无偏的随机数发生器 [英] unbiased random number generator

问题描述

我使用以下代码在域中生成随机数。当我绘制它们，他们看起来分组在右边。我可以告诉你我的情节，但我不知道如何上传它。基本上，我将一些数据值关联到相应的点。你能告诉我怎么能纠正它吗？我的完整代码是

I am generating the random numbers in a domain by using the following code. When I plot them they look grouped at the right. I could show you my plot but I do not know how to upload it. Basically I associate a some data value to the respective point. May you tell me how can I correct it please? My complete code is

#include <iostream>
#include <cmath>
#include <fstream>
#include <sstream>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <time.h>

using namespace std;
string int2string1( int l );
string int2string2( int m );
int main ()
{
    ofstream outFile;
    ofstream myimp;
    string filename;
    srand((unsigned)time(0));
    int nx = 400;
    int ny = 200;
    int i,j,ix,iy,xm,ym,nimp,nfam[nx][ny];
    float vo,rnd,rr,rad,sig,vimp[nx][ny];
    for (i=0; i<nx; i++)
    {
        for (j=0; j<ny; j++)
        {
            vimp[i][j] = 0.0;
        }
    }
    rad = 5.0;
    xm = 0;
    ym = 0;
    vo = 0.08;
    sig = 4.0;
    myimp.open("imp.dat");
    for(i=1; i<nx-1; i++)
    {
        for(j=1; j<ny-1; j++)
        {
            rnd = (random() %1000 + 1)*1.0/1000.0;
            if(rnd>0.99)
            {
                xm = random() % 398 + 1;              /***1 through 399 ***/
                ym = random() % 198 + 1;              /***1 through 199 ***/
                for(ix=xm-5; ix<=xm+5; ix++)
                {
                    for(iy=ym-5; iy<=ym+5; iy++)
                    {
                        rr = sqrt(pow(ix-xm,2.)+pow(iy-ym,2.));
                        if(rr<=rad)
                        {
                            vimp[ix][iy] = vo*1.6e-19;
                        }
                    }
                }
            }
            myimp<<i<<"\t\t"<<j<<"\t\t"<<xm<<"\t\t"<<ym<<"\t\t"<<nfam[i][j]<<"\t\t"<<vimp[i][j]*6.23e18<<"\n";
        }
    }
    myimp.close();
    return 0;
}

推荐答案

基本上， rand（）％N 如果 RAND_MAX 不是 N 。它以不均匀的方式将 [0，RAND_MAX] 中的数字投影到范围 [0，N]

Basically, the rand() % N expression introduces a bias if RAND_MAX is not a multiple of N. It projects numbers in [0,RAND_MAX] onto the range [0,N] with a non-uniform fashion.

假设 RAND_MAX = 4 和 N = 2 。然后，有3个数字产生 0 （ 0 ， 2 和 4 ）和2个产生 1 （ 1 和 3 ）。因此，您获得 0 的变化率为60％，40％的机会获得 1 。

Suppose that RAND_MAX=4 and N=2. Then, there are 3 numbers that produce 0 (0, 2 and 4) and 2 numbers that produce 1 (1 and 3). Thus, you have 60% change of getting 0 and 40% chance of getting 1.

正确的方式来实现从 [0，RAND_MAX] 到 [0，N] code>是通过重复调用 rand（），直到随机值在期望的间隔。请参见 Random.nextInt（） 在Java中（点击Oli Charlesworth的链接）。

The correct way to implement an unbiased projection from [0,RAND_MAX] onto [0,N] is by invoking rand() repeatedly until the random value is in the desired interval. See the documentation for Random.nextInt() in Java (Credits to Oli Charlesworth for the link).

，对于纯粹的执行速度，你想避免多次调用 rand（），生成最小偏差的方法是使用中间 double 数字，例如：

Assuming that, for sheer execution speed, you want to avoid calling rand() multiple times, the way to generate the least bias possible is to use an intermediate double number, such as:

double myrand ()
{
    return double(rand()) / double(RAND_MAX);
}

int myrand ( int max )
{
    return int(myrand() * double(max));
}

编辑：这里有一个简单的类， rand（）函数的输出到 [0，N] $ c> rand（）。

Here's a simple class that will project outputs of the rand() function in to a range [0,N] with no less bias than rand().

class BoundedRandom
{
    const int m_maximum;
    const int m_divisor;
public:
    BoundedRandom ( int maximum )
        : m_maximum(maximum),
          m_divisor(RAND_MAX/(maximum+1))
    {}

    int operator() ()
    {
        int result = rand() / m_divisor;
        while ( result > m_maximum ) {
            result = rand() / m_divisor;
        }
        return (result);
    }
};

警告：未经测试或调试。

Caution: not tested or debugged.

您可以像这样使用这个生成器：

You can use this generator like this:

BoundedRandom randomx(398);
BoundedRandom randomy(198);
// ...
 xm = randomx() + 1; // 1 through 399
 ym = randomy() + 1; // 1 through 199

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