Optimal subgroups

Given a full p-limit temperament of rank r, we may define a subgroup temperament of rank s<r by finding a set of generators of minimal maximum Benedetti height. We may also do the same thing with the added proviso that 2 must be in the subgroup. Presumably most of the time at least these will be uniquely determined. It would be nice to have an algorithm for finding them.

On Mon, Jul 23, 2012 at 7:31 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> Given a full p-limit temperament of rank r, we may define a subgroup
> temperament of rank s<r by finding a set of generators of minimal maximum
> Benedetti height. We may also do the same thing with the added proviso that
> 2 must be in the subgroup. Presumably most of the time at least these will
> be uniquely determined. It would be nice to have an algorithm for finding
> them.

I assume that you're only looking for integer monzos, right? e.g. no
fancy schmancy intervals that are defined as "the interval bisecting
3/2"?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> I assume that you're only looking for integer monzos, right? e.g. no
> fancy schmancy intervals that are defined as "the interval bisecting
> 3/2"?

Right. No cheating.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@> wrote:
>
> > I assume that you're only looking for integer monzos, right? e.g. no
> > fancy schmancy intervals that are defined as "the interval bisecting
> > 3/2"?
>
> Right. No cheating.
>

Another approach would be via the complexity of the multimonzo for the group.

On Tue, Jul 24, 2012 at 1:43 AM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> Another approach would be via the complexity of the multimonzo for the
> group.

OK, so you want to solve this problem for every s < r?

What other restrictions are there? Must the original temperament map
down to the new subgroup one? IOW, does 2.3 1/1 count as an invalid
solution for 2.3.11 243/242?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> What other restrictions are there? Must the original temperament map
> down to the new subgroup one?

Of course--as noted, they have the same commas.

IOW, does 2.3 1/1 count as an invalid
> solution for 2.3.11 243/242?

Yes, because 243/242 must be in the group.

I think I might make some progress on the rank two case by using something akin to the Fokker group; I'll report if that works out but I am busy this AM.

On Tue, Jul 24, 2012 at 1:00 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> IOW, does 2.3 1/1 count as an invalid
> > solution for 2.3.11 243/242?
>
> Yes, because 243/242 must be in the group.

OK. I didn't see that they have to contain the same commas. So does
2.9.5 81/80 count as a valid solution to 2.3.5 81/80? Does machine
count as a valid solution for supra?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> OK. I didn't see that they have to contain the same commas. So does
> 2.9.5 81/80 count as a valid solution to 2.3.5 81/80? Does machine
> count as a valid solution for supra?

Sure, those are fine.

I just tried my first test case for the Fokker-group-like stuff, and that part worked fine. But my idea that Benedetti height will usually be enough to ensure uniqueness ran into a problem. I get two normal lists with the same max height, 165, from 2.7/3.15/11 and 2.9/7.33/5. Do either one have a basis with smaller max height? Also, I don't like 2.9/7.33/5 much but there it is anyway. Of course, put them together and we get 2.3.7.11/5, which however is rank 3.

On Tue, Jul 24, 2012 at 8:20 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> I just tried my first test case for the Fokker-group-like stuff, and that
> part worked fine. But my idea that Benedetti height will usually be enough
> to ensure uniqueness ran into a problem. I get two normal lists with the
> same max height, 165, from 2.7/3.15/11 and 2.9/7.33/5. Do either one have a
> basis with smaller max height? Also, I don't like 2.9/7.33/5 much but there
> it is anyway. Of course, put them together and we get 2.3.7.11/5, which
> however is rank 3.

I'm starting to fall behind here; I was hoping to address some of this
in my post about Lp complexity for temperaments. But what's really
needed here is a rigorous way to address the complexity of a subgroup.
We've been talking about the norm of the multimonzo the subgroup
defines, but that leads to problems in comparing subgroups of
different ranks (or well, just one problem, which is how to normalize
the ranks against one another). This max Benedetti height of the
normal form basis is another idea.

What was your Fokker group idea?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> What was your Fokker group idea?

The idea is to get something which works for the most interesting case, which is a rank two subgroup temperament containing 2 in the group. Starting with a higher rank full p-limit temperament, look at the first row of the HNF map to see if the octave is divided. For instance, the rank four temperament from 540/539, it isn't, but for the rank three temperament from 250047/250000 it is divided into thirds with generator 63/50. The multimonzo deriving from 2 or the generator for 2, such as 63/50, plus a basis for the commas we may call V., Then the Fokker-like group consists of V^A for A a monzo of rank p. Find a basis for this, and LLL reduce it. The LLL reduction uses unweighted complexity but is a great start for a search for the minimal weighted complexity, or minimal Benedetti height of the normal list, or what have you.

In the case of 540/539, the minimal complexity trimonzo is |||1 -2 -1 0 0 0 0 0 0 0>>>, corresponding to the subgroup 2.3.539/5, which is nothing to write home about, and really counts as 2.3 anyway. Clearly the Benedetti height minimum idea is working better. The 2.7/3.15/11 group is the most interesting one; it is defined by |||1 1 -1 1 0 1 0 0 0 0>>>. 2.9/7.33/5 and even worse, 2.15/7.55/3 have more complex trimonzos but the same Benedetti height, which means Benedetti+L2 complexity seems to work better than either one separately.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

>which means Benedetti+L2 complexity seems to work better than either one separately.

It looks like perhaps L-inf complexity is what is really wanted.

On Tue, Jul 24, 2012 at 10:40 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> The multimonzo
> deriving from 2 or the generator for 2, such as 63/50, plus a basis for the
> commas we may call V., Then the Fokker-like group consists of V^A for A a
> monzo of rank p. Find a basis for this, and LLL reduce it. The LLL reduction
> uses unweighted complexity but is a great start for a search for the minimal
> weighted complexity, or minimal Benedetti height of the normal list, or what
> have you.

This is dual to something I was looking at, which is looking at
Fokker-groups of tuning maps (like the fokker group for the JIP). I
note that a non-trivial consequence of this fact is that you're
creating Fokker blocks in val space, whatever that means. I've thought
about them a lot but I have no idea wtf they are yet.

> In the case of 540/539, the minimal complexity trimonzo is |||1 -2 -1 0 0
> 0 0 0 0 0>>>, corresponding to the subgroup 2.3.539/5, which is nothing to
> write home about, and really counts as 2.3 anyway.

Auugh, that's terrible. Yeah, It makes sense that unweighted
complexity would fail miserably for multimonzos. It'll just wantonly
throw combinations of 7 and 11 around and smash everything.

Might be better to start from the weighted pseudoinverse. In fact, you
can probably sideswipe the whole Fokker group thing by just taking the
left null space of your V-map and then taking the pseudoinverse of it.
So I'd expect the generators with lowest L2 complexity will be the
ones which are also you can write as vals that temper out your
subgroup. Then maybe you can quantize from there to find the nearest
subgroup which only uses integer monzos.

> Clearly the Benedetti
> height minimum idea is working better. The 2.7/3.15/11 group is the most
> interesting one; it is defined by |||1 1 -1 1 0 1 0 0 0 0>>>. 2.9/7.33/5 and
> even worse, 2.15/7.55/3 have more complex trimonzos but the same Benedetti
> height, which means Benedetti+L2 complexity seems to work better than either
> one separately.

Or Linf complexity. This is another thing I was going to talk about in
my post which never materialized, but I might as well write it here:
there are two notions of complexity for any Lp norm other than L2.
There's the Lp on the multimonzo representing the kernel, which is
what Paul's using for TOP complexity, and then there's the dual Lp* on
the dual multival to that multimonzo. The coefficients will be the
same, but the choice of norm changes.

If you use Linf complexity, that's the same as just taking the Lp*
complexity of the multival tempering out your subgroup, rather than
the subgroup itself. This is interesting because it's totally dual to
what Paul's doing to calculate TOP: he takes the Lp complexity of the
dual of the multival defined by the temperament, so now you're going
to take the Lp* complexity of the dual of the multimonzo defined by
the subgroup.

I had a few conjectures about this, all of which are relevant, but I'm
too fried to think; I've been programming for hours and hours and
hours and hours. I need some time to clear my head. Maybe this gives
you some ideas though.

-Mike

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>
>
> --- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@> wrote:
>
> >which means Benedetti+L2 complexity seems to work better than either one separately.
>
> It looks like perhaps L-inf complexity is what is really wanted.
>

Or who the hell knows. Check out the 176/175 rank two subgroups. From |||0 0 0 1 1 1 0 0 0 0>>> we get 2.35.11/5, and from |||0 0 0 1 -2 0 0 0 0 0>>> we get 2.5.11/7. Clearly 11/5 beats 11/7, but clearly also the 11/7 group works out better with L2 complexity and I think makes for the preferable subgroup.

On Wed, Jul 25, 2012 at 1:47 AM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> Or who the hell knows. Check out the 176/175 rank two subgroups. From |||0
> 0 0 1 1 1 0 0 0 0>>> we get 2.35.11/5, and from |||0 0 0 1 -2 0 0 0 0 0>>>
> we get 2.5.11/7. Clearly 11/5 beats 11/7, but clearly also the 11/7 group
> works out better with L2 complexity and I think makes for the preferable
> subgroup.

Are these minimal in weighted Lp complexity or unweighted?

-Mike

Gene wrote:
>Given a full p-limit temperament of rank r, we may define a subgroup
>temperament of rank s<r by finding a set of generators of minimal
>maximum Benedetti height. We may also do the same thing with the added
>proviso that 2 must be in the subgroup. Presumably most of the time at
>least these will be uniquely determined. It would be nice to have an
>algorithm for finding them.

What's the goal here?

From the wiki:
>By the Hahn–Banach theorem, can be extended to an element of the
>full p-limit tuning space

Doesn't have to be the full p-limit... couldn't it be the subgroup
generated by the primes needed to factor the kernel?

-Carl

On Wed, Jul 25, 2012 at 3:19 AM, Carl Lumma <carl@lumma.org> wrote:
>
> From the wiki:
> >By the Hahn–Banach theorem, can be extended to an element of the
> >full p-limit tuning space
>
> Doesn't have to be the full p-limit... couldn't it be the subgroup
> generated by the primes needed to factor the kernel?

If you mean that you want to extend the Lp-optimal tuning map for a
temperament to the subgroup defined by the kernel of the temperament -
the temperament is in a larger subgroup than its kernel by definition,
so to do this would actually be a reduction, and the resulting tuning
map would be the scalar "0".

This would correspond to taking the Lp-optimal tuning map for 7-limit
meantone, putting it on the 81/80.126/125 subgroup, and then tempering
out 81/80 and 126/125.

-Mike

Mike wrote:

>> From the wiki:
>> >By the Hahn­Banach theorem, can be extended to an element of the
>> >full p-limit tuning space
>>
>> Doesn't have to be the full p-limit... couldn't it be the subgroup
>> generated by the primes needed to factor the kernel?
>
>If you mean that you want to extend the Lp-optimal tuning map for a
>temperament to the subgroup defined by the kernel of the temperament -
>the temperament is in a larger subgroup than its kernel by definition,

To factor the kernel. E.g., 5-limit meantone has kernel 81/80,
factored by primes 2, 3, 5. -Carl

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

<snip>
> a monzo of rank p. Find a basis for this, and LLL reduce
> it. The LLL reduction uses unweighted complexity but is a
> great start for a search for the minimal weighted
> complexity, or minimal Benedetti height of the normal
> list, or what have you.

Why unweighted? Weighted LLL is easy enough to
implement. You can see it here in a general purpose
language:

https://bitbucket.org/x31eq/regular/src/fac18c45e130/regutils.py

and here in a computer algebra system:

http://x31eq.com/parametric.gp

RMS TE complexity of intervals should be enough to give you
a unique basis.

TLLL as I have it working gives [99/100, 224/225,
540/539] as the TLLL-reduced set of kernel ratios for
11-limit magic. There's nothing to say they should be the
simplest set, and these are close enough to find 245:243 and
385:384. But it isn't always trivial to find the simplest
unison vectors from the TLLL set. In higher limits you
might find cases where my website misses some. Remember,
lattice basis reduction is a hard problem, but a well
studied one.

Graham

It seems to me what we really want to know is how many members of a given consonance set, such as an appropriate q-limit diamond, belong to the temperament. Why not use that as the measure? I'll give it a try.

On Wed, Jul 25, 2012 at 5:02 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> It seems to me what we really want to know is how many members of a given
> consonance set, such as an appropriate q-limit diamond, belong to the
> temperament. Why not use that as the measure? I'll give it a try.

If we're trying to get back to theoretical foundations, then what we
actually care about are how many members of a given consonance set of
chords belong to the temperament, where "chords" ranges over sets of
all cardinalities and includes individual intervals.

-Mike

On Wed, Jul 25, 2012 at 2:24 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
> On Wed, Jul 25, 2012 at 1:47 AM, genewardsmith <genewardsmith@sbcglobal.net>
> wrote:
>>
>> Or who the hell knows. Check out the 176/175 rank two subgroups. From |||0
>> 0 0 1 1 1 0 0 0 0>>> we get 2.35.11/5, and from |||0 0 0 1 -2 0 0 0 0 0>>>
>> we get 2.5.11/7. Clearly 11/5 beats 11/7, but clearly also the 11/7 group
>> works out better with L2 complexity and I think makes for the preferable
>> subgroup.
>
> Are these minimal in weighted Lp complexity or unweighted?

Still waiting for an answer to this. Are you getting all the way to
the end of the process using unweighted Lp complexity, or do you end
up weighting it at any point?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> > Are these minimal in weighted Lp complexity or unweighted?
>
> Still waiting for an answer to this.

IIRC, both.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
> It seems to me what we really want to know is how many members of a given consonance set, such as an appropriate q-limit diamond, belong to the temperament. Why not use that as the measure? I'll give it a try.
>

This works. If I put in the commas for the various subgroup temperaments on the chromatic pairs page, I seem to usually get them back; so far I've not found an exception to that rule. On the other hand, while I once called marvel the king of planar temperaments, it does not lend itself to subgroups, and you get a bunch of them of the same lame-ass number of 11-limit diamond consonances (4, or 5 counting the octave.) These are:

2.15.7.11/9, with consonances {8/7, 11/9, 18/11, 7/4}
2.5.9/7.33, with consonances {5/4, 9/7, 14/9, 8/5}
2.3.25/7.1375, with consonances {9/8, 4/3, 3/2, 16/9}

A sorry collection, but note the three-way tie.

On Wed, Jul 25, 2012 at 7:18 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
>
> > > Are these minimal in weighted Lp complexity or unweighted?
> >
> > Still waiting for an answer to this.
>
> IIRC, both.

I don't see how this could possibly be the case. For instance, in the
11-limit, the trimonzo |||2 0 0 0 0 0 0 0 0 0>>> could refer to the
4.3.5, 2.9.5, 2.3.25, 4.6.5, etc, subgroups.

On the other hand, the trimonzo |||0 0 0 0 0 0 0 0 0 2>>> could refer
to the 25.7.11, 5.49.11, 5.7.121, 5.11/7.121 subgroups.

They have the same unweighted complexity, and yet couldn't be further
apart for weighted complexity.

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> I don't see how this could possibly be the case. For instance, in the
> 11-limit, the trimonzo |||2 0 0 0 0 0 0 0 0 0>>> could refer to the
> 4.3.5, 2.9.5, 2.3.25, 4.6.5, etc, subgroups.

I'm not using any of that.

On Wed, Jul 25, 2012 at 8:07 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
>
> > I don't see how this could possibly be the case. For instance, in the
> > 11-limit, the trimonzo |||2 0 0 0 0 0 0 0 0 0>>> could refer to the
> > 4.3.5, 2.9.5, 2.3.25, 4.6.5, etc, subgroups.
>
> I'm not using any of that.

Not using any of "what"?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> On Wed, Jul 25, 2012 at 8:07 PM, genewardsmith <genewardsmith@...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@>
> > wrote:
> >
> > > I don't see how this could possibly be the case. For instance, in the
> > > 11-limit, the trimonzo |||2 0 0 0 0 0 0 0 0 0>>> could refer to the
> > > 4.3.5, 2.9.5, 2.3.25, 4.6.5, etc, subgroups.
> >
> > I'm not using any of that.
>
> Not using any of "what"?

What you might call contorted subgroups. While there is much to be said for temperaments like baldy or stacks, the method using multimonzos won't find them.