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Author Topic: Buddhabrot Creative Gallery  (Read 878 times)
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richardrosenman
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« on: August 02, 2010, 09:27:52 PM »


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cbuchner1
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« Reply #1 on: August 02, 2010, 09:36:17 PM »


amazing how some of these look just like Iterated Function Systems. A very creative project - can it be run outside of Photoshop as well?
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richardrosenman
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« Reply #2 on: August 02, 2010, 09:41:58 PM »

Hi cbuchner1;

Thanks for your reply. It runs as a plugin for Photoshop but you can run it various other graphics applications (some free ones as well) such as Gimp. Take a look here to see some free ones and also for a complete compatibity list of other numerous other apps that will run it:
http://www.richardrosenman.com/software/downloads/

-Richard
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BrutalToad
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« Reply #3 on: August 03, 2010, 12:17:54 AM »

Those are really great Richard.  I have downloaded some of your other plug-ins, and I really enjoy them.  Thank you for them and I can't wait until I can try out the buddhabrot plug-in. smiley

Edit:
I just downloaded Buddhabrot and it seems to be very good.
« Last Edit: August 03, 2010, 01:11:17 AM by BrutalToad » Logged

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kram1032
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« Reply #4 on: August 03, 2010, 11:23:26 PM »

you have a lot of nice renders going on there smiley

I suggest you to try tan(z)*z+c - that one turned out to be incredibly detailed.

The diamond one, is that based on split complex numbers?
(a+b*j where j²=1 but j!=1)

Because it looks a lot like that...

If so, you might as well want to try dual numbers:
a+b\epsilon where \epsilon^2=0 but \epsilon\neq0
that's (a+b\epsilon)^2=a^2+2ab\epsilon - seems incredibly simple as the real part is not at all dependend on the previous imaginary part but it certainly looks nice smiley - I actually prefer it over the split complex one.

That one turns out to have a beautiful fan-like structure smiley

Also very interesting is the set of polynomials of degree 5 (aribitary choice) where you could try either random parameters between -1 and 1 (bigger will basically shrink the features) or maybe some user input boxes with sliders so you can have nice comparisons smiley

Just some ideas I tried myself already

for the z*tan(z) one, you'll need very deep iterations. It will heavily oscillate between 0 and infinity but it does produce very nice images smiley - seriously, that's the Beauty and the Beast in a single function.
« Last Edit: August 03, 2010, 11:26:16 PM by kram1032 » Logged
richardrosenman
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« Reply #5 on: August 04, 2010, 07:45:22 AM »

you have a lot of nice renders going on there smiley

I suggest you to try tan(z)*z+c - that one turned out to be incredibly detailed.

The diamond one, is that based on split complex numbers?
(a+b*j where j²=1 but j!=1)

Because it looks a lot like that...

If so, you might as well want to try dual numbers:
<Quoted Image Removed> where <Quoted Image Removed> but <Quoted Image Removed>
that's <Quoted Image Removed> - seems incredibly simple as the real part is not at all dependend on the previous imaginary part but it certainly looks nice smiley - I actually prefer it over the split complex one.

That one turns out to have a beautiful fan-like structure smiley

Also very interesting is the set of polynomials of degree 5 (aribitary choice) where you could try either random parameters between -1 and 1 (bigger will basically shrink the features) or maybe some user input boxes with sliders so you can have nice comparisons smiley

Just some ideas I tried myself already

for the z*tan(z) one, you'll need very deep iterations. It will heavily oscillate between 0 and infinity but it does produce very nice images smiley - seriously, that's the Beauty and the Beast in a single function.

Hi Kram1032;

Thanks again for your reply. I'm very interested in trying out your suggestions. One thing I should mention straight up: I'm primarily an artist and only secondly a programmer so it takes quite a bit of extra effort for me to understand some of these equations. (I flunked algebra and geometry). I tend to learn best though example and some of the cryptic algorithms tend to lose me. wink

My main algorithm for my program is the following:

// Mandelbrot
xb=zx*zx-zy*zy+xa;   // real
yb=a*zx*zy+ya;                // imaginary

It would help if you could show me how your equation relates to my code so that I can understand it better.

The diamond equation is a simple variation on the above Mandelbrot but with negatives:

// Diamond
xb=-zx*zx-zy*zy+xa;   // real
yb=-a*zx*zy+ya;               // imaginary

I'm really curious to try your examples out (and especially the Tan function) so if you have some spare time (and patience) to show me how your equations relate that would be great. wink

-Richard

P.S. I had implemented proper powers in my Fractus 2D plugin but I found the computations far too slow to add to my Buddhabrot program. Perhaps one of these days if I find an optimized cheat... wink
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kram1032
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« Reply #6 on: August 05, 2010, 12:52:29 PM »

I see. so a is a free parameter?

you wont be able to keep this formula and change your program to allow tan...

complex squaring usually is like this:

(zx+i(zy))^2+ax+i(ay)=(zx^2+zy^2+ax)+i(-2 zx zy+ay)

what your diamont variation does is similar to the following:

(zx+j(zy))^2+ax+j(ay)=(zx^2+zy^2+ax)+j(2 zx zy+ay) - note that all that changed is the sign which is basically what you did. It's the split complex or double number Mandelbrot smiley

As said before, there also is the dual number Mandelbrot
(zx+\epsilon(zy))^2+ax+\epsilon(ay)=zx^2+ax+\epsilon(2 zx zy + ay)

The part in ()s with a letter in front are the corresponding,uh, unreal parts. You split them into the real and the other part as two seperate equations as usual.

Now I'm totally not sure how tan would be in those alternative algebras (they all are so-called clifford algebras) but for the complex part, that's easy:

\tan(zx+i(zy))(zx+i(zy))+ax+i(ay)={{zx \sin(2 zx)-zy \sinh(2 zy)}\over{\cos(2 zx)+\cosh(2 zy)}}+ax+i ({{zy \sin(2 zx) + zx \sinh(2 zy)}\over{\cos(2 zx) + \cosh(2 zy)}}+ay)

sinh and cosh are the so called hyperbolic functions. You can describe them as
\sinh(x) = {{1}\over{2}} (-e^{-x}+e^x) and
\cosh(x) = {{1}\over{2}} (e^{-x}+e^x)
I hope, this description is helpful...
« Last Edit: August 05, 2010, 12:55:38 PM by kram1032 » Logged
fengda2870
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« Reply #7 on: March 25, 2012, 03:55:32 PM »

very beautiful pic!!!!
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