Difference between revisions of "Spliced treble-dodging minor - 2"

From Changeringing Wiki
Jump to: navigation, search
(Formatting improvements)
(Corrections noted in third email applied.)
 
Line 16: Line 16:
 
Thu Sep 30 03:59:37 BST 2010
 
Thu Sep 30 03:59:37 BST 2010
  
This is the second email cataloguing the plans and this  
+
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.
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  
+
There will be a third (and hopefully shorter) email covering extents that can be described in terms of a mixture of  
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  
types of simple splice.  For example, extents such as the  
+
course and a three-lead splice (as well as lead splices and Parker splices for the 6ths place lead end variants).
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====
 
====MULTIPLE COURSE SPLICES====
  
Two of the lines in the course splice table from the  
+
Two of the lines in the course splice table from the previous email indicated a set of three mutually course  
previous email indicated a set of three mutually course  
 
 
splicing methods.
 
splicing methods.
  
Line 38: Line 30:
 
   [Ba, Cs, Fg, Sk], Do / [Bg, Kn, Rs, Wl], Ey
 
   [Ba, Cs, Fg, Sk], Do / [Bg, Kn, Rs, Wl], Ey
  
In these, the two sets of four bracketed lead splicers are  
+
In these, the two sets of four bracketed lead splicers are lead end variants of each other, and the two single methods  
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  
(Ox and Ms, or Do and Ey) both have course splices with the  
+
instead of looking at 2^6 plans, we have 3^6 plans. However, the removal of rotations complicates this.
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  
+
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  
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.)
considered the possibilities which have no leads of one of  
 
the methods.)
 
  
We know from earlier that, up to rotation, there's only one  
+
We know from earlier that, up to rotation, there's only one way of choosing four courses and the other two courses are  
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  
equivalent under rotation.  So the 4:1:1 method distribution  
+
courses).  With 3:2:1, we have two ways of choosing three courses, and in either case, the remaining three courses are  
gives 3 plans (one per choice of method for the four  
+
equivalent.  As there are six ways of assigning the methods, that gives 12 = 6*2 plans.
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  
+
Finally, there's the 2:2:2 method distribution.  Up to rotation, there's one way of picking two courses for the  
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  
first method.  How many ways are there of picking the  
+
discussion that given two courses, there are two ways of choosing a third couse -- two of the four unchosen courses  
courses for the second method?  We know from the earlier  
+
share a coursing pair with the two chosen courses, and two do not.  So if we want to choose two courses for the second  
discussion that given two courses, there are two ways of  
+
method, there are three ways of doing this, depending on whether 0, 1 or 2 of those courses share a coursing pair  
choosing a third couse -- two of the four unchosen courses  
+
with the first method's courses.  That gives another 3 plans.
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  
+
We had two sets of methods that shared three mutual course splices, so that gives 2*(3+12+3) = 36 plans that can be  
splices, so that gives 2*(3+12+3) = 36 plans that can be  
 
 
explained in terms of multiple course splices.
 
explained in terms of multiple course splices.
  
Unfortunately, it turns out that none of the plans actually  
+
Unfortunately, it turns out that none of the plans actually work particularly well.  The two extra methods (Ox and Ms,  
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  
or Do and Ey) are one 2nds and 6th place lead ends, and  
+
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  
because the remaining lead splice methods are all J/M lead  
+
composition with only, say, 2nds and 4th place lead ends, for example, by having a bob after every lead of Ox or Do.)  
ends, it's not possible to join the plan up with a plain  
+
This isn't a general problem with this type of composition -- it just happens that the only two sets of methods from  
lead of each method.  (In some cases it is possible to get a  
+
the 147 that this applies to have G/J/M/O lead ends which is particularly difficult to work with.
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====
 
====MULTIPLE SIX-LEAD SPLICES====
  
In the same way that we can apply two (or more, potentially)  
+
In the same way that we can apply two (or more, potentially) course splices, we can do the same with six-lead splices.  
course splices, we can do the same with six-lead splices.  
+
The following four sets of six-lead splices are candidates for this.
The following four sets of six-lead splices are candidates  
 
for this.
 
  
 
   [Bk, He], Pr, Wa / Bs, [Bv, Su], Cm                  3 [3]
 
   [Bk, He], Pr, Wa / Bs, [Bv, Su], Cm                  3 [3]
Line 103: Line 71:
 
   [Ch, Mu], Cl, Gl                                    6 [3]
 
   [Ch, Mu], Cl, Gl                                    6 [3]
  
Fortunately these are easier to enumerate than the multiple  
+
Fortunately these are easier to enumerate than the multiple course splices.  With five working bells, we can choose a  
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  
method for each pivot bell.  Two methods has already been  
+
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  
dealt with, with three methods the method balance can either  
+
in this case we get a chiral pair of plans).  We then just need to working out the combinatorical factors.  These  
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.
 
are tabulated below.
  
Line 121: Line 85:
 
   6            20*3    20*3    15*4      6*2          192
 
   6            20*3    20*3    15*4      6*2          192
  
Of the four sets of methods (above), two have three methods,  
+
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.
one five and one six.  That gives a total of 2*6+82+182 =  
 
276 plans.
 
  
 
====MULTIPLE THREE-LEAD SPLICES====
 
====MULTIPLE THREE-LEAD SPLICES====
  
The case of multiple three-lead splices is somewhat  
+
The case of multiple three-lead splices is somewhat different from the case of multiple six-lead splices or  
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  
multiple course splices.  In either of the latter, we have  
+
each pair.  There are no sets of three methods each of which have three-lead splices between them.  However, there are  
three methods, X, Y and Z, and there exists a splice between  
+
methods that have two *different* three-lead splices -- one between X and Y, and a different one between Y and Z.
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
 
   X                Y                Z
Line 144: Line 102:
 
   Gl      (2&3)  Ca        (4&5)  Av
 
   Gl      (2&3)  Ca        (4&5)  Av
  
Conceptually these work by starting with Y (e.g. Di) and  
+
Conceptually these work by starting with Y (e.g. Di) and then splicing some of X and Z in.  However, there's a  
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.  
subtlety.  Suppose I start with Di, and want to ring Ws when  
+
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  
bells (a,b) are in 2&3, and Ms when bells (c,d) are in 4&5.  
+
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  
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.
 
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  
+
Imagine we start with method Y and splice in just 3 leads (the minimal unit) of method X when (a,b) are in the  
(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:
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)
 
   (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  
+
So we cannot get any more than 21 leads of Z (which is borne out by the search results).
out by the search results).
 
  
Counting up the possibilities here is going to get tedious  
+
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:
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,b) + (a,c)
Line 175: Line 122:
 
   (a,c) + (b,d)  [comes in l. and r. handed versions]
 
   (a,c) + (b,d)  [comes in l. and r. handed versions]
  
We can see that only the last configuration exhibs  
+
We can see that only the last configuration exhibs chirality.  The first two are invariant under relabelling d  
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  
and e (as neither are used).  The third is invariant under  
+
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  
relabelling c and d.  However, in the fourth, if we swap the  
+
diagram (as introduced in the first email cataloguing the simple splices).  Here the four configurations listed above  
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.
 
are depicted in the same order from left to right.
  
Line 193: Line 136:
 
         b    e        b    e        b    e        b    e
 
         b    e        b    e        b    e        b    e
  
(The dotted vertical line is representing the X-Y splice  
+
(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  
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.)
(a,b) and makes bells a and b special.  Bell e is never  
 
involved.)
 
  
With three Y-Z splices there are eight choices (including  
+
With three Y-Z splices there are eight choices (including left and right handed versions of chiral pairs):
left and right handed versions of chiral pairs):
 
  
 
   (a,b) + (a,c) + (b,c)
 
   (a,b) + (a,c) + (b,c)
Line 208: Line 148:
 
   (a,c) + (a,d) + (a,e)
 
   (a,c) + (a,d) + (a,e)
  
The number of plans (up to rotation) with four, five, six or  
+
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,  
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.
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  
+
This gives the number of plans with one application of the X-Y splice and at least one application of the Y-Z splice  
X-Y splice and at least one application of the Y-Z splice  
 
 
as: 2+5+8+8+5+2+1 = 31.
 
as: 2+5+8+8+5+2+1 = 31.
  
Now we need to think about two applications of the X-Y  
+
Now we need to think about two applications of the X-Y splice.  (I did say this was going to get tedious!)  There  
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.  
are two ways (up to rotation) of choosing two three-lead  
+
Bearing in mind every Y-Z splice must share a bell with every X-Y splice, this leaves the following Y-Z splice slots  
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.
 
viable.
 
<pre>
 
<pre>
Line 230: Line 164:
 
   (a,b) + (c,d)      (a,c), (a,d), (b,c), (c,d)
 
   (a,b) + (c,d)      (a,c), (a,d), (b,c), (c,d)
 
</pre>
 
</pre>
(Semicolons separate splice slots that are not equivalent  
+
(Semicolons separate splice slots that are not equivalent under rotation.)  We only need to consider ways of choosing  
under rotation.)  We only need to consider ways of choosing  
 
 
one or two Y-Z splices.
 
one or two Y-Z splices.
 
<pre>
 
<pre>
Line 251: Line 184:
 
   (a,b) + (c,d)    (a,c) + (a,d)
 
   (a,b) + (c,d)    (a,c) + (a,d)
 
</pre>
 
</pre>
This gives 27 = 3+6+6+3+1 + 2+3+2+1 plans with two  
+
This gives 27 = 3+6+6+3+1 + 2+3+2+1 plans with two applications of X-Y.
applications of X-Y.
 
  
Three applications of X-Y.  I catalogued the four ways of  
+
Three applications of X-Y.  I catalogued the four ways of choosing three three-lead slots in the previous email.
choosing three three-lead slots in the previous email.
 
  
 
   X-Y splices                    Viable Y-Z splice slots
 
   X-Y splices                    Viable Y-Z splice slots
Line 263: Line 194:
 
   (3.4)  (a,b) + (b,c) + (a,c)    (a,b), (a,c), (b,c)
 
   (3.4)  (a,b) + (b,c) + (a,c)    (a,b), (a,c), (b,c)
  
We've already established that (3.2) is chiral.  This  
+
We've already established that (3.2) is chiral.  This results from a symmetry breaking in the choice of X-Y  
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  
splices.  We cannot restore that symmetry by careful choice  
+
such thing as a "doubly chiral" configuration.  (How could there be?  Chirality happens when the automorphism group of  
of Y-Z splices.  Nor can we break it further -- there's no  
+
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  
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.
 
chiral.
  
A bit of thought show that the number of plans with three  
+
A bit of thought show that the number of plans with three X-Y splices will be:
X-Y splices will be:
 
  
 
   (3.1):  1+1    = 2
 
   (3.1):  1+1    = 2
Line 283: Line 209:
 
Which gives a total of 22 = 2+2*5+7+3 plans.
 
Which gives a total of 22 = 2+2*5+7+3 plans.
  
Fortunately the remaining cases -- of four or more  
+
Fortunately the remaining cases -- of four or more applications of the X-Y splice -- require little additional  
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  
thought.  It's clear that as the number of applications of  
+
possible increase.  Once we've handled the case of 4 X-Y applications and 4 Y-Z applications, then we already have  
X-Y increases, the number of viable Y-Z slots cannot  
+
the remaining numbers simply by reversing X and Z.  (Both are three lead splices and the ordering was arbitrary.)
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  
+
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  
based on (3.3) as this is the only one with four viable Y-Z  
 
 
slots.
 
slots.
  
Line 299: Line 220:
 
   (3.3)  (a,b) + (b,c) + (b,d)    (a,b), (b,c), (b,d); (b,e)
 
   (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  
+
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  
getting 4 X-Ys and 4 Y-Zs:  by choosing the same four slots  
 
 
for both splices.
 
for both splices.
  
Now we just need to revisit the previous calculations  
+
Now we just need to revisit the previous calculations extracting the number of plans with four or more Y-Zs.
extracting the number of plans with four or more Y-Zs.
 
  
 
   With 1 X-Y:  8+5+2+1
 
   With 1 X-Y:  8+5+2+1
Line 312: Line 231:
 
                 22
 
                 22
  
There were four sets of methods that offered two three-lead  
+
There were four sets of methods that offered two three-lead splices.  So the total number of multiple three-lead splice  
splices.  So the total number of multiple three-lead splice  
 
 
plans is 412 = 4 * (31+27+22+1+22).  Phew!
 
plans is 412 = 4 * (31+27+22+1+22).  Phew!
  
 
====SUMMARY====
 
====SUMMARY====
  
The total number of extent plans explained so far is as  
+
The total number of extent plans explained so far is as follows.
follows.
 
  
 
   Single method plans .  . . . . . . . . .  75 \
 
   Single method plans .  . . . . . . . . .  75 \
Line 326: Line 243:
 
   Three-lead splices . . . . . . . . . . .  798 /
 
   Three-lead splices . . . . . . . . . . .  798 /
 
   Multiple course splices  . . . . . . . .  36 \
 
   Multiple course splices  . . . . . . . .  36 \
   Multiple six-lead splices  . . . . . . .  276 | This email
+
   Multiple six-lead splices  . . . . . . .  286 | This email
 
   Multiple three-lead splices  . . . . . .  412 /
 
   Multiple three-lead splices  . . . . . .  412 /
 
   ---------------------------------------------
 
   ---------------------------------------------
   TOTAL  . . . . . . . . . . . . . . . . . 1881
+
   TOTAL  . . . . . . . . . . . . . . . . . 1891
  
We now know that the total number of extent plans that can  
+
We now know that the total number of extent plans that can be explained solely in terms of simple splices is 2280.  
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  
(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.
 
there are 399 left to go.

Latest revision as of 18:48, 6 November 2010

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

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 [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+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!

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  . . . . . . .  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.