# Spliced treble-dodging minor - 3

Clusters of plans | Plans 1 | Plans 2 | Plans 3 | Plans 4 | Plans 5 | Plans 6

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

Back to the analysis ...

## Contents

#### 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

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