mirror of
https://github.com/RPCS3/llvm-mirror.git
synced 2024-11-24 11:42:57 +01:00
ae65e281f3
to reflect the new license. We understand that people may be surprised that we're moving the header entirely to discuss the new license. We checked this carefully with the Foundation's lawyer and we believe this is the correct approach. Essentially, all code in the project is now made available by the LLVM project under our new license, so you will see that the license headers include that license only. Some of our contributors have contributed code under our old license, and accordingly, we have retained a copy of our old license notice in the top-level files in each project and repository. llvm-svn: 351636
69 lines
2.1 KiB
C++
69 lines
2.1 KiB
C++
//===- 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?";
|
|
}
|