Python/maths/pi_monte_carlo_estimation.py
Christian Clauss ab3400bfad
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Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
2020-03-14 23:55:13 +01:00

64 lines
2.0 KiB
Python

import random
class Point:
def __init__(self, x: float, y: float) -> None:
self.x = x
self.y = y
def is_in_unit_circle(self) -> bool:
"""
True, if the point lies in the unit circle
False, otherwise
"""
return (self.x ** 2 + self.y ** 2) <= 1
@classmethod
def random_unit_square(cls):
"""
Generates a point randomly drawn from the unit square [0, 1) x [0, 1).
"""
return cls(x=random.random(), y=random.random())
def estimate_pi(number_of_simulations: int) -> float:
"""
Generates an estimate of the mathematical constant PI (see https://en.wikipedia.org/wiki/Monte_Carlo_method#Overview).
The estimate is generated by Monte Carlo simulations. Let U be uniformly drawn from the unit square [0, 1) x [0, 1). The probability that U lies in the unit circle is:
P[U in unit circle] = 1/4 PI
and therefore
PI = 4 * P[U in unit circle]
We can get an estimate of the probability P[U in unit circle] (see https://en.wikipedia.org/wiki/Empirical_probability) by:
1. Draw a point uniformly from the unit square.
2. Repeat the first step n times and count the number of points in the unit circle, which is called m.
3. An estimate of P[U in unit circle] is m/n
"""
if number_of_simulations < 1:
raise ValueError("At least one simulation is necessary to estimate PI.")
number_in_unit_circle = 0
for simulation_index in range(number_of_simulations):
random_point = Point.random_unit_square()
if random_point.is_in_unit_circle():
number_in_unit_circle += 1
return 4 * number_in_unit_circle / number_of_simulations
if __name__ == "__main__":
# import doctest
# doctest.testmod()
from math import pi
prompt = "Please enter the desired number of Monte Carlo simulations: "
my_pi = estimate_pi(int(input(prompt).strip()))
print(f"An estimate of PI is {my_pi} with an error of {abs(my_pi - pi)}")