From: "Gerald K. Dobiasovsky" <gerald.dob@aon.at> Reply-To: fractint@mailman.xmission.com To: <fractint@mailman.xmission.com> Subject: Re: [Fractint] Triternions Date: Fri, 31 Jan 2003 02:01:14 +0100

But why shouldn't it be possible for the fractal's cross- section for c2=real to look identical to the cross-section for c2=imag?

Gerald, Indeed, why not? There are, never the less, some strange things going on here, but I’m starting to get a handle on a few of them… E.g., MrT: MrT { x=real(pixel),y=imag(pixel),v=p1 x1=y1=v1=0: a=x1^2+2*y1*v1 b= v1^2+2*x1*y1 c= y1^2+2*x1*v1 x1=a+x,y1=b+y,v1=c+v z=sqrt(a^2+b^2+c^2) z < 8 } It was y1=b –y*y that turned MrT into the TMan, and it’s obvious that squaring all of the Y-axis values creates symmetry across the X-axis. y=sqrt(imag(pixel))^2 does this with no distortion. Then flipping the y values into the x positions gives symmetry across Y, whence TMan1. The 3-space we’re working in here is ruled by the X,Y,V axes, but actually defined as the intersection of three planes, as is borne out in the 3D effects seen in many of the fractal images. Yes, it seems that y=imag(pixel)*(0,1), which I once perceived as removing an imaginary component, is in fact introducing a 4th degree root which is not otherwise present, and hence expanding the space to 6D… (Now I see that earlier discussions regarding this make sense…) So it feels like I’m gradually getting my mind wrapped around this stuff… I’ve also come upon a general template, a pattern by which in theory all of the proper mathematical groups are uniformly configured as fractal generators … The extended version of MrT is below. MrTe { x=real(pixel), y=imag(pixel), v=p1 x1=x2=y1=y2=v1=v2=0: a1=x1^2+x2^2+2*y1*v1+2*y2*v2 a2=2*x1*x2+2*y1*v2+2*y2*v1 b1=v1^2+v2^2+2*x1*y1+2*x2*y2 b2=2*x1*y2+2*x2*y1+2*v1*v2 c1=y1^2+y2^2+2*x1*v1+2*x2*v2 c2=2*x1*v2+2*x2*v1+2*y1*y2 x1=a1+x,x2=a2-x,y1=b1+y,y2=b2-y,v1=c1+v,v2=c2-v z=sqrt((a1-a2)^2+(b1-b2)^2+(c1-c2)^2) z < 8 } This inclusive pattern is not always necessary; e.g., MrT vis-à-vis MrTe: When certain symmetries are present in the group table, just the upper left quadrant suffices. This wasn’t enough though for the order 8 cyclic group in the earlier CP8 form; it’s expressed as CP8e below. CP8e { x=real(pixel),y=imag(pixel),v=p1,w=p2 x1=x2=y1=y2=v1=v2=w1=w2=0: a1=x1^2+x2^2+2*y1*y2+2*v1*v2+2*w1*w2 a2=y1^2+y2^2+2*x1*x2+2*v1*w1+2*v2*w2 b1=v1^2+w2^2+2*x1*y1+2*x2*y2+2*v2*w1 b2=v2^2+w2^2+2*x1*y2+2*x2*y1+2*v1*w2 c1=2*x1*v1+2*x2*w2+2*y1*v2+2*y2*w1 c2=2*x1*v2+2*x2*w1+2*y1*w2+2*y2*v1 d1=2*x1*w1+2*x2*v2+2*y1*v1+2*y2*w2 d2=2*x1*w2+2*x2*v1+2*y1*w1+2*y2*v2 x1=a1+x,x2=a2-x,y1=b1+y,y2=b2-y, v1=c1+v,v2=c2-v,w1=d1+w,w2=d2-w z=sqrt((a1-a2)^2+(b1-b2)^2+(c1-c2)^2+(d1-d2)^2) z <= 16 } If a symmetrical fractal is desired, one needs to ensure that roots of equal degree rule each half of an axis, as was done here (but while evidently always necessary, this is not sufficient). C4 is a subgroup of C8, and the CP8e frm above generates the Mset. So, progress of a sort, apparently, and I’ll post some more on this later… meantime, is either of these MrTs configurable into one of your rotational formulas, the better for us to continue this investigation? Ciao, Russ _________________________________________________________________ Tired of spam? Get advanced junk mail protection with MSN 8. http://join.msn.com/?page=features/junkmail