Question

Here is the output for the below program.

value is : 2.7755575615628914E-17
Double.compare with zero : 1
isEqual with zero : true

My question is, what should be an epsilon value? Is there any robust way to obtain the value, instead of picking a number out from the sky.

---

package sandbox;

/**
 *
 * @author yccheok
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        double zero = 1.0/5.0 + 1.0/5.0 - 1.0/10.0 - 1.0/10.0 - 1.0/10.0 - 1.0/10.0;
        System.out.println("value is : " + zero);
        System.out.println("Double.compare with zero : " + Double.compare(zero, 0.0));
        System.out.println("isEqual with zero : " + isEqual(zero, 0.0));
    }

    public static boolean isEqual(double d0, double d1) {
        final double epsilon = 0.0000001;
        return d0 == d1 ? true : Math.abs(d0 - d1) < epsilon;
    }
}

Answer 1

The answer to your second question is no. The magnitude of finite-machine precision error can be arbitrarily large:

public static void main(String[] args) {
    double z = 0.0;
    double x = 0.23;
    double y = 1.0 / x;
    int N = 50000;
    for (int i = 0; i < N; i++) {
        z += x * y - 1.0;
    }
    System.out.println("z should be zero, is " + z);
}

This gives ~5.55E-12, but if you increase N you can get just about any level of error you desire.

There is a vast amount of past and current research on how to write numerically stable algorithms. It is a hard problem.