More Buffon’s Needle

In my previous blog entry, I discussed Buffon’s needle (a method of measuring \pi using random numbers). I tried to recreate what would occur in physical reality without using trig functions in the measurement of \pi. The tricky part was trying to make a uniform probability distribution along the angle \theta, which is what you expect with physical needles being dropped. In this entry, I will do something more exact, but less physical.

In reality, the probability density in \theta is uniform due to there being no preferential direction for a needle to point. To simulate this with uniform random numbers without using trigonometric functions (to avoid just measuring python’s \pi) is tricky. The way I did this was assign uniform random numbers (uniform probability density) to a projection of a needle, either perpendicular or parallel to the lines (with a probability of 0.5 assigned to each case). This results in a probability distribution in \theta that isn’t quite uniform, but it’s close enough to approximate \pi after a large number of needles are dropped:
2 \frac{N_T}{N_C} = \pi
where N_T and N_C are the total number of needles dropped and number of needles crossing a line, respectively.

If instead of assigning a uniform probability to either projection fairly, what if we assign it to just one? This was shown in the previous post. As the number of needles dropped increases, the ratio 2 \frac{N_T}{N_C} approaches some number around 2.52.6, shown in the figure below (\pi shown in red). Why is this the case?

The reason why has to do with a uniform probability density along a projection (q) does not translate into a uniform probability density along \theta, see figure below. This means needles have a preferred direction. It turns out this can still be used to extract a measurement of \pi!
The uniform probability density is assigned to the projection q:
\rho_q = 1/L, for 0 \leq q \leq L, and \rho_q=0 elsewhere.
To get this in terms of \theta, some substitutions are made in the probability density integral:
\int_0^L \rho_q dq = 1
The variables q and \theta are related by:
\text{cos}(\theta) = q/L
so by taking the derivative we get the substitution:
dq = -L\text{sin}(\theta) d \theta

So now the probability density integral can be written as:
\int_0^{\pi/2} \rho_q L \text{sin}(\theta) d \theta
which is also:
= \int_0^{\pi/2} \rho_{\theta} d \theta
So it gives the form of the probability density for \theta in terms of the one for q:
\rho_{\theta} = \rho_q L\text{sin}(\theta)

Plug in \rho_q = 1/L to get:
\rho_{\theta} = \text{sin}(\theta)
and it is obviously NOT uniform, it’s just the sine function!
Plotting a histogram of the number of needles per angle \theta will give the shape of the sine function:

Following the same steps as in the previous entry, we can find out why when a uniform probability is assigned to one projection, 2 N_T/N_C approaches 2.5 instead of \pi.
For simplicity W=L, and again for a needle to be crossing a line it has to be below the upper bound:
x \leq \frac{L}{2}\text{sin}(\theta)
and the probability of a needle crossing a line is:
P = \int_0^{\frac{L}{2}\text{sin}(\theta)} \int_0^{\pi/2} \rho_x \rho_{\theta} d \theta d x
Plugging in the probability densities: \rho_x = 2/L and \rho_\theta = \text{sin}(\theta):
P = \int_0^{\pi/2} \text{sin}^2(\theta) d \theta = \pi/4

P = \frac{N_C}{N_T} when the number of needles dropped is large, so by combining these equations:
2 \frac{N_T}{N_C} = \frac{8}{\pi} = 2.52
This is the number that is being approached!
But wait, \pi is there, so we can extract it by:
4 \frac{N_C}{N_T} = \pi

By plotting this ratio as a function of number of needles dropped, we see it approaching \pi. See the figure below.

Even though this isn’t how needles behave in reality, it’s a simpler way to measure \pi by using random numbers in a simulation. The code used to make these plots is shown below.

By the last drop:
2 \frac{N_T}{N_C} = 2.54
4 \frac{N_C}{N_T} = 3.14

import os
import sys
import math
import numpy
import pylab
import random

CL = 0
NN = 0

PIa = []
PIb = []
NNa = []
PIc = []

for ii in range(100000):

    PPy = random.uniform(0.0,1.0) #A-end y position (lines separated by unit distance)

    RRx = random.uniform(0.0,1.0) #x projection
    RRy = numpy.sqrt(1.0 - RRx**2) #y projection

    QQy = PPy + RRy #B-end y position
    QQyf = math.floor(QQy)
    PPyf = math.floor(PPy)

    NN = NN + 1
    if (QQyf-PPyf > 0.0): #if crossing line
        CL = CL + 1

    RAT = (1.0*CL)/(1.0*NN)
    if (CL==0):



print PIa[-1]
print PIb[-1]

pylab.ylabel("2 NT/NC")

pylab.hist(ANGLES, bins=10)
pylab.xlabel("Acute angle in factors of pi")
pylab.ylabel("Needles dropped")

pylab.ylabel("4 NC/NT")

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