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 Author Topic: Formulas easily Convertable to 3d, but not happened yet ...  (Read 4662 times) Description: 0 Members and 1 Guest are viewing this topic.
cKleinhuis
Fractal Senior

Posts: 7044

formerly known as 'Trifox'

 « on: September 26, 2010, 02:15:32 PM »

hello, all, i started this thread to collect formulas which appear to be easily converted to a triplex representation, but
not happened yet really ....

i use ultrafractal as base, and fractint names:

1. All 3 Fractint Barnsley Formulas, they consist just of an "if" alternation, and a "+" and a "*"
2. Phoenix Formula, a large formula but consisting only of "^","+","*", and the last "z" value
3. lambda formula, also consisting only of "^","*","+"
3. Burning Ship formula, just a slight modification of mandelbrot
4. Mandelbrot  "z^z+z^n"
5. Manowar

correct me if those have been implemented, or add more nice convertible formulas to the list as a review for implementors
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divide and conquer - iterate and rule - chaos is No random!
blob
Strange Attractor

Posts: 272

 « Reply #1 on: September 26, 2010, 02:41:57 PM »

I was wondering if there could be something like a 3D Newton???
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cKleinhuis
Fractal Senior

Posts: 7044

formerly known as 'Trifox'

 « Reply #2 on: September 26, 2010, 03:36:38 PM »

i just took formulas without a division operator, but a division is defined...
other than that the bailout condition is other than the mandelbrot condition, you check for for change in iteration values rather than absolute values
but this could also be implemented ...
i put in the list only escape time formulas, because they could be easily integrated into an existing rendering system
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divide and conquer - iterate and rule - chaos is No random!
Softology
Conqueror

Posts: 120

 « Reply #3 on: September 27, 2010, 02:09:43 AM »

I have tried;

Phoenix formula z=z^p+c*lastz

Polynomial1 formula z=z^p+z+c

Polynomial2 formula z=z^p-z+c

Manowar formula z=z^p+c+lastz

Ikenaga formula z=z^p+(c-1)*z-c

The more complex formulas do lead to more disconnected/dusty results.

More examples images here
http://softology.com.au/gallery/gallerymandelbulb.htm

Jason.
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David Makin
Global Moderator
Fractal Senior

Posts: 2286

 « Reply #4 on: September 27, 2010, 03:24:15 AM »

I don't think it's possible as things stand to do number 4 - as far as I'm aware z^z i.e. z^power where power is full triplex is as yet either completely undefined or only defined in terms of "possibly" correct log and exp functions that would take aeons to calculate.
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M Benesi
Fractal Schemer

Posts: 1075

 « Reply #5 on: November 02, 2010, 07:45:08 PM »

I don't think it's possible as things stand to do number 4 - as far as I'm aware z^z i.e. z^power where power is full triplex is as yet either completely undefined or only defined in terms of "possibly" correct log and exp functions that would take aeons to calculate.

Interesting.  I never noticed this thread or your statement until now, but I think it's going to be a LOT easier than you imagined (I was wrong... see below):  use my double complex formulas to "visualize" the mathematics.  We can easily do complex to a complex power...

The double complex uses 2 complex number groupings to the n power, so the only change would be to have it be to the z power... simplicity in itself.  I will try and post the results in thread.  I never thought of doing that...
Update:  not too pretty.  Discontinuous (although I only made a complex_1^complex_2 version, not complex_1^complex_2 + complex_1^n version).   In other words, looks like a major earthquake hit a normal fractal.  I'm not sure that the implementation I'm using is incorrect- it might simply be impossible too create a continuous version of this fractal <-- that might get someone working on it.

As to the OP, I've made the Burning Ship fractal, which is particularly promising in 3d (threads in the theory subforum 'beneath' this forum).  Extrapolating for the successful BS fractal, I realized it was simply sign management* that was the problem for z^2 Mandelbrots and corrected that (although perhaps the sign assignment method without singularities is NOT the correct one, as I've found other ones that produce nearly the amount of fractal detail).

*BS fractals using the absolute values of the real and imaginary components of the complex^n.
 « Last Edit: November 02, 2010, 08:08:23 PM by M Benesi » Logged

DarkBeam
Global Moderator
Fractal Senior

Posts: 2512

Fragments of the fractal -like the tip of it

 « Reply #6 on: October 30, 2011, 07:32:20 PM »

Code:
1. All 3 Fractint Barnsley Formulas, they consist just of an "if" alternation, and a "+" and a "*"
2. Phoenix Formula, a large formula but consisting only of "^","+","*", and the last "z" value
3. lambda formula, also consisting only of "^","*","+"
3. Burning Ship formula, just a slight modification of mandelbrot
4. Mandelbrot  "z^z+z^n"
5. Manowar

I converted Barnsley1 and 2 in 4D and nobody used it (almost)
Lambda... DONE 4D
Manowar ... Not so easy. And z^z = craziness!!!!!
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Power 8
Conqueror

Posts: 147

 « Reply #7 on: October 31, 2011, 01:06:59 PM »

Phoenix gives interesting results. An example I made a while ago using Softology's VOC:
I called it King Mandelbulb

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DarkBeam
Global Moderator
Fractal Senior

Posts: 2512

Fragments of the fractal -like the tip of it

 « Reply #8 on: November 03, 2011, 10:00:40 AM »

I wanted to translate Phoenix but it requires 3 extra slots that are so hard to find in MB3D! I can do this overwriting a non-used memory space but it seems to me a junky practice...
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No sweat, guardian of wisdom!
M Benesi
Fractal Schemer

Posts: 1075

 « Reply #9 on: November 18, 2011, 03:22:16 AM »

And z^z = craziness!!!!!
Maybe.  The following might work.  I might try it out later tonight, or after the weekend.     note: some mimetex notice is popping up with the latex...  ignore it unless you are the bossman

We can take the real and imaginary components of $e^{z\log{z}}$.  We'd have to do it for each of the individual "complex" numbers.   Note that log refers to the natural log in all these equations!

$whiskey=x + i \sqrt{y^2+z^2}$
So we would end up with:
$whiskey=whiskey^{whiskey}=e^{\left[x + i \sqrt{y^2+z^2}\right]\log\left[x + i \sqrt{y^2+z^2}\right]$
The we tackle our next complex variable:
$tango=y + iz$
And this other part is a bit more complicated, as we need to divide out by its "complex magnitude".  It's new magnitude would be (y and iz are negative because we are dividing!):
$tmag=e^{\left(-y - i z\right)\log\sqrt{y^2+z^2}}$
Note that the above latex is the magnitude of the y and z component to the -e to the [......]
Now we need to take tango to the tango:
$tango=tango^{tango}=e^{\left[y + iz\right]\log\left[y + i z\right]$
And multiply tango by tmag:
$tango=tango*tmag$
Now we sort out our components:
new_x component= the real part of whiskey.
new y component = the imaginary part of whiskey * the real part of tango
new z component = the imaginary part of whiskey * the imaginary part of tango

You might want to add in the regular z^n components as well... notice that the above is simply z^z.

Glad to see some good ideas bouncing around still!  I'll post an interesting modification for all trigonometric type fractals in a bit (I've already used it in several formulas, but it works in a general sense with cosine formulas with n=2,6,10,14...  and n=3,5,7,9.... (odd n)).
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Alef
Fractal Supremo

Posts: 1174

 « Reply #10 on: November 30, 2011, 10:37:55 PM »

hello, all, i started this thread to collect formulas which appear to be easily converted to a triplex representation, but
not happened yet really ....
3. Burning Ship formula, just a slight modification of mandelbrot

I just did it. mmm. Result is so so .
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fractal catalisator
DarkBeam
Global Moderator
Fractal Senior

Posts: 2512

Fragments of the fractal -like the tip of it

 « Reply #11 on: December 25, 2011, 12:08:45 PM »

And z^z = craziness!!!!!
Maybe.  The following might work.  I might try it out later tonight, or after the weekend.     note: some mimetex notice is popping up with the latex...  ignore it unless you are the bossman

We can take the real and imaginary components of $e^{z\log{z}}$.  We'd have to do it for each of the individual "complex" numbers.   Note that log refers to the natural log in all these equations!

$whiskey=x + i \sqrt{y^2+z^2}$
So we would end up with:
$whiskey=whiskey^{whiskey}=e^{\left[x + i \sqrt{y^2+z^2}\right]\log\left[x + i \sqrt{y^2+z^2}\right]$
The we tackle our next complex variable:
$tango=y + iz$
And this other part is a bit more complicated, as we need to divide out by its "complex magnitude".  It's new magnitude would be (y and iz are negative because we are dividing!):
$tmag=e^{\left(-y - i z\right)\log\sqrt{y^2+z^2}}$
Note that the above latex is the magnitude of the y and z component to the -e to the [......]
Now we need to take tango to the tango:
$tango=tango^{tango}=e^{\left[y + iz\right]\log\left[y + i z\right]$
And multiply tango by tmag:
$tango=tango*tmag$
Now we sort out our components:
new_x component= the real part of whiskey.
new y component = the imaginary part of whiskey * the real part of tango
new z component = the imaginary part of whiskey * the imaginary part of tango

You might want to add in the regular z^n components as well... notice that the above is simply z^z.

Glad to see some good ideas bouncing around still!  I'll post an interesting modification for all trigonometric type fractals in a bit (I've already used it in several formulas, but it works in a general sense with cosine formulas with n=2,6,10,14...  and n=3,5,7,9.... (odd n)).

tex is not working here
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DarkBeam
Global Moderator
Fractal Senior

Posts: 2512

Fragments of the fractal -like the tip of it

 « Reply #12 on: January 20, 2012, 09:45:44 AM »

I think that z^8 - .5 oldz + 0.5 z + c is good! A combination of Manowar and Phoenix with strength 0.5

( now I notice that it is a combo of Polynomial and Manowar, because effectively Phoenix is "+ z*c" ... )
Fractals are tricky

An image - I made it some days ago, Jesse told me that it should not be done but in my computer it works flawlessly

Juliabulbs are not symmetric though

Renders of variations of strength 1 are weird