Java中BigDecimal的平方根 [英] Square root of BigDecimal in Java

问题描述

我们是否可以仅使用Java API而不是定制的100行算法来计算Java中 BigDecimal 的平方根？

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?

推荐答案

我已经用过这个了，效果很好。
以下是算法如何在高级别运行的示例。

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

编辑：我很想知道下面定义的准确程度。以下是来自官方来源的sqrt（2）：

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

和这里使用我在下面概述的方法， SQRT_DIG 等于150：

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

(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206086685

在第一偏差后的 195位精度<发生/ strong>即可。如果您需要如此高的精确度，请自担风险。

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.

将 SQRT_DIG 更改为1000 1570位精度。

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));
}

请务必查看barwnikk的答案。它更简洁，看似提供更好或更好的精度。

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

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