[r-t] 147 TDMM
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
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.
- 123564 Lo
- 145236 Cx
- 145362 Lo
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
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)
a --- b a --- b a --- b
/ \ / / \
/ \ / / \
c --- d c d c d
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
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
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
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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