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# louismullie/pi-monte-carlo.py

Created Sep 23, 2012
Monte Carlo Estimation of PI in Python
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 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)

### dandrewmyers commented Apr 5, 2018

https://github.com/dandrewmyers/numerical/blob/master/mc_pi.py

### laffra commented Mar 13, 2019

I made a visualization for this code on PyAlgoViz.

### peduajo commented Oct 15, 2019

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

### AndreaPasqualini commented Feb 8, 2020 • edited

@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.

### vwang0 commented Apr 24, 2020 • edited

• 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.

### ifuchs commented Apr 27, 2020 • edited

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)
``````

### nisarkhanatwork commented Mar 22, 2021

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

### mfriar2 commented Apr 1, 2021

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

### srdinsek commented Feb 17, 2022

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
``````