[r-t] Emergency peal composition
holroyd at math.ubc.ca
Mon Dec 17 21:59:30 UTC 2012
Indeed. Mark's proof seems right, but assumes a plain hunt lead end if I
understand correctly. Can all 13 leads be got by mixing plain lead ends
(and lead end orders), and/or using half-lead calls (in methods with
regular half leads)?
On Sun, 16 Dec 2012, Mark Davies wrote:
> PABS writes,
>> However, if we move 3 then 2, adding a lead in between, it is possible
>> to get all 13 leads:
> But then we have all 13 leads rung forwards, with a 14th lead which is not
> part of the course. That was not at all Ander's idea!
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