Difference between revisions of "Spliced treble-dodging minor - 6"

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[[Spliced treble-dodging minor - clusters|Clusters of plans]]
[[Spliced treble-dodging minor - 1|Plans 1]]
[[Spliced treble-dodging minor - 2|Plans 2]]
[[Spliced treble-dodging minor - 3|Plans 3]]
[[Spliced treble-dodging minor - 4|Plans 4]]
[[Spliced treble-dodging minor - 5|Plans 5]]
[[Spliced treble-dodging minor - 6|Plans 6]]
Richard Smith richard at ex-parrot.com
Richard Smith richard at ex-parrot.com
Thu Oct 21 04:22:35 BST 2010
Thu Oct 21 04:22:35 BST 2010

Revision as of 18:18, 6 November 2010

Clusters of plans | Plans 1 | Plans 2 | Plans 3 | Plans 4 | Plans 5 | Plans 6

Richard Smith richard at ex-parrot.com Thu Oct 21 04:22:35 BST 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 Ipswich.


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 following:

  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 Yo.

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 considered.

  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 course.

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 variants.)


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 overworks.

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 methods.


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 plans.

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 chirality.

  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 Ol.

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.