holroyd at math.ubc.ca
Fri Sep 8 03:02:44 UTC 2006
This example raises a puzzling (to me) general issue. The method I give
here clearly has 15 plain courses which make up the extent, and the bob
gives q-sets of size 3 which enable one to reach any course. Since 15 is
odd, there is no parity obstacle to joining all the courses together, and
indeed this turns out to be possible (after some tedious searching).
But does anyone know general conditions under which such a thing is
possible (where "such a thing" means roughly: joining an odd number of
blocks to form an extent using q-sets of size 3)? Was it obvious
immediately that it would work in this case? And does anyone know an easy
way to find such a joining other than lots of trial and error?
(Sorry if this isn't very clear, but I don't know exactly what the correct
On Thu, 7 Sep 2006, Alexander Holroyd wrote:
> Here is the composition again in case the columns got messed up:
> On Thu, 7 Sep 2006, Alexander Holroyd wrote:
>> Very impressive, Richard. It _is_ possible on 6 (and I would guess on all
>> higher numbers). Here is an example (which I think it is safe to say
>> no-one in their right mind would ring).
>> p = 234.3188.8.131.52.3184.108.40.206.345.234.5
>> - = 234.3220.127.116.11.318.104.22.168.345.234.5
More information about the ringing-theory