Calculating nth root in Java using power method

Question

I was trying to get a cubic root in java using Math.pow(n, 1.0/3) but because it divides doubles, it doesn\'t return the exact answer. For example, with 125, this g

Answer 1

Since it is not possible to have arbitrary-precision calculus with double, you have three choices:

1. Define a precision for which you decide whether a double value is an integer or not.
2. Test whether the rounded value of the double you have is a correct result.
3. Do calculus on a BigDecimal object, which supports arbitrary-precision double values.

Option 1

private static boolean isNthRoot(int value, int n, double precision) {
    double a = Math.pow(value, 1.0 / n);
    return Math.abs(a - Math.round(a)) < precision; // if a and round(a) are "close enough" then we're good
}

The problem with this approach is how to define "close enough". This is a subjective question and it depends on your requirements.

Option 2

private static boolean isNthRoot(int value, int n) {
    double a = Math.pow(value, 1.0 / n);
    return Math.pow(Math.round(a), n) == value;
}

The advantage of this method is that there is no need to define a precision. However, we need to perform another pow operation so this will affect performance.

Option 3

There is no built-in method to calculate a double power of a BigDecimal. This question will give you insight on how to do it.

Answer 2

I wrote this method to compute floor(x^(1/n)) where x is a non-negative BigInteger and n is a positive integer. It was a while ago now so I can't explain why it works, but I'm reasonably confident that when I wrote it I was happy that it's guaranteed to give the correct answer reasonably quickly.

To see if x is an exact n-th power you can check if the result raised to the power n gives you exactly x back again.

public static BigInteger floorOfNthRoot(BigInteger x, int n) {
    int sign = x.signum();
    if (n <= 0 || (sign < 0))
        throw new IllegalArgumentException();
    if (sign == 0)
        return BigInteger.ZERO;
    if (n == 1)
        return x;
    BigInteger a;
    BigInteger bigN = BigInteger.valueOf(n);
    BigInteger bigNMinusOne = BigInteger.valueOf(n - 1);
    BigInteger b = BigInteger.ZERO.setBit(1 + x.bitLength() / n);
    do {
        a = b;
        b = a.multiply(bigNMinusOne).add(x.divide(a.pow(n - 1))).divide(bigN);
    } while (b.compareTo(a) == -1);
    return a;
}

To use it:

System.out.println(floorOfNthRoot(new BigInteger("125"), 3));

Edit Having read the comments above I now remember that this is the Newton-Raphson method for n-th roots. The Newton-Raphson method has quadratic convergence (which in everyday language means it's fast). You can try it on numbers which have dozens of digits and you should get the answer in a fraction of a second.

You can adapt the method to work with other number types, but double and BigDecimal are in my view not suited for this kind of thing.

Answer 3

I'd go for implementing my own function to do this, possibly based on this method.