[r-t] 147 TDMM

Richard Smith richard at ex-parrot.com
Thu Oct 21 03:22:35 UTC 2010

This is where is starts to get interesting!  Some of the 
compositions in this email are known, and some I expect are 
new.  However, I've never seen a good explanation of how 
these compositions work in general or how to find other 
similar compositions, and I hope this email goes someway to 
rectifying that.


I'm going to start by looking at a particular composition of 
surprise that hasn't yet been covered.

   Comp #1:  720 Spliced Surprise Minor (3m)

     123456 Yo     - 134625 Yo     - 153462 Cm
   - 123564 Ip       142356 Ip       162345 Cm
   - 145236 Cm     - 163425 Ip     - 124536 Ip
     136524 Cm     - 154632 Ip       152643 Ip
     124653 Cm       165243 Yo       165324 Cm
   - 145362 Ip     - 165432 Ip     - 152436 Cm
     134256 Ip       146253 Cm       136245 Ip
     123645 Cm       153624 Ip     - 152364 Yo
   - 134562 Cm     - 146532 Yo       126543 Ip
     162453 Ip     - 146325 Ip     - 135264 Ip
     ---------       ---------       ---------
   - 134625        - 153462          123456

   65s at back.  Contains a plain lead of each method.  ATW.
   360 Ipswich (Ip), 240 Cambridge (Cm) and 120 York (Yo).

This composition isn't new -- a trivial variant of it is on 
John Warboys' website, for example.  If anyone knows who 
first produced an extent on this plan, I'd be interested to 
know.  John Warboys also has a version incorporating Norfolk 
and Primrose too and thereby avoiding 65s at back which is 
worth repeating here:

   Comp #2:  720 Spliced Surprise Minor (5m)   Arr. JSW

     123456 Nf       142563 Ip       156234 Cm
     164523 Ip     - 135426 Cm     - 163425 Cm
     156342 Yo       126543 Pr     - 132546 Cm
   - 156423 Nf       164235 Pr     - 124653 Cm
     134256 Nf       143652 Pr     - 145362 Cm
   - 162345 Ip     - 135264 Ip       162534 Ip
     136524 Yo     - 142356 Yo     - 145623 Yo
   - 136245 Ip       125463 Nf       152436 Nf
   - 152364 Cm       134625 Ip       164352 Nf
   - 126435 Ip       163542 Ip     - 123645 Yo
     ---------       ---------       ---------
     142563          156234        - 123456

   No 65s at back.  Contains a plain lead of each method.
   216 Ipswich (Ip), 168 Cambridge (Cm), 144 Norfolk (Nf),
   120 York (Yo) and 72 Primrose (Pr).

But how does it work?  York doesn't splice with either 
Ipswich or Cambridge individually, so this isn't a simple 
extent.  (Clearly, or it would have been discussed 
elsewhere.)  However, it turns out that there is an 
irregular five-lead splice between York and King Edward 
which is the thirds-place half-lead variant of Cambridge and 


Irregular five-lead splices are odd things, occuring fairly 
rarely and generally being hard to work into an elegant 
composition.  (The example above is hardly elegant, though 
it's about as good as I can do.)  The familiar form of the 
five-lead splice is the course splice.  However, 
occasionally the five leads involved in the splice are not 
the same five leads of the plain course -- the splice 
between York and King Edward is an example of this.

The lead-heads and lead-ends involved in a splice for a 
group (or more generally, a left coset).  In the case of a 
five-lead splice, this is a dihedral group.  And for this 
particular splice, the lead-heads and -ends are as follows:

              /--- 123456 \
              |  / 132546 /
              |  \ 146532 \
              |  / 164352 /
  Leads of    |  \ 152364 \   Leads of
  King Edward |  / 125634 /   York
              |  \ 134625 \
              |  / 143265 /
              |  \ 165243 \
              \--- 156423 /

Sixth-place bell is pivot bell in York, and so the lead-ends 
and -heads pair up in leads as shown on the right; in King 
Edward, fourth-place bell is pivot, and they pair up 
differently, as shown on the left.

It's easy enough to put together a composition of spliced 
York and King Edward using this splice.  The Cm-Ip-Yo 
composition, above, is based on an arragement like the 

   Comp #3:  720 Spliced Surprise Minor (2m)

     123456 Yo       132654 KE       153624 KE
     135264 KE       156423 Yo       126435 KE
     162453 KE     - 156234 KE       134562 KE
     154326 KE       132465 KE     - 126543 KE
     123645 KE       164523 KE       145362 KE
     146532 Yo       125346 KE       163254 KE
     163425 KE       143652 KE       152436 KE
     124536 KE     - 125634 Yo       134625 Yo
   - 163542 KE       153246 KE       142356 KE
     145236 KE       142635 KE     - 165324 KE
     ---------       ---------       ---------
     132654        - 153624          123456

   Has 65s at back.  Contains a plain lead of each method.
   Methods: 600 King Edward (KE) and 120 York (Yo).

(With a better method balance, we could easiy get a 
three-part composition.)

We now have a 720 with the correct five leads of York and 
the rest as King Edward.  So it must be possible to cut up 
the leads of King Edward and reform them as Cambridge and 
Ipswich to get comp #1.

On the face of it, this seems easy -- any method with a 36 
half-lead has a three-lead splice with its 16 h.l. variant, 
and another three-lead splice with its 56 h.l. variant.  In 
this case, Cm-KE has a three-lead splice on 2,5, and Ip-KE 
has a three-lead splice on 3,6.

However we can quickly rule out the three-lead splices 
because the number of leads of King Edward, 25, is not 
divisible by 3, so we clearly can't get rid of more than 
24 of the leads of King Edward.  (In fact, we can't get rid 
of more than 15 of them.)


So how do we convert comp #3 into comp #1?  The leads of 
York are the same in both, and King Edward, Cambridge and 
Ipswich are all half-lead variants.  That means it must be 
possible to take the 25 leads of King Edward, cut them into 
50 half-leads and reassemble them as 15 leads of Ipswich and 
10 of Cambridge.

York has a G lead end, Ipswich is K, and Cambridge is H. 
Each of the five leads of York is in a different course, so 
we want to find a composite course that contains one G lead 
and the rest K or H.  I gave a table of all possible 
composite courses in the fourth email, and unfortunately the 
only composite course using just G, K and H is GHKHG which 
has two Gs.

Fortunately there is another type of composite course 
available: the fragmented composite course.  Instead of 
having a round block of five leads, the course fragments 
into a round block of two leads and a second round block of 
three leads.  It turns out that, up to rotation, there are 
only four ways of doing this with seconds place lead ends, 
and a further four with sixths place lead ends.  (The 
two-lead section needs to be a lead-head and its inverse: so 
HJ or GK.  Up to rotation, the only remaining choice is 
whether we visit the three remaining lead heads clockwise 
or anticlockwise.  That's 2*2 choices.)  The possible 
fragmented courses are:

   GK + GJG        LO + LNL
   GK + KHK        LO + OMO
   HJ + JKJ        NM + MLM
   HJ + HGH        NM + NON

The GK + KHK fragmented course is just what is needed to get 
one lead of Yo (G) in a course with Ip (K) and Cm (H). 
That's all we need to show -- it's necessarily the case that 
the King Edward can be reassembled into the appropriate 
leads of Ip and Cm.

So how many plans with Yo, Ip and Cm are there?  With five 
leads of Yo, the five courses involving Yo are totally 
constrained.  The remaining course can be either Ip or Cm. 
That gives two plans.  With ten leads of Yo, four of the 
courses contain two leads of Yo.  The only possible 
composite course for them is GHKHG.  Unfortunately two of 
the four courses contain consecutive leads of Yo, and two 
contain non-consecutive leads of Yo.  This composite course 
only copes with the former.  Therefore ten leads of Yo is 
not possible.  Similar arguments rule out larger amounts of 

In this case, we cannot extend the plan with furter methods. 
The only simple splices that Ip or Cm have (beyond the 
mutual course splice) are Cm's six-lead splices, and that's 
incompatible with the five-lead splice to York.  So this 
splice itself is only responsible for two further plans.


How would we find similar splices?  First we need to 
identify the key properties of the Yo-Ip-Cm splice that 
makes it work.  We might start by saying: (1) Ip and Cm are 
16 and 56 half-lead variants; and (2) Yo has a five-lead 
splice with the 36 half-lead variant of Ip and Cm.

Are there any other sets of methods where X and Y are 16/56 
h.l. variants, and Z has a 5-lead splice with their 
(necessarily irregular) 36 h.l. variant (let's call it W)? 
That turns out to be pretty restrictive, and the only other 
methods so related are the Carlisle-over equivalents:

   X   Y   Z
   Ip  Cm  Yo
   Nb  Cl  Ak

Can we generalise the idea?  For a start, does the W-Z 
splice need to be a five-lead splice?  If the splice is a 
six-lead splice then we've got a grid splice -- something 
we've already considered.  Because of this similarity, I've 
chosen to call the Ip-Cm-Yo splice a five-lead grid splice 
in contrast to the usual six-lead grid splice.

   X --(3)-- [W] --(3)-- Y       X --(3)-- [W] --(3)-- Y
              |                             |
             (5)                           (6)
              |                             |
              Z                             Z

   Five-lead grid splice         (Six-lead) grid splice

What if W-Z is a three-lead splice?  W has two crossing 
pairs.  W-X is a three-lead splice fixing one of them, and 
W-Y is a three-lead splice fixing the other.  If W-Z is a 
three-lead splice, then there must also be a three-lead 
splice betwen Y-Z or between X-Z.  That means, in either 
case, that X-Y-Z is just a combined course and three-lead 
splice plan and so already considered.

So does that mean that's it?  That this is a dead-end 
contributing four (already known) plans?  No.  With a bit of 
thought, it generalises considerably.


The mistake was to restrict ourselves to W being the 36 
place half-lead variant of X and Y.  Why can't W have a jump 
change at the half-lead?  The important thing is that W has 
two pairs swapping and one making a place at the half-lead. 
Just as when we generalised the grid splice to produce the 
triple-pivot grid splice, there was no requirement for the 
swapping bells to be adjacent.  Allowing such methods opens 
the door to a W-Z three-lead splice that hasn't already been 

   X   Y   Z
   Du  Su  Bo/Ki
   Nw  Mu  Sa/Te
   C1  Mp  C2/C3
   Ma  Ol  No
   Ma  Ol  Ms

Where there are two methods in the Z column it's because 
they share the same 3-lead splice as the W-Z splice. 
(That's why No and Ms are not combined on to a single line.)

The first two lines are moderately well known, with John 
Leary's compositions of 10 Cambridge-over surprise methods 
and 8 Carlisle-over surprise methods being notable examples 
of their use.  I've not seen any extents on the other plans.


Let's consider the Cambridge-over splice first as this is 
the best known one.  Bo/Ki has a J lead end, Du and Su are G 
and H respectively, so we'll be looking to work with the 
JGHGJ composite course or the HJ + HGH fragmented composite 

With one slot taken by Bo (i.e. three leads of Bo), we need 
to use the fragmented course three times, and we have three 
leads where we can choose either Du or Su, giving four 
plans.  An example is given below:

   123456 Su       134256 Su       142356 Su +
   156342 Du       156423 Du       156234 Du
   164523 Su       162534 Su +     163542 Su
   ---------       ---------       ---------
   123456          134256          142356

   135264 Bo       145362 Bo       125463 Bo
   142635 Su +     123645 Su       134625 Su
   ---------       ---------       ---------
   135264          145362          125463

   152436 Su +     153246 Su       154326 Su
   136245 Su       146325 Su       126435 Su
   145623 Su       125634 Su       135642 Su
   123564 Su       134562 Su       142563 Su +
   164352 Su       162453 Su +     163254 Su
   ---------       ---------       ---------
   152436          153246          154326

(The + denotes one of the three six-lead splice slot that 
could be used to introduce Cm into the plan.  This is 
discussed later.)

With two Bo slots (i.e. 6 leads), the slots can either share 
a fixed bell [(a,b) + (a,c)], or not [(a,b) + (c,d)].  In 
the former case, the Bo leads are scattered through five 
courses; in the latter, they're only present in four 
courses.  The question is, can those courses with two leads 
of Bo be arranged to fit the JGHGJ composite course?

The fixed place bells for the Bo splice are 3rds and 5ths 
place bells.  The J lead head is 164523 meaning that if 3,5 
are fixed in the first lead of Bo, then 2,4 are fixed for 
the next one.  That means that we can only fit two leads of 
Bo into the composite course if they don't share a fixed 
bell.  These plans have two courses with Bo where there is a 
free choice between a course of Du or a course of Su.  That 
results in three plans.  An example is given below:

   123456 Bo       134256 Su       142356 Du
   164523 Bo       156423 Su +     125463 Su
   135264 Du       123645 Su       163542 Su
   156342 Su +     145362 Su       ---------
   142635 Du       162534 Su       142356
   ---------       ---------
   123456          134256          156234 Su +
                                   134625 Bo
   154326 Bo       152436 Su       ---------
   163254 Bo       136245 Su +     156234
   142563 Du       145623 Su
   126435 Su +     123564 Su       153246 Du
   135642 Du       164352 Su       134562 Su
   ---------       ---------       162453 Su
   154326          152436          ---------

                                   146325 Su +
                                   125634 Bo

(As above, the + denotes the six-lead splice slot that could 
be used to introduce Cm into the plan.)

By adding a third slot of Bo (i.e. 9 leads), it becomes 
impossible to find composite courses to make it all fit 
together, and it might seem that we're stuck with just those 
4+3 = 7 plans.  But actually there's one more case that 
works.  If we choose one course to be entirely Bo, then this 
course shares two Bo splice slots with each other course. 
These slots fall adjacently allowing the JGHGJ composite 
course to be used in them.  This adds one further plan with 
15 leads of Bo, 10 of Du and 5 of Su.  The brings the total 
to 8.

This latter plan is of passing interest as it has a 
five-part structure.  Unfortunately it's not possible to 
join the parts together while retaining a five-part 
structure, though the following Relfe-like block gives rise 
to five mutually true blocks.

     123456 Du
     135264 Hu
     164523 Bk
     142635 Bo
     156342 Du
   - 156423 Bo
   - 123456

Clearly we can include both Bo and Ki in many of these 
plans.  Obviously with one Bo/Ki slot, we can have one or 
the other, but not both: 4*2 = 8 plans.  The three plans 
with 2 Bo/Ki slots can have Bo, Ki, or both: 3*3 = 9 plans. 
The plan with 5 Bo/Ki slots can have various combinations: 
1+1+2+2+1+1 = 8 plans.  (The terms are for 0,1,2,3,4,5 slots 
taken with Bo.  With 2 Bo, they can either be consecutive 
leads in the whole course of Bo, or separated.)  That brings 
us up to 8+9+8 = 25 plans.

What else can we get in the extent?  Su has a six-lead 
splice with Cm or Bs.  With one Bo/Ki slot, say with (a,b) 
fixed, if all of the spare courses are Su then there are 
three Su/Cm/Bs six-lead slots free: with c, d and e 
pivoting.  With no Su slots, we can have 0, 1, 2 or 3 Cm 
slots; with one Su slots, we can have 0, 1 or 2 Cm slots; 
and so on.  That gives a further 2*(4+3+2) = 18 plans. 
(The case of no Cm/Bs has already been counted.)  With two 
Bo/Ki slots, there's one Su/Cm/Bs six-lead slot, getting 
another 3*2 = 6 plans, bringing the total up to 49. This 
means that in an extent we can get any combination of Bo, 
Ki, Cm and Bs except all four together.

Also, Du has a three-lead splice with Yo with the fixed 
bells in 2,3.  With one Bo/Ki slot we have three HJ + HGH 
fragmented courses and thus three leads of Du there.  As can 
be seen from the fragments written out above, these three 
leads of Du form a Du/Yo three-lead splice slot, and so we 
can replace Du with Yo in any of those plans.

What if we have some whole courses of Du too?  That might be 
expected to provide additional Du/Yo splice slots, but it 
doesn't.  We know that there are two ways of choosing three 
courses depending on whether the courses share a coursing 
pair of bells.  The three fragmented courses necessarily 
share a coursing pair (because the three leads of Bo/Ki and 
the three leads of Du/Yo use it), that means the three free 
courses cannot share coursing pair, and so together they 
have no Du/Yo splice slots.  That means the 26 plans with 
three leads of Bo/Ki can have three leads of Du replaced 
with Yo (adding a further 26 plans), but nothing further. 
That tells us that if we want both Bs and Cm in the plan, we 
cannot have both Yo and Du.

What if we have two Bo/Ki slots?  For the same reason as 
above, only the two Du in the composite courses can be used 
for splicing in Yo.  With two slots, we would expect this to 
treble the number of plans depending on whether we want 
these six leads to be all Du, all Yo, or half and half.

Actually, it's a little more complicated.  Let's consider 
the 5 plans which contain both Bo and Ki.  (That's three 
from the choice of 0, 1 or 2 extra courses of Du, and two 
from the six-lead splices with Cm/Bs.)  Let's say that (a,b) 
are fixed for Bo and (c,d) for Ki.  If we choose (a,b) for 
Yo and (c,d) for Du, that's distinct from choosing (a,b) for 
Du and (c,d) for Yo.  That gives us an extra five plans.

With a little thought, we can see that it's not possible to 
insert Yo into the plan with 15 leads fo Bo/Ki.  In total 
that gives us 110 plans using this recipe:

    3 Bo/Ki:  (8+18)*2     =  52
    6 Bo/Ki:  (9+6)*3 + 5  =  50
   15 Bo/Ki:  8            =   8

John Leary's 1987 RW article summarises well what can be 
achieved with this plan:


There are 14 methods that are potentially available:

   Bs/Wa Cm/Pr Ki/Lv Bo/Hu [Su,Bv]/[He,Bk] Yo Du

We've also established two constraints on what's possible:

   (i)  we can get any combination of Bo/Hu, Ki/Lv, Cm/Pr and
        Bs/Wa except all four together; and

   (ii) if we want both Bs/Wa and Cm/Pr in the plan, we
        cannot have both Yo and Du.

John Leary's Cambridge 10 loses Bs/Wa to allow both Yo and 
Du to be included.

   Comp #4:  720 Spliced Surprise Minor (10m)   Comp. JRL

     123456 Cm       123645 Pr       123564 Cm
     156342 Yo       134256 Bv     - 136452 Bo
     164523 Bo*      156423 Yo       124536 Du
     135264 Bo     - 156234 Yo     - 124365 Du
   - 164235 He       163542 Pr     - 124653 Du
     143652 Cm       134625 Bo*      145236 Bv
     152364 Bk     - 125634 Bv       136524 He
     126543 Bk     - 153462 Cm       162345 Hu*
   - 164352 Su     - 136245 Su     - 145362 Hu
     152436 Su       145623 Su       162534 Bv
     ---------       ---------       ---------
   - 123645          123564        - 123456

   Can ring Bs for Cm and Wa for Pr throughout; also can ring
   Ki for Bo and Lv for Hu throughout.

It's tempting to ask whether this can be pushed to its 
logical maximum -- a twelve method extent, only omitting 
Bs/Wa or Cm/Pr?  It's almost possible to do it with comp #4 
by changing the leads marked * to Ki/Lv, though this leaves 
us without a plain lead of Lv.  It turns out that it's not 
quite possible to get plain leads of all twelve methods. 
(It's not even possible if you accept that you don't need 
plain leads of Yo and Du as they don't have lead-end 


The Nw-Mu-Sa/Te splice is essentially the same as the 
Du-Su-Bo/Ki one, and I'm not going to discuss it in much 
detail.  The methods have J, G and K lead heads 
respectively, which means we'll be using the KJGJK and GK + 
GJG composite courses.  With just three leads of Sa/Te, this 
will clearly work the same as the Cambridge-over methods.

What about two Sa/Te slots (six leads)?  We need to use two 
KJGJK courses, which puts the Sa leads adjacent.  Sa brings 
up the lead head 142635, and the two fixed bells in the Sa 
splice are 3rds and 6ths.  So if we have 3,6 in one lead we 
have 2,5 in the next -- disjoint pairs as with the Cambridge 
methods.  So everything works exactly the same as for the 
Cambridge methods and we have 25 plans just involving the 
four base methods (Nw, Mu, Sa, Te).

What else can we add to the plan?  Mu has six-lead splices 
with Cl and Gl, exactly as Su did with Cm and Bs; similarly, 
Nw has a three-lead splice with Ak, exactly as Du did with 
Yo.  None of that should be a surprise given the known 
duality between the Cambridge and Carlisle overworks.  The 
accounts of 110 plans, exactly as for the Cambridge 

However, there are two further splices to consider this 
time.  Gl has a three-lead splice with Ca and Cl with Cu. 
Fortunately these are straightforward to consider.  Cl and 
Gl are both introduced via six-lead splices into Ch.  To get 
one Gl/Ca or Cl/Cu slot we need 18 leads of Gl or Cl.  That 
means we can only have one Sa/Te slot and the three 
spare courses must be Ch.  That gives us 2*2*2 = 8 plans. 
(The factors of two come from the choices of Gl/Ca vs Cl/Cu, 
Ak vs Nw, and Sa vs Te.)  That gives us 118 plans all told.

Unfortunately, because of the large number of G-group 
methods (which do not have 6ths place variants), it's 
difficult to get compositions incorporating 6ths place 
methods.  The eight 'extra' plans (involving Ca or Cu) 
cannot be joined up while incorporating any 6ths place 
methods at all, though we can get a pleasing three-part with 
six 2nds place methods:

   Comp #5:  720 Spliced Surprise Minor (6m)

     123456 Mu
     135264 Cl
     156342 Cu
     164523 Cl
   - 164235 Nw
     152364 Cl
     126543 Cl
   - 126435 Sa
     142563 Cl
   - 142635 Ch
   - 142356

   Twice repeated.  Can ring Ak for Nw throughout; also can
   ring Ca and Gl for Cu and Cl throughout; also can ring Te
   for Sa throughout  For plain lead of Ch, swap Mu and Ch in
   one part.

John Leary's 1987 RW article discusses the the Carlisle 
methods too.  He provides an 8 method extent involving 
Nw/Mo, Ak/Ct, Ch/Mu, Sa/Wo, which is key to ringing the 41 
in twelve true extents.  It is possible to include Cl in 
this (using an analagous plan to comp #4), but again it's 
then not possible to include any 6ths place methods.  This 
is a more general problem with this particular recipe -- 
none of the plans that include Cl or Gl can include 6ths 
place methods in a round block.  Also like the Cm methods, 
in principle we should be able to add Te/Ev to JRL's 8 
spliced composition, though in practice it's not possible to 
do it in such a way that there is a plain lead of all ten 


Finally we're into territory that John Leary didn't explore 
in his 1987 RW article.  The main splice here is between 
Cotswold (C1), Mendip (Mp), Chiltern (C2) an Cheviot (C3). 
(There are more than 26 methods beginning with C in the 
147.)  C1 has a G lead head, Mp is H, and C2/C3 is J, so the 
JGHGJ and HJ + HGH composite courses will be relevant.

As with the previous two examples, we can have one, two or
five C2/C3 slots giving 25 basic plans.  We can extend this 
in three ways.  First (and simplest) C3 has a course splice 
with Pn.  This can only be effected when we have a whole 
course of C3 which requires the plan with five C3 slots; it 
is responsible for just one new plan.

   Comp #6:  720 Spliced Treble Bob Minor (4m)

     123456 C3       134256 C3       125463 Mp
     164523 C3       162534 C1       163542 C1
   - 123564 C3       123645 Mp       134625 C3
     145623 C1     - 134562 Mp       156234 C3
   - 145236 Pn       162453 C1       142356 C1
     124653 Pn       125634 C3     - 142563 Pn
     162345 Pn       146325 C3     - 135426 C1
     136524 Pn     - 125346 C3     - 135264 C1
   - 145362 C1       163425 C1       156342 Mp
     156423 C3       132654 Mp       142635 C1
     ---------       ---------       ---------
     134256        - 125463          123456

The second way of extending the basic extent is with the 
six-lead splices between Mp/Pv/Cc/Bh/By/Bw.  With one C2/C3 
slot and no whole courses of C1, there are three six-lead 
splice slots.  If they're all different that's 6*5*4/3! = 20 
plans; if two are the same, that's 6*5 = 30 plans; if 
they're all the same, that's 6 plans, one of which (all Mp) 
is already counted.  That's 55 extra plans.  Double that 
because we can have C2 or C3 in the C2/C3 slot.  With two 
C2/C3 slots, we've only one six-lead splice slot, which 
gives 5 extra plans.  Treble that for choice of methods in 
the C2/C3 slots.

Finally Pv has a three-lead splice with Li.  So if we use 
all three six-lead splice slots on Pv, we can add Li. 
That's responsible for two more plans: one with C2, one with 
C3; however, neither plan can be made to join up.  (This is 
an unusual situation where having 6ths place methods would 
help.)  All told, that results in 25+1+55*2+5*3+2 = 153 

Alarm bells should now be ringing.  I've just said that the 
Li plans cannot be joined up, but Li, Pv, Cc, Bh, Bw, By and 
Mp all have H lead heads.  If Li doesn't work, none of the 
others will too.  That's true, and because of that a large 
number of the plans cannot be joined up.  However, the 
situation isn't as bad as it might seem.  The extent can be 
salvaged in any of three ways:

   (i) including a Kent-over method (i.e. Bh, Bw or By) so
   that its 6ths place l.e. variant can be rung;

   (ii) including one or more whole courses of C1; or

   (iii) including more than one C2/C3 slot.

The first strategy mixes backworks up, which, depending on 
your perspective, might not be desirable.  It's not possible 
to get the 6ths place versions of all three Kent-over 
methods (Os, Wf, Kh), but one or two is easy enough.

   Comp #6:  720 Spliced Treble Dodging Minor (12m)

     123456 Cc       145362 C3       135426 Mp
     156342 C1     - 162345 By       126543 Mp
     164523 Pv       145236 Mp     - 164352 Pv
   - 142356 Pm       136524 Mp       152436 Ed
     156234 C1       124653 Md       123564 Cc
     163542 Cc       153462 Cc     - 136452 C3
   - 134256 Le     - 136245 Mp       124536 Le
     156423 C1       145623 Mp     - 143652 Cc
     162534 Kh     - 152364 Pm     - 135264 C3
     123645 Cc       164235 Bt       142635 Kh
     ---------       ---------       ---------
     145362          135426          123456

I think that's the most methods you can get in a single 
extent with this recipe.  (If we swap Cc for a Kent-over 
method with the aim of getting both lead-end variants, we 
need to introduce more bobs to do it and as a result lose 
the plain lead of at least one other method.)


The methods, Ma, Ol, No have J/M, K/N and O lead heads. 
That means we're going to be using the MONOM and NM + NON 
courses.  As before, we can have 3, 6 or 15 leads of No 
which gives us 4+3+1 = 8 basic plans (depending on the 
number of whole courses of Ma/Ol that we include).

Ma three-lead splices with Ta (exactly like Du does with 
Yo), and Ol six-lead splices with Bm (exactly like Su does 
with Cm or Bs).  That accounts for (4+3)*2 + (3+1)*3 + 1 = 
27 plans.  While we can easily get Ma, Ta and Bm in a single 
extent, it turns out not to be possible to get plain leads 
of all of the 2nds and 6ths place versions in a 15 method 
extent.  The following example gains all bar Ta:

   Comp #7:  720 Spliced Treble Dodging Minor (14m)

     123456 Bc       146325 Ns       154326 Ol
     164523 No       153246 No     - 163542 Wi
   - 156423 No     - 125346 No       156234 Cw
     145362 Ol       132654 Bc       142356 Bm
     134256 Bm       146532 No     - 163425 Hm
     123645 Cb       154263 Ma     - 125463 Wr
     162534 Hm     - 163254 Bc       134625 Cb
   - 134562 Ma       142563 Sl     - 156342 Br
     125634 Ng       135642 Cw       142635 Hm
     162453 Bm       126435 Wr       135264 No
     ---------       ---------       ---------
     146325          154326          123456

The other method that can be incorporated into this is El 
which has a three-lead splice with Ol (fixed bells: 2,4). 
To be able to incorporate some El we need the Ol/El fixed 
bells to have a bell in common with every No slot.  That 
means with one No slot, say (a,b), and no whole courses of 
Ma, there are seven El slots: (a,b), (a,c), (a,d), (a,e), 
(b,c), (b,d) and (b,e).  I counted the ways of selecting 
from these (modulo rotation) in the second email -- there 
are 31 ways of selecting a non-zero number of El slots.  If 
we want both El and Bm, the Bm splice must use the same 
fixed bell as the El splice(s), so if Bm uses c, only (a,c) 
and (b,c) can be used for El.  That adds a further 2 plans 
(for 3 or 6 leads of El).  Double that for plans with Ta 
instead of Ma: that gives 66 further plans.

What if we have some whole courses of Ma?  With one course 
of Ma, we lose four El slots.  (In that course, bell a must 
course two bells, neither of which can be b, say c+d; as 
must bell b, say d+e.  The slots with those coursing pairs 
-- (a,c), (a,d), (b,d), (b,e) -- include one lead in the 
course of Ma and so are no longer viable.)  That leaves 
three slots, say (a,b), (a,c), (b,d).  There are 5 ways of 
choosing a non-zero number of those El slots.  Doubling to 
allow for Ta gives another 10 plans.  With two or three 
courses of Ma, we can only use the (a,b) slot.  That adds a 
further 2*2 = 4 plans.

And what if we have two No slots, say (a,b)+(c,d), and no 
whole courses of Ma.  This leaves four El slots: (a,c), 
(a,d), (b,c), (b,d).  There's one way of choosing one slot 
and two of choosing two, but each are doubled because of 

   a --- c     a --- c     a --- c
   :     :     : \   :     :     :
   :     :     :  \  :     :     :
   :     :     :   \ :     :     :
   b     d     b     d     b --- d

   [chiral]               [chiral]

That gives 2+3+2+1 = 8 ways of choosing some El.  We need to 
treble that to account for six, three or zero leads of Ta. 
But we also need to revisit the number of patterns when one 
of (a,b) is Ma and the other is Ta: the middle diagram gets 
left-right reflected version because we are no longer free 
to relabel (a,b) as (c,d).  That means 8+9+8 = 25 plans.

Finally, we can add one whole course of Ma.  That will 
remove three of the four El slots, so if we want El, the 
only choice is whether we have six, three or zero leads of 
Ta -- 3 more plans.  That brings the total number of plans 
to 27+66+10+4+25+3 = 135.

   Comp #8:  720 Spliced Treble Dodging Minor (15m)

     123456 El       132546 No       163254 Cb
     142635 Ol     - 153246 No       126435 Cr
     164523 Bm       125634 Cw       154326 Bm
     156342 Ta       146325 Bm     - 163542 Wr
   - 142356 Cb       134562 Hm       125463 El
   - 163425 Bc       162453 Ng     - 134256 Ng
     154263 Ta     - 134625 Bc       123645 Wi
     132654 Ns       156234 Wi       162534 Ol
     146532 Sl     - 142563 Wr       156423 Bm
     125346 No       135642 Ol       145362 Cb
     ---------       ---------       ---------
   - 132546          163254        - 123456

This is the most methods we can get with this recipe.  There 
are no plans that include all four of El, Bm, Ma and Ta, so 
more than 15 methods clearly cannot be achieved using the 
plans listed here.


This is the last of the three-lead grid splices.

The methods, Ma, Ol, Ms have J, K and G lead heads which 
means using which means we'll be using the KJGJK and GK + 
GJG composite courses.  Ma and Ol are basically lead-end 
variants and Ms is the three-lead splice method (with fixed 
bells in 2-3), but if we look at the courses, G occurs once 
in the unfragmented course and three times in the fragmented 
course.  That means we can't just apply what we've done with 
all the previous three-lead grid splices.

Clearly one Ms slot (three leads) will work fine: we have 
three KJGJK courses and three whole courses of Ma or Ol. 
Equally clearly two Ms slots won't work because there will 
be a course with two leads of Ms and we have no courses with 
two G leads that we can use there.

Three Ms slots can be selected in four ways:

   1) a --- b    2) a --- b --- d    3) a --- b --- c --- d
       \   /              |
        \ /               |
         c                c      4) a --- b --- c     d --- e

In the course where a,b,c course in that order (or that 
order backwards), the first arrangement will need two G 
leads in that course.  The second arrangement will also need 
two G leads (because if b is coursing between a and c, it 
cannot be coursing d).  In the third arrangement, if abcde 
is an in-course coursing order (forwards or backwards), then 
that course contains three G leads.  But the badce course 
contains two slots: (a,b) and (d,c).  And in the fourth 
arrangement, the bcade course contains two slots.  So three 
Ms slots cannot work.

Four Ms slots?  With similar logic we can show that none of 
the arrangements work except for the following one:

   a --- b
    \   /
     \ /    d --- e

In the courses abcde, bcade, cabde it has three Ms slots, 
and in the courses acdbe, badce, cbdae it has one Ms slot. 
That gives us an plan with 12 leads of Ms, 9 each of Ma and 

The final working arrangement has five Ms slots.  It is the 
opposite to the arrangement used in the other three-lead 
grid splices.  In those arrangements we had a whole course 
of the Ms-equivalent method and two leads in each of the 
other courses; in this arrangement one we select the 
other five slots getting us a course with no Ms and five 
courses each with five.  The course without Ms can be chosen 
as Ol or Ma.

That accounts for 4+1+2 = 7 basic plans involving just Ma, 
Ol and Ms.  As with the Norwich-based three-lead grid, we 
can splice Ta for Ma.  Because the Ms course has two leads 
of Ma, it works slightly differently -- there are three 
splice slots, each having two leads in the Ma/Ol/Ms 
composite courses and one lead in the single-method courses. 
So if we have three whole courses of Ma, we can have 3, 6 or 
9 leads of Ta; if we have two whole courses of Ma, we can 
have 3 or 6 leads of Ta; and if we have one coures of Ma, we 
can have 3 leads of T.  That generates a 3+2+1 = 6 plans. 
With four Ms slots, the (d,e) component in the diagram above 
provides a single Ma/Ta slot, making one more plan.  It's 
fairly clear that there's no opportunity to get Ta into the 
five Ms slot plan.  That's 7+6+1 = 14 plans.

There are a few other methods that can be incorportated via 
simple splices.  Ol has a six-lead splice with Bm, so if we 
have three wholecourses of Ma, we can include one or two 
six-lead splices with Bm using the bells fixed in the Ms 
splice.  Similarly Ma has a six-lead splice with Ki.  If the 
single-method courses are all Ma, then six or twelve leads 
of Ki can be added.  Between them, that's four more plans.

Next there's the three-lead splice between El and Ol which 
works exactly the same as the Ta/Ma three lead splice. 
This means that if a single-method course is Ma, it provides 
a Ma/Ta splice slot, and if the course is Ol, it provides a 
Ol/El splice slot.  With one course of Ol, there are 3 
existing plans due to the choice of Ta; we get a further 3 
plans with El in them.  With two courses of Ol, there are 2 
existing plans; we get a further 4 by allowing 3 or 6 leads 
of El.  And with three courses of Ol, there's one existing 
plan to which we add 3 more.  That's 3+4+3 = 10 extra plans.

What about El when there's more than one Ms slot?  Again, 
the El/Ol <-> Ta/Ma duality holds, and we can add a single 
El slot.  Combined with the choice of Ta/Ma, that gives two 
new plans.

Finally, there's the three-lead splice between Ms and Di. 
Any of the plans with four Ms slot can incorporate three 
leads of Di.  That's 2*2 = 4 more plans.  (One factor of 2 
comes from Ta/Ma, the other comes from including or not 
including El.)  In total that's 14+4+10+4+2 = 34 plans.

The following composition uses a four Ms slot plan.  I've 
not included any of Old Oxford's lead splices as there's 
only one spare plain lead of Ns and none of Ol.

   Comp #9:  720 Spliced Treble Dodging Minor (10m)

     123456 El       164235 Ol       163425 Di
     142635 Ms     - 152643 Hm     - 163254 Ms
   - 142356 Ms       143265 El       135642 Ms
   - 142563 Ms     - 152436 Ol     - 135426 Di
     126435 Cr     - 164523 Ta       152364 Ma
     154326 Ta       135264 Ms     - 164352 Ma
   - 126354 Ns       156342 Ms       123564 Ns
     143526 Ma     - 156423 Di       145623 Ma
   - 126543 Br     - 156234 Ms     - 123645 Br
     143652 Ol       163542 Ms       145362 Ol
     164235        - 163425        - 123456


All of the plans in this email can be thought of as a 
generalisation of the grid splice, perhaps embellished 
through the addition of additional three- or six-lead 
splices.  It's been a long(!) email, and in total we've 
covered 110+118+153+135+34 = 550 plans here, including some 
with quite a lot of methods.

   Single method plans .  . . . . . . . . .   75   [1]
   Course splices . . . . . . . . . . . . .  108   [1]
   Six-lead splices . . . . . . . . . . . .  176   [1]
   Three-lead splices . . . . . . . . . . .  798   [1]
   Multiple course splices  . . . . . . . .   36   [2]
   Multiple six-lead splices  . . . . . . .  286   [2]
   Multiple three-lead splices  . . . . . .  412   [2]
   Combined course & three-lead splices . .  198   [3]
   Combined six- & three-lead splices . . .  163   [3]
   Other simple extents with four methods .   28   [3]
   Grid splices . . . . . . . . . . . . . .  124   [4]
   Triple-pivot grid splices  . . . . . . .  253   [4]
   Hidden triple-pivot grid splices . . . .    6   [4]
   Splice squares . . . . . . . . . . . . . 1224   [5]
   Five-lead grid splices . . . . . . . . .    4   [6]
   Three-lead grid splices  . . . . . . . .  550   [6]
   TOTAL  . . . . . . . . . . . . . . . . . 4441

This leaves us with just 173 unexplained plans.  Some of 
these build on the plans in this email in a fairly logical 
manner, for example by combining the Ma/Ol/No and Ma/Ol/Ms 
splices into a single framework.  But that's a topic for 
another email.

Apologies for a record-length email.


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