[r-t] 147 TDMM
Richard Smith
richard at ex-parrot.com
Sat Jun 9 23:25:45 UTC 2012
After an eighteen month break, it must be time for me to
write the next installment, the eighth in all, of my
analysis of bobs-only extents of the 'standard' 147
treble-dodging minor methods. An index to the parts so far
is given here:
http://ex-parrot.com/~richard/minor/147/emails.html
The analysis has largely been done in terms of 'plans'. A
plan is a table listing each lead head and lead end with the
associated method. The existance of a plan does not imply
that all of the leads can be joined together to form a
single round block (the extent), though more often than not,
it can be. Modulo rotation and lead splices, there are 4614
in-course plans.
The seventh part began with re-evaluation of how the grid
splice works; it continued with a look at composite courses,
noting that the established list of composite courses was
incomplete, and enumerated the missing composite courses;
finally, this was used to explain a few compositions
involving Marple (Ma), Old Oxford (Ol), Norwich (No) and
Morning Star (Ms). As of the end of the seventh part, we
had explained all but 136 of the plans.
Previously I had said that this part would investigate how
to extend the Ma-Ol-No-Ms system with Bedford (Be), Bourne
(Bo), Kirkstall (Ki) and Disley (Di). However, I've
discovered some more basic material that I think needs to be
covered first; indeed, ideally would have been covered
before much of the material in the last email.
Consequentially, extending Ma-Ol-No-Ms is deferred until a
later email. This email begins by looking again at the
usual grid splice and then considers a special case of it
which I shall call the two-lead grid splice. I also look at
a similar special case of the triple-pivot grid splice.
By the end of this email, I will finally have covered all
plans involving three methods.
MORE THAN JUST A GRID SPLICE
In the fourth email, we said that the grid splice could be
explained in terms of the following diagram:
X --(3)-- [G] --(3)-- Y
|
(6)
|
Z
G is the grid method, and X, Y and Z are the three methods
that appear in the plan. X and Y both have three-lead
splices with G; Z has a six-lead splice with G. Given three
methods, X, Y, Z, the grid method, G, can readily be
computed and the rows in a lead of G are uniquely defined
(though their order is not).
However, given three methods that have a grid splice, are
any plans possible besides the basic grid splice -- that is,
the plan with six identical composite courses?
In the third email, we met a set of methods that have a
splice very much like a grid splice:
Ki --(3)-- Bo --(3)-- Wl
|
(6)
|
Ne
I prefer not to think of this as a grid splice as all four
methods have the same lead head order, but this is just a
matter of definition
(One can imagine a situation where all four methods have the
same lead head order, but where the grid method (Bo here)
has some other undesirable property, such as more blows in
one place. This is in principle possible if the other three
methods have different parity structures and so the grid
method isn't simply formed by tracing the path of the grid
bell through the course. Whether it can happen in practice,
I'm unsure. In such a case, there is perhaps merit to
considering it a grid splice.)
In any case, as we've already demonstrated in the third
email, the only way to entirely eliminate the Bo is by
having every course the same -- the standard grid splice
arrangement. At no point during that analysis did we say
anything about the lead end orders of the methods: all we
said was that they shared three- or six-lead splices. With
different lead heads, we would have to make sure that any
plan had suitable composite courses available for each
course, but this is an additional restriction that might
rule out further plans -- it cannot allow additional plans.
So that analysis serves as a general proof that a grid
splice only produces a single plan.
But in that case, what about the grid splice between No, Ms
and Ma? In the seventh email we found plans using these
three methods that were not in the straightforward grid
splice arrangement. For example, we found a plan with three
leads each of No and Ms with the remainder as Ma. Haven't
we just proved that such a touch should not exist?
TWO-LEAD GRID SPLICES
We can deduce the grid method for this splice must have the
place notation &-34-4-2-1-34-1 (the lead end change is
irrelevent), and we can double-check readily enough that
this does indeed have three-lead splices with No and Ms; but
the grid method has a *two-lead* splice with Ma:
Ms --(3)-- [G] --(3)-- No
|
(2)
|
Ma
The fact that the grid method and Ma have a two-lead splice
is the critical point. The two-lead splice is special case
of a six-lead splice, and as expected for a six-lead splice,
the pivot bells for both methods (Ma and G) have the path.
(This isn't a firm requirement, and exceptions exist. The
actual rule is that if we shuffle the rows about so both
methods have the same parity structure, then the pivot bells
must have the same path.)
Marple Grid method
234165 234165
243615 243615
423651 426351
246315 246315
426351 423651
462531 432561
642513 345216
465231 435261
645213 342516
654123 324156
But let's look more carefully at the two methods. The only
difference is that Marple has Kent place under the treble,
where the grid method has Oxford places. Changing a 34-34
place notation to a -34- notation simply swaps the bells in
3-4. We do this twice in the lead, once just before the
half-lead and once just after. The first swaps 3 and 6, and
the second swaps 2 and 5. This tells us that if we want to
replace the grid method with Marple, we only do it in blocks
of two leads where the lead heads and ends are related as:
123456 132465
156423 165432
(These rows form a mathematical group known as the Klein
group. In Brian Price's paper, 'The Composition of Peals in
Parts', it is group [4.04].)
This means that while any plans that works for a normal grid
splice will work for these three methods, other plans may
also exist that use incomplete six-lead slots, and as we've
already seen, other plans for No-Ms-Ma do exist.
The obvious next question is do any of the other grid
splices have this property? We can find out by looking
through the 53 grid splices given in the fourth email,
calculating the grid method, and seeing how it splices with
the third method. And it turns out that precisely two sets
of methods have two-lead grid splices:
X Y Z
-----------
Dn Yo Cm
No Ms Ma
Plans involving No, Ms and Ma were covered in the seventh
email. (In that email, I failed to mention that I was
ignoring the possibility of four No slots because it just
gives rise to the standard grid splice. I think I thought
it obvious at the time, but when I re-read the text, it took
me some time to realise that this was the reason for not
considering that case. Clearly it was less obvious than I
thought.)
In a moment, I will look at the case of Dn, Yo and Cm, but
first a quick diversion to explain what might look like (but
isn't) an inconsistency with the previous (seventh) email.
TWO WAYS OF LOOKING AT GRID SPLICES
I gave in the seventh email a diagram showing the
No-Ms-Ma-Ol splice; if I omit Ol and flip it upside-down,
it looks like this:
Ms --(3)-- [G-Ma] --(hlv)-- [O-Ma] --(3)-- No
\ /
(hlv) (hlv)
\ /
Ma
This is rather different to the earlier diagram. Does that
mean one is wrong? No, they're just different ways of
looking at the same thing. The grid method, G-Marple and
O-Marple are all distinct methods: one difference is that
the grid method has Oxford places under the treble, where G-
and O-Marple have Kent places; another is that the grid
method has an irregular lead head whereas G- and O-Marple
are regular. We could show all these methods in one diagram
if we wanted, though the result is rather confused:
[G-Ma] --(hlv)-- [O-Ma]
/ | \ / | \
/ | (hlv) (hlv) | \
(3) | \ / | (3)
/ (3) Ma (3) \
/ \ | / \
Ms \ (2) / No
\ \ | / /
(3)-------- [G] -----------(3)
One way of looking at the splice is in terms of an extent of
the grid method, G, which we carefully remove though
judicious use of two- and three- lead splicing, and these
are then fitted into composite courses. Another way is to
consider the starting point as an extent of Marple, cut each
course up into composite courses of Marple, and G- and
O-Marple, and then splice the latter two methods out using
the three-lead splices. Both pictures are valid, both will
result in the same compositions, and it would be wrong to
consider one of these to be the "correct" model at the
expense of the other. (That said, later in this email we
will see an obscure case where the two models are not
equivalent.)
DUNEDIN, YORK AND CAMBRIDGE
As mentioned above, Dn-Yo-Cm also form a grid
splice system using a two-lead splice.
Dn --(3)-- [G] --(3)-- Yo
|
(2)
|
Cm
The lead heads and ends involved in a Dn (three-lead) splice
slot are given in the first column:
Dn slot Cm slot
123456 143265
163542 153624
124536 154263
164352 134625
125346 135264
165432 145623
Each of these rows will also be part of a Cm (two-lead)
splice slot, the other half of which is shown in the second
column. So if we ring the first column as Dn then we cannot
ring Cm for lead ends and heads in the second column.
Fortunately, the second column is precisely a Yo
(three-lead) slot. This demonstrates that any Dn-Yo-Cm
plans will contain the same number of leads of Dn and Yo.
We should also notice that each lead in the first column is
in the same course as exactly one lead from the second
column, meaning that each course must have equal numbers of
leads of Dn and Yo. Dn, Yo and Cm have lead ends J, G and
H, respectively, so we'll be interested in the HJ + HGH and
JGHGJ composite courses.
123456 J 123456 J
164523 G 164523 H
142635 H --------
135264 G 123456
156342 J
--------- 135264 G
123456 156342 H
142635 H
--------
135264
Exactly as we found with No-Ms-Ma, the leads of J and G in
the fragmented course correspond to slots of Dn and Yo that
together use up a whole Cm two-lead splice slot.
This gives one obvious composition straight away: with one
Dn slot and one Yo slot, we can use the HJ + HGH fragmented
course three times and then have three plain courses of Cm.
Of course, Cm has course splices with Ip which we can use
anything from zero to three times. That's 4 new plans.
Cm also has six-lead splices with Su and Bs, and whenever
either of the fixed bells of the Dn and Yo splices pivots,
we can ring Su or Bs. That gives two six-lead splice slots,
giving in turn a further 5 plans (one or two Su slots, one
or two Bs slots, or one of each). These are necessarily
incompatible with the Ip course splice. No other simple
splices can be worked into this plan. The example
composition below uses one Su and one Bs six-lead slot, and
includes all of the lead-end variants and lead splices.
720 Spliced Treble Dodging Minor (11m)
123456 Dn - 145362 Pr - 125463 Cm
164523 Cm 156423 He 163542 Dn
- 142356 Bs 162534 Wa - 142563 Bs
156234 Su 123645 Cm 163254 Cm
134625 Yo - 134562 Cm - 135426 Cm
- 134256 Cm 162453 Bs 126543 Su
- 145623 Yo - 125346 Pe 143652 Cm
- 145236 Cm 146532 Bk - 135264 Yo
136524 Bv 163425 Pr 156342 Bv
124653 Cm 132654 Bs 142635 Wa
--------- --------- ---------
145362 125463 123456
Again, just as with No-Ms-Ma, in the JGHGJ course one of the
J's is in the same Cm slot as one of the G's, and the other
J is in the same Cm slot as the other G. As both types of
composite courses (JGHGJ and HJ + HGH) have equal numbers of
Dn and Yo slots (which complement each other in using up a
Cm slot), both types of composite courses can be used in a
single plan. And so long as we avoid needing more than two
leads of Dn per course and avoid overlapping Dn and Yo
slots, we have a completely free choice of which Dn slots to
use.
The first email enumerated ways of selecting multiple
three-lead slots: we assigned them numbers like 2.1.
Because each Dn slot prescribes a corresponding Yo slot, and
we must avoid overlapping any Dn and Yo slots, only a small
number of arrangments are viable. These were mentioned
briefly in the seventh email in the context of No-Ms-Ma;
they are: 1.1, 2.1, 3.3, 3.4 and 4.6. 1.1 is the case we've
just discussed (with just three leads each of Dn and Yo),
and 4.6 is the familiar grid splice.
With two Dn slots (i.e. arrangement 2.1), there must be
exactly one course with two leads of Dn. This means we have
one JGHGJ course, four fragmented courses, and one plain
course of Cm. The plain course gives us the opportunity to
course-splice in Ip; and the fact that the Dn (and Yo) slots
in 2.1 share a common fixed bell means that there is
opportunity for one six-lead splice, either with Su or Bs.
That give 4 plans -- the basic one, and one each with Ip, Su
and Bs.
With three Dn slots, we no longer have a free course for Ip,
and we only have a six-lead splice opportunity if the Dn
slots share a common fixed bell. That's only true in 3.3,
so 3.3 has three plans depending how the six-lead slot is
used for Bs or Su, or left as Cm. In 3.4, we cannot use a
six-lead splice, but because we have three Yo slots with
fixed bells a,b, a,c, and b,c, we do have the opportunity to
use apply the Yo-Du three-lead splice with bells d,e. This
gives two more plans, depending on whether Du is included.
We've now explained a further 18 plans -- all involving Dn,
Yo and Cm.
Although they've already been counted in the seventh email,
the 21 plans of No, Ms and Ma, that do not include Ol, ought
really to be classified as two-lead grid splices.
MORE ON THE GRID SPLICE
In generating the short list of two-lead grid splices
earlier in this email, I simply ran down the list grid
splices from the fourth email checking whether any of them
have two-lead splices instead of six-lead splices. And in
the fourth email, I generated the list of all grid splices
by looking for ways of assigning methods palindromically in
a course such that all the courses were the same and the
plan true. So surely I ought to have found all the two-lead
grid splices there?
Not so. In an ordinary grid splice, the methods involved
splice as folows:
X --(3)-- [G] --(3)-- Y
|
(6)
|
Z
But there's an additional constraint that isn't shown in the
diagram -- we also require that the fixed place bells for
each splice are distinct. For example, the X-G fixed bells
might be 2,5, the Y-G fixed bells might be 3,6, and the Z-G
fixed bell might be 4.
Initially this requirement might seem tautologous. It might
seem that the grid method must have two disjoint pairs of
bells swapping at the half-lead giving rise to the
three-lead splices, and a different bell making a place,
giving rise to the six-lead splice. But let's stop thinking
in terms of the grid method, and instead think in terms of
half-lead variants.
X --(3)-- X' --(hlv)-- Y' --(3)-- Y
\ /
(hlv) (hlv)
\ /
Z'
|
(6)
|
Z
If the X-X' splice has a,b as its fixed bells, then X' can
have bells a,b each making a place (not swapping as usual),
and the other three bells making a three-cycle. Using this
model, there is no reason why the fixed bells in the X, Y
and Z splices have to be distinct. Let's explore this
possibility a bit further.
We're not interested in the case where X-X' and Y-Y' share
both fixed bells, as that implies X-Y have a simple
three-lead splice. So let's consider the case in which the
X-X' and Y-Y' splices share one fixed place bell. This
means every X slot used must share all or no fixed bell
with every Y slot used, otherwise we get leads that
must be both X and Y. For example, if we use the a,b slot
for X-X' and the a,c slot for Y-Y', we'll get a lead where
bell a is the common place bell to both splices, b is the
other fixed place bell for X-X', and c is the other fixed
place bell for Y-Y', and the lead would have to be both X
and Y which is impossible.
X slots used Y slots available
----------------------------------------------
1.1) a,b a,b; c,d, c,e, d,e
2.1) a,b, b,c d,e
2.2) a,b, c,d a,b, c,d
3.1) a,b, a,c, d,e d,e
3.4) a,b, a,c, b,c d,e
4.4) a,b, a,c, b,c, d,e d,e
(The numbers at the left are the configuration of the X
slots in the terminology of the first email.)
Z-Z' is a six-lead splice, and we showed back in the third
email that when combining three- and six-lead splices, the
use of even one three-lead splice slot excludes all but two
six-lead slots. Therefore, if we're to include all three
methods, we need can have at most 12 leads of Z, and
therefore we need at least 18 leads, or six slots, of X and
Y in some combination. However none of the lines in the
table above include six slots of X and Y, and therefore a
plan on these lines is not possible.
What if X-X' and Y-Y' do not share a common fixed bell? If
Z-Z' also does not share a common fixed bell with either
X-X' or Y-Y' we have the familiar grid splice that we've
already considered. The remaining case is that Z-Z' has a
common fixed bell with one of the three-lead splices, say
X-X'. With a bit of thought, we can see that every X slot
must overlap all Y slots, every Y slot must overlap with
every Z slot, but that the X and Z slots must not overlap.
Applying these rules, we find that the only possible
arrangements are as follows:
Z slots used X slots used Y slots available
------------------------------------------------
a b,c a,b, a,c
a b,c, b,d a,b
a b,c b,d, b,e a,b
However, these arrangement allow for at most 18 leads of X,
Y and Z, leaving at least 12 leads of the grid method
unsubstituted.
We've therefore demonstrated that the fixed bells for the
X-X', Y-Y' and Z-Z' splices have to be distinct in a
(six-lead) grid splice. This has been an assumption
implicit throughout this series of emails, but it's good to
prove it. It also reinforces the earlier statement that the
grid method model and the half-lead variants model are
equivalent.
This proof, however, assumes that the Z-Z' splice is a
six-lead splice and only a six-lead splice. What if it's a
two-lead splice or if Z and Z' are the same method? Our
proof no longer holds in these cases, and it turns out that
the 147 contains an example of such a splice.
MENDIP, ABBEYVILLE, CANTERBURY DELIGHT AND SOUTHWARK
The only remaining unexplained plans with just three methods
are a set of four plans involving Mp, Av, Ca and So. These
splices bear a strong resemblance to a grid splice: Ca/Av
have a three-lead splice with 4,5 fixed with a half-lead
variant of So denoted X'; Mp has a three-lead splice with
2,5 fixed with another half-lead variant of So denoted Y'.
Ca --(3)-- X' --(hlv)-- Y' --(3)-- Mp
\ / \ /
(3) (3) (hlv) (hlv)
\ / \ /
Av So
But for the fact that the 5 is fixed in both three-lead
splices, this would be a standard grid splice. As it is,
I'm not quite sure how best to name it.
As we said in the previous section, each X slot must either
share all or none of its fixed bells with each Y slot, and
we enumerated the possibilities in the first ofthe two
tables in that section. The next challenge is to find
suitable courses to join the half-lead variants together.
Ca/Av, So and Mp have lead-ends G, K and H respectively, so
we'll be using the GHKHG composite course, together with the
GK+KHK fragmented composite course.
GK+KHK GHKHG
123456 Ca 123456 Ca
135264 So 135264 Mp
--------- 164523 So
123456 156342 Mp
142635 Ca
164523 So ---------
156342 Mp 123456
142635 So
---------
164523
Exactly as we found with No-Ms-Ma (in the seventh email) and
Dn-Yo-Cm (above), we find here that in the fragmented
course, the same pair of bells is in 4,5 in Ca and in 2,5 in
Mp, and in the (unfragmented) composite course, we have the
same two disjoint pairs in 4,5 in Ca and in 2,5 in Mp. As
these courses (together with the plain course of So) are the
only courses we can use, we must use the same Ca/Av slots as
Mp slots, and fixed bells in each must be indentical.
This gives us two possibilities: we can have one Ca/Av slot
(say a,b) and one Mp slot (also a,b); or we can have two
Ca/Av slots (a,b and c,d) and two Mp slots (a,b and c,d).
In the former case, we have three fragmented courses and
three whole courses of So; in the latter, we have one GHKHG
course, two fragmented courses and one whole course of So.
This gives five plans straight off: one or two Ca slots, one
or two Av slots, or one of each.
The only simple splice that we can add to this is the So-Pn
six-lead splice, which has 5 as its fixed place bell. With
one Av/Ca slot (say a,b), there are three Pn slots available
(c, d or e); with two Av/Ca slots used (a,b and c,d),
there's still one Pn slot available (e). If we use three Pn
slots, we can then introduce a single Pn-Nm three-lead
splice to get three leads of Nm. This brings the total
number of plans up to 2*(4+1) + 3*2 = 16.
By the end of the third email we'd described all the plans
formed from simple splices. Now, at the end of the eighth
email, we've now described all plans formed from a compound
splice of three methods, together with any further simple
splices that can be added to them. 102 plans still remain
to be explained which I will start doing in the next email.
Hopefully it won't take me eighteen months to write it!
RAS
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