[r-t] Extents of Minor (was: Bobs only Erin Triples)

Philip Saddleton pabs at cantab.net
Sat Dec 5 11:04:56 UTC 2015

On 04/12/2015 20:02, Matthew Frye wrote:
>> On 4 Dec 2015, at 19:07, Ian Broster <ian at broster.co.uk 
>> <mailto:ian at broster.co.uk>> wrote:
>> > There are 26, excluding rotations.
>> I think that there are 140. I made a list once, but lost it long ago. 
>> Luckily the internet archive to the rescue!
>> https://web.archive.org/web/20050426201814/http://www-users.cs.york.ac.uk/~ianb/solid/plainbobminorbobs.text 
>> <https://web.archive.org/web/20050426201814/http://www-users.cs.york.ac.uk/%7Eianb/solid/plainbobminorbobs.text> 
> OK, taking that list as definitive, a little dab of python finds three 
> unique 360’s:
> Final list:  ['bbpbbpbpbppbbpbppbbppbbpbbpppp', 
> 'bbpbppbbppbpbppppbbpppbpbbpppp', 'bbpppbppppbbpppbppppbbpppbpppp’]
> The one I didn’t give previously can be thought of as WHW with an 
> extra Q-set, giving, eg. WFWHWWOHWIHW, where 7 leads (and HW) have 
> been removed from the first part and inserted in the third part by the 
> F/O/I Q-set. I *think* I knew about this, but forgot it because it’s 
> not a very nice compared to the other two, imho.
> Point is, I have no idea how extensive a tree-search is needed for 
> this problem, but such a strategy is probably beyond pencil and paper 
> in reasonable time? Whereas I was able to do it as a teenager in a 
> couple of hours with some basic Q-set considerations.
Here is a recipe that works for all symmetrical Plain methods with two 
pairs crossing at half-lead and lead head, and a bob affecting three bells:

Draw a graph with five vertices numbered 2-6, and join each pair of 
bells that are unaffected at a bobbed Q-set. Certain sub-graphs are 
ruled out of a 360, either in this graph or its complement (i.e. the 
plain Q-sets), corresponding to a plain course or a bob course. In the 
case of Plain Bob that means the graph cannot contain a triangle, and 
must contain at least one edge from each possible 5-cycle. Also any 
symmetry where swapping two pairs of vertices leaves the graph unchanged 
is ruled out as this would lead to a palindromic 360, which is impossible
(proof left as an exercise for the reader).

I think that, at least for methods with a 5-lead course, any remaining 
possibilities do give a 360.

A rotation or reversal corresponds to a renumbering of the vertices, and 
vice versa (this needs a bit of extra proof as the parity of the rows 
needs to be considered, but I think that all of the graphs that have not 
been ruled out have a symmetry where a single pair of vertices can be 
swapped), so find all of the graphs modulo isomorphism and you have the 
unique extents.

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