/*

 * Copyright (c) 2013, Oracle and/or its affiliates. All rights reserved.

 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.

 *

 * This code is free software; you can redistribute it and/or modify it

 * under the terms of the GNU General Public License version 2 only, as

 * published by the Free Software Foundation.

 *

 * This code 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

 * version 2 for more details (a copy is included in the LICENSE file that

 * accompanied this code).

 *

 * You should have received a copy of the GNU General Public License version

 * 2 along with this work; if not, write to the Free Software Foundation,

 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.

 *

 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA

 * or visit www.oracle.com if you need additional information or have any

 * questions.

 */

import java.util.*;

import java.util.function.*;

import java.util.stream.*;

/*

 * @test

 * @summary Test for use of non-naive summation in stream-related sum and average operations.

 */

public class TestDoubleSumAverage {

    public static void main(String... args) {

        int failures = 0;

        failures += testForCompenstation();

        if (failures > 0) {

            throw new RuntimeException("Found " + failures + " numerical failure(s).");

        }

    }

    /**

     * Compute the sum and average of a sequence of double values in

     * various ways and report an error if naive summation is used.

     */

    private static int testForCompenstation() {

        int failures = 0;

        /*

         * The exact sum of the test stream is 1 + 1e6*ulp(1.0) but a

         * naive summation algorithm will return 1.0 since (1.0 +

         * ulp(1.0)/2) will round to 1.0 again.

         */

        double base = 1.0;

        double increment = Math.ulp(base)/2.0;

        int count = 1_000_001;

        double expectedSum = base + (increment * (count - 1));

        double expectedAvg = expectedSum / count;

        // Factory for double a stream of [base, increment, ..., increment] limited to a size of count

        Supplier<DoubleStream> ds = () -> DoubleStream.iterate(base, e -> increment).limit(count);

        DoubleSummaryStatistics stats = ds.get().collect(DoubleSummaryStatistics::new,

                                                         DoubleSummaryStatistics::accept,

                                                         DoubleSummaryStatistics::combine);

        failures += compareUlpDifference(expectedSum, stats.getSum(), 3);

        failures += compareUlpDifference(expectedAvg, stats.getAverage(), 3);

        failures += compareUlpDifference(expectedSum,

                                         ds.get().sum(), 3);

        failures += compareUlpDifference(expectedAvg,

                                         ds.get().average().getAsDouble(), 3);

        failures += compareUlpDifference(expectedSum,

                                         ds.get().boxed().collect(Collectors.summingDouble(d -> d)), 3);

        failures += compareUlpDifference(expectedAvg,

                                         ds.get().boxed().collect(Collectors.averagingDouble(d -> d)),3);

        return failures;

    }

    }

    /**

     * Compute the ulp difference of two double values and compare against an error threshold.

     */

    private static int compareUlpDifference(double expected, double computed, double threshold) {

        double ulpDifference = Math.abs(expected - computed) / Math.ulp(expected);

        if (ulpDifference > threshold) {

            System.err.printf("Numerical summation error too large, %g ulps rather than %g.%n",

                              ulpDifference, threshold);

            return 1;

        } else

            return 0;

    }

}