改进蒙特卡洛代码以找到高维球体的体积。 （PYTHON）

Question

Considering Gaussian shape we can find the volume of n dimensional volume of a sphere. My intention is to find the volume using Monte Carlo method.

Using the gaussian integral I have found the formula

What I understand that the ratio of the points inside the n-dimensional sphere to the total number of points will then be roughly the same as the ratio of the volume of the ball to that of the cube. I mean the mass density would never change, whatever dimension I use.

Therefore I assume I should follow the same technique what I used to find the value of pi using Monte Carlo method.

I don't understand how to follow the code which I evaluated to find the value of pi.

import random

TIMES_TO_REPEAT = 10**5
LENGTH = 10**5

def in_circle(x, y):
    return x**2 + y**2 < LENGTH**2

inside_count = 0
for _ in range(TIMES_TO_REPEAT):
    point = random.randint(0,LENGTH), random.randint(0,LENGTH)
    if in_circle(*point):
        inside_count += 1

pi = (inside_count / TIMES_TO_REPEAT) * 4

print(pi) 

How can I apply the inequality condition in the code I have mentioned so the mass density would be same and I can find the value of volume in Higher dimension.?

Answer 1

import random

N = 10**5 # number of trials (ie, number of points to sample)
R = 10**5 # circle radius

def in_sphere(x, y, z):
    return x**2 + y**2 + z**2 < R**2

c = 0
for _ in range(N):
    p = random.randint(0,R), random.randint(0,R), random.randint(0,R)
    if in_sphere(*p):
        c += 1

pi = 6 * c / N
print(pi)

Answer 2

The basic idea: if a circle (radius R ) is inscribed inside a square (then, its edge must be 2R ), then the ratio (the area of circle over the area of square) will be π/4 . So, if you pick N points at random inside the square, approximately N * π/4 of those points should fall inside the circle.

hope the annotated/revised code help you understand the MC logic

import random

N = 10**5 # number of trials (ie, number of points to sample)
R = 10**5 # circle radius

# whether p(x,y) is inside a circle
def in_circle(x, y):
    return x**2 + y**2 < R**2

# use integer ops as much as possible for speed
c = 0
for i in range(N):
    x, y = random.randint(0,R), random.randint(0,R)
    if in_circle(x, y):
        c += 1

pi = 4 * c / N
print(pi) # pi-> 3.14