Recursion

Recursion is:

• A way of thinking about problems.
• A method for solving problems.
• Related to mathematical induction.

void f() {
   ... f() ... 
}

void f() {
    ... g() ...
}
	
void g() {
   ... f() ...
}

You might wonder about the following issues:

Q: Does using recursion usually make your code faster?
A: No.

Q: Does using recursion usually use less memory?
A: No.

Q: Then why use recursion?
A: It sometimes makes your code much simpler!

One way to think about recursion:

• When a recursive call is made, the method clones itself, making new copies of:
  • the code
  • the local variables (with their initial values),
  • the parameters
• Each copy of the code includes a marker indicating the current position. When a recursive call is made, the marker in the old copy of the code is just after the call; the marker in the "cloned" copy is at the beginning of the method.
• When the method returns, that clone goes away, but the previous ones are still there, and know what to execute next because their current position in the code was saved (indicated by the marker).

Here's an example of a simple recursive method:

   void printInt( int k ) {
1.     if (k == 0) {
2.         return;
3.     }
4.     System.out.println( k );
5.     printInt( k - 1 );
6.     System.out.println("all done");
}

The original call causes 2 to be output, and then a recursive call is made, creating a clone with k == 1. That clone executes line 1: the if condition is false; line 4: prints 1; and line 5: makes another recursive call, creating a clone with k == 0. That clone just returns (goes away) because the "if" condition is true. The previous clone executes line 6 (the line after its "marker") then returns, and similarly for the original clone.

Now let's think about what we have to do to make sure that recursive methods work correctly. First, consider the following recursive method:

void badPrint( int k ) {
    System.out.println( k );
    badPrint( k + 1 );
}

Every recursive method must have a base case -- a condition under which no recursive call is made -- to prevent infinite recursion.

Here's another example; this version does have a base case, but the call badPrint2(2) will still cause an infinite recursion:

void badPrint2( int k ) {
    if (k < 0) return;
    System.out.println(k);
    badPrint2( k+1 );
}

Every recursive method must make progress toward the base case to prevent infinite recursion.

This is how method calls (recursive and non-recursive) really work:

• At runtime, a stack of activation records (ARs) is maintained: one AR for each active method, where "active" means: has been called, has not yet returned.
• Each AR includes space for:
  • the method's parameters,
  • the method's local variables,
  • the return address -- where (in the code) to start executing after the method returns.
• When a method is called, its AR is pushed onto the stack. The return address in that AR is the place in the code just after the call (so the return address is the "marker" for the previous "clone").
• When a method is about to return, the return address in its AR is saved, its AR is popped from the stack, and control is transferred to the place in the code referred to by the return address.

Example (no recursion):

1.  void printChar( char c ) {
2.      System.out.print(c);
3.  }
4.
5.  void main (...) {
6.      char ch = 'a';
7.      printChar(ch);
8.      ch = 'b';
9.      printChar(ch);
10. }

When the first call to printChar returns, the top activation record is popped from the stack and the main method begins executing again at line 8. After executing the assignment at line 8, a second call to printChar is made (line 9), as illustrated below.

The two calls to printChar return to different places in main because different return addresses are stored in the ARs that are pushed onto the stack when the calls are made.

Now let's consider recursive calls:

1. void printInt( int k ) {
2.     if (k <= 0) return;
3.     System.out.println( k );
4.     printInt( k - 1 );
5. }
6. 
7. void main(...) {
8.     printInt(2);
9.}  

Note that all of the recursive calls have the same return address -- after a recursive call returns, the previous call to printInt starts executing again at line 5. In this case, there is nothing more to do at that point, so the method would, in turn, return.

Note also that each call to printInt causes the current value of k to be printed, so the output of the program is: 2 1.

Now consider a slightly different version of the printInt method:

1.  void printInt (int k) {
2.      if (k<=0) return;
3.      printInt (k-1);
4.      System.out.println(k)
5.  }

Sometimes a method has more to do following a recursive call; it gets done only after the recursive call (and all calls it makes) are finished.

Now let's think about when it is a good idea to use recursion and why. In many cases there will be a choice: many methods can be written either with or without using recursion.

Q: Is the recursive version usually faster?
A: No -- it's usually slower (due to the overhead of maintaining the stack)

Q: Does the recursive version usually use less memory?
A: No -- it usually uses more memory (for the stack).

Q: Then why use recursion??
A: Sometimes it is much simpler to write the recursive version.

Here are a few examples where recursion makes things a little bit clearer, though in the second case, the efficiency problem makes it a bad choice.

Factorial can be defined as follows:

• factorial of 0 is 1
• factorial of N (for N>0) is N*N-1* ... *3*2*1

• factorial of 0 is 1
• factorial of N (for N > 0) is N * factorial (N-1)

The first definition leads to an iterative version of factorial:

int factorial (int N) {
  if (N == 0) return 1;
  int tmp = N;
  for (int k=N-1; k>=1; k--) tmp = tmp*k;
  return (tmp);
}

int factorial (int N) {
  if (N == 0) return 1;
  else return (N*factorial(N-1));
}

• a little shorter,
• a little clearer (closer to the mathematical definition),
• a little slower

Fibonacci can be defined as follows:

• fibonacci of 1 or 2 is 1
• fibonacci of N (for N>2) is fibonacci of (N-1) + fibonacci of (N-2)

// iterative version
int fib (int N) {
  int k1, k2, k3;
  k1 = k2 = k3 = 1;
  for (int j = 3; j <= N; j++) {
      k3 = k1 + k2;
      k1 = k2;
      k2 = k3;
  }
  return k3;
}

// recursive version
int fib (int N) {
  if ((N == 1) || (N == 2)) return 1;
  else return (fib (N-1) + fib (N-2));
}

• shorter,
• clearer,
• but much slower

Given that the recursive version of fibonacci is slower than the iterative version, is there a good reason for using it? The answer may be yes: because the recursive solution is so much simpler, it is likely to take much less time to write, debug, and maintain. If those costs (the cost for programming time) are more important than the cost of having a slow program, then the advantages of using the recursive solution outweigh the disadvantage, and you should use it! If the speed of the final code is of vital importance, then you should not use the recursive fibonacci.

So far, we've been doing all our recursion based on the value of some int parameter; e.g., for factorial, we start with some value N, and each recursive call passes a smaller value (N - 1) until we get down to 0. A more common way to use recursion in programming is to base the recursion on a recursive data structure. A data structure is recursive if you can define it in terms of itself. For example, we can give a recursive definition of a string:

A string is either empty, or it is a character followed by a string.

We can give a similar, recursive definition of a linked list:

A linked list is either empty, or it is a list-node followed by a linked list.

Below are two methods that use recursive data structures; the first one prints each character of its String parameter, and the second prints each value in its linked-list parameter.

Example 1: Use a String parameter

printStr( String S ) {
    if (S.equals("")) return;
    System.out.print(S.substring(0,1) + " ");
    printStr(S.substring(1));
}

Example 2: Use a linked-list parameter

printList( Listnode L ) {
    if (L == null) return;
    System.out.println(L.getData());
    printList(L.getNext());
}

Now suppose we want to print the values in a linked list backwards; i.e., if the list looks like this:

we want to print:

itemN
itemN-1
 ...
item2
item1

It is easy to do this using recursion; all we have to do is swap the order of the print statement and the recursive call in the Example 2 code given above. Here's the new version:

printList( Listnode L ) {
    if (L == null) return;
    printList(L.getNext());
    System.out.println(L.getData());
}

There are two ways to determine how much time is required for a call to a recursive method:

1. Informal Reasoning
2. Using Recurrence Equations

Informal Reasoning

To reason about the time required for a call to a recursive method, you must figure out how many times total the method will be called, and how much work (in terms of the problem size, N) is done on each call in addition to the recursive calls. The total time is the product of those two values.

For example, consider recursive method printInt:

public static void printInt( int k ) {
  if (k==0) return;
  System.out.println(k);
  printInt( k-1 );
}

Using Recurrence Equations

A more formal way to analyze the running time of a recursive method is to use recurrence equations. The recurrence equations for a method include an equation for the base case (or several equations if there are several base cases), and an equation for the recursive case.

For example, for the printInt method above, the recurrence equations are:

T(0) = 1
T(N) = 1 + T(N-1)

To find the worst-case time for a problem of size N, we need to find a closed-form solution to these equations; i.e., a solution for T(N) that doesn't involve any T(...) on the right-hand side. To do that we perform three steps:

1. Expand: Find the times required for a problem of size 1, 2, 3, etc.
2. Look for a pattern, and guess a solution for T(N).
3. Plug in your solution and see if you got it right.

Step 1: Expand

T(0) = 1

T(1) = 1 + T(0) = 1 + 1 = 2

T(2) = 1 + T(1) = 1 + 2 = 3

T(3) = 1 + T(2) = 1 + 3 = 4

Step 2: Look for a pattern and guess a solution

We'll guess that T(N) = N+1

Step 3: Verify the guessed solution

T(N)	=	1 + T(N-1)	// given
N+1	=?=	1 + ((N-1)+1)	// replace T(...) on both sides with the guessed solution (N+1) in which each N is replaced by ...
N+1	=?=	1 + N - 1 + 1	// simplify
N+1	=	1 + N	// yes the two sides are equal!

If you like math, you may enjoy thinking about how to use mathematical induction to prove that recursive code is correct. If not, feel free to skip this section.

When using induction to prove a theorem, you need to show:

1. that the base case (usually n=0 or n=1) is true
2. that case k implies case k+1

It is sometimes straightforward to use induction to prove that recursive code is correct. Let's consider how to do that for the recursive version of factorial. First we need to prove the base case: factorial(0) = 0! (which, by definition is 1). The correctness of the factorial method for N=0 is obvious from the code: when N==0 it returns 1.

Now we need to show that if factorial(k) = k!, then factorial(k+1) = (k+1)!. Looking at the code we see that for N != 0, factorial(N) = (N)*factorial(N-1). So factorial(k+1) = (k+1)*factorial(k). By assumption, factorial(k) = k!, so factorial(k+1) = (k+1)*(k!). By definition, k! = k*(k-1)*(k-2)*...*3*2*1. So (k+1)*(k!) = (k+1)*k*(k-1)*(k-2)*...*3*2*1, which is (by definition) (k+1)!, and the proof is done.

Note that we've shown that the factorial method is correct for all values of N greater than or equal to zero (because our base case was N=0). We haven't shown anything about factorial for negative numbers.

• Use recursion for clarity, and (sometimes) for a reduction in the time needed to write and debug code, not for space savings or speed of execution.
• Remember that every recursive method must have a base case (rule #1).
• Also remember that every recursive method must make progress towards its base case (rule #2).
• Sometimes a recursive method has more to do following a recursive call. It gets done only after the recursive call (and all calls it makes) finishes.
• Use informal reasoning or recurrence equations to determine how much time a recursive call will require.
• Recursion is often simple and elegant, can be efficient, and tends to be underutilized. Consider using recursion when solving a problem!