Tail Recursion Example

Tail recursive saves stack memory.

A recursive function is said to be tail-recursive if there is nothing to do after the function returns except returning its value. In this case, instead of allocating a stack frame for each call, the compiler can rework the code to simply reuse the current stack frame, meaning a tail-recursive function will only use a single stack frame as opposed to hundreds or even thousands of stack frame as in the case of regular recursive calls.

Also, in a tail-recursive case, with each evaluation of the recursive call, the solution (e.g., a running total) is updated through parameter. A smart compiler can make Tail Recursion as efficient as iteration normally is.

We'll see an tail-recursion implementation of Fibonacci at the end of this post. Using Fibonacci as an example, a Fibonacci sequence: [f(0)...f(n)]=[0, 1, 1, 2, 3, 5, 8, 13, ...]

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A regular recursive implementation of Fibonacci:

public double fib(int n)  {
  if(n == 0)
    return 0d;
  else if(n == 1)
    return 1d;
  else
    return fib(n - 1) + fib(n - 2);
}

The run-time breakdown of the above intuit but inefficient recursive algorithm which takes a lot of stack space:

  fib(100) 
= fib(99) + fib(98)
= fib(98) + fib(97) + fib(97) + fib(96)
= fib(97) + fib(96) + fib(96) + fib(95) + fib(96) + fib(95) + fib(95) + fib(94)
= ...

You may notice the above  algorithm will be terribly performed with O(2^n). For performance's sake, here's the non-recursive version using a loop:

double f(int n){
    double prev_prev=0d, prev=1d, result=0d;
    for (int i = 1; i <n; i++) {
        result=prev_prev+prev;
        prev_prev=prev;
        prev=result;
    }
    return result;
}

With the above non-recursive code, to get fib(n), simply call f(n-1).

Can't we stay with the recursive way of implementation? Sure. I view tail recursion as a way of introducing a running total which is updated through parameter of the recursive calls. In the Fabonacci example below, 'running tatal' is updated and maintained through 'val' parameter, and the previous value to be calculated toward Fabonocci is passed through the 'prev' parameter:

double fib(int term, double val = 1d, double prev = 0d)

Consider the following tail recursion way of implementation using Groovy's default parameter (not supported by Java):

double fib(int term, double val = 1d, double prev = 0d){
   if(term == 0) return prev;
   if(term == 1) return val;
   return fib(term - 1, val+prev, val);
}

references

• What is tail-recursion?
  
• Bad Code Leads to Bad Conclusions
  
• Tail Recursion