[r-t] Methods as polyhedra
King, Peter R
peter.king at imperial.ac.uk
Wed Apr 9 15:31:56 UTC 2008
I remembered the name of the solid - it is a cuboctahedron.
> -----Original Message-----
> From: ringing-theory-bounces at bellringers.net
> [mailto:ringing-theory-bounces at bellringers.net] On Behalf Of
> King, Peter R
> Sent: 09 April 2008 15:53
> To: ringing-theory at bellringers.net
> Subject: Re: [r-t] Methods as polyhedra
> You can also represent touches as polyhedra. For simplicity,
> if you just allow bobs at W & H then there are 12 coursing
> orders. 3 H's or 3 W's can be represented by triangles.
> Attached is a picture of the flattened out polyhedron
> resulting (I think this is called a Schlegel diagram).
> I'm not entirely sure what the polyhedron is called, it has
> 12 vertices (the coursing orders), 8 triangles (4 lots of 3Hs
> and 4 lots of 3Ws) and
> 6 squares (blocks of WHWH). So I think it is a truncated
> cube, but I think it has some special name which I can't
> remember. The Hs are the blue traingles and the Ws the arrows.
> Then the problem of composition is to find closed paths
> through the vertices. For example on the second picture the
> red line is 3(H2W).
> Presumably some kind of search algorithm related to the
> travelling salesman problem could be used to select such
> paths. In the presence of falseness vsiting certain vertices
> would then preclude other vertices from the proposed path,
> but I haven't thought through how to implement this.
> Also I don't think the full polyhedron for tenors together
> compositions should be that complicated. There are 60 in
> course coursing orders which are the vertices. The faces
> would then be triangles (3Hs 3 Ws or 3Ms) and pentagons (5
> Bs) and other simple polygons for WHWH, MHMH etc. I'm sure
> some bright spark out there could construct it, perhaqps it
> has already been done.
> > -----Original Message-----
> > From: ringing-theory-bounces at bellringers.net
> > [mailto:ringing-theory-bounces at bellringers.net] On Behalf Of Mark
> > Davies
> > Sent: 08 April 2008 22:19
> > To: ringing-theory at bellringers.net
> > Subject: [r-t] Methods as polyhedra
> > Hugh Pumphrey:
> > > After making the original post I discovered that double
> > court minor also
> > > makes a truncated isosahedron.
> > Truncated icosahedron, presumably!
> > It's interesting that all of these you've named so far are
> > solids, presumably this is because these solid have the
> property that
> > all vertices radiate the same number of edges: in ringing
> terms, every
> > leadhead has the same number of connections via plains and
> calls. Can
> > we construct methods to match other such solids? According to
> > Wikipedia, ones with sensible numbers of vertices (factorials or
> > factorials/2) are:
> > Truncated cube - 3 edges per vertex, 24 vertices - but has
> an 8-sided
> > face, is this too long? Would need PBPBPBPB to be a touch.
> > Snub cube - 5 edges per vertex, 24 vertices, each vertex has two
> > 3-sided faces and one 4-sided face. So would need e.g. a
> plain course
> > of 4 leads and a bob and a single course of 3 leads each,
> where a bob
> > and a single are reverse transpositions. Is this possible? Sounds
> > unlikely. If it is, do the two chiral forms give rise to
> two different
> > but related methods?
> > Truncated dodecahedron - 3 edges per vertex, 60 vertices - contains
> > 10-sided faces, so e.g. PB*5 must be a touch. Hmm.
> > Rhombicosidodecahedron - 4 edges per vertex, 60 vertices,
> faces of 3,
> > 4 and
> > 5 sides. Sounds all right. Method?
> > Truncated rhombicosidodecahedron - 3 edges per vertex, 120
> > faces of 4, 6 and 10 sides. Hmm again.
> > Snub dodecahedron - 5 edges per vertex, 60 vertices, faces of
> > 3 and 5 sides.
> > Would need two types of call, both in-course, so not very sensible.
> > If none of these work, why not? Do any non-convex polyhedra with
> > regular vertices work? What about the Great disnub
> > dirhombidodecahedron??
> > MBD
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