/*
 * SPDX-License-Identifier: Apache-2.0
 *
 * The OpenSearch Contributors require contributions made to
 * this file be licensed under the Apache-2.0 license or a
 * compatible open source license.
 */

/*
 * Licensed to Elasticsearch under one or more contributor
 * license agreements. See the NOTICE file distributed with
 * this work for additional information regarding copyright
 * ownership. Elasticsearch licenses this file to you under
 * the Apache License, Version 2.0 (the "License"); you may
 * not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *    http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing,
 * software distributed under the License is distributed on an
 * "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
 * KIND, either express or implied.  See the License for the
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/*
 * Modifications Copyright OpenSearch Contributors. See
 * GitHub history for details.
 */

package org.opensearch.search.aggregations.metrics;

import org.opensearch.test.OpenSearchTestCase;
import org.junit.Assert;

public class CompensatedSumTests extends OpenSearchTestCase {

    /**
     * When adding a series of numbers the order of the numbers should not impact the results.
     *
     * <p>This test shows that a naive summation comes up with a different result than Kahan
     * Summation when you start with either a smaller or larger number in some cases and
     * helps prove our Kahan Summation is working.
     */
    public void testAdd() {
        final CompensatedSum smallSum = new CompensatedSum(0.001, 0.0);
        final CompensatedSum largeSum = new CompensatedSum(1000, 0.0);

        CompensatedSum compensatedResult1 = new CompensatedSum(0.001, 0.0);
        CompensatedSum compensatedResult2 = new CompensatedSum(1000, 0.0);
        double naiveResult1 = smallSum.value();
        double naiveResult2 = largeSum.value();

        for (int i = 0; i < 10; i++) {
            compensatedResult1.add(smallSum.value());
            compensatedResult2.add(smallSum.value());
            naiveResult1 += smallSum.value();
            naiveResult2 += smallSum.value();
        }

        compensatedResult1.add(largeSum.value());
        compensatedResult2.add(smallSum.value());
        naiveResult1 += largeSum.value();
        naiveResult2 += smallSum.value();

        // Kahan summation gave the same result no matter what order we added
        Assert.assertEquals(1000.011, compensatedResult1.value(), 0.0);
        Assert.assertEquals(1000.011, compensatedResult2.value(), 0.0);

        // naive addition gave a small floating point error
        Assert.assertEquals(1000.011, naiveResult1, 0.0);
        Assert.assertEquals(1000.0109999999997, naiveResult2, 0.0);

        Assert.assertEquals(compensatedResult1.value(), compensatedResult2.value(), 0.0);
        Assert.assertEquals(naiveResult1, naiveResult2, 0.0001);
        Assert.assertNotEquals(naiveResult1, naiveResult2, 0.0);
    }

    public void testDelta() {
        CompensatedSum compensatedResult1 = new CompensatedSum(0.001, 0.0);
        for (int i = 0; i < 10; i++) {
            compensatedResult1.add(0.001);
        }

        Assert.assertEquals(0.011, compensatedResult1.value(), 0.0);
        Assert.assertEquals(Double.parseDouble("8.673617379884035E-19"), compensatedResult1.delta(), 0.0);
    }

    public void testInfiniteAndNaN() {
        CompensatedSum compensatedResult1 = new CompensatedSum(0, 0);
        double[] doubles = { Double.NEGATIVE_INFINITY, Double.POSITIVE_INFINITY, Double.NaN };
        for (double d : doubles) {
            compensatedResult1.add(d);

        }

        Assert.assertTrue(Double.isNaN(compensatedResult1.value()));
    }
}