[r-t] Methods as polyhedra

King, Peter R peter.king at imperial.ac.uk
Wed Apr 9 14:52:39 UTC 2008

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

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 Archimedean
> 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 
> vertices, 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??
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