# Spliced treble-dodging minor - 7

Richard Smith richard at ex-parrot.com Thu Dec 30 17:30:17 GMT 2010

After a lengthy delay I have finally written the seventh part of my analysis of extents of the 147 TDMM.

I'm going to start by returning to the topic of the last email and looking at the underlying explanation in a bit more detail. There's also quite a long detour looking at different composite courses and noting that our list of fragmented courses is incomplete. This will allow us to generalise some of the extents in the sixth email in a few new ways, and I use the Marple - Old Oxford - Norwich - Morning Star system to illustrate these.

A SECOND LOOK AT THE THREE-LEAD GRID SPLICE

In my previous email I said that this splice involves three methods, X, Y and Z, that splice in the following manner:

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

This diagram can be interpreted to mean the extent starts as a single-method extent of W (a method that has some undesirable property), and X, Y, Z are introduced via three-lead splices to remove all of the W.

It turns out that this model is not exactly true. Let's say that the fixed bells for the W-X splice are a,b. That means that at the half-lead of both methods, a must cross with b. (The fact that W might have a jump change at the half-lead doesn't change that, though it may mean a and b are not adjacent at the half-lead of W.)

Similarly, the fixed bells for the W-Y splice much swap at the half-leads of those methods. That gives two possibilities: either the fixed bells are also a,b, or both are not a,b. Were the fixed bells to be a,b, that would imply a simple three-lead splice between X-Y which is not the case. (In all the cases considered they actually shared a course splice.) So the W-Y fixed bells must be different bells -- let's call them c,d.

We now know that at the half-lead of W, a,b cross, as do c,d, and therefore e is the pivot. If W has a three-lead splice with Z, it must either use a,b or c,d as fixed bells, in which case it must share a three-lead splice with X or Y, which is not the case. So what is actually happening? And if the model's wrong, why did it do such a good job explaining the touches found in the previous email (including correctly predicting the number of touches found)?

It turns out that W is a half-lead variant of X and Y, but

- not* a three-lead splice -- contrary to what I said in the

sixth email. We can see this by looking at the Du/Su/Bo example in more detail. Du and Su are half-lead variant having a 16 and 56 half-lead change, respectively. Let's create the 'W' method by putting a jump change at the half-lead such that it produces a J lead head. Using Michael Fould's naming convention we can call this method J-Durham (or, equally, J-Surfleet). The middle part of the lead is shown below in the left-hand column:

J-Durham reordered

264153 + 264153 + 624513 + 624513 + 642531 + 645231 - 465213 - 465213 - 645231 - 642531 + 243651 - 246351 + 423615 - 423615 - 246351 + 243651 - 264315 + 264315 + 624135 + 624135 +

The right-hand column contains the same rows, but re-ordered so that it has a right-place parity structure. We can see that 3 and 5 just ring Bourne in the right hand column, and therefore J-Durham has a three-lead splice with Bourne (with 3,5 as the fixed bells). It would be more accurate to depict the splice as follows (where 'hlv' stands for half-lead variant):

X --(hlv)-- [W] --(hlv)-- Y | (3) | Z

In fact, a similar diagram would be more relevant for five-lead grid splices (such as between Ip/Cm/Yo) too -- although King Edward (the grid method) does have a three-lead splice with Ipswich and Cambridge, the relevant fact is that it is a half-lead variant. This is clear because some of the compositions (including the very first one in the email) did not have a multiple of three leads of Cm.

With this in mind, I'm minded to rename the three- and five-lead grid splices, but as I can't immediately think of an alternative name, I'll stick with it for the time being.

OLD OXFORD, NORWICH AND MORNING STAR

The last two three-lead grid splices I discussed in the last email were both involved Ma and Ol which are half-lead variants. (Technically, it's actually Marple and Willesden that are half-lead variants, but Willesden and Old Oxford are lead splices, and I've chosen Old Oxford -- with the D1 underwork -- as the canonical lead splice.)

X Y Z ---------- Ma Ol No Ma Ol Ms

So Norwich must be a three-lead splice with O-Marple, and Morning Star with G-Marple. We can draw this as:

Ma / | \ / | \ / | \ Ms --(3)-- G-Ma -(hlvs)- O-Ma --(3)-- No \ | / \ | / \ | / Ol

A composite course (whether fragmented or otherwise) needs to have at least three different lead-ends. (At least, because the GNJLO and OHMKG composites involve five lead-ends.) The sixth email dealt with compositions with Ma, Ol and No, and also those with Ma, Ol and Ms. So can we get compositions with both Ms and No?

Let's start by thinking about just three methods: Ol, Ms, No. These have K/N, G and O lead ends. The GNOKG composite course is the only one going. (None of the fragmented ones have both G and O.) With only that composite course available, it's clear we can have at most three leads of No as the No must form three-lead splice slots, and if we had more than one slots, one course would have more than one lead of No.

But with three leads of No, do the six G leads form splice slots for Ms? The No three-lead splice has fixed bells in 5-6, so lets consider the three tenors together courses:

135264 Ms 145362 Ms 125463 Ms 156342 Ns 156423 Ns 156234 Ns 123456 No 134256 No 142356 No 142635 Ol 123645 Ol 134625 Ol 164523 Ms 162534 Ms 163542 Ms --------- --------- --------- - 164235 - 162345 - 163425

The Ms three-lead splice has fixed bells in 2-3; it's fairly clear that the six leads of Ms don't form three-lead slots, so this is false. So it seems we can't get a composition with just Ol, Ms and No. This is confirmed by the fact that the list of unexplained plans (or, indeed, explained plans) contains none with just these methods.

ENUMERATING COMPOSITE COURSES

The proof that there were no extents of Ol, Ms and No relies on our list of composite courses (and fragmented composite courses) being exhaustive. Can we be sure that it is?

As we're considering plans, we don't care what lead ends are used to join the leads up. So at this stage, the plain course of an H method (e.g. Cambridge) is the same as the plain course of an H metheod (e.g. Primrose). In both, the 23456 lead head is joined to the 53624 lead end, the 56342 lead head to the 46253 lead head, and so on.

So how many courses are there? That's easy enough to work out. A course contains ten lead ends/heads. We can depict these at the vertices of a decagon -- let's say ordered as they appear in Plain Bob. In the Plain Bob, the verticies are alternately lead heads and lead ends. However, if we were to draw a Parker course on to our decagon, we'd find some of the PB lead heads occured as lead ends in the Parker course and vice versa. So instead of refering to them as lead heads and lead ends, let's just call them red and green rows. Our decagon now consists of alternate red and green vertices.

Let's pick a red vertex to start -- any one will do. We can associate that with any of the five green vertex. In our course they'll form (in some order) the head and end of a lead. The five choices correspond to G, H/L, J/M, K/N and O lead ends. Now consider the next red vertex around the decagon. We can associate this with any of the remaining four green vertices. Continuing around the decagon we find there must be 5! = 120 courses. Is that consistent with what we've already found?

To start with, let's think about single-method courses. There are five of these:

GGGGG HHHHH / LLLLL JJJJJ / MMMMM KKKKK / NNNNN OOOOO

In the fourth email, we identified the standard composite courses, such as HKJKH.

Base Composite Parker Base Composite Parker ---------------------- ---------------------- S HKJKH NLJKH V NLMLN NLJKL HLJNK HLJNN HKMLK HKMLN NKMHH NKMHL

P KJGJK NJGMK T LMOML HMOJL NJNJG HMHMO GMKMK OJLJL

Q GHKHG NLHGG W ONLNO OOKKL GGLLK HKNOO

R JGHGJ GMLJG U MONOM OJKMO

In the same way that we didn't count both JJJJJ and MMMMM (which only differ by whether seconds or sixths are made at the lead end), we cannot count any of the Parker courses, above, as they are all derived from the corresponding composite course by changing some lead ends.

Can we count all of the eight composites? Or is there duplication there too? It's trivial to see that there's no duplication with a column -- if there were, the course would simply be listed twice which none are. And we can see fairly easily that the bottommost six cannot involve duplication between columns -- the left-hand ones include G and the right-hand ones O.

That just leaves the top two which are duplicates of each other. This is perhaps most obviously apparent by looking at two examples of them:

123456 Cm 123456 Nf 156342 Ip 164523 Pr 135264 Bo 142635 Hu 142635 Ip 135264 Pr 164523 Cm 156342 Nf --------- --------- 123456 123456

Alternatively we could observe that the irregular S lead-end type is simply the seconds place version of the V lead-end type. That leaves us with 7 course. However we can start them at any of five different places giving 5*7 = 35 courses.

Also in the fourth email, we identified 24 Parker courses. These were the 2 basic ones, another 16 (above) derived from the standard composite courses, and finally 6 miscellaneous ones. The first 2+16 have already been accounted for -- they only differ from single-method courses and standard composite courses in that they have a mixture of 2nds, 4ths and 6ths place lead ends. But the last six are new. These courses were:

OHGLO GGMOO OOJGG GNJLO OHMKG GNOKG

Is there any duplication here? Yes, though it's subtle. >From a quick inspection, it's clear that the only possibilities for duplication are between GGMOO and OOJGG, and between GNJLO and OHMKG. As these are Parker courses, they're already a mixture of seconds and sixths place lead ends -- the sequence being fixed by the choice of observation bell, which in turn is fixed by where the course starts -- so on the face of it they can't be duplicates. But there's another way they can be duplicates: by reversal. GGMOO is the reverse of OOJGG. (M turns to J because the symbol represents the lead and the lead end change following it, but not the preceding lead end change.) Similarly GNJLO is the reverse of OHMKG.

How is that relevant? When we calculated that there were 120 courses we did it by thinking about pairs of red and green vertices around a decagon. But at no point did we say which order the pairs were in. The decagon for GGMOO is shown here:

35264 32546 / \ 53624 23456

56342 ----------------------- 24365

65432 42635 \ / 64523 46253

It looks exactly the same as OOJGG because the diagram is symmetrical -- the symmetry is only broken when we decide how to join the leads up. (Note the duplication is not is just because the two courses, GGMOO and OOJGG, are reverses -- the fact that the underlying decagon pattern is symmetrical is also important.) The same is true of GNJLO and OHMKG:

35264 --- 32546

53624 ------------------- 23456

56342 ----------------------- 24365

65432 ------------------- 42635

64523 --- 46253

That gives another 5*4 = 20 courses. (We might question whether there really are five rotations because the Parker courses must start and end with bobs. That's true, but any bell can be selected to make the bob twice.)

The sixth email discussed fragmented composite courses of which 8 were listed:

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

But again, this reduces to 7 because HJ + JKJ and NM + MLM only differ by having 2nds or 6ths place lead ends. That gives another 5*7=35 courses.

We should briefly consider whether there might be any overlap between the 5 single-method courses, the 35 standard composite courses, the 20 miscellaneous Parker courses, and the 35 fragmented composite courses. We can quickly see that there isn't. The fragmented composites all have one lead-end order represented three times -- that happens in none of the others. Obviously the single-methods courses have a lead-end order present five times which none of the others have. None of the standard composites do not have both G, whereas all of the miscellaneous Parkers do.

But that only gives 5+35+20+35 = 95 courses. That means there are another 25 somewhere that we've not yet considered.

MISSING FRAGMENTED COMPOSITE COURSES

The fragmented composite courses we looked at all divided the course into round block of three leads and another round block of two leads. Can we instead divide it into a four and a one? On the face of it this sounds impossible. How do we get a one-lead round block? The answer is to use a O or G group method, but with a 2nds or 6ths place lead end to give a one-lead course. Obviously we're not actually interested in such methods, but if we're just considering extent plans, then we have to consider the possibility.

With seconds place lead ends, there must be six ways of choosing the leads in the four-lead fragments.

Four-lead Corresponding sixths fragments place version --------------------------------------- 2O + GJJJ 6G + OMMM 2O + HGGG 6G + 6G + 6G + OL 2O + JKKK NM + NON 2O + KHHH LO + LNL 2O + KHGJ 6G + OMLN 2O + HKJG 6G + ONLM

2O denotes an O-group method with a seconds place lead-end and 6G denotes a G-group method with a sixths place lead-end. The third and fourth lines turn out to be equivalent to fragmented courses already considered. What of the last two, 2O+KHGJ and 2O+HKJG? These are reflections of each other. The decagon for 2O+KHGJ is shown below.

35264 32546 / \ 53624 / 23456 \ / \ / 56342 ----------------------- 24365 / \ / \ 65432 \ 42635 \ / 64523 46253

This is clearly symmetrical. But there is a difference: whereas the two previous diagrams examined had two planes of symmetry, this only has one. An asymetric pattern inscribed on a decagon has 20 distinct rotations / reflections; the one here has 10 rotations (the three lines can cross just by any of the ten verticies); but the previous diagrams (e.g. GGMOO) only have 5 distinct rotations because of the second symmetry plane. So the five rotations from starting the composition at an arbitrary point are enough to get the full set of GGMOO-like diagrams, but they are not enough to get the 10 rotations of the 2O+KHGJ diagram. For that reason, we need to include both 2O+KHGJ and 2O+HKJG.

Similarly, there are six sixths place four-lead fragments.

Four-lead Corresponding seconds fragments place version --------------------------------------- 6G + LNNN GK + KHK 6G + MLLL HJ + HGH 6G + NOOO 2O + 2O + 2O + GK 6G + OMMM 2O + GJJJ 6G + OMLN 2O + KHGJ 6G + ONLM 2O + HKJG

That gives one new four-lead fragment (6G + NOOO). Altogether we now have five new four-lead fragments giving us 5*5 = 25 new courses:

2O + HGGG 6G + NOOO 2O + GJJJ 6G + OMMM 2O + KHGJ 6G + OMLN 2O + HKJG 6G + ONLM

Are these 25 courses all new? Or is there duplication with the 95 we already knew about? We've already looked at the correspoding seconds / sixths place versions of them, so the only scope for duplication is with the six miscellaneous Parker courses.

The first three lines in the four-course fragment table all include three indentical leads, something that doesn't happen in any of the miscellaneous Parker courses. That just leaves 2O+KHGJ / 2O+HKJG. Could this be the same as GNJLO / OHMKG? No. We've already drawn the diagrams for these and they're not remotely similar.

We've now identified all possible ways of forming a course. Sure, we haven't looked at possible ways of combining different lead heads; nor have we looked at ways of moving between courses. In a true extent, every course must be one of the 120 enumerated here -- even if the course is chopped up, rung with a mixture of lead heads, or similar.

MARPLE, OLD OXFORD, NORWICH AND MORNING STAR

As noted at the start of this detour into composite courses, the proof that there were no extents of Ol, Ms and No relies on our list of composite courses (and fragmented composite courses) being exhaustive, and as we've just shown, it wasn't. The methods have K/N, G and O lead ends, and so, in addition to the GNOKG composite course, we also have the 6G + NOOO fragmented course to play with.

We can't have exactly three No splice slots because that gives us nine leads of No -- we can put three in one course, but that leaves six leads amongst the other five, which would mean one had to have two leads. Adding a fourth No slot can fix the course with only two leads of No, but produces other courses with the same problem. And enumerating all the further options shows that they too all have similar problems. So we were correct in stating it's not possible to get Ol, Ms and No in an extent (and, indeed, the search results confirm that).

What about Ma, Ms and No? Ma/Br is J/M and this gives us more options -- this time we have both the GGMOO and OOJGG Parker courses, as well as the 2O+GJJJ and 6G+OMMM fragmented courses. That gives more scope -- unlike the previous example where we could have 0, 1, 3 or 5 leads of No per course, this time we can have 0, 1, 2 or 5. That's an important difference.

If we have one No three lead slot, we have three (fragmented) courses:

156342 Ms 156423 Ms 156234 Ms 164523 Ma 162534 Ma 163542 Ma 135264 Ma 145362 Ma 125463 Ma 142635 Ma 123645 Ma 134625 Ma --------- --------- --------- 156342 156423 156234

123456 No 134256 No 142356 No --------- --------- --------- x 123456 x 134256 x 142356

... where 'x' represents the seconds place 'call' that brings one lead of Norwich into a round block. The three leads of No immediately form a three-lead splice (fixed bells: 5-6); but so too do the leads of Ms (fixed bells: 2-3). This, together with three courses of Ma gives us a working plan.

In the same way that the 2O+GJJJ course has the same pair of fixed bells in 2-3 for G as in 5-6 for O, the OOJGG Parker course has the same two pairs in 2-3 for G as in 5-6 for O:

123456 O 142635 O 164523 J 135264 G 156342 G -------- - 156423

That means we can have two No slots on coursing pairs (2.1 in the notation of the first email in the series), and also three No slots in either the 3.3 or 3.4 configurations. This gives four plans using just Ma, Ms and No. The following composition is an example with three No slots in the 3.3 position:

720 Spliced Treble Dodging Minor (4m)

123456 Ms 145362 Ma 165243 Ma 135264 Ma - 162345 Ma - 143265 Br 142635 Ms 153462 Ma 165324 No - 142356 Ms 124653 Ms 136452 No 125463 Ms 145236 Br - 143652 No - 125634 Ms 136524 No - 164352 No 153246 Ma - 153624 No 136245 Ma 162453 Ma 165432 No 152436 Ms 134562 Ma - 146532 No 123564 Ms - 162534 Ma - 154632 No - 123645 Ms --------- --------- --------- 145362 165243 - 123456

As discussed in the sixth email, Ma has a three-lead splice with Ta with 3-5 fixed. This partially overlaps the No splice slot which has 5-6 fixed. With one No slot, say with (a,b) fixed, we either want neither of a,b involved in Ta splice, or both. (If one of a,b is in 5ths place, the other one must be guaranteed not to be in 6ths place -- fixing it in 3rds does that.) That gives four slots for Ta: (a,b), (c,d), (d,e), (c,e). Modulo rotation, the latter three are the same, and so we have the following ways of choosing Ta slots:

None (a,b) (c,d) (a,b), (c,d) (c,d), (c,e) (a,b), (c,d), (c,e) (c,d), (c,e), (d,e) (a,b), (c,d), (c,e), (d,e)

With more than one No slot, our only choice is to avoid all the fixed bells for the No. So wth two slots, (a,b), (b,c), we must choose (d,e) if we're to have any Ta; with three slots, (a,b), (a,c), (b,c), we must choose (d,e); and with three slots, (a,b), (a,c), (a,d), there is no viable slot for Ta. The number of plans including Ta is 8+2+2+1 = 13.

Ma has a six-lead splice with Ki, with fixed bell in 4ths. To use this we need to make sure a fixed bell from No is in 4ths; with one No slot, we can use either or both No fixed bells; with two No slots, we must use the fixed bell involved in both; and with three, it is only viable in the configuration 3.3: (a,b), (a,c), (a,d) -- i.e. the configuration which does not allow Ta. That gives 4 plans.

How does Ki overlap with including Ta? With more than one No slot, Ta can only be included by avoiding the No splice bells; but we can only include Ki if any splice with Ta uses the fixed bell from the Ki splice (which is also a fixed bell from No). So we cannot have both Ta and Ki with more than one No splice. With one No splice on (a,b), we can have Ki whenever a and/or b pivots, and still ring Ta when (a,b) are in 3-5. That increases the plans by two depending on whether there's six or twelve leads of Ki.

Next, Ms has a three-lead splice with Di fixing 4,5. If three No slots are used of the form (a,b), (a,c), (b,c), then two of (a,b,c) are always in 2-3 during the Ms, and so we can ring Di whenever (d,e) are in 4-5. The plan has a single Ta slot available.

Finally, being half-lead variants, Ma and Ol have a course splice. If we don't use the Ki six-lead splice, we might have a course of Ma which we can swap for Ol. With only three leads of No with (a,b) fixed, the (a,b) Ta slot falls in the same three courses as the No and so it does not affect how much Ol can be included. Ignoring the (a,b) Ta slot, if there are no further Ta slots used, there will be three whole courses of Ma of which we can convert 0, 1, 2 or 3 to Ol; if there's one further Ta slot, there are two free courses, and with two further Ta slots, just one free course; if all three further Ta slots are used, there are no courses available for converting to Ol. So with three leads of No and no Ki, there are 2*(4+3+2+1) = 20 plans.

The following example is a composition with only three leads of No, three whole courses of Ol, and three leads of Ta spliced in. This arrangment allows us to include all the lead splices and lead-end variants of the methods:

720 Spliced Treble Dodging Minor (14m)

123456 Ms 154326 Ol 135642 Sl - 123564 Ms - 163542 Ns 126435 Ol - 123645 Ms 125463 Sl 142563 Cb 134256 Br 134625 Cw - 135426 Cb 156423 No 156234 Wr 143652 Ng - 145623 No 142356 Ns 164235 Wi 164352 Ma - 163425 Ta 126543 Wi - 152364 Ng 154263 Ma - 135264 Br - 143526 Hm - 163254 Cw 164523 No 126354 Ma - 142635 Ta 156342 Ma --------- --------- --------- - 154326 - 135642 123456

With two No slots, we have a similar situation. No is rung when (a,b) or (b,c) are in 5-6. That gives two courses of GGMOO, two of 2O+GJJJ, and two whole courses of Ma of which 0, 1 or 2 can be replaced by Ol. If the sole Ta slot (d,e) is taken, only one course can be changed to Ol. That gives 3+2=5 plans with six leads of No and no Ki.

Without Ki or Di, that gives 20+5+2+1=28 plans; add the 4+2=6 Ki plans and 2 Di plans, that gives 36 plans in total.

COMPLEX SPLICES WITH OLD OXFORD

In the diagram at the top of this email showing how Ms, Ma, Ol and No splice, we looked at ways of splicing Ms-Ma-Ol and Ma-Ol-No in the sixth email, and we've just finished looking at ways of splicing Ms-Ol-No and Ms-Ma-No. But we haven't looked at ways of including all four methods. Yes, a lot of the Ms-Ma-No plans were extended to include whole courses of Ol, but is it possible to use composite courses to include Ol other than in whole courses?

With four lead end orders to play with, there's long list of composite courses available. With so many possibilities, how do we start to think about putting them together? Let's choose one method -- No, say -- and think how it would fit with the various courses here. We can arrange the possible courses by the number of O leads (for No) they contain:

0 leads of O 1 lead of O 2 leads of O More ------------ --------------- ------------ -------

GGGGG GNOKG! MONOM 6G+NOOO ! JJJJJ/MMMMM * NM+NON GGMOO/OOJGG *! OOOOO KKKKK/NNNNN * 2O+GJJJ/6G+OMMM *! KJGJK GK+GJG

The courses marked with a * are those with equal numbers of leads of No (O) and Ms (G). Because these only involve Ol (K/N) in whole courses, we've already considered extents made solely from those courses. We need at least one three-lead splice with No and one with Ms (because extents without both have already been considered). The splice with No involves a coursing pair (5-6), as does the one with Ms (2-3). That means that it's not possible to choose the Ms and No splices so that no course contains both Ms and No. Therefore we need at least one course with both No and Ms. Such courses are marked with an !.

Let's start by considering 6G+NOOO. That needs at least three No slots using (a,b), (b,c), (c,d), and one Ms slot using (b,c). The other leads of No and Ms are distributed as follows:

Coursing order O slots G slots -------------- -------- ------- abcde ab bc cd bc abdec ab abecd ab cd acbed bc bc adbcd bc bc aebdc cd

Can we make MONOM fit around the two No leads in the abecd course? The course is written out below:

123456 M 156342 O 135264 N 142635 O 164523 M -------- 123456

The plain course coursing order is 53246, and the O leads have 42 and 35 in 5-6. That's exactly what's wanted to have ab and cd in 5-6 when the coursing order is abecd (or equivalently cdabe). So that course works. The courses with both O and G can easily be done with 2O+GJJJ (we've done exactly that with the earlier plans in this email), and the courses with just one O with NM+NON. As we've found composite courses that fit with the requirements of the No and Ms three-lead splices, we have a valid plan. An example composition produced from it is given below.

720 Spliced Treble Dodging Minor (5m)

123456 Ms 125463 Ol 164235 No - 123564 Ms 142356 Ns 126543 Ns - 123645 Ms 163542 No 135426 Ol 134256 Br 156234 Ol - 164352 Br 156423 No - 142563 Ns 152436 Br 145362 Br 135642 No 136245 Br 162534 Ma 163254 Br 145623 No - 134562 Ns 154326 Br - 164523 No 125634 Ma 126435 No 156342 No - 134625 Br - 142635 No 135264 Ol --------- --------- --------- 125463 - 164235 123456

There's no further scope for splicing into this plan. There are no complete courses of Ol or Ma to allow course splicing with the other, there's not enough Ms to include Di, there's too much No to allow Ki to be spliced into Ma, and examination of the courses shows there are no splice slots to add Ta into Ma.

Can we find further similar plans? We started by looking at the 6G+NOOO course. This immediately fixed told us we needed No when (a,b), (b,c) and (c,d) were in 5-6, and that we mustn't have No when (d,e) or (e,a) were in 5-6. What about the other five No slots -- can we use any? The requirement to have Ws when (b,c) are in 2-3 means b or c must be involved in every No slot. That just rules out (a,d).

What of the other four slots? Looking at the abecd course (which already contains two No slots -- see table above), this tells us we cannot have (b,e) or (c,e) as the only course with three Os is 6G+NOOO which requires three consecutive coursing pairs. The acbed course tells us we cannot have (a,c) because that would require a course with two Os and one G, and there is no such course. A similar argument eliminates (b,d). So, no, we cannot add any futher No.

The only possible slots for Ms are those that involved fixed bells from each of the three No slots: (a,c), (b,c) or (b,d). Can we add (a,c)? No, for the same reason we can't add a No slot of (a,c). But can we do both simultaneously, and ring GGMOO for the acbed course? No, because it causes problems with the abdec course which cannot be fixed by adding furter No or Ms.

That rules out any other plans using the 6G+NOOO course. A similar analysis with the GNOKG course shows that it's not possible to do anything with that either.

We already know there are lots of plans using the GGMOO/OOJGG or 2O+GJJJ/6G+OMMM courses -- the question is, is there anything else? It's possible to show that those plans require the same choice of fixed bells for the No (O) splice slots as Ms (G) slots. Any attempt to do so would involve a course that included both No and Ms in unequal amounts. (This is slightly awkward to prove, but intuitively obvious.) But the only two such courses are 6G+NOOO and GNOKG and we've already explored all the plans involving them. That only leaves courses without K/N methods (or the whole course of K/N), and they were covered earlier in this email. So there's nothing more to find.

SUMMARY

In total we've found 37 new plans using the methods Ma, Ol, No and Ms, augmented with simple simples to Ta, Di and Ki. These should really be considered togeter with the 169 similar plans found in the previous email.

Quite a few (16) of the plans cannot be made to join up. Fragmented courses impose quite a constraint on how you can try to join the courses together, and that's particularly true when one of fragments only contains one lead. Mixing G and O lead ends is also tricky, especially if you want plain leads of both.

The plans and compositions in this email will be of very niche interest indeed, as they cover an obscure set of methods that can be included in plenty of other, better compositions, but the ideas behind them are hopefully of more general applicability.

The list of unexplained plans is now down to 136 plans. The next email in this series will, in some ways, be an addendum to this one. My plan is to cover two further ways of extending the Ma-Ol-No-Ms system: one by adding Bedford (Be); the second by adding Bourne (Bo), Kirkstall (Ki) and Disley (Di). These are slightly more complex than the simple splices (e.g. the three-lead splice to Taxal) that have been covered here, and I've elected to hold them over until the next email. In any case, this email was quite long enough of without them!