[r-t] Extents in half leads - was Colin Wyld's extent of Yorkshire Major

Mike Ovenden mike.ovenden at homecall.co.uk
Fri Jun 17 23:27:48 UTC 2005

>   1. Colin Wyld's extent of Yorkshire Major (Richard Smith)
> I think I might take a look at some of these issues in a
> further email.  I've also taken a look at some other methods
> where this, or a similar, idea applies -- there are
> surprisingly many of them.  Again, more in a later email.

Excellent post; I look forward to the next instalment.  My 2 pence ...

There are two aspects: the existence of sets of half leads that cover the
extent, and joining them up.  The former has a connection with groups that's
perhaps worth setting out in more detail.  It applies to other stages too,
but I'll use Minor for a concrete example (some of us might not fancy
ringing extents of Major).

Consider a half lead in which the treble falls into each position precisely
twice.  Each treble position gives rise to two false half lead heads
(FHLHs), one being the inverse of the other.

Here are three half leads illustrating the two FHLHs arising with the treble
in 3rds place.
1   123456  f1=a.inv(b)  f2=b.inv(a)=inv(f1)
2   -1----  -1----       -1----
3   1-----  1-----       1-----
4   -1----  -1----       -1----
5   a       a.inv(b).a   b
6   ---1--  ---1--       ---1--
7   b       a            b.inv(a).b
8   ---1--  ---1--       ---1--
9   ----1-  ----1-       ----1-
10  -----1  -----1       -----1
11  ----1-  ----1-       ----1-
12  -----1  -----1       -----1

Suppose a half lead head x appears in an extent (it could actually be a lead
head or lead end).  Then x.f1 and x.inv(f1) are precluded.  But if x.f1
can't occur, what about its fifth row x.f1.a?  That MUST occur as the
seventh row of the half lead x.f1.a.inv(b) = x.f1^2.  The same thing is true
of any FHLH fi:  if x occurs as a half lead head, x.f cannot occur, and
x.f^2 must occur.

Now construct two sets, L and R, based on that observation.  L will be what
must be present; R will be what can't be.  Start L off by including rounds.
Calculate some elements of R by transposing each element of L (initially
just rounds) by each FHLH in turn.  Now calculate some new elements of L by
transposing each element of R by each FHLH in turn.  And so on, stopping
when no new elements are found.

L comprises all distinct products of an even number of FHLHs.  It clearly
satisfies the required closure property for a group, and it contains the
identity, so yes, it's a permutation group.  R comprises all distinct
products of an odd number of FHLHs, so it can be viewed as either a left or
right coset of L (in whatever), or it will turn out to be identical to L.
If L and R are identical, then the method won't yield an extent in half
leads (the same half lead heads are both required and not allowed);
otherwise there is hope.

If L and R are not identical, then their union is also a group, say G,
comprising all the distinct products of any number of FHLHs.  The group of
possible half lead heads is S(n-1), i.e. keeping treble fixed and permuting
the rest.  S(n-1) now neatly divides up into G and its cosets.  Each coset
of G can be split into two cosets of L that are mutually exclusive.  Hey
presto, a recipe for generating sets of half leads that cover the extent.

In the case of Yorkshire S Major, L is a representation of A6.  I suppose
there are likely to be Major methods where L is considerably smaller that
maybe afford the composer more latitude?

Back to treble dodging or treble place Minor.  The problem of getting an
extent is already solved, trivially, for methods with conventional parity
structure.  If you're willing to consider extents in half leads, it's also
possible to obtain extents for a few other methods.

Here's a delightful example : x34x14.56.12x16x12x36 lh16

In this case L is {123456,123465,134256,134265,142356,142365}

s s 163254
s   123465
s s 125643
s s 156324
  s 134256
Half lead singles are in the first column
All singles are 1236

The problems with these unconventional methods are
- some of the possibilities are invalid due to falseness in the plain course
- sometimes the half leads can't actually be joined in any reasonable way
- most have the notation 56 (for treble dodging anyway)
- difficult compositions

But there are some half decent possibilities amongst the dross; Double
Oxford Treble Place Minor for example (IMO!)


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