[r-t] 147 TDMM

Richard Smith richard at ex-parrot.com
Wed Oct 6 01:05:43 UTC 2010

Richard Smith wrote:

> 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).

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 ...


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).


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 

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


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.


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

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

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

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

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 

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.


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.


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