C++ Monte Carlo 集成：如何在不对结果求和的情况下多次运行编码？

Question

这是我更改代码之前的代码，它包含多次运行的功能：

#include <iostream>
#include <fstream>
#include <stdlib.h>
#include <math.h>
#include <random>
#include <iomanip>
#include <vector>

using namespace std; 

int main () {
    double vol;
    double hit;
    int samples;
    int i, j;
    double sum;
    double pt;
    double actual_vol;
    const double PI = 2.0*atan2(1.0,0.0);
    double abs_err;
    double rel_err;

    random_device dev;
    default_random_engine e{ dev() };
    uniform_real_distribution<double> u{0.0,1.0};

    samples = 1000000 * dim;

    actual_vol = pow(PI, double(dim/2.0)) / exp(lgamma(double(dim/2.0)+1.0));

    for (i = 0; i < samples; i++) {
            sum = 0;
            for (j = 0; j < dim; j++) {
                    pt = 2*(u(e)-0.5);
                    sum += pt*pt;
            }
            if (sqrt(sum) < 1) {
                    hit += 1;
            }
    }

    vol = ( pow(2,dim) * hit ) / samples;
    abs_err = fabs( actual_vol - vol);
    rel_err = abs_err / actual_vol;

    cout << "Average volume of your sphere: " << setprecision(7) << vol << endl;
    cout << "Actual volume: " << setprecision(7) << actual_vol << endl;
    cout << "Absolute Error: " << setprecision(7) << abs_err << endl;
    cout << "Relative Error: " << setprecision(7) << rel_err << endl;
}

我会得到如下所示的正确输出：

Average volume of your sphere: 3.140924
Actual volume: 3.141593
Absolute Error: 0.0006686536
Relative Error: 0.000212839

现在，当我更改它以便我可以多次调用该函数并运行它时，使用以下代码：

#include <iostream>
#include <fstream>
#include <stdlib.h>
#include <math.h>
#include <random>
#include <iomanip>
#include <vector>

using namespace std;

double monte_carlo (int dim) {
    double vol;
    double hit;
    int samples;
    int i, j;
    double sum;
    double pt;
    double actual_vol;
    const double PI = 2.0*atan2(1.0,0.0);
    double abs_err;
    double rel_err;

    random_device dev;
    default_random_engine e{ dev() };
    uniform_real_distribution<double> u{0.0,1.0};

    samples = 1000000 * dim;

    actual_vol = pow(PI, double(dim/2.0)) / exp(lgamma(double(dim/2.0)+1.0));

    for (i = 0; i < samples; i++) {
            sum = 0;
            for (j = 0; j < dim; j++) {
                    pt = 2*(u(e)-0.5);
                    sum += pt*pt;
            }
            if (sqrt(sum) < 1) {
                    hit += 1;
            }
    }

    vol = ( pow(2,dim) * hit ) / samples;
    abs_err = fabs( actual_vol - vol);
    rel_err = abs_err / actual_vol;

    cout << "Average volume of your sphere: " << setprecision(7) << vol << endl;
    cout << "Actual volume: " << setprecision(7) << actual_vol << endl;
    cout << "Absolute Error: " << setprecision(7) << abs_err << endl;
    cout << "Relative Error: " << setprecision(7) << rel_err << endl;
}

int main (int argc, char* argv[]) {

    int dim = 0;
    int runs = 0;
    int i;

    dim =  atoi(argv[1]);
    runs = atoi(argv[2]);

    for (i = 0; i < runs; i++) {
            monte_carlo(dim);
    }

    return 0;
}

我得到这些结果，现在将以前的值与当前值相加，这不是我想要的：

Average volume of your sphere: 3.141764
Actual volume: 3.141593
Absolute Error: 0.0001713464
Relative Error: 5.454126e-05
Average volume of your sphere: 6.283674
Actual volume: 3.141593
Absolute Error: 3.142081
Relative Error: 1.000156
Average volume of your sphere: 9.427502
Actual volume: 3.141593
Absolute Error: 6.285909
Relative Error: 2.000867
Average volume of your sphere: 12.56937
Actual volume: 3.141593
Absolute Error: 9.427775
Relative Error: 3.000954
Average volume of your sphere: 15.71272
Actual volume: 3.141593
Absolute Error: 12.57113
Relative Error: 4.001515
Average volume of your sphere: 18.85378
Actual volume: 3.141593
Absolute Error: 15.71219
Relative Error: 5.001345
Average volume of your sphere: 21.99504
Actual volume: 3.141593
Absolute Error: 18.85345
Relative Error: 6.001239

你会注意到球体平均体积的第一个值大约是 3.14，然后是它的第二个实例，现在是 6.28（或者是第一个的两倍），第三个实例是 9.42（大约是第一个的三倍）第一个）等等

它应该做的是运行对每个运行进行一次新计算，并且每个值都应该徘徊在 3.14 左右。如何让它停止对前一个运行的值求和？

谢谢！！

Answer 1

这可能是因为您从未重新初始化变量。

你也有强烈的 "old" C 偏见（C headers，atoi，fabs...的用法），在你需要的时候声明你的变量，而且你的路径总是相似的，因为你使用相同的随机数生成器和相同的种子（默认构造）。

仍然，对于你的问题：

double hit = 0;
double samples = 0;

以此类推

同样对于 PI，如果你有提升，使用它的常量，而不是用低于它所能提供的精度重新计算它。

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