ReservoirSamplerTest.cpp
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//===- 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?";
}