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Author Topic: Jerusalem cube - can be made via KIFS?  (Read 15872 times)
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David Makin
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« Reply #45 on: December 08, 2012, 12:52:39 PM »

BTW!
the only way IMHO to have a correct DE is to go for (classic) IFS (and LRIFS) or a mix between KIFS and IFS. Obtaining the DE for IFS requires a priority queue. I have successfully adapted Hart's -IFS raytracing- method to return DE instead. My implementation is experimental and use a simple array as priority queue therefore it is slow.

Have you managed to get the Hart+DE method to work with non-affine transforms ?
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knighty
Fractal Iambus
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Posts: 819


« Reply #46 on: December 08, 2012, 03:08:11 PM »

Not yet but I believe it's possible.

@M Benessi: See the attached fragmentarium script. afro

* JC-NonDE.frag (3.43 KB - downloaded 272 times.)
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M Benesi
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« Reply #47 on: December 10, 2012, 07:47:04 AM »

  DOH!@!#!@$   

 Thanks Knighty.  cheesy   hahahhaa...  that cracked me up.  Still grinnin. 
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DarkBeam
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Fragments of the fractal -like the tip of it


« Reply #48 on: December 10, 2012, 11:38:26 AM »

I modded your formula this way;

Code:
JC4(x,y,z,alpha,beta){
   t=beta-1;
   scl=(alpha+t);
   scl1=scl/alpha;
   a=1/scl1;
   r=x*x+y*y+z*z;dd=1;
   for(i=0;i<MI && r<100;i++){
      x=abs(x);y=abs(y);z=abs(z);
      if(x<y){t=x;x=y;y=t;}
      if(z<x){t=z;z=x;x=t;}
      if(x<y){t=x;x=y;y=t;}

      if(y<a && x>1-beta*a+y){
         x-=1;z-=1;
         x*=scl1;y*=scl1;z*=scl1;dd*=scl1;
         x+=1;z+=1;
      }else{
         x-=1;y-=1;z-=1;
         x*=scl;y*=scl;z*=scl;dd*=scl;
         x+=1;y+=1;z+=1;
      }
      r=x*x+y*y+z*z;
   }
   (sqrt(r)-1.75)/dd
}
It is nice but even more discontinue embarrass
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knighty
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« Reply #49 on: December 10, 2012, 08:02:08 PM »

It is always difficult to get continuous DE when the transform contains 'if' statments (if not possible at all).  hurt
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