Question

Answer in Python please:

Compose a function mc_pi( n ) to estimate the value of π using the Buffon's Needle method. n describes the number of points to be used in the simulation. mc_pi should return its estimate of the value of π as a float. Your process should look like the following:

Prepare an array of coordinate pairs xy. This should be of shape ( n,2 ) selected from an appropriate distribution (see notes 1 and 2 below).

Calculate the number of coordinate pairs inside the circle's radius. (How would you do this mathematically? Can you do this in NumPy without a loop?—although a loop is okay.)

Calculate the ratio ncirclensquare=AcircleAsquare, which implies (following the development above), π≈4ncirclensquare.

Return this estimate of π.

You may find it edifying to try the following values of n, and compare each result to the value of math.pi: 10, 100, 1000, 1e4, 1e5, 1e6, 1e7, 1e8. How does the computational time vary? How about the accuracy of the estimate of π?

You will need to consider the following notes:

Which kind of distribution is most appropriate for randomly sampling the entire area? (Hint: if we could aim, it would be the normal distribution—but we shouldn't aim in this problem!)

Since numpy.random distributions accept sizes as arguments, you could use something like np.random.distribution( n,2 ) to generate coordinate pairs (in the range [0,1) which you'll then need to transform)—but use the right distribution! Given a distribution from [0,1), how would you transform it to encompass the range [−1,1)? (You can do this to the entire array at once since addition and multiplication are vectorized operations.)

Your submission should include a function mc_pi( n ).

Answer 1

`Hey,

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Note: Brother sometimes while uploading on HomeworkLib the indentations change. So, I request you to verify it with screenshot once. This is the link where I have saved the code too

https://trinket.io/python3/e0a2ee1763

import numpy as np

def sample_cosine():
rr=2.
while rr > 1.:
u1=np.random.uniform(0,1.)
u2=np.random.uniform(0,1.)
v1=2*u1-1.
rr=v1*v1+u2*u2
cc=(v1*v1-u2*u2)/rr
return cc

class mc_pi:
def __init__(self,x,y,n,m):
self.x = float(x) #width of the needle
self.y = float(y) #witdh of the space
self.r = [] #coordinated of the centre of the needle
self.z = [] #measure of the alignment of the needle
self.n = n #no of throws
self.m = m #no of simulations
self.p = self.x/self.y
self.pi_approx = []

def samples(self):
# throwing the needles
for i in range(self.n):
self.r.append(np.random.uniform(0,self.y))
C=sample_cosine()
self.z.append(C*self.x/2.)
return [self.r,self.z]

def simulation(self):
# m simulation
for j in range(self.m):
self.r=[]
self.z=[]
self.samples()
# n throw
hits = 0 #setting the success to 0
for i in range(self.n):
#condition for a hit
if self.r[i]+self.z[i]>=self.y or self.r[i]-self.z[i]<0.:
hits += 1
else:
continue
est =self.p*float(self.n)/float(hits)
self.pi_approx.append(est)
return self.pi_approx

y = mc_pi(1,2,80000,5)

print (y.simulation())

Note: Brother According to Chegg's policy we are only allowed to answer first part if there are many. So, I request you to post other part as separate posts

Kindly revert for any queries

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