[r-t] irregular leadheads
Richard Smith
richard at ex-parrot.com
Thu Dec 2 15:15:25 UTC 2004
Peter King:
> How many leadheads are there on n bells which (keep the
> treble fixed) and lead to courses of n-1 leads
Me:
> [...] So there's the answer: (n-2)!.
Peter King:
> That's what I thought and is true for n-1 prime but for Royal (for
> example) there are only 6 regular lheads (and presumably each family of
> irregular ones has only 6 members) to avoid the cycles of 3. Or am I
> wrong?
Your right that there are only 6 regular lead heads for
Royal, but this does not mean the answer is not still
(n-2)!. Try writing down the place bell order for one of
the short-course regular Royal lead heads. It will be
something like
2 -> 8 -> 7 -> 2,
3 -> 6 -> 9 -> 3,
4 -> 0 -> 5 -> 4.
That is, it is (obviously) not a single list sequence. Any
9-cycle can be written as a single sequence of place bells
starting say with the tenor. Tenths place bell can become
any of the 9 other place bells, which can become any of 8
remaining place bells, until only one remains. That place
bell becomes 10ths place bell. So the number of 9-cycle
lead heads is still (n-2)! even though there are few regular
lead heads than might be expected.
Richard
More information about the ringing-theory
mailing list