Tuning-Math Digests messages 10575 - 10599

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Message: 10575

Date: Mon, 08 Mar 2004 22:49:31

Subject: Re: Dual L1 norm deep hole scales

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> 
> > P.S. Why aren't they simply the "L1 deep hole" results? Why dual? 
I 
> > tend to think of the monzo space or lattice of notes itself as 
> > the 'standard' space, while the space of linear functionals on it 
> > (breeds) as its dual. Or am I misunderstanding?
> 
> Presumably L1 would mean 
> 
> ||3^a 5^b 7^c|| = |a|+|b|+|c|
> 
> This has a different geometry from that, certainly.

If that's your concern, I would say "Triangular L1 norm" or something 
similar, for this.

> "Dual" comes in
> because if you use an L1 norm when measuring errors of tunings,

What do errors have to do with any of this? These are simply 
symmetrical JI scales . . . 

> dual
> to that is the "Dual L1 norm" on note-classes.

Since the dual of L1 is L-inf, is this actually a "Triangular L-inf 
norm" on note-classes?


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Message: 10578

Date: Mon, 08 Mar 2004 23:17:01

Subject: Re: Dual L1 norm deep hole scales

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" 
<gwsmith@s...> 
> > wrote:
> > > --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" 
<perlich@a...> 
> > wrote:
> > > > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" 
> > <gwsmith@s...> 
> > > > wrote:
> > > > 
> > > > > Ball 2 radius 6 14 notes
> > > > > [1, 21/20, 7/6, 6/5, 49/40, 5/4, 21/16, 7/5, 3/2, 8/5, 
42/25, 
> > > > > 12/7, 7/4,9/5]
> > > > 
> > > > Just as I suspected. We've finally constructed the Stellated 
> > Hexany, 
> > > > aka Mandala!
> > > 
> > > Eh, I think that's turned up already. :)
> > 
> > Did it?
> 
> Sure. It came up as the scale of the unit cube in connection with
> chord-lattice scales, and again as the union of the first two
> Euclidean shells around a deep hole.

Oops -- missed that.


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Message: 10579

Date: Mon, 08 Mar 2004 23:23:44

Subject: Re: Dual L1 norm deep hole scales

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> 
> > > "Dual" comes in
> > > because if you use an L1 norm when measuring errors of tunings,
> > 
> > What do errors have to do with any of this? These are simply 
> > symmetrical JI scales . . . 
> 
> Quite a lot, as these correspond to the ways we measure error.

The raison d'etre for these scales would seem to be if someone's 
looking for a good JI scale to use, without regard to its melodic 
structure. What does he or she care how we measure error?

> The
> Hahn norm corresponds to the minimax error,

I thought the "Hahn scales" we were coming up with were minimax in 
terms of note-classes. So wouldn't the dual of that be L1 in terms of 
error?

> the Euclidean norm to the
> rms error, and this "dual L1" gadget to the L1 error measurement 
which
> you were advocating.

Don't get it.

> > > dual
> > > to that is the "Dual L1 norm" on note-classes.
> > 
> > Since the dual of L1 is L-inf, is this actually a "Triangular L-
inf 
> > norm" on note-classes?
> 
> No, because "L-inf" is an L-inf on errors of consonances, and hence
> has the symmetry properties deriving from that.

Don't get it. But either way, dual or not, wouldn't "L1" indicate a 
rectangular, rather than triangular, lattice or metric? You were the 
one to bring this up but I don't see how duality ameliorates it.


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Message: 10580

Date: Tue, 09 Mar 2004 08:37:28

Subject: Re: Octave equivalent calculations (Was: Hanzos

From: Graham Breed

Gene Ward Smith wrote:

> If you wedge <21 33 49 59| with <41 65 95 115| you get
> <12 -14 -4 -50 -40 30|; dividing this through by two gives you the
> wedgie for 7-limit miracle, <6 -7 -2 -25 -20 15|. If you wedge the
> monzo for 16805/16087 with that for 1029/1024 and take the compliment,
> once again you get twice the wedgie for miracle, and hence miracle.
> These are the results you should be getting by any correct method, but
> don't seem to be.

Oh, right.  Yes, I get a different result in the first case if I don't 
use wedge products.  I think the non-wedge version is correct.  As 
there's no way of knowing where a wedge product comes from I have to 
treat contorsion the same way as torsion.  If I wanted to remove the 
contorsion, that'd be easy enough by looking at the mapping, and much 
easier than using wedge products.  So, despite your arrogance in telling 
me that I'm incorrect (still with no reason) this can't be a reason for 
using wedge products.

I can't factorize 16805/16087, and I still wouldn't be able to factorize 
it if I weren't using wedge products.  I get both 3361 and 16087 to be 
primes.


              Graham


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Message: 10581

Date: Tue, 09 Mar 2004 20:39:02

Subject: Re: Graham on contorsion

From: Paul Erlich

My rambling thoughts:

If the wedge product of the monzos (i.e., bimonzo) of two commas lead 
to torsion, they lead to torsion; one should make a big deal about 
this fact but not sweep it under the rug. Similarly, if the wedge 
product of the breeds (i.e., cross-breed) of two ETs lead to 
contorsion, they lead to contorsion; one should make a big deal about 
this fact but not sweep it under the rug. It's fine to then define 
the "wedgie" (why don't we call this the smith) as either of these 
wedge products with the (con)torsion removed by dividing through by 
gcd, and then insist that true temperaments correspond to a 
wedgie/smith.


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Message: 10588

Date: Tue, 09 Mar 2004 10:52:07

Subject: Re: Graham on contorsion

From: Graham Breed

Gene Ward Smith wrote:
> This is what Graham says on his web site:
> 
> "Whether temperaments with contorsion should even be thought of as
> temperaments is a matter of debate. They're really a way of
> constructing a scale with a simpler temperament. One problem is that
> there's always more than one qualitatively different generator for a
> given linear temperament with contorsion. So the temperament isn't
> uniquely determined by the mapping by period and generator."
> 
> Is there some reason why this is no longer applicable? 

It looks good to me, and shows why wedgies are inadequate for generating 
such scales.


                   Graham


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Message: 10592

Date: Tue, 09 Mar 2004 11:07:16

Subject: Re: Octave equivalent calculations (Was: Hanzos

From: Graham Breed

Gene Ward Smith wrote:

> Why is it correct? Is there a reason you want to give, or is this what
> we call in the States a faith-based initiative?

An equal scale with 21 notes should combine with an equal scale of 41 
notes to give an equal scale of 62 notes with maximally even subsets of 
21 and 41 notes.  Alternatively, an MOS of 62 notes with a maximally 
even sub-MOS of either 21 or 41 notes.  All the scales I give for 21&41 
comply with this, but miracle does not.

> That there is no way to tell where a wedge product comes from seems to
> be to be a major advantage.

Why on earth would that be?  It loses information.

> I don't know why it is arrogant to assume we remove contorsion; this
> is what we've been doing all along. If you are not doing so, I'd
> suggest people take both your results and your temperament finder with
> a grain of salt. 

It's not what I've been doing, and I've been not doing it for longer 
than you've been here.

> It should have been clear that this was a typo for 16875/16807, as
> that is a well-enough known seven-limit comma. In any case, that is
> the one you could try.

Then it's the one I tried on Sunday, and it gives the same result today 
-- miracle, but with a warning about the period not being what was 
expected (indicating some kind of torsion) for the matrix method.


                     Graham


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Message: 10593

Date: Tue, 09 Mar 2004 11:30:30

Subject: Re: Graham on contorsion

From: Graham Breed

Gene Ward Smith wrote:

> It doesn't mention wedgies, so it can hardly do that. A wedgie is a
> wedge product which has deliberately had the common factor taken out
> if the gcd is greater than one; if you wanted to leave it in and
> proceed on that basis you could, of course.

It doesn't mention wedgies, but does mention what wedgies are used to 
derive -- the mapping by (period and) generator.

But if you want wedge products, you can have wedge products.  the 5&19 
example comes from these two vals:

<5 8 12]
<19 30 44]

whose wedge product is

<<-2 -8 -8]]

The "scale with tempering" (if you don't want to call it a temperament) is:

5/24, 251.7 cent generator

basis:
(1.0, 0.20975898813907931)

mapping by period and generator:
[(1, 0), (2, -2), (4, -8)]

mapping by steps:
[(19, 5), (30, 8), (44, 12)]

highest interval width: 8
complexity measure: 8  (9 for smallest MOS)
highest error: 0.004480  (5.377 cents)
unique

You get a different MOS from the same family if you do 19&24 or 24&43.

A very similar scale can be found by combining 24 and 31:
16/55, 348.3 cent generator

basis:
(1.0, 0.29024101186092066)

mapping by period and generator:
[(1, 0), (1, 2), (0, 8)]

mapping by steps:
[(31, 24), (49, 38), (72, 56)]

highest interval width: 8
complexity measure: 8  (10 for smallest MOS)
highest error: 0.004480  (5.377 cents)
unique

Count back and this is the same as 17&7 with best 5-limit approximations:

7/24, 348.3 cent generator

basis:
(1.0, 0.29024101186092066)

mapping by period and generator:
[(1, 0), (1, 2), (0, 8)]

mapping by steps:
[(17, 7), (27, 11), (40, 16)]

highest interval width: 8
complexity measure: 8  (10 for smallest MOS)
highest error: 0.004480  (5.377 cents)
unique

That's generator of 7/24, which is different to the 5/24 of the previous 
scale.  The original vals are

<24 38 56]
<31 49 72]

and their wedge product is

<<-2 -8 -8]]

which is exactly the same as the 5/24 scale above.  So given a wedge 
product of <<-2 -8 -8]], how do you know if it should give the 5/24 
family or the 7/24 family?


                       Graham


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Message: 10596

Date: Tue, 09 Mar 2004 01:01:13

Subject: Re: Dual L1 norm deep hole scales

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> 
> > The raison d'etre for these scales would seem to be if someone's 
> > looking for a good JI scale to use, without regard to its melodic 
> > structure. What does he or she care how we measure error?
> 
> It seems to me the duality provides another and convincing raison 
d'etre.

Can you outline this raison d'etre, please? I have no idea what it 
could be.

> > > The
> > > Hahn norm corresponds to the minimax error,
> > 
> > I thought the "Hahn scales" we were coming up with were minimax 
in 
> > terms of note-classes. So wouldn't the dual of that be L1 in 
terms of 
> > error?
> 
> The situation is confusing, because we've got three different (at
> minimum) norms to contend with. We have a norm which is Linf 
(minimax)
> or Euclidean (rms.) This leads to *another* norm by applying it to 
the
> consonances.
> 
> Starting with the Euclidean norm (norm #1) (x^2+y^2+z^2)^(1/2),

How is this used below?

> we
> apply it to 3,5,7,5/3,7/3,7/5 and get the following:
> 
> x^2+y^2+z^2+(y-x)^2+(z-x)^2+(z-y)^2 = 3(x^2+y^2+z^2)-2(xy+xz+yz)
> 
> Taking the square root of this is norm #2.

This is the rms error criterion (or rms 'loss function'), right?

> Now form the symmetric
> matrix for the above, and take the inverse:
> 
> [[3,-1,-1],[-1,3,-1],[-1,-1,3]]^(-1) = 
> [[1/2,1/4,1/4],[1/4,1/2,1/4],[1/4,1/4,1/2]]
> 
> The quadratic form for this last is
> 
> (a^2+b^2+c^2+ab+ac+bc)/2, which gives us norm #2; of course we can
> rescale by multiplying by 2.

So you're saying the dual of rms error is euclidean norm in the 
symmetric oct-tet lattice, yes?

> The same situation, three *different* norms, we find if we start 
with
> norm #1 being the L_inf norm. Then norm #2 has a unit ball which is
> the convex hull of the twelve consonances--ie, a cuboctahedron. From
> the 14 faces of this we get norm #3.

I guess I must have been wrong above. What's the difference between 
this latter #2 and #3? And what about duality and the fact that on 
your Tenney page, you say that the dual of L1 is L_inf, but today you 
seem to be saying something different (is it the triangularity of the 
lattice that alters the situation)?

Gene, it takes a huge amount of guesswork to navigate your posts. 
Being concise is all well and good, but taking things step-by-step 
will reap many rewards in human understanding, I predict.



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Message: 10597

Date: Wed, 10 Mar 2004 15:31:03

Subject: Re: Octave equivalent calculations (Was: Hanzos

From: Graham Breed

Gene:
>>>Why is it correct? Is there a reason you want to give, or is this what
>>>we call in the States a faith-based initiative?

Me:
>>An equal scale with 21 notes should combine with an equal scale of 41 
>>notes to give an equal scale of 62 notes with maximally even subsets of 
>>21 and 41 notes.  Alternatively, an MOS of 62 notes with a maximally 
>>even sub-MOS of either 21 or 41 notes.  All the scales I give for 21&41 
>>comply with this, but miracle does not.

Gene:
> Since the question concerns temperaments, not scales, this is hardly
> relevant. I agree that programs which take two ets and produce scales
> out of them are a good thing, but they aren't temperament programs.

If the question concerns temperaments, why didn't the question contain 
come hint as to this fact?  Like maybe the word "temperament" along with 
a pointer to a definition so that we can all see what you're taking 
"temperament" to mean today.

I was certainly talking about the results of my program, which include a 
definition of a temperament (most of the time, depending on how you 
define "temperament") but also (at least implicitly) a generic MOS 
(excluding the precise tuning, although this was supplied for the 
results I showed here) with a particular number of notes to the octave. 
  And, definitely by implication this time, a family of MOS scales 
consistent with the mapping from JI.

Well, nice to see that you now think my program is "a good thing".  But 
how does that suddenly stop it from being a temperament program?  It 
produces, as far as I can tell, a homomorphic mapping from what may be 
rational numbers to a tone group with a smaller rank, in line with your 
definition here:

Regular Temperaments *

That's somewhat vague as to whether contorsion is allowed.  First it 
says that an "icon" has to be epimorphic.  But then "given an icon" we 
have to "find a homomorphic mapping".  Well, you can certainly find the 
mapping for Vicentino's 7&31 system using the icon for meantone, so I 
suppose it must be a temperament.

Still if you've now decided that temperaments don't have contorsion, 
that's fine, we'll find another name for the things like temperaments 
that may have contorsion.

Gene:
>>>That there is no way to tell where a wedge product comes from seems to
>>>be to be a major advantage.

Me:
>>Why on earth would that be?  It loses information.

Gene:
> Why should you care if the temperament comes from putting together
> vals or putting together monzos? As for the information lost in
> reduction of the wedge product to a wedgie, that is deliberately
> eliminated. The sign involves ordering of the product, and I would be
> interested to hear what use this precious information is. The rest
> could be used to define quasi-temperaments with contorsion if you
> wanted to, but this strikes me as a pointless proceedure. If you want
> that, just take a temperament and contort it in all the various ways
> that suit you.

If you don't care, then it can hardly be a "major advantage" not to be 
told.  If not being told is a "major advantage" then you must deeply 
care about not knowing, which is bizarre.  At the minimum, reasonable 
people wouldn't care (not about whether the result came from vals or 
monzos, which is a straw man, but about which vals it came from) in 
which case it doesn't matter how they do the calculation.  If people do 
care about getting the correct answer to the question they asked, then a 
method that throws away some of the information required for the answer 
and then tries to reconstruct it looks to be at a disadvantage compared 
to a method that goes straight to the correct answer.

The information may be deliberately lost, depending on who is doing the 
deliberating.  I don't think the sign carries any information.  Why did 
you mention it?

Yes, producing contorted quasi-temperaments from wedge products may be 
pointless.  That could be why nobody, except you, has ever advocated 
doing so.


Anyway, those who want contorsion-free temperaments will be pleased to 
learn that the module at

http://microtonal.co.uk/temper.py *

has been updated with a method to remove any contorsion from a 
"LinearTemperament" object.  So, to get a contorsion-free temperament 
consistent with the best 5-limit mappings of 19- and 5-equal, you can do:

 >>> temper.Temperament(19,5,temper.limit5).uncontort()

5/12, 503.4 cent generator

basis:
(1.0, 0.41951797627815951)

mapping by period and generator:
[(1, 0), (2, -1), (4, -4)]

mapping by steps:
[(7, 5), (11, 8), (16, 12)]

highest interval width: 4
complexity measure: 4  (5 for smallest MOS)
highest error: 0.004480  (5.377 cents)
unique


                Graham


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Message: 10598

Date: Wed, 10 Mar 2004 15:40:40

Subject: Re: Graham on contorsion

From: Graham Breed

Me:
>>which is exactly the same as the 5/24 scale above.  So given a wedge 
>>product of <<-2 -8 -8]], how do you know if it should give the 5/24 
>>family or the 7/24 family?

Gene:
> The question is meaningless. It should give both if you want
> contorsion, and neither if (like me) you don't.

The question is perfectly meaningful, and if you don't understand it you 
can't have been paying attention.  Go back and re-read the message in 
which I asked it.  I give a particular result for a pair of vals, and a 
different result for a different pair of vals.  It isn't a mattor of 
"want contorsion" or "don't want contorsion" but wanting a *particular 
scale* with tempering, which happens to contain contorsion.  Being given 
a list of scales that may be correct isn't good enough.

If you don't want contorsion, then you shouldn't be following this 
thread, which is about contorsion (and clearly says so in the subject line).


                  Graham


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Message: 10599

Date: Wed, 10 Mar 2004 16:32:48

Subject: Re: Graham on contorsion

From: Graham Breed

Paul Erlich wrote:
> My rambling thoughts:
> 
> If the wedge product of the monzos (i.e., bimonzo) of two commas lead 
> to torsion, they lead to torsion; one should make a big deal about 
> this fact but not sweep it under the rug. Similarly, if the wedge 
> product of the breeds (i.e., cross-breed) of two ETs lead to 
> contorsion, they lead to contorsion; one should make a big deal about 
> this fact but not sweep it under the rug. It's fine to then define 
> the "wedgie" (why don't we call this the smith) as either of these 
> wedge products with the (con)torsion removed by dividing through by 
> gcd, and then insist that true temperaments correspond to a 
> wedgie/smith.

But the two cases are different.  There is at least one historically 
important instance of contorsion: Vicentino's enharmonic of 1555. 
Although he gives higher limit ratios (sometimes incorrect) the harmony 
is all 5-limit, but with an equal division of the chromatic semitone.  I 
expect other "quartertone" composers work on similar principles.  My 
program will give this system of any two 7, 24 and 31-equal (7 and 31 
are 5-limit consistent, 24 is twice 12).  These numbers are important 
because:

- 7 is the number of notes to the octave in staff notation.  The 
alternative 5-limit contorted-meantone can't be written in staff 
notation with "quartertones".

- 24 is the number of notes to a quartertone scale.  If you think in 
terms of a 12 note chromatic, this is the simplest subdivision that 
gives you quartertones, or neutral thirds, or whatever.  Vicentino never 
invokes such a scale, but never gives examples that lie outside it. 
Also Example 48.3 of Book III (p.205 of the Maniates translation) 
includes a fourth descending through all quartertones and Example 46.1 
(p.199) includes a note that is incorrect given his definition of the 
enharmonic, but belongs to the correct "quartertone" scale.

- 31 is the number of notes in Vicentino's tuning.

If you asked for 7&24 or 24&31 than you should expect to get Vicentino's 
system.  If the program automatically removed contorsion, you would be 
rightly surprised to get a result that wasn't consistent with 24-equal.

If you asked for 7&31 in some other context, you may be surprised to get 
the quartertones coming back.  The program can't really be sure which 
question you meant to ask.  It could give you an uncontorted scale if 
neither of the inputs were contorted, but that means it won't give the 
right answer if you did want a microtonal system with 7 nominals 
consistent with 31-equal.  So it currently returns the contorted result 
and leaves you to explicitly remove the contorsion (explicit is better 
than implicit).


Torsion of unison vectors is a different matter.  I don't know of any 
cases, theoretical or otherwise, in which torsion is desired in a 
tempered MOS.  I'm not even sure what it would mean.  A periodicity 
block with torsion certainly doesn't correspond to an MOS with 
contorsion.  Where you don't supply a chromatic unison vector, it's even 
more likely that you simply wanted a torsion free scale that tempers out 
all the commatic unison vectors, and that's what you get.  Maybe if you 
supplied the chromatic unison vector, you would want to be warned of 
torsion.  But you aren't.  So there.


                     Graham


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