# Pinhole Photographs MMXVII

2017-05-06, post № 169

art, #light, #nature, #photography, #picture, #tulip, #World-Wide Pinhole Day, #WWPD

# Multibrot Set

2017-04-22, post № 168

Java, mathematics, programming, #animated gif, #animation, #Cartesian, #complex, #complex arithmetic, #fractal, #generalization, #gif, #Mandelbrot set, #multi-threading, #polar, #reals, #threading

The Mandelbrot Set is typically defined as the set of all numbers $c\in\mathbb{C}$ for which — with $z_0=0$, $z_{n+1}=f_c(z_n)$ and $f_c(z)=z^2+c$ — the limit $\lim\limits_{n\to\infty}z_n$ converges. Visualizations of this standard Mandelbrot Set can be seen in three of my posts (Mandelbrot Set, Mandelbrot Set Miscalculations and Mandelbrot Set II).

However, one can extend the fractal’s definition beyond only having the exponent 𝟤 in the function to be $f_c(z)=z^\text{exp}+c$ with $\text{exp}\in\mathbb{R}$ . The third post I mentioned actually has some generalization as it allows for $\text{exp}\in\{2,3,4,5\}$, although the approach used cannot be extended to real or even rational numbers.

The method I used in the aforementioned post consists of manually expanding $(a+b\cdot i)^n$ for each 𝑛. The polynomial $(a+b\cdot i)^3$, for example, would be expanded to $(a^3-3\cdot a\cdot b^2)+(3\cdot a^2\cdot b-b^3)\cdot i$.
This method is not only tedious, error-prone and has to be done for every exponent (of which there are many), it also only works for whole-number exponents. To visualize real Multibrots, I had to come up with an algorithm for complex number exponentiation.

Luckily enough, there are two main ways to represent a complex number, Cartesian form $z=a+b\cdot i$ and polar form $z=k\cdot e^{\alpha\cdot i}$. Converting from Cartesian to polar form is simply done by finding the number’s vector’s magnitude $k=\sqrt{a^2+b^2}$ and its angle to the 𝑥-axis $\alpha=\mbox{atan2}(\frac{a}{b})$. (The function $\mbox{atan2}$ is used in favor of $\arctan$ to avoid having to divide by zero. View this Wikipedia article for more on the function and its definition.)
Once having converted the number to polar form, exponentiation becomes easy, as With the exponentiated $z^\text{exp}$ in polar form, it can be converted back in Cartesian form with Using this method, converting the complex number to perform exponentiation, I wrote a Java program which visualizes the Multibrot for a given range of exponents and a number of frames.
Additionally, I added a new strategy for coloring the Multibrot Set, which consists of choosing a few anchor colors and then linearly interpolating the red, green and blue values. The resulting images have a reproducible (in contrast to randomly choosing colors) and more interesting (in contrast to only varying brightness) look.

The family of Multibrot Sets can also be visualized as an animation, showing the fractal with an increasing exponent. The animated gif shown below was created using ImageMagick’s convert -delay <ms> *.png multibrot.gif command to stitch together the various .png files the Java application creates. To speed up the rendering, a separate thread is created for each frame, often resulting in 𝟣𝟢𝟢％ CPU-usage. (Be aware of this should you render your own Multibrot Sets!)

To use the program on your own, either copy the source code listed below or download the .java file. The sections to change parameters or the color palette are clearly highlighted using block comments (simply search for /*).
To compile and execute the Java application, run (on Linux or MacOS) the command javac multibrot.java; java -Xmx4096m multibrot in the source code’s directory (-Xmx4096m tag optional, though for many frames at high quality it may be necessary as it allows Java  to use more memory).
If you are a sole Windows user, I recommend installing the Windows 10 Bash Shell.

Source code: multibrot.java

# Easter MMXVII

2017-04-16, post № 167

art, ascii, haiku, poetry, #ascii egg, #celebration, #easter egg, #egg

Winds swirl through the air,
Water sloshes at the shore;
A peaceful island.

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A fairly well-known way to approximate 𝜋 is to randomly choose points in a square (often thought of as throwing darts at a square piece of cardboard), determine their distance to a circle’s center and do a division, as I did in my 𝜋 Generator post.

However, 𝜋 does not only appear in the formula for a circle’s area, $A=\pi\,r^2$, yet  also in the formula for a sphere’s volume, $V=\frac{4}{3}\,\pi\,r^3$, and for all the infinite hyperspheres above dimension three (view this Wikipedia article for more about volumes of higher-dimensional spheres).

In particular, the formula for the hypervolume of a hypersphere in four dimensions is defined as being $V=\frac{\pi^2}{2} \cdot r^4$. Using this formula, my Python script randomly chooses four-dimensional points (each in the interval $\left[0,1\right)$), calculates their distance to the point $\left(0.5,0.5,0.5,0.5\right)$ and determines if they are in the hypersphere around that point with radius 𝟢.𝟧.
By dividing the number of random points which lie in the hypersphere by the number of iterations used ( $10^6$ in the example below), the script approximates the hypersphere’s hypervolume. By then rearranging the equation $V=\frac{\pi^2}{2}\cdot r^4$ with 𝑟 = 𝟢.𝟧 to $\pi=\sqrt{V\cdot 32}$, the desired constant can be approximated.

\$ python pi.py
3.14196371717
Source code: pi-day-mmxvii.py

# A278328

2017-03-11, post № 162

mathematics, programming, Python, #2016, #decimal reverse, #difference, #integer, #OEIS, #On-Line Encyclopedia of Integer Sequences, #palindrome, #reverse, #sequence, #square

The On-Line Encyclopedia of Integer Sequences (also known by its acronym, OEIS) is a database hosting hundreds of thousands of — as the name implies — integer sequences. Yet, despite the massive number of entries, I contributed a new integer sequence, A278328.

A278328 describes numbers whose absolute difference to their decimal reverse are square. An example would be 𝟣𝟤 or 𝟤𝟣 (both are the decimal reverse to each other), since $\left|12-21\right|=9$ and $9=3^2$.

Not a whole lot is known about the sequence , partly due to its definition only resulting in the sequence when using the decimal system, though it is known that there are infinitely many numbers with said property. Since there are infinitely many palindromes (numbers whose reverse is the number itself), $\left|n-n\right|=0$ and $0=0^2$.

Due to there — to my knowledge — not being a direct formula for those numbers, I wrote a Python script to generate them. On the sequence’s page, I posted a program which endlessly spews them out, though I later wrote a Python two-liner, which only calculates those members of the sequence in the range from 𝟢 to 𝟫𝟪 (shown below entered in a Python shell).

>>> import math
>>> filter(lambda n:math.sqrt(abs(n-int(str(n)[::-1])))%1 == 0, range(99))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 21, 22, 23, 26, 32, 33, 34, 37, 40, 43, 44, 45, 48, 51, 54, 55, 56, 59, 62, 65, 66, 67, 73, 76, 77, 78, 84, 87, 88, 89, 90, 95, 98]`
Jonathan Frech's blog; built 2021/10/02 17:36:09 CEST