[rt] 147 TDMM
Richard Smith
richard at exparrot.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.
A SECOND LOOK AT THE THREELEAD 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 singlemethod extent of W (a method that has some
undesirable property), and X, Y, Z are introduced via
threelead 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 WX splice are a,b. That means
that at the halflead of both methods, a must cross with b.
(The fact that W might have a jump change at the halflead
doesn't change that, though it may mean a and b are not
adjacent at the halflead of W.)
Similarly, the fixed bells for the WY splice much swap at
the halfleads 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 threelead splice between XY which is not
the case. (In all the cases considered they actually shared
a course splice.) So the WY fixed bells must be different
bells  let's call them c,d.
We now know that at the halflead of W, a,b cross, as do
c,d, and therefore e is the pivot. If W has a threelead
splice with Z, it must either use a,b or c,d as fixed bells,
in which case it must share a threelead 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 halflead variant of X and Y, but
*not* a threelead 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 halflead variant
having a 16 and 56 halflead change, respectively. Let's
create the 'W' method by putting a jump change at the
halflead such that it produces a J lead head. Using
Michael Fould's naming convention we can call this method
JDurham (or, equally, JSurfleet). The middle part of the
lead is shown below in the lefthand column:
JDurham reordered
264153 + 264153 +
624513 + 624513 +
642531 + 645231 
465213  465213 
645231  642531 +
243651  246351 +
423615  423615 
246351 + 243651 
264315 + 264315 +
624135 + 624135 +
The righthand column contains the same rows, but reordered
so that it has a rightplace parity structure. We can see
that 3 and 5 just ring Bourne in the right hand column, and
therefore JDurham has a threelead 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
halflead variant):
X (hlv) [W] (hlv) Y

(3)

Z
In fact, a similar diagram would be more relevant for
fivelead grid splices (such as between Ip/Cm/Yo) too 
although King Edward (the grid method) does have a
threelead splice with Ipswich and Cambridge, the relevant
fact is that it is a halflead 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
fivelead 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 threelead grid splices I discussed in the last
email were both involved Ma and Ol which are halflead
variants. (Technically, it's actually Marple and Willesden
that are halflead 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 threelead splice with OMarple, and
Morning Star with GMarple. We can draw this as:
Ma
/  \
/  \
/  \
Ms (3) GMa (hlvs) OMa (3) No
\  /
\  /
\  /
Ol
A composite course (whether fragmented or otherwise) needs
to have at least three different leadends. (At least,
because the GNJLO and OHMKG composites involve five
leadends.) 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 threelead 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 threelead splice has fixed bells in
56, 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 threelead splice has fixed bells in 23; it's fairly
clear that the six leads of Ms don't form threelead 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 singlemethod 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 lefthand ones include G
and the righthand 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 leadend
type is simply the seconds place version of the V leadend
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 singlemethod 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 singlemethod 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
leadend order represented three times  that happens in
none of the others. Obviously the singlemethods courses
have a leadend 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 onelead 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 onelead 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 fourlead fragments.
Fourlead 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 Ogroup method with a seconds place leadend
and 6G denotes a Ggroup method with a sixths place
leadend. 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 GGMOOlike 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 fourlead fragments.
Fourlead 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 fourlead fragment (6G + NOOO).
Altogether we now have five new fourlead 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 fourcourse 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 threelead splice (fixed
bells: 56); but so too do the leads of Ms (fixed bells:
23). 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 23 for G as in 56 for O, the OOJGG Parker
course has the same two pairs in 23 for G as in 56 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 threelead splice
with Ta with 35 fixed. This partially overlaps the No
splice slot which has 56 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 sixlead 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 35. That increases the plans by two depending on
whether there's six or twelve leads of Ki.
Next, Ms has a threelead 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 23 during the Ms, and so
we can ring Di whenever (d,e) are in 45. The plan has a
single Ta slot available.
Finally, being halflead variants, Ma and Ol have a course
splice. If we don't use the Ki sixlead 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 leadend 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 56. 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 MsMaOl and
MaOlNo in the sixth email, and we've just finished looking
at ways of splicing MsOlNo and MsMaNo. But we haven't
looked at ways of including all four methods. Yes, a lot of
the MsMaNo 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
threelead splice with No and one with Ms (because extents
without both have already been considered). The splice with
No involves a coursing pair (56), as does the one with Ms
(23). 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 56. That's exactly what's wanted to have
ab and cd in 56 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 threelead 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 56, and that
we mustn't have No when (d,e) or (e,a) were in 56. What
about the other five No slots  can we use any? The
requirement to have Ws when (b,c) are in 23 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 MaOlNoMs 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 threelead 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!
RAS
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