How does the recursion here work?

Asked 2010-03-09T05:14:43.120

Code 1:

public static int fibonacci (int n){ 
    if (n == 0 || n == 1) { 
        return 1; 
    } else { 
        return fibonacci (n-1) + fibonacci (n-2); 
    }        
}

How can you use fibonacci if you haven't gotten done explaining what it is yet? I've been able to understand using recursion in other cases like this:

Code 2:

class two 
{
    public static void two (int n) 
    {
        if (n>0) 
        {
            System.out.println (n) ;
            two (n-1) ;
        }
        else
        {
            return ;
        }
    } 

    public static void main (String[] arg) 
    {
        two (12) ;
    }
}

In the case of code 2, though, n will eventually reach a point at which it doesn't satisfy n>0 and the method will stop calling itself recursively. In the case of code 2, though, I don't see how it would be able to get itself from 1 if n=1 was the starting point to 2 and 3 and 5 and so on. Also, I don't see how the line return fibonacci (n-1) + fibonacci (n-2) would work since fibonacci (n-2) has to contain in some sense fibonacci (n-1) in order to work, but it isn't there yet.

The book I'm looking at says it will work. How does it work?

java recursion fibonacci

Edit Report

4 Answers

In the case of code 2 though n will eventualy reach a point at which it doesnt satisfy n>0 and the method will stop calling itself recursivly

to make it look similar you can replace condition if (n == 0 || n == 1) with if (n < 2)

Also i don't see how the line `return fibonacci (n-1) + fibonacci (n-2) would work since fibbonacci n-2 has to contain in some sense fibonacci n-1 in order to wrok but it isn't there yet.

I suspect you wanted to write: "since fibbonacci n-1 has to contain in some sense fibonacci n-2"
If I'm right, then you will see from the example below, that actually fibonacci (n-2) will be called twice for every recursion level (fibonacci(1) in the example):
1. when executing fibonacci (n-2) on the current step
2. when executing fibonacci ((n-1)-1) on the next step

(Also take a closer look at the Spike's comment)

Suppose you call fibonacci(3), then call stack for fibonacci will be like this:
(Veer provided more detailed explanation)

n=3. fibonacci(3)  
n=3. fibonacci(2) // call to fibonacci(n-1)
   n=2. fibonacci(1) // call to fibonacci(n-1)
      n=1. returns 1
   n=2. fibonacci(0) // call to fibonacci(n-2)
      n=0. returns 1
   n=2. add up, returns 2
n=3. fibonacci(1) //call to fibonacci(n-2)
   n=1. returns 1
n=3. add up, returns 2 + 1

Note, that adding up in fibonacci(n) takes place only after all functions for smaller args return (i.e. fibonacci(n-1), fibonacci(n-2)..

answered 2010-03-09T07:13:58.160

"How can you use Fibonacci if you haven't gotten done explaining what it is yet?"

This is an interesting way to question recursion. Here's part of an answer: While you're defining Fibonacci, it hasn't been defined yet, but it has been declared. The compiler knows that there is a thing called Fibonacci, and that it will be a function of type int -> int and that it will be defined whenever the program runs.

In fact, this is how all identifiers in C programs work, not just recursive ones. The compiler determines what things have been declared, and then goes through the program pointing uses of those things to where the things actually are (gross oversimplification).

answered 2010-03-09T05:28:14.307

Try debugging and use watches to know the state of the variable

answered 2010-03-09T05:22:45.197

Understanding recursion requires also knowing how the call stack works i.e. how functions call each other.
If the function didn't have the condition to stop if n==0 or n==1, then the function would call itself recursively forever. It works because eventually, the function is going to petter out and return 1. at that point, the return fibonacci (n-1) + fibonacci (n-2) will also return with a value, and the call stack gets cleaned up really quickly.

answered 2010-03-09T05:28:31.040

How does the recursion here work?

4 Answers

Your Answer