Im almost used to the routine by now, which is that every time I believe I have the Tset formula written down properly, I see a way to refine it. With this latest and last little lick though, I think it's done My first efforts to resolve the set in 2D, were mostly to demonstrate its existence, and hopefully inspire someone to render it in 3D. I used an HP SX48 calculator, and it was a simple matter to construct a program with the 3D vectors that are built-in. The skewed, truncated Msets that this method produced seemed a reasonable result, although not quite what Id hoped for. Then, a few weeks ago, Tim Wegner directed my attention to the Fractint list and the Morgan Owens formula posted on 12/11/02. At the time, I was still focused on a full 3D rendering, so I wasnt that excited about seeing a 2D set in Fractint. Besides, the Owens formula looked about right, but it produces images quite distinct from those on the HP, and I thus suspected that the Fractint parser wasnt up to the job. But just for drill I guess, I tweaked the formula to bring it a little closer to the HP version and voila the images Id found earlier were there. Well that must be right, I thought, reckoning that if one machine confirmed the other, how could it be otherwise but in a short while I saw that the imaginary value for c2 (which the HP's vectors had snuck in on me) was not appropriate in this case. The simple reason is that triternions are based on the order six cyclic group, and there are no 4th roots of unity in C6 (or for that matter, in the other order six group, D3). I saw that simply squaring c2 would rectify the imaginary term, and the images that this change produced were far more intricate and fun to work with. Moreover, Jim Muths flip of c1 produced a nice two-fold symmetry (and incidentally z1=t1-c1*c1 works well with that too), so it looked like wed arrived. Upon further reflection, however, it became clear that in c2*c2, the term also acts upon itself as a scalar, and that this disrupts its proportionality with c1. The proper solution then is c2*(0,1) or (better) c2=imag(pixel) *(0,1) in the definition line, to minimize calculations. There it is then: Weve watched the Tset go from embryo to fetus to now a newborn, and guess what Cuzzins, Its A Girl! You can go to p.12 of my paper, where shes posing for a picture, dressed all pretty in pink: http://fibonacci-arrays.com/Triternions.pdf Or you can render her yourself with TGirl { c1=real(pixel),c2=imag(pixel)*(0,-1),c3=p1 z1=z2=z3=0: t1=z1*z1+2*z2*z3, t2=z3*z3+2*z1*z2, t3=z2*z2+2*z3*z1 z1=t1+c1,z2=t2+c2,z3=t3+c3 z=z1+z2+z3 z < 64 } TGirl is easily persuaded to produce Julia sets: TJul { z1=real(pixel),z2=imag(pixel)*(0,1),z3=p3: t1=z1^p1+2*z2*z3, t2=z3^p1+2*z1*z2, t3=z2^p1+2*z3*z1 z1=t1+p1,z2=t2+p2,z3=t3+p3 z=z1*z2+z2*z3+z3*z1 z < 64 } Try, say, z1=-0.75 with z2 and/or z3 at or near 0.1. Looking forward to seeing where this goes Ciao, Russell _____________________________________________________________ Get 25MB, POP3, Spam Filtering with LYCOS MAIL PLUS for $19.95/year. http://login.mail.lycos.com/brandPage.shtml?pageId=plus&ref=lmtplus