[seqfan] Re: a question about arrangements of pennies
Jonathan Post
jvospost3 at gmail.com
Mon Dec 14 08:08:49 CET 2009
I'd commented in a 3-D geometry enumeration sequence: "Note that a
'floppy' polytetrahedron with only e2e {edge-to-edge} connections can
actually be mechanically rigid."
A119602 Number of nonisomorphic polytetrahedra with n identical
regular tetrahedra connected face-to-face or edge-to-edge (chiral
shapes counted twice).
I am still working on the associated paper, which includes 2-D, 3-D,
hyperbolic (as in A119611), and 4-D analogues. Hundreds of pages of
hand-drawn illustrations, scores of typed pages of text. "Rigidity"
is a subtle concept.
-- Prod. Jonathan Vos Post
On Sun, Dec 13, 2009 at 8:56 PM, William Keith <wjk26 at drexel.edu> wrote:
>> There are several ways to enforce rigidity. Perhaps
>> the best is to add a further constraint to the first
>> problem. Form a second graph with nodes = triangles,
>> edges = two triangles that share an edge. Then this second graph
>> must be connected.
>
> Counterexample!
>
> ****
> *****
> ** **
> * *
> ** **
> *****
> ****
>
> If that doesn't come across well graphically, picture a "mostly
> hexagon" with 4 in the first row, 5 in the second, the third row being
> 2-blank-2, the fourth row being two pennies, just one in the middle of
> each pair, and the bottom half a vertical reflection.
>
> This is mechanically rigid -- I think that's the intuition you sought
> to capture -- but the associated graph as you describe would be two
> disconnected arcs.
>
> (Which is not to say that your sequence isn't also interesting; it
> would be counting such graphs with a weight of 3 per node, minus 2 per
> edge, plus a somewhat more complicated summand accounting for the
> possibility of cycles and part-cycles of various sizes. It would be a
> "stronger" subset of rigid.)
>
> William Keith
>
>
>
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