C：随机游走(布朗运动)程序不返回预期的理论值.为什么？

发布时间：2020-12-16 05:00:01 所属栏目：百科来源：网络整理

导读：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).

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 ( "%gn",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 ( "%gn",expected_distance );
  return EXIT_SUCCESS;
}

（编辑：李大同）

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