|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!|
Mar 13, 2019
I made a visualization for this code on PyAlgoViz.
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
Feb 8, 2020
@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
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.
Apr 24, 2020
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.
Apr 27, 2020
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)
Mar 22, 2021
Thank you. Here is a translated SNAP! program version...https://snap.berkeley.edu/project?user=nisar&project=monte_carlo_pi_calculation
Apr 1, 2021
Thank you! This helped me understand how to run it with x and y.
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
Made a version of your Monte Carlo pi code with a plot. Here is link: