pi can be approximated by throwing darts at a circle in a square, and calculating the ratio of darts landing on the circle vs the square.



In [200]:

    
import math
import random

simple way

First, we generate a bunch of throws by using random.random()

We are throwing darts in only the top right quadrant of the square, so the origin is the center of the square, and the diameter of the circle is 2.

We figure out if the points are in the circle by using the pythogeras theorem to calculate the distance of each point from the origin (0,0). If the distance is greater than the radius of the circle, it's outside.



In [209]:

    
darths_thrown = 10000
throws = [[random.random(), random.random()] for i in range(darths_thrown)]

in_circle=0
out_circle=0

for throw in throws:
    if math.sqrt(throw[0]**2 + throw[1]**2) <= 1:
        in_circle +=1
    else:
        out_circle += 1

pi_estimate = in_circle/len(throws) * 4

print(f'estimate: {pi_estimate}')
print(f'actual: {math.pi}')
print(f'difference is {abs(pi_estimate-math.pi)}')









    



estimate: 3.146
actual: 3.141592653589793
difference is 0.004407346410206792

now to jazz this up visually



In [210]:

    
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
%matplotlib inline

df = pd.DataFrame(np.random.rand(darths_thrown,2)*2, columns=["x", "y"])
df.head()



In [211]:

    
def is_in_circle(row):
    x = row['x']
    y = row['y']
    if math.sqrt((x-1)**2 + (y-1)**2) <= 1:
        return True
    else:
        return False

df['In_Circle'] = df.apply(is_in_circle, axis=1)
df.head()



In [242]:

    
plt.scatter(df['x'], df['y'], c=[within and 'blue' or 'red' for within in df['In_Circle']], alpha=0.4,s= 25)

fig = plt.gcf()
fig.set_size_inches(9,9);

the ratio of the area of the circle divided by the area of the square gives us pi/4:



In [205]:

    
pi_estimate = df['In_Circle'].sum() / len(df) * 4
pi_estimate









    Out[205]:





3.1453333333333333



In [206]:

    
pi_estimate - math.pi









    Out[206]:





0.0037406797435401984

So the estimate is pretty close! Neat, since all it took was some very simple maths and a random number generator.



In [ ]:

	x	y
0	0.946924	0.950627
1	1.053205	0.041109
2	1.365198	1.888242
3	1.806928	0.279787
4	0.162452	0.148584

	x	y	In_Circle
0	0.946924	0.950627	True
1	1.053205	0.041109	True
2	1.365198	1.888242	True
3	1.806928	0.279787	False
4	0.162452	0.148584	False