edward.w.martin at gmail.com
Fri Aug 4 20:05:30 UTC 2006
On 8/4/06, Alexander Holroyd <holroyd at math.ubc.ca> wrote:
> On Fri, 4 Aug 2006, edward martin wrote:
> > I tried in vain to put myself in FS's shoes and figure out what
> > percievable logic was behind the principle Crambo
> I recently realised that there is a beautiful and mysterious logic behind
> Crambo. Start with an course of Stedman - this has all 60 in-course rows.
> Now, to turn this into an extent, one option is to try to _insert_ the 60
> out-of-course rows, each one between two of the existing in-course rows.
> To put it another way, expand each double change in the stedman into two
> single changes that have the same effect:
> 1 becomes 123.145 or 145.123
> 3 becomes 123.345 or 345.123
> 5 becomes 125.345 or 345.125
> It doesn't seem obvious that this can be done in a way that produces a
> true 120, but in fact it can, in exactly one way (up to symmetries, if we
> require a 5-part). The result is Crambo (reversed and started in a
> different place):
> The same trick works for some other in-course extents - Carter:
> Gransire (p-*3):
> Spliced grandsire and double grandsire :
> (unfortunately these 3-part ones all have 5 blows in one place).
> All of them are strange and unsymmetrical, so what is really going one
> here? I don't know. E.g. can you prove or disprove that any in-course
> extent of doubles can be treated in this way?
I wpuld have said yes in that if you have an extent of pure doubles
(say Grandsire or Rev Grandsire called P B P B P B or the plain course
of Stedman or Rev Stedman or Carter's or Rev Carters) then, since each
row is produced by a double then two singles are bound to accomplish
the same thing. Thus if we compare the backstroke rows of Crambo we
21435 5 via 345 125 = say A
24153 1 via 145 123 = say B
42135 3 via 345 123 = say C
24315 5 via 125 345 = say D
42351 3 via 123 345 = say E
24531 5 via 125 345 = D
42513 3 via 123 345 = E
45231 1 via 145 123 = B
54321 5 via 345 125 = A
45312 3 via 345 123 = C
54132 5 via 345 125 = A
45123 3 via 345 123 = C
Thus on the backstrokes we have the figures of Reverse Stedman Doubles
However, If we compare the handstroke rows of Crambo we have:
24135 = jump 4 to 2
42153 = 3 via E
42315 = jump 5 to 3
24351 = 3 via C
42531 = 5 via A
24513 = 3 via C
45213 = jump 1 to 3
54231 = 3 via E
45321 = 5 via D
54312 = 3 via E
45132 = 5 via D
54123 = 3 via E
we get something reminiscent of Rev Stedman but with three jump changes
> Did FS have this in mind? Who can say?
I don't think so but if not then what?
> Can you ring Crambo by thinking of it as slowed-down stedman? Errrr.....
Remember (A B C D E D E B A C A C)"5 and you've got it made!
More information about the ringing-theory