[r-t] Long lengths of Bristol Surprise Major
holroyd at math.ubc.ca
Thu Jul 14 00:39:12 UTC 2016
On Wed, 13 Jul 2016, Joe Norton wrote:
> Given such a set of mutually true leads of a given method, would it be sufficient
> to find a Hamiltonian cycle through them to get a composition containing all the
> given leads? That is, if they were joined in a graph by edges representing
> possible lead-ends.
> Also, if it could be shown that no such cycle exits then would it follow that no
> composition can exist for the set of leads?
Basically yes. If you allow singles (i.e. calls that change the parity of
the lead end) then there is the additional complication that a lead can be
rung backwards. Such a composition corresponds to a cycle in a larger
graph that has a pair of vertices for each lead (corresponding to ringing
it forwards or backwards). The required condition on the cycle is not
Hamiltonicity but that it visit exactly one vertex from each pair.
And could anything be inferred as to
> whether another composition with the same number of leads could or could not be
No. There could be other sets of mutually true leads of the same or
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