[r-t] 147 TDMM

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


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

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 

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

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 

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.


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:

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

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.


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 

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:


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 

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:


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 


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.


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 

   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 

   (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.


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 !

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

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 

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 

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.


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!

Happy New Year!


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