Tuesday, 27 November 2018

Tower of Hanoi



The Tower of Hanoi is a classic mathematical puzzle consisting of three pegs and disks of different sizes. The aim of the puzzle is to move the stack of increasing size disks from one peg to the other, by moving only the upper disk of any peg at a time. On each movement, no larger disk may be placed above another smaller disk.

The problem was first described by the French mathematician Édouard Lucas in 1883 and it is a  problem where recursion can be easily used to solve the game. Along with the Fibonacci series, it is the classic exercise of a 101 programming course where students are asked to apply recursion.

Recently I built three wooden play sets of Tower of Hanoi and thought it would be interesting to increase my knowledge on the problem.






As a first approach of the problem, I wrote the classical recursive solution, which I post here. I will tackle the problem from a different perspective, probably graph theory, in the coming posts. 

The Python code can be found on my github repository. I have splitted the solution in three different Python classes, Rectangle, which deals with the graphic representation of the objects using Pyglet library, Disk, which represents a disk in the Tower of Hanoi problem, keeping track of the position of itself at any moment, and finally the Hanoi class, which solves the problem using the recursive algorithm and show a graphical representation of the step by step solution.

Following the instructions of the README.md file at the git hub repository, you can get the following solution video.



Stay tuned for upcoming post on the subject with different solution approaches.


Sunday, 28 October 2018

Human Hearing Frequency Range Test


The other day I came across an article on human hearing which sparked my curiosity on what my audible frequency range was.

Human ear can usually detect sounds ranging from frequencies from 20Hz to 20KHz. However, as we grow up, the auditory organ loses capabilities and the dynamic range gets narrower. I wondered what my current dynamic range was, and I wrote this simple Python script to test it.

The script makes use of PyAudio, a Python Package which gives us access to the sound card. The code saves a number of num_samples of sound samples into a NumPy matrix. Each sample is generated from a sinusoidal signal of a different frequency, specified in the frequencies list. 

Once the different waves have been saved, each frequency in a row of the matrix waves, we play the sounds by writing into the stream object created from PyAudio module.

The script starts by playing a sinusoidal sound of 5Hz, and increases step by step the frequency of the samples. Each frequency is played for 5secs. In my case, I start hearing sounds after the 20Hz tone. As the frequency increases beyond the 20Hz, even if the intensity is kept constant in the script, my ear response to the different frequencies is not the same, giving us the illusion of a different intensity for each tone. After the 14KHz barrier, I was unable to hear any signal.

If you have pets around, try to test this scripts by increasing the frequency beyond the 22KHz. Observe however that if you increase the frequency beyond 22KHz, you should increase as well the sample rate, to satisfy the requirements imposed by Nyquist sampling theorem.



Saturday, 23 July 2016

Generating QR Code Personal Business Cards

It's been a long time since my last post, time flies, and I am embarked in a very important project, one in my life Master Plan, eheheh, that is preventing me from publishing posts in both blogs.

These days I have had the need to print some personal cards and I thought It would be a good idea to embed a QR code with my vCard, so that other people I meet at conferences and events can easily add me to their contact list with all information I am willing to share with them. Also, it might be interesting to add in a techy resume.

It was interesting to read a bit about how QR were invented, and the different formats and versions there exists. Reading about its encoding and error correction transported me to my days of Electrical Engineering student.

On the other hand, vCard format, described on RFC6350, is a standard format to share contact details commonly used by phones and internet applications. Like many internet formats, it is human readable, so it is easy to craft one's vCard with the fields you want to include. Skimming RFC6350 you might get surprised how many interesting fields you can include in your vCard, from your private encryption keys to an URL pointing to the most updated vCard of you, in case you change contact details such as address or phones frequently. Not sure how many contacts application make use of the last one.

To generate a QR-Code I have used qrcode Python package, which enable to create a QRcode image in just three lines of code. Since it is pip compliant, one can easily install the package with the following command:

>sudo -H pip install qrcode 

Once we have qrcode up and running on our python distribution, we can use the code below, included in my github repository to generate the QR code. The vCard file is downloaded from my website, you can easily craft yours with a text editor and then feed this text to the qr.add_data() function.

As usual you can get the code at my github profile: https://github.com/kokomero/miscelanea/tree/master/vCard_QRcode

The result is a QR-code that can be inserted in your business cards or in your resume. The more information you add to your vCard, the more characters the QRcode needs to encode and thus, the bigger it gets. Hence, I strongly encourage to add only the important information and add a field such as your personal website where more information can be found if needed. Another option is encoding instead an URL pointing to a more complete vCard stored in your website.



Wednesday, 30 December 2015

Buffon Needle Problem



Buffon Needle problem is a classical mathematical problem first suggested on 18th century by the French naturalist Georges-Louis Leclerc, Comte de Buffon. Although it was not the original aim of Buffon, the problem turned out to be a method to estimate the number π.

The problem is as follows: we have a set of evenly distanced parallel bars on a board. If we drop a needle on the board, what is the probability that the needle will touch one of the bars?

Assume the distance between the bars is t, and the length of the needle is l. Assume also that t is greater than l. If that is not the case, the problem can still be easily solved but the solution is different.

This problem is supposed to the very first problem of geometrical probability, and can be easily solved using simple integral calculus. At the same time, it turns out that the ubiquitous number π is in the solution of the problem, so, it can be equally used as a Monte-Carlo estimation of this number. Although it is not the most efficient way to estimate the digits of π, we will solve the problem and write a Python program to check it, just for the fun of it.

Assume $x$ is the distance from the middle point of the needle to the closest bar, then assume, between any two bars, $x \in [0, \frac{t}{2}]$. Assume also that $\theta$ is the angle of the needle with the bar. Then, $x, \theta$ will be uniform random variables for each of the random needle throw. The p.d.f of each variable will be $f_X(x) = \frac{2}{t} = , f_{\Theta}(\theta) = \frac{2}{\pi}$. Since both variables will be independent, the joint distribution density probability function will be just the product of the two, i.e. $f_{X,\Theta}(x, \theta) = \frac{4}{\pi t}$.

Then, once we have drawn the $x, \theta$ random variables at each trial, the needle will touch the bar if the condition $x < \frac{l}{2}sin ( \theta )$.

Thus, the probability that the needle touches a bar will be $P = \int_{\theta = 0}^{\frac{\pi}{2}} \int_{x=0}^{\frac{l}{2}sin(\theta)}  f_{X, \Theta}(x, \theta) dx d \theta$. With a little of calculus, we get $  P = \frac{4}{\pi t} \int_{\theta = 0}^{\frac{\pi}{2}} \frac{l}{2}sin(\theta) d \theta = \frac{2 l}{\pi t}$.

So, the probability of the needle touching one bar is surprisingly related to number $\pi$ through the expression $P = \frac{2 l}{\pi t}$.

Assume that we throw indeed $n$ needles in a simulation, then, let $p$ be the number of trials in which the needle will touch one of the bar. An estimate of the probability above, will be $\frac{p}{n}$ , i.e. $\frac{p}{n} \approx \frac{2 l}{\pi t}$.

Now we have a relation between number $\pi$ and a probabilistic simulation, we can write a snippet of Python code to run simulations and see how good the estimates of number $\pi$ are using this method.

The Python code is included at the end of this blog post. The code will simulate $m$ times the Buffon problem, each problem throwing $n$ needles in a three bars board. The image resulting from the Python code gives us an idea of the needles touching the bars, drawn in red, and the ones not touching any bar, drawn in green.


Example of simulation with n = 100, command line argument:
python BuffonNeedle.py -n 150 -m 1


Executing the program with the parameter $m = 1000$, we get the mean and standard deviation of this estimator for number $\pi$.

n Mean Std
100 3.21196453418 0.502342987432
1000 3.16177668689 0.152438640479
10000 3.15099821931 0.0462329285661

As for the result, we see that it might not be the most efficient estimator for number $\pi$, anyway, we had a lot of fun thinking and solving the problem, which was the point of all this.

The code, as usual, can be found on my GitHub page.

References:


Sunday, 20 September 2015

Math Art Work with just segments and circles


This post was inspired by the article in the CNN in the following link.

Using a big number of segments or circles we can create nice art work like the images below. When a number of segments under a certain parametrization intersect at some points, the set of intersection points define a curve on the plane called the evolute.

For example, imagine a fixed length bar which stands vertical on the Y-axis. Then, we take the extreme of the bar at (0,0) and move it through the X-axis, while letting the other extreme descend through the Y-axis. The intersection of all this segment will define a curve, and the segments are tangent to the curve at each of the intersection points. Many curves such as parabola, hyperbola can be constructed in this way.


python mathart.py -c black -C black -n 100 -o "artwork10.png" -e line -X1 "0.0" -Y1 "sin(pi/2.0*k/n)*(-1.0)" -X2 "cos(pi/2.0*k/n)" -Y2 "0.0"

Using more complex parametrizations as the one described in the origina article and in the script.sh file, and letting our imagination fly, we can come up with nice geometrical figures.

When I first tackled this problem with the very first lines of code at the early stage of the program, I observed that the images I obtained where fuzzy. If I increased the number of elements to get more defined curves, what I got was large areas of pixels of the same color.  This is due to the fact that computers image are a set of discrete points instead of a continuum of points. When drawing many elements the non-integer coordenates will be plotted at the nearest pixel, creating an overlapping, as we can appreciate on the picture below. This effect is known in signal processing as the Aliasing problem.



Example of figure without using anti-aliasing technique.

The solution to this problem is obviously to increase the resolution of the image.  In order to increase the resolution while preserving the same image size, what we do is to create a first image larger than our final image. All calculations are done in this larger image, larger by a factor specified on the command line of the program as super-sampling. Once we have this larger image, we resize or down-size the image to the size we originally wanted. In order to down-size the image, we can choose amongst several algorithms to apply: Nearest, Bilinear, Bicubic, Antialias. In this case we use the Antialias filtering which is the one resulting in higher quality images in this case.

Using this super-sampling technique, we come up with much better quality pictures. The pictures below have been created with the Python script at the end of this post. The legend of each pictures include the command-line used to create the picture. All these examples are under the script.sh file on the same directory. 

You can get the code from my github repository at the following link.



python mathart.py -c red -C blue -n 25000 -o "artwork1.png" -e line -X1 "sin(108.0*pi*k/n)*sin(4.0*pi*k/n)" -Y1 "cos(106.0*pi*k/n)*sin(4.0*pi*k/n)" -X2 "sin(104.0*pi*k/n)*sin(4.0*pi*k/n)" -Y2 "cos(102.0*pi*k/n)*sin(4.0*pi*k/n)"

python mathart.py -c black -C blue -n 25000 -o "artwork2.png" -e line -X1 "3.0/4.0*cos(86.0*pi*k/n)" -Y1 "sin(84.0*pi*k/n)**5" -X2 "sin(82.0*pi*k/n)**5" -Y2 "3.0/4.0*cos(80.0*pi*k/n)"

python mathart.py -c red -C blue -n 25000 -o "artwork3.png" -e circle -Xc "sin(14.0*pi*k/n)" -Yc "cos(26.0*pi*k/n)**3" -R "1.0/4.0*cos(40.0*pi*k/n)**2"

python mathart.py -c red -C blue -n 25000 -o "artwork4.png" -e circle -Xc "cos(6.0*pi*k/n)" -Yc "sin(20.0*pi*k/n)**3" -R "1.0/4.0*cos(42.0*pi*k/n)**2"

python mathart.py -c red -C blue -n 25000 -o "artwork5.png" -e circle -Xc "cos(6.0*pi*k/n)" -Yc "sin(14.0*pi*k/n)**3" -R "1.0/4.0*cos(66.0*pi*k/n)**2"

python mathart.py -c black -C blue -n 25000 -o "artwork6.png" -e circle -Xc "cos(14.0*pi*k/n)**3" -Yc "sin(24.0*pi*k/n)**3" -R "1.0/3.0*cos(44.0*pi*k/n)**4"

python mathart.py -c red -C green -n 25000 -o "artwork7.png" -e circle -Xc "sin(22.0*pi*k/n)**3" -Yc "cos(6.0*pi*k/n)" -R "1.0/5.0*cos(58.0*pi*k/n)**2"

python mathart.py -c yellow -C blue -n 25000 -o "artwork8.png" -e circle -Xc "cos(2.0*pi*k/n)" -Yc "sin(18.0*pi*k/n)**3" -R "1.0/4.0*cos(42.0*pi*k/n)**2"

The Python code is include below.

Wednesday, 16 September 2015

Diffusion Limited Aggregation and Brownian Trees

Diffusion limited aggregation (DLA) is a type of stochastic process in which particles move following a Brownian motion until they stick together with other particles.

Imagine a square in which we have put some glue on the bottom edge. Then, we release one particle at a time on that square. The particles follow a random-walk within the square until they stick to the bottom edge or to a particle already fixed or rooted. Then, that particle becomes itself attached to the fixed structure. This fixed structure so formed is commonly called Brownian trees, because it recalls the shape of a tree.

Many natural phenomena, such as electrodeposition, dielectric breakdown (lightings), coral growth, snowflake formation and other not so natural such as urban areas growth can be simulated using DLA. To the basic diffusion process, we can add other physical features such as gravitation, external forces, attraction between particles, fluids, different frames where particles can be attached... to achieve different results.

The snippet of code below is an example of a simple two dimensional DLA simulation written in Python.




The following images were created using the script above, for different type of frames.






On the bit-player.org blog, we can find interesting examples of such processes. There are even several programming libraries to simulate such processes. As usual, adding a litte bit of imagination we can easily hack the script above to generate new wonderful images.



Thursday, 10 September 2015

Benford Law in Stock Market Index



The Benford's Law is an empirical law about the probability distribution of the first digit of each number. In everyday datasets, especially ones with values with different order of magnitude,  such as bill, number in addresses, populations, stock market prices and so on, the probability of the first digit is not uniform as one could think. Instead, the probability of the first digit d is proportional to the space between d and d+1 in logarithmic space. Thus, the probability of the digit d, being d a digit in [1,9] is a decreasing function which can be calcualted as:

For a more detailed description of the Law, I strongly recommend to visit the wikipedia entry for the Benford's Law.

Being a little curious about this law, I wrote a short R script to check this law against the prices of stocks in the different stock market indexes, such as S&P 500, DJ EUROSTOXX50E, FTSE 100, IBEX 35, RUSSELL 2000 and NIKKEI 225. In order to run this script, you will need a computer with a Bloomberg connection. I guess it can be easily hacked so that we can get the stock prices from the Yahoo! finance website, but I haven't tried.


library(Rbbg) #Bloomberg Connection
 
#Connect to Bloomberg API
conn <- blpConnect()
 
indexes <- c("RTY Index", "SX5E Index", "IBEX Index", "SPX Index", "NKY Index", "UKX Index")
 
#Get members for index and their RICS
memb_codes <- bds(conn, indexes[6], c("INDX_MEMBERS") )
rics <- paste(memb_codes[ ,1], "Equity", sep= " ")
 
#Get spots for each member and their first digit
spots <- bdp(conn, rics, c("PX_LAST") )
firstDigit <- as.numeric( substring( spots[ , 1], 1, 1 ) )
 
#Get histogram for first Digit
histogram <- hist( firstDigit, breaks = 0:9 )
 
#Benford Law Theoretical Probability
benfordTheoretical <- log10( 1 + 1/(1:9) )
 
#Result dataframe
results_UKX <- data.frame(FirstDigit = 1:9, Probability = histogram$density, Theoretical = benfordTheoretical)
 
 
#For those not having a bloomberg connection, we can load the data
load("BenfordLaw.RData")
 
#Plot results
plot(results_UKX$Probability, main='Benford Law for stock prices in FTSE100 Index', 
     xlab="First Digit", ylab="Probability", col="green", pch=1, xaxt="n")
axis(1, at=seq(1,9))
lines(results_UKX$Theoretical, col="red", pch=0)
legend("topright", c("Results", "Theoretical"), lty=c(0,1), pch=c(1,0), col=c("green", "red"))



I attach the results for each of the index to see how well the distribution matches the theoretical Benford distribution.