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Author Topic: 4th AND 5th Order ABS Fractals Are Here!!!  (Read 4614 times)
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Kalles Fraktaler
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« Reply #15 on: April 16, 2016, 12:58:03 PM »

Thanks for your extensive work stardust, however I am sorry but splitting up the Z^4 to (Z^2)^2 does unfortunately not work with perturbation, that uses the (reference+delta) - (reference) construction.

I found out that the problem is not the imaginary part, but the real part.
The expression I showed actually contains r^4-r^4 and i^4-i^4.
With the help of this page
http://quickmath.com/webMathematica3/quickmath/algebra/simplify/basic.jsp
I reduced the real part, and got it to work! 
joy dancing banana dancing chilli joy dancing banana dancing chilli


* bs4.jpg (96.37 KB, 660x445 - viewed 180 times.)
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stardust4ever
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« Reply #16 on: April 17, 2016, 04:45:44 AM »

Thanks for your extensive work stardust, however I am sorry but splitting up the Z^4 to (Z^2)^2 does unfortunately not work with perturbation, that uses the (reference+delta) - (reference) construction.

I found out that the problem is not the imaginary part, but the real part.
The expression I showed actually contains r^4-r^4 and i^4-i^4.
With the help of this page
http://quickmath.com/webMathematica3/quickmath/algebra/simplify/basic.jsp
I reduced the real part, and got it to work!  
joy dancing banana dancing chilli joy dancing banana dancing chilli
So you found an "island" somewhere around e44 in the 4th order burning ship? Neato! I was only thinking maybe the equations could be factored into two second order half-iterations (which you claimed won't work) but kudos to you regardless!
joy dancing banana dancing chilli joy dancing banana dancing chilli

You said it contained r^4-r^4 and i^4-i^4. If any of these expressions resulted in the denominator of a fraction, it would invalidate the equation.

There was a fairly famous algebraic "proof" that I remember reading from an old algebra textbook in high school. Or was it eighth grade; I cannot remember exactly. It starts with A=B, then manipulates both sides of the equation extensively before resulting in B=A+A and substituting A for B and factoring A out of the equation with the final erroneous result 1=2. Upon further examination of the equation, you have both sides of the equal sign with (A-B)/(A-B) or similar expression in both the numerator and denominator of a fraction, resulting in 0/0 which is undefined. Anything times zero equals zero, so working backwards, any value could be applied to an invalid fraction such as 0/0, such a conundrum which allowed the equation to be manipulated to 1=2 as a result. tongue stuck out

My calculus sucks, but there is a great deal of math involved in determining limits, which a curve or infinite series may approach but never actually reach a specific value with finite terms. Discovering the precise limit to a problem, or even whether a series escapes to infinity or has a defined limit, is not trivial. For fractals, we sometimes work out the equation to millions of iterations to determine if a point falls within a set. If the deltas involved with perturbation use infinitesimal values to compute "limits", a zero term anywhere in the equation, especially under a denominator, could really foul the result.

Kudos to you again regardless.
joy dancing banana dancing chilli joy dancing banana dancing chilli

I look forward to future Kalles Fraktaler updates. I would also love to see 5th order B.S. [Z=|Z|^5+C] and Buffalo [Z=|Z^5|+C] at some point if they are possible to implement, since the odd order versions of both these fractals have that beautiful diagonal symmetry that is so intriguing.
« Last Edit: April 17, 2016, 04:52:04 AM by stardust4ever » Logged
Kalles Fraktaler
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« Reply #17 on: April 17, 2016, 10:40:54 AM »

You said it contained r^4-r^4 and i^4-i^4. If any of these expressions resulted in the denominator of a fraction, it would invalidate the equation.
This is actually the main magic of perturbation, reducing the larger term to be able to calculate with low precision.
We calculate the delta, i.e. the reference (x) plus an offset (d) and then subtract the reference.
An easy example, (x+d)^2 - x^2. Expaning the parentesis yiels (x^2 + 2xd + d^2) - x^2.
The largest term of the reference, x^2, can be reduced, and left is only 2xd + d^2, which fortunately doesn't require high precision to be calculated accurately.

I managed to forget the importance of this since last time I implemented perturbation formulas cheesy

In 4th Power Burning Ship there were also a term r*r*i*i that needed to be reduced in order to get it working.
5th order requires some good analysis I assume.
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stardust4ever
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« Reply #18 on: April 17, 2016, 11:47:48 AM »

This is actually the main magic of perturbation, reducing the larger term to be able to calculate with low precision.
We calculate the delta, i.e. the reference (x) plus an offset (d) and then subtract the reference.
An easy example, (x+d)^2 - x^2. Expaning the parentesis yiels (x^2 + 2xd + d^2) - x^2.
The largest term of the reference, x^2, can be reduced, and left is only 2xd + d^2, which fortunately doesn't require high precision to be calculated accurately.

I managed to forget the importance of this since last time I implemented perturbation formulas cheesy

In 4th Power Burning Ship there were also a term r*r*i*i that needed to be reduced in order to get it working.
5th order requires some good analysis I assume.
I get it now. Working out the polynomials and reducing them is a pain for higher orders, but it's still just basic algebra. The 2zΔ + Δ^2 would start out infintessimally small, considering the deltas are the equivalent of a few pixel widths onscreen. For a distance in the neighborhood of e-44, the Δ^2 term would be on the order of e-88, about twice the current arbitrary precision depth. As a result, the Δ^2 could be safely discarded. Additionally, if floats or scaled integers are used, the required precision decreases as the deltas drift apart with every iteration.

But I see how the equations get ugly for higher orders. (z+Δ)^5 would be
Code:
z^5 + 5*z^4*Δ + 10*z^3*Δ^2 + 10*z^2*Δ^3 + 5*z*Δ^4 + Δ^5
Going into complex domain, we have
Code:
((zr + zi*i) + (Δr + Δi*i))^5 - (zr + zi*i)^5
with the sum of four terms inside the brackets to which an exponent is raised, which I won't even attempt to factor out.

But essentially we are dealing with powers of sums (a + b + c + d)^N which seems really nasty to expand, even if a bunch of terms from -(a + b)^N ultimately either get cancelled out or are so infintessimally small we can safely ignore them. I've only ever done polynomial powers of sums with two variables each (for which one can use the horizontal rows of Pascal's triangle as a cheat sheet). Four variables (necessary for summation of two complex numbers) would be a PITA. Only the terms containing only Zr and Zi can be algebraically eliminated.

Edit: (a+b*i+c+d*i)^5-(a+b*i)^5 is plain nasty, with 50 terms! hurt
http://www.wolframalpha.com/input/?i=expand+(a%2Bb*i%2Bc%2Bd*i)^5+-+(a%2Bb*i)^5
Code:
5 a^4 c+5 i a^4 d+20 i a^3 b c-20 a^3 b d+10 a^3 c^2+20 i a^3 c d-10 a^3 d^2-30 a^2 b^2 c-30 i a^2 b^2 d+30 i a^2 b c^2-60 a^2 b c d-30 i a^2 b d^2+10 a^2 c^3+30 i a^2 c^2 d-30 a^2 c d^2-10 i a^2 d^3-20 i a b^3 c+20 a b^3 d-30 a b^2 c^2-60 i a b^2 c d+30 a b^2 d^2+20 i a b c^3-60 a b c^2 d-60 i a b c d^2+20 a b d^3+5 a c^4+20 i a c^3 d-30 a c^2 d^2-20 i a c d^3+5 a d^4+5 b^4 c+5 i b^4 d-10 i b^3 c^2+20 b^3 c d+10 i b^3 d^2-10 b^2 c^3-30 i b^2 c^2 d+30 b^2 c d^2+10 i b^2 d^3+5 i b c^4-20 b c^3 d-30 i b c^2 d^2+20 b c d^3+5 i b d^4+c^5+5 i c^4 d-10 c^3 d^2-10 i c^2 d^3+5 c d^4+i d^5
(50 terms)
« Last Edit: April 17, 2016, 12:47:40 PM by stardust4ever » Logged
Kalles Fraktaler
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« Reply #19 on: April 18, 2016, 09:32:20 AM »

5th Burning Ship is unfortunately even harder.
Because delta and reference are encapsulated by the abs function for the real value, so there is no reducing easily available.

Edit: No, I just made some typos, it will work

Anyway, here is a test zoom of the 4th power:
<a href="https://www.youtube.com/v/z-HbUHNhbGg&rel=1&fs=1&hd=1" target="_blank">https://www.youtube.com/v/z-HbUHNhbGg&rel=1&fs=1&hd=1</a>


* bs5.jpg (194.61 KB, 659x497 - viewed 167 times.)
« Last Edit: April 18, 2016, 10:01:04 AM by Kalles Fraktaler » Logged

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stardust4ever
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« Reply #20 on: April 18, 2016, 07:32:00 PM »

Holy cow! Thanks dude... wink
Wow

EDIT:
In other news, here's a short 9th Order B.S. (!) zoom video I made in Fractal Extreme yesterday using my own compact float point code:
<a href="https://www.youtube.com/v/TeEEyeuBA-U&rel=1&fs=1&hd=1" target="_blank">https://www.youtube.com/v/TeEEyeuBA-U&rel=1&fs=1&hd=1</a>

Code:
9th Order Burning Ship [nested]
        zi= ((zrsqr * 3) -zisqr) * abs(zi);
        zr= (zrsqr - (zisqr * 3)) * abs(zr);
        zisqr = zi * zi;
        zrsqr = zr * zr;
        zi= ((zrsqr * 3) -zisqr) * zi + JuliaI;
        zr= (zrsqr - (zisqr * 3)) * zr + JuliaR;
        zisqr = zi * zi;
        zrsqr = zr * zr;
« Last Edit: April 18, 2016, 09:01:34 PM by stardust4ever » Logged
Kalles Fraktaler
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kallesfraktaler
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« Reply #21 on: April 18, 2016, 09:38:46 PM »

Holy cow! Thanks dude... wink
Wow

EDIT:
In other news, here's a short 9th Order B.S. (!) zoom video I made in Fractal Extreme yesterday using my own compact float point code:
<a href="https://www.youtube.com/v/TeEEyeuBA-U&rel=1&fs=1&hd=1" target="_blank">https://www.youtube.com/v/TeEEyeuBA-U&rel=1&fs=1&hd=1</a>

Code:
9th Order Burning Ship [nested]
        zi= ((zrsqr * 3) -zisqr) * abs(zi);
        zr= (zrsqr - (zisqr * 3)) * abs(zr);
        zisqr = zi * zi;
        zrsqr = zr * zr;
        zi= ((zrsqr * 3) -zisqr) * zi + JuliaI;
        zr= (zrsqr - (zisqr * 3)) * zr + JuliaR;
        zisqr = zi * zi;
        zrsqr = zr * zr;
Cool, looks like an owl.
Outzoomed it looks like almost the 9th power mandelbrot smiley
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Kalles Fraktaler
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« Reply #22 on: April 19, 2016, 11:13:47 PM »

<a href="http://www.youtube.com/v/R9yme1WAGW8&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/R9yme1WAGW8&rel=1&fs=1&hd=1</a>
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stardust4ever
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« Reply #23 on: April 20, 2016, 12:51:07 AM »

 Repeating Zooming Self-Silimilar Thumb Up, by Craig
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quaz0r
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« Reply #24 on: April 20, 2016, 08:13:56 AM »

 Repeating Zooming Self-Silimilar Thumb Up, by Craig
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stardust4ever
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« Reply #25 on: April 21, 2016, 02:51:45 AM »

Some side by side large details I rendered of the West Miniships in 4th order and 5th order Burning Ship / Quasi Burning Ship fractals. The Quasi and standard Burning Ship fractal variants share a striking similarity along the west needle, regarding placement of dendrites and minis.

High resolution zoomed in areas of the west needle in the 4th and 5th B.S. and Quasi B.S.:
http://sta.sh/01e6u2eank9n

The 3rd order version of the Quasi Burning Ship for comparison. Notice how more tightly compacted the dendrites are in the 3rd Quasi B.S mini:
http://sta.sh/0aiensqbdgu

In the 3rd order version of the Quasi Burning Ship, there also exists a northward facing needle that strikingly resembles that same region in the Buffalo 3rd. You can see the northward facing needle in this high resolution zoomed out capture of the entire set:
http://sta.sh/01hrnckb4dvc

Also for comparison, I have done zoomed in renders of the west needle for the Buffalo and Celtic fractals in the 4th and 5th orders:
http://sta.sh/029v7h94cu9p

Note that the 4th order fractals exhibits 8-fold symmetry and the 5th order fractals exhibit 10-fold symmetries with regards to the dendrites. The Buffalo and Celtics are indeed very "bulby" fractals when getting into higher orders! cheesy


Generalized abs() formula for any power (arbitrary exponent):

Burning Ship: Absolutes the value of Zr and Zi before the exponent.
Buffalo: Absolute the value of Zr and Zi after exponent.
Celtic: Absolutes the value of Zr after exponent.
Quasi Burning Ship: Absolutes the value of Zr and Zi before the exponent, AND absolutes the value of Zi after the exponent.*

*For Quasi Burning Ship, it is not always necessary to absolute Zi or Zr before the exponent, if Zi or Zr only appears in the imaginary portion of the expanded equation. For even powers, the Quasi Burning Ship is the Partial Buffalo (Imaginary Only), which takes the imaginary side of the Buffalo and the real side of the standard Mandelbrot. For odd Powers, the Quasi Burning Ship is a hybrid combining the real side of the Burning Ship and the Imaginary side of the Buffalo formula. For 2nd order only, the Burning Ship and Quasi Burning Ship are the same fractal.
« Last Edit: April 21, 2016, 02:59:05 AM by stardust4ever » Logged
Kalles Fraktaler
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« Reply #26 on: April 22, 2016, 04:42:41 PM »

If you can put up with not being able to go deeper than e300, I can upload a new version with these new fractals included?
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TheRedshiftRider
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« Reply #27 on: April 22, 2016, 10:10:43 PM »

Great, I guess depth will not be an issue, just being able to explore them would already be nice.
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stardust4ever
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« Reply #28 on: April 23, 2016, 04:13:06 AM »

If you can put up with not being able to go deeper than e300, I can upload a new version with these new fractals included?
Go for it. Well, e300 is a still a hella lot better than the 43 zooms I'm currently getting with my Fractal Extreme sampleplugin hacks! bubble gum

IS there a reason why does it fail beyond e300?
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Kalles Fraktaler
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« Reply #29 on: April 23, 2016, 09:35:25 AM »

Go for it. Well, e300 is a still a hella lot better than the 43 zooms I'm currently getting with my Fractal Extreme sampleplugin hacks! bubble gum

IS there a reason why does it fail beyond e300?
The reason is that e300 is the limit of the exponent of the hardware datatype 64-bit double.
One can extend that to e600 by multiplying and dividing the values, "scaling", but that gets too complicated for anything beyond cubic Mandelbrot, at least for me.
Beyond e600 I use the dll that has the formulas in 80-bit long double compiled with gcc. But I have to uninstall and install to switch between 32 and 64 bits. I will eventually do it though to be able to make also the new formulas beyond e300.
Beyond e4900 (which also could be extended for plain Mandelbrot with scaling though), the limit of the exponent of long double, the much slower custom datatype floatexp is used, i.e. a double and integer pair where the integer replaces the exponent of the double. The limit of the exponent is then some 2 billion  cheesy

Ok the exponent limit numbers above are not the exact limits but what I use, since there is some margin
Anyway, get the new formulas from http://www.fractalforums.com/index.php?topic=23602.msg92220#new
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