# -*- coding: utf-8 -*-
"""
volume of an n-sphere, relative to an n-cube
@author: neil 2022
"""

import numpy as np

print('numerical integration to achieve 2 figure accuracy for an 8-sphere requires >> 10^17 calculations')

# Analytic soln
print('Analytic solution of volume on an 8 dimensional sphere: {:.5f}'.format( 4.057*(.5)**8 ))
  
# Our approach to MC integration is to throw darts (random guesses)
# at an n-sphere inscribed in a n-cube and calculate the number of hits vs the total throws, 
# which will approx the ratio of the volumes.
# For this we use a radius of 0.5, that way we know easily the volume of the n-cube (=1!)
# Define number of samples
N = 100000
d = 8  # define number of dimensions
x = np.random.rand(N,d)-.5  # a 'pythonic' method to make an N by d array of uniform random numbers between -.5 and +.5

hits=0
for j in np.arange(N):
    norm=0
    for s in range(d):
        norm+= x[j,s]**2         # calc the square of the pythagorean distance from the origin in n-dimensions
    # np.linalg.norm(x,axis=1) replaces much of the above loop, returns the sum of the squares of a row of the x=array
    if np.sqrt(norm) <= .5:      # if the distance is less than the radius, we have a HIT!
        hits += 1
     
# after the above we have the sum of chosen points in n-space that were inside the n-sphere
# Calculate integral by taking the ratio of points in and out to the area of the unit square
unit_vol  = 1**d                 # volume of unit cube in all dimensions is 1!
integral = (hits / N)
print("Monte Carlo Integral estimate of 8-sphere volume:       {:.5f}".format(integral/unit_vol))