std :: normal_distribution< double>导致错误的顺序窗口对比linux? [英] std::normal_distribution<double> results in wrong order windows versus linux?
问题描述
任何人访问此问题?根据 1 ,实施不需要生成相同的数据。
在实践中,在arm,x86,自由和商业编译器之间的STL实现有很多不同吗?
Anyone came access this issue? Per 1 the implementations are not required to produce same data. What about in practice - are there many differences in STL implementations among arm, x86, free and commercial compilers?
// g++ --std=c++11 -o a minimal.cpp && ./a
#include <iostream>
#include <random>
using namespace std;
int
main()
{
std::mt19937_64 gen;
gen.seed(17);
cout << "\nNormal\n";
normal_distribution<double> distr1;
for (int i = 0; i < 2; i++) {
double delay = distr1(gen);
printf(" Value = %15.15g\n", delay);
}
return(0);
}
/*
Results
(1) gcc-4.8.0 linux 64b version result
Normal
Value = 1.03167351251536
Value = 1.21967569130525
(2) Microsoft Visual Studio Community 2015 Version 14.0.23107.0 D14REL
or
Microsoft Visual Studio Professional 2012 Version 11.0.60610.01 Update 3
Normal
Value = 1.21967569130525
Value = 1.03167351251536 // same values in wrong (different) order
*/
我可以理解在一些特殊的硬件平台上使用不同的算法来生成生成器或分布,但是这种差异似乎更像是一个错误。
I could understand use of a different algorithm for either the generator or the distribution on some special HW platform but this difference seems more as a bug.
这里是一些更多的代码,我用来诊断差异来自哪里,并解决它:
- 生成器和统一分布匹配win和linux。
- 除了成对顺序以外,正态分布数字匹配
Here is some more code I used to diagnose where does the difference come from and workaround it: - Generator and uniform distribution match on win and linux. - The normal distribution matches numerically except for pair-wise order
// g++ --std=c++11 -o a workaround.cpp && ./a
#include <iostream>
#include <random>
#include <stack>
using namespace std;
typedef std::mt19937_64 RandGenLowType;
// Helper wrapper - it did confirm that the differences
// do NOT come from the generator
class RandGenType : public RandGenLowType {
public:
result_type operator()() {
result_type val = RandGenLowType::operator()();
printf(" Gen pulled %20llu\n", val);
return(val);
}
};
typedef normal_distribution<double> NormalDistrLowType;
// Workaround wrapper to swap the output data stream pairwise
class NormalDistrType : NormalDistrLowType {
public:
result_type operator()(RandGenType &pGen) {
// Keep single flow (used variables, includes) same for all platforms
if (win64WaStack.empty()) {
win64WaStack.push(NormalDistrLowType::operator()(pGen));
#ifdef _MSC_VER
win64WaStack.push(NormalDistrLowType::operator()(pGen));
#endif
}
result_type lResult = win64WaStack.top();
win64WaStack.pop();
return(lResult);
}
private:
std::stack<result_type> win64WaStack;
};
int
main()
{
RandGenType gen;
gen.seed(17);
// No platform issue, no workaround used
cout << "\nUniform\n";
uniform_real_distribution<double> distr;
for (int i = 0; i < 4; i++) {
double delay = distr(gen);
printf(" Delay = %15.15g\n", delay);
}
// Requires the workaround
#ifdef _MSC_VER
cout << "Workaround code is active, swapping the output stream pairwise\n";
#endif
cout << "\nNormal\n";
//normal_distribution<float> distr1;
NormalDistrType distr1;
for (int i = 0; i < 10; i++) {
double delay = distr1(gen);
printf(" Value = %15.15g\n", delay);
}
return(0);
}
推荐答案
分布的随机数,例如 Box-Muller变换和 Marsaglia极坐标法,一次生成两个随机数。使用这些方法之一的分发对象将生成两个随机数,返回其中一个,并在下次调用时保存另一个。
Several common methods of generating normally distributed random numbers, such as the Box-Muller transform and the Marsaglia polar method, generate two random numbers at once. A distribution object using one of those methods would generate two random numbers, return one of them, and save the other one for the next time it's called.
返回哪一个并且哪一个被存储是当然完全取决于图书馆作者。它看起来像libstdc ++和MSVC使用相同的算法,但发生选择不同。
Which one is returned and which one is stored is, of course, completely up to the library author. It looks like libstdc++ and MSVC use the same algorithm, but happen to choose differently.
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