/* 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); } }