[r-t] (no subject)

Robin Woolley robin at robinw.org.uk
Fri Sep 24 08:56:50 UTC 2004

Philip Earis asked me to post this - any errors are a consequence of not
having had it checked, but other people have seen it and not commented.

There are three things proved:

i) A symmetric section only has one falseness group associated with it,
ii) The inversion of a symmetric section (e.g., 38x38 for x38x) has the same
falseness group as the original,
iii) An asymmetric section may also have just one falseness group.

I have been told that (i) was printed in the comic in the sixties. This
is where your 'electronic journal' idea would fill a huge gap. Most
subscribers to the list were not ringing nor yet born in the 60's.

Here goes then - the formulae look alright in Courier New

Conventional single treble dodging methods are considered so all numerical
examples are 'without loss of generality'. In any method name, 'Surprise
Major' is assumed.

Consider the first four changes of Cambridge suitably labelled:

A 12345678
B 21436587
C 12463857
D 21648375

Falseness occurs between corresponding rows with respect to treble position,
so A & C may be false and B & D also.

The basic ratio is one of the pair multiplied by the inverse of the other.
(The inverse is that permutation resulting in rounds). So:

 X(AC) = inv(C).A  X(BD) = inv(D).B

(In the example, inv(C) = 12537486, inv(D) = 21648375, X(AC) = 12537486
(since A = rounds = 'e' in common notation), X(BD) = 12537486)

It remains to show that it is generally true the X(AC) = X(BD) for a
symmetric section, i.e., of the form x38x rather than 38x56.

An interesting result, due entirely to the fact that a bell moves only one
position at a time, is that a place notation particle can be written as a
self-inverse permutation, e.g., 'x' = 21436587, '34'= 21346587, '38' =
21354768, etc. Therefore, we can write change B (supra) as pA, C as qB = qpA
and D as pqpA. To find the inverse of an algebraic expression, invert each
term and write down backwards. Since p is self inverse, inv(p) = p and p x p
= e.

Straight away, X(AC) = inv(C).A = inv(A).pqA
and            X(BD) = inv(D).B = inv(D).pA = inv(A)pqp.pA
                     = inv(A).pqA = X(AC)

Inverting the place notation, e.g., 38x38 for x38x gives C = pqA, so
              X'(AC) = inv(C).A = inv(A).qpA = inv(inv(A)pqA)
                     = inv(X(AC))
which is the inverse of the original ratio, thus giving the same falseness.

Finally, an asymmetric section has place notation of the form pqr, so D =
rpqA. X(AC) is only one of eight ratios giving the complete falseness.
X(AC).J (where J = 12436587) is another.
We look at X(AC)= X(BD).J:

X(AC) = inv(A).pqA = X(BD).J = inv(A)pqr.pA.J
       => pq = pqrpJ (on letting A = e (rounds))
       =>  e = rpJ => rp = J

An example of such a method is one starting 34x5678 - if you'd ever want to
ring it.

Best wishes

More information about the ringing-theory mailing list