Title: 3d Hypercomplex Fractal
Post by: jhazelden on June 19, 2016, 03:32:39 PM
Hi all. I would like to outline a fractal I thought up recently that, as far as I know, has not been discovered. The fractal is generated using a hypercomplex number system with dimension of 3, allowing for a 3d fractal. In this system, numbers are represented in terms of three basis vectors 1, i, and j. I defined mutiplication of these basis vectors in the following table:
Using this definition of multiplication, here's what the product of two number in this system, say a+bi+cj and d+ei+fj, would look like:
(d+ei+fj) = (ad+bf+ce) + (ae+bd+be)i + (af+cd+cf)j$)
.
It is easy to prove that this system is distributive, making it hypercomplex, along with also being associative and commutative.
Finally, I applied the mandelbrot transformation:
, where c was the current coordinate in 3d, and colored it based on how long it took to grow outside of a box, just like the mandelbrot.
The end result is in this video:
https://www.youtube.com/watch?v=gbqG23NxaAY
I'm rendering it in openGL by cutting out planes so it doesn't look so nice right now, but I would like to, eventually, render it with volume rendering. If anyone feels so inclined: I would love seeing what the fractal looks like
rendered in a professional program. Anyways, I just thought I would share this fractal because I haven't seen one generated this way, but please correct me if I'm wrong. The method I used could be changed tons
because of all the different permutations possible with the multiplication definition and the method of determining the c constant, just like with the mandelbrot.
Please tell me what you think, or provide some better images of the fractal, if you like :).
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It is easy to prove that this system is distributive, making it hypercomplex, along with also being associative and commutative.
Finally, I applied the mandelbrot transformation:
The end result is in this video:
https://www.youtube.com/watch?v=gbqG23NxaAY
I'm rendering it in openGL by cutting out planes so it doesn't look so nice right now, but I would like to, eventually, render it with volume rendering. If anyone feels so inclined: I would love seeing what the fractal looks like
rendered in a professional program. Anyways, I just thought I would share this fractal because I haven't seen one generated this way, but please correct me if I'm wrong. The method I used could be changed tons
because of all the different permutations possible with the multiplication definition and the method of determining the c constant, just like with the mandelbrot.
Please tell me what you think, or provide some better images of the fractal, if you like :).
Title: Re: 3d Hypercomplex Fractal
Post by: lkmitch on June 20, 2016, 04:37:37 PM
Hi,
This idea has been explored before, but I'm not aware of any videos of it. If I remember correctly, it resembled other 3d systems and wasn't all that interesting for the regular Mandelbrot set. But then again, I don't think it was explored that much.
Kerry
This idea has been explored before, but I'm not aware of any videos of it. If I remember correctly, it resembled other 3d systems and wasn't all that interesting for the regular Mandelbrot set. But then again, I don't think it was explored that much.
Kerry
Title: Re: 3d Hypercomplex Fractal
Post by: hgjf2 on June 21, 2016, 10:08:56 AM
This fractal don't look as fractal. This function is simmilar with the Henon map.
Answer: for the hypercomplex a+bi+cj if (a+bi+cj)*(d+ei+fj)=(ad+bf+ce)+(bd+ae+be)i+(cd+af+cf)j, the plane section for j=0.
This function is newx=x*x+cx and newy=y*y+2*x*y+cy with x=newx and y=newy for each iteration, but this 2D transformation isn't holomorph.
Also the Henon map isn't holomorph. :spork:
Neither for plane section for i=0 that having same 2D transformation. But for real number give 0 (a=0 and d=0) having bf+ce+bei+cfj more complicated due bi+cj isn't algebrical subgroup.
Answer: for the hypercomplex a+bi+cj if (a+bi+cj)*(d+ei+fj)=(ad+bf+ce)+(bd+ae+be)i+(cd+af+cf)j, the plane section for j=0.
This function is newx=x*x+cx and newy=y*y+2*x*y+cy with x=newx and y=newy for each iteration, but this 2D transformation isn't holomorph.
Also the Henon map isn't holomorph. :spork:
Neither for plane section for i=0 that having same 2D transformation. But for real number give 0 (a=0 and d=0) having bf+ce+bei+cfj more complicated due bi+cj isn't algebrical subgroup.