Question

Theres an exercise from a Java book I’m reading that has me confused:

A Fibonacci sequence is the sequence of numbers 1, 1, 2, 3, 5, 8,
13, 21, 34, etc., where each number (from the third on) is the sum
of the previous two. Create a method that takes an integer as an
argument and displays that many Fibonacci numbers starting from the
beginning. If, e.g., you run java Fibonacci 5 (where Fibonacci is
the name of the class) the output will be: 1, 1, 2, 3, 5.

I could have swore that it would need an array or some way of storing the previous numbers but when I saw the answer it wasn’t the case:

import java.util.*; 

public class Fibonacci { 

    static int fib(int n) {
         if (n <= 2) 
              return 1;

         return fib(n-1) + fib(n-2);
    } 
    public static void main(String[] args) {
    // Get the max value from the command line: 
    int n = Integer.parseInt(args[0]); 
    if(n < 0) {
        System.out.println("Cannot use negative numbers"); return;
    } 
    for(int i = 1; i <= n; i++)
        System.out.print(fib(i) + ", ");
    }
}

Would someone be able to explain how using a function within itself produces this?

Answer 1

The code you gave is an example of a recursive solution. When the function is called for the first time, it executes until it calls itself. Its state is stored on the stack, and the code begins executing again with new input data. This process repeats until the input is less than 2, at which point 1 is returned, and the answer returns to the previous caller.

For example, calling fib(5) results in the following execution:

fib(5):
    fib(4):
        fib(3):
            fib(2): 1
            fib(1): 1
        fib(2): 1
    fib(3):
        fib(2): 1
        fib(1): 1

Note that you are partially correct. Intermediate results are stored on the stack in this solution. That is one of the reasons why the recursive solution is so expensive. The other is its O(2^n) complexity. However, if is possible to compute Fibonacci(n) iteratively and without storing all previous results. All you really need is to store the last to results and count from 1 up to n.