robin at robinw.org.uk
Thu Apr 23 14:49:36 UTC 2015
Alan Reading said "I think I am correct in saying that if you start with
any FCH and apply these techniques for all plain bob lead heads and ends
then you end up with all other FCHs that *must* also be present (i.e.
all the other ones that fall under the same letter)."
Most like, because it is certainly true that every FCH of a group can be
generated from just the pair of corresponding rows in the 1st half-lead
of a symmetric method. Take Cambridge, the first corresponding rows
12345678 and 12463857. Invert the second and multiply this on the left
of the first - trivially giving 12537486 - call this X. If J = 12436587,
form JX, XJ and JXJ and each inverse. Multiply these on the left of the
seven lead ends gives 56 false leads which can be rotated to lead-heads
(multiply succesively by the method's 1st lead head). For each of these,
transpose by 13254768 to give 56 lead-heads in total.
These resolve themselves into distinct sets of 14, 28 or 56 of which
half are in course and half out of course. As we know, the first FCH of
Cambridge we generate is 2436578 and this is the only one of the set
with in-course tenors-together falseness. The out-of-course falseness
has 4 tenors-together elements. Group E has 14 distinct in-course
members but just two are tenors-together, for example.
PS "Multiply .. on the left of the seven leads" means "operate on the
seven leads by ..." in my notation.
PPS - I think this is right - I haven't thought about it for 10 years or
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the ringing-theory