Testing for block status

Does anyone have an idea for an algorithm for testing if a scale is a Fokker block? I'd like to know if I could get an example of a hobbit which was not a tempering of a Fokker block.

Gene wrote:

>Does anyone have an idea for an algorithm for testing if a scale is a
>Fokker block? I'd like to know if I could get an example of a hobbit
>which was not a tempering of a Fokker block.

Didn't you propose that any convex hull which has the
epimorphic property should be a Fokker block?

-Carl

It might be helpful if you defined what a Fokker block is in mathematical terms, what a tempering of a Fokker block is in mathematical terms, what a hobbit is in mathematical terms and what a scale is in mathematical terms. This might seem naive or even facetious to you but I think that you might need to go back to the initial conditions of your problem.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
> Does anyone have an idea for an algorithm for testing if a scale is a Fokker block? I'd like to know if I could get an example of a hobbit which was not a tempering of a Fokker block.
>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> Gene wrote:
>
> >Does anyone have an idea for an algorithm for testing if a scale is a
> >Fokker block? I'd like to know if I could get an example of a hobbit
> >which was not a tempering of a Fokker block.
>
> Didn't you propose that any convex hull which has the
> epimorphic property should be a Fokker block?

Good catch! In fact, the hobbit business can be reformulated in terms of Fokker blocks: there will be a minimal comma basis in terms of the metric on pitch classes derived from the Euclidean (weighted) norm, and the hobbit will be the image of one of the Fokker blocks derived from these commas. For instance, for 12et in marvel, the minimal 5-limit commas are 2048/2025 (smallest) and 128/125 (second smallest), leading to the diadie blocks. By my definition, diadie2 is the preimage for the hobbit, but we could just as well use diadie1 (aka lumma5.)

Gene wrote:

>> Didn't you propose that any convex hull which has the
>> epimorphic property should be a Fokker block?
>
>Good catch! In fact, the hobbit business can be reformulated in terms
>of Fokker blocks: there will be a minimal comma basis in terms of the
>metric on pitch classes derived from the Euclidean (weighted) norm,
>and the hobbit will be the image of one of the Fokker blocks derived
>from these commas. For instance, for 12et in marvel, the minimal
>5-limit commas are 2048/2025 (smallest) and 128/125 (second smallest),
>leading to the diadie blocks. By my definition, diadie2 is the
>preimage for the hobbit, but we could just as well use
>diadie1 (aka lumma5.)

It sounds very much like this

/tuning-math/message/18248

except with a Euclidean norm instead of a Tenney norm.

Probably there's a way to do my steps 5 & 6 algebraically, to
avoid having to identify and remove duplicate pitch classes, but
I doubt I'd be able to understand it well enough to trust it.

-Carl

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

> It sounds very much like this
>
> /tuning-math/message/18248
>
> except with a Euclidean norm instead of a Tenney norm.

I'd say some small similarly rather than very much alike. The biggest difference, perhaps, is that I'm using a norm derived from the temperament.

Let's define a "hobbit block" for a regular temperament as a tempering of a Fokker block for a minimal set of generators, with minimal not Tenney height or anything like it, but the Euclidean distance from the origin for commas projected onto octave-equivalent temperament subspace (that is, the subspace define by adding 2 as a comma to the comma list of the temperament.) While I singled out hobbits in order to give a unique scale, really all the hobbit blocks have an equal claim to consideration. The family of blocks can be defined by a set of commas in a JI subgroup being used to represent notes of the temperament. Here are three examples, the first two in 5-limit JI for "happy" temperaments, and the third in the {2,3,7} subgroup. The Fokker blocks are constructed as usual, and the hobbit blocks are simply the temperings. The point of using these sets of commas is that they are designed to give good results for the temperament--in this 7-limit case, that means to give 7-limit harmony.

marvel (225/224) comma bases

8: 16/15, 250/243
9: 135/128, 128/125
10: 28/27, 64/63 {2,3,7} subgroup
11: 135/128, 2048/1875
12: 2048/2025, 128/125
15: 128/125, 32768/30375
17: 25/24, 2278125/2097152
19: 16875/16384, 81/80
21: 128/125, 273375/262144
22: 2048/2025, 3125/3072
29: 16875/16384, 32805/32768
31: 81/80, 34171875/33554432

starling (126/125) comma bases

7: 25/24, 81/80
8: 16/15, 648/625
9: 27/25, 128/125
11: 144/125, 135/128
12: 128/125, 628/625
15: 128/125, 250/243
16: 648/625, 3125/3072
17: 25/24, 20480/19683
19: 81/80, 3125/3072
27: 128/125, 78732/78125
28: 648/625, 16875/16384
31: 81/80, 1990656/1953125
34: 15625/15552, 2048/2025

sensamagic (245/243) comma bases in {2,3,7} subgroup

12: 729/686, 64/63
17: 64/63, 19683/19208
19: 49/48, 177147/175616
22: 64/63, 537824/531441
24: 64/63, 15059072/14348907

Gene wrote:

>Let's define a "hobbit block" for a regular temperament as a tempering
>of a Fokker block for a minimal set of generators, with minimal not
>Tenney height or anything like it, but the Euclidean distance from the
>origin for commas projected onto octave-equivalent temperament
>subspace (that is, the subspace define by adding 2 as a comma to the
>comma list of the temperament.) While I singled out hobbits in order
>to give a unique scale, really all the hobbit blocks have an equal
>claim to consideration. The family of blocks can be defined by a set
>of commas in a JI subgroup being used to represent notes of the
>temperament. Here are three examples, the first two in 5-limit JI for
>"happy" temperaments, and the third in the {2,3,7} subgroup. The
>Fokker blocks are constructed as usual, and the hobbit blocks are
>simply the temperings. The point of using these sets of commas is that
>they are designed to give good results for the temperament--in this
>7-limit case, that means to give 7-limit harmony.
>
>marvel (225/224) comma bases
>8: 16/15, 250/243

These are the chromatic UVs of the block then? That would
fit if these are 7-limit examples. You say they're 5-limit but
225/224 isn't in the 5-limit. Can you produce scl files for a
few of them?

>9: 135/128, 128/125
>10: 28/27, 64/63 {2,3,7} subgroup

Is the "{2,3,7} subgroup" here a clipboard error?

-Carl

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

> >marvel (225/224) comma bases
> >8: 16/15, 250/243
>
> These are the chromatic UVs of the block then? That would
> fit if these are 7-limit examples.

Chromas are not very important; it's the commas which are interesting. These are commas for 8et.

You say they're 5-limit but
> 225/224 isn't in the 5-limit.

Indeed. The block produce what I've been calling "transversals"; 5-limit (in this case) preimages.

Can you produce scl files for a
> few of them?

You'll find tons of them from the links on the Xenwiki pages for "Marvel family" or "Scalesmith". You can also find some of them among the 5-limit Fokker block surveys I've done: 16/15 and 250/243 give the eight semipor scales; 135/128 and 128/125 the three mavdie scales; 25/24 and 2048/2025 the five diachrome scales; 2048/2025 and 128/125 the two diadie scales. These can be found in

/tuning-math/files/gene/fokker/

> >9: 135/128, 128/125
> >10: 28/27, 64/63 {2,3,7} subgroup
>
> Is the "{2,3,7} subgroup" here a clipboard error?

Sorry; that should have been

10: 25/24, 2048/2025

Gene wrote:

>> >marvel (225/224) comma bases
>> >8: 16/15, 250/243
[snip]
>These are commas for 8et.

Huh? You mean 16/15, 250/243 & 225/224 are a basis for
8-ET in the 7-limit? Fine, but we don't use 8-ET, we use
a scale containing 16/15 & 250/243 and tempering out
225/224.

>> Can you produce scl files for a
>> few of them?
>
>You'll find tons of them from the links on the Xenwiki pages for
>"Marvel family" or "Scalesmith". You can also find some of them among
>the 5-limit Fokker block surveys I've done:

Yeah yeah; the problem is to find a single scale.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> Gene wrote:
>
> >> >marvel (225/224) comma bases
> >> >8: 16/15, 250/243
> [snip]
> >These are commas for 8et.
>
> Huh? You mean 16/15, 250/243 & 225/224 are a basis for
> 8-ET in the 7-limit?

No, I mean 16/15 and 250/243 are a basis for the commas of 8et in the 5-limit, and in fact the Minkowski basis, in terms of the norm defined by restricting the seminorm derived from 225/224 to the 5-limit.

> Yeah yeah; the problem is to find a single scale.

That problem can be solved readily enough. I thought you wanted a single scale with some unusual properties, like GMP.

To find a single scale starting with a p-limit regular temperament and an et val supporting the temperament

(1) Find a mapping, or ikon, or whatever you like to call it. Could be normalized by Hermite reduction.

(2) Find associated generators, which are rational numbers mapped to 0 by all of the vals except for one, which maps it to 1.

(3) Reduce the list of generators to a normal form by reducing them to a normal interval list. This defines a subgroup of the p-limit group.

(4) Form a val-like thingie by applying v to each element of the normal list.

(5) Find a dual set of monzo-like thingies. You can do this by the rational projection map method: if A is the matrix whose rows are the val-like thingies, P = A`A, where A` is the pseudoinverse, is a projection map, and Q = I - P is the dual projection map. Applying a saturation algorithm to the rows (or columns, it's symmetric) of Q gives us the monzo-like thingies.

(6) Convert the monzo-like thingies to rational numbers by using the coefficients as exponents for the elements of the normal list.

(7) These are now a set of lattice basis elements for the lattice of commas of v restricted to the subgroup defined by the normal list. Now find the Minkowski basis for this lattice under the norm defined by the temperament commas plus octaves.

(8) Calling these c2, c2, ..., ck, find c1 in the subgroup with the property that v(c1) = 1.

(9) Convert c1,...,ck to a square matrix of monzos, invert, and find multiplicites e1,...,ek from the first row of the inverse matrix.

(10) form the scale

scale[i] = c1^i * c2^round(e2*i/n) * ... * ck^round(ek^i/n)

where v(2) = n.

(11) Temper the scale in your chosen tuning for the temperamant.

Gene wrote:

>No, I mean 16/15 and 250/243 are a basis for the commas of 8et in the
>5-limit, and in fact the Minkowski basis, in terms of the norm defined
>by restricting the seminorm derived from 225/224 to the 5-limit.

It sounds like I might approve of this if I fully understood it.

>> Yeah yeah; the problem is to find a single scale.
>
>That problem can be solved readily enough. I thought you wanted a
>single scale with some unusual properties, like GMP.

Well, yes. It looks like you're telling us which commas, but
it'd be nice to get a canonical block.

Do you always get GMP? Doesn't seem like it. If you apply your
procedure to a rank 2 temperament, what do you get for non-MOS
cardinalities?

>To find a single scale starting with a p-limit regular temperament and
>an et val supporting the temperament
>
>(1) Find a mapping, or ikon, or whatever you like to call it. Could be
>normalized by Hermite reduction.
>
>(2) Find associated generators, which are rational numbers mapped to 0
>by all of the vals except for one, which maps it to 1.
>
>(3) Reduce the list of generators to a normal form by reducing them to
>a normal interval list. This defines a subgroup of the p-limit group.
>
>(4) Form a val-like thingie by applying v to each element of the normal list.

Is v the original ET val supporting the temperament?

>(5) Find a dual set of monzo-like thingies. You can do this by the
>rational projection map method: if A is the matrix whose rows are the
>val-like thingies,

You lost me here.

>(7) These are now a set of lattice basis elements for the lattice of
>commas of v restricted to the subgroup defined by the normal list. Now
>find the Minkowski basis for this lattice under the norm defined by
>the temperament commas plus octaves.
>
>(8) Calling these c2, c2, ..., ck, find c1 in the subgroup with the
>property that v(c1) = 1.
>
>(9) Convert c1,...,ck to a square matrix of monzos, invert, and find
>multiplicites e1,...,ek from the first row of the inverse matrix.
>
>(10) form the scale
>
>scale[i] = c1^i * c2^round(e2*i/n) * ... * ck^round(ek^i/n)
>
>where v(2) = n.
>
>(11) Temper the scale in your chosen tuning for the temperamant.

I mean, wow. I know some of this is explained on the xenwiki,
but, dzam.

-Carl

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

> Well, yes. It looks like you're telling us which commas, but
> it'd be nice to get a canonical block.

I'm finding that the hobbit is not always in the block defined by the Minkowski basis. I'm beginning to think that the best solution would be to come with a better method for computing hobbits, but neither you nor Graham seems interested in them as a solution to this problem. Why is that?

> Do you always get GMP? Doesn't seem like it.

No. Those seem to be rare.

If you apply your
> procedure to a rank 2 temperament, what do you get for non-MOS
> cardinalities?

A non-MOS generated scale.

>> Well, yes. It looks like you're telling us which commas, but
>> it'd be nice to get a canonical block.
>
>I'm finding that the hobbit is not always in the block defined by the
>Minkowski basis. I'm beginning to think that the best solution would
>be to come with a better method for computing hobbits, but neither you
>nor Graham seems interested in them as a solution to this problem.
>Why is that?

I'm interested, just not likely to be of much use. I'm also
interested in tackling the problem (of MOS generalization) in
ways I feel I understand better.

>> Do you always get GMP? Doesn't seem like it.
>
>No. Those seem to be rare.

Well, do they follow a pattern??

-Carl

On 26 October 2010 08:50, genewardsmith <genewardsmith@sbcglobal.net> wrote:

> I'm finding that the hobbit is not always in the block defined by the Minkowski basis. I'm beginning to think that the best solution would be to come with a better method for computing hobbits, but neither you nor Graham seems interested in them as a solution to this problem. Why is that?

I might be interested if I knew what they were. I can see the point
in scales made up of notes of minimal complexity. But maybe my
generalization of that is simpler than yours. I still don't take
mathematically generated scales that seriously, though. I'd rather
arrange the intervals how I want them.

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:

> I might be interested if I knew what they were. I can see the point
> in scales made up of notes of minimal complexity. But maybe my
> generalization of that is simpler than yours.

Hobbits are constructed using notes of minimal complexity relative to the temperament and under the assumption we are constructing scales with octave repetition. How do you define "minimal complexity"?

I still don't take
> mathematically generated scales that seriously, though. I'd rather
> arrange the intervals how I want them.

Except that's getting pretty hard by 46 notes. And if mathematically generated scales are so uninteresting, why all the interest in MOS?

On 26 October 2010 23:07, genewardsmith <genewardsmith@sbcglobal.net> wrote:

> Hobbits are constructed using notes of minimal complexity relative to the temperament and under the assumption we are constructing scales with octave repetition. How do you define "minimal complexity"?

There's no obvious way of measuring the complexity of intervals in a
higher-rank temperament. What I'll suggest is first taking some
octave-orthogonal, near-self-orthogonal basis. The octave-equivalent
part of the Hermite basis will probably do. Better would be the
Tenny-weighted LLL reduction of this.

Tenney-weighted LLL reduction is, in fact, easy to do with a standard
LLL function. What you do is weight your vectors in the usual way
(divide by logs of primes for mappings, multiply by logs of primes for
intervals) and feed them into the LLL function. It works because the
basis is still orthogonal. Reverse the weighting and round off to get
your results if necessary. If your LLL function requires integers,
multiply the weights by an arbitrary large integer and round off.
(Don't round off the weighted vectors because you might end up in the
wrong lattice.)

So we have a new, near-orthogonal basis with low scalar complexity for
the temperament. You can calculate the scalar complexity of tempered
intervals using a Euclidean norm (dot or scalar product) with a
suitable weighting. Probably conserving the scalar complexity of the
basis vectors is the way to go.

To go from complexity of intervals to complexity of notes, measure the
intervals relative to the tonic. The obvious way is to take the
simplest interval that maps to the correct number of scale degrees. I
guess that'll work, but maybe it's balls. The generalization of
dwarves would be to only consider positive steps of the
octave-orthogonal vectors. That would entail defining which direction
counts as positive to get the correct otonal bias. There's probably
some function of the mapping that would do this.

For rank 2, this would give generated scales, and so MOS with the
right number of notes.

>  I still don't take
>> mathematically generated scales that seriously, though.  I'd rather
>> arrange the intervals how I want them.
>
> Except that's getting pretty hard by 46 notes. And if mathematically generated scales are so uninteresting, why all the interest in MOS?

If you want to argue about this, find somebody who disagrees with you.

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:
>
> On 26 October 2010 23:07, genewardsmith <genewardsmith@...> wrote:
>
> > Hobbits are constructed using notes of minimal complexity relative to the temperament and under the assumption we are constructing scales with octave repetition. How do you define "minimal complexity"?
>
> There's no obvious way of measuring the complexity of intervals in a
> higher-rank temperament.

Whether or not it's obvious, Euclidean distance on the subspace the TOP-RMS projection projects orthogonally onto is precisely a way of measuring complexity for higher-rank temperaments. In fact, it even works for rank one, where it gives something proportional to the count of scale steps. What's less obvious is whether it's legitimate to get rid of octaves by adding 2 to the commas, but it seems to work, so I think my characterization is accurate.

"genewardsmith" <genewardsmith@sbcglobal.net> wrote:

> Whether or not it's obvious, Euclidean distance on the
> subspace the TOP-RMS projection projects orthogonally
> onto is precisely a way of measuring complexity for
> higher-rank temperaments. In fact, it even works for rank
> one, where it gives something proportional to the count
> of scale steps. What's less obvious is whether it's
> legitimate to get rid of octaves by adding 2 to the
> commas, but it seems to work, so I think my
> characterization is accurate.

I'd call I-P the TOP-RMS projection, but anyway...

In another message, I showed your complexity is the same as
a quadratic form on intervals in the temperament given by:

K = inverse(mapping~*W*W*mapping)

So the complexity is:

comp(interval) = interval*K*interval~

One thing I forgot to mention in that other message is it's
much better to minimize intervals in the temperament than
using the norm on ratio space intervals. In the
temperament, there are fewer dimensions, and the defining
matrix is positive definite, which makes it easier to work
with.

Anyway, this does indeed work with equal temperaments, as K
becomes a scalar. It also works with just intonation,
because the mapping becomes the identity matrix:

K = inverse(I~*W*W*I)
= inverse(W*W)

This is the Tenney-weighted Euclidean metric equivalent, or
dual, to scalar complexity.

I'm certainly looking for some kind of quadratic form like
this. I've concluded that what it's doing is transforming
to an ideal basis where everything's perfectly orthogonal
(which is impossible on the lattice) and weighting the
basis intervals according to their size. That leads to
strange consequences, like secors being much simpler than
octaves in Miracle. But then, that makes sense, because
you're likely to pile up more secors in any given scale
than octaves.

I've been through various examples that I don't have time
to give now. For well behaved temperaments, it does make
sense. But I think it fails on mystery:

? mystery = [29, 46, 67, 81, 100, 107; 58, 92, 135, 163,
201, 215]~;
? qfminim(inverse(mystery~ * W*W * mystery),
0.03, 20, 2)[3]

[1 2 3 4 5 6 7]

[2 4 6 8 10 12 14]

These are the simplest column vectors in the lattice shown
as steps of 29 (top) and 58 (bottom) note scales. All it's
doing is following 29-equal, and it keeps doing that for
over two octaves before it does anything more interesting.

I don't, currently, have any better ideas for metrics
though. Well, except that you can transform to some
standard reduction and define a diagonal matrix.

Graham

Graham Breed <gbreed@gmail.com> wrote:

> I've been through various examples that I don't have time
> to give now. For well behaved temperaments, it does make
> sense. But I think it fails on mystery:
>
> [1 2 3 4 5 6 7]
>
> [2 4 6 8 10 12 14]
>
> These are the simplest column vectors in the lattice shown
> as steps of 29 (top) and 58 (bottom) note scales. All
> it's doing is following 29-equal, and it keeps doing that
> for over two octaves before it does anything more
> interesting.

After sleeping on it, I think I was wrong. What it's doing
is defining the 29-equal scale first, which is the first
MOS. Then it extends it to higher octaves before starting
on the second, 58-note MOS. But there are still 29 notes
per octave, which is what the odd-limit complexity tells us
we should have before non-Pythagorean intervals come in.
Then, it'll start defining the 58 note MOS to get 58 notes
per octave. So that's correct.

Given how simple it is, let's assume this is the valid
metric for tempered interval complexity until we have a
good reason to think otherwise. Well done Gene!

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:

> Given how simple it is, let's assume this is the valid
> metric for tempered interval complexity until we have a
> good reason to think otherwise. Well done Gene!

Thanks. It's got a quadratic form, so it's got to be good. :)