Fractals Graphics Math Programming Python

Visualizing Data Using the Hilbert Curve

Some time ago, a coworker asked me to help him visualize some data. He had a very long series (many millions) of data points, and he thought that plotting a pixel for each one would visualize it well, so he asked for my help.
I installed Python & PIL on his machine, and not too long after, he had the image plotted. The script looked something like:

data_points = get_data_points()
n =  int((len(data_points)**0.5) + 0.5)
image = Image('1', (n, n))
for idx, pt in enumerate(data_points):
	image.putpixel(pt, (idx/n, idx%n))'bla.png', 'png')

Easy enough to do. Well, easy enough if you have enough memory to handle very large data sets. Luckily enough, we had just enough memory for this script & data series, and we were happy. The image was generated, and everything worked fine.

Still, we wanted to improve on that. One problem with this visualization is that two horizontally adjacent pixels don’t have anything to do with each other. Remembering xkcd’s “Map of the Internet“, I decided to use the Hilbert Curve. I started with wikipedia’s version of the code for the Python turtle and changed it to generate a string of instructions of where to put pixels. On the way I improved the time complexity by changing it to have only two recursion calls instead of four. (It can probably be taken down to one by the way, I leave that as a challenge to the reader :)

Unfortunately, at this point we didn’t have enough memory to hold all of those instructions, so I changed it into a generator. Now it was too slow. I cached the lower levels of the recursion, and now it worked in reasonable time (about 3-5 minutes), with reasonable memory requirements (no OutOfMemory exceptions). Of course, I’m skipping a bit of exceptions and debugging along the way. Still, it was relatively straightforward.

Writing the generator wasn’t enough – there were still pixels to draw! It took a few more minutes to write a simple “turtle”, that walks the generated hilbert curve.
Now, we were ready:

hilbert = Hilbert(int(math.log(len(data_points), 4) + 0.5))
for pt in data_points:
    x,y =
    image.putpixel(pt, (x,y))

A few minutes later, the image was generated.

computer science Fractals Python

Fractals in 10 minutes No. 6: Turtle Snowflake

I didn’t write this one, but I found it’s simplicity and availability so compelling, I couldn’t just not write about it.
In a reddit post from a while ago, some commenter named jfedor left the following comment:

A little known fact is that you can do the following on any standard Python installation:

from turtle import *
def f(length, depth):
   if depth == 0:
     f(length/3, depth-1)
     f(length/3, depth-1)
     f(length/3, depth-1)
     f(length/3, depth-1)
f(500, 4)

If you copy paste, it’s a fractal in less than a minute. If you type it yourself, it’s still less than 10. And it’s something you can show a kid. I really liked this one.

Fractals Graphics Programming Python

A second look at the dragon fractal

At first when I drew the (twin)dragon fractal, I had a small bug. I used the base 1+i instead of 1-i. This also generated a very similar looking beast. Thinking about that for a while, made me curious. Just like the Mandelbrot set, maybe other interesting fractals could be generated using exactly the same method, with a different complex ‘seed’.

So I patched up my code, and made it output a series of images for complex numbers on a path. I thought the reasonable choice would be a spiral, so using t*(cost+isint) as a base I wrote a spiral that would go around the origin several times and finally land on 1-i.
It might seem obvious to you to try and make this interactive instead – why not move the mouse around and watch the different fractals that emerge? Well, I wanted a video.

I also wanted the fractal to convey a little more information. So each point in the set was given a color according to its generation. I decided after some trial and error that white was better for the older generation, and red for younger generations.
After some PIL and ffmpeg work, it was ready.

While working on the video, I witnessed some very interesting fractals with different seeds.
I was very curious as to why certain shapes emerged. For example, a square pattern was very common:
Squares fractal
This turned out to be not that surprising. Its generator number is of the shape 0+ia. So it made sense. I still didn’t figure out how come hexagon shapes were so common:
Hexagons1 Hexagons2 Hexagons1

Maybe it had something to do with pi/6, I’m not too sure about that. If it did, I would expect to see many more regular polygons.
Here is another curious one:

Another interesting phenomenon was what I started to call ‘the dragon’s heart’. You see, in the final dragon, the starting generations were spread about pretty evenly. However, with other bases, even ones generating something which is pretty similar to the dragon, the starting generations are clustered together – sometimes at the side, sometimes at the middle.

I’ve got a feeling (not proven or anything) that to be space filling, a fractal’s starting generations should be spread about. What do you think?

Fractals Graphics Programming Python

Fractals in 10 minutes No. 2 – Recursive Spring

Imagine a straight wire. Bend this wire until its ends meet. You get a ring. Next stage.
Take another straight wire, bend it as before, but this time, don’t let the ends meet, instead continue bending the wire more and more, until you get a spring. Now, Think of this spring as the first wire, and bend it until the spring’s ends meet. We get a spring-ring. (See pictures below)

At this point it should be obvious – instead of making the ends of the spring meet, we can continue bending the spring into a meta-spring, and again make this meta-spring’s ends meet. Rinse and repeat ad-infinitum.

At first I had a hard time picturing it, and I thought it would be tough to write about it without drawing it. So I set with pen and paper, and tried to draw it. That was even harder than picturing it. So I thought I could write a script to draw it, and that turned out to be pretty easy.

Here are the first four stages. The second and third are shown from the side. Click for bigger versions.

To implement this, I wrote a function that draws any track. A track is defined to be a function takes a float argument between 0 and 1, and returns the appropriate 3d point in between, where 0 is one end of the track and 1 is the other. (Actually, I also decided that tracks should return a “right hand” vector as well). Then I wrote a ring:

def ring(t):
    pi = math.pi
    cos = math.cos
    sin = math.sin
    x = numpy.array((2*cos(2*pi*t),2*sin(2*pi*t),0))
    return (x,-x/3)

Lastly I wrote a wrapper function, that takes any track, and springifies it. It was called springify_track. At this point I could do this:
(The constants are the radius of the spring loops, the sampling delta, and the number loops.)

t1 = ring
t2 = springify_track(t1, 0.25, 0.1, 50)
t3 = springify_track(t2, 0.06, 0.0011, 5000)
t4 = springify_track(t3, 0.011, 0.000012, 500000)
Algorithms Fractals Math

Riddle of the Week – LStrings

So I wrote about lstrings. I intend to write about them again in a short while – I already finished the basic script a few days ago, but I’m waiting, until I will be satisfied with it.

In the meantime, here is a curious riddle (That I came up with):

Version 1 (not final)

Assume you are given some iteration of an lstring – S. How do you discover the original lstring used to create S?


Well, those who are quick to answer will first try to discover the length of the original string – it could be computed directly from S. At this point you should come to the conclusion that without being given the exact number of iterations – the riddle is not that interesting… Iterating on an lstring is an associative operation. If we denote the substitution of a string A into a string B as multiplication, when both are iterations on an original string T , we will get:

A*B =T^n * T^k = T^(n+k) = T^k* T^n

This will result in the obvious answer to the original riddle – “Why, the lstring S is the original string, with zero iterations!”. This answer is correct – and obviously useless :)

So, let us rephrase the riddle:

Version 2 (final)

Assume you are given some iteration of an lstring – S. How do you discover the minimum possible lstring that could be used to create S?

Discussion – But not a solution

This is the proper riddle – have at it! I will be glad to read your thoughts…

note: The acceptable solution should be either an algorithm, or a constructive mathematical proof. I don’t like reading only statements of existence :)

Fractals Programming Python

Fractals in 10 minutes No. 1 – LStrings

Lately it started to be quite fun to find some fun mathematical formula, and just code it in python, and plot the result. Fractals are such fun. LStrings is an example of that. For those who don’t know – lstrings are simple strings with three letters, ‘f’ for forward, ‘l’ for left, ‘r’ for right. An iteration of an lstring is replacing all occurrences of ‘f’ with the original string.

During high-school I recall long lessons when I would draw complicated iterations of some simple lstring just to see the result. Sometimes I still draw them. A few days ago I just wanted to see how long it would take me to program one. It actually was quite short. What really took me time is blogging about it ;)

The source and the results are available here.

This post is going to be the first in a series of posts about some fractals. I will continue to write other examples of simple code that (hopefully) generates beautiful results. Stay tuned for more.

Do you know any elegant\short\simple pieces of code that generate fractals?