[r-t] Emergency peal composition
mark at snowtiger.net
Sat Dec 15 19:24:30 UTC 2012
> Good work, Mark. Do you have a proof that all 13 leads aren't possible?
Well, I only considered the two possible compositions using two big bobs
and no other calls. One of these moves the 2 and 3 amongst the back
bells to give a 4567890ETAB part end, the other moves A and B to the
front to give the part end used in my final composition.
Since the composition is primarily made up of plain leads, but some of
these are rung backwards, it is convenient to count up the pairs of full
treble leads. I numbered these from 0 to 12, so that, in the plain
course, 0 is the pair (132547698E0BTA, 1234567890ETAB), 1 is
(1537294E6A8B0T, 13527496E8A0BT) and so on. You can then count up the
pairs used in the composition regardless of which method is being rung
(leadhead order doesn't matter since we are effectively counting the
changes on either side of a plain lead) and regardless of whether we're
in a forward or reverse bit.
Of course the 18 calls do split the pairs up. In the "AB" arrangement,
the first call splits pair 3, so that only the handstroke row is rung,
but fortunately the second call brings us back to the same position in
the course, so the pair 3 backstroke is collected. But, interestingly,
the same thing does not occur with the backstroke/handstroke treble
leads on the other side of the call. The first big bob gives us a
backstroke lead which is in fact the handstroke lead of pair 9; just
before the second bob we get the backstroke lead of pair 10.
Since pairs 9 and 10 are split, without introducing further calls, we
can never ring the other half of them. So one full lead must be missed
out of the course.
The same reasoning applies to the "23" arrangement. Here the reversed
section between the calls does begin and end with the same lead pair,
but the forward-hunting bit doesn't: the first big bob splits a 9 lead,
and the second splits a 10. So the same reasoning applies: 12 leads out
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