[r-t] 147 TDMM

Richard Smith richard at ex-parrot.com
Mon Oct 18 17:37:06 UTC 2010

Richard Smith wrote:

> 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.

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 


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:

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

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

   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

   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 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) 

    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

   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

   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

   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

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 

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.


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.


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