How does the f function end up with a value of 1024?

I’m trying to solve a sample test question and I’m having trouble understanding the last part of the problem. For the first print statement, I get 32. Using that same process I get a completely different value than the one returned by my compiler. The compiler returns 32 then 1024. Using the same logic I used to get 32, I’m confused on how the compiler gets 1024. Could anyone clearly explain what I’m doing wrong?

public static void main(String[] args) {
    System.out.println(f(2, 5));
    System.out.println(f(2, 10));
}
public static long f(long x, long y) {
    if (y == 0) {
        return 1;
    } 
    else if (y == 1) {
        return x;
    } 
    else {
        long z = y / 2;
        long e = f(x, z);
        if (y % 2 == 0) {
            return e * e;
        } 
        else {
            return x * e * e;
        }
    }
}

>Solution :

Just.. walk through it.

f(2, 10)

y is neither 0 nor 1, but is divisible by 2 so the else clause triggers, and the y%2 == 0 clause. e*e is returned, where e is f(2, 5).

f(2, 5)

y is neither 0 nor 1, and is not divisible by 2 so both else clauses trigger. x*e*e is returned, where x is 2 and e is f(2, 2).

f(2, 2)

y is neither 0 nor 1, but is divisible by 2 so the else clause triggers, and the y%2 == 0 clause. e*e is returned, where e is f(2, 1).

f(2, 1)

y is 1, therefore x is returned, which is 2.

Now go backwards

• f(2, 1) = 2
• f(2, 2) = e*e = 4
• f(2, 5) = 2*e*e = 32
• f(2, 10) = e*e = 1024