/*
This file is part of Cyclos (www.cyclos.org).
A project of the Social Trade Organisation (www.socialtrade.org).
Cyclos is free software; you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation; either version 2 of the License, or
(at your option) any later version.
Cyclos is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with Cyclos; if not, write to the Free Software
Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
*/
package nl.strohalm.cyclos.utils;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.MathContext;
import nl.strohalm.cyclos.entities.settings.LocalSettings;
/**
* Helper class for handling <code>BigDecimal</code>s Contains several useful BigDecimal mathematical functions taken from:
* "Java Number Cruncher, the java programmer's guide to numerical computing" by Ronald Mak, Prentice Hall PTR, 2003. pages 330 & 331
*
* @author luis
* @author Rinke
* @author Ronald Mak
*/
public class BigDecimalHelper {
public static final BigDecimal ONE_HUNDRED = new BigDecimal(100.0);
/**
* Compute the arctangent of x to a given scale, |x| < 1
* @param x the value of x
* @param scale the desired scale of the result
* @return the result value
*/
public static BigDecimal arctan(final BigDecimal x, final int scale) {
// Check that |x| < 1.
if (x.abs().compareTo(BigDecimal.valueOf(1)) >= 0) {
throw new IllegalArgumentException("|x| >= 1");
}
// If x is negative, return -arctan(-x).
if (x.signum() == -1) {
return arctan(x.negate(), scale).negate();
} else {
return arctanTaylor(x, scale);
}
}
/**
* returns amount as a percentage of total. Example: if amount is 5 and total is 50, then BigDecimalHelper.asPercentageOf(amount, total) equals 10
* (%).
* @param amount the BigDecimal to be written as a percentage.
* @param total the BigDecimal representing the total amount of which amount is a percentage.
* @return amount as a BigDecimal percentage of total.
*/
public static BigDecimal asPercentageOf(final BigDecimal amount, final BigDecimal total) {
final MathContext mathContext = new MathContext(LocalSettings.MAX_PRECISION);
final BigDecimal asFractionOf = amount.divide(total, mathContext);
return asFractionOf.multiply(ONE_HUNDRED, mathContext);
}
/**
* returns a nominal percentage (like 5.2%) as a fraction (0.052).
* @return the BigDecimal divided by 100. In the above example this would be 0.052.
*/
public static BigDecimal asPercentFraction(final BigDecimal bigDecimal) {
final MathContext mathContext = new MathContext(LocalSettings.MAX_PRECISION);
return bigDecimal.divide(ONE_HUNDRED, mathContext);
}
/**
* 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(final BigDecimal x, final 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.
final BigDecimal xFraction = x.subtract(xWhole);
// z = 1 + fraction/whole
final BigDecimal z = BigDecimal.valueOf(1).add(xFraction.divide(xWhole, scale, BigDecimal.ROUND_HALF_EVEN));
// t = e^z
final BigDecimal t = expTaylor(z, scale);
final 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);
}
/**
* Compute x^exponent to a given scale.
* @param x the value x
* @param exponent the exponent value
* @param scale the desired scale of the result
* @return the result value
*/
public static BigDecimal intPower(BigDecimal x, long exponent, final int scale) {
// If the exponent is negative, compute 1/(x^-exponent).
if (exponent < 0) {
return BigDecimal.valueOf(1).divide(intPower(x, -exponent, scale), scale, BigDecimal.ROUND_HALF_EVEN);
}
BigDecimal power = BigDecimal.valueOf(1);
// Loop to compute value^exponent.
while (exponent > 0) {
// Is the rightmost bit a 1?
if ((exponent & 1) == 1) {
power = power.multiply(x).setScale(scale, BigDecimal.ROUND_HALF_EVEN);
}
// Square x and shift exponent 1 bit to the right.
x = x.multiply(x).setScale(scale, BigDecimal.ROUND_HALF_EVEN);
exponent >>= 1;
Thread.yield();
}
return power;
}
/**
* 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, final long index, final int scale) {
// Check that x >= 0.
if (x.signum() < 0) {
throw new IllegalArgumentException("x < 0");
}
final int sp1 = scale + 1;
final BigDecimal n = x;
final BigDecimal i = BigDecimal.valueOf(index);
final BigDecimal im1 = BigDecimal.valueOf(index - 1);
final 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)
final BigDecimal xToIm1 = intPower(x, index - 1, sp1);
// x^index
final BigDecimal xToI = x.multiply(xToIm1).setScale(sp1, BigDecimal.ROUND_HALF_EVEN);
// n + (index-1)*(x^index)
final BigDecimal numerator = n.add(im1.multiply(xToI)).setScale(sp1, BigDecimal.ROUND_HALF_EVEN);
// (index*(x^(index-1))
final 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 the natural logarithm of x to a given scale, x > 0.
*/
public static BigDecimal ln(final BigDecimal x, final 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.
final 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)
final BigDecimal root = intRoot(x, magnitude, scale);
// ln(x^(1/magnitude))
final BigDecimal lnRoot = lnNewton(root, scale);
// magnitude*ln(x^(1/magnitude))
return BigDecimal.valueOf(magnitude).multiply(lnRoot).setScale(scale, BigDecimal.ROUND_HALF_EVEN);
}
}
/**
* Returns zero when the given BigDecimal is null
*/
public static BigDecimal nvl(final BigDecimal bigDecimal) {
return bigDecimal == null ? BigDecimal.ZERO : bigDecimal;
}
/**
* calculates x^exponent, that is, raise x to the poser of exponent. (by Rinke)
* @param x the BigDecimal to be "powered".
* @param scale the desired scale of the result
* @param exponent the double which will be used to raise x to its power.
* @return the result value
*/
public static BigDecimal pow(final BigDecimal x, final int scale, final BigDecimal exponent) {
final BigDecimal lnX = BigDecimalHelper.ln(x, scale);
final BigDecimal newExponent = lnX.multiply(exponent);
return BigDecimalHelper.exp(newExponent, scale);
}
/**
* extremely simple version natural logarithm, using no algorith tricks, but just via doubles and recreating BigDecimal Might give less accurate
* results, but this is to be tested. As long as this hasn't happened it is deprecated.
*/
@Deprecated
public static BigDecimal simpleLn(final BigDecimal x, final int scale) {
final long unscaledValue = x.unscaledValue().longValue();
final int scalevalue = x.scale();
final double result = Math.log(unscaledValue) - (scalevalue * Math.log(10.0));
return new BigDecimal(result, new MathContext(scale));
}
/**
* Compute the square root of x to a given scale, x >= 0. Use Newton's algorithm.
* @param x the value of x
* @param scale the desired scale of the result
* @return the result value
*/
public static BigDecimal sqrt(final BigDecimal x, final int scale) {
// Check that x >= 0.
if (x.signum() < 0) {
throw new IllegalArgumentException("x < 0");
}
// n = x*(10^(2*scale))
final BigInteger n = x.movePointRight(scale << 1).toBigInteger();
// The first approximation is the upper half of n.
final int bits = (n.bitLength() + 1) >> 1;
BigInteger ix = n.shiftRight(bits);
BigInteger ixPrev;
// Loop until the approximations converge
// (two successive approximations are equal after rounding).
do {
ixPrev = ix;
// x = (x + n/x)/2
ix = ix.add(n.divide(ix)).shiftRight(1);
Thread.yield();
} while (ix.compareTo(ixPrev) != 0);
return new BigDecimal(ix, scale);
}
/**
* Compute the arctangent of x to a given scale by the Taylor series, |x| < 1
* @param x the value of x
* @param scale the desired scale of the result
* @return the result value
* @author Ronald Mak: "Java Number Cruncher, the java programmer's guide to numerical computing" Prentice Hall PTR, 2003. pages 330 & 331
*/
private static BigDecimal arctanTaylor(final BigDecimal x, final int scale) {
final int sp1 = scale + 1;
int i = 3;
boolean addFlag = false;
BigDecimal power = x;
BigDecimal sum = x;
BigDecimal term;
// Convergence tolerance = 5*(10^-(scale+1))
final BigDecimal tolerance = BigDecimal.valueOf(5).movePointLeft(sp1);
// Loop until the approximations converge
// (two successive approximations are within the tolerance).
do {
// x^i
power = power.multiply(x).multiply(x).setScale(sp1, BigDecimal.ROUND_HALF_EVEN);
// (x^i)/i
term = power.divide(BigDecimal.valueOf(i), sp1, BigDecimal.ROUND_HALF_EVEN);
// sum = sum +- (x^i)/i
sum = addFlag ? sum.add(term) : sum.subtract(term);
i += 2;
addFlag = !addFlag;
Thread.yield();
} while (term.compareTo(tolerance) > 0);
return sum;
}
/**
* Compute e^x to a given scale by the Taylor series.
* @param x the value of x
* @param scale the desired scale of the result
* @return the result value
* @author Ronald Mak: "Java Number Cruncher, the java programmer's guide to numerical computing" Prentice Hall PTR, 2003. pages 330 & 331
*/
private static BigDecimal expTaylor(final BigDecimal x, final int scale) {
BigDecimal factorial = BigDecimal.valueOf(1);
BigDecimal xPower = x;
BigDecimal sumPrev;
// 1 + x
BigDecimal sum = x.add(BigDecimal.valueOf(1));
// Loop until the sums converge
// (two successive sums are equal after rounding).
int i = 2;
do {
// x^i
xPower = xPower.multiply(x).setScale(scale, BigDecimal.ROUND_HALF_EVEN);
// i!
factorial = factorial.multiply(BigDecimal.valueOf(i));
// x^i/i!
final BigDecimal term = xPower.divide(factorial, scale, BigDecimal.ROUND_HALF_EVEN);
// sum = sum + x^i/i!
sumPrev = sum;
sum = sum.add(term);
++i;
Thread.yield();
} while (sum.compareTo(sumPrev) != 0);
return sum;
}
/**
* Compute the natural logarithm of x to a given scale, x > 0. Use Newton's algorithm.
* @author Ronald Mak: "Java Number Cruncher, the java programmer's guide to numerical computing" Prentice Hall PTR, 2003. pages 330 & 331
*/
private static BigDecimal lnNewton(BigDecimal x, final int scale) {
final int sp1 = scale + 1;
final BigDecimal n = x;
BigDecimal term;
// Convergence tolerance = 5*(10^-(scale+1))
final BigDecimal tolerance = BigDecimal.valueOf(5).movePointLeft(sp1);
// Loop until the approximations converge
// (two successive approximations are within the tolerance).
do {
// e^x
final 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);
}
}