# [r-t] Colin Wyld's extent of Yorkshire Major

Richard Smith richard at ex-parrot.com
Wed Jun 8 13:51:25 UTC 2005

```Last month I made some comments on Colin Wyld's composition
of Yorkshire, and said that as it has a single at every
half-lead, it not really Yorkshire.  I'm not sure where I
got this idea from, because it's not true.  (It does have a
call at every lead end, and perhaps I was thinking of this.)

Robin Woolley kindly sent me a copy of Colin's 1974 Ringing
World article on the composition, and I was struck by the
elegance of the underlying idea.

The key to the composition lies with an observation
half-lead of the plain course.  These are:-

12463857
12537486
12843657
12547683
12365478
12365847
12375486

These all have odd parity (which is unsurprising as
Yorkshire has a conventional parity structure -- that is,
all divisions either the place notation -A- or B-C, where A,
B and C are triple changes); more importantly, the 2nd is
fixed across all of the false half-leads.

For methods with a conventional parity structure (which
almost all commonly-rung methods, except Cambridge, have),
if you ring every half-lead in-course, you get enough
mutually true leads to build an extent.  Unfortunately, as
convention methods have a triple change at the half-lead and
lead-end, this requires some sort of parity-swapping call at

In Yorkshire you can avoid this.  Because each false
half-lead has the 2nd fixed, the extent divides into seven
sets of half-leads depending on which bell is 2nds place
bell.  The parity of each of these sets can be chosen
independently.  Colin chooses to ring those half-leads with
2,3 as 2nds place bell in-course, and the remaining ones
out-of-course.

The next question is how to join the half-leads together.
In Colin's words,

| If we examine the last full lead of a plain course of
| Yorkshire we find two changes, both HLEs [meaning lead
|
|   12436587
|   12345678
|
| The first is out of course and the second in course. Only
| one of these is allowed by our previous rule about the
| can not be used.  At the full lead we use either a bob or
| a single.  At the half lead we use either a plain or a
| single.  A single can be called anywhere since it returns
| the 2nds place belt of the previous half lead to that
| position.  In the next half lead and produces the correct
| type of HLE.  Singles can only be omitted where the 2nds
| place bell of the half lead produced is of opposite type
| to the 2nds place bell of the previous half lead (i.e.,
| their HLEs are of opposite nature: one is in course, the
| other out of course).

Assuming we choose to omit half-lead singles whereever
possible, this gives a set of rules for generating round
blocks starting from a given row.  It turns out that there
are three distinct types of block depending on the position
of 2,3.

The A block     The B block     The C block

12345678        12436578        12453678
------------    ------------    ------------
- 13578264      s 16478253    s - 14278365
s 18564327    s s 16453827    s - 12465837
s s 18527436      - 14327685      s 15437286
s s 18536742    s - 13485762    s s 15486723
s s 18542673      s 15462378    s s 15423678
- 15273864      - 14278536      - 14378562
s 13264587    s - 12436857    s - 13462857
------------    ------------    ------------
x 6             x 3             x 4

10 A blocks, 20 B blocks and 15 C blocks contain all the
rows in the extent.  These are joined with singles into a
five-part arrangement.  (Bobs cannot be used as the Q-set
has three different bells as 2nds place bell.  As a bob has
even parity, you would need to make the sure that each pair
of bells running in and out together were of opposite
parities, and this clearly is not possible.)

Given the numbers of each type of block, a five-part
composition seems obvious.  The next question is how to
produce a five-part structure from the six-, three- and
four-part blocks available.  Colin's composition does this
by joining five A blocks into a single five-part structure
(which I've called an A' block).  This he does using ten
extra half lead singles (marked with capital letters,
below).

The A' Block

1   2   3   4   5   6   7  12345678
-------------------------------------
-   s s s s s s s   -   s  13264587
S s   -   s s s s s s s   -  17352648
s   -   s s s s s s s   -  18263574
s   -   s s s s s s s   -  14352687
s   -   s s s s s s s   -  17263548
S -   s s s s s s s   -   s  12356784
-------------------------------------
x 5

Unsurprisingly, the ten joining singles form five Q-sets:

46573821  64758321
64753821  46578321

57683421  75864321
75863421  57684321

68743521  86475321
86473521  68745321

74853621  47586321
47583621  74856321

85463721  58647321
58643721  85467321

Choosing the five Q-sets is simple.  Choose your part end
(12356784 in this case) and choose a Q-set (e.g. the first
one above).  Then work out what this Q-set becomes in the
other four parts.

Finally, in each part, single in the remaining A block, the
four B blocks and the two C blocks and you have an extent.
(Colin's composition actually starts with an A block and
singles one part from the A' block into it, but the idea is
the same.)

Here is his composition.  I have enclosed the various
singled-in blocks in paretheses and capitalised the singles
used to insert these blocks.  I've also added a comment
saying what the singled-in block actually is and whether the
single replaces a bob or a plain.

40,320  Yorkshire Surprise Major
Comp.  Colin J.E. Wyld

1   2   3   4   5   6   7  12345678
-------------------------------------
S(s -   s s s s s   - s -  12563847   C, s for b
s - s -   s s s s s   - s -  13582764
s - s -   s s s s s   - s -  12573486
s - s -   s s s s s   - s -  13542678
s)S S(- s -   s s s   - s -  12564873   B, s for p
s   - s -   s s s   - s -  12586473
s   - s -   s s s   - s -  12548673
s)S s s s s s s s   -   s  13264587
-   s s s s s s s   -   s  12356478
-   s s s s s s s   -   s  13245687
S(s -   s s s s s   - s -  13562748   B, s for b
s - s -   s s s s s   - s -  12573864
s S(S(- s -   s s s   - s -  13587462   A', s for b;  B, s for p
s   - s -   s s s   - s -  13548762
s   - s -   s s s   - s -  13574862
s)S s s s s s s s   -   s  12387546
s s   -   s s s s s s s   -  16253874
s   -   s s s s s s s S(s  15482367   B, s for p
- s -   s s s   - s -   s  17482356
- s -   s s s   - s -   s  16482375
- s -   s s s   - s -)S -  14382567
s   -   s s s s s s s   -  17253846
s   -   s s s s s s s   -  16382574
s S(s -   s s s s s   - s -  12653487   C, s for b
s - s -   s s s s s   - s -  13642758
s - s -   s s s s s   - s -  12673845
s - s -   s s s s s   - s -  13682574
s)S   s s s s s s s   -   s  13258647
)S s -   s s s s s   - s -  13862754   [end of A']
s - s -   s s s s s   - s -  12873465
s)S S(- s -   s s s   - s -  13847562   B, s for p
s   - s -   s s s   - s -  13854762
s   - s -   s s s   - s -  13875462
s)S s s s s s s s   -   s  12347856
-   s s s s s s s   -   s  13284765
-   s s s s s s s   -   s  12378456
-------------------------------------

Repeat four times

In the conclusion to Colin's article, he says,

| Some people may not consider this a fair extent of
| Yorkshire since it does not contain a plain full lead.  It
| is impossible to produce this extent in half leads with a
| plain full lead.  It is possible, however, to produce the
| extent of Woodstock, the 8ths place version of Yorkshire,
| with a plain full lead.
|
| This is, I think, the nearest that we can approach to the
| ideal extent of Yorkshire Surprise Major.

I'm not entirely sure I agree with this last point.  His
composition contains 87s at back, and, whilst some people
profess to like such things, I personally prefer to avoid
them.  I can see no reason why this shouldn't be possible.

Secondly, it is possible to get a composition with fewer
half-lead singles.  In his building blocks (A, B and C)
11/21sts of half-leads have a single.  It is certainly
possible to reduce this to 3/7ths at the expense of
introducting two more types of block and of loosing the
five-part structure.

Finally, it might be possible to get a six-part composition.
Clearly this can't be the standard sort with, say, 234
cycling and 56 swapping, but it might be possible to have 23
swapping, 45 swapping and 678 cycling.  (Of course, as this
looses the observation bells, it might not be considered an
improvement.)  This arrangement ought to still be possible
using the blocks with fewer half-lead singles mentioned in
the previous paragraph.

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.

RAS

```