Example usage for org.apache.commons.math3.fraction BigFraction ONE_HALF

Introduction

In this page you can find the example usage for org.apache.commons.math3.fraction BigFraction ONE_HALF.

Prototype

BigFraction ONE_HALF

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Document

A fraction representing "1/2".

Usage

From source file:cz.cuni.mff.spl.evaluator.statistics.KolmogorovSmirnovTestFlag.java

/***
 * Creates {@code H} of size {@code m x m} as described in [1] (see above).
 *
 * @param d statistic//  w  ww. j  a  v  a2s.com
 * @param n sample size
 * @return H matrix
 * @throws NumberIsTooLargeException if fractional part is greater than 1
 * @throws FractionConversionException if algorithm fails to convert {@code h} to a
 *         {@link org.apache.commons.math3.fraction.BigFraction} in expressing {@code d} as \((k
 *         - h) / m\) for integer {@code k, m} and \(0 <= h < 1\).
 */
private FieldMatrix<BigFraction> createExactH(double d, int n)
        throws NumberIsTooLargeException, FractionConversionException {

    final int k = (int) Math.ceil(n * d);
    final int m = 2 * k - 1;
    final double hDouble = k - n * d;
    if (hDouble >= 1) {
        throw new NumberIsTooLargeException(hDouble, 1.0, false);
    }
    BigFraction h = null;
    try {
        h = new BigFraction(hDouble, 1.0e-20, 10000);
    } catch (final FractionConversionException e1) {
        try {
            h = new BigFraction(hDouble, 1.0e-10, 10000);
        } catch (final FractionConversionException e2) {
            h = new BigFraction(hDouble, 1.0e-5, 10000);
        }
    }
    final BigFraction[][] Hdata = new BigFraction[m][m];

    /*
     * Start by filling everything with either 0 or 1.
     */
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < m; ++j) {
            if (i - j + 1 < 0) {
                Hdata[i][j] = BigFraction.ZERO;
            } else {
                Hdata[i][j] = BigFraction.ONE;
            }
        }
    }

    /*
     * Setting up power-array to avoid calculating the same value twice: hPowers[0] = h^1 ...
     * hPowers[m-1] = h^m
     */
    final BigFraction[] hPowers = new BigFraction[m];
    hPowers[0] = h;
    for (int i = 1; i < m; ++i) {
        hPowers[i] = h.multiply(hPowers[i - 1]);
    }

    /*
     * First column and last row has special values (each other reversed).
     */
    for (int i = 0; i < m; ++i) {
        Hdata[i][0] = Hdata[i][0].subtract(hPowers[i]);
        Hdata[m - 1][i] = Hdata[m - 1][i].subtract(hPowers[m - i - 1]);
    }

    /*
     * [1] states: "For 1/2 < h < 1 the bottom left element of the matrix should be (1 - 2*h^m +
     * (2h - 1)^m )/m!" Since 0 <= h < 1, then if h > 1/2 is sufficient to check:
     */
    if (h.compareTo(BigFraction.ONE_HALF) == 1) {
        Hdata[m - 1][0] = Hdata[m - 1][0].add(h.multiply(2).subtract(1).pow(m));
    }

    /*
     * Aside from the first column and last row, the (i, j)-th element is 1/(i - j + 1)! if i -
     * j + 1 >= 0, else 0. 1's and 0's are already put, so only division with (i - j + 1)! is
     * needed in the elements that have 1's. There is no need to calculate (i - j + 1)! and then
     * divide - small steps avoid overflows. Note that i - j + 1 > 0 <=> i + 1 > j instead of
     * j'ing all the way to m. Also note that it is started at g = 2 because dividing by 1 isn't
     * really necessary.
     */
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < i + 1; ++j) {
            if (i - j + 1 > 0) {
                for (int g = 2; g <= i - j + 1; ++g) {
                    Hdata[i][j] = Hdata[i][j].divide(g);
                }
            }
        }
    }
    return new Array2DRowFieldMatrix<BigFraction>(BigFractionField.getInstance(), Hdata);
}

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