Very simple formula for fractal patterns

[lkmitch]

Can you give me another tip on how this coloring works?

Keeping track of the average imag(z)/real(z) is easy enough.  Total imag(z)/real(z) during the iterations, and then divide total by the iteration count after all the iterations for the pixel have been calculated.

How do you then index a color palette entry for the resulting average floating point value?  For the curent color methods I use colval=trunc(avmag*colscale)mod 255 to get the floating point value into the 0-255 color palette range, but when using the imag(z)/real(z) value I am not getting those awesome appolonian-like shapes you are?  The colscale is needed (I found) otherwise if you have a very smooth palette, you don't see all the details.

Thanks for any tips.

Jason.

Hi Jason,

It's simply the arithmetic mean of imag(z)/real(z), over all of the iterations.  One "trick" is the clamping: Ultra Fractal clamps negative values to 0 and I (in this image, not in the coloring) clamp values larger than 1 to 1.  Values between 0 and 1 are mapped to a ramp between white and black.  But I see that you've got it figured out already.  :-)

Kerry

[Softology]

Softology, Reply #73 looks like a solution of a special circle packing problem.

The are some sphere packing fractal ideas (i.e. some indra pearls), but most with a complicatd generation algorithms.


Cool finding,

Trafassel

Thanks, it was Kerry that found it.  The julia coords just happen to have that sphere packing result.  The rest of the julia sets don't seem to have the same patterns.  Still a very cool find.

I have Indra's Pearls on my to read pile to implement one of these days.

Jason.

[fracmonk]

Really fantastic renders, Kali!  And growing fast!  Kudos!

I'm going to try a FractInt implementation, since I'm so used to doing that...(& I'm old & in the way...)... 1st chance I get...

Later!

[trafassel]

The corresponding video to reply #64 with changing julia parameters.

<a href="http://www.youtube.com/v/1q45CN-3SuA&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/1q45CN-3SuA&rel=1&fs=1&hd=1</a>

[Kali]

Nice and interesting video, Per! thanks.

PS: I'll be trying the new version of Gestaltlupe, thanks for the advice

[Pauldelbrot]

I'd like to see support for this in Mandelbulber.

[Pauldelbrot]

Some more good variations Kali.

A couple of samples I found when exploring.

z=abs(z*c+1)+1/abs(z*c+1) images

Jason, I can't reproduce these using the formula provided. I mostly don't get space-filling patterns with exponential smoothing and 1000 iterations, and when I do they seem to be much noisier and less structured. What were the c parameters for those images?

[Softology]

Some more good variations Kali.

A couple of samples I found when exploring.

z=abs(z*c+1)+1/abs(z*c+1) images

Jason, I can't reproduce these using the formula provided. I mostly don't get space-filling patterns with exponential smoothing and 1000 iterations, and when I do they seem to be much noisier and less structured. What were the c parameters for those images?

OK, here are the images (I realised I messed up the order of images linked to with the formulas in the other post) and formula details.


Formula z=abs(z*c+1)+1/abs(z*c+1)
The C values are 0.288235294117647 real 0.842261306532663 imaginary.
This one uses "Dual Real<0" symmetry.  Before doing the usual complex math calcs each iteration mirror z by using if z.x<0 then z.r:=-z.r


Formula z=abs(z*c+1)+1/abs(z*c+1)
The C values are -0.370588235294118 real 0.872613065326633 imaginary.
This one uses "Dual Real>0" symmetry.  Before doing the usual complex math calcs each iteration mirror z by using if z.x>0 then z.r:=-z.r


Formula z=(z*c+1)+1/(z*c+1)
The C values are -0.347058823529412 real -0.508391959798995 imaginary.
This one uses "Full Tetrahedral" symmetry.  Before doing the usual complex math calcs each iteration perform the following code (which was an experiment with using the KIFS 3D fold techniques in 2D)
                 if(z.x+z.y<0) then
                 begin
                      x1:=-z.y;
                      z.y:=-z.x;
                      z.x:=x1;
                 end;
                 if(z.x-z.y<0) then
                 begin
                      x1:=z.y;
                      z.y:=z.x;
                      z.x:=x1;
                 end;


Formula z=(z*c+1)+1/(z*c+1)
The C values are -0.323529411764706 real -0.0834673366834171 imaginary.
This one uses "Octahedral" symmetry.  Before doing the usual complex math calcs each iteration perform the following code (which was an experiment with using the KIFS 3D fold techniques in 2D)
                 //octahedral
                 z.x:=abs(z.x);
                 z.y:=abs(z.y);
                 if (z.x-z.y<0) then
                 begin
                      x1:=z.y;
                      z.y:=z.x;
                      z.x:=x1;
                 end;
Also I have an option to raise z to a specified power each iteration.  In the above image the power is 1.11.  The others are power 1, so no change is needed for them.

Hope that helps.

Jason.

[Pauldelbrot]

Thanks for the information. I'll check this out.

[Kali]

@Paul: you are doing too much iterations, try with <50
@Jason: Nice variations and images. And thanks for the mention in your blog 

I have tried a lot of variations with good results, like adding transformations, foldings, rotations, translations, and so on...

But this is a more simple and still nice add-on to original formula, just multiplying by a constant complex number:

z=abs(1/z)*m+c

The attached example is for c=-1-2i and m=1+2i

Also try z=abs(1/z)*c+m in mandelbrot mode for some given m value. Good for a "preview" of the patterns to pick up Julia values.

[Kali]

another apollonian packing

c=1
m=i

[fracmonk]

Kali-  Love your last one- it's a classic subject...wish I had time to investigate it.  Very frustrating for me...hardly have time for my my own fave.  What you're doing here looks very fruitful, though.  I'm v. happy 4 U.  I'll be in touch.

[Pauldelbrot]

Formula z=abs(z*c+1)+1/abs(z*c+1)
The C values are -0.370588235294118 real 0.872613065326633 imaginary.
This one uses "Dual Real>0" symmetry.  Before doing the usual complex math calcs each iteration mirror z by using if z.x>0 then z.r:=-z.r

Ah, the original post didn't mention this step, only the z=abs(z*c+1)+1/abs(z*c+1) part.

However, though I now come closer with my own code, I still can't quite reproduce the above. Here's the closest I've got:



That's with 3x3 AA and no serious attempt to replicate the exact color scheme. And I don't know what center coordinates you used.

There's a resemblance to your image now, and space-filling structure, but don't see the "over/under bridges" aspect that is very pleasing in your image, so I'm not apparently getting the same exact Julia set. Which means something is still not exact.

The loop is

Code:

complex one = COMPLEX(1,0);
...
double r = REAL(z);
if (r > 0) {
        z = COMPLEX(-r,IMAG(z));
}
complex t = CABS(CADD(CMUL(z,c),one);
z = CADD(t,CINV(t));

and the various complex manipulation macros are all definitely correct (a normal Mandelbrot comes out correctly, for starters).

Any more information you might have would be useful.

[Pauldelbrot]

Here's another one, on the Y axis this time and with a slightly closer color scheme. I still see a resemblance, but not the over/under thing. I wonder if there's still a mirroring step missing somehow.

[Softology]

Your code does look right.

This is how I do it... the references to .x and .y are for the real and imaginary components of the complex number.
cone is initialised to real=1, imag=0

if z.x>0 then z.x:=-z.x;
tmpc:=cadd(cmult(z,c),cone);
tmpc.x:=abs(tmpc.x);
tmpc.y:=abs(tmpc.y);
z:=cadd(tmpc,cdiv(cone,tmpc));
CSet(z.x,z.y,cfrectangular);

The real (x axis) coords for Z go from approx -4 to +4
The imaginary (Y axis) coords go from approx -7 to -2

Jason.