# [CM] cmn factorize

René Bastian rbastian@free.fr
Fri, 7 Apr 2006 14:21:01 +0200

```Le Vendredi  7 Avril 2006 13:38, Bill Schottstaedt a écrit :
> The "logic" in the 1/9 case
> goes something like: if you ask for (say) triplets (i.e. n/3), you' re
> putting 3 (say 8ths) in a quarter, so you get, for example, q + e = q.
> If you ask for 5 divisions in a quarter, that's q + s = q, (or is it h + e
> = q?) -- since the beat is closer to a 16th, this gets two beams, but the
> problem for cmn is that in nested n-lets so to speak, you need to keep the
> beams
> logical (say a triplet within a triplet = 1/9) so the outer triplet, the
> unbroken beam, has one beam, and the inner triplet is considered
> 3 16ths in a triplet 1/8, so it ought to have 2 beams, but that means
> a quarter at the outer level = h + s at the innermost level.  You'll
> notice in the code (just below the point you mention) a long comment
> about the problem, all brought about by rqq.lisp and its fancy nested
> groupings.  I suppose we could add a flag to choose which style is
> desired.  (The factorization was trying to catch these nested cases --
> as you can see in the code I originally just chose the closest match
> between flags and actual duration, but the experts disagree even on
> this).
>
There is another simple rule :
wholes : 1
halfs : 2 3
quarters : 4 5 6 7
8-th-s : 8 ... 15 [== 1 beam]
16-th-s : 16 ... 31 [== 2 beams]

=> 9 (equ. 18) == 2 beams

Hm, but if the divisor is near 31, I would prefer 3 beams.
Where is the limit ? ["chacun pour soi" :-)]
One can find such "strange beam attractors" in Stockhausen's "Gruppen"

--
René Bastian
« Le progrès consiste plus que jamais dans l'apprentissage de l'abstraction
et dans la rupture avec les images. »  Jean-Michel Besnier (La croisée des
sciences)

```