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.

Answer 2

public static BigDecimal sqrt(BigDecimal A, final int SCALE) {     BigDecimal x0 = new BigDecimal("0");     BigDecimal x1 = new BigDecimal(Math.sqrt(A.doubleValue()));     while (!x0.equals(x1)) {         x0 = x1;         x1 = A.divide(x0, SCALE, ROUND_HALF_UP);         x1 = x1.add(x0);         x1 = x1.divide(TWO, SCALE, ROUND_HALF_UP);      }     return x1; } 

This work perfect! Very fast for more than 65536 digits!