AT ONE TIME OR ANOTHER, you've probably been told that you can't define something in terms of itself. Nevertheless, if it's done right, defining something at least partially in terms of itself can be a very powerful technique. A recursive definition is one that uses the concept or thing that is being defined as part of the definition. For example:
An "ancestor" is either a parent or an ancestor of a parent.
A "sentence" can be, among other things, two sentences joined by a conjunction such as "and."
A "directory" is a part of a disk drive that can hold files and directories.
In mathematics, a "set" is a collection of elements, which can be other sets.
A "statement" in C# can be a while statement, which is made up of the word "while", a boolean-valued condition, and a statement.
Recursive definitions can describe very complex situations with just a few words. A definition of the term "ancestor" without using recursion might go something like "a parent, or a grandparent, or a great-grandparent, or a great-great-grandparent, and so on." But saying "and so on" is not very rigorous. (I've often thought that recursion is really just a rigorous way of saying "and so on.") You run into the same problem if you try to define a "directory" as "a file that is a list of files, where some of the files can be lists of files, where some of those files can be lists of files, and so on." Trying to describe what a C# statement can look like, without using recursion in the definition, would be difficult and probably pretty comical.
Recursion can be used as a programming technique. A recursive method is one that calls itself, either directly or indirectly. To say that a method calls itself directly means that its definition contains a method call statement that calls the method that is being defined. To say that a method calls itself indirectly means that it calls a second method which in turn calls the first method (either directly or indirectly). A recursive method can define a complex task in just a few lines of code. In the rest of this section, we'll look at a variety of examples, and we'll see other examples in the remaining sections of this chapter.
Let's start with an example that you've seen before: the binary search algorithm from Section 8.4. Binary search is used to find a specified value in a sorted list of items (or, if it does not occur in the list, to determine that fact). The idea is to test the element in the middle of the list. If that element is equal to the specified value, you are done. If the specified value is less than the middle element of the list, then you should search for the value in the first half of the list. Otherwise, you should search for the value in the second half of the list. The method used to search for the value in the first or second half of the list is binary search. That is, you look at the middle element in the half of the list that is still under consideration, and either you've found the value you are looking for, or you have to apply binary search to one half of the remaining elements. And so on! This is a recursive description, and we can write a recursive method to implement it.
Before we can do that, though, there are two considerations that we need to take into account. Each of these illustrates an important general fact about recursive methods. First of all, the binary search algorithm begins by looking at the "middle element of the list." But what if the list is empty? If there are no elements in the list, then it is impossible to look at the middle element. Having a non-empty list is a "precondition" for looking at the middle element, and this is a clue that we have to modify the algorithm to take this precondition into account. What should we do if we find ourselves searching for a specified value in an empty list?
The answer is easy: We can say immediately that the value does not occur in the list. An empty list is a base case for the binary search algorithm. A base case for a recursive algorithm is a case that is handled directly, rather than by applying the algorithm recursively. The binary search algorithm actually has another type of base case: If we find the element we are looking for in the middle of the list, we are done. There is no need for further recursion.
The second consideration has to do with the parameters to the method. The problem is phrased in terms of searching for a value in a list. In the original, non-recursive binary search method, the list was given as an array. However, in the recursive approach, we have to able to apply the method recursively to just a part of the original list. Where the original method was designed to search an entire array, the recursive method must be able to search part of an array. The parameters to the method must tell it what part of the array to search. This illustrates a general fact that in order to solve a problem recursively, it is often necessary to generalize the problem slightly.
Here is a recursive binary search algorithm that searches for a given value in part of an array of integers:
static int BinarySearch(int[] A, int loIndex, int hiIndex, int value) {
// Search in the array A in positions from loIndex to hiIndex,
// inclusive, for the specified value. It is assumed that the
// array is sorted into increasing order. If the value is
// found, return the index in the array where it occurs.
// If the value is not found, return -1.
if (loIndex > hiIndex) {
// The starting position comes after the final index,
// so there are actually no elements in the specified
// range. The value does not occur in this empty list!
return -1;
}
else {
// Look at the middle position in the list. If the
// value occurs at that position, return that position.
// Otherwise, search recursively in either the first
// half or the second half of the list.
int middle = (loIndex + hiIndex) / 2;
if (value == A[middle])
return middle;
else if (value < A[middle])
return BinarySearch(A, loIndex, middle - 1, value);
else // value must be > A[middle]
return BinarySearch(A, middle + 1, hiIndex, value);
}
} // end BinarySearch()
In this routine, the parameters loIndex and hiIndex specify the part of the array that is to be searched. To search an entire array, it is only necessary to call BinarySearch(A, 0, A.length - 1, value). In the two base cases -- where there are no elements in the specified range of indices and when the value is found in the middle of the range -- the method can return an answer immediately, without using recursion. In the other cases, it uses a recursive call to compute the answer and returns that answer.
Most people find it difficult at first to convince themselves that recursion actually works. The key is to note two things that must be true for recursion to work properly: There must be one or more base cases, which can be handled without using recursion. And when recursion is applied during the solution of a problem, it must be applied to a problem that is in some sense smaller -- that is, closer to the base cases -- than the original problem. The idea is that if you can solve small problems and if you can reduce big problems to smaller problems, then you can solve problems of any size. Ultimately, of course, the big problems have to be reduced, possibly in many, many steps, to the very smallest problems (the base cases). Doing so might involve an immense amount of detailed bookkeeping. But the computer does that bookkeeping, not you! As a programmer, you lay out the big picture: the base cases and the reduction of big problems to smaller problems. The computer takes care of the details involved in reducing a big problem, in many steps, all the way down to base cases. Trying to think through this reduction in detail is likely to drive you crazy, and will probably make you think that recursion is hard. Whereas in fact, recursion is an elegant and powerful method that is often the simplest approach to solving a complex problem.
A common error in writing recursive methods is to violate one of the two rules: There must be one or more base cases, and when the method is applied recursively, it must be applied to a problem that is smaller than the original problem. If these rules are violated, the result can be an infinite recursion, where the method keeps calling itself over and over, without ever reaching a base case. Infinite recursion is similar to an infinite loop. However, since each recursive call to the method uses up some of the computer's memory, a program that is stuck in an infinite recursion will run out of memory and crash before long. (In C#, the program will crash with an exception of type StackOverflowError.)
Binary search can be implemented with a while loop, instead of with recursion. Next, we turn to a problem that is easy to solve with recursion but difficult to solve without it. This is a standard example known as "The Towers of Hanoi". The problem involves a stack of various-sized disks, piled up on a base in order of decreasing size. The object is to move the stack from one base to another, subject to two rules: Only one disk can be moved at a time, and no disk can ever be placed on top of a smaller disk. There is a third base that can be used as a "spare". The situation for a stack of ten disks is shown in the top half of the following picture. The situation after a number of moves have been made is shown in the bottom half of the picture.
The problem is to move ten disks from Stack 0 to Stack 1, subject to certain rules. Stack 2 can be used a spare location. Can we reduce this to smaller problems of the same type, possibly generalizing the problem a bit to make this possible? It seems natural to consider the size of the problem to be the number of disks to be moved. If there are N disks in Stack 0, we know that we will eventually have to move the bottom disk from Stack 0 to Stack 1. But before we can do that, according to the rules, the first N-1 disks must be on Stack 2. Once we've moved the N-th disk to Stack 1, we must move the other N-1 disks from Stack 2 to Stack 1 to complete the solution. But moving N-1 disks is the same type of problem as moving N disks, except that it's a smaller version of the problem. This is exactly what we need to do recursion! The problem has to be generalized a bit, because the smaller problems involve moving disks from Stack 0 to Stack 2 or from Stack 2 to Stack 1, instead of from Stack 0 to Stack 1. In the recursive method that solves the problem, the stacks that serve as the source and destination of the disks have to be specified. It's also convenient to specify the stack that is to be used as a spare, even though we could figure that out from the other two parameters. The base case is when there is only one disk to be moved. The solution in this case is trivial: Just move the disk in one step. Here is a version of the method that will print out step-by-step instructions for solving the problem:
void TowersOfHanoi(int disks, int from, int to, int spare) {
// Solve the problem of moving the number of disks specified
// by the first parameter from the stack specified by the
// second parameter to the stack specified by the third
// parameter. The stack specified by the fourth parameter
// is available for use as a spare.
if (disks == 1) {
// There is only one disk to be moved. Just move it.
Console.WriteLine("Move a disk from stack number "
+ from + " to stack number " + to);
}
else {
// Move all but one disk to the spare stack, then
// move the bottom disk, then put all the other
// disks on top of it.
TowersOfHanoi(disks-1, from, spare, to);
Console.WriteLine("Move a disk from stack number "
+ from + " to stack number " + to);
TowersOfHanoi(disks-1, spare, to, from);
}
}
This method just expresses the natural recursive solution. The recursion works because each recursive call involves a smaller number of disks, and the problem is trivial to solve in the base case, when there is only one disk. To solve the "top level" problem of moving N disks from Stack 0 to Stack 1, it should be called with the command TowersOfHanoi(N,0,1,2).
There is, by the way, a story that explains the name of this problem. According to this story, on the first day of creation, a group of monks in an isolated tower near Hanoi were given a stack of 64 disks and were assigned the task of moving one disk every day, according to the rules of the Towers of Hanoi problem. On the day that they complete their task of moving all the disks from one stack to another, the universe will come to an end. But don't worry. The number of steps required to solve the problem for N disks is 2N - 1, and 264 - 1 days is over 50,000,000,000,000 years. We have a long way to go.
Turning next to an application that is perhaps more practical, we'll look at a recursive algorithm for sorting an array. The selection sort and insertion sort algorithm algorithms, which were covered in Section 8.4, are fairly simple, but they are rather slow when applied to large arrays. Faster sorting algorithms are available. One of these is Quicksort, a recursive algorithm which turns out to be the fastest sorting algorithm in most situations.
The Quicksort algorithm is based on a simple but clever idea: Given a list of items, select any item from the list. This item is called the pivot. (In practice, I'll just use the first item in the list.) Move all the items that are smaller than the pivot to the beginning of the list, and move all the items that are larger than the pivot to the end of the list. Now, put the pivot between the two groups of items. This puts the pivot in the position that it will occupy in the final, completely sorted array. It will not have to be moved again. We'll refer to this procedure as QuicksortStep.
QuicksortStep is not recursive. It is used as a method by Quicksort. The speed of Quicksort depends on having a fast implementation of QuicksortStep. Since it's not the main point of this discussion, I present one without much comment.
static int QuicksortStep(int[] A, int lo, int hi) {
// Apply QuicksortStep to the list of items in
// locations lo through hi in the array A. The value
// returned by this routine is the final position
// of the pivot item in the array.
int pivot = A[lo]; // Get the pivot value.
// The numbers hi and lo mark the endpoints of a range
// of numbers that have not yet been tested. Decrease hi
// and increase lo until they become equal, moving numbers
// bigger than pivot so that they lie above hi and moving
// numbers less than the pivot so that they lie below lo.
// When we begin, A[lo] is an available space, since it used
// to hold the pivot.
while (hi > lo) {
while (hi > lo && A[hi] > pivot) {
// Move hi down past numbers greater than pivot.
// These numbers do not have to be moved.
hi--;
}
if (hi == lo)
break;
// The number A[hi] is less than pivot. Move it into
// the available space at A[lo], leaving an available
// space at A[hi].
A[lo] = A[hi];
lo++;
while (hi > lo && A[lo] < pivot) {
// Move lo up past numbers less than pivot.
// These numbers do not have to be moved.
lo++;
}
if (hi == lo)
break;
// The number A[lo] is greater than pivot. Move it into
// the available space at A[hi], leaving an available
// space at A[lo].
A[hi] = A[lo];
hi--;
} // end while
// At this point, lo has become equal to hi, and there is
// an available space at that position. This position lies
// between numbers less than pivot and numbers greater than
// pivot. Put pivot in this space and return its location.
A[lo] = pivot;
return lo;
} // end QuicksortStep
With this method in hand, Quicksort is easy. The Quicksort algorithm for sorting a list consists of applying QuicksortStep to the list, then applying Quicksort recursively to the items that lie to the left of the pivot and to the items that lie to the right of the pivot. Of course, we need base cases. If the list has only one item, or no items, then the list is already as sorted as it can ever be, so Quicksort doesn't have to do anything in these cases.
static void Quicksort(int[] A, int lo, int hi) {
// Apply quicksort to put the array elements between
// position lo and position hi into increasing order.
if (hi <= lo) {
// The list has length one or zero. Nothing needs
// to be done, so just return from the method.
return;
}
else {
// Apply quicksortStep and get the pivot position.
// Then apply quicksort to sort the items that
// precede the pivot and the items that follow it.
int pivotPosition = QuicksortStep(A, lo, hi);
Quicksort(A, lo, pivotPosition - 1);
Quicksort(A, pivotPosition + 1, hi);
}
}
As usual, we had to generalize the problem. The original problem was to sort an array, but the recursive algorithm is set up to sort a specified part of an array. To sort an entire array, A, using the quickSort() method, you would call quicksort(A, 0, A.length - 1).