algorithm. Quadratic sorts are the slowest practical algorithms you'll see in the real world. Finally, selection sort is very efficient in terms of data movement. If comparisons are significantly cheaper than swaps, selection sort performs well.

As presented above, the straight selection sort is not stable. That is, if the items have more data attached to them that aren't a part of the comparison, the relative order of duplicate items is unspecified by the algorithm. To make the sort stable such that duplicate items retain their relative order, we can select the last of any duplicates and shift instead of swap Of course, this removes the primary benefit of selection sort, lack of data movement, because we're moving items around in the shift:

1 void jsw_selection ( int a[], int n ) 2 { 3 while ( --n > 0 ) { 4 int i, max = n; 5 6 for ( i = 0; i < n; i++ ) { 7 if ( a[i] >= a[max] ) 8 max = i; 9 } 10 11 if ( max != n ) { 12 int save = a[max]; 13 14 for ( i = max; i < n; i++ ) 15 a[i] = a[i + 1]; 16 17 a[n] = save; 18 } 19 } 20 }

Let's look at how that works with a quick trace by adding some extra data to each item and throwing in a few duplicates. Notice how the secondary numbers are in the same order after the sort as before the sort. For example, 69,4 is in front of 69,5 both before and after the algorithm touches them:





Heap Sort

The straight selection sort is probably the simplest sorting algorithm, but that simplicity is really the only benefit. Selection sort has quadratic time complexity, which is the worst complexity of the practical sorting algorithms that you'll see in the real world. There are a handful of quadratic sorting algorithms in common use, and while selection sort isn't the worst of them, you're really comparing the lowest common denominators with each other. Maybe we should work smarter instead of trying to pick the best of the worst. ;-)

The very name of selection sort suggests a way to improve the algorithm. Since the biggest problem with selection sort is the linear time it takes to select the largest item, we can find the most improvement by solving the selection problem more efficiently. In real code, we would do that with a priority queue, where items are ordered so that extracting the largest item is quick and easy. The code to do that if we assume an existing priority queue library is simplicity itself:

1 void jsw_selection ( int a[], int n ) 2 { 3 pqueue pq; 4 int i; 5 6 for ( i = 0; i < n; i++ ) 7 pq_insert ( pq, a[i] ); 8 9 while ( !pq_empty ( pq ) ) 10 a[--n] = pq_max ( pq ); 11 }

Okay, but what if that nifty little library doesn't already exist? Well, we need to do it manually. The best known priority queue implementation is implemented with a max heap. The heap data structure has two very simple rules. It's a complete binary tree where the children of a node cannot be larger (or smaller if you want a min heap) than their parent. If you have a valid max heap then the largest item is the root of the tree. This structure can be easily converted into an array because the tree is complete, so there are no gaps:


 is equivalent to


                    /              \

                500                   724

           /           \            /     \

       467               464    334         478

     /     \           /

 358         41    169

By converting an array into a heap structure, we can then easily extract the largest item from index 0. After doing that, the heap needs to be fixed so that it's still a valid heap again, and the process is repeated. This method of sorting is called heap sort, and it's one of the faster sorting algorithms with a worst case that has the same time complexity as the average case! By working smarter with the selection problem, we've turned one of the worst sorting algorithms into one of the best. Not bad, eh? :-)

Cool, so how do we do all of that junk? Well, naturally we need to figure out how to force an array to have the heap property. Basically, we can say that for each index i, the child nodes are at i * 2 + 1 and i * 2 + 2. In the above tree, we can see that 467 is a child of 500. In the array, 500 is at index 1 and 1 * 2 + 1 is index 3, which is where 467 is in the array. Try this with any of the indices and you'll see that this property holds for all of them. Also notice that the nodes with children are roughly the first half of the array and the nodes without children are the last half. So to build a heap, all we need to do is force the heap property from the lowest nodes to the root. Let's look at an example:



                 /              \

              67                  34

         /          \           /    \

       0              69     24        78

     /   \          /

 58        62    64

To build a heap out of this array (I'll also include a tree representation to make things more obvious), we need to work with subtrees first. We start with 69 and its one child. Since 64 is smaller than 69, the heap property holds. Then we move to 0 and its children. 62 is the largest of the three, so we swap 62 and 0. Note that the relative order of the children is irrelevant:



                 /              \

              67                  34

         /          \           /    \

       62             69     24        78

     /    \         /

 58         0    64

Now we move up a level and all the way to the right. In the array we're just decrementing the parent index though. At 34, we test the 3 nodes and find that 78 is the largest. So we swap 78 and 34 to restore the heap property:



                 /              \

              67                  78

         /          \           /    \

       62             69     24        34

     /    \         /

 58         0    64

Now we go left to 67 and its two children. Since 69 is the greatest of the children and is greater than 67, we swap 67 and 69:



                 /              \

              69                  78

         /          \           /    \

       62             67     24        34

     /    \         /

 58         0    64

Finally, at the root of the tree, testing 1 and its children we find that 78 is greater than both 1 and 69, so we make 78 the new root of the tree. Uh oh, this isn't a valid heap though. 78 is greater than 1, but 1 isn't greater than either 24 or 34, so we've got a problem. It appears that we can't just do local changes to a parent and its two children, we need to push the parent down as long as it violates the heap property. We can do this because we build a heap from the bottom up, so the lower subtrees are always valid heaps unless a change higher up the tree violates it. Let's do the final step by fixing 78's right subtree:



                 /             \

              69                 34

         /          \          /    \

       62             67    24        1

     /    \         /

 58         0    64

Now we have a valid heap! Let's write a jsw_do_heap function that walks down the tree, making sure that the heap property is met. This is a fairly obvious method. The outer loop of jsw_do_heap handles cases where multiple parents need to be fixed down the tree. The first step inside the loop finds the largest child, the second step swaps the largest child and the parent if the child is larger. Notice how the i * 2 + 1 trick is used to find the first child. jsw_heapsort is just a translation of the priority queue pseudocode:

1 void jsw_do_heap ( int a[], int i, int n ) 2 { 3 int k = i * 2 + 1; 4 5 while ( k < n ) { 6 if ( k + 1 < n && a[k] < a[k + 1] ) 7 ++k; 8 9 if ( a[i] < a[k] ) 10 jsw_swap ( &a[i], &a[k] ); 11 12 k = ++i * 2 + 1; 13 } 14 } 15 16 void jsw_heapsort ( int a[], int n ) 17 { 18 int i = n / 2; 19 20 while ( i-- > 0 ) 21 jsw_do_heap ( a, i, n ); 22 23 while ( --n > 0 ) { 24 jsw_swap ( &a[0], &a[n] ); 25 jsw_do_heap ( a, 0, n ); 26 } 27 }

This is a surprisingly simple algorithm for how good heap sort is supposed to be, but it's not very fast, and we can do much better with only minor changes to jsw_do_heap. The current algorithm walks down every node in the tree until the end of the array. That means we're doing a lot of unnecessary work if only one push down is required in an array of 1,000,000 items. We also don't need to call jsw_swap in the loop. We can simply save the first root that we're pushing down, move children into its place, and after the loop is finished, put that saved node into its final resting place:

1 void jsw_do_heap ( int a[], int i, int n ) 2 { 3 int k = i * 2 + 1; 4 int save = a[i]; 5 6 while ( k < n ) { 7 if ( k + 1 < n && a[k] < a[k + 1] ) 8 ++k; 9 10 if ( save >= a[k] ) 11 break; 12 13 a[i] = a[k]; 14 k = ++i * 2 + 1; 15 } 16 17 a[i] = save; 18 }

That's much faster than our last attempt, but because we're only incrementing i by 1 each time, we're basically fixing the entire heap when we only need to fix the subtree that we swapped with, not both subtrees. So by making the simple change of setting i to k instead of incrementing it, we improve the speed of the algorithm by several orders of magnitude. jsw_heapsort doesn't change throughout all of these improvements, and we come to the final (and fastest) incarnation of heapsort for this tutorial:

1 void jsw_do_heap ( int a[], int i, int n ) 2 { 3 int k = i * 2 + 1; 4 int save = a[i]; 5 6 while ( k < n ) { 7 if ( k + 1 < n && a[k] < a[k + 1] ) 8 ++k; 9 10 if ( save >= a[k] ) 11 break; 12 13 a[i] = a[k]; 14 i = k; 15 k = i * 2 + 1; 16 } 17 18 a[i] = save; 19 }

Okay, I'm a liar. :-) We'll look at another alternative before moving on. A lot of people are surprised to learn that heap sort can be implemented any number of ways. The heap property is a very flexible rule when it comes to how you want to enforce it, so there can be a lot of variation in how heap sort is implemented. The best way to find alternative implementations is to think about how heaps work and look for patterns. One pattern in particular that we will look at is the relationship between each index and its child indices. If you think about the simple calculation for finding a child, you'll come up with the following pattern:

    |67|  |0|69|
       |34|    |24|78|
          |0|        |58|62|
            |69|           |64|

This shows the parents and their direct children. You can verify that the right children go with the right parents by using the i * 2 + 1 trick. 41 is index 0, 0 * 2 + 1 is 1 and 0 * 2 + 2 is 2. Okay, that's an easy one, how about the last parent and child? 69 is index 4, 4 * 2 + 1 is 9, which is the index for 64. So we know that the pattern matches what we were doing by finding k with i * 2 + 1. But since the parent index increments by 1 each time and the index of the first child increments by 2, we can work directly with the pattern by using two indices (once again, jsw_heapsort remains unchanged):

1 #define jmax(p,q) ( *(p) > *(q) ? (p) : (q) ) 2 #define maxof3(p,q,r) jmax ( (p), jmax ( (q), (r) ) ) 3 4 void jsw_do_heap ( int a[], int i, int n ) 5 { 6 int k; 7 8 for ( k = i + 1; k < n; i++, k += 2 ) { 9 if ( k == n - 1 ) { 10 if ( a[k] > a[i] ) 11 jsw_swap ( &a[i], &a[k] ); 12 } 13 else 14 jsw_swap ( &a[i], maxof3 ( &a[i], &a[k], &a[k + 1] ) ); 15 } 16 }

But this is really two steps backward since it's even slower than our first attempt at heapsort, and implementing the same optimizations that we used before would make the code much more complicated, so this is really a substandard solution even though it's intellectually interesting to play around with alternatives. :-) I like to think that if you understand the variations we've looked at, you'll have a good feel for heap sort.

One problem that heap sort has (though you won't see it here) is that the traditional algorithm assumes 1-based array indices, while C uses 0-based arrays. That's proven to be a very frustrating problem because getting a 1-based algorithm to work with 0-based arrays can be tricky. As such, you'll probably see different tricks that avoid the issue, like ignoring the 0th index, or shifting the array so that the 0th index actually refers to one element in front of the array. The former is awkward and the latter is undefined by C. Other attempts are just wrong in subtle ways, and while they may seem to work most of the time, they'll break, usually at the worst possible time. ;-) No worries here though, all of my implementations are written specifically with 0-based arrays in mind. You're welcome.


Insertion Sort

The other most obvious way to sort a collection of items is by insertion. Insertion takes each new item, then locates the current ordered position in the collection and inserts the item there. The common example is sorting a bridge hand, but I don't play bridge and I don't know anyone that does, so an execution trace on our list of numbers as an example would fare better:

  1)  |41|
  2)  |41|67|
  3)  |34|41|67|
  4)  |00|34|41|67|
  5)  |00|34|41|67|69|
  6)  |00|24|34|41|67|69|
  7)  |00|24|34|41|67|69|78|
  8)  |00|24|34|41|58|67|69|78|
  9)  |00|24|34|41|58|62|67|69|78|
 10)  |00|24|34|41|58|62|64|67|69|78|

Each new item is moved into the correct position as it arrives. Unlike sorting by selection, this suggests that sorting by insertion is well suited to sorting an unknown number of items, and that the items don't need to be present before sorting can begin. So acquiring items through a stream or a generator is perfectly acceptable when sorting by insertion.

However, that very sequential nature suggests that sorting by insertion has potential performance issues when used with arrays. Insertion into the middle of an array requires that a hole be created by shifting all items after the hole to the right. For a large array, that can be incredibly slow and expensive. On the other hand, sorting by insertion into a linked list can be far more efficient in theory because the shifting step is not required.

Note that sorting by insertion can be done in-place without using any extra memory. Using insertion sort on an existing array of items would result in the following execution:

  1)  |41>67|34|00|69|24|78|58|62|64|
  2)  |41|67>34|00|69|24|78|58|62|64|
  3)  |34|41|67>00|69|24|78|58|62|64|
  4)  |00|34|41|67>69|24|78|58|62|64|
  5)  |00|34|41|67|69>24|78|58|62|64|
  6)  |00|24|34|41|67|69>78|58|62|64|
  7)  |00|24|34|41|67|69|78>58|62|64|
  8)  |00|24|34|41|58|67|69|78>62|64|
  9)  |00|24|34|41|58|62|67|69|78>64|
 10)  |00|24|34|41|58|62|64|67|69|78|

The code to do this is pretty simple. All we need to do is pick each item in the array, then make a hole where it needs to go and insert it. In fact, insertion sort is considered to be one of the simplest sorting algorithms both in concept and in implementation:

1 void jsw_insertion ( int a[], int n ) 2 { 3 int i; 4 5 for ( i = 1; i < n; i++ ) { 6 int j, save = a[i]; 7 8 for ( j = i; j >= 1 && a[j - 1] > save; j-- ) 9 a[j] = a[j - 1]; 10 11 a[j] = save; 12 } 13 }

Unfortunately, being one of the simplest sorting algorithms also means that insertion sort is also one of the least efficient. It's in the class of sorting algorithms that have quadratic time complexity, which means that it doesn't scale well to large arrays. However, insertion sort has one really awesome feature that makes it useful as a part of more sophisticated algorithms: insertion sort is blazing fast on arrays that are sorted or almost sorted. This makes it ideal for finishing up a partial quicksort, which we'll see shortly.

Insertion sort on a linked list is almost as simple as insertion sort on an array. With linked lists we may not be able to go backward, but that's hardly a stumbling block since we can easily save the sorted items in a temporary list by extracting the nodes from the original list, inserting them into the temporary list, then pointing the original list pointer back to the temporary list before returning. The temporary list only has the overhead of an extra variable, so it's space efficient:

1 void jsw_insertion ( struct jsw_list *list ) 2 { 3 struct jsw_node result = {0}; 4 struct jsw_node *i, *step, *next; 5 6 for ( step = list->head; step != NULL; step = next ) { 7 next = step->next; 8 9 for ( i = &result; i->next != NULL; i = i->next ) { 10 if ( i->next->data > step->data ) 11 break; 12 } 13 14 step->next = i->next; 15 i->next = step; 16 } 17 18 list->head =; 19 }


Shell Sort

While straight insertion is particularly efficient on already sorted or almost sorted arrays or lists, it's still a quadratic sort because the overall result only sorts the items by one position at a time. Now, while this may sound like a given, it's possible to force the algorithm to move items over wider ranges by dividing the array into multiple interleaved groups. For example, dividing the insertion sort example into two groups of five, then sorting each corresponding value in those groups, then combining the groups and sorting the one group results in the following execution:

  1)  |41|67|34|00|69|   |24|78|58|62|64|
  2)  |24|67|34|00|69|   |41|78|58|62|64|
  3)  |24|67|34|00|69|   |41|78|58|62|64|
  4)  |24|67|34|00|69|   |41|78|58|62|64|
  5)  |24|67|34|00|64|   |41|78|58|62|69|
  6)  |00|24|67|34|64|41|78|58|62|69|
  7)  |00|24|34|64|67|41|78|58|62|69|
  8)  |00|24|34|41|64|67|78|58|62|69|
  9)  |00|24|34|41|58|64|67|78|62|69|
 10)  |00|24|34|41|58|62|64|67|78|69|
 11)  |00|24|34|41|58|62|64|67|69|78|

Hmm, not much better, is it? How about we divide the array into four groups instead of two? That might be better:

  1)  |41|67|   |34|00|69|   |24|78|   |58|62|64|
  2)  |24|67|   |34|00|69|   |41|78|   |58|62|64|
  3)  |24|00|   |34|62|69|   |41|67|   |58|78|64|
  4)  |24|00|   |34|62|64|   |41|67|   |58|78|69|
  5)  |00|24|34|62|64|41|67|58|78|69|
  6)  |00|24|34|41|62|64|67|58|78|69|
  7)  |00|24|34|41|58|62|64|67|78|69|
  8)  |00|24|34|41|58|62|64|67|69|78|

Sorting interleaved groups is the idea behind the diminishing increment sort, better known by the popular name Shell sort. Donald Shell capitalized on the fact that sorting by insertion is very fast for small arrays that are almost sorted. He then introduced the concept of breaking the array into increasing groups of smaller numbers (usually called the h-increment) of items. As long as the groups eventually shrink down to N groups of 1, the final step is a straight insertion and the algorithm terminates successfully:

1 void jsw_shellsort ( int a[], int n ) 2 { 3 int i, h = 1; 4 5 while ( h <= n / 9 ) 6 h = 3 * h + 1; 7 8 while ( h > 0 ) { 9 for ( i = h; i < n; i++ ) { 10 int j, save = a[i]; 11 12 for ( j = i; j >= h && a[j - h] > save; j -= h ) 13 a[j] = a[j - h]; 14 15 a[j] = save; 16 } 17 18 h /= 3; 19 } 20 }

The only difference between this code and the insertion sort is that we've added an extra loop around the insertion sort that handles the step increments and replaced references to 1 with h. By moving items that are separated by a large number of items, they get closer to their final resting place faster. The examples above don't do the algorithm justice, but until quicksort was developed, Shell sort was the fastest sorting algorithm available.

A careless choice of values for the h-increments, such as powers of two, could result in strikingly bad performance, while a seemingly arbitrary sequence can result in very good performance. To date, nobody has determined what the very best sequence is. Two sequences known to be good are the square of the Fibonacci sequence (1,4,9,25,64,169,441,...), and a sequence recommended by Donald Knuth that's fast and easy to calculate (1,4,13,40,121,364,1093,3280,9841...). While there are better sequences, these two are not substantially worse, and they have the benefit of being easy to create with minimal cost. The function above uses Knuth's sequence.

The Shell sort is simple to understand, easy to implement, and has good performance for small to medium arrays, but its time complexity is difficult to analyze and depends heavily on the sequence of h-increments chosen. In general, Shell sort should be the first sort used because it's easier to implement than the higher order sorting algorithms (with the possible exception of heap sort) and has good performance. If Shell sort turns out to be too slow, then some of the faster algorithms can be considered.


Bubble Sort

Sorting by exchanging is just that. Pairs of items that are out of order are exchanged repeatedly until no more pairs exist. When no more pairs exist that are out of order, the array must be completely sorted. Sorting by exchanging is particularly dangerous because the efficient solutions aren't nearly as obvious as the hideously inefficient ones. This is true for all sorting categories, but sorting by exchanging has the great honor of boasting the worst sorting algorithm that you might see in the real world (if you're unlucky). Here is a look at the most obvious algorithm:

  1)  |41|67|34|00|69|24|78|58|62|64|
  2)  |41>34|67|00|69|24|78|58|62|64|
  3)  |41|34>00|67|69|24|78|58|62|64|
  4)  |41|34|00|67>24|69|78|58|62|64|
  5)  |41|34|00|67|24|69>58|78|62|64|
  6)  |41|34|00|67|24|69|58>62|78|64|
  7)  |41|34|00|67|24|69|58|62>64|78|
  8)  >34|41|00|67|24|69|58|62|64|78|
  9)  |34>00|41|67|24|69|58|62|64|78|
 10)  |34|00|41>24|67|69|58|62|64|78|
 11)  |34|00|41|24|67>58|69|62|64|78|
 12)  |34|00|41|24|67|58>62|69|64|78|
 13)  |34|00|41|24|67|58|62>64|69|78|
 14)  >00|34|41|24|67|58|62|64|69|78|
 15)  |00|34>24|41|67|58|62|64|69|78|
 16)  |00|34|24|41>58|67|62|64|69|78|
 17)  |00|34|24|41|58>62|67|64|69|78|
 18)  |00|34|24|41|58|62>64|67|69|78|
 19)  |00>24|34|41|58|62|64|67|69|78|

As you can clearly see, this particular exchange sort is exceptionally inefficient, even when compared to straight selection and insertion, which have similar time complexity properties, even when the execution left out any of the steps that didn't result in a swap. Hey, I don't want to spend half my life typing out those executions, so I made it shorter. In the process I made the algorithm look better than it is, and it STILL looks bad. ;-)

Unfortunately, because the algorithm is simple and easy to understand this so-called “bubble sort” is commonly taught to beginning programmers as the first sorting algorithm, and often as a first algorithm entirely. I think that's a shame, because bubble sort only has practical use in very specific and unlikely circumstances, and for everything else it just sucks too much. For comparison and completeness, I'll provide an implementation of the bubble sort as it's usually taught:

1 void jsw_bubblesort ( int a[], int n ) 2 { 3 int i; 4 5 for ( i = 0; i < n; i++ ) { 6 int j; 7 8 for ( j = n - 1; j > 0; j-- ) { 9 if ( a[j] < a[j - 1] ) 10 jsw_swap ( &a[j], &a[j - 1] ); 11 } 12 } 13 }

Well, you have to admit that it's simple. Of course, once you optimize it a bit, that simplicity starts to go away. A flag for testing whether a swap has been made is critical for bubble sort, because otherwise it would make several unnecessary passes over the collection before terminating. While strictly optional, a test for early completion is considered a mandatory optimization to the bubble sort algorithm. No further optimizations are necessary as they won't improve the performance significantly, and simpler algorithms, such as insertion sort, will almost always outperform it. Here's the “optimized” bubble sort:

1 void jsw_bubblesort ( int a[], int n ) 2 { 3 int i; 4 5 for ( i = 0; i < n; i++ ) { 6 int j, done = 1; 7 8 for ( j = n - 1; j > 0; j-- ) { 9 if ( a[j] < a[j - 1] ) { 10 jsw_swap ( &a[j], &a[j - 1] ); 11 done = 0; 12 } 13 } 14 15 if ( done ) 16 break; 17 } 18 }

Bubble sort is a quadratic algorithm, but it's frighteningly inefficient even if you only compare it with other quadratic algorithms. Words can't describe my loathing for this algorithm, and I've gotten into very heated debates about whether it should ever be used or not, even as an introductory algorithm. My recommendation: forget bubble sort exists. You'll be better off for it. :-)



Bubble sort is unmercifully slow. It's the worst possible sorting algorithm that you'll see in the real world, and even then only because the author of the code probably doesn't know any better. Fortunately, we can improve bubble sort in much the same way that we improved the straight selection sort. But since we don't have a neat and tidy concept to improve, along with a way to improve it (ie. the selection problem), we need another way to improve the performance of exchanging.

The good news is that we can use recursion to split the array at a median value at each step by moving the median value to its sorted position, then recursively sort each subarray the same way in a very good algorithm called quicksort. By dividing the problem at the median each time, we can guarantee that this part of the algorithm is O(log N). Now it's just a simple matter of figuring out a good way to sort the subarrays that doesn't take longer than O(N) and we'll have an optimal comparison sort.

Okay, maybe it's not that simple. :-) We have two immediate problems. First we need a way to partition the array. Partitioning is forcing a median value to its sorted position while making sure that all items less than the median come before it and all items greater than or equal to the median come after it. Then we need a way to find the median value without taking too much time. To really find the median requires at least an O(N) algorithm on top of the partitioning which will at best be O(N). Now, O(N) + O(N) is still O(N), but the algorithm will run more slowly than if we only had one O(N) part, and we want this to be as fast as possible.

Rather than waste cycles trying to find the true median, we'll just guess. Our first attempt at guessing is the simplest. We just take the first item in the subarray and pretend that it's the median value. :-) This could be a serious problem if the first item is the smallest item in the array, but we'll look at that a bit later. One problem at a time. Assuming a partition function, we can easily write a recursive function to partition, then recursively do the same thing on each subarray as long as the subarray has more than one item:

1 void jsw_quicksort_r ( int a[], int first, int last ) 2 { 3 if ( first < last ) { 4 int pivot = jsw_partition ( a, first, last ); 5 jsw_quicksort_r ( a, first, pivot - 1 ); 6 jsw_quicksort_r ( a, pivot + 1, last ); 7 } 8 } 9 10 void jsw_quicksort ( int a[], int n ) 11 { 12 jsw_quicksort_r ( a, 0, n - 1 ); 13 }

Let's look at a sample execution to see how this function could possibly work. It's not obvious at first unless you're familiar with recursion and can see how partitioning forces an item to its sorted position recursively results in a sorted file. In this example, we take the first item in each subarray and use it as the median and ignore cases where the subarray is too small to partition:


The base case for recursion is when the indices first and last cross. So we stop the recursion if we get to a subarray with only a single item. jsw_quicksort passes an inclusive array to jsw_quicksort_r, which basically means that both a[first] and a[last] are valid items. This is different from most of our other algorithms (such as heap sort) where last is the first index after the last element of the array. This is for the benefit of jsw_partition, which is more easily written if both first and last refer to valid indices. The pivot variable refers to the index of the median value so there's no need to include it in the recursive calls as it's already sorted after partitioning. Here's the code that performs the partition magic:

1 int jsw_partition ( int a[], int first, int last ) 2 { 3 int pivot = first; 4 5 while ( first < last ) { 6 if ( a[first] <= a[last] ) 7 jsw_swap ( &a[pivot++], &a[first] ); 8 9 ++first; 10 } 11 12 jsw_swap ( &a[pivot], &a[last] ); 13 14 return pivot; 15 }

Partitioning an array is surprisingly difficult to get right. The algorithm above is a relatively safe, simple, and efficient one that gets the job done in O(N) time. The idea is to walk across the array, swapping the pivot with items that are less than the pivot as we go, and resetting the pivot to the next item when we swap. When we get to the end of the array, the current pivot is the median value that's in its sorted position. Note that the median isn't necessarily the pivot that we chose at the beginning of the partition.

That's all there is to quicksort! Once you have a partition algorithm that works, writing the code to do a quicksort is trivial, and then you have the fastest non-hybrid sorting algorithm at your disposal. But we're not done yet because we can improve the algorithm. There are a few problems that need to be fixed and optimizations that can make quicksort even faster.

The first problem is the biggest. If the pivot value chosen in jsw_partition is always the smallest item in the array, such as when the array is already sorted, no swaps will be made and the median will be the first item in the array. This results in a vastly unbalanced split where the left half has one item and the right half has the rest of the array. This is a degenerate case where quicksort really isn't that quick, with a quadratic time complexity.

There are two widely accepted ways to minimize the chances of hitting a degenerate case. First, we can use a random number generator to pick which value in the array to use as a pivot. Then we guarantee that nothing untoward can happen by swapping that value with the first item in the array because the partition algorithm assumes that pivot starts at first. Otherwise we would end up walking off the end of the array, and that's never a good thing. ;-) The changes are minimal:

1 int jsw_partition ( int a[], int first, int last ) 2 { 3 int pivot = first + rand() % ( last - first + 1 ); 4 5 jsw_swap ( &a[pivot], &a[first] ); 6 pivot = first; 7 8 while ( first < last ) { 9 if ( a[first] <= a[last] ) 10 jsw_swap ( &a[pivot++], &a[first] ); 11 12 ++first; 13 } 14 15 jsw_swap ( &a[pivot], &a[last] ); 16 17 return pivot; 18 }

Notice that the code to find a random index within the range is somewhat complicated. We need to take into account that first is not always zero, so we use last - first + 1 to find the number of items in the subarray (we add 1 because last is the last valid item, not one beyond the last valid item), then use that as the modulus for rand (ignoring the issues with rand). Finally, because the resulting number is ranged from zero to the number of elements, we add first to fit it within the range of [first,last] as a valid index.

The second, and preferred, option is to take the median of three items picked at random from the array (though the first, last, and middle are most common), and use that as the pivot. The process is the same, except instead of picking a random index and swapping its item with a[first], we pick the median of a[first], a[mid], and a[last], then swap that median with a[first]:

1 int jsw_partition ( int a[], int first, int last ) 2 { 3 int pivot, mid = ( first + last ) / 2; 4 5 /* Largest of two */ 6 pivot = a[first] > a[mid] ? first : mid; 7 8 /* Smallest of remaining */ 9 if ( a[pivot] > a[last] ) 10 pivot = last; 11 12 jsw_swap ( &a[pivot], &a[first] ); 13 pivot = first; 14 15 while ( first < last ) { 16 if ( a[first] <= a[last] ) 17 jsw_swap ( &a[pivot++], &a[first] ); 18 19 ++first; 20 } 21 22 jsw_swap ( &a[pivot], &a[last] ); 23 24 return pivot; 25 }

The next improvement is an optimization that doesn't touch jsw_partition (yay!). Since it's really a waste to recurse on small subarrays, we can set a threshold for when to stop calling jsw_quicksort_r recursively. As it is, we're recursing until first and last cross, which is equivalent to a threshold of 0. But if we set a cutoff of, say, 15 or 20, we can mostly sort a large array, then hand it off to insertion sort, which is much faster for small or nearly sorted arrays. This suggests two possible variations. First, we could call insertion sort inside jsw_quicksort_r, if we get to the cutoff. Robert Sedgewick studied this thoroughly and discovered that it's slightly more efficient to call insertion sort after quicksort finishes, which brings us to this (where THRESHOLD is a macro with whatever cutoff value you choose):

1 void jsw_quicksort_r ( int a[], int first, int last ) 2 { 3 if ( last - first > THRESHOLD ) { 4 int pivot = jsw_partition ( a, first, last ); 5 jsw_quicksort_r ( a, first, pivot - 1 ); 6 jsw_quicksort_r ( a, pivot + 1, last ); 7 } 8 } 9 10 void jsw_quicksort ( int a[], int n ) 11 { 12 jsw_quicksort_r ( a, 0, n - 1 ); 13 jsw_insertion ( a, n ); 14 }

The next optimization that we'll look at handles the problem of already sorted subarrays. If the subarrays are already sorted or have a large number of duplicates, quicksort will take its sweet time going through them all. There's no point in using a sophisticated sort on an already sorted subarray, and we can make quicksort consistently fast by testing if the subarray is already sorted and doing nothing if it is. The change is made to jsw_quicksort_r, and we can use the is_sorted function given earlier:

1 void jsw_quicksort_r ( int a[], int first, int last ) 2 { 3 if ( last - first > THRESHOLD ) { 4 int pivot; 5 6 if ( is_sorted ( a, first, last ) ) 7 return; 8 9 pivot = jsw_partition ( a, first, last ); 10 jsw_quicksort_r ( a, first, pivot - 1 ); 11 jsw_quicksort_r ( a, pivot + 1, last ); 12 } 13 }

You might be thinking that calling a linear function might have performance issues, but in practice it actually makes quicksort faster. But that's still not the best implementation! It's fast, yes. It's definitely a very good quicksort, and a good compiler will give you efficient machine code for it. But a not so good compiler probably won't take care of the tail recursion in jsw_quicksort_r. Removing recursion is the last optimization for quicksort, and I'll leave it to you because I'm evil. >:)

As an aside, the worst case of quicksort is still O(N2) even though we've shrunk the possibility of encountering it to virtually nothing. However, by putting together several sorting algorithms into a hybrid “super-algorithm”, we can have our cake and eat it too. Here are the common players:

Heap sort: O(Nlog N) in the average and worst cases, but still slower than quicksort due to more comparisons.

Quicksort: O(Nlog N) in the average case, but O(N2) in the worst case. Fast as greased lightning if the worst case isn't encountered though.

Insertion sort: O(N2) in the average case, but O(N) if the array is already sorted or almost sorted.

Here's what we're going to do. We'll use quicksort until it's obvious that we're nearing a degenerate case. We can test for that by figuring out the depth of recursion that qualifies as O(log N). When we get to that depth, we're starting to leave the boundaries of quicksort's average case. Once we hit that cutoff, we then switch to heap sort because it's guaranteed to be O(Nlog N). By doing this, we never leave a complexity of O(Nlog N). We can also implement all of the optimizations for quicksort such as the threshold and insertion sort at the end, as well as already sorted subarrays. The resulting algorithm is called introsort, short for introspective sort because it's smart enough to change how it works when things start to get hairy:

1 void jsw_introsort_r ( int a[], int first, int last, int depth ) 2 { 3 if ( last - first > THRESHOLD ) { 4 if ( depth == 0 ) 5 jsw_heapsort ( &a[first], last - first + 1 ); 6 else { 7 int pivot; 8 9 if ( is_sorted ( a, first, last ) ) 10 return; 11 12 pivot = jsw_partition ( a, first, last ); 13 jsw_introsort_r ( a, first, pivot - 1, depth - 1 ); 14 jsw_introsort_r ( a, pivot + 1, last, depth - 1 ); 15 } 16 } 17 } 18 19 void jsw_introsort ( int a[], int n ) 20 { 21 jsw_introsort_r ( a, 0, n - 1, (int)( 2 * log ( n ) ) ); 22 jsw_insertion ( a, n ); 23 }

Heap sort and quicksort remain unchanged, as does insertion sort, but we need to be careful to only call heapsort on the correct subarray, which means passing an array slice. Introsort only puts them together into an algorithm that guarantees good performance in all cases. But we can improve even this beautiful algorithm by paring it down to trim the fat. Since there's a logarithmic cutoff on the depth of the recursion, we can remove the left side recursion in favor of a loop. This change helps to negate any overhead that introsort might have so that it's that much closer to a plain quicksort in the average case:

1 void jsw_introsort_r ( int a[], int first, int last, int depth ) 2 { 3 while ( last - first > THRESHOLD ) { 4 if ( depth == 0 ) 5 jsw_heapsort ( &a[first], last - first + 1 ); 6 else { 7 int pivot; 8 9 if ( is_sorted ( a, first, last ) ) 10 return; 11 12 pivot = jsw_partition ( a, first, last ); 13 jsw_introsort_r ( a, pivot + 1, last, depth - 1 ); 14 last = pivot - 1; 15 } 16 } 17 }

Note that heap sort isn't a requirement. You can use any sorting algorithm that guarantees at least O(Nlog N) performance, such as merge sort or radix sort. However, heap sort has an added benefit that it sorts in-place while merge sort and radix sort require extra memory to temporarily store subarrays. The same goes for insertion sort. Any sort that's most efficient with almost sorted arrays will work just as well.


Merge Sort

One of the first sorting methods suggested was by John Von Neumann. The concept is simple: split an array into two parts, sort those two parts, then merge them together. This is generally done recursively, where each subarray is then separated into two smaller subarrays, those subarrays are separated into four smaller subarrays, and so on until you have N subarrays of 1 item. At that point, you merge all of the subarrays together and the array is sorted. The basic code for this process is similar to quicksort:

1 void jsw_mergesort_r ( int a[], int first, int last ) 2 { 3 if ( first < last - 1 ) { 4 int mid = ( first + last ) / 2; 5 jsw_mergesort_r ( a, first, mid ); 6 jsw_mergesort_r ( a, mid, last ); 7 jsw_merge ( a, first, mid, last ); 8 } 9 } 10 11 void jsw_mergesort ( int a[], int n ) 12 { 13 jsw_mergesort_r ( a, 0, n ); 14 }

It seems as if merge sort is the inverse of quicksort. Where quicksort first partitions and then recurses, merge sort recurses and then merges, which is a reasonable observation. As with quicksort, if the merge operation is assumed to work properly, the algorithm is easy to visualize, even if it's difficult to describe. Also, while the recursive scheme works for arrays in fast memory, merge sort's claim to fame is that it can easily be modified to handle sequential data such as from a stream or a generator. As such, merge sort is excellent for sorting large linked lists, and it's often the algorithm of choice for external sorting of files that are too large to fit in fast memory all at one time.

It's a good idea, as with quicksort's partition, to choose the simplest merge algorithm available. The process of merging is one of the deceptively simple algorithms in computer science. Given two arrays, compare each item with the other as they come and assign the smallest to the final array. When one of the sequences is exhausted, copy the other in bulk to the final array. Here is one algorithm, called straight merging, that's easy to understand and works well in many cases:

1 void jsw_merge ( int a[], int first, int mid, int last ) 2 { 3 int i = first, j = mid, k = 0; 4 int *save = malloc ( ( last - first ) * sizeof *save ); 5 6 while ( i < mid && j < last ) { 7 if ( a[i] <= a[j] ) 8 save[k++] = a[i++]; 9 else 10 save[k++] = a[j++]; 11 } 12 13 while ( i < mid ) 14 save[k++] = a[i++]; 15 16 while ( j < last ) 17 save[k++] = a[j++]; 18 19 for ( i = 0; i < ( last - first ); i++ ) 20 a[first + i] = save[i]; 21 22 free ( save ); 23 }

The merge simply works with two subarrays: a left subarray from first to mid, and a right subarray from mid to last. As long as there are items in both subarrays, we copy the smallest of them to the temporary buffer and move only the index for the buffer and the subarray forward. After one of the subarrays is exhausted, we copy the remaining subarray to the buffer, then copy the buffer to the original array and terminate (but not before freeing the memory we allocated). Here's an execution trace of the entire sorting algorithm (leaving out the base cases because they do nothing):


It should be immediately obvious that for the merge algorithm to work, extra space equal to the size of the array is required. If we didn't use a temporary buffer, we might copy over the current item in one of the subarrays and lose it forever. That's not conducive to a correct sorting algorithm, so we live with the extra memory for a buffer.

The extra space requirement can be removed so that the merge runs in-place, but the resulting algorithm is both complicated and slower, so extra space is a necessary evil for an efficient merge sort using arrays. This is why the merge sort is well suited to linked lists; an item can simply be spliced in and out of a list representing the merged halves without using any extra space, aside from a pointer or two.

The running time of a merge sort is largely insensitive to the order of the items, so it has guaranteed performance properties. This is in contrast to, for example, quicksort, where certain degenerate orders can cause the running time of the algorithm to drop from O(Nlog N) to O(N2). On the other hand, because guaranteed performance is a double edged sword, a straight merge sort can't take advantage of any inherent order already in the array to improve its running time. Insertion sort is very fast, approaching and sometimes reaching O(N), if the array is almost sorted, and other algorithms can take advantage of the distribution of items to run very quickly.

Of course, like quicksort we can add a test to see if the subarray is already sorted and avoid recursing on it. This counts as an introspective algorithm that takes advantage of existing order in the subarray. It also could be educational to try the threshold trick we got from quicksort on merge sort, where we stop at a certain cutoff to make the array almost sorted and finish the algorithm with an insertion sort. Remember that tricks from one sorting algorithm can sometimes be reused in other algorithms.

You might notice that while I said that merge sort is ideal for data that can only be accessed sequentially, the merge sort algorithm above actually relies heavily on random access for its speed and requires that all of the items be present before the algorithm can be run so that it can find the middle item. The same algorithm with a linked list would be much slower because finding the middle of a list if you are only given the head is somewhat expensive. The best solution is to walk two links down the list, one moving twice as fast. When the faster link reaches the end of the list, the slower link is at the middle. The code to split a list in this way and then sort it by merging is relatively simple. The split is easy to write, the merge is far easier to write and far shorter, and the sort really doesn't change that much:

1 struct jsw_node *jsw_split ( struct jsw_node *a ) 2 { 3 struct jsw_node *b = a->next; 4 5 while ( b != NULL && b->next != NULL ) { 6 a = a->next; 7 b = b->next->next; 8 } 9 10 b = a->next; 11 a->next = NULL; 12 13 return b; 14 } 15 16 struct jsw_node *jsw_merge ( struct jsw_node *a, struct jsw_node *b ) 17 { 18 struct jsw_node result, *it = &result; 19 struct jsw_node **half; 20 21 while ( a != NULL && b != NULL ) { 22 half = ( a->data < b->data ) ? &a : &b; 23 it->next = *half; 24 it = *half; 25 *half = (*half)->next; 26 } 27 28 it->next = ( a == NULL ) ? b : a; 29 30 return; 31 } 32 33 struct jsw_node *jsw_mergesort_r ( struct jsw_node *a ) 34 { 35 if ( a != NULL && a->next != NULL ) { 36 struct jsw_node *b = jsw_split ( a ); 37 38 a = jsw_mergesort_r ( a ); 39 b = jsw_mergesort_r ( b ); 40 41 return jsw_merge ( a, b ); 42 } 43 44 return a; 45 } 46 47 void jsw_mergesort ( struct jsw_list *list ) 48 { 49 list->head = jsw_mergesort_r ( list->head ); 50 }

However, as I've already said several times, jsw_split slows down the algorithm by a substantial amount, so this algorithm isn't very practical. A much better option is to take items sequentially from the beginning, do some magic, and then merge the magical result into a sorted array or list. This is much better suited to a sequential delivery method such as a stream or generator, and it improves the performance of the sort on linked lists considerably.

Because we are accepting items as they exist in the collection, we can guarantee that the algorithm is stable. We can also take advantage of the current order of the collection, unlike with straight merge sorting. Such an algorithm is called a “natural” merge, because it uses the natural order of the collection to improve its performance. The worst case of a natural merge sort is O(Nlog N), but the best case is O(N). This is an improvement over guaranteed O(Nlog N) performance in all cases because if the array is already sorted or almost sorted, the performance is as good as running insertion sort from the beginning.

The algorithms for a natural merge sort can vary wildly, but a simple algorithm that performs well uses two queues to act as each half of the merge. The first item goes into queue A, and for each new item, if the item is smaller than the top of the current queue, the algorithm switches queues. Because we switch queues only when the order of the sequence changes, the number of items in each queue won't be balanced. In fact, all of the items in the array could be in one queue, in which case the array is completely sorted. Here's the code to do a natural merge sort with arrays and queues:

1 void jsw_merge ( int a[], int *q[2], int top[2] ) 2 { 3 int i[2] = {0}, k = 0; 4 int dir = q[1][0] < q[0][0]; 5 6 a[k++] = q[dir][i[dir]++]; 7 8 while ( i[dir] < top[dir] ) { 9 dir = q[1][i[1]] < q[0][i[0]]; 10 11 /* Special case: Use larger item */ 12 if ( q[dir][i[dir]] < a[k - 1] 13 && a[k - 1] < q[!dir][i[!dir]] ) 14 { 15 dir = !dir; 16 } 17 18 a[k++] = q[dir][i[dir]++]; 19 } 20 21 while ( i[0] < top[0] ) 22 a[k++] = q[0][i[0]++]; 23 while ( i[1] < top[1] ) 24 a[k++] = q[1][i[1]++]; 25 } 26 27 void jsw_mergesort ( int a[], int n ) 28 { 29 int *q[2], top[2]; 30 int half, i; 31 32 if ( n < 2 ) 33 return; 34 35 q[0] = malloc ( n * sizeof *q[0] ); 36 q[1] = malloc ( n * sizeof *q[1] ); 37 38 for ( ; ; ) { 39 top[0] = top[1] = 0; 40 q[0][top[0]++] = a[0]; 41 half = 0; 42 43 for ( i = 1; i < n; i++ ) { 44 if ( a[i] < q[half][top[half] - 1] ) 45 half = !half; 46 47 q[half][top[half]++] = a[i]; 48 } 49 50 if ( top[0] == 0 || top[1] == 0 ) 51 break; 52 53 jsw_merge ( a, q, top ); 54 } 55 56 half = top[0] == 0; 57 58 for ( i = 0; i < n; i++ ) 59 a[i] = q[half][i]; 60 61 free ( q[0] ); 62 free ( q[1] ); 63 }

Notice that jsw_merge no longer needs to allocate memory because that memory is allocated in jsw_mergesort to the queues, and then the queues are passed to jsw_merge. This makes the merge simpler and faster because we're not constantly calling an expensive memory manager. A simple boolean flag is used to switch between queues. This works because we only have two queues, so the only valid indices are 0 and 1. That coincides perfectly with the result of a boolean test. However, we need to be careful to reset the queues on each iteration of the outer loop (by setting the tops to 0 and giving the first queue the first item in the array).

Let's look at an example, because this algorithm could easily be confusing. :-) The following execution depicts the worst case for a natural merge sort (where the algorithm has O(Nlog N) performance). The worst case is an already sorted array in descending order, which is basically the opposite of what we want (in this tutorial we're sorting in ascending order). Pay careful attention to how the two queues work together to sort the array:

          Result              Queue A             Queue B
 1)  |8|7|6|5|4|3|2|1|   | | | | | | | | |   | | | | | | | | |
 2)  |7|5|3|1|8|6|4|2|   |8|6|4|2| | | | |   |7|5|3|1| | | | |
 3)  |5|1|7|3|6|2|8|4|   |7|3|6|2| | | | |   |5|1|8|4| | | | |
 4)  |1|5|3|6|4|7|2|8|   |5|3|6|4| | | | |   |1|7|2|8| | | | |
 5)  |1|3|5|4|6|2|7|8|   |1|5|4|7| | | | |   |3|6|2|8| | | | |
 6)  |1|3|4|5|2|6|7|8|   |1|3|5|2|7|8| | |   |4|6| | | | | | |
 7)  |1|2|3|4|5|6|7|8|   |1|3|4|5| | | | |   |2|6|7|8| | | | |
 8)  |1|2|3|4|5|6|7|8|   |1|2|3|4|5|6|7|8|   | | | | | | | | |

It's possible to modify the algorithm to eliminate the need for two queues, but the result is dreadfully complicated and only serves to confuse most of the people that look at it. :-) The only problem with using two queues implemented as arrays is that both have to be at least as big as the array itself, so you are looking at an extra space cost of 2N instead of N. However, using a dynamic queue can decrease that cost while causing the algorithm to potentially run slower.

The extra extra space for a natural merge sort on arrays isn't such a bad thing because you shouldn't use a natural merge sort on arrays in the first place. Even though a natural merge sort on an array is theoretically better than a straight merge sort, in practice it's much slower due to excessive copying of data in and out of the queues. However, the natural merge sort is ideal for linked lists because by splicing nodes, you avoid the problem of copying data all over the place. Here's a tuned natural merge sort for linked lists:

1 /* Remove a node from a list and push it onto a tail pointer */ 2 #define split_push(head,tail,item) do { \ 3 struct jsw_node *save = item->next; \ 4 item->next = NULL; \ 5 if ( tail == NULL ) \ 6 head = tail = item; \ 7 else { \ 8 tail->next = item; \ 9 tail = tail->next; \ 10 } \ 11 item = save; \ 12 }while ( 0 ) 13 14 struct jsw_node *jsw_merge ( struct jsw_node *a, struct jsw_node *q[2] ) 15 { 16 int dir = q[0]->data > q[1]->data; 17 struct jsw_node *save = a = q[dir]; 18 19 /* Intended assignment */ 20 while ( q[dir] = q[dir]->next ) { 21 dir = q[0]->data > q[1]->data; 22 23 if ( q[dir]->data < a->data && a->data < q[!dir]->data ) 24 dir = !dir; 25 26 a->next = q[dir]; 27 a = a->next; 28 } 29 30 dir = q[0] == NULL; 31 32 while ( q[dir] != NULL ) { 33 a->next = q[dir]; 34 q[dir] = q[dir]->next; 35 a = a->next; 36 } 37 38 return save; 39 } 40 41 void jsw_mergesort ( struct jsw_list *list ) 42 { 43 struct jsw_node *q[2] = {0}; 44 struct jsw_node *a = list->head; 45 46 for ( ; ; ) { 47 struct jsw_node *end[2] = {0}; 48 int half = 0; 49 50 split_push ( q[0], end[0], a ); 51 52 while ( a != NULL ) { 53 if ( a->data < end[half]->data ) 54 half = !half; 55 split_push ( q[half], end[half], a ); 56 } 57 58 if ( q[0] == NULL || q[1] == NULL ) 59 break; 60 61 a = jsw_merge ( a, q ); 62 } 63 64 list->head = q[q[0] == NULL]; 65 }

Not only is the code simpler (assuming you're familiar with linked lists), it's also much faster than the array version. The extra space cost has virtually been eliminated because the queues are nothing more than pointers to nodes that already exist, so no extra memory needs to be allocated. jsw_merge is just a translation of the natural merge algorithm used for arrays to linked lists, and split_push is nothing more than a specialized way to extract a node from the list and push it onto a queue.

The performance properties of merge sort are very good. It has guaranteed O(Nlog N) performance, like heap sort, and when a natural merge sort is used it can reach O(N) performance, but the performance of merge sort comes at the price of extra space equivalent to the number of items being sorted. This cost is eliminated when sorting linked lists, which is why merge sort is a common list sorting algorithm. When the merge algorithm is chosen carefully, mergesort is one of the few stable sorting algorithms. Most other high speed sorting algorithms that can handle large arrays or lists require extra work and extra space to make them stable, so in light of the stability question, mergesort fares very well. Another huge benefit is that when written carefully, merge sort does not require that all items be present. It can sort an unknown number of items coming in from a stream or generator, which is a very useful property.


Radix Sort

Up to this point, we've looked mostly at algorithms that sort by comparison. That is, we figure out the relative order of two items by testing which is less than the other and place them accordingly. This section will cover a different method, where we use the distribution of items and place them in buckets, then put the buckets together to form a final sorted array. The first algorithm that uses distribution is called counting sort. The concept is to map each item to an array, where the item's value is the array index. Then you count how many items are smaller than that item, each time incrementing the index. Then you use the count as a location index to move the item to its final location.

Here's the process on an array of 10 items ranged from [0,10) using a 1-based index algorithm described in Donald Knuth's “The Art of Computer Programming”, volume 3:


 Get counts:


 Count smaller items:


 Use counts as indices:


It's a nifty algorithm, but when I first read about it, I wasn't able to completely understand what was going on until I walked through it on paper, so I won't try to explain in detail. I'll leave you with the basic concept, the code, and let you do the same thing (probably because I'm evil). There's a lot to be said for working out algorithms on paper though. :-) Here's the code that does a counting sort. Be sure to figure it out before moving on, because this will be a fundamental part of the coming radix sorts. Be happy that it's in C, I had to translate it from MIX...

1 void jsw_countsort ( int a[], int m, int n ) 2 { 3 int *aux = malloc ( n * sizeof *aux ); 4 int *count = calloc ( m + 1, sizeof *count ); 5 int i; 6 7 for ( i = 0; i < n; i++ ) 8 ++count[a[i] + 1]; 9 10 for ( i = 1; i < m; i++ ) 11 count[i] += count[i - 1]; 12 13 for ( i = 0; i < n; i++ ) 14 aux[count[a[i]]++] = a[i]; 15 16 for ( i = 0; i < n; i++ ) 17 a[i] = aux[i]; 18 19 free ( count ); 20 free ( aux ); 21 }

As we discovered previously in the tutorial, the lower bound for sorting algorithms that use comparisons to find the relative order of items is O(Nlog N). However, it's possible to improve that to O(N) by eschewing comparisons and using distribution as the counting sort did. Counting sort is a linear algorithm because it makes a set number of linear passes through the array, regardless of pre-existing order.

However, the biggest problem with counting sort is that it requires extra memory equivalent to the range of items being sorted. So if you have 10 items in the range of [0,1,000,000), you need to allocate an extra array to hold the counts for one million items. That's a ridiculous requirement, and it makes counting sort far less practical than other sorting algorithms. If only we could figure out a way to use counting sort but minimize the extra space we need to allocate for the counts.

This idea is exactly what a class of algorithms called radix sort do. Instead of using the full range of each item, they use the range of a single digit, for some definition of “digit”, and then use a distribution sort on each digit either from the most significant digit to the least significant digit or vice versa. The two algorithms called radix sort are LSD (for “least significant digit”, you druggies!), and MSD (for most significant digit). The LSD algorithm is what was used to sort punch cards back in the early days of computing.

Before we get into the code, let's look at an example execution where we start at the least significant digit of several 3 digit numbers and use counting sort until we get to the most significant digit:

   1     2     3     4
 |341| |100| |100| |100|
 |267| |341| |224| |178|
 |534| |762| |534| |224|
 |100| |534| |341| |267|
 |369| |224| |958| |341|
 |224| |564| |762| |369|
 |178| |267| |564| |534|
 |958| |178| |267| |567|
 |762| |958| |369| |762|
 |564| |369| |178| |958|

There are several things to remember about LSD radix sort. Each digit needs to be sorted with a stable algorithm. The most common choice is a counting sort because counting sort is an algorithm that can reach O(N) performance. If you have short keys then LSD radix sort can also be O(N), whereas if we used another stable sort like a straight shift select, each digit would be sorted with an O(N2) sort, and the entire algorithm would become quadratic. Next, each key has to be the same length. That's why most radix sorts will use the number of bytes as the radix, so 4 byte integers would be sorted from the least significant byte to the most significant byte, where each byte is a digit. The performance of any radix sort is dependent on the length of the items.

With this in mind, LSD radix sort is relatively easy to implement. You simply modify a counting sort to use digits, then call the counting sort on each digit of the array. The auxiliary array that counting sort uses can be allocated in the function that calls the counting sort, so we avoid multiple calls to the memory manager. The nice thing about using counting sort for a radix sort is that the count array can be small. In our case, we only need to count all of the possible bytes, so the count array can be as small as ( 1U << CHAR_BIT ) + 1, which is exactly what we do. :-)

1 #define RANGE ( ( 1U << CHAR_BIT ) + 1 ) 2 #define digit(x,i) ( (x) >> ( (i) * CHAR_BIT ) & UCHAR_MAX ) 3 4 void jsw_radix_pass ( int a[], int aux[], int n, int radix ) 5 { 6 int i; 7 int count[RANGE] = {0}; 8 9 for ( i = 0; i < n; i++ ) 10 ++count[digit ( a[i], radix ) + 1]; 11 12 for ( i = 1; i < RANGE; i++ ) 13 count[i] += count[i - 1]; 14 15 for ( i = 0; i < n; i++ ) 16 aux[count[digit ( a[i], radix )]++] = a[i]; 17 18 for ( i = 0; i < n; i++ ) 19 a[i] = aux[i]; 20 } 21 22 void jsw_radixsort ( int a[], int n ) 23 { 24 int i; 25 int *aux = malloc ( n * sizeof *aux ); 26 27 for ( i = 0; i < sizeof *a; i++ ) 28 jsw_radix_pass ( a, aux, n, i ); 29 30 free ( aux ); 31 }

The biggest problem with LSD radix sort is that it starts at the digits that make the least difference. If we could start with the most significant digits, the first pass would go a long way toward sorting the entire range, and each pass after that would simply handle the details. The idea of MSD radix sorting is to divide all of the digits with an equal value into their own bucket, then do the same thing with all of the buckets until the array is sorted. Naturally, this suggests a recursive algorithm, but it also means that we can now sort variable length items and we don't have to touch all of the digits to get a sorted array. This makes MSD radix sort considerably faster and more useful.



 First Pass:

   1     2     3     4     5     6
 |100| |267| |369| |564| |762| |958|
 |178| |224| |341|

 Second Pass:

   1     2     3     4     5     6     7     8     9    10
 |100| |178| |224| |267| |334| |341| |369| |564| |762| |958|



Notice that because all of the second digits are unique, no further sorting is needed beyond the second pass. Each of the buckets are separated into their own buckets containing items that have duplicate digits. So in the first pass, every item that started with 3 was put into a bucket, every item that started with 2 was put in a bucket, and so on. This is a good example of MSD radix sort being more efficient than LSD radix sort, which would have to sort all digits before the array would be sorted.

The bad news is that MSD radix sort is going to be recursive in this tutorial. It looks surprisingly similar to quicksort though, which is a good thing because we can use the threshold trick to stop before spending too much time on small buckets and finish the almost sorted array with insertion sort. jsw_radix_pass is pretty much the same thing, but for variety and to show the variable length item strength of MSD radix sort, we'll implement this one to sort strings instead of integers:

1 /* Extract ith MSD byte */ 2 #define digit(x,i) ( (x)[i] ) 3 #define bin(x) ( first + count[x] ) 4 5 void jsw_radix_pass ( char *a[], char *aux[], int count[], int first, int last, int radix ) 6 { 7 int i; 8 9 for ( i = first; i < last; i++ ) 10 ++count[digit ( a[i], radix ) + 1]; 11 12 for ( i = 1; i < UCHAR_MAX; i++ ) 13 count[i] += count[i - 1]; 14 15 for ( i = first; i < last; i++ ) 16 aux[count[digit ( a[i], radix )]++] = a[i]; 17 18 for ( i = first; i < last; i++ ) 19 a[i] = aux[i - first]; 20 } 21 22 void jsw_radixsort_r ( char *a[], char *aux[], int first, int last, int radix ) 23 { 24 if ( radix <= UCHAR_MAX && last - first > THRESHOLD ) { 25 int count[UCHAR_MAX + 1] = {0}; 26 int i; 27 28 jsw_radix_pass ( a, aux, count, first, last, radix ); 29 30 for ( i = 0; i < UCHAR_MAX - 1; i++ ) { 31 int left = bin ( i ); 32 int right = bin ( i + 1 ) - 1; 33 jsw_radixsort_r ( a, aux, left, right, radix + 1 ); 34 } 35 } 36 } 37 38 void jsw_radixsort ( char *a[], int n ) 39 { 40 char **aux = malloc ( ( n + 1 ) * sizeof *aux ); 41 42 jsw_radixsort_r ( a, aux, 0, n, 0 ); 43 jsw_insertion ( a, n ); 44 45 free ( aux ); 46 }

MSD radix sort doesn't require a stable sort as the workhorse algorithm, even though we still used counting sort because it's fast. So MSD radix sort is considerably more flexible than LSD radix sort. One thing to be careful of is that in both the LSD and MSD algorithms, I've implemented 0 as “start here”, so LSD sees digits in 5,4,3,2,1,0 order while MSD sees digits in 0,1,2,3,4,5 order. In other words, be careful when mixing and matching because these algorithms can be very fragile.



The art of sorting is unbelievably diverse, and one single source can't possibly teach you everything there is to know about it. I regret that this tutorial can't be more comprehensive, but it was already too long when I felt that I included the bare minimum of information. I hope that this article was helpful, and I encourage you to read Donald Knuth's “The Art of Computer Programming” volume 3, which is considered the definitive text on sorting. However, keep in mind that it is somewhat out of date and that more recent discoveries on sorting will require more intensive searching to uncover...ironically enough. :-)


From the twisted mind of Julienne Walker
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