BigDecimal的对数

Java Number Cruncher: The Java Programmer's Guide to Numerical Computing 提供使用 Newton's Method . 书中的源代码可用 here . 以下内容摘自第章 12.5大除数函数 (第330页和第331页):

/**
 * Compute the natural logarithm of x to a given scale, x > 0.
 */
public static BigDecimal ln(BigDecimal x, int scale)
{
    // Check that x > 0.
    if (x.signum() <= 0) {
        throw new IllegalArgumentException("x <= 0");
    }

    // The number of digits to the left of the decimal point.
    int magnitude = x.toString().length() - x.scale() - 1;

    if (magnitude < 3) {
        return lnNewton(x, scale);
    }

    // Compute magnitude*ln(x^(1/magnitude)).
    else {

        // x^(1/magnitude)
        BigDecimal root = intRoot(x, magnitude, scale);

        // ln(x^(1/magnitude))
        BigDecimal lnRoot = lnNewton(root, scale);

        // magnitude*ln(x^(1/magnitude))
        return BigDecimal.valueOf(magnitude).multiply(lnRoot)
                    .setScale(scale, BigDecimal.ROUND_HALF_EVEN);
    }
}

/**
 * Compute the natural logarithm of x to a given scale, x > 0.
 * Use Newton's algorithm.
 */
private static BigDecimal lnNewton(BigDecimal x, int scale)
{
    int        sp1 = scale + 1;
    BigDecimal n   = x;
    BigDecimal term;

    // Convergence tolerance = 5*(10^-(scale+1))
    BigDecimal tolerance = BigDecimal.valueOf(5)
                                        .movePointLeft(sp1);

    // Loop until the approximations converge
    // (two successive approximations are within the tolerance).
    do {

        // e^x
        BigDecimal eToX = exp(x, sp1);

        // (e^x - n)/e^x
        term = eToX.subtract(n)
                    .divide(eToX, sp1, BigDecimal.ROUND_DOWN);

        // x - (e^x - n)/e^x
        x = x.subtract(term);

        Thread.yield();
    } while (term.compareTo(tolerance) > 0);

    return x.setScale(scale, BigDecimal.ROUND_HALF_EVEN);
}

/**
 * Compute the integral root of x to a given scale, x >= 0.
 * Use Newton's algorithm.
 * @param x the value of x
 * @param index the integral root value
 * @param scale the desired scale of the result
 * @return the result value
 */
public static BigDecimal intRoot(BigDecimal x, long index,
                                 int scale)
{
    // Check that x >= 0.
    if (x.signum() < 0) {
        throw new IllegalArgumentException("x < 0");
    }

    int        sp1 = scale + 1;
    BigDecimal n   = x;
    BigDecimal i   = BigDecimal.valueOf(index);
    BigDecimal im1 = BigDecimal.valueOf(index-1);
    BigDecimal tolerance = BigDecimal.valueOf(5)
                                        .movePointLeft(sp1);
    BigDecimal xPrev;

    // The initial approximation is x/index.
    x = x.divide(i, scale, BigDecimal.ROUND_HALF_EVEN);

    // Loop until the approximations converge
    // (two successive approximations are equal after rounding).
    do {
        // x^(index-1)
        BigDecimal xToIm1 = intPower(x, index-1, sp1);

        // x^index
        BigDecimal xToI =
                x.multiply(xToIm1)
                    .setScale(sp1, BigDecimal.ROUND_HALF_EVEN);

        // n + (index-1)*(x^index)
        BigDecimal numerator =
                n.add(im1.multiply(xToI))
                    .setScale(sp1, BigDecimal.ROUND_HALF_EVEN);

        // (index*(x^(index-1))
        BigDecimal denominator =
                i.multiply(xToIm1)
                    .setScale(sp1, BigDecimal.ROUND_HALF_EVEN);

        // x = (n + (index-1)*(x^index)) / (index*(x^(index-1)))
        xPrev = x;
        x = numerator
                .divide(denominator, sp1, BigDecimal.ROUND_DOWN);

        Thread.yield();
    } while (x.subtract(xPrev).abs().compareTo(tolerance) > 0);

    return x;
}

/**
 * Compute e^x to a given scale.
 * Break x into its whole and fraction parts and
 * compute (e^(1 + fraction/whole))^whole using Taylor's formula.
 * @param x the value of x
 * @param scale the desired scale of the result
 * @return the result value
 */
public static BigDecimal exp(BigDecimal x, int scale)
{
    // e^0 = 1
    if (x.signum() == 0) {
        return BigDecimal.valueOf(1);
    }

    // If x is negative, return 1/(e^-x).
    else if (x.signum() == -1) {
        return BigDecimal.valueOf(1)
                    .divide(exp(x.negate(), scale), scale,
                            BigDecimal.ROUND_HALF_EVEN);
    }

    // Compute the whole part of x.
    BigDecimal xWhole = x.setScale(0, BigDecimal.ROUND_DOWN);

    // If there isn't a whole part, compute and return e^x.
    if (xWhole.signum() == 0) return expTaylor(x, scale);

    // Compute the fraction part of x.
    BigDecimal xFraction = x.subtract(xWhole);

    // z = 1 + fraction/whole
    BigDecimal z = BigDecimal.valueOf(1)
                        .add(xFraction.divide(
                                xWhole, scale,
                                BigDecimal.ROUND_HALF_EVEN));

    // t = e^z
    BigDecimal t = expTaylor(z, scale);

    BigDecimal maxLong = BigDecimal.valueOf(Long.MAX_VALUE);
    BigDecimal result  = BigDecimal.valueOf(1);

    // Compute and return t^whole using intPower().
    // If whole > Long.MAX_VALUE, then first compute products
    // of e^Long.MAX_VALUE.
    while (xWhole.compareTo(maxLong) >= 0) {
        result = result.multiply(
                            intPower(t, Long.MAX_VALUE, scale))
                    .setScale(scale, BigDecimal.ROUND_HALF_EVEN);
        xWhole = xWhole.subtract(maxLong);

        Thread.yield();
    }
    return result.multiply(intPower(t, xWhole.longValue(), scale))
                    .setScale(scale, BigDecimal.ROUND_HALF_EVEN);
}

一个适用于大数字的黑客小算法使用关系式 log(AB) = log(A) + log(B) . 下面介绍如何在底10中进行转换(您可以将其转换为任何其他对数底):

1. 计算答案中的小数位数。这是对数的整数部分, 加一 . 例子: floor(log10(123456)) + 1 是6,因为123456有6个数字。

2. 如果您只需要对数的整数部分,就可以在这里停止:只需从步骤1的结果中减去1。

3. 要得到对数的小数部分,请将该数字除以 10^(number of digits) ,然后使用 math.log10() (或其他;如果没有其他可用的值,则使用简单的序列近似值),并将其添加到整数部分。示例:获取 log10(123456) 计算 math.log10(0.123456) = -0.908... ,并将其添加到步骤1的结果中: 6 + -0.908 = 5.092 ,这就是 Log10(123456) . 请注意,您基本上只是将小数点附加到大数字的前面;在您的用例中,可能有一个很好的方法来优化这个问题,对于非常大的数字,您甚至不需要费心获取所有的数字。-- log10(0.123) 是一个伟大的近似 log10(0.123456789) .

这是一个超级快,因为:

• 不 toString()
• 不 BigInteger 数学(牛顿分数/连分数)
• 甚至连一个新的 双整数
• 只使用固定数量的非常快的操作

一次通话大约需要20微秒(大约每秒5万次通话)

但是:

• 只适用于 双整数

解决问题 BigDecimal (未测试速度):

• 移动小数点,直到值为2^53
• 使用 toBigInteger() (使用一个) div 内部)

该算法利用对数可以作为指数和尾数的对数之和来计算。如:

12345有5个数字,因此以10为基数的日志介于4和5之间。 对数(12345)=4+对数(1.2345)=4.09149…(基10日志)

此函数计算基2日志,因为查找占用位的数目很简单。

public double log(BigInteger val)
{
    // Get the minimum number of bits necessary to hold this value.
    int n = val.bitLength();

    // Calculate the double-precision fraction of this number; as if the
    // binary point was left of the most significant '1' bit.
    // (Get the most significant 53 bits and divide by 2^53)
    long mask = 1L << 52; // mantissa is 53 bits (including hidden bit)
    long mantissa = 0;
    int j = 0;
    for (int i = 1; i < 54; i++)
    {
        j = n - i;
        if (j < 0) break;

        if (val.testBit(j)) mantissa |= mask;
        mask >>>= 1;
    }
    // Round up if next bit is 1.
    if (j > 0 && val.testBit(j - 1)) mantissa++;

    double f = mantissa / (double)(1L << 52);

    // Add the logarithm to the number of bits, and subtract 1 because the
    // number of bits is always higher than necessary for a number
    // (ie. log2(val)<n for every val).
    return (n - 1 + Math.log(f) * 1.44269504088896340735992468100189213742664595415298D);
    // Magic number converts from base e to base 2 before adding. For other
    // bases, correct the result, NOT this number!
}

你可以用

log(a * 10^b) = log(a) + b * log(10)

基本上 b+1 将是数字中的位数,并且 a 将是一个介于0和1之间的值,您可以使用正则表达式计算的对数 double 算术。

或者你可以使用一些数学技巧-例如,接近1的数字的对数可以通过级数展开计算出来。

ln(x + 1) = x - x^2/2 + x^3/3 - x^4/4 + ...

根据你试图取对数的类型,可能会有这样的数字你可以使用。

编辑 :要获得以10为底的对数,可以将自然对数除以 ln(10) 或类似的任何其他基础。

这就是我想到的:

//http://everything2.com/index.pl?node_id=946812        
public BigDecimal log10(BigDecimal b, int dp)
{
    final int NUM_OF_DIGITS = dp+2; // need to add one to get the right number of dp
                                    //  and then add one again to get the next number
                                    //  so I can round it correctly.

    MathContext mc = new MathContext(NUM_OF_DIGITS, RoundingMode.HALF_EVEN);

    //special conditions:
    // log(-x) -> exception
    // log(1) == 0 exactly;
    // log of a number lessthan one = -log(1/x)
    if(b.signum() <= 0)
        throw new ArithmeticException("log of a negative number! (or zero)");
    else if(b.compareTo(BigDecimal.ONE) == 0)
        return BigDecimal.ZERO;
    else if(b.compareTo(BigDecimal.ONE) < 0)
        return (log10((BigDecimal.ONE).divide(b,mc),dp)).negate();

    StringBuffer sb = new StringBuffer();
    //number of digits on the left of the decimal point
    int leftDigits = b.precision() - b.scale();

    //so, the first digits of the log10 are:
    sb.append(leftDigits - 1).append(".");

    //this is the algorithm outlined in the webpage
    int n = 0;
    while(n < NUM_OF_DIGITS)
    {
        b = (b.movePointLeft(leftDigits - 1)).pow(10, mc);
        leftDigits = b.precision() - b.scale();
        sb.append(leftDigits - 1);
        n++;
    }

    BigDecimal ans = new BigDecimal(sb.toString());

    //Round the number to the correct number of decimal places.
    ans = ans.round(new MathContext(ans.precision() - ans.scale() + dp, RoundingMode.HALF_EVEN));
    return ans;
}

一个Java实现的MeWOR68伪代码,我用几个数字测试:

public static BigDecimal log(int base_int, BigDecimal x) {
        BigDecimal result = BigDecimal.ZERO;

        BigDecimal input = new BigDecimal(x.toString());
        int decimalPlaces = 100;
        int scale = input.precision() + decimalPlaces;

        int maxite = 10000;
        int ite = 0;
        BigDecimal maxError_BigDecimal = new BigDecimal(BigInteger.ONE,decimalPlaces + 1);
        System.out.println("maxError_BigDecimal " + maxError_BigDecimal);
        System.out.println("scale " + scale);

        RoundingMode a_RoundingMode = RoundingMode.UP;

        BigDecimal two_BigDecimal = new BigDecimal("2");
        BigDecimal base_BigDecimal = new BigDecimal(base_int);

        while (input.compareTo(base_BigDecimal) == 1) {
            result = result.add(BigDecimal.ONE);
            input = input.divide(base_BigDecimal, scale, a_RoundingMode);
        }

        BigDecimal fraction = new BigDecimal("0.5");
        input = input.multiply(input);
        BigDecimal resultplusfraction = result.add(fraction);
        while (((resultplusfraction).compareTo(result) == 1)
                && (input.compareTo(BigDecimal.ONE) == 1)) {
            if (input.compareTo(base_BigDecimal) == 1) {
                input = input.divide(base_BigDecimal, scale, a_RoundingMode);
                result = result.add(fraction);
            }
            input = input.multiply(input);
            fraction = fraction.divide(two_BigDecimal, scale, a_RoundingMode);
            resultplusfraction = result.add(fraction);
            if (fraction.abs().compareTo(maxError_BigDecimal) == -1){
                break;
            }
            if (maxite == ite){
                break;
            }
            ite ++;
        }

        MathContext a_MathContext = new MathContext(((decimalPlaces - 1) + (result.precision() - result.scale())),RoundingMode.HALF_UP);
        BigDecimal roundedResult = result.round(a_MathContext);
        BigDecimal strippedRoundedResult = roundedResult.stripTrailingZeros();
        //return result;
        //return result.round(a_MathContext);
        return strippedRoundedResult;
    }

做对数的伪代码算法。

假设我们想要x的对数

result = 0;
base = n;
input = x;

while (input > base)
  result++;
  input /= base;

fraction = 1/2;
input *= input;   

while (((result + fraction) > result) && (input > 1))
  if (input > base)
    input /= base;
    result += fraction;
  input *= input;
  fraction /= 2.0;

大的while循环可能看起来有点混乱。

在每次传递中,您可以将输入平方,也可以取基数的平方根;无论哪种方法,您都必须将分数除以2。我发现将输入平方化,而不使用基数,这样更准确。

如果输入值变为1,我们就通过了。对于任何基,1的日志都是0,这意味着我们不需要再添加任何内容。

如果(结果+分数)不大于结果,那么我们的编号系统就达到了精度的极限。我们可以停下来。

显然,如果您使用的是一个精度任意多个数字的系统,那么您需要在其中放入其他东西来限制循环。

如果您只需要在您可以使用的数字中找到10的幂:

public int calculatePowersOf10(BigDecimal value)
{
    return value.round(new MathContext(1)).scale() * -1;
}

我在寻找这个确切的东西,最后用一个连续的分数方法。连分数可以在 here 或 here

代码:

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

public static long ITER = 1000;
public static MathContext context = new MathContext( 100 );
public static BigDecimal ln(BigDecimal x) {
    if (x.equals(BigDecimal.ONE)) {
        return BigDecimal.ZERO;
    }

    x = x.subtract(BigDecimal.ONE);
    BigDecimal ret = new BigDecimal(ITER + 1);
    for (long i = ITER; i >= 0; i--) {
    BigDecimal N = new BigDecimal(i / 2 + 1).pow(2);
        N = N.multiply(x, context);
        ret = N.divide(ret, context);

        N = new BigDecimal(i + 1);
        ret = ret.add(N, context);

    }

    ret = x.divide(ret, context);
    return ret;
}

老问题,但我认为这个答案更可取。它具有良好的精度,支持几乎任何大小的参数。

private static final double LOG10 = Math.log(10.0);

/**
 * Computes the natural logarithm of a BigDecimal 
 * 
 * @param val Argument: a positive BigDecimal
 * @return Natural logarithm, as in Math.log()
 */
public static double logBigDecimal(BigDecimal val) {
    return logBigInteger(val.unscaledValue()) + val.scale() * Math.log(10.0);
}

private static final double LOG2 = Math.log(2.0);

/**
 * Computes the natural logarithm of a BigInteger. Works for really big
 * integers (practically unlimited)
 * 
 * @param val Argument, positive integer
 * @return Natural logarithm, as in <tt>Math.log()</tt>
 */
public static double logBigInteger(BigInteger val) {
    int blex = val.bitLength() - 1022; // any value in 60..1023 is ok
    if (blex > 0)
        val = val.shiftRight(blex);
    double res = Math.log(val.doubleValue());
    return blex > 0 ? res + blex * LOG2 : res;
}

核心逻辑( logBigInteger 方法)复制自 this other answer 我的。

我为biginteger创建了一个函数,但是可以很容易地为bigdecimal修改它。分解日志并使用日志的一些属性是我所做的,但我只得到双倍精度。但它适用于任何基地。:)

public double BigIntLog(BigInteger bi, double base) {
    // Convert the BigInteger to BigDecimal
    BigDecimal bd = new BigDecimal(bi);
    // Calculate the exponent 10^exp
    BigDecimal diviser = new BigDecimal(10);
    diviser = diviser.pow(bi.toString().length()-1);
    // Convert the BigDecimal from Integer to a decimal value
    bd = bd.divide(diviser);
    // Convert the BigDecimal to double
    double bd_dbl = bd.doubleValue();
    // return the log value
    return (Math.log10(bd_dbl)+bi.toString().length()-1)/Math.log10(base);
}