[r-t] Rapid Rap revisited

[Mike Ovenden]

Sorry Mark, no 720s here - indeed no compositions at all.  This post is
just to look at some _possibilities_ for multiple extent compositions of RR.

For completeness, start with what's already been done...

The simplest approach is to produce a composition in which every possible
permutation occurs exactly once as a lead head.  No need to worry about the
internal details of the method - the result is an 8-fold extent 8*720=5760.
Each perm also turns up at every other point in the lead somewhere in the
composition.  (If this isn't evident, read up on Lagrange's theorem?)  PJE
has supplied a composition of this sort (at least, I assume that's what it
is).

For a symmetrical method, a composition of half that length can be had if
every possible perm occurs exactly once as either a lead head or lead end.
RR is not symmetrical.

Now look at parity:
   123456 +1 612345 -1 etc
x  214365 -1 163254 +1
36 124635 -1 613524 +1
x  216453 +1 165342 -1
34 126435 +1 615324 -1
16 162345 +1 651234 -1
34 612354 +1 561243 -1
x  163245 -1 652134 +1
   ------  ------
56 612345 -1 561234 +1

The distribution is uneven within the lead, but we can sort that out by
rotation:
   123456 +1 261345 -1 etc
34 213465 +1 621354 -1
x  124356 -1 263145 +1
56 213456 -1 621345 +1
x  124365 +1 263154 -1
36 214635 +1 623514 -1
x  126453 -1 265341 +1
34 216435 -1 625314 +1
   ------  ------
16 261345 -1 652134 +1

A composition in which every possible even perm, but no other, occurs
exactly once as a lead head will be a 4-fold extent (2880) of this rotated
method.  One problem - the lh-to-lh transposition for a plain lead is odd -
so no plain leads would be possible.  (And maybe you'd then need 3 types of
call.)  So that's useless (except that you might manage to join the pieces
up using an additional extent of something well-behaved, for a 3600 of
spliced).  Try something different ...

Sticking with the rotated form of the method, consider the 24 lead heads
permuting 3,4,5,6 with 1,2 fixed.  These leads must in aggregate contain the
expected 48 rows 4 times each.  The required composition would now be a 2880
joining 360 distinct lead heads, constrained by the requirement that any
pair in 1,2 must always appear the same way.

Is there a weaker constraint?  Yes, using parity.  Consider the 12 lead
heads comprising even permutations of 3,4,5,6 with 1,2 fixed.
   123456 +1 ... 12{even permutation of 3456}
34 213465 +1 ... 21{odd permutation of 3456}
x  124356 -1 ... 12{odd permutation of 3456}
56 213456 -1 ... 21{even permutation of 3456}
x  124365 +1 ... 12{even permutation of 3456}
36 214635 +1 ... 21{odd permutation of 3456}
x  126453 -1 ... 12{odd permutation of 3456}
34 216435 -1 ... 21{even permutation of 3456}
   ------        ----------------------------
This set of 12 leads clearly contains the expected 48 perms belonging to
S2xS4 (mathematicians please correct me if this is not the correct way of
describing it) twice each.  It is therefore conceivable that a 1440 might be
obtainable.

For any pair fixed on the front, there are four such sets, equivalent in
terms of the rows they contain.  A 1440 must contain all 12 leads from one
of them (and none from the others).  Here they are for 1, 2 on the front ...

Set 1A
123456 123564 123645 124365 124536 124653
125346 125463 125634 126354 126435 126543

Set 1B
213456 213564 213645 214365 214536 214653
215346 215463 215634 216354 216435 216543

Set 1C
123465 123546 123654 124356 124563 124635
125364 125436 125643 126345 126453 126534

Set 1D
213465 213546 213654 214356 214563 214635
215364 215436 215643 216345 216453 216534

So there's a way of selecting 'true' leads of RR for a 1440.  Whether it's
possible to join them up in any reasonable way is another matter - an
exercise for the reader ;-|

Other useful(?) facts ...

1.  If a plain course is rung from some lead head, all other plain courses
from lead heads in the same set of 12 are excluded within the constraints of
a 1440 as set out above.

2.  The call x will join 3 leads from the same set of 12.  The entire set
can be arranged into 4 x-courses.

3.  The call 3456 will join 3 leads from the same set of 12.  The entire set
can be arranged into 4 3456-courses.

4.  An entire set of 12 can be linked into a round block by calling
xxssxxssxxss or xxsxssxxsxss or their reversals/rotations, where s=3456.

5.  The directed graph for the above sits nicely on the edges of a
cuboctahedron.

6.  Various place notations are unhelpful as calls for a 1440 because they
join leads from incompatible sets. That goes for: 12, 34, 36, 56, 1234,
1236, 1256.

Can a 1440 actually be realised?????


Mike