refer to: http://www.stealthcopter.com/blog/2009/09/python-calculating-pi-using-random-numbers/

During my undergraduate degree I wrote a program in fortran 95 to calculate pi using random numbers. My aim is to rewrite it efficiently in python. I know its a terrible way to calculate pi, and there are much better ways to do it but its fun!

First I’ll explain the maths so you can visualise what’s going on. As we should know _pi_ is the ratio of circle’s radius to its circumference, which is conveniently the same as the ratio of a circle’s area to the square of its radius (wiki…)

So what we are going to be doing is picking lots of random coordinates in an x-y grid and calculating if they are within the circle or the square.

We will assign the radius to be 1, because that makes it easy to work with. By default a random number in python ( random() ) will return a floating point number between 0 and 1. To test if a point is within a circle we simply use Pythagoras.

So if the sqrt(a**2+b**2)<=1 then the point lies inside the circle’s radius. In the diagram above we see that point A lies within the circle, and point B lies outside the circle.

We can really don’t need to use the whole circle as it has symmetry, so we can just take a quartre, which makes the generating of random numbers easier as you only need to use a random number for x and y between 0 and 1, rather than -1 and 1. It will look like the diagram below.

Now for a confusing bit of maths. We are calculating the ratio of the area of a circle to the area of a square.

# Area of circle
A=pi*r**2
# where r = 1
A = pi
# Area of square
A = l ** 2
# in this case (see diagram) our square's length is twice the radius
l=2*r
A=(1+1)**2 = 4 #Therefore our ratio will be pi : 4.
# Which means we must multiply our result by four to get pi.

Final version (efficient for using)

from random import *
from math import sqrt
inside=0
n=1000
for i in range(0,n):
x=random()
y=random()
if sqrt(x*x+y*y)<=1:
inside+=1
pi=4*inside/n
print pi

Below we can see the values it creates

n	calc	error
1 4.00000000 0.73686317
10 3.60000000 0.45840735
100 3.24000000 0.09840735
1000 3.06400000 -0.07759265
10000 3.16160000 0.02000735
100000 3.14140000 -0.00019265
1000000 3.14293600 0.00134335
10000000 3.14117920 -0.00041345

So we can see that the program quickly solves pi to about two decimal places, but it is a terribly inefficient method and will struggle to get much more accuracy than this.

Resources to check out:

Summary: How to calculate PI? Based on Monte Carlo method的更多相关文章

  1. (转)Monte Carlo method 蒙特卡洛方法

    转载自:维基百科  蒙特卡洛方法 https://zh.wikipedia.org/wiki/%E8%92%99%E5%9C%B0%E5%8D%A1%E7%BE%85%E6%96%B9%E6%B3%9 ...

  2. Monte Carlo Method(蒙特·卡罗方法)

    0-故事: 蒙特卡罗方法是计算模拟的基础,其名字来源于世界著名的赌城——摩纳哥的蒙特卡罗. 蒙特卡罗一词来源于意大利语,是为了纪念王子摩纳哥查理三世.蒙特卡罗(MonteCarlo)虽然是个赌城,但很 ...

  3. 蒙特·卡罗方法(Monte Carlo method)

    蒙特·卡罗方法(Monte Carlo method),也称统计模拟方法,是二十世纪四十年代中期由于科学技术的发展和电子计算机的发明,而被提出的一种以概率统计理论为指导的一类非常重要的数值计算方法.是 ...

  4. 蒙特卡罗算法(Monte Carlo method)

    蒙特卡罗方法概述 蒙特卡罗方法又称统计模拟法.随机抽样技术,是一种随机模拟方法,以概率和统计理论方法为基础的一种计算方法,是使用随机数(或更常见的伪随机数)来解决很多计算问题的方法.将所求解的问题同一 ...

  5. (转)Markov Chain Monte Carlo

    Nice R Code Punning code better since 2013 RSS Blog Archives Guides Modules About Markov Chain Monte ...

  6. Introduction To Monte Carlo Methods

    Introduction To Monte Carlo Methods I’m going to keep this tutorial light on math, because the goal ...

  7. 蒙特卡罗方法、蒙特卡洛树搜索(Monte Carlo Tree Search,MCTS)初探

    1. 蒙特卡罗方法(Monte Carlo method) 0x1:从布丰投针实验说起 - 只要实验次数够多,我就能直到上帝的意图 18世纪,布丰提出以下问题:设我们有一个以平行且等距木纹铺成的地板( ...

  8. History of Monte Carlo Methods - Part 1

    History of Monte Carlo Methods - Part 1 Some time ago in June 2013 I gave a lab tutorial on Monte Ca ...

  9. Monte Carlo Policy Evaluation

    Model-Based and Model-Free In the previous several posts, we mainly talked about Model-Based Reinfor ...

随机推荐

  1. 单片机 C 语言模块化编程

    好的开始是成功的一半 通过上一章的学习,我想你已经掌握了如何在程序中释放CPU了.希望能够继续坚持下去.一个良好的开始是成功的一半.我们今天所做的一切都是为了在单片机编程上做的更好. 在谈论今天的主题 ...

  2. 【Xamarin开发 Android 系列 6】 Android 结构基础(上)

    原文:[Xamarin开发 Android 系列 6] Android 结构基础(上) 前面大家已经熟悉了什么是Android,而且在 [Xamarin开发 Android 系列 4] Android ...

  3. 从C#到Python —— 4 类及面向对象

    http://www.cnblogs.com/yanxy/archive/2010/04/04/c2p_4.html 如果你熟悉C#,那么对类(Class)和面向对象(Object Oriented) ...

  4. git图示所有分支的历史

    1.第一种方法 git gui 菜单栏上 repository-->visual all branch history 或者直接使用命令gitk --all 2.在git bash中,使用命令查 ...

  5. NOI2005瑰丽华尔兹

    1499: [NOI2005]瑰丽华尔兹 Time Limit: 3 Sec  Memory Limit: 64 MBSubmit: 893  Solved: 508[Submit][Status] ...

  6. Master Nginx(1) - Installing Nginx and Third-Party Modules

    Installing NGINX and Third-Party Modules Installing Nginx using a package manager Linux(deb-based) s ...

  7. 微软Sharepoint的一些缺点

    转:http://bbs.tianya.cn/post-144-566491-1.shtml 微软Sharepoint的一些缺点(一) 微软Sharepoint的一些缺点 关于SharePoint,它 ...

  8. Python生成器 yield

    迭代器与list相比较,就for in句型循环拿数据而言: 用list写很简洁,但如果list数据过大,会很消耗资源. 用iteration 迭代器写,则不会消耗那么多资源.他会随用随取,用一个拿一个 ...

  9. Win7,8的上帝模式文件夹 GodMode.{ED7BA470-8E54-465E-825C-99712043E01C}

    新建一文件夹,命名为:“GodMode.{ED7BA470-8E54-465E-825C-99712043E01C}”, 创建后双击进入,可以看到计算机所有的配置选项信息.

  10. visual studio 资源视图 空白 解决方案

    visual studio 资源视图打开后显示空白的解决方案步骤: 在解决方案view下,右键点击工程 1 unload projects 完成第一步后仍然在解决方案view下,右键点击工程 2 re ...