recursive method to comput psF(n)
can someone explain to me how recursive methods work
we went over it in class but i was not listening and now i'am to scared to tell the teacher. :) can't focus for more than 50 mins at a time
how can you use a method within it self
static int psF(int n) {
if(n<2)
return n;
else
return 2*psF(n-1)+psF(n-2);
}
how can he use a method call with in a method
[427 byte] By [
Mcody2] at [2007-11-11 9:56:34]

# 2 Re: recursive method to comput psF(n)
can someone explain to me how recursive methods work
we went over it in class but i was not listening and now i'am to scared to tell the teacher. :) can't focus for more than 50 mins at a time
how can you use a method within it self
static int psF(int n) {
if(n<2)
return n;
else
return 2*psF(n-1)+psF(n-2);
}
how can he use a method call with in a method
If you don't understand some code, try putting some System.out.println() statements in it to see what's happening to the variables in your code.
The previous' posters code can even be shortened with only one if-statement:
public class Fibo {
public static int fibonacci(int n) {
System.out.println("********************************************");
System.out.println("Begin of the method, n = "+n);
if(n < 2) {
System.out.println(" - returning: "+n);
return n;
} else {
System.out.println(" - calling fibonacci("+n+"-1)+fibonacci("+n+"-2)");
return fibonacci(n-1)+fibonacci(n-2);
}
}
public static void main(String[] args) {
int N = 4;
System.out.println("\n\nAnswer fibonacci("+N+") = "+fibonacci(N));
}
}
When you run the code the following will be printed:
********************************************
Begin of the method, n = 4,
- calling fibonacci(4-1)+fibonacci(4-2) [1]-+
******************************************** |
Begin of the method, n = 3, |
- calling fibonacci(3-1)+fibonacci(3-2) <--+ [2]-+
******************************************** | |
Begin of the method, n = 2, | |
- calling fibonacci(2-1)+fibonacci(2-2) | <--+ [3]-+
******************************************** | | |
Begin of the method, n = 1, | | |
- returning: 1 | | <--+
******************************************** | | |
Begin of the method, n = 0, | | |
- returning: 0 | | <--+
******************************************** | |
Begin of the method, n = 1, | |
- returning: 1 | <--+
******************************************** |
Begin of the method, n = 2, |
- calling fibonacci(2-1)+fibonacci(2-2) <--+ [4]-+
******************************************** |
Begin of the method, n = 1, |
- returning: 1 <--+
******************************************** |
Begin of the method, n = 0, |
- returning: 0 <--+
Answer fibonacci(4) = 3
I've drawn some arrows to show how the flow of the recursive calls work. As you can see, the number 1 gets returned three times, which is the answer.
Hope it's clear to you now.
Good luck.