city block (1)

>I see, I have changed the algorithm and decided to follow the shortest
route >in the triangular lattice but at each step taking the log of the
highest >prime instead of the lowest one. Otherwise one would get the
strange result >of 7/6 and 6/5 having a lower complexity than 5/4.

Do you mean that, in the case of (say) 7/6, you'd use log(7) instead of
log(3) to weight the number of 7/6's?

This doesn't seem to fix anything, because now the 5/4 and 6/5 are weighted
the same. What about weighting by the log of the sum of the odd limits of
the numerator and denominator? So 7/6=log(10), 6/5=log(8), 5/4=log(5)?

Carl

On Fri, 12 Mar 1999, Carl Lumma wrote:
> Do you mean that, in the case of (say) 7/6, you'd use log(7) instead of
> log(3) to weight the number of 7/6's?

Yes.

> This doesn't seem to fix anything, because now the 5/4 and 6/5 are weighted
> the same.

That's as it should be, as this metric ignores factors of 2.

--pH <manynote@lib-rary.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

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> That's as it should be, as this metric ignores factors of 2.

Up to a point, I'd agree that it makes sense to ignore factors of two. But
along related lines, I don't think that the overall size of the interval is not
important. For example, if the interval between two adjacent tones in a chord
is on the order of, say, 3 to 4 octaves (especially in a two-note chord), the
two pitches start to become so distant that they hardly interact with each other
at all. I would take that to default to sounding consonant, or at least not
dissonant.