Learning Java - Do not fully understand how this sequence is calculated (Fibonacci) [duplicate]

Question

This question already has an answer here:

• Java recursive Fibonacci sequence 36 answers

I am learning Java and I have this code from the internet and running it in Eclipse:

public class Fibonacci {

    public static void main (String [] args) { 
        for (int counter = 0; counter <= 3; counter++){
            System.out.printf("Fibonacci of %d is: %d\n",  counter, fibonacci(counter));

    }

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

I've tried to understand it but cannot get it. So I run through the code and counter gets passed in through the fibonacci method. As counter starts at 0 and this is what gets passed first, then 1 and I understand the method passes back 0 and then 1.

When it reaches 2: it will return 2-1 + 2-2 = 2 and it does return this.
When it reaches 3: it will return 3-1 + 3-2 = 3 but it does not return 3 it returns 2.

Please can someone explain to me why as I cannot figure this out?

Thanks

Answer 1

First, I have to tell you that this recursive version has a dramatic exponential cost. Once you understand how it works, my advice for you would be to learn about tail recursivity, write a tail-recursive solution, an iterative solution, and compare them to your current method for high values of "number".

Then, your function basically uses the mathematical definition of the Fibonacci sequence :

f0 = 1, f1 = 1, fn = fn-1 + fn-2 for all n >= 2

For example if we call fibonacci(3), this will return fibonacci(2) + fibonacci(1). fibonacci(2) will be executed first and will return fibonacci(1) + fibonnacci(0). Then fibonacci(1) will return immediately 1 since it is a terminal case. It happens the same thing with fibonnacci(0), so now we have computed fibonnacci(2) = 1 + 0 = 1. Let's go back to fibonacci(3) which has been partially evaluated at this point : 1 + fibonnacci(1). We just have to compute fibonnacci(1) and we can finally return 1 + 1 = 2.

Even in this little example, you can see that we evaluated twice fibonacci(1), that is why this version is so slow, it computes many times the same values of the sequence, and it gets worth when "number" is high.