I'm a programmer that stumbled across an interesting fractal. Can anyone shed some light on what I've done?

GoldyOrNugget · 3 months

Neeaaaaat. I just wrote it recursively and although it doesn't have the same transparency effect, the general black/white regions are preserved. Here's a side-by-side comparison of traditional sierpinski and this: http://i.imgur.com/2FYhL.png (colours inverted for consistency).

CODE:

from turtle import *

ratio = 1 / 1.6180339887
up()
goto(-800,-400)
down()

tracer(1000)

def sierpinski(length):
    if length >= 3:
        sierpinski(length * ratio)

        left(60)
        forward(length * ratio)
        right(60)
        sierpinski(length * ratio)

        right(60)
        forward(length * ratio)
        left(60)
        sierpinski(length * ratio)

        backward(length * ratio)


sierpinski(500)

ratio = 0.5
up()
goto(0, -400)
down()

sierpinski(800)

raw_input()

harrisonbeaker · 3 months

What you've done is built an iterated function system and let it run. It's a nice and relatively easy fact to prove that as long as each individual function is a contraction (it brings points closer together), the system will result in an attractor with some self-similarity.

Many times this will end up ugly or boring, but occasionally you can hit on some nice patterns.

There isn't such a thing as an 'established' fractal. There are the handful of famous ones, but then there are also the thousands of other beautiful images produced by hordes of comp sci and math students every semester.

To get a better intuitive idea of why yours looks like that, note that a chaos game is really just a way to approximate a standard iterated function system. Your system has 3 functions:

f1(x,y) = (1/phi * x, 1/phi * y)

f2(x,y) = (1/phi * x + (1-1/phi), 1/phi * y)

f3(x,y) = (1/phi * x + (1/2 * 1/phi), 1/phi * y +sqrt(3) * 1/phi)

Now start with an equilateral triangle T0 with one corner at the origin and side lengths one. Each of these functions shrinks the triangle (by a factor of phi in your case) and shifts it to re-approximate the original triangle. We say that F(T0) = T1 is the union of each of these three 'images' of T0. Now we do the same thing, using T1 as our starting point.

Since these triangles overlap you end up with your transparency and if you play through the first few iterates you start to see your pattern develop. Note that if you replace phi by 2, you get the sierpinski triangle.

EDIT: I think I messed up the functions a bit. Rather than do all the algebra to correct them, it's easier and more informative to describe what they do. f1 shrinks the entire triangle by the desired factor (in this case 1/phi) and leaves the bottom left corner on the origin. f2 shrinks, then scoots small triangle over so that the right corner is the point (1,0). And finally f3 shrinks and scoots the small triangle up and over so the top is at (1/2,1). The reason for these translations is to keep the iterations bounded inside the original triangle. Since you're using a factor larger than 1/2, there will be plenty of overlapping.

coveritwithgas · 3 months

What did 1/3 and 1/5 yield? Without them, I'm on the fence as to whether this is something or not.

aurath · 3 months

Wow! Thanks for the feedback guys, interesting stuff. I thought I'd post the processing code as it currently stands in case anyone wants to screw with it. Keep in mind processing isn't the fastest thing in the world and it will take a while to run. Press any key to save a screenshot.

int s = 800; //Size of one side of the triangle in pixels
float[][] mp = new float[3][2]; // The triangles main points
float[] p = new float[2];

void setup(){
  int w = s; int h = round(.866 * w); //Sets the width and height based on the triangle edge length
  size(w, h);
  background(0);
  stroke(255);
  frameRate(3000);

  //Set the main points of the triangle
  mp[0][0] = w / 2; mp[0][1] = 0;
  mp[1][0] = 0;     mp[1][1] = h;
  mp[2][0] = w;     mp[2][1] = h;

  p[0] = random(w); p[1] = random(h); //picks a starting point  
}

void draw(){
  int rp = floor(random(mp.length)); //picks an index for a random main point to jump towards.

  //Interpolate between the two points
  float xdiff = p[0] - mp[rp][0];
  float ydiff = p[1] - mp[rp][1];
  p[0] -= xdiff * 0.381966;
  p[1] -= ydiff * 0.381966;

  //Draw the new point
  int x = int(p[0]); int y = int(p[1]);
  color c = get(x, y);
  c = lerpColor(c, color(255), 0.03);
  set(x, y, c);
}

void keyPressed(){
  saveFrame();
}

ScallopPusher · 3 months

There is a post here in r/math that should help you: http://www.reddit.com/r/math/comments/p2297/interactive_chaos_game_path_plotter/ or direkt: http://koozdra.ca/chaosgame/pathPlotter5.html

try the Triangle fixed 1.618 setting (Golden Ratio)

koozdra · 3 months

If you use a hexagon, the magnitudes 0.5 and 1.5 draw the same image: Hexagon chaos game inner zoom

FarDareisMai · 3 months

I'm not much of a programmer, but I do have a lot of experience with fractal art, so I opened up Apophysis and ran with the idea until I made this. Sorry I can't answer any of your questions but I thought you might be interested nonetheless.

zserf · 3 months

You will get fractals like this if you do anything in the correct range. Anything larger than 1/2 gets you overlapping triangles, and anything smaller than 2/3 gets you empty space. It just so happens that phi is in the right range to have both of these properties be easily observable. Values between .6 and .63 look the best to me.

I'm having trouble with imgur, but if you have java you can play around with this code, which goes quite quickly.

public class Serpenski{  
public static void main(String[] args){  
    double x=Double.parseDouble (args [0]);  
    double y=Double.parseDouble (args [1]);  
    double a=Double.parseDouble (args [2]);  
    double b=Double.parseDouble (args [3]);  
    double point1x=0;  
    double point1y=0;  
    double point2x=1;  
    double point2y=0;  
    double point3x=.5;  
    double point3y=.86603;  
    StdDraw.setXscale(0.0, 1.0);  
    StdDraw.setYscale(0.0, 1.0);  
    StdDraw.show (0);  
    for (double i=1.0; i<=100000; i++){  
        double r=Math.random();  
        if (r<(.333)){                   
            StdDraw.point(((a*point1x+b*x)/(a+b)),((a*point1y+b*y)/(a+b)));  
            x=(a*point1x+b*x)/(a+b);  
            y=(a*point1y+b*y)/(a+b);  
        }  
        else if (r<(.667)){  
            StdDraw.point(((a*point2x+b*x)/(a+b)),((a*point2y+b*y)/(a+b)));  
            x=(a*point2x+b*x)/(a+b);  
            y=(a*point2y+b*y)/(a+b);}  
        else {  
            StdDraw.point(((a*point3x+b*x)/(a+b)),((a*point3y+b*y)/(a+b)));  
            x=(a*point3x+b*x)/(a+b);  
            y=(a*point3y+b*y)/(a+b);}  
    }  
    StdDraw.show (0);  
}

}

Change all the &lt to < . I'm not sure why reddit is switching that.

Scorcherr · 3 months

I did fool around with this as a kid (hahh.. TurboPascal, good times, good times!). Try 4 or more vertices (don't remember if I stayed at 1/2 jumps), arranged as a cross, a star, ... funny things will happen. :)

Platz · 3 months

Also, works nicely as my smartphone wallpaper. Thanks!

de_la_Seoul · 3 months

greatest triforce ever.

newton1996 · 3 months

First, a question: Have you tried starting from all three vertices with ratio 1/4 or 1/5 and overlaying the results?

Now, comments: Serpinski gasket can be considered as the following set:

Every point in the triangle has 3 barycentric coordinates between 0 and 1, with sum equal to 1, see http://en.wikipedia.org/wiki/Barycentric_coordinate_system_%28mathematics%29 We write this [a,b,c] (with a+b+c=1).

Now write those in binary. What can the first bits of the three coordinates be? Well, the coordinates add up to 1=1/2+1/4+1/8+.., which means that the most significant digits have to add up to 1. If there is a carry from the second most significant digit, then the first bits are (0,0,0) - with the carry providing the 1. If there is no carry then the first bits are (1,0,0) or (0,1,0) or (0,0,1). So we get 4 possibilities. These options divide the triangle into 4 smaller pieces - one for each possibility. The first one is the middle triangle of teh Serpinski construction, that we take out at the first stage, and the other 3 are the three "corner" triangle that remain. Now let's assume we are in one of the 3 later cases, and let's consider the second most significant bits (the quarters). Sine we know we get no carry, we again have to have those add up to 0 if there is a carry from the eights, and to 1 if there is no such carry. So each of our three triangles is again divided into 4 parts, and 3 that remain in Serpinski construction are the ones that have exactly one non-zero bit in the 1/4's place. And so on. The conclusion is that Serpinski gasket is the set of those points in th triangle, for which the binary expansion of the barycentric coordinates has exactly one 1 in each position (i.e. out of 3 bits in each 1/2^k place exactly one is 1).

The plotting construction starts with [1, 0, 0] which has 1=.11111111..., 0=.000000...., 0=.0000000 which indeed has exactly one 1 in each position. At each stage the barycentric coordinates [a, b, c] go to [a/2, b/2, c/2+1] (if we move towards the third vertex). In binary, all coordinates are shifted once to the right and the .1 is added to the third coordinate. This clearly preserves the property of having exactly one 1 in each position. And, i fact, you can see that you can get to any triple of coordinates with this property by repeating these moves. That's why the program plots the serpinsky gasket.

Now for the other ratios you could write the barycentric coordinates in base 3 or 5 and do similar analysis to see which sets you would get. I think you still get nice sets if you seed from all three vertices.

For the golden ratio, I think you could use the golden-ratio based number system http://en.wikipedia.org/wiki/Golden_ratio_base to see what the map does....

dietaryfiber0 · 3 months

I was messing around with this after this post and wrote a Java program for a Sieprinski Hexagon (same thing, but with a hexagon, as you probably could have guessed).

You can do the same thing with one over the golden ratio, and you get this: http://i.imgur.com/nI0Ge.png As opposed to the standard 2/3 ratio, which gives: http://i.imgur.com/GTU0m.png

I don't know much about fractals to understand the difference, but it's certainly cool!

WarzoneOfDefecation · 3 months

mid point from a random point within the triangle to its closest vertex. We did this in highschool, its a neat little fractal.

mycophile · 3 months

This is what is see when I close my eyes.

3 months

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