[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 

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:


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 

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.


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