这些随机数我做错了什么?

时间:2016-11-08 08:10:36

标签: c++ random benchmarking uniform-distribution

我被告知rand()mod n产生有偏差的结果,所以我试着让这段代码来检查它。它会生成从{1}到s的{​​{1}}个数字,然后按出现次数进行排序。

l

运行此代码后,我可以发现rand()的预期偏差:

#include <iostream>
#include <random>

using namespace std;

struct vec_struct{
    int num;
    int count;
    double ratio;
};

void num_sort(vec_struct v[], int n){
    for (int i = 0; i < n-1; i++){
        for (int k = 0; k < n-1-i; k++){
            if (v[k].num > v[k+1].num) swap(v[k], v[k+1]);
        }
    }
}

void count_sort(vec_struct v[], int n){
    for (int i = 0; i < n-1; i++){
        for (int k = 0; k < n-1-i; k++){
            if (v[k].count < v[k+1].count) swap(v[k], v[k+1]);
        }
    }
}

int main(){

    srand(time(0));

    random_device rnd;

    int s, l, b, c = 1;

    cout << "How many numbers to generate? ";
    cin >> s;

    cout << "Generate " << s << " numbers ranging from 1 to? ";
    cin >> l;

    cout << "Use rand or mt19937? [1/2] ";
    cin >> b;

    vec_struct * vec = new vec_struct[s];

    mt19937 engine(rnd());
    uniform_int_distribution <int> dist(1, l);

    if (b == 1){
        for (int i = 0; i < s; i++){
            vec[i].num = (rand() % l) + 1;
        }
    } else if (b == 2){
        for (int i = 0; i < s; i++){
            vec[i].num = dist(engine);
        }   
    }
    num_sort(vec, s);

    for (int i = 0, j = 0; i < s; i++){
        if (vec[i].num == vec[i+1].num){
            c++;
        } else {
            vec[j].num = vec[i].num;
            vec[j].count = c;
            vec[j].ratio = ((double)c/s)*100;
            j++;
            c = 1;  
        }
    }
    count_sort(vec, l);

    if (l >= 20){

        cout << endl << "Showing the 10 most common numbers" << endl;
        for (int i = 0; i < 10; i++){
            cout << vec[i].num << "\t" << vec[i].count << "\t" << vec[i].ratio << "%" << endl;
        }

        cout << endl << "Showing the 10 least common numbers" << endl;
        for (int i = l-10; i < l; i++){
            cout << vec[i].num << "\t" << vec[i].count << "\t" << vec[i].ratio << "%" << endl;
        }
    } else {

        for (int i = 0; i < l; i++){
            cout << vec[i].num << "\t" << vec[i].count << "\t" << vec[i].ratio << "%" << endl;
        }
    }
}

Hoover我与$ ./rnd_test How many numbers to generate? 10000 Generate 10000 numbers ranging from 1 to? 50 Use rand or mt19937? [1/2] 1 Showing the 10 most common numbers 17 230 2.3% 32 227 2.27% 26 225 2.25% 25 222 2.22% 3 221 2.21% 10 220 2.2% 35 218 2.18% 5 217 2.17% 13 215 2.15% 12 213 2.13% Showing the 10 least common numbers 40 187 1.87% 7 186 1.86% 39 185 1.85% 42 184 1.84% 43 184 1.84% 34 182 1.82% 21 175 1.75% 22 175 1.75% 18 173 1.73% 44 164 1.64% mt19937获得了相同的结果!这有什么不对?不应该是统一的,还是测试没用?

3 个答案:

答案 0 :(得分:1)

不,它不应该是完全一致的。因此,上述内容并非任何错误的证据。

它们是随机的,因此应该相当统一,但不完全一致。

特别是你会期望每个数字出现大约10000/50 = 200次 - 大致标准偏差为sqrt(200),大约为14 - 对于50个数字,你可以预期大约2个标准偏差 - 是+ - / 28。

使用RAND_MAX模数引起的偏差小于此值;所以你需要更多的样本来检测偏见。

答案 1 :(得分:0)

您必须使用更多样本进行此类随机数测试。我用你的代码尝试了50000,结果是:

要生成多少个数字? 50000

生成500到1的数字? 50

使用rand还是mt19937? [1/2] 2

显示10个最常见的数字

36 1054 2.108%

14 1051 2.102%

11 1048 2.096%

27 1045 2.09%

2 1044 2.088%

33 1035 2.07%

21 1034 2.068%

48 1034 2.068%

34 1030 2.06%

39 1030 2.06%

显示10个最不常见的数字

47 966 1.932%

16 961 1.922%

38 960 1.92%

28 959 1.918%

8 958 1.916%

10 958 1.916%

30 958 1.916%

32 958 1.916%

18 953 1.906%

23 953 1.906%

答案 2 :(得分:-1)

据我所知 http://www.cplusplus.com/reference/random/mersenne_twister_engine/ mt19937将遭受与rand()

相同的偏见

偏差是由于rand()在某个范围[0-MAX_RAND]中生成无符号整数,当你取模数时它会更小一些(除非你的除数是MAX_RAND的整数除数)

考虑:

Range [0-74]:
0 % 50 = 0
40 % 50 = 40
50 % 50 = 0
74 % 50 = 24
(numbers less than 25 occur twice)
相关问题