IMHO reason behind no 3D mandelbrot.

[Alef]

In XY plane there are nice 2D mandelbrot. In 3D, using canonical number system - quaternions, there is just revolution surface. But if you cut in any direction throught X axis you 'll get nice mandelbrot. It is fractal, but becouse Y=Z, quaternionic mandelbrot is fractal just in 2 dimensions. If you cut it perpendicular to X axis, you 'll get circle. This should not have something to do with k part being left, becouse i, j and k behaves similar, so it would be revolution surface in 4 dimensions, too.

1. So I tought that first requirement are X><Y><Z><X.

Code:

i^2=-1, j^2=-i, i*j=1

zzx = sqr(zx) - sqr(zy) + 2*zy*zz + Cx;
zzy = 2* zx * zy - sqr(zz)  + Cy;
zzz = 2*zx*zz  + Cz;

Fractal above (i posted somewhere) is a bitt better, but even with formula having X><Y><Z><X the fractal is extruded. That means, it curves in all 3 dimensions, but shows fractal features only in 2D cutouts. (Well, this formula have nice curved insides.)

2. So, relation between all X, Y and Z must be fractal, that is, all cutouts must be fractal.
(This sounds as common sense, but it isn't so common.)

==================

http://www.renderosity.com/mod/gallery/index.php?image_id=1308487&member
A little bitt of water:
Idealised imagined 3D mandelbrot have stalks all around its body, so in cutout perpendicular to X axis it would have a higher degree mandelbrot shape (5 degree mandelbrot at least). But I think, quadratic 3D mandelbrot must not have higher degree shape in any cutouts. Quadratic 3D mandelbrot must have only quadratic shapes. Actualy there are just a few cannonical power 2 shapes: mandelbrot and tricorn. Alsou an ideal square. (Burning ship hardly could be included) Maybe alsou z=z*z - c*c shape as it is symmetrical along X and Y axis, so idealy would fitt as cutout .

These polynomials could generate all these shapes. Coefficients 1,-1,0,0,0,2 generates Mandelbrot set, -1,1,0,0,0,2 are Tricorn fractal and 1,1,0,0,0,2 is square.

X = K*X*X + L*Y*Y + M*X*Y  + Cx
Y = O*X*X + P*Y*Y + Q*X*Y   + Cy

IMHO, so I think,
1. In 3D there must be some hidden coeffiсients or polynomial parts whatsoever and formula should be longer than above. Just multiplication don't generate enought polynomial parts;)
2. Cutouts perpendicular to X axis should have some fractal shape, maybe tricorn or z=z^2 - C^2 shape.

[Aexion]

Let me add something more to your post:

Most people forgot that the Julia Sets of the 3D Mandelbrot set are as important as the 3D Mandelbrot itself, such as in this simple example:



Mostly because they share the 3D Mandelbrot dynamics (a very biological looking structures)..

(BTW: the formula is missing, I stored it away and forgot on where..    )

[DarkBeam]

(BTW: the formula is missing, I stored it away and forgot on where..    )

Oh you!

 

[Alef]

Nice julia, indeed looks biological, like octupus Paul.
 

Interesting, that boring quaternion mandelbrot set do have nice julia sets beeing real fractal in all 3 dimensions. IMHO this should have something to do with julia seed. If all (quaternionic mandelbrot) julia seed parts would be a same (0.2, 0.2, 0.2, 0.2), probably julia set would be some oval ball. Throught none realy knows, how julia 3D shapes correspondents to 3D mandelbrot set shape;)

So this could give some hints about properties of "real 3D mandelbrot set", but this too would not give exact formula.

[cKleinhuis]

just thinking ... when using those nice 3d julias, the iteration loop for the mandelbrot could be modified, e.g. it is a member of the 3d mandelbrot if and only if black areas in 3d quaternion julia are connected, could be a bit hard to implement, but perhaps could lead to a hint on how the 3d mandy would look like...

[Alef]

just thinking ... when using those nice 3d julias, the iteration loop for the mandelbrot could be modified, e.g. it is a member of the 3d mandelbrot if and only if black areas in 3d quaternion julia are connected, could be a bit hard to implement, but perhaps could lead to a hint on how the 3d mandy would look like...

If I 'm correct, it sounds as supermandelbrot formula in IKM2.upr. I hadn't looked at code, probably it tests, if julia set is connected if julia set in point (0,0) is "inside" or something like that. In 3D it would bee bitt harder, but it could alsou lead back to quaternion shape.


Here I made few illustrations.
3D mandelbrot cutout in Z=0 (2D mandelbrot), possible cutout in X=0, and 3D examples with cutouts.

It seems, that Z=Z^2+C^2 fits as possible shape of cutout parpendicular to X axis. It have two small bulbs corresponding to similar bulbs of mandelbrot, 3D mandelbulb could be symmetrical along Z axis, and the two valleys would be extrapolation of ridge starting from elephant valley. IMHO this is slightly julia like set coz pixel^2. Just that it is too big, coresponding bulbs are outside of mandelbrot range, so it is similar, but not what could be called cutout;)

I was thinking, if there would be formula for cutout in X=0, (where X=0), it would be quite an easy to get 3D formula. Problem with this example is that the highest and the most recognisible shape are somewhere around X=-0.1 making everything much harder.

Open question is, if X unit squared is 1, Y unit squared is -1, then what would be some special value of Z unit squared and what would correspond z^2-+c^2?

[Tglad]

There is no 3d mandelbrot because unlike 2d, in 3d (or higher) there are only 4 transforms that preserve shape (translate, rotate, scale, inverse+reflect). If you make a fractal using any other transform what-so-ever it will have stretch in it. So there is nothing remotely equivalent to the mandebrot set in 3d.

Even if there did exist a conformal transform like the square operator, the core feature of the mandelbrot is that it is universal; you can move it, rotate it or scale it each iteration and you still end up with a mandelbrot. That would be even more impossible in 3d

[hgjf2]

Where is image!!???
Wow: the image file is crashed

[hgjf2]

Let me add something more to your post:

Most people forgot that the Julia Sets of the 3D Mandelbrot set are as important as the 3D Mandelbrot itself, such as in this simple example:

<Quoted Image Removed>

Mostly because they share the 3D Mandelbrot dynamics (a very biological looking structures)..

(BTW: the formula is missing, I stored it away and forgot on where..    )

where is image?

[Alef]

There is no 3d mandelbrot because unlike 2d, in 3d (or higher) there are only 4 transforms that preserve shape (translate, rotate, scale, inverse+reflect).

Even if there did exist a conformal transform like the square operator, the core feature of the mandelbrot is that it is universal;

Alsou in 2D many different formulas at certain zooms do have mandelbrot islands. In 3D this is not the case. So this means, there are no universal cannonical 3D mandelbrot, just a few ~mandelbrots, having more or less expected 3D mandelbrot features.

[Alef]

With these ideas in mind, I found one formula with perfectly matching mandelbrot in X=0, mirrored only positive x m-brot;) Looks like X=0 line touches upper bulb.

Code:

zzx=sqr(zx) - sqr(zy)  + abs (real (c))
zzy=2*zx*zy    + imag (c)

x of this would be z of 3D mandebrot, so that x is a real numbers, y is an imaginary numbers, and z is well a double positive numbers. Z must have only a positive value, hence allways modulus of Z by abs. It left open question x*z=? and y*y=?, solved using intuition. 

Code:

zx=real(z);
zy=imag(z);
zz=abs(part_j(z));

zzx = zx*zx - zy*zy + 2*zz*zx  + Cx;
zzy = 2* zx * zy + 2* zz * zy   + Cy;
zzz= zz*zz - zy*zy  + abs(Cz);

z= quaternion(zzx, zzy, zzz, 0);

Not quite a "true 3D mandelbrot" with expected features, but I think, pretty perspective. Right angles should be becouse of abs function dividing + and - orbits.


Funny shape of a rear;) Cutting along X=0 and with the same colour method reveals smomething like 2D, but with some mirroring in the middle. Didn't expected more precise resemblance.


Julias are much more angular. It seems, that formulas having visual m-brots have simpler stretched in one dimension julias, and julias curved in all 3 dimensions correspond to solid mass m-brots. So probably 3D m-brot features are seen only when they are not too dense.






However, insides of this fractal is interesting. Zooming inside always shows something like this, realy an unexplored world. So maybe some of 3D mandelbrots would have endlesly curved fractal insides coresponding to 2D spirals, but using only outside rendering would leave them unnoticed.
Larger pics:
http://www.fractalforums.com/images-showcase-(rate-my-fractal)/inside-of-m-brot-set/

p.s
Uploaded this as Slonofractal (coined from saxifragales and slavic name for an elephant) throught Chaos pro are a bitt of past generation renderer http://www.chaospro.de/formulas/display.php?fileid=222

[cKleinhuis]

so, what exactly are you doing there ?!?
nice candidate, can you create a power8 one ?!

[Alef]

Power 3 allredy is very hard;) To get formula, each new power polynimial must be  multiplied by all 3 number parts, so it gets longer and harder. With another more simpler idea for 3D mandelbrot I achieved maximum of 4th power, and that took lot of time, patience and error check. But here is no multiplication rules, more like try and error.

[Alef]

Another inside.


But in 2D formula produced something not so mandelbrot like. Probably becouse some of fractalness goes to Z axis, throught it have interesting angles.


Here I uploaded Chaos Pro parameter file with these fractals (mandelbrot insides).

[Alef]

Using as rigid as possible algebraic order multiplication was kind of not so hard. Eurocodes is harder. Square root factor worked nicely only with power 2, probably coz it is mathematical uncorrect, so I switched it with factor 1, this alsou makes formula more easy.
Multiplication rules are:  r=1 (r is plain real number), i*i=-r -j, j=|j| (j>0), i*r=i, i*j=i, r*r=r, j*r=r, j*j=j .

Multiplaying power 2 formula with with (Xr, Yi, Zj) I got 3rd power slonofractal formula. z is allways positive, so abs must be used:

Code:

zzx= zx^3 - 4*zy^2*zx + 3*zx^2*zz + 3*zx*zy^2 - 3*zy^2*zz + Cx;
zzy= 3*zx^2*zy -2*zy^3 + 6*zx*zy*zz + 3*zy*zz^2 + Cy;
zzz= zz^3 - 2*zx*zy^2 -3*zy^2*zz + abs(Cz);

zx= zzx;
zy= zzy;
zz= abs(zzz);

Power 3 slonofractal is symmetric in Y axis, but not in Z and X axis. In X>0 side fractal have common features with 2th power shape, but dissapointingly lacks side bulbs. Well didn't found mandelbrot shape in inside rendering as I expected. But the spot showing mandelbrot shapes, if switched to cube power, shows something similar in shape. (Colour mathod is my version of exponent smoothing.) 



In next multiplication step I got 4th power slonofractal:

Code:

zzx= zx^4 +2*zy^4 -9*zx^2*zy^2 +4*zx^3*zz +6*zx^2*zz^2 -16*zx*zy^2*zz +4*zx*zy^3 -6*zy^2*zz^2 +Cx;
zzy= 4*zx^3*zy +12*zx^2*zy*zz -8*zx*zy^3 +12*zx*zy*zz^2 -8*zy^3*zz +4*zy*zz^3 +Cy;
zzz= zz^4 +2*zy^4 -3*zx^2*zy^2 -8*zx*zy^2*zz +6*zy^2*zz^2 +abs(Cz)

zx= zzx;
zy= zzy;
zz= abs(zzz);

Power 4 fractal is pretty funny, reminds some thing, well but this maybe is just my imagination. Outer bulbs are curved in Z axis, each in different direction. Fractal looks particulary long, but in middle it is much shorter. IMHO too smooth.



Then I got 5th power slonofractal formula:

Code:

zzx= zx^5 +12*zx*zy^4 -16*zx^3*zy^2 +5*zx^4*zz +10*zx^3*zz^2 -45*zx^2*zy^2*zz +10*zx^2*zz^3 -28*zx*zy^2*zz^2 +5*zx*zz^4 +10*zy^4*zz -10*zy^2*zz^3 + Cx;
zzy= 4*zy^5+5*zy^4*zy +20*zx^3*zy*zz -20*zx^2*zy^3 +30*zx^2*zy*zz^2 -40*zx*zy^3*zz +20*zx*zy*zz^3 -8*zy^3*zz^2 +5*zy*zz^4 +Cy;
zzz= zz^5 -4*zx^3*zy^2 -15*zx^2*zy^2*zz +8*zx*zy^4 -20*zx*zy^2*zz^2 +10*zy^4*zz +2*zy^2*zz^3 +abs(Cz);

zx= zzx;
zy= zzy;
zz= abs(zzz);

Power 5 fractal is much more interesting if compared to previous, the most symmetrical of all, a bitt like power 2 and power 4 shapes, but very unlike of power 3 shape. This is kind of interesting, as if there were any mistake in multiplication, all following powers would contain that mistake and distorted shapes, but here it is even more symmetric than power 2, and do have side bulbs.


Inside render of julia set:



And power 6 slonofractal:

Code:

zzx= zx^6 -4*zy^6 +40*zx^2*zy^4 -25*zx&4*zy^2 +6*zx^5*zz +15*zx^4*zy^2 -96*zx^3*zy^2*zz +20*zx^3*zy^3 -123*zx^2*zy^2*zz^2 +15*zx^2*zz^4 +72*zx*zy^4*zz -56*zx*zy^2*zz^3 +6*zx*zz^5 +18*zy^4*zz^2 -15*zy^2*zz^4 +Cx;
zzy= 24*zx*zy^5 +6*zx^5*zy +30*zx^4*zy*zz -40*zx^3*zy^3 +60*zx^3*zy*zz^2 -120*zx^2*zy^3*zz +60*zx^2*zy*zz^3 -96*zx*zy^3*zz^2 +30*zx*zy*zz^4 +24*zy^5*zz -16*zy^3*zz^3 +6*zy*zz^5 +Cy;
zzz= zz^6 -4*zy^6 -5*zx^4*zy^2 -24*zx^3*zy^2*zz +20*zx^2*zy^4 -45*zx^2*zy^2*zz^2 +48*zx*zy^4*zz -40*zx*zy^2*zz^3 +18*zy^4*zz^2 -3*zy^2*zz^4 +abs(Cz);

zx= zzx;
zy= zzy;
zz= abs(zzz);

Two or three antennas, irregular as is power 4.


Power 7 is overkill:

Code:

zzx= zx^7 +100*zx^3*zy^4 -36*zx^5*zy^2 +7*zx^6*zz +21*zx^5*zy^2 -175*zx^4*zy^2*zz +35*zx^4*zy^3 -324*zx^3*zy^2*zz^2 +35*zx^3*zz^4 +280*zx^2*zy^4*zz -279*zx^2*zy^2*zz^3 +21*zx^2*zz^5 -32*zx*zy^6 +204*zx*zy^4*zz^2 -104*zx*zy^2*zz^4 +7*zx*zz^6 -28*zy^6*zz +34*zy^4*zz^3 -21*zy^2*zz^5 +Cx;
zzy= 84*zx^2*zy^5 -8*zy^7 +7*zx^6*zy +42*zx^5*zy*zz -70*zx^4*zy^3 +105*zx^4*zy*zz^2 -280*zx^3*zy^3*zy*zz +140*zx^3*zy*zz^3 -376*zx^2*zy^3*zz^2 +105*zx^2*zy*zz^4 +168*zx*zy^5*zz -208*zx*zy^3*zz^3 +42*zx*zy*zz^5 +60*zy^5*zz^2 -34*zy^3*zz^4 +7*zy*zz^6 +Cy;
zzz= zz^7 -24*zx*zy^6 -6*zx^5*zy^2 -35*zx^4*zy^2*zz +40*zx^3*zy^4 -84*zx^3*zy^2*zz^2 +140*zx^2*zy^4*zz -105*zx^2*zy^2*zz^3 +144*zx*zy^4*zz^2 +70*zx*zy^2*zz^4 -28*zy^6*zz +34*zy^4*zz^3 -11*zy^2*zz^5 + abs(Cz);

zx= zzx;
zy= zzy;
zz= abs(zzz);

In power 7 fractal is not very symmetric as I expected with symmetric power 5. There could be some algebraic mistakes. (EDITED: I found the mistake, and now it is a bitt symmetrical.) Cutout in z=-0.15 do shows mandelbrot like fractal with lesser bulbs around main set and indeed fractal is different in all cutouts even outside don't look anywhere like mandelbrot set.




https://sites.google.com/site/3dfractals/discussion
p.s.
Added Chaos pro savefile with these spots.