Buffon's needle simulation in python

Question

import numpy as np
import matplotlib.pylab as plt

class Buffon_needle_problem:

    def __init__(self,x,y,n,m):
        self.x = x #width of the needle
        self.y =         


        

Answer 1

Looks like you were committing a simple rounding error. The code below works, though the results are not very close to pi...

import numpy as np
import matplotlib.pylab as plt

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

def samples(self):
    # throwing the needles
    for i in range(self.n):
        self.r.append(np.random.uniform(0,self.y))
        self.z.append(np.random.uniform(0,self.x/2.0))
    return [self.r,self.z]

def simulation(self):
    #self.samples()
    # m simulations
    for j in range(self.m):
        self.r=[]
        self.z=[]
        for i in range(self.n):
            self.r.append(np.random.uniform(0,self.y))
            self.z.append(np.random.uniform(0,self.x/2.0))
        # n throws
        hits = 0 # setting the succes 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] <= 0.0:
                hits += 1
            else:
                continue
        hits = 2.0*(float(self.x)/self.y)*float(self.n)/float(hits)
        self.pi_approx.append(hits)

    return self.pi_approx

y = Buffon_needle_problem(1,2,40000,5)
print (y.simulation())

Also note that you were using the same sample for all simulations!