split-complex ternary algebra xD

[kujonai]

With a triplex  a + qb + pc   ←→  (a,b,c)   with a,b,c belong to z (complex)

      q^1=+q      q^2=$p      q^3=$1
      q^4=$q      q^5=#p      q^6=#1
      q^7=#q      q^8=+p      q^9=+1

and  $1=(-1+i*sqrt(3))/2     #1=(-1-i*sqrt(3))/2  

..example for the simple iteration F(n+1)=F(n)^3+k  
with F(0)=0 and k belongs to z ....
or how and in what program can i to make this 3d fractal? some suggestions?

good luck to this new year

[David Makin]

With a triplex  a + qb + pc   ←→  (a,b,c)   with a,b,c belong to z (complex)

      q^1=+q      q^2=$p      q^3=$1
      q^4=$q      q^5=#p      q^6=#1
      q^7=#q      q^8=+p      q^9=+1

and  $1=(-1+i*sqrt(3))/2     #1=(-1-i*sqrt(3))/2  

..example for the simple iteration F(n+1)=F(n)^3+k  
with F(0)=0 and k belongs to z ....
or how and in what program can i to make this 3d fractal? some suggestions?

good luck to this new year

If you can write code yourself and use Windows then a free solution is write a formula for either Fractint or ChaosPro, alternatively use Ultra Fractal.
Fractint is also a solution under Linux, I'm not sure what you could use on a Mac.

Note that there are 3D/4D formulas publicly available for all three programs (for say Julibrots or quaternions etc.) that you should be able to adapt.

[jehovajah]

The problem as i see it kujonai is that 3d fractal programmes are based on quaternions or some related set of algebra. So for example to use quaternions you will need to define your p in terms of j. This then may alter your mod (9) ring for q. Unless you can design a programme using a 3d plotter, you will have to adapt your vectors to quaternions with the appropriate scale factors.

[Timeroot]

With a triplex  a + qb + pc   ←→  (a,b,c)   with a,b,c belong to z (complex)

      q^1=+q      q^2=$p      q^3=$1
      q^4=$q      q^5=#p      q^6=#1
      q^7=#q      q^8=+p      q^9=+1

and  $1=(-1+i*sqrt(3))/2     #1=(-1-i*sqrt(3))/2  

..example for the simple iteration F(n+1)=F(n)^3+k  
with F(0)=0 and k belongs to z ....
or how and in what program can i to make this 3d fractal? some suggestions?

good luck to this new year

If you can write code yourself and use Windows then a free solution is write a formula for either Fractint or ChaosPro, alternatively use Ultra Fractal.
Fractint is also a solution under Linux, I'm not sure what you could use on a Mac.

Note that there are 3D/4D formulas publicly available for all three programs (for say Julibrots or quaternions etc.) that you should be able to adapt.

Although there are built-in formulas for some 3D/4D fractals in FractInt, you cannot make your own 3D images of them - only 2D slices at a time.

[David Makin]

With a triplex  a + qb + pc   ←→  (a,b,c)   with a,b,c belong to z (complex)

      q^1=+q      q^2=$p      q^3=$1
      q^4=$q      q^5=#p      q^6=#1
      q^7=#q      q^8=+p      q^9=+1

and  $1=(-1+i*sqrt(3))/2     #1=(-1-i*sqrt(3))/2  

..example for the simple iteration F(n+1)=F(n)^3+k  
with F(0)=0 and k belongs to z ....
or how and in what program can i to make this 3d fractal? some suggestions?

good luck to this new year

If you can write code yourself and use Windows then a free solution is write a formula for either Fractint or ChaosPro, alternatively use Ultra Fractal.
Fractint is also a solution under Linux, I'm not sure what you could use on a Mac.

Note that there are 3D/4D formulas publicly available for all three programs (for say Julibrots or quaternions etc.) that you should be able to adapt.

Although there are built-in formulas for some 3D/4D fractals in FractInt, you cannot make your own 3D images of them - only 2D slices at a time.

I thought Fractint's compiler was sophisticated enough to write your own 3D ray-tracing code, but it is a while since I used Fractint

[Timeroot]

Nope, the Fractint compiler is pretty weak. Doesn't even allow you to write your own ray-tracing code, because you can't include loops  It's too bad, because it calculates pretty fast.

[jehovajah]

I wonder if Terry Gintz idynamasz might do the job?

[jehovajah]

(a+q*b+c*p)^3  i get

a^3+4*a*b*c +2*b*c^2+b^3*(-1+i*sqrt(3))/2+c^3*(-1-i*sqrt(3))/2

  (3*(a^2)*b+2*a*b*c +b*c^2)*p

   (3*(a^2)*c +b^2*c+3*a*c^2*(-1-i*sqrt(3))/2+(3*(a)*b^2+2*b^2*c)*(-1+i*sqrt(3))/2)*q

Could somebody, or Kujonai check this please?