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Spliced treble-dodging minor - 7

2011-06-27T06:55:26Z

Holroyd: Created page with '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'…'

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!

Spliced treble-dodging minor - Introduction

2011-06-27T06:54:28Z

Holroyd:

[[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]]
|
[[Spliced treble-dodging minor - 7|Plans 7]]

Richard Smith richard at ex-parrot.com
Tue Sep 28 04:19:50 BST 2010

I've spent quite a lot of the last month looking at spliced
extents of treble dodging minor.

Thanks to a cunning algorithm (which I shall describe in a
moment) designed by Ander which we've been fine-tuning it
turns out to be possible to do exhaustive searches over
search spaces that I had previously thought were impossibly
large.

As a demonstration, I have just done a search for all true
extents of minor using just methods from the standard 147
treble dodging minor methods rung with 4ths place lead-end
bobs. I will do some further verification of this result
over the next few days, but I believe the number of extents
of this form is

5,862,727,200,079,423,275,554

To put this number into perspective, if I were to produce a
booklet listing these in a similar format to that used in
the CC's spliced minor collection, then the resulting
booklet would be about 5 light-years thick.

THE ALGORITHM

There are five main stages to the search algorithm.

First we remove lead splices and lead-end variants from the
list of methods. So, for example, we only want to include
one of Beverley, Surfleet, Berwick and Hexham. This reduces
the list of methods from 147 to 75.

The second stage is to associate each lead end or lead head
row with a method. Start with a list of the 60 in-course
rows with the treble leading -- these will all appear as a
l.e. or a l.h., and we need to choose a method for each one,
and doing so will join a l.h. to the subsequent l.e.

Suppose some l.e./l.h. rows already have methods chosen. Of
the remaining rows, we call a method 'possible' if

(i) the l.e. that would be reached by ringing a lead of
the method starting at the given l.h. row is not
associated with a method; and

(ii) the lead would be true against all other chosen
leads.

Take the row that has the fewest possible methods and, in
sequence, try each of its possible methods, recursing.
This gives an exhaustive tree search. The result of this is
a 'plan' -- a list of which method is rung from each lead,
but with no information on how to join the leads up.

Stage two can be speeded up significantly by implementing a
form of rotational pruning. Put the methods in some
arbitrary order. Any method (other than the first one
chosen) must not be before the first one chosen in the
ordering. This will remove some but not all rotations and
reflections. If you want an accurate count, it's a good
idea to check whether a plan is in its canonical rotation
and only output it if it is.

The third stage is to do an exhaustive search of ways to
join the 30 leads in each plan using 12, 14 or 16 lead end
changes. An normal tree search for compositions will do
this fine. There's no need to check for truth beyond
checking for repetition of lead heads and lead ends as this
was dealt with in stage two. For each plan you then have a
list of compositions that produce the extent.

Fourth, we remove compositions that include 16 lead ends in
London (3-3.4) or Hills (3-34.6) backworks. This is a
little subtle for plans that include one of these backworks
and another one -- as 16 lead ends are fine as long as they
only occur in the non-London, non-Hills backworks.

A further subtlety arises if rotational pruning was done in
stage two. Because there is no clear distinction between
rotation and reflection of a plan (because we don't yet know
which rows will become a l.h. and which a l.e.), pruning
removes both rotations and reflections. However, going from
Carlisle-over to London-over with a 16 l.e. is fine; but
going the other way is not.

This gives the complete set of extents.

Fifth, and assuming we want to count them, for each plan,
the number of extents is the product of three terms: the
number of distinct rotations / reflections (assuming
rotational pruning); the number of lead splices (N^n where N
is the number of methods in the lead splice set -- 2 or 4
for everything in the 147 -- and n the number of leads of
it); and the number of compositions for each plan. Adding
the values for each plan gives the overall total.

For the 147, the five stages took: 4s, 4h 1m, 1h 7m, 16m 44,
and 1m 18s. So the total search time was just under 6h.
I've only made an effort to optimise stages two and three
(stage five in particular is woefully suboptimal), but given
that's where most of the time is spent, that seems
reasonable. I reckon that without too much work the search
could be reduced to under 4h -- maybe even under 3h.

THE EXTENTS

Because the search first finds plans, and the number of
plans (modulo rotation) is a fairly managable 4614, it's
fairly easy to get a good idea of what's there. And a quick
scan through the list of plans shows that there are some
interesting plans that are new (at least to me). I'll give
a breakdown of what's there in a later email.

RAS

File:Bananabob.png

2011-01-16T09:48:29Z

Holroyd:

Template:Navigation

2010-10-28T06:56:08Z

Holroyd:

{| class="noprint toccolours" style="clear: both; width: {{{width|100%}}}; margin: 0.5em auto;" cellpadding="2"
|align="center" style="background-color:#336699; color:white; font-weight:bold;"|<div style="position:relative; width:100%; white-space:nowrap;">Contents</div>
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'''[[:Category:Articles|Miscellaneous Articles]]''' 
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|valign="top" width=33% |
'''[[:Category:Methods|Methods]]''' 
*[[Falsest Method]]
*[[Popular Surprise Major Methods]]
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'''[[:Category:Composition|Composition]]''' 
*[[Spliced treble-dodging minor - Introduction|Spliced treble-dodging minor]]
*[[Compositions of the Decade]]
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'''[[Central Council Decisions]]''' 
*[[Calls that pass to another part of the same course]]
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Spliced treble-dodging minor - 6

2010-10-28T06:53:08Z

Holroyd: Created page with '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 ex…'

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.

ISPWICH, CAMBRIDGE AND YORK

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

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.)

FRAGMENTED COMPOSITE COURSES

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.

FIVE-LEAD GRID SPLICES

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.

THREE-LEAD GRID SPLICE

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.

DURHAM, SURFLEET AND BOURNE

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 ---------
153246

146325 Su +
125634 Bo
---------
146325

(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
-----
110

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

http://www.rrhorton.net/learycomps.html#learycomps.comp00

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.)

NEWCASTLE, MUNDEN AND SANDIACRE

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.

COTSWOLD, MENDIP, CHILTERN AND CHEVIOT

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.)

MARPLE, OLD OXFORD AND NORWICH

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.

MARPLE, OLD OXFORD AND MORNING STAR

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
c

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

SUMMARY

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.

Email
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.

Spliced treble-dodging minor - 5

2010-10-28T06:52:11Z

Holroyd: Created page with 'Richard Smith richard at ex-parrot.com Mon Oct 18 18:37:06 BST 2010 And here's the next installment. This email is relatively short as it only discusses a single type of splic…'

Richard Smith richard at ex-parrot.com
Mon Oct 18 18:37:06 BST 2010

And here's the next installment. This email is relatively
short as it only discusses a single type of splice. This
splice accounts for 1224 of the 1951 as-yet unexplained
plans.

SPLICE SQUARES

A lot of three-lead splices work by similarly to the one
between London and Wells -- by swapping 34.16.34 for
14.36.14 at the half-lead. (In the first email in this
series, these three-lead splices were marked with an
asterisk.) Methods with the London and Wells underworks are
one example of this; the Canterbury and Abbeyville
underworks are another example, as are the Bucknall and
Castleton underworks. I shall refer to these as London- or
Wells-like underworks.

Any method with a London-like underwork has a three-lead
splice with the corresponding Wells-like method. What if
the London-like method also has another splice (e.g. a
six-lead splice) with a different London-like method? That
method will also have a Wells-like variant which will have a
splice back with the first Wells-like method. The result is
a square of splices:

3-lead
L1 ---------- W1
| |
| e.g. | e.g.
| 6-lead | 6-lead
| |
L2 ---------- W2

Amongst the 147, there are two sets of methods with these
splices:

W1 L1 L2 W2
--------------
Nw Ak Cz Ww
We Lo Bn Cx

For the rest of this discussion, I shall assume the second
splice is a six-lead splice as this is the case for both of
these sets of methods. With a larger set of methods, we
might find examples where L1 and L2 shared, say, a course
splice or another type of three-lead splice.

Let imagine we start with an extent of L1 and splice in some
L2. As discussed in the third email, if there are 18, 24 or
30 leads of L1, we can splice in some W1; likewise if there
are only 6 or 12 leads of L1, we can add some W2. That
email also explained why it was not possible to get all four
methods using simple splices in either of:

W --(6)-- X --(3)-- Y --(6)-- Z

W --(3)-- X --(6)-- Y --(3)-- Z

However, in the case we're currently considering, the
diagram is now a square (W and Z are connected). This
allows us to go beyond the realm of simple splices.

Imagine we have 12 leads of L1 (when a or b pivot) and 18
leads of L2. The only L2-W2 splice slot available is the
one with (a,b) as the the fixed bells. But why can't we use
an arbitrary splice slot? All of the leads have a
London-like underwork. Why can't we just swap the
London-like bit for the corresponding Wells-like bit without
concerning ourselves about what's happening elsewhere in the
method? The answer is that we can and it will result in us
changing some of each of L1 and W1 into L2 and W2.

For example, the following three-part works by ringing Lo/We
when 5 or 6 pivots and Bn/Cx otherwise. The Wells-like
underwork is rung when 6 and any other bell crosses on the
front, which happens in 2nds & 4ths place bells Lo/We, and
3rds & 6ths place bells Bn/Cx.

123456 Lo
142635 Bn
164523 Bn
156342 Cx
- 123564 Lo
- 145236 Cx
124653 We
162345 Bn
136524 Cx
- 145362 Lo
------
134256

Twice repeated.

Unfortunately neither set of methods allows a mixture of
2nds and 6th place methods from the 147 -- the Lo/We/Bn/Cx
one because the 6ths place variants all have four blows
behind and are not included in the 147; the Nw/Ak/Cz/Ww one
because they have J/M lead-ends, and it's not possible to
mix both lead ends in a round block without adding a non-J/M
lead-end method.

The idea is very simple and it must surely have been
discovered before. I can't check Michael Foulds' books as
I've leant them to someone and not got them back :-/

In principle it would be possible to extend the plan further
if any of the four methods had a another three- or six-lead
splice (but not a course splice). However it turns out that
none of the methods in question have such a splice.

COUNTING THE CORRESPONDING PLANS

Counting how many plans this is responsible is very long,
tedious and not especially elucidating. Skip to 'SUMMARY'
if you don't care about this. The reason I've been
carefully counting these up is two-fold: (i) it's a good way
of checking I understand the limitations of what can be done
with the plan; and, more importantly, (ii) it allows me to
verify there are no plans hiding amongst them that need
explaining in some simpler way.

We're only interested here in plans that include all four
methods. Plans with one or two methods were covered in the
first email, and those with three in the third.

Lets start by considering the case where we have 6 leads of
L1 (when bell a pivots) and 24 of L2. The splice slots
(a,b), (a,c), (a,d) and (a,e) are all equivalent under
rotation and do not introduce any W1. The other six splice
slots introduce both W1 and W2, and are all equivalent:
(b,c), (b,d), (b,e), (c,d), (c,e), (d,e).

We've already established (see fourth email) that there are
twelve ways of choosing slots from these six. We can choose
none, but we're not interested in that case (as it results
in no W1). We can choose one slot from the six -- say
(b,c). How many ways of choosing (a,x) slots are there?
Two of the four slots involve b or c, and two do not. That
gives two ways of choosing one, and seemingly three ways of
choosing three (depending on whether zero, one or two
involve b or c). However, the choice (a,b)+(a,d) is chiral
as each of the five bells is in some way unique. That gives
1+2+4+2+1 = 10 ways of choosing from (a,x).

With two slots from the six, they can overlap (b,c)+(c,d) or
not (b,c)+(d,e). In the former case, there are three types
of (a,x) slot: (a,b) and (a,d) are equivalent, the other two
are both unique. That seemingly gives 1+3+4+3+1 = 12 ways
of choosing (a,x), except that the choices involving
only one of (a,b) and (a,d) choice split because of
chirality. That increases it to 1+4+6+4+1 = 16 plans.

In the latter case (two non-overlapping slots), all the
(a,x) slots start indistinguishable, but once one, say
(a,b), is chosen, one of the remaining slots (a,c) is now
distinct because of the selected (b,c) slot. If we don't
select that slot, the choice (a,b)+(a,d) splits on chiral
grounds. That gives 1+1+3+1+1 = 7 plans.

There are three ways of choosing three of the six non-(a,x)
slots:

a --- b a --- b a --- b
/ \ / / \
/ \ / / \
c --- d c d c d

[chiral pair]

In the first case, we have 1+2+4+2+1 = 10 plans, doubled
to 20 because of chirality. In diagrams for the the second
and third cases are (modulo rotation) complementary graphs
(if an edge is present in one, it's not in the other and
vice versa), so the number of plans from each will be the
same. In both cases we have 1+2+2+2+1 = 8 plans, doubled
makes 16.

Four or five of the non-(a,x) slots are the same as two or
one. And we don't want all six of them because otherwise
there's no L1.

That gives a total of 10+(16+7)+(20+16)+(16+7)+10 = 102
plans with six leads of L1/W1 which agrees with what was
actually found.

What about when there are 12 leads of L1, say when a and b
pivot. Only the (a,b) slot will give just W2, so we need at
least one of the other nine slots to be present. I'm going
to take a slightly different approach to counting these.
I've already enumerated (in the first email) the 38 ways of
choosing 1 to 9 three-lead slots. I'm going to look at
each of these in turn and count the ways of assign a, b to
two of the nodes in the graph.

With one slot:

(1.1) A --- B C D E

there are three ways of assigning (a,b) to these: (A,B),
(A,C) or (C,D). In the case (a,b) = (A,B) we have no
non-(a,b) slots so we're not interested in it. That leaves
two relevant ways.

I'm not going to repeat all the diagrams here -- see the
first email for them. I'm just going to enumerate the
ways of assigning (a,b) for each plan. An asterisk denotes
a chiral pair.

1.1 (A,C) (C,D) = 2

2.1 (A,B) (A,C) (A,D)* (B,D) (D,E) = 6
2.2 (A,B) (A,C)* (A,E) = 4
----
10

3.1 (A,B) (A,C) (A,D)* (B,D) (D,E) = 6
3.2* (A,B) (A,C) (A,D) (A,E) (B,C) (B,E) = 12
3.3 (A,B) (A,C) (A,E) (B,E) = 4
3.4 (A,B) (A,D) (D,E) = 3
----
25

4.1* (A,B) (A,C) (A,D) (A,E) (B,C) (B,D) = 12
4.2 (A,B) (A,C) (A,D)* (B,C) (B,D)* (C,D)* (D,E) = 10
4.3 (A,B)* (A,C) (A,D)* (A,E)* (B,D) (B,E) (D,E) = 10
4.4 (A,B) (A,D) (D,E) = 3
4.5 (A,B)* (A,D) (A,E) = 4
4.6 (A,B) (A,C) = 2
----
41

5.1* (A,B) (A,C) = 4
5.2 (A,B)* (A,C)* (A,D)* (A,E) (B,C) (B,D) (C,D) = 10
5.3 (A,B)* (A,C)* (A,D)* (A,E) (B,C) (B,D) (C,D) = 10
5.4* (A,B) (A,C) (A,D) (A,E) (B,C) (B,E) = 12
5.5 (A,B)* (A,C) (A,D) (B,C) (B,E) = 6
5.6 (A,B)* (A,D) (A,E) (B,C) (B,E) = 6
----
48

By symmetry we can write down the number of plans with 6, 7
or 8 slots: 41, 25 and 10. With 9 slots, as with 1 slot,
one of the three choices is irrelevant because it leaves us
with no L1. That gives 2+10+25+41+48+41+25+10+2 = 204
plans.

Clearly with 18 or 24 leads of L1 there are another 204+102
plans. That gives 612 in total for one set of four methods.
The 147 TDMM contains two sets of four methods in this
arrangement, so that explains a total of 1224 further plans.

SUMMARY

This one type of extent accounts for just over a quarter of
all the plans (modulo rotation) involving methods from the
147 TDMM. Having four separate methods breaks the symmetry
of the plan quite a lot meaning that rotational pruning
doesn't remove all that many plans. But the two involved
splices work together well so that there are lots of
possible plans.

Of the 506 clusters of plans, 14 were explained by simple
splices (in the first three emails), 26 by grid splices, 4
by triple-pivot grid splices, 1 by the hidden triple-pivot
grid splice (all in the fourth email), and a further 388 by
the splices squares described here.

That means that we now have 73 clusters containing, in
total, 727 plans left to explain.

Spliced treble-dodging minor - 4

2010-10-28T06:50:53Z

Holroyd: Created page with 'Richard Smith richard at ex-parrot.com Sat Oct 16 02:10:49 BST 2010 Despite impressions, I haven't yet got bored of this. I've now analysed all of the composition plans that ca…'

Richard Smith richard at ex-parrot.com
Sat Oct 16 02:10:49 BST 2010

Despite impressions, I haven't yet got bored of this.

I've now analysed all of the composition plans that can be
explained solely in terms of simple splices -- that is,
course, three- and six-lead splices. This was described
in the following series of emails:

http://ex-parrot.com/~richard/r-t/2010/09/003650.html
http://ex-parrot.com/~richard/r-t/2010/09/003660.html
http://ex-parrot.com/~richard/r-t/2010/10/003675.html

That explained 2280 of the 4614 plans. The remaining 2334
plans are listed here:

http://ex-parrot.com/~richard/minor/147/compound-plans.txt

I've noted in another thread that these can be partitioned
into 492 clusters of plans where each cluster contains plans
related by simple plans. These plans are interesting
because this is where any potential new compositions will be
found. So let's start to look at these. This email covers
grid splices, the related (though often overlooked)
triple-pivot grid splice, and a new (if rather useless)
generalisation which I've termed the hidden triple-pivot
grid splice.

GRID SPLICES

Grid splices are the best-understood splice that is not a
simple splice. A grid splice involves three methods with
different lead-ends in the ratio 2:2:1, and the choice of
method depends solely on the position of the observation
bell for the splice. (That is sufficient for a definition.)
The most rung example of a grid splice must be the one
between Cambridge, Ipswich and Bourne which have H, K and J
lead ends, respectively:

123456 Ip
142635 Bo
156342 Ip
- 123564 Cm
164352 Cm
- 145236 Ip
124653 Bo
136524 Ip
- 145362 Cm
162534 Cm
---------
134256

Twice repeated

By using a Parker splice to get both 2nds and 6ths place
lead ends, A.G. Driver produced a three-part arrangement
with six methods -- the so-called 'Cambridge six'.
(Incidentally, I was slightly surprised to see no mention of
Driver's work composing spliced minor in his obituary in
the RW a fortnight ago [RW 2010, p1000].)

Grid splices were mentioned in passing in the 'OTHER EXTENTS
WITH FOUR METHODS' section of my third email looking at
simple splice plans, when I said

> X --(3)-- G --(3)-- Y [diagram relabeled]
> |
> (6)
> |
> Z
>
> This arrangement of splices is the one that makes a grid
> splice work, except that for a regular grid splice, G is
> an irregular method and entirely removed.

When all four methods are regular methods, they all must
have the same lead-end order, and so under the definition of
a grid splice that I've adopted, an arrangement of X, Y and
Z doesn't count as a grid splice.

However, if method G is irregular then X, Y and Z must each
have different lead-end orders if they are regular. This
means that, because we're only searching for regular
methods, we will see three-method plans for X, Y and Z in
the list of compound (i.e. non-simple) plans. More
generally, G might have other undesirable properties such as
multiple consecutive blows in one place or jump changes.

ENUMERATING GRID SPLICES

This raises an interesting question. To enumerate simple
splices, we simply look at each pair of methods in turn and
ask what splice they have. Even for a fairly long list of
methods, that's quite efficient. But how do we efficiently
enumerate grid splices?

One possibility is to take a list of methods that include
irregular methods, and look at the simple splices between
all pairs of methods. Then whenever we have a set of of
four methods G,X,Y,Z with different lead end orders that
share splices as marked shown in the diagram above, we know
that X,Y,Z have a grid splice. The problem with this is
that the list of methods needs to be very long so as to
include G even when it contains jump changes or some other
undesirable property.

Another possibility is to look at all choices of three
method, X,Y,Z, put them into a grid splice and see if it's
true. The strategy I've used is a variation on this. I've
taken my code for searching for extents of the 147 and
modified it to search for plans with some part-end group.
The relevant part-end group is the 12 in-course rows of the
form 1....6 -- these are the course heads and course ends of
the composition, and by having both, we're taking into
account the palindromic nature of the grid splice.

This loses some of the search's efficiency as it means
droppping rotational pruning; it also complicates the
inter-method falseness handling. (Mathematically, one way
of thinking about the latter is as a consequence of the fact
that, unlike Cayley graphs, Schreier graphs are not vertex
transitive.)

That search turns up 53 grid splices listed below:

X Y Z course #plans
--------------------------------------

Cc Lo Ke/Ce (S) HKJKH 2
Mp So Ke/Ce (S) HKJKH 2
Li/Pv Pn Sg (S) HKJKH 5
Fo Sa/Te Ti/Tr (S) HKJKH 10
Cm Ip Bo/Ne (S) HKJKH 2

C1 Mp So/Pn (Q) GHKHG 2
Cu/Cl Nb Sa (Q) GHKHG 5
Di/Ws Es Po (Q) GHKHG 5

Dn Yo Cm/Su/Bs (R) JGHGJ 3
Wl/Bo Ey Cj (R) JGHGJ 5
Bp Bu Cm/Su/Bs (R) JGHGJ 3
Dk Di/Ms Be (R) JGHGJ 5

Ip Bo/Ki Ey (P) KJGJK 5
Rc Bp Bu (P) KJGJK 1
El/Ol Bo/Ki Bu (P) KJGJK 36
Te Tr Ms (P) KJGJK 1

Qu/Tr Kt Po (U) MONOM 5
Dk Ox Po (U) MONOM 1

No El/Ol Be (W) ONLNO 5
No Ip Es (W) ONLNO 1

Do Fr Ey (-) OHGLO 1
Do Cj Bu (-) OHGLO 1

C1 Pn Kt (-) GNOKG 1
Di/Ms Rc Ox (-) GNOKG 5

No Ms Ki/Ma (-) OOJGG 2 } Cannot be made
No Di/Ws Bo/Ne (-) OOJGG 10 } to join up

In each grid splice, there are 12 leads of each of the
methods listed in the first two columns (headed X and Y),
and 6 of the method in the last (Z) column. Where several
grid splices just differ by a simple splice (i.e. a three-
or six-lead splice), they're listed on the same line above.
With two (or more) methods in the Z column, it's not
possible to get more than one of them in the composition,
because they share a six-lead splice.

But when there are two methods in the X or Y columns, the
methods share a three-lead splice and both methods can be
present in composition. There are four 3-lead splice slots
for X or Y (with the observation bell each each other bell
as the fixed bells). This gives rise to five plans (up to
rotation and reflection) depending on whether 0, 1, 2, 3
or 4 of the slots are used.

In one case, both columns X and Y have two methods. We can
label the four X splice slots a, b, c and d, and there are
four corresponding Y splice slots with the same fixed bells.
There are two X methods: lets call them X1 and X2. If we
have no X1 or no X2 then, we have five different ways of
applying the Y splice (solely depending on the ratio of the
two Y1 methods). With one splice slot used to get X1
(say slot a), we have eight ways of choosing Y:

0, a, b, a+b, b+c, a+b+c, b+c+d, a+b+c+d

With equal amounts of X1 and X2 (say by having X1 at a and
b) we have nine ways of choosing Y:

0, a, c, a+b, a+c, c+d, a+b+c, a+c+d, a+b+c+d

However we need to think about chirality. This is relevant
in one case -- when a+b are X1 and a+c are Y1:

Y1
a ------ c
| |
X1 | | X2
| |
b ------ d
Y2

If we relabel (say) a and b, we also need to relabel c and
d. That's an even parity relabeling, so we've got two
versions of that plan. That a total of gives 5+8+10+8+5 =
36 plans.

The table above shows the number of plans for each grid
splice. Adding them all up gives 124 plans. With a few
moments thought, we can see that it's not possible to build
on a grid splice by adding further methods.

COMPOSITE COURSES

[This section is a digression from the analysis of the
extent plans found in the search.]

The fourth column of the table of grid splices shows, in
parentheses, the lead end order of the grid method -- the
method G in the diagram at the top of this email. Of the
eight irregular lead-end codes, only six are represented
above. S and V are just lead-end variants of each other;
the lead-end code that's really missing is T. With a larger
selection of methods to play with, it's possible to get grid
splices were the grid method is T-group method; however, it
turns out that there are no suitable methods in the 147.

The other thing in the fourth column is the composite course
-- that is, the sequence of lead-end codes that make up the
course. There are eight of these corresponding to the eight
possible irregular lead-end codes.

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

The left hand set of columns corresponds to grid splices
with seconds place lead ends; the right hand to sixths place
ones. Only the S / V line corresponds to the same methods,
because that is the only irregular lead end that produces a
five-lead method with both 2nds and 6ths place lead ends.

It's easy enough to see that, with 2nds place lead ends,
there ought to be 24 ways of ordering the lead heads in the
course. (The first, rounds is fixed, the remaining four can
be in any order.) 20 of these are the composite courses
shown in the second column above (five rotations of each of
the four courses); the remaining four are single method
courses (GGGGG, HHHHH, JJJJJ and KKKKK). Similarly for 6ths
place lead ends.

The 'Parker' column corresponds to courses with mixed lead
heads. A Parker course is not a round block. Just as it
would go false, a bob is called to bring up the course
head 156423 -- the 4th there is observation, and if the 4th
is at the back, a 12 l.e. is rung, and if the 4th is at the
front, a 16 l.e. is rung. As with the 2nds and 6ths place
courses, the lead end/head pairs can crop up in 24 different
orders. 16 are in the table above, and are derived from the
composite courses above.

What are other eight? Six are miscellaneous courses that do
not correspond to a 2nds or 6ths place course because they
have both G and O group methods:

OHGLO GGMOO OOJGG GNJLO OHMKG GNOKG

The remaining two are derived from single-method courses,
instead of composite courses. These are are standard Parker
splices for, say, Cambridge/Primrose and Ipswich/Norfolk.

HLLHH NKKNN

TRIPLE-PIVOT GRID SPLICES

This neglected splice is closely related to the grid splice.
I believe Michael Foulds mentions it in passing in the
fourth of his excellent series of books, but I've lent my
copies to someone and so can't check. (If whoever has them
is reading this, can I have them back?)

When I generated the list of grid splices, I asked my
computer to generate a list of all plans with a
twelve-element part-end group (A_4). This is effectively
looking for palindromic courses that can be rung in each of
the six courses to give the extent.

Obviously this produced all the single-method plans, as well
as all the three-lead splices (showing up with methods in
the ratio 3:2, with the splice applied solely based on the
position of a single observation) and all the six-lead
splices (with methods in the ratio 4:1). Grid splices
turned up with a method ratio of 2:2:1; plans with both
three- and six-lead splices and plans with two three-lead
splices also had a 2:2:1 ratio, looking much like grid
splices, except that all the methods had the same lead end
group. That was all I anticipated finding.

In fact I found a further 51 palindromic course plans, many
of which are triple-pivot grid splices.

In grid splices, the grid method has a lead end that swap
two pairs of bells. For example, the grid method to the
Cm/Ip/Bo grid splice is King Edward which has lead end
156423, swapping 2-5 and 3-6; 4 is the pivot bell. 2-5 are
then used as fixed bells for the three-lead splice with Cm,
3-6 for the three-lead splice with Ip, and 4 for the
six-lead splice with Bo. That means the half lead change in
the grid method must be in the 2,2,1,1 equivalence class --
by which I mean it has two pairs of bells swapping and two
fixed bells.

But what if the half-lead change in the grid method is in
the 3,1,1,1 equivalence class? Clearly that's not possible
for an ordinary change, but there's no reason why the grid
method shouldn't have a jump change at the half lead. For
example, Norwich with the following underwork:

234165 +
243615 +
423651 +
246315 -
426351 - } jump
642351 - } change
462315 -
643251 +
463215 +
436125 +

It's fairly straightforward to see that this 'method' has a
six-lead splice with Bedford (with 3 fixed), and also with
Old Oxford (with 5 fixed). (The 3 just rings pivot bell
Bedford, and the 5 pivot bell Old Oxford.) Perhaps less
clearly, it also has a three-lead splice with with Marple
with 3 and 5 fixed. That's because we can relabel 3 and 5
at the half-lead and make a corresponding relabelling to 2,
4 or 6 to preserve parity

This allows us to take an extent of this 'method', do a
six-lead splice with Be when the observation is 3rds place
bell, do another six-lead splice with Ol when the
observation is 5ths place bell, and do three-lead splices
with Ma whenever the observation is not 3rds or 6ths place
bell. The reason for the name (triple-pivot grid splice) is
that the observation bell rings the pivot bell in all three
methods.

Joining the parts up can be a little delicate because each
course fragments into two bits meaning lots of bobs are
required, but it's often possible. For example,

123456 Ta
- 156423 Ol
- 134562 Ta
125634 Ta
- 134625 Ol
163542 Ta
- 142563 Be
163254 Ta
- 154263 Ta
132654 Be
---------
- 125463

Twice repeated;
no 65s at back

In total, there are 37 triple-pivot grid splices using
methods from the 147:

X Y Z Additional splices
-----------------------------------------------------
Av/Ca So/Pn Ke/Ce [3-lead: Ca/Gl] *
C3/C2 So/Pn Cc/Pv/Mp/By/Bh/Bw
Ma/Ta Bm/Ol Be [6-lead: Ma/Ki]
Bs Bu Ta [6-lead: Bs/Cm/Su]

[* = these plans cannot be joined up]

The method(s) in the X column are the three-lead splice of
which there are 18 leads -- the six splice-slots involving
the non-observation bells (a,b), (a,c), (a,d), (b,c), (b,d)
and (c,d). This means that when there are two methods (X1
and X2) in the X column, we can incorporate both.

Clearly there's one way of having just X1, and one way of
having one slot of X2. Two X2 slots: either they overlap or
they don't: (a,b), (a,c) or (a,b), (c,d). And there are
three ways of chosing three slots:

a --- b a --- b a --- b
/ \ / / \
/ \ / / \
c --- d c d c d

The first of these is a chiral pair. That gives twelve
plans when there are two methods in the X column. So there
are 144 = 12 * 2 * 6 plans in the cluster containing C2/C3
as X.

But with 18 leads of an X method, if X has a six-lead
splice, it's possible to incorporate six leads of that
method too -- for example Cm or Su splices with Bs. These
are indicated in the table above, and generate one more
plan for each choice of Y and Z. That means the Ma/Ta
cluster contains 26 = (12+1)*2 plans, and the Bs cluster
contains just 3.

It's also possible to include a three-lead splice into X.
As there's only one set of methods where this applies, we
may as well be concrete about it. The six X slots can each
be either Av or Ca, and if we have enough Ca we can splice
Gl in using Ca's other 3-lead splice. If bell e is the
observation for the triple-pivot grid splice, the six slots
for Ca or Av are (a,b), (a,c), (a,d), (b,c), (b,d) and
(c,d).

If all of these are Ca (i.e. we have no Av), then we have
four Ca-Gl splice slots: (a,e), (b,e), (c,e) and (d,e).
(Bell e must be involved in the Gl splice, because otherwise
some of the Gl leads will fall in the Y or Z methods.) This
gives four extra plans depending on whether 1, 2, 3 or 4 of
these slots are used. It's worth noting that if all four
slots are used, the only leads of Ca remaining are when
the observation is pivot bell.

If one slot is Av -- say (a,b) -- then either a or b must
also be involved in the Gl splice leaving Gl two slots:
(a,e) and (b,e). That gives another two plans. With two
overlapping slots of Av, (a,b) and (a,c), then there's just
one Gl slot: (a,e). The same is true when there's three
mutually overlapping Av slots: (a,b), (a,c), (a,d).

In all, that gives a total of 8 extra plans involving Gl.
So the number of plans in the Av/Ca cluster is (12+8)*2*2 =
80.

Adding these all up, we have 80+144+26+3 = 253 plans.

HIDDEN TRIPLE-PIVOT GRID SPLICES

There's one final developement to the triple-pivot splice
that warrants discussion. The triple-pivot splice works by
having an imaginary method G which has two different
six-lead splices (methods Y and Z, with fixed bells a and b,
respectively) and a three-lead splice (with X when a,b are
fixed). Sometimes X has a six-lead splice with another
method, W, allowing the pivot leads of X to be removed. And
sometimes X has a different three-lead splice with a method,
V, which allows some or all of the non-pivot leads of X to
be removed.

W Y
\ /
\ /
X ------ G
/ \
/ \
V Z

In principle, with suitable methods, this means we might be
able to remove all of X, just leaving V, W, Y and Z. There
are no suitable choices of V,W,X,Y,Z amongst the 147 to make
this possible, but why does X need to be one of the 147?
We've already accepted that the grid method, G, can be
outside of the 147 (e.g. by having jump changes) -- the same
can be true of X.

Any examples of this will have been found by my search for
grid splices, and the only such plans are given below:

V W Y Z
-----------------------
Av Mu/Cl/Gl Te Ti/Tr

This accounts for 6 = 3*2 further plans. It's fairly clear
that we cannot apply any further simple splices to this to
add additional methods.

Sadly none of these plans can be joined up to give a working
extent.

The labeling of W, Y and Z is somewhat arbitrary. Despite
the diagram above, it's not the case that W has a different
status to Y and Z in the splice by virtue of the fact that W
splices with X, and Y and Z with G. As X and G are really
just arbitrary sets of rows, we can recombine them
differently to get two other methods X' and G' such that
it's Y that splices with X' and W and Z with G'.

I wonder whether further investigation of this kind of use
of imaginary methods might yield a general theory of
splicing that would allow us to understand splicing of three
or more methods as well as we currently understand the
splicing of two methods.

SUMMARY

My initial intention in this email was to look just at grid
splices. To create an exhaustive list of grid splices I ran
a plan search using the twelve in-course 1....6 rows as the
part-end group. This found all plans where each course was
the same and also palindromic. As well as finding grid
splices, this turned up some triple-pivot grid splices
(about which I had forgotten), and a generalisation of this.
We've now enumerated all plans related to these by simple
(i.e. course, three- or six-lead) splices.

To update the running count of plans, this shows:

Single method plans . . . . . . . . . . 75 \
Course splices . . . . . . . . . . . . . 108 | See first
Six-lead splices . . . . . . . . . . . . 176 | email
Three-lead splices . . . . . . . . . . . 798 /
Multiple course splices . . . . . . . . 36 \ See second
Multiple six-lead splices . . . . . . . 286 | email
Multiple three-lead splices . . . . . . 412 /
Combined course & three-lead splices . . 198 \ See third
Combined six- & three-lead splices . . . 163 | email
Other simple extents with four methods . 28 /
Grid splices . . . . . . . . . . . . . . 124 \ See this
Triple-pivot grid splices . . . . . . . 253 | email
Hidden triple-pivot grid splices . . . . 6 /
---------------------------------------------
TOTAL . . . . . . . . . . . . . . . . . 2663

Only 1951 plans left to explain, and these promise to be
particularly interesting as we've now basically exhausted
the standard splicing recipes. Stay tuned.

Spliced treble-dodging minor - Introduction

2010-10-28T06:48:40Z

Holroyd:

[[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
Tue Sep 28 04:19:50 BST 2010

I've spent quite a lot of the last month looking at spliced
extents of treble dodging minor.

Thanks to a cunning algorithm (which I shall describe in a
moment) designed by Ander which we've been fine-tuning it
turns out to be possible to do exhaustive searches over
search spaces that I had previously thought were impossibly
large.

As a demonstration, I have just done a search for all true
extents of minor using just methods from the standard 147
treble dodging minor methods rung with 4ths place lead-end
bobs. I will do some further verification of this result
over the next few days, but I believe the number of extents
of this form is

5,862,727,200,079,423,275,554

To put this number into perspective, if I were to produce a
booklet listing these in a similar format to that used in
the CC's spliced minor collection, then the resulting
booklet would be about 5 light-years thick.

THE ALGORITHM

There are five main stages to the search algorithm.

First we remove lead splices and lead-end variants from the
list of methods. So, for example, we only want to include
one of Beverley, Surfleet, Berwick and Hexham. This reduces
the list of methods from 147 to 75.

The second stage is to associate each lead end or lead head
row with a method. Start with a list of the 60 in-course
rows with the treble leading -- these will all appear as a
l.e. or a l.h., and we need to choose a method for each one,
and doing so will join a l.h. to the subsequent l.e.

Suppose some l.e./l.h. rows already have methods chosen. Of
the remaining rows, we call a method 'possible' if

(i) the l.e. that would be reached by ringing a lead of
the method starting at the given l.h. row is not
associated with a method; and

(ii) the lead would be true against all other chosen
leads.

Take the row that has the fewest possible methods and, in
sequence, try each of its possible methods, recursing.
This gives an exhaustive tree search. The result of this is
a 'plan' -- a list of which method is rung from each lead,
but with no information on how to join the leads up.

Stage two can be speeded up significantly by implementing a
form of rotational pruning. Put the methods in some
arbitrary order. Any method (other than the first one
chosen) must not be before the first one chosen in the
ordering. This will remove some but not all rotations and
reflections. If you want an accurate count, it's a good
idea to check whether a plan is in its canonical rotation
and only output it if it is.

The third stage is to do an exhaustive search of ways to
join the 30 leads in each plan using 12, 14 or 16 lead end
changes. An normal tree search for compositions will do
this fine. There's no need to check for truth beyond
checking for repetition of lead heads and lead ends as this
was dealt with in stage two. For each plan you then have a
list of compositions that produce the extent.

Fourth, we remove compositions that include 16 lead ends in
London (3-3.4) or Hills (3-34.6) backworks. This is a
little subtle for plans that include one of these backworks
and another one -- as 16 lead ends are fine as long as they
only occur in the non-London, non-Hills backworks.

A further subtlety arises if rotational pruning was done in
stage two. Because there is no clear distinction between
rotation and reflection of a plan (because we don't yet know
which rows will become a l.h. and which a l.e.), pruning
removes both rotations and reflections. However, going from
Carlisle-over to London-over with a 16 l.e. is fine; but
going the other way is not.

This gives the complete set of extents.

Fifth, and assuming we want to count them, for each plan,
the number of extents is the product of three terms: the
number of distinct rotations / reflections (assuming
rotational pruning); the number of lead splices (N^n where N
is the number of methods in the lead splice set -- 2 or 4
for everything in the 147 -- and n the number of leads of
it); and the number of compositions for each plan. Adding
the values for each plan gives the overall total.

For the 147, the five stages took: 4s, 4h 1m, 1h 7m, 16m 44,
and 1m 18s. So the total search time was just under 6h.
I've only made an effort to optimise stages two and three
(stage five in particular is woefully suboptimal), but given
that's where most of the time is spent, that seems
reasonable. I reckon that without too much work the search
could be reduced to under 4h -- maybe even under 3h.

THE EXTENTS

Because the search first finds plans, and the number of
plans (modulo rotation) is a fairly managable 4614, it's
fairly easy to get a good idea of what's there. And a quick
scan through the list of plans shows that there are some
interesting plans that are new (at least to me). I'll give
a breakdown of what's there in a later email.

RAS

Spliced treble-dodging minor - 3

2010-10-28T06:47:50Z

Holroyd: Created page with 'Richard Smith richard at ex-parrot.com Wed Oct 6 02:05:43 BST 2010 This is that third email. But first to correct a typo in the second email. At the end of the 'MULTIPLE SIX-…'

Richard Smith richard at ex-parrot.com
Wed Oct 6 02:05:43 BST 2010

This is that third email.

But first to correct a typo in the second email. At the end
of the 'MULTIPLE SIX-LEAD SPLICES' section, I said:

> [...] That gives a total of 2*6+82+182 = 276 plans.

This should have said 2*6+82+192 = 286 plans. (The 192
terms was correct in the previous table.) The table at the
end of email needs updating accordingly; but this is
repeated (and extended) at the end of this email.

Back to the analysis ...

COMBINING COURSE AND SIX-LEAD SPLICES

Each bell pivots once during a course (of a single method)
so it is not possible to combine course and six-lead splices
in a single extent using simple splices. (It might be
possible to do some cunning cross-splice type thing with
suitable methods, though I'm not aware of any. But it would
then no longer be a simple splice and so is beyond the scope
of this calculation).

COMBINING COURSE AND THREE-LEAD SPLICES

Course splices do combine with three-lead splices, as the
extents of the six wrong-place Cambridge-over surprise
methods demonstrate. The table below shows all methods
where X-Y have a course splice and Y-Z have a three-lead
splice (the fixed bells for which are marked).

X Y Z
----------------------
Ol Ma Ta (3&5)
Ne Lf Wm (2&5)
Dk/Ox Ms Di (4&5) [see below]
Po Ws Di (2&3)
Ma Ol El (2&4)

Su Du Yo (2&3)
Ey/Do Wl Bo (2&6) [see below]
Ws Po Sa (2&4)
Mu Nw Ak (2&6)
C3 Pn Nm (2&4)

Pn C3 C2 (3&5)
Pv Cx Bn (3&6)
Ce Av Ca (4&5)
Cx Pv Li (2&5)
Lo Cu Cl (2&3)

Nb Cl Cu (2&3)
Cu Lo We (2&4)

(Lead splices and lead-end variants have been excluded from
the table for reasons of brevity.)

Lets start with method X and add courses of Y. Clearly we
need at least three courses of Y before we can
exploit the Y-Z three-lead splice. And if we have six
courses of Y, there's no X left and the splice has been
covered elsewhere.

Up to rotation, there are two ways of selecting three
courses to make Y. In one way, the three Y courses share a
coursing pair; in the other way they don't. If the
three-lead splice involves a coursing pair (e.g. 3&5 for
Ma-Ta) then the former choice of three courses allows a
single application of the three-lead splice and the latter
none; if the three-lead splice involves a non-coursing pair
(e.g. 2&5 for Lf-Wm) then it's other choice of three courses
that allows the three-lead splice to be applied. Either
way, that gives us one plan (up to rotation).

There is just one way of select four courses of Y. We know
that the two courses of X share two coursing pairs, which
means that 8 = 5*2-2 coursing pairs have used in the X
leaving two that can be used for the three-lead splice. (If
the three-lead splice involves non-coursing pairs, change
'coursing' for 'non-coursing' in the preceding sentence.)
That contributes two plans depending on whether we have one
or two applications of the three-lead splice.

Finally, five courses of Y which can be chosen in just one
way. Only five coursing pairs are involved in the course of
X leaving five viable three-lead splice slots. These are
(2,3), (3,5), (5,6), (6,4), (4,2) if the splice involves a
coursing pair or (2,5), (5,4), (4,3), (3,6), (6,2)
otherwise. Either way, we can label the slots

(a,b), (b,c), (c,d), (d,e), (e,a)

There's one way of choosing one slot, two of choosing two
(together or separate), two of choosing three (all together
or one separate), one of choosing four, and one of choosing
five. That gives seven plans.

So for each set of method (X,Y,Z), we have 10 = 1+2+7 plans.

There are 15 ordinary sets of methods in the table above,
plus a further two with two methods in the X column. In
these, Y course-splices with both X methods. If we want to
include plans with either (or both) X methods, this gives us
4*1 + 3*2 + 2*7 = 24 plans (up to rotation) for those two
lines.

All in all, that gives us 15*10 + 2*24 = 198 plans.

COMBINING SIX-LEAD AND THREE-LEAD SPLICES

We can also combine three-lead and six-lead splices in a
single extent. The table below shows all methods where X-Y
have a six-lead splice and Y-Z have a three-lead splice (the
fixed bells for which are marked).

X Y Z
--------------------------------
Bm Ol El (2&4)
Bp/Cn/Dk/Dn Wl Bo (2&6)
Ki Ma Ta (3&5)
Ma Ki Bo (3&5)
Bh/Bw/By/Cc/Mp Pv Li (2&5)

Ti Tr Qu (2&6)
Cl/Mu Gl Ca (2&3)
Gl/Mu Cl Cu (2&3)
Ak Cz Ww (3&5)
Cz Ak Nw (2&6)

Nw Ww Cz (3&5)
Ww Nw Ak (2&6)
So Pn Nm (2&4)
Fo Li Pv (2&5)
Bn Lo We (2&4)

Lo Bn Cx (3&6)
Cx We Lo (2&4)
We Cx Bn (3&6)
Ne Bo Wl (2&6)
Ne Bo Ki (3&5)

Let's start with an extent of Y and apply the X-Y six-lead
splice once when bell a pivots. These six leads each rule
out a different three-lead splice slot leaving just the four
slots involving bell a: (a,b), (a,c), (a,d), (a,e). That
gives four plans (depending on whether we have 1, 2, 3 or 4
applications of the Y-Z splice).

If we have a two applications of the X-Y splice -- using
pivots a and b, there's only one three-lead slot available:
(a,b). This gives one more plan giving five in total.

There are sixteen sets of methods (X,Y,Z) with a single
method in the X column -- that gives 80 = 16*5 plans.
There's a further (4+5+2+2)*4 + (10+15+3+3)*1 = 83
plans from the entries with multiple X methods.

All together, that gives us 163 plans.

OTHER EXTENTS WITH FOUR METHODS

We've now covered all possible simple extents using three
methods. As we know that a simple extent cannot involve
both course and six-lead splices, this leaves four possible
types of three-method extent:

X --(5)-- Y --(5)-- Z
X --(6)-- Y --(6)-- Z --(3)-- denotes a 3-lead splice
X --(3)-- Y --(3)-- Z --(5)-- denotes a course splice
X --(5)-- Y --(3)-- Z --(6)-- denotes a 6-lead splice
X --(6)-- Y --(3)-- Z

What about extents with four methods? Quite a lot of these
have been covered too. The course and six-lead splices are
both transitive -- that is, if X and Y have a course (or
six-lead) splice, and so do Y and Z, then X and Z do too.
Whenever the X-Y splice is transitive, the possibility of
multiple X methods has already been considered.

This only leaves a few more possibilities to consider.

W --(3)-- X --(5)-- Y --(3)-- Z

With three courses of X and three courses of Y, if the
W-X splice uses a coursing pair and Y-Z uses a
non-coursing pair then there's exactly one plan with all
four methods. However there are no sets of methods in the
147 that have suitable splices to make this work.

W --(5)-- X --(3)-- Y --(3)-- Z

If we want a single application of Y-Z splice on (a,b), we
know we can have at most seven applications of X-Y using:
(a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e). Do these
provide enough X to get a W-X course splice? No. Because
we know that the pairs in a course splice are of the form
(p,q), (q,r), (r,s), (s,t), (t,p). So we cannot get four
methods in this way.

W --(5)-- X --(3)-- Y --(5)-- Z

If we one course of W and five of X, then the five pairs
that course / don't course in W can be used in the X-Y
splice (depending with it uses a non-coursing or coursing
pair). However if we want to add a course of Z we would
need the courses of W and Z not to share any coursing pairs
and that isn't possible. So we cannot get four methods this
way either.

W --(3)-- X --(6)-- Y --(3)-- Z

This cannot work as we know that we need at least 2/5 of the
extent on the three-lead splice side of the six-lead splice.
As this has three-lead splices on both sides of the six-lead
splice, it cannot work.

W --(6)-- X --(3)-- Y --(3)-- Z

With only six leads of W when bell a pivots, we can get up
to twelve leads of Y whenever bell a is in the fixed
position for the X-Y splice. However, this leaves no
opportunity for Y-Z. All four methods have the same
lead-end order, and the pivot bell for W-X, the two fixed
bells for X-Y and the two fixed bells for Y-Z are all
different place bells. If the Y-Z splice doesn't have a
as a fixed bell, then two of the leads will fall in the W.
If it does have a as a fixed bell then all of the leads are
in the X. Either way, no Z can be included.

W --(6)-- X --(3)-- Y --(6)-- Z

All the methods must have same lead-end order which means
W-X and Y-Z have the same fixed (pivot) place bell. If we
ring W when bell a pivots, we can only ring Y when a is
fixed in the three-lead splice. Clearly a can't pivot in Z,
but neither can anything else because only those leads with
bell a in the fixed position for Y-Z are present. So this
doesn't work.

W --(3)-- X --(3)-- Y --(3)-- Z

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

These plans can both be made to work, but there are no
methods in the 147 that have these particular arrangements
of three-lead splices.

W --(3)-- X --(3)-- Y
|
(5)
|
Z

This plan cannot work with regular methods. If X has two
three-lead splices, then one must involve a coursing pair
and one must involve a non-coursing pair. If we have a
course of Z, then only pairs that do not course in Z are
available for three-lead splicing in X

W --(3)-- X --(3)-- Y
|
(6)
|
Z

This arrangement of splices is the one that makes a grid
splice work, except that for a regular grid splice, X is an
irregular method and entirely removed. So we know that it
works. Exactly one set of methods in the 147 exists that
has splices in this particular arrangement:

Ki --(3&5)-- Bo --(2&6)-- Wl
|
(4)
|
Ne

Let's start with an extent of Bo. We know that we can apply
the Bo-Ne six-lead splice at most twice if we want to be
able to have retain a three-lead splice slot for Ki or Wl.
However, we've already counted those plans with only one of
Ki and Wl. Can we get both methods while also including
twelve leads of Ne? Yes. If we ring Ne well bells a or b
pivot, then we can also ring Ki when (a,b) are in 3&5 and Wl
when (a,b) are in 2&6. That's one plan up to rotation.

What about if we only have six leads of Ne, rung when bell a
pivots? That leaves four slots for Ki: (a,b), (a,c), (a,d),
(a,e); and four slots of Wl (with the same fixed bells).

Ignoring Wl, we know there are four ways of choosing Ki, up
to rotation, depending on whether there are 3, 6, 9 or 12
leads of Ki. Adding Wl is more complicated because the
leads of Ki mean the slots are no longer equivalent under
rotation. With one application of Bo-Ki, there are 2+2+2+1
= 7 ways of choosing Wl (depending whether we share the
Bo-Ki fixed pair); with two application of Bo-Ki, there are
2+3+2+1=8 ways of choosing Wl; and by symetry, with three
applications of Bo-Ki there are 7, and with four there are
4.

Finally we need to think about whether chirality is relevant
to any of them. This will only happen if each bell is in
some way unique. The pivot bell in Ne is a, which makes
that unique. If one bell (say e) is not fixed in either Ki
or Wl, that makes that unique. If one bell (b) is fixed in
both Ki and Wl, that can be unique. Which leaves c and d
which can be fixed in Ki and Wl respectively. So the only
plan that splits due to chirality is the plan with two
applications each of Bo-Ki and Bo-Wl, where one pair of
fixed bells is common to the two splices.

That gives 1+7+9+7+4 = 28 plans.

It's worth mentioning in passing that one of these plans
(the one with four applications of Bo-Ki and four of Bo-Wl)
contains no Bo -- it has twelve leads of Ki, twelve of Wl
and six of Ne. What's unusual in this case is that we have
a three-method plan in which none of the methods share a
splice, yet it can be explained in terms of simple
splices by introducing a fourth method. This turns out to
be common with grid splices, though most of the time, the
introduced method (the grid method) is not one of the
methods being considered. For example, with the Cm-Ip-Bo
grid splice, the grid method is King Edward which is not one
of the 147.

SUMMARY

That brings to an end the analysis of all plans that can be
explained in terms of just simple splices. They can be
grouped as follows:

Single method plans . . . . . . . . . . 75 \
Course splices . . . . . . . . . . . . . 108 | See first
Six-lead splices . . . . . . . . . . . . 176 | email
Three-lead splices . . . . . . . . . . . 798 /
Multiple course splices . . . . . . . . 36 \ See second
Multiple six-lead splices . . . . . . . 286*| email
Multiple three-lead splices . . . . . . 412 /
Combined course & three-lead splices . . 198 \ This
Combined six- & three-lead splices . . . 163 / email
Other extents with four methods . . . . 28
---------------------------------------------
TOTAL . . . . . . . . . . . . . . . . . 2280

[* = corrected from previous email; see note at top]

It comes as something of a relief that the total of 2280
plans calculated over the three emails in this analysis is
the same as the total number of simple plans counted
automatically by getting a computer to compare plans to each
other, and locating connected components which contain
single method plans.

RAS

Spliced treble-dodging minor - Introduction

2010-10-28T06:44:09Z

Holroyd:

[[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]]

Richard Smith richard at ex-parrot.com
Tue Sep 28 04:19:50 BST 2010

I've spent quite a lot of the last month looking at spliced
extents of treble dodging minor.

Thanks to a cunning algorithm (which I shall describe in a
moment) designed by Ander which we've been fine-tuning it
turns out to be possible to do exhaustive searches over
search spaces that I had previously thought were impossibly
large.

As a demonstration, I have just done a search for all true
extents of minor using just methods from the standard 147
treble dodging minor methods rung with 4ths place lead-end
bobs. I will do some further verification of this result
over the next few days, but I believe the number of extents
of this form is

5,862,727,200,079,423,275,554

To put this number into perspective, if I were to produce a
booklet listing these in a similar format to that used in
the CC's spliced minor collection, then the resulting
booklet would be about 5 light-years thick.

THE ALGORITHM

There are five main stages to the search algorithm.

First we remove lead splices and lead-end variants from the
list of methods. So, for example, we only want to include
one of Beverley, Surfleet, Berwick and Hexham. This reduces
the list of methods from 147 to 75.

The second stage is to associate each lead end or lead head
row with a method. Start with a list of the 60 in-course
rows with the treble leading -- these will all appear as a
l.e. or a l.h., and we need to choose a method for each one,
and doing so will join a l.h. to the subsequent l.e.

Suppose some l.e./l.h. rows already have methods chosen. Of
the remaining rows, we call a method 'possible' if

(i) the l.e. that would be reached by ringing a lead of
the method starting at the given l.h. row is not
associated with a method; and

(ii) the lead would be true against all other chosen
leads.

Take the row that has the fewest possible methods and, in
sequence, try each of its possible methods, recursing.
This gives an exhaustive tree search. The result of this is
a 'plan' -- a list of which method is rung from each lead,
but with no information on how to join the leads up.

Stage two can be speeded up significantly by implementing a
form of rotational pruning. Put the methods in some
arbitrary order. Any method (other than the first one
chosen) must not be before the first one chosen in the
ordering. This will remove some but not all rotations and
reflections. If you want an accurate count, it's a good
idea to check whether a plan is in its canonical rotation
and only output it if it is.

The third stage is to do an exhaustive search of ways to
join the 30 leads in each plan using 12, 14 or 16 lead end
changes. An normal tree search for compositions will do
this fine. There's no need to check for truth beyond
checking for repetition of lead heads and lead ends as this
was dealt with in stage two. For each plan you then have a
list of compositions that produce the extent.

Fourth, we remove compositions that include 16 lead ends in
London (3-3.4) or Hills (3-34.6) backworks. This is a
little subtle for plans that include one of these backworks
and another one -- as 16 lead ends are fine as long as they
only occur in the non-London, non-Hills backworks.

A further subtlety arises if rotational pruning was done in
stage two. Because there is no clear distinction between
rotation and reflection of a plan (because we don't yet know
which rows will become a l.h. and which a l.e.), pruning
removes both rotations and reflections. However, going from
Carlisle-over to London-over with a 16 l.e. is fine; but
going the other way is not.

This gives the complete set of extents.

Fifth, and assuming we want to count them, for each plan,
the number of extents is the product of three terms: the
number of distinct rotations / reflections (assuming
rotational pruning); the number of lead splices (N^n where N
is the number of methods in the lead splice set -- 2 or 4
for everything in the 147 -- and n the number of leads of
it); and the number of compositions for each plan. Adding
the values for each plan gives the overall total.

For the 147, the five stages took: 4s, 4h 1m, 1h 7m, 16m 44,
and 1m 18s. So the total search time was just under 6h.
I've only made an effort to optimise stages two and three
(stage five in particular is woefully suboptimal), but given
that's where most of the time is spent, that seems
reasonable. I reckon that without too much work the search
could be reduced to under 4h -- maybe even under 3h.

THE EXTENTS

Because the search first finds plans, and the number of
plans (modulo rotation) is a fairly managable 4614, it's
fairly easy to get a good idea of what's there. And a quick
scan through the list of plans shows that there are some
interesting plans that are new (at least to me). I'll give
a breakdown of what's there in a later email.

RAS

Spliced treble-dodging minor - Introduction

2010-10-28T06:41:53Z

Holroyd:

[[Spliced treble-dodging minor - clusters|Clusters of plans]]

Richard Smith richard at ex-parrot.com
Tue Sep 28 04:19:50 BST 2010

I've spent quite a lot of the last month looking at spliced
extents of treble dodging minor.

Thanks to a cunning algorithm (which I shall describe in a
moment) designed by Ander which we've been fine-tuning it
turns out to be possible to do exhaustive searches over
search spaces that I had previously thought were impossibly
large.

As a demonstration, I have just done a search for all true
extents of minor using just methods from the standard 147
treble dodging minor methods rung with 4ths place lead-end
bobs. I will do some further verification of this result
over the next few days, but I believe the number of extents
of this form is

5,862,727,200,079,423,275,554

To put this number into perspective, if I were to produce a
booklet listing these in a similar format to that used in
the CC's spliced minor collection, then the resulting
booklet would be about 5 light-years thick.

THE ALGORITHM

There are five main stages to the search algorithm.

First we remove lead splices and lead-end variants from the
list of methods. So, for example, we only want to include
one of Beverley, Surfleet, Berwick and Hexham. This reduces
the list of methods from 147 to 75.

The second stage is to associate each lead end or lead head
row with a method. Start with a list of the 60 in-course
rows with the treble leading -- these will all appear as a
l.e. or a l.h., and we need to choose a method for each one,
and doing so will join a l.h. to the subsequent l.e.

Suppose some l.e./l.h. rows already have methods chosen. Of
the remaining rows, we call a method 'possible' if

(i) the l.e. that would be reached by ringing a lead of
the method starting at the given l.h. row is not
associated with a method; and

(ii) the lead would be true against all other chosen
leads.

Take the row that has the fewest possible methods and, in
sequence, try each of its possible methods, recursing.
This gives an exhaustive tree search. The result of this is
a 'plan' -- a list of which method is rung from each lead,
but with no information on how to join the leads up.

Stage two can be speeded up significantly by implementing a
form of rotational pruning. Put the methods in some
arbitrary order. Any method (other than the first one
chosen) must not be before the first one chosen in the
ordering. This will remove some but not all rotations and
reflections. If you want an accurate count, it's a good
idea to check whether a plan is in its canonical rotation
and only output it if it is.

The third stage is to do an exhaustive search of ways to
join the 30 leads in each plan using 12, 14 or 16 lead end
changes. An normal tree search for compositions will do
this fine. There's no need to check for truth beyond
checking for repetition of lead heads and lead ends as this
was dealt with in stage two. For each plan you then have a
list of compositions that produce the extent.

Fourth, we remove compositions that include 16 lead ends in
London (3-3.4) or Hills (3-34.6) backworks. This is a
little subtle for plans that include one of these backworks
and another one -- as 16 lead ends are fine as long as they
only occur in the non-London, non-Hills backworks.

A further subtlety arises if rotational pruning was done in
stage two. Because there is no clear distinction between
rotation and reflection of a plan (because we don't yet know
which rows will become a l.h. and which a l.e.), pruning
removes both rotations and reflections. However, going from
Carlisle-over to London-over with a 16 l.e. is fine; but
going the other way is not.

This gives the complete set of extents.

Fifth, and assuming we want to count them, for each plan,
the number of extents is the product of three terms: the
number of distinct rotations / reflections (assuming
rotational pruning); the number of lead splices (N^n where N
is the number of methods in the lead splice set -- 2 or 4
for everything in the 147 -- and n the number of leads of
it); and the number of compositions for each plan. Adding
the values for each plan gives the overall total.

For the 147, the five stages took: 4s, 4h 1m, 1h 7m, 16m 44,
and 1m 18s. So the total search time was just under 6h.
I've only made an effort to optimise stages two and three
(stage five in particular is woefully suboptimal), but given
that's where most of the time is spent, that seems
reasonable. I reckon that without too much work the search
could be reduced to under 4h -- maybe even under 3h.

THE EXTENTS

Because the search first finds plans, and the number of
plans (modulo rotation) is a fairly managable 4614, it's
fairly easy to get a good idea of what's there. And a quick
scan through the list of plans shows that there are some
interesting plans that are new (at least to me). I'll give
a breakdown of what's there in a later email.

RAS

Banana Doubles

2010-10-27T06:50:24Z

Holroyd:

[[Image:banana.png|right]]

Banana doubles is an interesting and elegant method, highly recommended for a 5-bell band looking for something new. The music and the "feel" are very different from the more familiar methods. The plain course is easier to learn than Stedman. The calls require some getting used to, but once learned they are not much harder than Grandsire.

'''Tips for the plain course'''

* It's a principle (like Stedman). That is, all the bells do the same work, and there is no hunt bell.
* It's a double method (like Double Bob and Bristol Surprise). That is, in addition to the usual symmetry (the blue line is the same forwards as backwards), there is also front-back symmetry (the back-work is the front-work upside-down). But note that the start is not at a symmetry point.
* All places in 1st, 2nd, 4th, and 5th are made right (hand-back), all places in 3rd are made wrong.
* Of the four places where a bell leads, the middle two (adjacent to the 3rds-from-the-front) have another bell making 2nds at the same time. Similarly, the middle two lies have a bell making 4ths under them. (This helps with keeping track of where you are).

'''Bobs'''

Bobs can come both at the lead-end and the half-lead. The place notation for a plain lead is:
3.125.3.125.3.145.3.145
A half-lead bob replaces the middle 125 with 145:
3.125.3.'''145'''.3.145.3.145
A lead-end bob replaces the last 145 with 125:
3.125.3.125.3.145.3.'''125'''

If that sounds too daunting, it can all be done by learning some simple rules. First, some terminology: we call "lead,2nds,lead,3rds,lead,2nds,lead" the frontwork, and "lie,4ths,lie,3rds,lie,4ths,lie" the backwork. So the plain course is: frontwork, 3rds, backwork, 3rds.

* All calls happen at backstroke, and take effect the next backstroke.

If a bob happens when you are:

* about to lead, then unaffected. (This can happen for the first two leads of the front work).
* about to make the middle 3rds of the front work, then make 3rds, 5ths, 3rds and restart the front work
* about to make the last 2nds of the front work, then make 3rds instead and restart the front work
* about to leave the front work, then make an extra 2nds, lead, 2nds, lead, then 3rds and backwork

Because the method and calls are double, exactly the same rules apply to the backwork, but upside down:

* lie: unaffected
* middle 3rds: 3rds, lead, 3rd, backwork
* last 4ths: 3rds, backwork
* end of backwork: 4ths, lie, 4ths, lie

'''Calling a 120'''

Call a bob every one-and-a-half leads (i.e. every 12 changes), 10 bobs in total. If you are covering or very good at counting, this may be enough, otherwise it's easy enough to learn the sequence of works at the bobs. E.g.

call the 4th: first lead (of frontwork), last 2nds,
middle 3rds, second lead, end of frontwork;
then the same for the backwork.

File:Banana.png

2010-10-27T06:47:47Z

Holroyd:

Method Tips

2010-10-27T06:29:52Z

Holroyd:

Banana Doubles

2010-10-27T06:25:43Z

Holroyd: Created page with 'Banana doubles is an interesting and elegant method, highly recommended for a 5-bell band looking for something new. The music and the "feel" are very different from the more fa…'

Banana doubles is an interesting and elegant method, highly recommended for a 5-bell band looking for something new. The music and the "feel" are very different from the more familiar methods. The plain course is easier to learn than Stedman. The calls require some getting used to, but once learned they are not much harder than Grandsire.

'''Tips for the plain course'''

* It's a principle (like Stedman). That is, all the bells do the same work, and there is no hunt bell.
* It's a double method (like Double Bob and Bristol Surprise). That is, in addition to the usual symmetry (the blue line is the same forwards as backwards), there is also front-back symmetry (the back-work is the front-work upside-down). But note that the start is not at a symmetry point.
* All places in 1st, 2nd, 4th, and 5th are made right (hand-back), all places in 3rd are made wrong.
* Of the four places where a bell leads, the middle two (adjacent to the 3rds-from-the-front) have another bell making 2nds at the same time. Similarly, the middle two lies have a bell making 4ths under them. (This helps with keeping track of where you are).

'''Bobs'''

Bobs can come both at the lead-end and the half-lead. The place notation for a plain lead is:
3.125.3.125.3.145.3.145
A half-lead bob replaces the middle 125 with 145:
3.125.3.'''145'''.3.145.3.145
A lead-end bob replaces the last 145 with 125:
3.125.3.125.3.145.3.'''125'''

If that sounds too daunting, it can all be done by learning some simple rules. First, some terminology: we call "lead,2nds,lead,3rds,lead,2nds,lead" the frontwork, and "lie,4ths,lie,3rds,lie,4ths,lie" the backwork. So the plain course is: frontwork, 3rds, backwork, 3rds.

* All calls happen at backstroke, and take effect the next backstroke.

If a bob happens when you are:

* about to lead, then unaffected. (This can happen for the first two leads of the front work).
* about to make the middle 3rds of the front work, then make 3rds, 5ths, 3rds and restart the front work
* about to make the last 2nds of the front work, then make 3rds instead and restart the front work
* about to leave the front work, then make an extra 2nds, lead, 2nds, lead, then 3rds and backwork

Because the method and calls are double, exactly the same rules apply to the backwork, but upside down:

* lie: unaffected
* middle 3rds: 3rds, lead, 3rd, backwork
* last 4ths: 3rds, backwork
* end of backwork: 4ths, lie, 4ths, lie

'''Calling a 120'''

Call a bob every one-and-a-half leads (i.e. every 12 changes), 10 bobs in total. If you are covering or very good at counting, this may be enough, otherwise it's easy enough to learn the sequence of works at the bobs. E.g.

call the 4th: first lead (of frontwork), last 2nds,
middle 3rds, second lead, end of frontwork;
then the same for the backwork.

Spliced treble-dodging minor - 2

2010-10-24T23:05:09Z

Holroyd: Created page with 'Richard Smith richard at ex-parrot.com Thu Sep 30 03:59:37 BST 2010 This is the second email cataloguing the plans and this email aims to cover all those plans with three or mo…'

Richard Smith richard at ex-parrot.com
Thu Sep 30 03:59:37 BST 2010

This is the second email cataloguing the plans and this
email aims to cover all those plans with three or more
methods that can be described solely in terms of a single
type of simple splices -- that is multiple course splices,
multiple six-lead splices or multiple three-lead splices.

There will be a third (and hopefully shorter) email covering
extents that can be described in terms of a mixture of
types of simple splice. For example, extents such as the
six wrong-place Cambridge-over methods which combine a
course and a three-lead splice (as well as lead splices and
Parker splices for the 6ths place lead end variants).

MULTIPLE COURSE SPLICES

Two of the lines in the course splice table from the
previous email indicated a set of three mutually course
splicing methods.

[Ci, Ks, Ls, Sd], Ox / [Cf, Dk, Ny, Oc], Ms
[Ba, Cs, Fg, Sk], Do / [Bg, Kn, Rs, Wl], Ey

In these, the two sets of four bracketed lead splicers are
lead end variants of each other, and the two single methods
(Ox and Ms, or Do and Ey) both have course splices with the
other eight methods and with each other. This means that
instead of looking at 2^6 plans, we have 3^6 plans.
However, the removal of rotations complicates this.

With three possible methods, the number of courses of each
method can be: 4:1:1, 3:2:1 or 2:2:2. (We've already
considered the possibilities which have no leads of one of
the methods.)

We know from earlier that, up to rotation, there's only one
way of choosing four courses and the other two courses are
equivalent under rotation. So the 4:1:1 method distribution
gives 3 plans (one per choice of method for the four
courses). With 3:2:1, we have two ways of choosing three
courses, and in either case, the remaining three courses are
equivalent. As there are six ways of assigning the methods,
that gives 12 = 6*2 plans.

Finally, there's the 2:2:2 method distribution. Up to
rotation, there's one way of picking two courses for the
first method. How many ways are there of picking the
courses for the second method? We know from the earlier
discussion that given two courses, there are two ways of
choosing a third couse -- two of the four unchosen courses
share a coursing pair with the two chosen courses, and two
do not. So if we want to choose two courses for the second
method, there are three ways of doing this, depending on
whether 0, 1 or 2 of those courses share a coursing pair
with the first method's courses. That gives another 3
plans.

We had two sets of methods that shared three mutual course
splices, so that gives 2*(3+12+3) = 36 plans that can be
explained in terms of multiple course splices.

Unfortunately, it turns out that none of the plans actually
work particularly well. The two extra methods (Ox and Ms,
or Do and Ey) are one 2nds and 6th place lead ends, and
because the remaining lead splice methods are all J/M lead
ends, it's not possible to join the plan up with a plain
lead of each method. (In some cases it is possible to get a
composition with only, say, 2nds and 4th place lead ends,
for example, by having a bob after every lead of Ox or Do.)
This isn't a general problem with this type of composition
-- it just happens that the only two sets of methods from
the 147 that this applies to have G/J/M/O lead ends which is
particularly difficult to work with.

MULTIPLE SIX-LEAD SPLICES

In the same way that we can apply two (or more, potentially)
course splices, we can do the same with six-lead splices.
The following four sets of six-lead splices are candidates
for this.

[Bk, He], Pr, Wa / Bs, [Bv, Su], Cm 3 [3]
[Ed, Kh], Os, Wf
/ Bh, [Bt, Le, Md, Pv], Bw, [By, Pm], Cc, Mp 3 [6]
[Ba, Cs, Fg, Sk], [Ci, Ks, Ls, Sd], Pe, Ri, Wv
/ [Bg, Kn, Rs, Wl], Bp, [Cf, Dk, Ny, Oc], Cn, Dn 4 [5]
[Ch, Mu], Cl, Gl 6 [3]

Fortunately these are easier to enumerate than the multiple
course splices. With five working bells, we can choose a
method for each pivot bell. Two methods has already been
dealt with, with three methods the method balance can either
be 3:1:1 or 2:2:1, with four methods the method balance has
to be 2:1:1:1, and with five it's always 1:1:1:1:1 (however
in this case we get a chiral pair of plans). We then
just need to working out the combinatorical factors. These
are tabulated below.

Number of /-------- Number of plans --------\
Methods 3:1:1 2:2:1 2:1:1:1 1:1:1:1:1 Total
--------------------------------------------------------
3 1*3 1*3 0*4 0*2 6
4 4*3 4*3 1*4 0*2 28
5 10*3 10*3 5*4 1*2 82
6 20*3 20*3 15*4 6*2 192

Of the four sets of methods (above), two have three methods,
one five and one six. That gives a total of 2*6+82+182 =
276 plans.

MULTIPLE THREE-LEAD SPLICES

The case of multiple three-lead splices is somewhat
different from the case of multiple six-lead splices or
multiple course splices. In either of the latter, we have
three methods, X, Y and Z, and there exists a splice between
each pair. There are no sets of three methods each of which
have three-lead splices between them. However, there are
methods that have two *different* three-lead splices -- one
between X and Y, and a different one between Y and Z.

X Y Z
-----------------------------------------------------------
Ms (4&5) Di (2&3) [Ws, Ad]
Lv / Ki (3&5) Hu / Bo (2&6) [Ba, Cs, Fg, Sk]
/ [Bg, Kn, Rs, Wl]
Ev / Te (3&6) Wo / Sa (2&4) [Ck, Wt] / [Dt, Po]
Gl (2&3) Ca (4&5) Av

Conceptually these work by starting with Y (e.g. Di) and
then splicing some of X and Z in. However, there's a
subtlety. Suppose I start with Di, and want to ring Ws when
bells (a,b) are in 2&3, and Ms when bells (c,d) are in 4&5.
This causes a problem with the l.h. 1abcde as it is part of
both splices. As a result, the bells fixed in each of the
splices with method 1 must overlap with the bells fixed in
each of the splices with method 2. E.g. Ws when (a,b) are
in 2&3, and Ms when (b,c) are in 4&5 is fine.

Imagine we start with method Y and splice in just 3 leads
(the minimal unit) of method X when (a,b) are in the
relevant position. If we want to add some Z, we can have
any or all of:

(a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e)

So we cannot get any more than 21 leads of Z (which is borne
out by the search results).

Counting up the possibilities here is going to get tedious
rapidly. We have two ways of choosing one 3-lead splice
with Z: (a,b) is not equivalent to the others under
rotation. If we want two Y-Z splices we have the following
choices:

(a,b) + (a,c)
(a,c) + (b,c)
(a,c) + (a,d)
(a,c) + (b,d) [comes in l. and r. handed versions]

We can see that only the last configuration exhibs
chirality. The first two are invariant under relabelling d
and e (as neither are used). The third is invariant under
relabelling c and d. However, in the fourth, if we swap the
labels on c and d we must also swap the labels on a and b,
hence the two variants. This can be easier to see on a
diagram (as introduced in the first email cataloguing the
simple splices). Here the four configurations listed above
are depicted in the same order from left to right.

a a a a
/ | / : / : \ / :
/ | / : / : \ / :
c | d c : d c : d c : d
| \ : : : /
| \ : : : /
b e b e b e b e

(The dotted vertical line is representing the X-Y splice
using (a,b) that exists even if there isn't a Y-Z splice on
(a,b) and makes bells a and b special. Bell e is never
involved.)

With three Y-Z splices there are eight choices (including
left and right handed versions of chiral pairs):

(a,b) + (a,c) + (b,c)
(a,b) + (a,c) + (a,d)
(a,b) + (a,c) + (b,d) [chiral]
(a,c) + (a,d) + (b,c) [chiral]
(a,c) + (a,d) + (b,e)
(a,c) + (a,d) + (a,e)

The number of plans (up to rotation) with four, five, six or
seven Y-Z splices must be the same as the number with three,
two, one or zero Y-Z splices, respectively, because there
are only seven viable splice slots.

This gives the number of plans with one application of the
X-Y splice and at least one application of the Y-Z splice
as: 2+5+8+8+5+2+1 = 31.

Now we need to think about two applications of the X-Y
splice. (I did say this was going to get tedious!) There
are two ways (up to rotation) of choosing two three-lead
splice slots depending on whether or not they share a bell.
Bearing in mind every Y-Z splice must share a bell with
every X-Y splice, this leaves the following Y-Z splice slots
viable.

X-Y splices Viable Y-Z splice slots

(a,b) + (b,c) (a,b), (b,c); (a,c); (b,d), (b,e)
(a,b) + (c,d) (a,c), (a,d), (b,c), (c,d)

(Semicolons separate splice slots that are not equivalent
under rotation.) We only need to consider ways of choosing
one or two Y-Z splices.

X-Y splices Y-Z splices

(a,b) + (b,c) (a,b)
(a,b) + (b,c) (a,c)
(a,b) + (b,c) (b,d)

(a,b) + (b,c) (a,b) + (b,c)
(a,b) + (b,c) (a,b) + (a,c)
(a,b) + (b,c) (a,b) + (b,d) [chiral]
(a,b) + (b,c) (a,c) + (b,d)
(a,b) + (b,c) (b,d) + (b,e)

(a,b) + (c,d) (a,c) [chiral]

(a,b) + (c,d) (a,c) + (b,d) [chiral]
(a,b) + (c,d) (a,c) + (a,d)

This gives 27 = 3+6+6+3+1 + 2+3+2+1 plans with two
applications of X-Y.

Three applications of X-Y. I catalogued the four ways of
choosing three three-lead slots in the previous email.

X-Y splices Viable Y-Z splice slots
(3.1) (a,b) + (b,c) + (d,e) (b,d), (b,e)
(3.2) (a,b) + (b,c) + (c,d) (b,c); (a,c), (b,d)
(3.3) (a,b) + (b,c) + (b,d) (a,b), (b,c), (b,d); (b,e)
(3.4) (a,b) + (b,c) + (a,c) (a,b), (a,c), (b,c)

We've already established that (3.2) is chiral. This
results from a symmetry breaking in the choice of X-Y
splices. We cannot restore that symmetry by careful choice
of Y-Z splices. Nor can we break it further -- there's no
such thing as a "doubly chiral" configuration. (How could
there be? Chirality happens when the automorphism group of
the configuration graph being a subgroup of A_5. Either it
is or it isn't.) So all plans derived form (3.2) will be
chiral.

A bit of thought show that the number of plans with three
X-Y splices will be:

(3.1): 1+1 = 2
(3.2): 2+2+1 = 5 [chiral]
(3.3): 2+2+2+1 = 7
(3.4): 1+1+1 = 3

Which gives a total of 22 = 2+2*5+7+3 plans.

Fortunately the remaining cases -- of four or more
applications of the X-Y splice -- require little additional
thought. It's clear that as the number of applications of
X-Y increases, the number of viable Y-Z slots cannot
possible increase. Once we've handled the case of 4 X-Y
applications and 4 Y-Z applications, then we already have
the remaining numbers simply by reversing X and Z. (Both
are three lead splices and the ordering was arbitrary.)

So can we get 4 X-Ys and 4 Y-Zs? If we can, it must be
based on (3.3) as this is the only one with four viable Y-Z
slots.

X-Y splices Viable Y-Z splice slots
(3.3) (a,b) + (b,c) + (b,d) (a,b), (b,c), (b,d); (b,e)

It's immediately apparent that there is precisely one way of
getting 4 X-Ys and 4 Y-Zs: by choosing the same four slots
for both splices.

Now we just need to revisit the previous calculations
extracting the number of plans with four or more Y-Zs.

With 1 X-Y: 8+5+2+1
With 2 X-Ys: 3+1 + 1
With 3 X-Ys: 1
-------
22

There were four sets of methods that offered two three-lead
splices. So the total number of multiple three-lead splice
plans is 412 = 4 * (31+27+22+1+22). Phew!

SUMMARY

The total number of extent plans explained so far is as
follows.

Single method plans . . . . . . . . . . 75 \
Course splices . . . . . . . . . . . . . 108 | See first
Six-lead splices . . . . . . . . . . . . 176 | email
Three-lead splices . . . . . . . . . . . 798 /
Multiple course splices . . . . . . . . 36 \
Multiple six-lead splices . . . . . . . 276 | This email
Multiple three-lead splices . . . . . . 412 /
---------------------------------------------
TOTAL . . . . . . . . . . . . . . . . . 1881

We now know that the total number of extent plans that can
be explained solely in terms of simple splices is 2280.
(This number comes from counting the number of extents in
each simple splice cluster -- see other emails.) This means
there are 399 left to go.

Spliced treble-dodging minor - 1

2010-10-24T23:03:48Z

Holroyd: Created page with 'Richard Smith richard at ex-parrot.com Tue Sep 28 17:21:54 BST 2010 I'm going to start by cataloguing those plans that can be explained simply in terms of well-understood splic…'

Richard Smith richard at ex-parrot.com
Tue Sep 28 17:21:54 BST 2010

I'm going to start by cataloguing those plans that can be
explained simply in terms of well-understood splices,
probably in two separate emails. This will then leave the
shorter list of plans that deserve further study.

This email covers all plans with one or two methods. That
means there's nothing new in this email as splicing two
methods (at least with a fixed treble) is well understood.

SINGLE METHOD PLANS

As we've got 75 methods (modulo lead splices and lead-end
variants), 75 of the 4614 plans contain just a single
method.

The fact that the plans only include a single method doesn't
mean that they cannot produce extents of spliced -- for
example, we can easily produce an 8-method extent of spliced
using Old Oxford's lead-splices and lead-end variants.
Similarly an extent of Beverley, Surfleet, Berwick and
Hexham is derived from one of these single method plans.

SIMPLE SPLICES

Let's call a splice 'simple' if it can involve just two
methods. So for example the three-lead splice between York
and Durham is a simple splice -- sure, we can continue by
combining, say, course of Beverley into the touch, but this
is optional -- the touch works with just York and Durham and
so the three-lead splice is 'simple'.

On the other hand, the grid splice with Ipswich, Bourne and
Cambridge is not simple as all three methods are an integral
part of the splice -- we cannot get an extent of just
Ipswich and Bourne, for example.

For the remainder of this email, I shall refer to methods
using their standard two-letter abbreviations. These are
give on John Warboy's website:

http://website.lineone.net/~jswcomps/comp06.htm#TD

It's well understood how to generate a complete list of
simple splices. I'm not sure an explanation of this has
ever been covered explicitly on this list, though it has
been mentioned in passing. But I'm not going to break this
discussion to explain how to do it -- though I might write
another email on it.

Splices are usually described in terms of the minimum number
of leads of the method that can be inserted. For TDMMs,
this number can be 1, 2, 3, 5 or 6. In practice 2-lead
splices are rare and do not occur amonst any of the standard
147. The most common form a of a 5-lead splice is the
course-splice where the five leads to be replaced form a
course. There are no non-course 5-lead splices using
methods from the 147. Let's take these types of splices one
by one.

LEAD SPLICES

The following lead splices exist amongst the standard 147.

(i) Using the D1, D2, D3 & D4 underworks

[Ci, Ks, Ls, Sd] / [Cf, Dk, Ny, Oc] J/M
[Cw, Ns, Sl, Wr] / [Cb, Ng, Ol, Wi] K/N
[Cd, Ce, Sw, Va] J

(ii) Using the S1, S2, S3 & S4 underworks

[Ba, Cs, Fg, Sk] / [Bg, Kn, Rs, Wl] J/M
[Bt, Le, Md, Pv] H

(iii) Using the Westminster & Allandale underworks

[Ad, Ws] G
[Co, Li] H
[Ck, Wt] / [Dt, Po] K/N

(iv) Using the Beverley & Surfleet underworks

[Bk, He] / [Bv, Su] H/L
[Ed, Kh] / [By, Pm] H/L
[Ch, Mu] G

A note on notation. [Bk, He] / [Bv, Su] means that Bk and
He are lead splices and that Bv and Su are their 2nds place
lead-end variants and which also form a pair of lead
splices. Whether you consider He and Bv to be lead splices
is simply a matter of definition and of no great relevance
here. The letters in the last column are the lead-end
orders.

Because lead splices were excluded when reducing the list of
methods to 75, they do not appear in the list of plans.

Unfortunately it is not possible to include both J and M
variants in an extent (without also including other lead-end
orders). This means that the Old Oxford group is the only
one of these that can give an eight method extent. This
plan is responsible for 55% of the 5.86 x 10^21 extents.
This is because there are 4^30 lead splices and 2796
possible callings (allowing 2nds, 4ths and 6ths lead ends).
Multiplying these together gives 3.2 x 10^21 extents.

COURSE SPLICES

The following is a table of all course splices using methods
from the 147. This table was calculated from first
principles (and is much the same as the one in Michael
Foulds' books on spliced TDMM) rather than extracted from
the results of the search.

Br, [Cw, Ns, Sl, Wr] / [Cb, Ng, Ol, Wi], Ma o
Ab, Ro / Lf, Ne
[Ci, Ks, Ls, Sd], Ox / [Cf, Dk, Ny, Oc], Ms o #
Nf, Pr / Cm, Ip o
[Bk, He] / [Bv, Su], Du o

[Ba, Cs, Fg, Sk], Do / [Bg, Kn, Rs, Wl], Ey o #
[Ck, Wt] / [Ad, Ws], [Dt, Po] +
Wh / Cl, Nb o
Mo / [Ch, Mu], Nw o
C1, Mp o

C3, Pn
[Bt, Le, Md, Pv], Cx o
Av, [Cd, Ce, Sw, Va] o
Cu, Lo +

Notation. As with lead splices, a slash separates 6th
place and 2nds place methods. Where a group of methods are
enclosed in square brackets, they are lead splices. An o
denotes that the course splice is just a half-lead variant,
often with a set of lead splices. A # notes that the line
contains three separate course splices, e.g. Ox, Ms and the
eight lead splices are three sets of course splices. A +
notes that multiple backworks are present.

It's worth calculating the number of plans that can be
accounted for solely in terms of course splices. This is
worthwhile because the easiest way of checking that there's
nothing interesting hidden amongst the list of
seemingly-ordinary plans is by checking that the search
found the predicted number.

With six courses, we would expect 2^6 = 64 plans. However,
our list of 4614 plans exclude rotations and reflections,
and many of the 64 plans will just be rotations of each
other.

If we apply the splice zero times, then we have a single
method plan (already considered above). All ways of
applying it once are equivalent -- we can always rotate /
reflect the plan so that the splice is applied to the 123456
l.h.

What about two applications? Put succinctly, are all
choices of two courses equivalent? We know that from the
plain course we can reach any other course using just one
bob -- therefore all pairs of (distinct) courses are related
by cycling three coursing bells and are thus equivalent.

By symmetry, four, five and six applications of the splice
will be the same as two, one and zero respectively. This
just leaves the case of three applications of the course
splice. Are all choices of three courses equivalent? No.
For example, we know that a block of three bobs can join the
three tenors-together courses, but the same is not true of
the three split-tenors courses.

We know that any two courses must share two coursing pairs.
Three distinct courses cannot all share two coursing pairs
because there are only 10 pairs in total and 3*(5-2)+2 > 10.
So they must either all share a single coursing pair (as the
tenors together courses do, which is what allows them to be
joined by a Q-set of bobs on this pair) or none (as the
split tenors courses do).

How many of each type of choice of three courses are there?
Once we've selected two courses, there are four remaining.
Two of the unselected courses each share a (different)
single coursing pair with the two courses, and therefore the
other two do not share any coursing pair with both the
already chosen courses. One way of looking at this is that
courses A,B,C can be joined with a block of three homes,
A,D,B with a block of three before, but A,B,E and A,B,F
cannot be joined in any order using a block of three calls.
So of the 20 ways of selecting three courses, 10 share a
single coursing pair, and 10 do not. Once rotations and
reflectins have been factored out, this just leaves two ways
of selecting three courses.

So we have 1+1+2+1+1 = 6 plans for course splices (excluding
those that none of one or other method). The table above
has 18 course splices (noting that the two lines marked with
a # each contain three pairs of course splices). This means
that 108 = 18 * 6 out of the 4614 plans can be explained
just in terms of a single course splices, perhaps applied
multiple times.

SIX-LEAD SPLICES

The following is a table of all 6-lead splices using methods
from the 147.

Do, No 2
Bl, Wk 2
[Bk, He], Pr, Wa / Bs, [Bv, Su], Cm 3 [3]
[Ed, Kh], Os, Wf
/ Bh, [Bt, Le, Md, Pv], Bw, [By, Pm], Cc, Mp 3 [6]
Ml / [Co, Li], Fo 3

[Ba, Cs, Fg, Sk], [Ci, Ks, Ls, Sd], Pe, Ri, Wv
/ [Bg, Kn, Rs, Wl], Bp, [Cf, Dk, Ny, Oc], Cn, Dn 4 [5]
Br, Lv / Ki, Ma 4
Ab, Hu / Bo, Ne 4
Km, Sh / Ti, Tr 4
Ct, Cy / Ak, Cz 4

Lu, Mo / Nw, Ww 4
[Cd, Ce, Sw, Va], Ke 4
Bc, [Cw, Ns, Sl, Wr] / Bm, [Cb, Ng, Ol, Wi] 5
Pn, So 5
Bn, Lo 5

Cx, We 5
[Ch, Mu], Cl, Gl 6 [3]

Notation. As above, a slash separates lead end variants,
and lead splices are enclosed in square brakcets. The
number in the right-hand column is the fixed (pivot) bell
for the splice. Where a number is given in square brackets
at the end of the line, this is number of groups of mutually
six-lead splicing methods on the line.

Counting the plans that these are responsible for is
trivial. Because the splice uses all six rows where a given
bell pivots, up to rotation, there is exactly one way of
applying the splice once, one way of applying it twice, one
way of applying three times, and one way of applying if four
times. (Zero or five applications results in a single
method extent, already considered above.)

The table above has 17 lines, but four rows list multiple
six-lead splices. With n six-lead splice clusters (i.e. a
row marked [n]), there are n(n-1)/2 separate pairs of
six-lead splicers. This gives 17-4 + 2*(3*2/2) + 5*4/2 +
6*5/2 = 44 six-lead splices.

This means that 176 = 44 * 4 out of the 4614 plans can be
explained just in terms of a single 6-lead splices, perhaps
applied multiple times.

THREE-LEAD SPLICES

[Ad, Ws], Di 2&3 +
Du, Yo 2&3 *
Ca, Gl 2&3 +
Cl, Cu 2&3 +
Cr, [Cw, Ns, Sl, Wr] / [Cb, Ng, Ol, Wi], El 2&4 +

[Ck, Wt], Wo / [Dt, Po], Sa 2&4 +
Nm, Pn 2&4
Lo, We 2&4 *
Ro, St / Lf, Wm 2&5 *
[Bt, Le, Md, Pv], [Co, Li] 2&5 +

[Ba, Cs, Fg, Sk], Hu / [Bg, Kn, Rs, Wl], Bo 2&6 +
Km, Sn / Qu, Tr 2&6
Ct, Mo / Ak, Nw 2&6 *
Br, Hm / Ma, Ta 3&5 *
Hu, Lv / Bo, Ki 3&5

Cy, Lu / Cz, Ww 3&5 *
C2, C3 3&5 *
Ev, Wo / Sa, Te 3&6
Bn, Cx 3&6 *
Di, Ms 4&5

Av, Ca 4&5 *

(Notation. As with lead splices, a slash separates 6th
place and 2nds place methods. Where a group of methods are
enclosed in square brackets, they are lead splices. The
numbers in the right-hand column are the fixed place bells
for the splice. A * notes that the splice works like
London and Wells by swapping 34.16.34 for 14.36.46 at the
half-lead. A + notes that multiple backworks are present.)

With 30 leads in the extent, we can apply the 3-lead splice
any number of times from 0 to 10. Another way of looking at
this is that there are ten ways of choosing a pair of bells
from the five working bells (10 = 5*4/2). This means that
there are 2^10 different plans for each extent. However,
our list of 4614 plans exclude rotations and reflections,
and many of the 1024 = 2^10 will just be rotations of each
other which complicates things a bit.

If we apply the splice zero times, then we have a single
method plan (already considered above). All ways of
applying it once are equivalent -- we can always rotate /
reflect the plan so that the splice is applied to the 123456
l.h. With two applications, either the two applications
share a fixed bell (e.g. 2&3 and 2&4) or they do not (e.g.
2&3 and 4&5). Up to rotation and reflection, that's the
only choice left. We can show these diagramatically with
letters A-E indicating the five working bells and a
representing each application of the splice by joining the
two fixed bells.

(1.1) A --- B C D E

(2.1) A --- B --- C D E

(2.2) A --- B C --- D E

With three applications, we apparently have four
possibilities.

(3.1) A --- B --- C D --- E

(3.2) A --- B --- C --- D E

(3.3) A --- B --- C E
|
|
D

(3.4) A --- B D E
\ /
\ /
C

However, this isn't what the search found. For example, it
found five plans (up to rotation and reflection) containing
21 leads of London and 6 of Wells -- (3.2) appeared twice.

The reason is to do with parity. Because the plan only uses
in-course l.h.s and l.e.s we can only rotate or reflect the
plan by an even permutation. In (3.1), A and C are
equivalent as are D and E. When rotating (3.1), if we find
we need an odd permutation, we simply swap the labels on A
and C and use an even permutation.

But with (3.2) we can't do that. Yes, A and D are
equivalent as are B and C. But we cannot indepdently swap
labels on one pair of these -- if we swap the labels on A
and D we also need to swap the labels on B and C for the
graph to remain unaltered. This means we cannot simply
relabel so that an odd permutation 'rotation' converts into
an even permutation. The result is that there are two
versions of (3.2) which we might term a right-handed and a
left-handed version.

What of four applications of the splice?

(4.1) A --- B --- C --- D --- E [has l+r versions]

(4.2) A --- B --- C --- D
|
|
E

(4.3) A --- B --- D E
\ /
\ /
C

(4.4) A --- B D --- E
\ /
\ /
C

(4.5) A --- B E
| |
| |
C --- D

(4.6) A
|
|
B --- C --- D
|
|
E

And finally, for five applications:

(5.1) A --- B --- C [l+r variants]
\ /
\ /
D --- E

(5.2) A --- B --- C
| |
| |
D --- E

(5.3) A --- B --- C --- D
\ /
\ /
E

(5.4) A --- B --- C --- D [l+r variants]
\ /
\ /
E

(5.5) A
|
|
B --- C --- D
\ /
\ /
E

(5.6) A --- B E
\ / \
\ / \
C --- D

Six or more applications of the splice are, by symmetry, the
same as four or fewer. This gives the total number of plans
for a 3-lead splice as: 1+2+5+7+8+7+5+2+1 = 38. There are
21 3-leads splices in the table above, so that means that
3-lead splices are responsible for 798 = 21 * 38 of the 4614
plans.

SUMMARY

Scanning through the results of the search, I find 1157
plans with one or two methods. If I add the numbers above,
I get:

Single method plans . . . . . . . . . . 75
Course splices . . . . . . . . . . . . . 108
Six-lead splices . . . . . . . . . . . . 176
Three-lead splices . . . . . . . . . . . 798
---------------------------------------------
TOTAL . . . . . . . . . . . . . . . . . 1157

This isn't surprising. As I noted at the beginning of the
email, the theory of spliced with just two methods is well
understood and we wouldn't expect to find anything new.
However, this has been a productive exercise on two counts.
First, it increases my confidence that the search results
are correct as it agrees with the already well-tested theory
on splicing two methods. Secondly, it has allowed me to
work out techniques for counting extents -- for example,
identifying the potential problem with chirality (handness)
of certain three-lead splices.

Of course, with 3459 plans left to study, there's still
plenty to do!

Spliced treble-dodging minor - clusters

2010-10-24T23:01:44Z

Holroyd: Created page with 'Alexander Holroyd holroyd at math.ubc.ca Wed Sep 29 12:35:12 BST 2010 Here is an idea for an automated method for analysing the plans. For each pair of plans in the list (ie a …'

Alexander Holroyd holroyd at math.ubc.ca
Wed Sep 29 12:35:12 BST 2010

Here is an idea for an automated method for analysing the plans.

For each pair of plans in the list (ie a few million pairs), do the
following. "Rotate" one of the pair through all 60 possible starting
rows. (Your plans are already in standard form, aren't they?) For each
such rotation, compare the two plans. Specifically, look to see whether
the two plans are identical except that one is obtained from the other by
replacing some set of leads all of method X with all method Y. If so, say
that there is a "simple splice" between the two plans. The simple splice
itself may be described by saying what methods X and Y are, and what the
set of leads is, in "standard form" (i.e. rotated to it's smallest version
in lexicographic order).

After this is done for all pairs, the set of simple splices that arise had
better be what exactly we expect, i.e. things like "2 copies of the
Cambride-Beverley 6-lead splice".

We now have a graph, whose vertices (nodes) are the plans, and whose edges
(links) are the simple splices. Break this graph up into its connected
components, or "clusters". Each cluster is a group of plans all of which
communicate with each other via simple splices.

For a start I would like to know how many clusters there are.

Now we want to analyse the clusters. Any cluster that contains a
single-method extent can be removed from the game, because these we
(hopefully) completely understand. For the others, it would be desirable
to nominate a "standard representative" plan from each cluster - the basic
plan which then gets embelished with various simple splices to form the
other members of the cluster. I'm not sure whether there is a canonical
way to do this in general, but one natural thing that springs to mind is
to list, for each cluster, the plans that have the smallest number of
methods, and see by hand which of these (if there is more than one) is the
most "natural".

Then one can summarize the whole thing by listing, for each cluster, its
standard representative, and the list of all simple splices it involves.
I suspect such a list would be quite manageable.

----

Richard Smith richard at ex-parrot.com
Wed Sep 29 19:57:24 BST 2010

There are 506 connected components of which 14 are simple.
The simple components contain the methods:

1. Sg
2. Nm So C3 C2 Pn
3. Nw Mo Ak Ct Mu Ch Av Ca Ke Ce Cd Sw Va Li Co Cc We Lo
Cu Cx Bn Pv Bt Le Md Mp C1 Cl Nb Wh Gl Fo Ml By Pm Ed
Kh Ww Lu Cz Cy Bh Os Bw Wf
4. Qu Sn Ti Sh Tr Km
5. Kt
6. Bl Wk
7. Fr Cg
8. Cj Nl
9. Du Yo Su He Bv Bk Cm Pr Ip Nf Bs Wa
10. Bu
11. Es Cv
12. Rc Bz
13. Bm Bc El Cr Wm St Lf Ro Sa Wo Te Ev Po Dt Ck Wt Di Ms
Do Dn Pe Wl Bg Kn Rs Ba Cs Fg Sk Ey Bp Wv Bo Hu Ki Lv
Cn Ri Dk Cf Ny Oc Ci Ks Ls Sd Ox No Ne Ab Ws Ad Ma Br
Ta Hm Ol Cb Ng Wi Cw Ns Sl Wr
14. Be Me

The full list of plans grouped into clusters is here:

http://ex-parrot.com/~richard/minor/147/plans-in-clusters.txt

Spliced treble-dodging minor - Introduction

2010-10-24T22:55:45Z

Holroyd: moved Spliced treble-dodging minor to Spliced treble-dodging minor - Introduction

Richard Smith richard at ex-parrot.com
Tue Sep 28 04:19:50 BST 2010

I've spent quite a lot of the last month looking at spliced
extents of treble dodging minor.

Thanks to a cunning algorithm (which I shall describe in a
moment) designed by Ander which we've been fine-tuning it
turns out to be possible to do exhaustive searches over
search spaces that I had previously thought were impossibly
large.

As a demonstration, I have just done a search for all true
extents of minor using just methods from the standard 147
treble dodging minor methods rung with 4ths place lead-end
bobs. I will do some further verification of this result
over the next few days, but I believe the number of extents
of this form is

5,862,727,200,079,423,275,554

To put this number into perspective, if I were to produce a
booklet listing these in a similar format to that used in
the CC's spliced minor collection, then the resulting
booklet would be about 5 light-years thick.

THE ALGORITHM

There are five main stages to the search algorithm.

First we remove lead splices and lead-end variants from the
list of methods. So, for example, we only want to include
one of Beverley, Surfleet, Berwick and Hexham. This reduces
the list of methods from 147 to 75.

The second stage is to associate each lead end or lead head
row with a method. Start with a list of the 60 in-course
rows with the treble leading -- these will all appear as a
l.e. or a l.h., and we need to choose a method for each one,
and doing so will join a l.h. to the subsequent l.e.

Suppose some l.e./l.h. rows already have methods chosen. Of
the remaining rows, we call a method 'possible' if

(i) the l.e. that would be reached by ringing a lead of
the method starting at the given l.h. row is not
associated with a method; and

(ii) the lead would be true against all other chosen
leads.

Take the row that has the fewest possible methods and, in
sequence, try each of its possible methods, recursing.
This gives an exhaustive tree search. The result of this is
a 'plan' -- a list of which method is rung from each lead,
but with no information on how to join the leads up.

Stage two can be speeded up significantly by implementing a
form of rotational pruning. Put the methods in some
arbitrary order. Any method (other than the first one
chosen) must not be before the first one chosen in the
ordering. This will remove some but not all rotations and
reflections. If you want an accurate count, it's a good
idea to check whether a plan is in its canonical rotation
and only output it if it is.

The third stage is to do an exhaustive search of ways to
join the 30 leads in each plan using 12, 14 or 16 lead end
changes. An normal tree search for compositions will do
this fine. There's no need to check for truth beyond
checking for repetition of lead heads and lead ends as this
was dealt with in stage two. For each plan you then have a
list of compositions that produce the extent.

Fourth, we remove compositions that include 16 lead ends in
London (3-3.4) or Hills (3-34.6) backworks. This is a
little subtle for plans that include one of these backworks
and another one -- as 16 lead ends are fine as long as they
only occur in the non-London, non-Hills backworks.

A further subtlety arises if rotational pruning was done in
stage two. Because there is no clear distinction between
rotation and reflection of a plan (because we don't yet know
which rows will become a l.h. and which a l.e.), pruning
removes both rotations and reflections. However, going from
Carlisle-over to London-over with a 16 l.e. is fine; but
going the other way is not.

This gives the complete set of extents.

Fifth, and assuming we want to count them, for each plan,
the number of extents is the product of three terms: the
number of distinct rotations / reflections (assuming
rotational pruning); the number of lead splices (N^n where N
is the number of methods in the lead splice set -- 2 or 4
for everything in the 147 -- and n the number of leads of
it); and the number of compositions for each plan. Adding
the values for each plan gives the overall total.

For the 147, the five stages took: 4s, 4h 1m, 1h 7m, 16m 44,
and 1m 18s. So the total search time was just under 6h.
I've only made an effort to optimise stages two and three
(stage five in particular is woefully suboptimal), but given
that's where most of the time is spent, that seems
reasonable. I reckon that without too much work the search
could be reduced to under 4h -- maybe even under 3h.

THE EXTENTS

Because the search first finds plans, and the number of
plans (modulo rotation) is a fairly managable 4614, it's
fairly easy to get a good idea of what's there. And a quick
scan through the list of plans shows that there are some
interesting plans that are new (at least to me). I'll give
a breakdown of what's there in a later email.

RAS

Spliced treble-dodging minor

2010-10-24T22:55:45Z

Holroyd: moved Spliced treble-dodging minor to Spliced treble-dodging minor - Introduction

#REDIRECT [[Spliced treble-dodging minor - Introduction]]

Spliced treble-dodging minor - Introduction

2010-10-24T22:50:08Z

Holroyd: Created page with 'Richard Smith richard at ex-parrot.com Tue Sep 28 04:19:50 BST 2010 I've spent quite a lot of the last month looking at spliced extents of treble dodging minor. Thanks to a cu…'