Richard Smith richard at ex-parrot.com
Fri Dec 3 18:36:07 UTC 2004

```Peter King wrote:

> things like 3254 (2 lead course) or 3542 (3 lead course)
> which leaves just two families (the labelling I am using
> is out of simplicity and is not related to the usual
>
> a 3524
> b 5432
> c 4253

Except that b has a 2-lead course -- it is a differential
hunter.

> and the cyclic ones
> d 3452
> e 4523
> f 5234

Likewise, e has a 2-lead course.

Because the lead ends form the whole of a conjugacy class
for the
> So there are 6 ie (n-2)! exactly as you said.

As PABS points out, you are missing

4532
5423

Removing b and e and replacing them with these two gives the
a non-differential (n-1)-lead course, there are (n-2)!
despite the fact that four is not prime.

> The question of symmetry is an interesting one I hadn't
> thought about previously but then if I restrict myself to
> certain symmetry types what does the answer become (so eg
> is irregular.

Let's first focus on single-hunt even bell methods with
palindromic symmetry (i.e. "normal" methods).  As per my
previous email, the number of lead ends is (n-1)(n-3)...3.
Let's assume that all of these can be obtained (this is not
always true: certainly for Little methods there is not space
to reach all of the lead ends).

We require the lead head to be an (n-1)-cycle.  This must be
reached from the lead end via a change containing exactly
two places, one of which is in first place (the treble).
There are n/2 such changes.  (This is including lead end
changes that would usually be though of as bobs, such as 14
in minor).  This gives (n/2).(n-1)(n-3)...3 lead-end /
(n-1)-cycle.

Concentrating on a specific lead end change (say 12), what
end?  Let's draw the change 12 (on six bells) as follows:

1  2  3  4  5  6
o  o  o--o  o--o

That is, with lines joining swapping pairs.  And let's
look at one possible lead end row, 153624, which we can draw

1  2  3  4  5  6
_____
|     |
o  o  o  o  o  o
|________|

If we combine these two pictorial representations, we get

1  2  3  4  5  6
_____
|     |
o  o  o--o  o--o
|________|

which is a line passing through all five working bells.
When we multiply the lead end row 153624 by the lead end
show that all (n-1)-cycles (where n is even) correspond to
lines like this.

So, given the change,

1  2  3  4  5  6
o  o  o--o  o--o

how many ways of turning this into a 5-cycle are there?
First we need a line from 2 to one of the other (n-2)
working bells; this bell will already have a line to another
bell.  We now must join this to one of the (n-4) remaining
working bells, and so on.  This gives (n-2)(n-4)...2

Allowing any lead end change, there are (n/2).(n-2)(n-4)...2
closed formula, 2^(n/2-1) (n/2)!.  Tabulating the first few
values:

Bells   Le/lhs
4            2
6           24
8          192
10        1920
12       23040
14      322560
16     5160960

> DoubleCambridge cyclic hasa different
> fall out). Once again Ipresume the answer you gave
> previously holds if n-1 is prime.

possible.

> Also is it possible to have an odd bell cyclic method with
> the usual symmetry

Only on five bells -- not on more.

> Finally there doesn't appear to be a classification
> beyond those you mentioned. So I can't find any standard
> other irregular ones).

As you can see from the table above, it wouldn't be feasible
to have codes for irregular methods at higher stages.  Yes,
you could have two-letter codes for Major, three-leter ones
for Royal, etc., but this doesn't strike me as particularly
useful.

When and if cyclic methods become more popular, we might
start to see the adoption of some naming convention for