Square root of BigDecimal in Java

Question

Can we compute the square root of a BigDecimal in Java by using only the Java API and not a custom-made 100-line algorithm?

Answer 1

I've used this and it works quite well. Here's an example of how the algorithm works at a high level.

Edit: I was curious to see just how accurate this was as defined below. Here is the sqrt(2) from an official source:

(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147

and here it is using the approach I outline below with SQRT_DIG equal to 150:

(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206086685

The first deviation occurs after 195 digits of precision. Use at your own risk if you need such a high level of precision as this.

Changing SQRT_DIG to 1000 yielded 1570 digits of precision.

private static final BigDecimal SQRT_DIG = new BigDecimal(150);
private static final BigDecimal SQRT_PRE = new BigDecimal(10).pow(SQRT_DIG.intValue());

/**
 * Private utility method used to compute the square root of a BigDecimal.
 * 
 * @author Luciano Culacciatti 
 * @url http://www.codeproject.com/Tips/257031/Implementing-SqrtRoot-in-BigDecimal
 */
private static BigDecimal sqrtNewtonRaphson  (BigDecimal c, BigDecimal xn, BigDecimal precision){
    BigDecimal fx = xn.pow(2).add(c.negate());
    BigDecimal fpx = xn.multiply(new BigDecimal(2));
    BigDecimal xn1 = fx.divide(fpx,2*SQRT_DIG.intValue(),RoundingMode.HALF_DOWN);
    xn1 = xn.add(xn1.negate());
    BigDecimal currentSquare = xn1.pow(2);
    BigDecimal currentPrecision = currentSquare.subtract(c);
    currentPrecision = currentPrecision.abs();
    if (currentPrecision.compareTo(precision) <= -1){
        return xn1;
    }
    return sqrtNewtonRaphson(c, xn1, precision);
}

/**
 * Uses Newton Raphson to compute the square root of a BigDecimal.
 * 
 * @author Luciano Culacciatti 
 * @url http://www.codeproject.com/Tips/257031/Implementing-SqrtRoot-in-BigDecimal
 */
public static BigDecimal bigSqrt(BigDecimal c){
    return sqrtNewtonRaphson(c,new BigDecimal(1),new BigDecimal(1).divide(SQRT_PRE));
}

be sure to check out barwnikk's answer. it's more concise and seemingly offers as good or better precision.