[r-t] Double Cambridge cyclic bob Minor
Richard Smith
richard at ex-parrot.com
Wed May 18 21:49:45 UTC 2011
Jonathan Agg wrote:
> I've been trying to get a touch of Double Cambridge Cyclic bob minor suitable
> for a quarter peal (I read an archive for this list that said an extent was
> impossible) so was instead trying to find a shorter length e.g. 540.
Sorry to be negative, but I think you'll be very lucky to
find something that long -- the method is very false.
> Can anyone prove why an extent is not possible?
That's an easier question. There's a simple algorithm for
determining whether you can get a fixed-treble extent of an
arbitrary asymetric plain method. (Yes, Double Cambridge
has glide symmetry, but that doesn't invalidate the
algorithm.)
On n bells, there are (n-1)! rows that have the treble
leading. For an extent to be possible, we need to find
(n-1)!/2 mutually true leads -- in other words, we need to
use half of the (n-1)! rows as lead heads and the other half
as lead ends.
First work out the false lead heads for the method. For
Double Cambridge they are:
123456
135426
136452
152436
162453
164352
We can ignore the first one as that's trivial falseness.
The second one is the anticipated "lead is false against
itself backwards" falseness. The others are the critical
ones. So for each lead, there are five others that are
false against it.
Imagine you have a large piece of paper and you write all
120 potential lead heads down on it. Now imagine drawing
lines between each pair of false rows. You would now have
300 lines joining the rows. The get two coloured pens --
say red and green. Colour one of the rows red. Now colour
green all of the rows that are linked to a red row. Then
vice versa: colouring red all of the rows linked to a green
row. Stop when either (i) you have two rows of the same
colour linked by a line, or (ii) you have no uncoloured
lines left.
If you've successfully coloured all the rows, then all of
the red rows (or, indeed, all of the green rows) are a
mutually true set of lead heads. If you've ended up with
adjacent rows of the same colour, you've proved it
impossible. In the case of Double Cambridge, it's the
latter.
Or if you prefer to think of it more mathematically, then
let M be the set of rows in the first lead (from lead head
[123456] to lead end [135426] inclusive). Define the set M'
(I usually write it M with an over-bar if the medium permits
it) to be the set of inverses of elements of M:
M' = { m^-1 : m in M }
And define set multiplication in the obvious way:
A B = { a b : a in A, b in B }
The set of false lead heads for the method, which I
typically notate F(M), can be expressed:
F(M) = M M'
Let S_{n-1} be the group of rows on n bells with the the
treble leading.
For a fixed-treble extent to be possible, it is necessary
for the Cayley graph, Cay(S_{n-1}, F(M)), to be bipartite.
RAS
More information about the ringing-theory
mailing list