|  | //===- ReservoirSampler.cpp - Tests for the ReservoirSampler --------------===// | 
|  | // | 
|  | // Part of the LLVM Project, under the Apache License v2.0 with LLVM Exceptions. | 
|  | // See https://llvm.org/LICENSE.txt for license information. | 
|  | // SPDX-License-Identifier: Apache-2.0 WITH LLVM-exception | 
|  | // | 
|  | //===----------------------------------------------------------------------===// | 
|  |  | 
|  | #include "llvm/FuzzMutate/Random.h" | 
|  | #include "gtest/gtest.h" | 
|  | #include <random> | 
|  |  | 
|  | using namespace llvm; | 
|  |  | 
|  | TEST(ReservoirSamplerTest, OneItem) { | 
|  | std::mt19937 Rand; | 
|  | auto Sampler = makeSampler(Rand, 7, 1); | 
|  | ASSERT_FALSE(Sampler.isEmpty()); | 
|  | ASSERT_EQ(7, Sampler.getSelection()); | 
|  | } | 
|  |  | 
|  | TEST(ReservoirSamplerTest, NoWeight) { | 
|  | std::mt19937 Rand; | 
|  | auto Sampler = makeSampler(Rand, 7, 0); | 
|  | ASSERT_TRUE(Sampler.isEmpty()); | 
|  | } | 
|  |  | 
|  | TEST(ReservoirSamplerTest, Uniform) { | 
|  | std::mt19937 Rand; | 
|  |  | 
|  | // Run three chi-squared tests to check that the distribution is reasonably | 
|  | // uniform. | 
|  | std::vector<int> Items = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; | 
|  |  | 
|  | int Failures = 0; | 
|  | for (int Run = 0; Run < 3; ++Run) { | 
|  | std::vector<int> Counts(Items.size(), 0); | 
|  |  | 
|  | // We need $np_s > 5$ at minimum, but we're better off going a couple of | 
|  | // orders of magnitude larger. | 
|  | int N = Items.size() * 5 * 100; | 
|  | for (int I = 0; I < N; ++I) { | 
|  | auto Sampler = makeSampler(Rand, Items); | 
|  | Counts[Sampler.getSelection()] += 1; | 
|  | } | 
|  |  | 
|  | // Knuth. TAOCP Vol. 2, 3.3.1 (8): | 
|  | // $V = \frac{1}{n} \sum_{s=1}^{k} \left(\frac{Y_s^2}{p_s}\right) - n$ | 
|  | double Ps = 1.0 / Items.size(); | 
|  | double Sum = 0.0; | 
|  | for (int Ys : Counts) | 
|  | Sum += Ys * Ys / Ps; | 
|  | double V = (Sum / N) - N; | 
|  |  | 
|  | assert(Items.size() == 10 && "Our chi-squared values assume 10 items"); | 
|  | // Since we have 10 items, there are 9 degrees of freedom and the table of | 
|  | // chi-squared values is as follows: | 
|  | // | 
|  | //     | p=1%  |   5%  |  25%  |  50%  |  75%  |  95%  |  99%  | | 
|  | // v=9 | 2.088 | 3.325 | 5.899 | 8.343 | 11.39 | 16.92 | 21.67 | | 
|  | // | 
|  | // Check that we're in the likely range of results. | 
|  | //if (V < 2.088 || V > 21.67) | 
|  | if (V < 2.088 || V > 21.67) | 
|  | ++Failures; | 
|  | } | 
|  | EXPECT_LT(Failures, 3) << "Non-uniform distribution?"; | 
|  | } |