Odd-limit vs octave-equivalent Tenney height

I think the point of odd-limits are to isolate a set of intervals
which tend to combine well to form chords. For instance, the
5-odd-limit gives you 1/1, 4/3, and 5/3, and 5/3 appears before 15/8.
Does this actually work, though?

For instance, you need to look at n*d <=15, ignoring factors of 2, to
get all of the intervals in the 5-odd-limit. Is there some
mathematically precise sense in which the average complexity of the
triads that you get by looking at the
15-octave-equivalent-Tenney-limit is much higher than the average
complexity of the triads you can get in the 5-odd-limit?

Or, to make it simpler, we can stop assuming octave equivalence and
just consider Tenney-limits and integer-limits instead, which place
bounds on Tenney height and Weil height. Does n*d <= 15 give crappier
triads on average than max(n,d) <= 5?

-Mike

At 11:44 PM 8/22/2012, you wrote:
>I think the point of odd-limits are to isolate a set of intervals
>which tend to combine well to form chords. For instance, the
>5-odd-limit gives you 1/1, 4/3, and 5/3, and 5/3 appears before 15/8.
>Does this actually work, though?
>
>For instance, you need to look at n*d <=15, ignoring factors of 2, to
>get all of the intervals in the 5-odd-limit. Is there some
>mathematically precise sense in which the average complexity of the
>triads that you get by looking at the
>15-octave-equivalent-Tenney-limit is much higher than the average
>complexity of the triads you can get in the 5-odd-limit?
>
>Or, to make it simpler, we can stop assuming octave equivalence and
>just consider Tenney-limits and integer-limits instead, which place
>bounds on Tenney height and Weil height. Does n*d <= 15 give crappier
>triads on average than max(n,d) <= 5?
>
>-Mike

What relationship to triads are you assuming?

I thought the point of odd limit was to be the best octave-
equivalent estimator of harmonic complexity for dyads.

We considered max(n,d) in 1999. It works fairly well but is
less discriminative than Tenney height. For instance, it gives
all of: 15/1, 15/2, 15/4, 15/7, 15/8, 15/11, 15/13 and 15/14 the
same complexity.
That seems worse than giving 5/3 and 15/1 the same complexity.

-Carl

On Thu, Aug 23, 2012 at 4:19 PM, Carl Lumma <carl@lumma.org> wrote:
>
> What relationship to triads are you assuming?

Say that we're in the p-prime-limit. If we are, then we don't only
care about intervals with only prime factors of p, but we also care
about triads and larger chords with only prime factors of p.
Specifically, care about those larger chords in the p-limit which
themselves aren't very complex, like 4:5:6 for the 5-limit, and maybe
4:5:6:7 for the 7-limit, etc. The set of all dyads that appears in the
chord 1:2:3:4:...:p is exactly the set of p-integer-limit dyads, and
if you assume octave equivalence you end up looking at the set of
p-odd-limit dyads. If instead you look at the smallest Tenney-limit
containing this entire integer limit, you end up getting a bunch of
extra crap that doesn't appear in the chord you're looking at, and the
same applies in the octave-equivalent case. So even though it ranks
individual intervals strangely, e.g. 5/3 and 5/4 and 5/1 are all
equal, it might make more sense when looking at larger chords.

Or so I thought, but now I'm having doubts about it. I guess it boils
down to how you want to rate chord complexity. For instance, I'm
implicitly assuming an integer-limit or odd-limit measure of chord
complexity above. But is that correct, or is Tenney height for chords
more correct? For instance, if one decides that 1:3:5 is just as
complex as 1:1:15, or 1:3:4 is just as complex as 1:2:6, then maybe
using octave-equivalent Tenney height makes more sense.

> I thought the point of odd limit was to be the best octave-
> equivalent estimator of harmonic complexity for dyads.

OK, but in line with the above, why is odd-limit better than
octave-equivalent Tenney height?

> We considered max(n,d) in 1999. It works fairly well but is
> less discriminative than Tenney height. For instance, it gives
> all of: 15/1, 15/2, 15/4, 15/7, 15/8, 15/11, 15/13 and 15/14 the
> same complexity.
> That seems worse than giving 5/3 and 15/1 the same complexity.

So why does that become OK if octave equivalence is assumed? For
instance, in the 9-limit, why is it good for 9/1, 9/5, and 9/7 to be
all rated the same?

-Mike

Mike wrote:

>> What relationship to triads are you assuming?
>
>Say that we're in the p-prime-limit. If we are, then we don't only
>care about intervals with only prime factors of p, but we also care
>about triads and larger chords with only prime factors of p.
>Specifically, care about those larger chords in the p-limit which
>themselves aren't very complex, like 4:5:6 for the 5-limit, and maybe
>4:5:6:7 for the 7-limit, etc. The set of all dyads that appears in the
>chord 1:2:3:4:...:p is exactly the set of p-integer-limit dyads, and
>if you assume octave equivalence you end up looking at the set of
>p-odd-limit dyads. If instead you look at the smallest Tenney-limit
>containing this entire integer limit, you end up getting a bunch of
>extra crap that doesn't appear in the chord you're looking at, and the
>same applies in the octave-equivalent case. So even though it ranks
>individual intervals strangely, e.g. 5/3 and 5/4 and 5/1 are all
>equal, it might make more sense when looking at larger chords.

Sure, odd limits make a lot of sense when dealing with complete
harmonic series segments -- that's how Partch came up with them.
They also "work" with subharmonic series segments though, and
Partch never properly dealt with that. One option is to disallow
fractions in the spelling of chords. Subharmonic segments can hit
large odds when expressed as harmonics. The minor triad would
already be 15-limit. That's what we do with generalized Tenney
height, which is also more expressive for chords that aren't
saturated (just like in the dyadic case). But it's not octave-
equivalent.

Prime limits are pretty much useless for talking about chords.

>> I thought the point of odd limit was to be the best octave-
>> equivalent estimator of harmonic complexity for dyads.
>
>OK, but in line with the above, why is odd-limit better than
>octave-equivalent Tenney height?

What's octave-equivalent Tenney height?

>> We considered max(n,d) in 1999. It works fairly well but is
>> less discriminative than Tenney height. For instance, it gives
>> all of: 15/1, 15/2, 15/4, 15/7, 15/8, 15/11, 15/13 and 15/14 the
>> same complexity.
>> That seems worse than giving 5/3 and 15/1 the same complexity.
>
>So why does that become OK if octave equivalence is assumed? For
>instance, in the 9-limit, why is it good for 9/1, 9/5, and 9/7 to be
>all rated the same?

It isn't necessarily OK -- what's the alternative?
It is true that any octave-equivalent estimator has to cope
with rating things like 16/15, 15/8 and 15/1 the same, so
maybe it just doesn't stick out as much.

-Carl

On Thu, Aug 23, 2012 at 5:37 PM, Carl Lumma <carl@lumma.org> wrote:
>
> Sure, odd limits make a lot of sense when dealing with complete
> harmonic series segments -- that's how Partch came up with them.
> They also "work" with subharmonic series segments though, and
> Partch never properly dealt with that.

Sure, and an even worse thing is that you can combine 1/1, 6/5, and
5/4 together to get 1/1:6/5:5/4 = 20:24:25, which is bad.

> One option is to disallow
> fractions in the spelling of chords. Subharmonic segments can hit
> large odds when expressed as harmonics. The minor triad would
> already be 15-limit. That's what we do with generalized Tenney
> height, which is also more expressive for chords that aren't
> saturated (just like in the dyadic case). But it's not octave-
> equivalent.

What does "not saturated" mean here?

> >> I thought the point of odd limit was to be the best octave-
> >> equivalent estimator of harmonic complexity for dyads.
> >
> >OK, but in line with the above, why is odd-limit better than
> >octave-equivalent Tenney height?
>
> What's octave-equivalent Tenney height?
//snip
> It isn't necessarily OK -- what's the alternative?

Any rational n/d can be expressed as 2^k * a/b, where k, a, and b are
integers, and a and b are coprime with each other and also with 2.
Then the octave-equivalent Tenney height of n/d is log(a*b).

Or another way to think about it is just to drop all factors of 2 and
then take a*b rather than max(a,b).

-Mike

Mike wrote:
>> Sure, odd limits make a lot of sense when dealing with complete
>> harmonic series segments -- that's how Partch came up with them.
>> They also "work" with subharmonic series segments though, and
>> Partch never properly dealt with that.
>
>Sure, and an even worse thing is that you can combine 1/1, 6/5, and
>5/4 together to get 1/1:6/5:5/4 = 20:24:25, which is bad.

That wouldn't be a 5-odd-limit chord, since the odd limit is
supposed to apply to all the dyads in the chord.

>> One option is to disallow
>> fractions in the spelling of chords. Subharmonic segments can hit
>> large odds when expressed as harmonics. The minor triad would
>> already be 15-limit. That's what we do with generalized Tenney
>> height, which is also more expressive for chords that aren't
>> saturated (just like in the dyadic case). But it's not octave-
>> equivalent.
>
>What does "not saturated" mean here?

Not having or nearly having all harmonics up to its limit.
Like 4:5:9 is not a saturated 9-limit chord. Generalized
Tenney height can distinguish it from 4:5:6:7:9 but odd limit
cannot, just as Tenney height can distinguish 15/8 and 15/11
while odd limit cannot.

>> What's octave-equivalent Tenney height?
>//snip
>> It isn't necessarily OK -- what's the alternative?
>
>Any rational n/d can be expressed as 2^k * a/b, where k, a, and b are
>integers, and a and b are coprime with each other and also with 2.
>Then the octave-equivalent Tenney height of n/d is log(a*b).
>Or another way to think about it is just to drop all factors of 2 and
>then take a*b rather than max(a,b).

Ok, sure. Keenan based an OEHE curve on it once
/tuning-math/message/19139
You can do it with chords too, by removing all factors of 2 from
each element of the harmonic series representation. I've played
with that a bit.

I wonder if Keenan ever resolved the OEHE issue to his
satisfaction. It would be nice if it had a fixed point with
respect to the basis, like Paul found for regular HE (where
the output always seemed to agree with Tenney height regardless
of basis).

-Carl

On Fri, Aug 24, 2012 at 1:12 AM, Carl Lumma <carl@lumma.org> wrote:
>
> Ok, sure. Keenan based an OEHE curve on it once
> /tuning-math/message/19139
> You can do it with chords too, by removing all factors of 2 from
> each element of the harmonic series representation. I've played
> with that a bit.

So why did everyone come to the conclusion that is odd-limit is ideal
for octave-equivalent dyadic complexity then if this is here? I have
my own reasons for liking it, but I'm curious what everyone else
thinks.

-Mike

At 12:32 AM 8/24/2012, you wrote:
>On Fri, Aug 24, 2012 at 1:12 AM, Carl Lumma <carl@lumma.org> wrote:
>>
>> Ok, sure. Keenan based an OEHE curve on it once
>> /tuning-math/message/19139
>> You can do it with chords too, by removing all factors of 2 from
>> each element of the harmonic series representation. I've played
>> with that a bit.
>
>So why did everyone come to the conclusion that is odd-limit is ideal
>for octave-equivalent dyadic complexity then if this is here? I have
>my own reasons for liking it, but I'm curious what everyone else
>thinks.

Paul was its main advocate. His OEHE agreed with odd-limit when
seeded with ratios under an odd-limit cutoff. It would be
interesting to see what happens when seeding with no-2s Tenney.

-Carl

On Fri, Aug 24, 2012 at 4:14 AM, Carl Lumma <carl@lumma.org> wrote:
>
> Paul was its main advocate. His OEHE agreed with odd-limit when
> seeded with ratios under an odd-limit cutoff. It would be
> interesting to see what happens when seeding with no-2s Tenney.
>
> -Carl

OK, so how do I actually do this? Do I just pick all the ratios lying
within a few octaves that are bounded by their no-2's Tenney height
and then run HE as normal?

-Mike

>> Paul was its main advocate. His OEHE agreed with odd-limit when
>> seeded with ratios under an odd-limit cutoff. It would be
>> interesting to see what happens when seeding with no-2s Tenney.
>>
>> -Carl
>
>OK, so how do I actually do this? Do I just pick all the ratios lying
>within a few octaves that are bounded by their no-2's Tenney height
>and then run HE as normal?

Yes.

There's some question about how to assign widths to each
seed ratio. Mediants probably shouldn't be used. Instead,
Paul used "limit-weighted midpoints"

/tuning/topicId_12722.html#12722

and Keenan apparently used plain midpoints (which Paul said
produce almost identical results).

-Carl

I wrote:
>There's some question about how to assign widths to each
>seed ratio. Mediants probably shouldn't be used. Instead,
>Paul used "limit-weighted midpoints"
>
> /tuning/topicId_12722.html#12722
>
>and Keenan apparently used plain midpoints (which Paul said
>produce almost identical results).

As long as the seed function cutoff is significantly higher
than the point where its relationship to concordance breaks down,
I suspect any reasonable midpoint should be OK. But it does
look like a 'no2s-TH-weighted midpoint' passes the test Paul
showed in that message, i.e.

log(3)*498.04 + log(35)*582.51 / log(3)+log(35) = 562.57

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>

> >OK, so how do I actually do this? Do I just pick all the ratios lying
> >within a few octaves that are bounded by their no-2's Tenney height
> >and then run HE as normal?
>
> Yes.

Or you could try the idea I proposed a while back, and use the theta function in place of the normal distribution.