[FIXED] How do you calculate log base 2 in Java for integers?

Issue

I use the following function to calculate log base 2 for integers:

public static int log2(int n){
    if(n <= 0) throw new IllegalArgumentException();
    return 31 - Integer.numberOfLeadingZeros(n);
}

Does it have optimal performance?

Does someone know ready J2SE API function for that purpose?

UPD1 Surprisingly for me, float point arithmetics appears to be faster than integer arithmetics.

UPD2 Due to comments I will conduct more detailed investigation.

UPD3 My integer arithmetic function is 10 times faster than Math.log(n)/Math.log(2).

Solution

If you are thinking about using floating-point to help with integer arithmetics, you have to be careful.

I usually try to avoid FP calculations whenever possible.

Floating-point operations are not exact. You can never know for sure what will (int)(Math.log(65536)/Math.log(2)) evaluate to. For example, Math.ceil(Math.log(1<<29) / Math.log(2)) is 30 on my PC where mathematically it should be exactly 29. I didn't find a value for x where (int)(Math.log(x)/Math.log(2)) fails (just because there are only 32 "dangerous" values), but it does not mean that it will work the same way on any PC.

The usual trick here is using "epsilon" when rounding. Like (int)(Math.log(x)/Math.log(2)+1e-10) should never fail. The choice of this "epsilon" is not a trivial task.

More demonstration, using a more general task - trying to implement int log(int x, int base):

The testing code:

static int pow(int base, int power) {
    int result = 1;
    for (int i = 0; i < power; i++)
        result *= base;
    return result;
}

private static void test(int base, int pow) {
    int x = pow(base, pow);
    if (pow != log(x, base))
        System.out.println(String.format("error at %d^%d", base, pow));
    if(pow!=0 && (pow-1) != log(x-1, base))
        System.out.println(String.format("error at %d^%d-1", base, pow));
}

public static void main(String[] args) {
    for (int base = 2; base < 500; base++) {
        int maxPow = (int) (Math.log(Integer.MAX_VALUE) / Math.log(base));
        for (int pow = 0; pow <= maxPow; pow++) {
            test(base, pow);
        }
    }
}

If we use the most straight-forward implementation of logarithm,

static int log(int x, int base)
{
    return (int) (Math.log(x) / Math.log(base));
}

this prints:

error at 3^5
error at 3^10
error at 3^13
error at 3^15
error at 3^17
error at 9^5
error at 10^3
error at 10^6
error at 10^9
error at 11^7
error at 12^7
...

To completely get rid of errors I had to add epsilon which is between 1e-11 and 1e-14. Could you have told this before testing? I definitely could not.

Answered By - Rotsor

Answer Checked By - Marilyn (PHPFixing Volunteer)