我如何计算一个BigDecimal的对数?有人知道可以使用哪些算法吗?
到目前为止,我的谷歌搜索只提供了(无用的)将BigDecimal转换为double并使用Math.log的想法。
我会提供所需答案的精度。
编辑:任何底数都可以。如果在x进制中更容易,我会采用x进制。
我如何计算一个BigDecimal的对数?有人知道可以使用哪些算法吗?
到目前为止,我的谷歌搜索只提供了(无用的)将BigDecimal转换为double并使用Math.log的想法。
我会提供所需答案的精度。
编辑:任何底数都可以。如果在x进制中更容易,我会采用x进制。
Java数字计算专家:Java程序员的数值计算指南 提供了使用牛顿法的解决方案,书中源代码在此处可获得。以下内容摘自第12.5大十进制函数章节(p330和p331):
/**
* 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);
}
Thread.yield()
应该不在这里。如果你的目的是让一个计算密集型线程成为一个“好公民”,那么你可以将其替换为一些代码来测试线程的“中断”标志并退出。但是调用 Thread.yield()
会干扰正常的线程调度,并且可能会使方法运行非常缓慢...这取决于其他事情发生了什么。 - Stephen Cexp()
和intRoot()
的代码。 - Maarten Bodewesexp()
和 intRoot()
https://github.com/javadev/calc/blob/master/src/main/java/com/github/calc/BigDecimalUtil.java - kanelog(AB) = log(A) + log(B)
。以下是如何在以10为底的情况下进行操作(你可以轻松地将其转换为任何其他对数基数):
统计答案中的十进制数字数量,这就是对数的整数部分,再加上1。例如:floor(log10(123456)) + 1
得到6,因为123456有6个数字。
如果只需要对数的整数部分,则可以在此处停止:只需从步骤1的结果中减去1即可。
要获取对数的小数部分,请将数字除以10^(数字的位数)
,然后使用math.log10()
(或其他任何方法;如果没有其他方法可用,则使用简单的级数近似)计算该数字的对数,并将其添加到整数部分。例如:要获取log10(123456)
的小数部分,请计算math.log10(0.123456) = -0.908...
,并将其加到步骤1的结果上:6 + -0.908 = 5.092
,这就是log10(123456)
。请注意,您基本上只是在大数前面加上一个小数点;在您的用例中可能有一种很好的优化方法,对于非常大的数字,您甚至不需要担心获取所有数字--log10(0.123)
是log10(0.123456789)
的一个很好的近似值。
toString()
BigInteger
函数(牛顿/连分数)BigInteger
BigInteger
BigDecimal
的解决办法(未经过速度测试):
toBigInteger()
(内部使用一个div
)12345 有5位数字,所以以10为底的对数在4和5之间。 log(12345) = 4 + log(1.2345) = 4.09149... (以10为底的对数)
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!
}
1.44269504088896340735992468100189213742664595415298D
这么长?Java 的浮点数只有 16 位精度,所以在 Java 中 1.44269504088896340735992468100189213742664595415298D == 1.4426950408889634
(大多数具有浮点精度的语言也是如此)。不过我还是可以确认它工作得非常好,所以给个赞。 - Kevin Cruijssenlog(a * 10^b) = log(a) + b * log(10)
b+1
是数字的位数,a
是 0 到 1 之间的一个值,你可以通过使用常规的 double
算术计算其对数。或者有一些数学技巧可以使用 - 例如,接近 1 的数的对数可以通过级数展开计算。ln(x + 1) = x - x^2/2 + x^3/3 - x^4/4 + ...
根据你要取对数的数字类型,可能会有类似这样的方法可用。
编辑: 要获取以10为底的对数,可以将自然对数除以ln(10)
,或者对于任何其他底数也可以使用类似的方法。
public int calculatePowersOf10(BigDecimal value)
{
return value.round(new MathContext(1)).scale() * -1;
}
//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;
}
这是一个Meower68伪代码的Java实现,我用几个数字进行了测试:
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;
}
我一直在寻找这个东西,最终采用了连分数的方法。 连分数可以在这里或这里找到。
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;
}
进行对数的伪代码算法。
假设我们想要计算 x 的 log_n。
double fraction, input;
int base;
double result;
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;
}
虽然这是一个旧问题,但我认为这个答案更可取。它具有很高的精度,并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;
}
logBigInteger
方法)是从我另一个答案中复制的。