[r-t] Methods as polyhedra

King, Peter R peter.king at imperial.ac.uk
Wed Apr 9 18:54:06 UTC 2008

Yes but given the limted number of calling positions/course I am not
convinced the polyhedron is actually that huge. My example used only Ws
& Hs so each node that I had is really a line. As you say for Major,
tenors together each of the 60 nodes is a square (ie a total of 240
vertices). For royal & above there are just 3 calling positions (so 180
vertices). These are certainly complicated polyhedra, but there is a
huge amount of symemtry so it can't be that hard to construct. And once
its done it is done. The only point I was trying to make is that you
don't a vertex for each lead end (except for minor or doubles). Whether
this is useful or not I don't know.

> -----Original Message-----
> From: ringing-theory-bounces at bellringers.net 
> [mailto:ringing-theory-bounces at bellringers.net] On Behalf Of 
> Mark Davies
> Sent: 09 April 2008 18:26
> To: ringing-theory at bellringers.net
> Subject: [r-t] Methods as polyhedra
> Peter King writes,
> > You can also represent touches as polyhedra.
> You mean, as paths along the edges of polyhedra? Yes, that's 
> what we've been
> doing. The polyhedron as a whole represents the "search 
> space" - i.e. all
> the lead ends or coursing order nodes that can be reached 
> (vertices) plus
> the calls that join them (edges). That's what Hugh meant when 
> he said put a
> method onto a polyhedron; it is really the search space of 
> the method plus
> calls.
> > There are 60 in course coursing orders which are the vertices.
> Not exactly - in Minor, there are 60 in-course leadheads, 
> however if you are
> looking at tenors-together compositions on high numbers, then 
> there are
> always 60 in-course coursing orders, but there are many more vertices
> required to represent a composition. Unless you are only 
> dealing with one
> calling position (e.g. 3H your only touch!) you have to look 
> at individual
> parts of the course.
> For example, the section from W to H in coursing order 53246 
> is distinct
> from that between H and W in 53246, and so more than one 
> vertex is required
> to represent 53246. You have to split the course into 
> individual nodes. For
> a simple case with calls at M, W and H you may have just 
> three nodes per
> course, but that is still 180 vertices not 60. If allowing Befores or 
> dealing with an nth's place method, there are more.
> And sadly, I don't think representing the search space as a 
> polyhedron or
> connected graph gives you any new composition search algorithms - tree
> searches are still all you have.
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