Re: [math-fun] Exact rational 3D spacefill
Jason>
I don't recall if someone already posted this sculpture of trilbert: http://blog.makezine.com/archive/2010/11/math_monday_3d_hilbert_curve_in_ste...
An 8^3 (not a 4^3, Steve) in plastic would be delicate, at best. The entire weight is everywhere supported by the thickness of a single pipe, often far offcenter. Neil Bickford built a 4^3 from construction cubes that you dasn't breathe on. And the illustrated 8^3 is a quivering mass of robot intestines. (Amazingly organic, for such a seemingly crunchy and artificial concept.) Neil was investigating the connection of Trilberts with Morse (= Stutter Free = "Square" Free) sequences, and found the spacefilling property enormously underconstraining. There seem to be 2^(n+1)-1 cubes of order n, corresponding to all the ways you can reflect a subcube in the symmetry plane that contains the edge it spans. In fact, Corey's first exact map had the inverse image of (1,1,1) at 85/126 instead of 19/28. I probably missed something, but am having difficulty finding a stutter-free (no straight angles) 8^4. It seems to require a lot of searching and manual reflecting instead of a nice, clean recursion. --rwg That's no hamster--it's an illegal gerbil! Get him!
On Thu, Jan 6, 2011 at 1:59 PM, Bill Gosper <billgosper@gmail.com> wrote:
Jason>
I don't recall if someone already posted this sculpture of trilbert:
http://blog.makezine.com/archive/2010/11/math_monday_3d_hilbert_curve_in_ste...
An 8^3 (not a 4^3, Steve) in plastic would be delicate, at best. The entire weight is everywhere supported by the thickness of a single pipe, often far offcenter. Neil Bickford built a 4^3 from construction cubes that you dasn't breathe on. And the illustrated 8^3 is a quivering mass of robot intestines. (Amazingly organic, for such a seemingly crunchy and artificial concept.)
Neil was investigating the connection of Trilberts with Morse (= Stutter Free = "Square" Free) sequences, and found the spacefilling property enormously underconstraining. There seem to be 2^(n+1)-1 cubes of order n, corresponding to all the ways you can reflect a subcube in the symmetry plane that contains the edge it spans. In fact, Corey's first exact map had the inverse image of (1,1,1) at 85/126 instead of 19/28. I probably missed something, but am having difficulty finding a stutter-free (no straight angles) 8^4. It seems to require a lot of searching and manual reflecting instead of a nice, clean recursion. --rwg That's no hamster--it's an illegal gerbil! Get him!
Corey strikes again, this time with "Spirilbert", a recursion I don't undertand, which he has tested through 8^6. I just determined that *all* kth differences of the order 3 are free of {0,0,0}. Here's the order 4<http://gosper.org/CTrilbert.png> . --rwg
Please define your notation here: eg. does "8^6" mean edge 2^8 in 6-D space? Several earlier such constructions of this nature had poor locality, compared to the classic Hilbert walk in 2-space. Have you looked at how well these curves perform in this respect? WFL On 1/9/11, Bill Gosper <billgosper@gmail.com> wrote:
On Thu, Jan 6, 2011 at 1:59 PM, Bill Gosper <billgosper@gmail.com> wrote:
Jason>
I don't recall if someone already posted this sculpture of trilbert:
http://blog.makezine.com/archive/2010/11/math_monday_3d_hilbert_curve_in_ste...
An 8^3 (not a 4^3, Steve) in plastic would be delicate, at best. The entire weight is everywhere supported by the thickness of a single pipe, often far offcenter. Neil Bickford built a 4^3 from construction cubes that you dasn't breathe on. And the illustrated 8^3 is a quivering mass of robot intestines. (Amazingly organic, for such a seemingly crunchy and artificial concept.)
Neil was investigating the connection of Trilberts with Morse (= Stutter Free = "Square" Free) sequences, and found the spacefilling property enormously underconstraining. There seem to be 2^(n+1)-1 cubes of order n, corresponding to all the ways you can reflect a subcube in the symmetry plane that contains the edge it spans. In fact, Corey's first exact map had the inverse image of (1,1,1) at 85/126 instead of 19/28. I probably missed something, but am having difficulty finding a stutter-free (no straight angles) 8^4. It seems to require a lot of searching and manual reflecting instead of a nice, clean recursion. --rwg That's no hamster--it's an illegal gerbil! Get him!
Corey strikes again, this time with "Spirilbert", a recursion I don't undertand, which he has tested through 8^6. I just determined that *all* kth differences of the order 3 are free of {0,0,0}. Here's the order 4<http://gosper.org/CTrilbert.png> . --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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