Monte Carlo Estimation of PI in Python

pi-monte-carlo.py

import random as r
import math as m

# Number of darts that land inside.
inside = 0
# Total number of darts to throw.
total = 1000

# Iterate for the number of darts.
for i in range(0, total):
  # Generate random x, y in [0, 1].
  x2 = r.random()**2
  y2 = r.random()**2
    # Increment if inside unit circle.
    if m.sqrt(x2 + y2) < 1.0:
      inside += 1

# inside / total = pi / 4
pi = (float(inside) / total) * 4

# It works!
print(pi)

hmmm the code doesn't give me the pi number. I think that the equation is x² + y² <= 1, and you are doing sqrt(x² + y²) <= 1. I have checked this equation with a large number of points and it is really close to the pi number, like ~3.1415

@peduajo: taking the square root in this specific case is not a problem. At worst, we mis-count points very much on the circle because of rounding error. This is because the circle (and the points inside it) is defined as the set of points (x, y) such that x^2 + y^2 <= r^2. If r=1, then the square root becomes irrelevant and we could as well check x^2 + y^2 <= 1. The problem is that the circle in the code is not the unit circle. By taking x and y from uniforms over [0, 1], we are using a circle inscribed in the unit square, which is not the unit circle but is the half-unit circle.

• I would suggest using random.uniform() instead of random.random() since some darts might be landed on the edge of the square, but random.random() only generates a random float uniformly in the semi-open range [0.0, 1.0), while random.uniform() generates a random float uniformly in the range [0.0, 1.0].

• no indentation needed in the if clause.

• no math.sqrt needed (reduce computational cost). if math.sqrt(x2 + y2) < 1.0 then x2+y2 < 1.0, vice versa.

Doesn't this do the same thing without dealing with the square or the x,y coords?

import random
iterations = 1000000
x = 0
for i in range(iterations):
	x+=(1-random.random()**2)**.5
print(4*x/iterations)

Thank you. Here is a translated SNAP! program version...https://snap.berkeley.edu/project?user=nisar&project=monte_carlo_pi_calculation

Thank you! This helped me understand how to run it with x and y.

Here is a slightly faster version.

def square_pi(N):
    # square_pi(100000000) is fast and gives 4 digits. 
    xy = np.random.random((N, 2)) ** 2
    counts = np.sum((xy[:, 0] + xy[:, 1]) <= 1)

    print("pi was approximated at ::",4*counts/N)
    return 4*counts/N