[r-t] ringing-theory Digest, Vol 52, Issue 7

Mark Davies mark at snowtiger.net
Thu Jan 15 20:31:29 UTC 2009

Tom writes,

> You might argue that A,B,C are all the same composition because A=B and 
> B=C...

Ah no - we're not saying the compositions are the same. We're saying they 
are variations, which is not at all the same thing. A can be a variation of 
B, and B a variation of C, without requiring A to be a variation of C.

Again, in practice I see this all the time. Composition B will come along, 
and I will judge it to be a variation of A. Then next year I get composition 
C, which is clearly closely related to B, but less so to C. So in the 
composition collection we have a chain:

  Comp A

  Comp B
  A variation of comp A

  Comp C
  A variation of comp B

After a bit you get a nice family of compositions, all related.

What happens if comp C arrives before comp B? Well I realise B is a 
variation of C (the operation is commutative if not associative) as well as 
spotting that B is related to A. So I might put something like:

  Comp A

  Comp C

  Comp B
  A variation of comp C; see also comp A.

In the end it amounts to much the same thing - the relationships between 
compositions can be seen. To be strictly fair, when this sort of thing 
happens someone has generally spotted the relationship between A and C, and 
maybe annotated as follows:

  Comp A

  Comp C
  See also comp A

  Comp B
  A variation of comp C.

The "see also" operator is like "a variation of", but a bit less strict. :-)


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