蒙特卡罗法近似求解圆周率π
作者:互联网
文章目录
1. 原理
- 给出 x∈[0,1),y∈[0,1) 的均匀分布随机点,模拟 t 次,落在以 (0,0) 为圆心,半径 r=1 的圆以内的次数为 c
- 当模拟次数足够大时,可以看成面积比 4π≈tc⇒π≈4c/t
2. 模拟代码
# -*- coding:utf-8 -*-
# @Python Version: 3.7
# @Time: 2020/5/2 9:02
# @Author: Michael Ming
# @Website: https://michael.blog.csdn.net/
# @File: monte_carlo_cal_pi.py
# @Reference:
import random
import matplotlib.pyplot as plt
simulations = [100, 1000, 10000]
plt.figure()
plt.rcParams['font.sans-serif'] = 'SimHei' # 消除中文乱码
plotid = 1
color = ['r', 'b']
for i in range(len(simulations)):
f = plt.subplot(1, 3, plotid)
plotid += 1
count = 0
t = simulations[i]
time = simulations[i]
while time > 0:
time -= 1
x = random.random() # [0,1)
y = random.random() # [0,1)
val = x ** 2 + y ** 2
pos = 1
if (val < 1):
pos = 0
count += 1
f.scatter(x, y, c=color[pos])
pi = 4 * count / t
f.set_title("模拟次数{},pi的值{:.4f}".format(t, pi))
plt.suptitle("蒙特卡罗法近似求解圆周率pi")
plt.show()
标签:plt,求解,圆周率,random,pos,蒙特卡罗,simulations,pi,模拟 来源: https://blog.csdn.net/qq_21201267/article/details/105886943