这只是基于“Feynman物理学讲座”第6-3节的实验：
In its simplest version,we imagine a “game” in which a “player”
starts at the point x=0 and at each “move” is required to take a step
either forward (toward +x) or backward (toward −x). The choice is to
be made randomly,determined,for example,by the toss of a coin.
资料来源：http://www.feynmanlectures.caltech.edu/I_06.html#Ch6-S3
我的目标是计算距离说明点的预期距离.所以,我想每一步都等于一个距离单位.我写了一个简单的C程序来模拟30个随机步骤,然后计算起点的最终距离.重复一百万次,程序平均距离以获得预期距离.
理论上,预期距离应该是步数的平方根.那应该是sqrt(30)= 5.48.
但是,该程序运行几次并保持返回接近4.33的值(更确切地说,4.33461,4.33453和4.34045).为什么它甚至不接近理论值约5.48？
这是我的代码：
- #include <time.h>
- #include <stdlib.h>
- #include <stdio.h>
- int main ( int argc,char *argv[] )
- {
- int number_of_steps = 30;
- int repetition = 1000000;
- int distance = 0;
- int total_distance = 0;
- double expected_distance;
- int i,j;
- srand(time(NULL));
- for ( i = 0; i < repetition; i++ ) {
- for ( j = 0; j < number_of_steps; j++) {
- distance += rand() & 1 ? -1 : 1;
- }
- total_distance += abs(distance);
- distance = 0;
- }
- expected_distance = (float) total_distance / i;
- printf ( "%g\n",expected_distance );
- return EXIT_SUCCESS;
- } /* ---------- end of function main ---------- */
解决方法
从您链接到的讲座中,您的理论期望基于
root mean square,这与您编码的
arithmetic mean不同.通过将算法从一个更改为另一个,the code现在可以为您提供预期的结果.
- for ( i = 0; i < repetition; i++ ) {
- for ( j = 0; j < number_of_steps; j++) {
- distance += rand() & 1 ? -1 : 1;
- }
- total_distance += distance * distance;
- distance = 0;
- }
- expected_distance = sqrt((float) total_distance / repetition);
- printf ( "%g\n",expected_distance );
- return EXIT_SUCCESS;
- }
