Java AllowedSolution类代码示例

import org.apache.commons.math3.analysis.solvers.AllowedSolution; //导入依赖的package包/类
/**
 * Solve for a zero in the given interval, start at {@code startValue}.
 * A solver may require that the interval brackets a single zero root.
 * Solvers that do require bracketing should be able to handle the case
 * where one of the endpoints is itself a root.
 *
 * @param maxEval Maximum number of evaluations.
 * @param f Function to solve.
 * @param min Lower bound for the interval.
 * @param max Upper bound for the interval.
 * @param startValue Start value to use.
 * @param allowedSolution The kind of solutions that the root-finding algorithm may
 * accept as solutions.
 * @return a value where the function is zero.
 * @exception NullArgumentException if f is null.
 * @exception NoBracketingException if root cannot be bracketed
 */
public Dfp solve(final int maxEval, final UnivariateDfpFunction f,
                 final Dfp min, final Dfp max, final Dfp startValue,
                 final AllowedSolution allowedSolution)
    throws NullArgumentException, NoBracketingException {

    // checks
    MathUtils.checkNotNull(f);

    // wrap the function
    RealFieldUnivariateFunction<Dfp> fieldF = new RealFieldUnivariateFunction<Dfp>() {

        /** {@inheritDoc} */
        public Dfp value(final Dfp x) {
            return f.value(x);
        }
    };

    // delegate to general field solver
    return solve(maxEval, fieldF, min, max, startValue, allowedSolution);

}
 

import org.apache.commons.math3.analysis.solvers.AllowedSolution; //导入依赖的package包/类
@Test
public void testConvergenceOnFunctionAccuracy() {
    BracketingNthOrderBrentSolverDFP solver =
            new BracketingNthOrderBrentSolverDFP(relativeAccuracy, absoluteAccuracy,
                                                 field.newDfp(1.0e-20), 20);
    UnivariateDfpFunction f = new UnivariateDfpFunction() {
        public Dfp value(Dfp x) {
            Dfp one     = field.getOne();
            Dfp oneHalf = one.divide(2);
            Dfp xMo     = x.subtract(one);
            Dfp xMh     = x.subtract(oneHalf);
            Dfp xPh     = x.add(oneHalf);
            Dfp xPo     = x.add(one);
            return xMo.multiply(xMh).multiply(x).multiply(xPh).multiply(xPo);
        }
    };

    Dfp result = solver.solve(20, f, field.newDfp(0.2), field.newDfp(0.9),
                              field.newDfp(0.4), AllowedSolution.BELOW_SIDE);
    Assert.assertTrue(f.value(result).abs().lessThan(solver.getFunctionValueAccuracy()));
    Assert.assertTrue(f.value(result).negativeOrNull());
    Assert.assertTrue(result.subtract(field.newDfp(0.5)).subtract(solver.getAbsoluteAccuracy()).positiveOrNull());
    result = solver.solve(20, f, field.newDfp(-0.9), field.newDfp(-0.2),
                          field.newDfp(-0.4), AllowedSolution.ABOVE_SIDE);
    Assert.assertTrue(f.value(result).abs().lessThan(solver.getFunctionValueAccuracy()));
    Assert.assertTrue(f.value(result).positiveOrNull());
    Assert.assertTrue(result.add(field.newDfp(0.5)).subtract(solver.getAbsoluteAccuracy()).negativeOrNull());
}
 

import org.apache.commons.math3.analysis.solvers.AllowedSolution; //导入依赖的package包/类
/**
 * Solve for a zero in the given interval.
 * A solver may require that the interval brackets a single zero root.
 * Solvers that do require bracketing should be able to handle the case
 * where one of the endpoints is itself a root.
 *
 * @param maxEval Maximum number of evaluations.
 * @param f Function to solve.
 * @param min Lower bound for the interval.
 * @param max Upper bound for the interval.
 * @param allowedSolution The kind of solutions that the root-finding algorithm may
 * accept as solutions.
 * @return a value where the function is zero.
 * @exception NullArgumentException if f is null.
 * @exception NoBracketingException if root cannot be bracketed
 */
public Dfp solve(final int maxEval, final UnivariateDfpFunction f,
                 final Dfp min, final Dfp max, final AllowedSolution allowedSolution)
    throws NullArgumentException, NoBracketingException {
    return solve(maxEval, f, min, max, min.add(max).divide(2), allowedSolution);
}