[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:


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.


In the fourth email, we said that the grid splice could be 
explained in terms of the following diagram:

   X --(3)-- [G] --(3)-- Y

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

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?


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

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 

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.


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)
                       \    /

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 


As mentioned above, Dn-Yo-Cm also form a grid 
splice system using a two-lead splice.

    Dn --(3)-- [G] --(3)-- Yo

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

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 

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.


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

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)
                   \    /

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

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 

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.


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

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!


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