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Author Topic: Not so fractal related - tube triangular DE patterns (solved in 2 ways)  (Read 9198 times)
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Alef
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« Reply #15 on: January 02, 2013, 10:00:07 AM »

Rotating trough z axis =  <Quoted Image Removed>Uploaded at ImageFra.me

Cool pattern, and realy nice animation. Becouse of colours it's somewhat not so Christmas and New Year, but like ornaments in Marrakesh.  IMHO, the further east the more sided simmetries they prefare. In Europe its squares, further east its hexagonal, and even further east in Asia it's octagonal;)
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DarkBeam
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Fragments of the fractal -like the tip of it


« Reply #16 on: January 02, 2013, 06:15:41 PM »

Lots of thanks my friend,
by the way octagonal tiling does not exist in regular euclidean geometry wink But you can do an hypertessellation.
It is very complicated btw Knighty was able to do it. cheesy
You can do some fiddling with octagons and squares.
into ultrafractal folders Sam made a "semiregular tessellation" and otherwild stuff like some knotted rings, my head hurts cheesy
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knighty
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« Reply #17 on: January 02, 2013, 06:50:38 PM »

Very nice animation.   the wave When will the formula be available?

BTW you can obtain semi-regular tilings: just take a point inside the fundamental domain (the triangle bounded by x,y axis and the 3rd folding lines) then draw from that point three lines prependicular to the 3 folding lines.
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DarkBeam
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Fragments of the fractal -like the tip of it


« Reply #18 on: January 02, 2013, 11:38:21 PM »

The formula is already available of course cheesy
Find it in difs transforms
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Alef
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« Reply #19 on: January 03, 2013, 09:37:00 AM »

It's not so much about math, but I think that 2, 4, 6, 8 simmetries are visualy appealing. But odd number simmetries rarely looks nice (none uses cube power mandelbrot). And 10, 12 etc are just too much;)
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knighty
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« Reply #20 on: January 03, 2013, 11:55:10 AM »

The formula is already available of course cheesy
Find it in difs transforms
You mean the attachment of 1st post in this thread: http://www.fractalforums.com/mandelbulb-3d/re-custom-formulas-and-transforms-release-t9810/

I haven't understood immediately.  hurt

Thank you.
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DarkBeam
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Fragments of the fractal -like the tip of it


« Reply #21 on: January 03, 2013, 12:19:25 PM »

Uh, I wonder if there is a x,y function to get the "colour" of every hexagon that you find after folding. It can be so very cool. smiley

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DarkBeam
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« Reply #22 on: January 04, 2013, 08:30:53 PM »

Btw I updated hexgrid & hextgrid3 with custom angles and center, it's awesome thanks for suggestions A Beer Cup


* hexgridIFS.JPG (133.23 KB, 962x742 - viewed 224 times.)
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knighty
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« Reply #23 on: January 09, 2013, 04:09:04 PM »

Thank you Darkbeam!

Uh, I wonder if there is a x,y function to get the "colour" of every hexagon that you find after folding. It can be so very cool. smiley

It's possible. You need to record the folding/transformation history of the transformed point.

The problem is that you'll need some (maybe) solid abstract math background (say group theory) to solve this if you want to make some particular effect. An example is the hexagonal tiling above. More details later... Just don't expect a full solution. I don't have the required math level   hurt.
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DarkBeam
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Fragments of the fractal -like the tip of it


« Reply #24 on: January 09, 2013, 05:44:20 PM »

oh I feared that. don't worry then. You already helped a lot, and hopefully you will continue! smiley
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knighty
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« Reply #25 on: January 16, 2013, 08:48:34 PM »

Hi,
Uh, I wonder if there is a x,y function to get the "colour" of every hexagon that you find after folding. It can be so very cool. smiley

In the case of hexagonal tiling, a more simple appreach is to convert to a sheared coordinates system then find the nearest integer coordinates point.
Code:
static a[2]={1,0};
static b[2]={0.5,0.5*sqrt(3)};
(x,y){
   inthex(&x,&y);
   abs(sin(3*x+y*10))//gen a color
}
inthex(&x,&y){//rouds to nearest hexagonal grid "integer" coordinates
   //convert to "sheared" coodinates (x,y) -> (l,m) by solving the system:
   //x=l*a[0]+m*b[0];
   //y=l*a[1]+m*b[1];

   idet=1/(a[0]*b[1]-b[0]*a[1]);
   m=(-x*a[1]+y*a[0])*idet;
   l=(x*b[1]-y*b[0])*idet;
   
   m0=floor(m);l0=floor(l);//coordinates of the lower left nearest "integer" corner
   
   //find nearest "integer" corner
   d2=(l0*a[0]+m0*b[0]-x)^2+(l0*a[1]+m0*b[1]-y)^2;
   nl=l0; nm=m0;nd=d2;
   d2=((l0+1)*a[0]+m0*b[0]-x)^2+((l0+1)*a[1]+m0*b[1]-y)^2;
   if(d2<nd) {nl=l0+1; nm=m0;nd=d2;}
   d2=(l0*a[0]+(m0+1)*b[0]-x)^2+(l0*a[1]+(m0+1)*b[1]-y)^2;
   if(d2<nd) {nl=l0; nm=m0+1;nd=d2;}
   d2=((l0+1)*a[0]+(m0+1)*b[0]-x)^2+((l0+1)*a[1]+(m0+1)*b[1]-y)^2;
   if(d2<nd) {nl=l0+1; nm=m0+1;nd=d2;}
   
   x=nl; y=nm;
}

In the picture below, I've drawn the fundamental domain of 3-3-3 triangle group in the right side. if you draw from the center of that fundamental domain three lines perpendicular to it's edges you will obtain an hexagonal grid. The three lines divide the fundamental domain into three regions with a different color each. Therefor, you get a 3 colored hexagonal tiling. This is basically what I used in the fragmentarium script presented here: http://www.fractalforums.com/fragmentarium/triangle-groups-tessellation/ .

The two triangles at the left are the fundamental domain of 2-3-6 triangle group and it's mirror image about the diagonal (I could also have mirrored it about any other two edges). The procedure is to count the number of folding (necessary to go inside the fundamental domain) then give a color dependent on the parity of folding number to the bigger part (the fundamental domain is divided in two parts by the line perpendicular to the diagonal).

In my previous post I was referring to a general (too general) method of using subgroups of the triangle group. In the euclidean case, Most of the wallpaper groups (and friese groups) are subgroups of triangle groups. I haven't worked out the math though.  hurt


* Tile_63.PNG (93.47 KB, 440x440 - viewed 222 times.)
« Last Edit: January 16, 2013, 08:50:17 PM by knighty » Logged
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