java - Java中BigDecimal的平方根

Question

BigDecimal我们可以只使用 Java API 而不是定制的 100 行算法来计算 Java 中 a 的平方根吗？

Answer 1

我用过这个，效果很好。 这是该算法如何在高层次上工作的示例。

编辑：我很想知道下面定义的准确度。这是来自官方来源的 sqrt(2) ：

(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147

这里使用的是我在下面概述的方法，SQRT_DIG等于 150：

(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206086685

第一个偏差发生在195 位精度之后。如果您需要如此高的精度，请自行承担风险。

更改SQRT_DIG为 1000 会产生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，似乎提供了良好或更好的精度。

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;
}

这项工作完美！超过 65536 位数字非常快！

Answer 3

从 Java 9 开始，您可以！见BigDecimal.sqrt()。

Answer 4

通过使用 Karp 的技巧，这可以在没有循环的情况下仅在两行中实现，精度为 32 位：

public static BigDecimal sqrt(BigDecimal value) {
    BigDecimal x = new BigDecimal(Math.sqrt(value.doubleValue()));
    return x.add(new BigDecimal(value.subtract(x.multiply(x)).doubleValue() / (x.doubleValue() * 2.0)));
}

Answer 5

如果您只需要找到整数平方根- 这是可以使用的两种方法。

牛顿法- 即使对于 1000 位 BigInteger 也非常快：

public static BigInteger sqrtN(BigInteger in) {
    final BigInteger TWO = BigInteger.valueOf(2);
    int c;

    // Significantly speed-up algorithm by proper select of initial approximation
    // As square root has 2 times less digits as original value
    // we can start with 2^(length of N1 / 2)
    BigInteger n0 = TWO.pow(in.bitLength() / 2);
    // Value of approximate value on previous step
    BigInteger np = in;

    do {
        // next approximation step: n0 = (n0 + in/n0) / 2
        n0 = n0.add(in.divide(n0)).divide(TWO);

        // compare current approximation with previous step
        c = np.compareTo(n0);

        // save value as previous approximation
        np = n0;

        // finish when previous step is equal to current
    }  while (c != 0);

    return n0;
}

二分法- 比牛顿法慢 50 倍 - 仅用于教育目的：

 public static BigInteger sqrtD(final BigInteger in) {
    final BigInteger TWO = BigInteger.valueOf(2);
    BigInteger n0, n1, m, m2, l;
    int c;

    // Init segment
    n0 = BigInteger.ZERO;
    n1 = in;

    do {
        // length of segment
        l = n1.subtract(n0);

        // middle of segment
        m = l.divide(TWO).add(n0);

        // compare m^2 with in
        c = m.pow(2).compareTo(in);

        if (c == 0) {
            // exact value is found
            break;
        }  else if (c > 0) {
            // m^2 is bigger than in - choose left half of segment
            n1 = m;
        } else {
            // m^2 is smaller than in - choose right half of segment
            n0 = m;
        }

        // finish if length of segment is 1, i.e. approximate value is found
    }  while (l.compareTo(BigInteger.ONE) > 0);

    return m;
}

Answer 6

java api中没有任何东西，所以如果double不够准确（如果没有，为什么要使用BigDecimal？）那么你需要类似下面的代码。）

import java.math.BigDecimal;

public class BigDSqrt {
  public static BigDecimal sqrt(BigDecimal n, int s) {
    BigDecimal TWO = BigDecimal.valueOf(2);

    // Obtain the first approximation
    BigDecimal x = n
        .divide(BigDecimal.valueOf(3), s, BigDecimal.ROUND_DOWN);
    BigDecimal lastX = BigDecimal.valueOf(0);

    // Proceed through 50 iterations
    for (int i = 0; i < 50; i++) {
      x = n.add(x.multiply(x)).divide(x.multiply(TWO), s,
          BigDecimal.ROUND_DOWN);
      if (x.compareTo(lastX) == 0)
        break;
      lastX = x;
    }
    return x;
  }
}

来源：http ://www.java2s.com/Code/Java/Language-Basics/DemonstrationofhighprecisionarithmeticwiththeBigDoubleclass.htm

Answer 7

正如之前所说：如果您不介意答案的精度，但只想在第 15 个仍然有效的数字之后生成随机数字，那么您为什么要使用 BigDecimal 呢？

这是应该使用浮点 BigDecimals 实现技巧的方法的代码：

    import java.math.BigDecimal;
    import java.math.BigInteger;
    import java.math.MathContext;



public BigDecimal bigSqrt(BigDecimal d, MathContext mc) {
    // 1. Make sure argument is non-negative and treat Argument 0
    int sign = d.signum();
    if(sign == -1)
      throw new ArithmeticException("Invalid (negative) argument of sqrt: "+d);
    else if(sign == 0)
      return BigDecimal.ZERO;
    // 2. Scaling:
    // factorize d = scaledD * scaleFactorD 
    //             = scaledD * (sqrtApproxD * sqrtApproxD)
    // such that scalefactorD is easy to take the square root
    // you use scale and bitlength for this, and if odd add or subtract a one
    BigInteger bigI=d.unscaledValue();
    int bigS=d.scale();
    int bigL = bigI.bitLength();
    BigInteger scaleFactorI;
    BigInteger sqrtApproxI;
    if ((bigL%2==0)){
       scaleFactorI=BigInteger.ONE.shiftLeft(bigL);
       sqrtApproxI=BigInteger.ONE.shiftLeft(bigL/2);           
    }else{
       scaleFactorI=BigInteger.ONE.shiftLeft(bigL-1);
       sqrtApproxI=BigInteger.ONE.shiftLeft((bigL-1)/2 );          
    }
    BigDecimal scaleFactorD;
    BigDecimal sqrtApproxD;
    if ((bigS%2==0)){
        scaleFactorD=new BigDecimal(scaleFactorI,bigS);
        sqrtApproxD=new BigDecimal(sqrtApproxI,bigS/2);
    }else{
        scaleFactorD=new BigDecimal(scaleFactorI,bigS+1);
        sqrtApproxD=new BigDecimal(sqrtApproxI,(bigS+1)/2);         
    }
    BigDecimal scaledD=d.divide(scaleFactorD);

    // 3. This is the core algorithm:
    //    Newton-Ralpson for scaledD : In case of f(x)=sqrt(x),
    //    Heron's Method or Babylonian Method are other names for the same thing.
    //    Since this is scaled we can be sure that scaledD.doubleValue() works 
    //    for the start value of the iteration without overflow or underflow
    System.out.println("ScaledD="+scaledD);
    double dbl = scaledD.doubleValue();
    double sqrtDbl = Math.sqrt(dbl);
    BigDecimal a = new BigDecimal(sqrtDbl, mc);

    BigDecimal HALF=BigDecimal.ONE.divide(BigDecimal.ONE.add(BigDecimal.ONE));
    BigDecimal h = new BigDecimal("0", mc);
    // when to stop iterating? You start with ~15 digits of precision, and Newton-Ralphson is quadratic
    // in approximation speed, so in roundabout doubles the number of valid digits with each step.
    // This fmay be safer than testing a BigDecifmal against zero.
    int prec = mc.getPrecision();
    int start = 15;
    do {
        h = scaledD.divide(a, mc);
        a = a.add(h).multiply(HALF);
        start *= 2;
    } while (start <= prec);        
    // 3. Return rescaled answer. sqrt(d)= sqrt(scaledD)*sqrtApproxD :          
    return (a.multiply(sqrtApproxD));
}

作为一个测试，尝试重复数次平方而不是重复平方根，看看你离开始的地方有多近。

Answer 8

如果要计算位数超过双精度数的数字的平方根（大比例的 BigDecimal）：

维基百科有一篇计算平方根的文章：http ://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method

这是我的实现：

public static BigDecimal sqrt(BigDecimal in, int scale){
    BigDecimal sqrt = new BigDecimal(1);
    sqrt.setScale(scale + 3, RoundingMode.FLOOR);
    BigDecimal store = new BigDecimal(in.toString());
    boolean first = true;
    do{
        if (!first){
            store = new BigDecimal(sqrt.toString());
        }
        else first = false;
        store.setScale(scale + 3, RoundingMode.FLOOR);
        sqrt = in.divide(store, scale + 3, RoundingMode.FLOOR).add(store).divide(
                BigDecimal.valueOf(2), scale + 3, RoundingMode.FLOOR);
    }while (!store.equals(sqrt));
    return sqrt.setScale(scale, RoundingMode.FLOOR);
}

setScale(scale + 3, RoundingMode.Floor)因为过度计算会提供更高的准确性。RoundingMode.Floor截断数字，进行RoundingMode.HALF_UP正常舍入。

Answer 9

这是非常准确和快速的解决方案，它基于我的BigIntSqRoot 解决方案和下一个观察：A^2B 的平方根 - 是 A 乘以 B 的根。使用这种方法，我可以轻松计算 2 的平方根的前 1000 位。

1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727350138462309122970249248360558507372126441214970999358314132226659275055927557999505011527820605714701095599716059702745345968620147285174186408891986095523292304843087143214508397626036279952514079896872533965463318088296406206152583523950547457502877599617298355752203375318570113543746034084988471603868999706990048150305440277903164542478230684929369186215805784631115966687130130156185689872372352885092648612494977154218334204285686060146824720771435854874155657069677653720226485447015858801620758474922657226002085584466521458398893944370926591800311388246468157082630100594858704003186480342194897278290641045072636881313739855256117322040245091227700226941127573627280495738108967504018369868368450725799364729060762996941380475654823728997180326802474420629269124859052181004459842150591120249441341728531478105803603371077309182869314710171111683916581726889419758716582152128229518488472

所以这里是源代码

public class BigIntSqRoot {
    private static final int PRECISION = 10000;
    private static BigInteger multiplier = BigInteger.valueOf(10).pow(PRECISION * 2);
    private static BigDecimal root = BigDecimal.valueOf(10).pow(PRECISION);
    private static BigInteger two = BigInteger.valueOf(2L);

    public static BigDecimal bigDecimalSqRootFloor(BigInteger x)
            throws IllegalArgumentException {
        BigInteger result = bigIntSqRootFloor(x.multiply(multiplier));
        //noinspection BigDecimalMethodWithoutRoundingCalled
        return new BigDecimal(result).divide(root);
    }

    public static BigInteger bigIntSqRootFloor(BigInteger x)
            throws IllegalArgumentException {
        if (checkTrivial(x)) {
            return x;
        }
        if (x.bitLength() < 64) { // Can be cast to long
            double sqrt = Math.sqrt(x.longValue());
            return BigInteger.valueOf(Math.round(sqrt));
        }
        // starting with y = x / 2 avoids magnitude issues with x squared
        BigInteger y = x.divide(two);
        BigInteger value = x.divide(y);
        while (y.compareTo(value) > 0) {
            y = value.add(y).divide(two);
            value = x.divide(y);
        }
        return y;
    }

    public static BigInteger bigIntSqRootCeil(BigInteger x)
            throws IllegalArgumentException {
        BigInteger y = bigIntSqRootFloor(x);
        if (x.compareTo(y.multiply(y)) == 0) {
            return y;
        }
        return y.add(BigInteger.ONE);
    }

    private static boolean checkTrivial(BigInteger x) {
        if (x == null) {
            throw new NullPointerException("x can't be null");
        }
        if (x.compareTo(BigInteger.ZERO) < 0) {
            throw new IllegalArgumentException("Negative argument.");
        }

        return x.equals(BigInteger.ZERO) || x.equals(BigInteger.ONE);
    }
}

Answer 10

我想出了一个算法，它不仅取平方根，而且取每个 BigDecimal 的整数以下的每个根。它的最大优势在于它不执行搜索算法，因此运行时间为 0,1ms - 1ms，速度非常快。

但是你得到的速度和多功能性，它缺乏准确性，它平均 5 个正确的数字，第五个数字的偏差为 3。（用一百万个随机数和根进行测试），虽然测试运行的根非常高，所以如果你将根保持在 10 以下，你可以期待更高的准确性。

结果只有 64 位精度，其余数字为零，因此如果您需要非常高的精度，请不要使用此函数。

它可以处理非常大的数字，非常大的根，而不是非常小的数字。

public static BigDecimal nrt(BigDecimal bd,int root) {
//if number is smaller then double_max_value it's faster to use the usual math 
//library
    if(bd.compareTo(BigDecimal.valueOf(Double.MAX_VALUE)) < 0) 
        return new BigDecimal( Math.pow(bd.doubleValue(), 1D / (double)root ));

    BigDecimal in = bd;
    int digits = bd.precision() - bd.scale() -1; //take digits to get the numbers power of ten
    in = in.scaleByPowerOfTen (- (digits - digits%root) ); //scale down to the lowest number with it's power of ten mod root is the same as initial number

    if(in.compareTo(BigDecimal.valueOf( Double.MAX_VALUE) ) > 0) { //if down scaled value is bigger then double_max_value, we find the answer by splitting the roots into factors and calculate them seperately and find the final result by multiplying the subresults
        int highestDenominator = highestDenominator(root);
        if(highestDenominator != 1) {
            return nrt( nrt(bd, root / highestDenominator),highestDenominator); // for example turns 1^(1/25) 1^(1/5)^1(1/5)
        }
        //hitting this point makes the runtime about 5-10 times higher,
        //but the alternative is crashing
        else return nrt(bd,root+1) //+1 to make the root even so it can be broken further down into factors
                    .add(nrt(bd,root-1),MathContext.DECIMAL128) //add the -1 root and take the average to deal with the inaccuracy created by this
                    .divide(BigDecimal.valueOf(2),MathContext.DECIMAL128); 
    } 
    double downScaledResult = Math.pow(in.doubleValue(), 1D /root); //do the calculation on the downscaled value
    BigDecimal BDResult =new BigDecimal(downScaledResult) // scale back up by the downscaled value divided by root
            .scaleByPowerOfTen( (digits - digits % root) / root );
    return BDResult;
}
private static int highestDenominator(int n) {
    for(int i = n-1; i>1;i--) {
        if(n % i == 0) {
            return i;
        }
    }
    return 1;
}

它通过使用一个数学属性来工作，该属性基本上说，当你做平方根时，你可以将 x^0.5 更改为 (x/100)^0,5 * 10，因此将底数除以 100 取幂并乘以 10。

将其推广为 x^(1/n) = (x / 10^n) ^ (1/n) * 10。

因此，对于立方根，您需要将底除以 10^3，对于四根，您需要除以 10^4，依此类推。

该算法使用该函数将输入缩小到数学库可以处理的范围，然后根据输入的缩小程度再次将其放大。

它还处理了一些输入无法足够缩小的边缘情况，正是这些边缘情况增加了很多准确性问题。

Answer 11

public static BigDecimal sqrt( final BigDecimal value )
{
    BigDecimal guess = value.multiply( DECIMAL_HALF ); 
    BigDecimal previousGuess;

    do
    {
        previousGuess = guess;
        guess = sqrtGuess( guess, value );
   } while ( guess.subtract( previousGuess ).abs().compareTo( EPSILON ) == 1 );

    return guess;
}

private static BigDecimal sqrtGuess( final BigDecimal guess,
                                     final BigDecimal value )
{
    return guess.subtract( guess.multiply( guess ).subtract( value ).divide( DECIMAL_TWO.multiply( guess ), SCALE, RoundingMode.HALF_UP ) );
}

private static BigDecimal epsilon()
{
    final StringBuilder builder = new StringBuilder( "0." );

    for ( int i = 0; i < SCALE - 1; ++i )
    {
        builder.append( "0" );
    }

    builder.append( "1" );

    return new BigDecimal( builder.toString() );
}

private static final int SCALE = 1024;
private static final BigDecimal EPSILON = epsilon();
public static final BigDecimal DECIMAL_HALF = new BigDecimal( "0.5" );
public static final BigDecimal DECIMAL_TWO = new BigDecimal( "2" );

Answer 12

BigDecimal.valueOf(Math.sqrt(myBigDecimal.doubleValue()));