Spliced treble-dodging minor - 5
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.
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
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
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.
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.