Spliced treble-dodging minor - 2
Richard Smith richard at ex-parrot.com Thu Sep 30 03:59:37 BST 2010
This is the second email cataloguing the plans and this email aims to cover all those plans with three or more methods that can be described solely in terms of a single type of simple splices -- that is multiple course splices, multiple six-lead splices or multiple three-lead splices.
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).
MULTIPLE COURSE SPLICES
Two of the lines in the course splice table from the previous email indicated a set of three mutually course splicing methods.
[Ci, Ks, Ls, Sd], Ox / [Cf, Dk, Ny, Oc], Ms [Ba, Cs, Fg, Sk], Do / [Bg, Kn, Rs, Wl], Ey
In these, the two sets of four bracketed lead splicers are lead end variants of each other, and the two single methods (Ox and Ms, or Do and Ey) both have course splices with the other eight methods and with each other. This means that instead of looking at 2^6 plans, we have 3^6 plans. However, the removal of rotations complicates this.
With three possible methods, the number of courses of each method can be: 4:1:1, 3:2:1 or 2:2:2. (We've already considered the possibilities which have no leads of one of the methods.)
We know from earlier that, up to rotation, there's only one way of choosing four courses and the other two courses are equivalent under rotation. So the 4:1:1 method distribution gives 3 plans (one per choice of method for the four courses). With 3:2:1, we have two ways of choosing three courses, and in either case, the remaining three courses are equivalent. As there are six ways of assigning the methods, that gives 12 = 6*2 plans.
Finally, there's the 2:2:2 method distribution. Up to rotation, there's one way of picking two courses for the first method. How many ways are there of picking the courses for the second method? We know from the earlier discussion that given two courses, there are two ways of choosing a third couse -- two of the four unchosen courses share a coursing pair with the two chosen courses, and two do not. So if we want to choose two courses for the second method, there are three ways of doing this, depending on whether 0, 1 or 2 of those courses share a coursing pair with the first method's courses. That gives another 3 plans.
We had two sets of methods that shared three mutual course splices, so that gives 2*(3+12+3) = 36 plans that can be explained in terms of multiple course splices.
Unfortunately, it turns out that none of the plans actually work particularly well. The two extra methods (Ox and Ms, or Do and Ey) are one 2nds and 6th place lead ends, and because the remaining lead splice methods are all J/M lead ends, it's not possible to join the plan up with a plain lead of each method. (In some cases it is possible to get a composition with only, say, 2nds and 4th place lead ends, for example, by having a bob after every lead of Ox or Do.) This isn't a general problem with this type of composition -- it just happens that the only two sets of methods from the 147 that this applies to have G/J/M/O lead ends which is particularly difficult to work with.
MULTIPLE SIX-LEAD SPLICES
In the same way that we can apply two (or more, potentially) course splices, we can do the same with six-lead splices. The following four sets of six-lead splices are candidates for this.
[Bk, He], Pr, Wa / Bs, [Bv, Su], Cm 3  [Ed, Kh], Os, Wf / Bh, [Bt, Le, Md, Pv], Bw, [By, Pm], Cc, Mp 3  [Ba, Cs, Fg, Sk], [Ci, Ks, Ls, Sd], Pe, Ri, Wv / [Bg, Kn, Rs, Wl], Bp, [Cf, Dk, Ny, Oc], Cn, Dn 4  [Ch, Mu], Cl, Gl 6 
Fortunately these are easier to enumerate than the multiple course splices. With five working bells, we can choose a method for each pivot bell. Two methods has already been dealt with, with three methods the method balance can either be 3:1:1 or 2:2:1, with four methods the method balance has to be 2:1:1:1, and with five it's always 1:1:1:1:1 (however in this case we get a chiral pair of plans). We then just need to working out the combinatorical factors. These are tabulated below.
Number of /-------- Number of plans --------\ Methods 3:1:1 2:2:1 2:1:1:1 1:1:1:1:1 Total -------------------------------------------------------- 3 1*3 1*3 0*4 0*2 6 4 4*3 4*3 1*4 0*2 28 5 10*3 10*3 5*4 1*2 82 6 20*3 20*3 15*4 6*2 192
Of the four sets of methods (above), two have three methods, one five and one six. That gives a total of 2*6+82+192 = 286 plans.
MULTIPLE THREE-LEAD SPLICES
The case of multiple three-lead splices is somewhat different from the case of multiple six-lead splices or multiple course splices. In either of the latter, we have three methods, X, Y and Z, and there exists a splice between each pair. There are no sets of three methods each of which have three-lead splices between them. However, there are methods that have two *different* three-lead splices -- one between X and Y, and a different one between Y and Z.
X Y Z ----------------------------------------------------------- Ms (4&5) Di (2&3) [Ws, Ad] Lv / Ki (3&5) Hu / Bo (2&6) [Ba, Cs, Fg, Sk] / [Bg, Kn, Rs, Wl] Ev / Te (3&6) Wo / Sa (2&4) [Ck, Wt] / [Dt, Po] Gl (2&3) Ca (4&5) Av
Conceptually these work by starting with Y (e.g. Di) and then splicing some of X and Z in. However, there's a subtlety. Suppose I start with Di, and want to ring Ws when bells (a,b) are in 2&3, and Ms when bells (c,d) are in 4&5. This causes a problem with the l.h. 1abcde as it is part of both splices. As a result, the bells fixed in each of the splices with method 1 must overlap with the bells fixed in each of the splices with method 2. E.g. Ws when (a,b) are in 2&3, and Ms when (b,c) are in 4&5 is fine.
Imagine we start with method Y and splice in just 3 leads (the minimal unit) of method X when (a,b) are in the relevant position. If we want to add some Z, we can have any or all of:
(a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e)
So we cannot get any more than 21 leads of Z (which is borne out by the search results).
Counting up the possibilities here is going to get tedious rapidly. We have two ways of choosing one 3-lead splice with Z: (a,b) is not equivalent to the others under rotation. If we want two Y-Z splices we have the following choices:
(a,b) + (a,c) (a,c) + (b,c) (a,c) + (a,d) (a,c) + (b,d) [comes in l. and r. handed versions]
We can see that only the last configuration exhibs chirality. The first two are invariant under relabelling d and e (as neither are used). The third is invariant under relabelling c and d. However, in the fourth, if we swap the labels on c and d we must also swap the labels on a and b, hence the two variants. This can be easier to see on a diagram (as introduced in the first email cataloguing the simple splices). Here the four configurations listed above are depicted in the same order from left to right.
a a a a / | / : / : \ / : / | / : / : \ / : c | d c : d c : d c : d | \ : : : / | \ : : : / b e b e b e b e
(The dotted vertical line is representing the X-Y splice using (a,b) that exists even if there isn't a Y-Z splice on (a,b) and makes bells a and b special. Bell e is never involved.)
With three Y-Z splices there are eight choices (including left and right handed versions of chiral pairs):
(a,b) + (a,c) + (b,c) (a,b) + (a,c) + (a,d) (a,b) + (a,c) + (b,d) [chiral] (a,c) + (a,d) + (b,c) [chiral] (a,c) + (a,d) + (b,e) (a,c) + (a,d) + (a,e)
The number of plans (up to rotation) with four, five, six or seven Y-Z splices must be the same as the number with three, two, one or zero Y-Z splices, respectively, because there are only seven viable splice slots.
This gives the number of plans with one application of the X-Y splice and at least one application of the Y-Z splice as: 2+5+8+8+5+2+1 = 31.
Now we need to think about two applications of the X-Y splice. (I did say this was going to get tedious!) There are two ways (up to rotation) of choosing two three-lead splice slots depending on whether or not they share a bell. Bearing in mind every Y-Z splice must share a bell with every X-Y splice, this leaves the following Y-Z splice slots viable.
X-Y splices Viable Y-Z splice slots (a,b) + (b,c) (a,b), (b,c); (a,c); (b,d), (b,e) (a,b) + (c,d) (a,c), (a,d), (b,c), (c,d)
(Semicolons separate splice slots that are not equivalent under rotation.) We only need to consider ways of choosing one or two Y-Z splices.
X-Y splices Y-Z splices (a,b) + (b,c) (a,b) (a,b) + (b,c) (a,c) (a,b) + (b,c) (b,d) (a,b) + (b,c) (a,b) + (b,c) (a,b) + (b,c) (a,b) + (a,c) (a,b) + (b,c) (a,b) + (b,d) [chiral] (a,b) + (b,c) (a,c) + (b,d) (a,b) + (b,c) (b,d) + (b,e) (a,b) + (c,d) (a,c) [chiral] (a,b) + (c,d) (a,c) + (b,d) [chiral] (a,b) + (c,d) (a,c) + (a,d)
This gives 27 = 3+6+6+3+1 + 2+3+2+1 plans with two applications of X-Y.
Three applications of X-Y. I catalogued the four ways of choosing three three-lead slots in the previous email.
X-Y splices Viable Y-Z splice slots (3.1) (a,b) + (b,c) + (d,e) (b,d), (b,e) (3.2) (a,b) + (b,c) + (c,d) (b,c); (a,c), (b,d) (3.3) (a,b) + (b,c) + (b,d) (a,b), (b,c), (b,d); (b,e) (3.4) (a,b) + (b,c) + (a,c) (a,b), (a,c), (b,c)
We've already established that (3.2) is chiral. This results from a symmetry breaking in the choice of X-Y splices. We cannot restore that symmetry by careful choice of Y-Z splices. Nor can we break it further -- there's no such thing as a "doubly chiral" configuration. (How could there be? Chirality happens when the automorphism group of the configuration graph being a subgroup of A_5. Either it is or it isn't.) So all plans derived form (3.2) will be chiral.
A bit of thought show that the number of plans with three X-Y splices will be:
(3.1): 1+1 = 2 (3.2): 2+2+1 = 5 [chiral] (3.3): 2+2+2+1 = 7 (3.4): 1+1+1 = 3
Which gives a total of 22 = 2+2*5+7+3 plans.
Fortunately the remaining cases -- of four or more applications of the X-Y splice -- require little additional thought. It's clear that as the number of applications of X-Y increases, the number of viable Y-Z slots cannot possible increase. Once we've handled the case of 4 X-Y applications and 4 Y-Z applications, then we already have the remaining numbers simply by reversing X and Z. (Both are three lead splices and the ordering was arbitrary.)
So can we get 4 X-Ys and 4 Y-Zs? If we can, it must be based on (3.3) as this is the only one with four viable Y-Z slots.
X-Y splices Viable Y-Z splice slots (3.3) (a,b) + (b,c) + (b,d) (a,b), (b,c), (b,d); (b,e)
It's immediately apparent that there is precisely one way of getting 4 X-Ys and 4 Y-Zs: by choosing the same four slots for both splices.
Now we just need to revisit the previous calculations extracting the number of plans with four or more Y-Zs.
With 1 X-Y: 8+5+2+1 With 2 X-Ys: 3+1 + 1 With 3 X-Ys: 1 ------- 22
There were four sets of methods that offered two three-lead splices. So the total number of multiple three-lead splice plans is 412 = 4 * (31+27+22+1+22). Phew!
The total number of extent plans explained so far is as follows.
Single method plans . . . . . . . . . . 75 \ Course splices . . . . . . . . . . . . . 108 | See first Six-lead splices . . . . . . . . . . . . 176 | email Three-lead splices . . . . . . . . . . . 798 / Multiple course splices . . . . . . . . 36 \ Multiple six-lead splices . . . . . . . 286 | This email Multiple three-lead splices . . . . . . 412 / --------------------------------------------- TOTAL . . . . . . . . . . . . . . . . . 1891
We now know that the total number of extent plans that can be explained solely in terms of simple splices is 2280. (This number comes from counting the number of extents in each simple splice cluster -- see other emails.) This means there are 399 left to go.