[r-t] 147 TDMM
Richard Smith
richard at ex-parrot.com
Thu Sep 30 02:59:37 UTC 2010
>> THE EXTENTS
>
> I'm going to start by cataloguing those plans that can be
> explained simply in terms of well-understood splices,
> probably in two separate emails. This will then leave the
> shorter list of plans that deserve further study.
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 [3]
[Ed, Kh], Os, Wf
/ Bh, [Bt, Le, Md, Pv], Bw, [By, Pm], Cc, Mp 3 [6]
[Ba, Cs, Fg, Sk], [Ci, Ks, Ls, Sd], Pe, Ri, Wv
/ [Bg, Kn, Rs, Wl], Bp, [Cf, Dk, Ny, Oc], Cn, Dn 4 [5]
[Ch, Mu], Cl, Gl 6 [3]
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+182 =
276 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!
SUMMARY
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 . . . . . . . 276 | This email
Multiple three-lead splices . . . . . . 412 /
---------------------------------------------
TOTAL . . . . . . . . . . . . . . . . . 1881
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.
RAS
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