Yahoo Tuning Groups Ultimate Backup tuning-math Harmonized melody in the 7-limit

Two 7-limit notes less than 2 apart in the symmetric lattice can be
harmonized by two tetrads sharing at least one common note, and notes
2 or more apart cannot be so harmonized. Hence, a scale consisting
only of such intervals has the property that stepwise progressions can
be harmonized by tetrads with common notes--though not, of course,
necessarily tetrads all of whose notes belong to the scale.

If list all notes reduced to an octave which are at a distance of less
than three from the unison and smaller than 200 cents in size, we
obtain this:

{28/25, 10/9, 35/32, 15/14, 16/15, 21/20, 25/24, 36/35, 49/48}

The possible types of JI scale with the above property, in terms of
the intervals and their multiplicities, with the above restriction on
step size are given below. We can obtain tempered versions of these by
temperaments which equate steps; we have (28/25)/(10/9) = 126/125,
(10/9)/(35/32) = 64/63, (35/32)/(15/14) = 49/48, (15/14)/(16/15) =
225/224, (16/15)/(21/20) = 64/63, (21/20)/(25/24) = 126/125,
(25/24)/(36/35) = 875/864, (36/35)/(49/48) = 1728/1715. Meantone,
magic, orwell, pajara, porcupine, blackwood, superpythagorean,
tripletone, kleismic or nonkleismic would all be reasonable linear
temperaments to try--or beep if you think that is reasonable.

[28/25, 10/9, 35/32, 36/35] [1, 3, 2, 3]
[28/25, 10/9, 25/24, 36/35] [3, 1, 4, 3]
[28/25, 15/14, 16/15, 49/48] [1, 5, 3, 2]
[28/25, 16/15, 25/24, 36/35] [3, 1, 5, 3]

[10/9, 35/32, 16/15, 36/35] [2, 3, 2, 3]
[10/9, 35/32, 36/35, 49/48] [4, 1, 5, 2]
[10/9, 35/32, 21/20, 36/35] [4, 1, 2, 3]
[10/9, 15/14, 16/15, 21/20] [2, 3, 2, 3]
[10/9, 15/14, 21/20, 36/35] [4, 1, 3, 2]
[10/9, 15/14, 36/35, 49/48] [4, 1, 5, 3]
[10/9, 21/20, 25/24, 36/35] [4, 3, 1, 3]
[10/9, 25/24, 36/35, 49/48] [4, 1, 6, 3]

[35/32, 15/14, 16/15, 36/35] [3, 2, 4, 1]
[35/32, 16/15, 25/24, 36/35] [3, 4, 2, 3]

[15/14, 16/15, 21/20, 25/24] [3, 4, 3, 2]
[15/14, 16/15, 21/20, 49/48] [5, 4, 1, 2]
[15/14, 16/15, 36/35, 49/48] [5, 4, 1, 3]

[16/15, 21/20, 25/24, 36/35] [4, 3, 5, 3]
[16/15, 25/24, 36/35, 49/48] [4, 5, 6, 3]

[28/25, 10/9, 15/14, 25/24] [3, 1, 3, 1]
[28/25, 15/14, 16/15, 25/24] [3, 3, 1, 2]
[28/25, 35/32, 15/14, 16/15] [1, 2, 3, 3]
[28/25, 10/9, 15/14, 21/20] [2, 2, 3, 1]

[35/32, 15/14, 16/15, 21/20] [2, 3, 4, 1]

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

>Two 7-limit notes less than 2 apart in the symmetric lattice can be
>harmonized by two tetrads sharing at least one common note, and notes
>2 or more apart cannot be so harmonized. Hence, a scale consisting
>only of such intervals has the property that stepwise progressions can
>be harmonized by tetrads with common notes--though not, of course,
>necessarily tetrads all of whose notes belong to the scale.
>
>If list all notes reduced to an octave which are at a distance of less
>than <correction>two</correction> from the unison and smaller than 200
>cents in size, we obtain this:
>
>{28/25, 10/9, 35/32, 15/14, 16/15, 21/20, 25/24, 36/35, 49/48}
>
>The possible types of JI scale with the above property, in terms of
>the intervals and their multiplicities, with the above restriction on
>step size are given below. We can obtain tempered versions of these by
>temperaments which equate steps; we have (28/25)/(10/9) = 126/125,
>(10/9)/(35/32) = 64/63, (35/32)/(15/14) = 49/48, (15/14)/(16/15) =
>225/224, (16/15)/(21/20) = 64/63, (21/20)/(25/24) = 126/125,
>(25/24)/(36/35) = 875/864, (36/35)/(49/48) = 1728/1715. Meantone,
>magic, orwell, pajara, porcupine, blackwood, superpythagorean,
>tripletone, kleismic or nonkleismic would all be reasonable linear
>temperaments to try--or beep if you think that is reasonable.

Rock!

>[28/25, 10/9, 35/32, 36/35] [1, 3, 2, 3]
>[28/25, 10/9, 25/24, 36/35] [3, 1, 4, 3]
>[28/25, 15/14, 16/15, 49/48] [1, 5, 3, 2]
>[28/25, 16/15, 25/24, 36/35] [3, 1, 5, 3]
>
>[10/9, 35/32, 16/15, 36/35] [2, 3, 2, 3]
>[10/9, 35/32, 36/35, 49/48] [4, 1, 5, 2]
>[10/9, 35/32, 21/20, 36/35] [4, 1, 2, 3]
>[10/9, 15/14, 16/15, 21/20] [2, 3, 2, 3]
>[10/9, 15/14, 21/20, 36/35] [4, 1, 3, 2]
>[10/9, 15/14, 36/35, 49/48] [4, 1, 5, 3]
>[10/9, 21/20, 25/24, 36/35] [4, 3, 1, 3]
>[10/9, 25/24, 36/35, 49/48] [4, 1, 6, 3]
>
>[35/32, 15/14, 16/15, 36/35] [3, 2, 4, 1]
>[35/32, 16/15, 25/24, 36/35] [3, 4, 2, 3]
>
>[15/14, 16/15, 21/20, 25/24] [3, 4, 3, 2]
>[15/14, 16/15, 21/20, 49/48] [5, 4, 1, 2]
>[15/14, 16/15, 36/35, 49/48] [5, 4, 1, 3]
>
>[16/15, 21/20, 25/24, 36/35] [4, 3, 5, 3]
>[16/15, 25/24, 36/35, 49/48] [4, 5, 6, 3]
>
>[28/25, 10/9, 15/14, 25/24] [3, 1, 3, 1]
>[28/25, 15/14, 16/15, 25/24] [3, 3, 1, 2]
>[28/25, 35/32, 15/14, 16/15] [1, 2, 3, 3]
>[28/25, 10/9, 15/14, 21/20] [2, 2, 3, 1]
>
>[35/32, 15/14, 16/15, 21/20] [2, 3, 4, 1]

Rad. No 6- or 7-toners, eh? I wonder about the "9-limit".
The 9-limit has the further advantage that you can hit more
fifths, and thus improve omnitetrachordality.

Also, could we screen based on which of the above
combinations, and which orderings of those, produce the
most low-numbered ratios in the scale? Or does such
an approach fail on the grounds that it ignores temperament
(aka TOLERANCE)?

-Carl

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

>> Rad. No 6- or 7-toners, eh? I wonder about the "9-limit".
>> The 9-limit has the further advantage that you can hit more
>> fifths, and thus improve omnitetrachordality.
>
>The 9-limit would be different, for sure. The simple symmetrical
>lattice criterion wouldn't work, but it would be easy enough to
>find what does.

Nobody ever answered me if symmetrical is synonymous with unweighted.

>> Also, could we screen based on which of the above
>> combinations, and which orderings of those, produce the
>> most low-numbered ratios in the scale? Or does such
>> an approach fail on the grounds that it ignores temperament
>> (aka TOLERANCE)?
>
>The main problem I see with that is that it is a huge computational
>job, since in each case you need to find the optimal version of the
>scale in question. I wouldn't want to try it using Maple.

The thing is to only store one permutation in memory at a time.
Alas, I haven't come up with an easy way to code these kinds of
evaluations in scheme. They're very natural in C, I think. The
present problem may still be hard on account of CPU cycles, tho.

-Carl

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

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

> One of us is still misunderstanding Paul Hahn's 9-limit approach.
> In the unweighted version 3, 5, 7 and 9 are all the same length.

In this system you don't exactly, have 7-limit notes and intervals.
You do have "hanzos", with basis 2,3,5,7,9. The hahnzo |0 -2 0 0 1> is
a comma, 9/3^2, which obviously would play a special role. Hahnzos map
onto 7-limit intervals, but not 1-1. Are you happy with the idea that
two scales could be different, since they have steps and notes which
are distinct as hahnzos, even though they have exactly the same steps
and notes in the 7-limit?

We've got three hahnzos corresponding to 81/80; if we take any two of
them and wedge, we get the planar wedgie <<<0 1 0 4 0 -2 4 0 0 8|||.
For 126/125 we get both |1 2 -3 1 0> and |1 0 -3 1 1> as a hahnzo.
Going through all six combinations, I get <<1 4 10 2 4 13 0 12 -8
-26|| as the wedgie, leading to [<1 2 4 7 4|, <0 -1 -4 -10 -2|] as the
mapping. The mechanism seems to work for hahnzos; here it is telling
us the generator is a fourth, and 16*(4/3)^(-2) = 9. I can also get
this by sticking in the dummy hahnzo <0 -2 0 0 1| as a comma. If I try
three hahnzo commas which don't have a dummy relationship, I get a
disguised et as a "linear" temperament.

🔗Carl Lumma <ekin@lumma.org>

>> One of us is still misunderstanding Paul Hahn's 9-limit approach.
>> In the unweighted version 3, 5, 7 and 9 are all the same length.
>
>In this system you don't exactly, have 7-limit notes and intervals.
>You do have "hanzos", with basis 2,3,5,7,9.

The usual point of odd-limit is to get octave equivalence, and
therefore I'd say the 2s should be dropped from the basis.

>The hahnzo |0 -2 0 0 1> is a comma, 9/3^2, which obviously would
>play a special role. Hahnzos map onto 7-limit intervals, but not
>1-1. Are you happy with the idea that two scales could be
>different, since they have steps and notes which are distinct as
>hahnzos, even though they have exactly the same steps and notes
>in the 7-limit?

I think the answer here is yes, though I'm at a loss for why
you're mapping hanzos to the 7-limit.

>We've got three hahnzos corresponding to 81/80;

My recollection is that Paul H.'s algorithm assigns a unique
lattice route (and therefore hanzo) to each 9-limit interval.
Certainly it can be used to find the set of lattice points
within distance <= 2 of a given point.

-Carl

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

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

> >> My recollection is that Paul H.'s algorithm assigns a unique
> >> lattice route (and therefore hanzo) to each 9-limit interval.
> >
> >So what? You still get an infinite number representing each
interval,
> >since you can multiply by arbitary powers of the dummy comma 9/3^2.
>
> An infinite number from where? If you look at the algorithm, that
> dummy comma has zero length.

If it has length zero then we are not talking about a lattice at all,
though a quotient of it (modding out the dummy comma) might be. In a
symmetrical lattice it necessarily has the same length as, for
example, 11/3^2, which is of length sqrt(1^2+2^2-1*2)=sqrt(3). It
does *not* have the same length as 11/9, which is of length one, of
course.

> >> Certainly it can be used to find the set of lattice points
> >> within distance <= 2 of a given point.
> >
> >Hahn's alogorithm can, or hahnzos can, or what?
>
> The algorithm can.

Why do you think Hahn's definition of "distance" would work for this
problem? If you tell me what it is, we could check and see. However,
you've just told me it does not have the basic properties of a
metric, so I'm inclined to object to calling it a "distance".

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> One of us is still misunderstanding Paul Hahn's 9-limit approach.
> >> In the unweighted version 3, 5, 7 and 9 are all the same length.
> >
> >In this system you don't exactly, have 7-limit notes and intervals.
> >You do have "hanzos", with basis 2,3,5,7,9.
>
> The usual point of odd-limit is to get octave equivalence, and
> therefore I'd say the 2s should be dropped from the basis.

Bad move -- for one thing, you can't detect torsion.

> >The hahnzo |0 -2 0 0 1> is a comma, 9/3^2, which obviously would
> >play a special role. Hahnzos map onto 7-limit intervals, but not
> >1-1. Are you happy with the idea that two scales could be
> >different, since they have steps and notes which are distinct as
> >hahnzos, even though they have exactly the same steps and notes
> >in the 7-limit?
>
> I think the answer here is yes, though I'm at a loss for why
> you're mapping hanzos to the 7-limit.

7-prime-limit.

> >We've got three hahnzos corresponding to 81/80;
>
> My recollection is that Paul H.'s algorithm assigns a unique
> lattice route (and therefore hanzo) to each 9-limit interval.

Unique "hanzo" but not unique lattice route -- even monzos don't
assign a unique lattice route.

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Carl Lumma <ekin@lumma.org>

>>>>My recollection is that Paul H.'s algorithm assigns a unique
>>>>lattice route (and therefore hanzo) to each 9-limit interval.
>>>
>>>So what? You still get an infinite number representing each
>>>interval, since you can multiply by arbitary powers of the dummy
>>>comma 9/3^2.
>>
>>An infinite number from where? If you look at the algorithm, that
>>dummy comma has zero length.
>
>If it has length zero then we are not talking about a lattice at all,
>though a quotient of it (modding out the dummy comma) might be. In a
>symmetrical lattice it necessarily has the same length as, for
>example, 11/3^2, which is of length sqrt(1^2+2^2-1*2)=sqrt(3). It
>does *not* have the same length as 11/9, which is of length one, of
>course.

Yes, you're onto something here. In the unweighted lattice there is
a point for 9/3^2, which lies on the diameter-1 hull.

>> >> Certainly it can be used to find the set of lattice points
>> >> within distance <= 2 of a given point.
>> >
>> >Hahn's alogorithm can, or hahnzos can, or what?
>>
>> The algorithm can.
>
>Why do you think Hahn's definition of "distance" would work for this
>problem? If you tell me what it is, we could check and see. However,
>you've just told me it does not have the basic properties of a
>metric, so I'm inclined to object to calling it a "distance".

I thought you acknowledged the receipt of the algorithm...

>Given a Fokker-style interval vector (I1, I2, . . . In):
>
>1. Go to the rightmost nonzero exponent; add its absolute value
>to the total.
>
>2. Use that exponent to cancel out as many exponents of the opposite
>sign as possible, starting to its immediate left and working right;
>discard anything remaining of that exponent.
>
>3. If any nonzero exponents remain, go back to step one, otherwise
>stop.

As for the problem, let's start over. Call the position occupied by
1/1 in 1/1,8/7,4/3,8/5 the root of utonal tetrads. Now 7-limit tetrads
sharing a common dyad (pair of pitches) with an otonal tetrad rooted
on 1/1 will have roots...

5/4, 3/2, 15/8, 7/4, 35/32, 21/16

...and those sharing a common tone (single pitch) will have roots...
ack, this is visually exhausting... am I correct that they are the
1- and 2-combinations of:

1, 3, 5, 7, 1/3, 1/5, 1/7

If so, can you say why adding 9 and 1/9 to this list will not produce
an equivalent 9-limit result?

-Carl

🔗Carl Lumma <ekin@lumma.org>

>Given a Fokker-style interval vector (I1, I2, . . . In):

[-2 0 0 1]

>1. Go to the rightmost nonzero exponent; add its absolute value
>to the total.

T=1

>2. Use that exponent to cancel out as many exponents of the opposite
>sign as possible, starting to its immediate left and working right;
>discard anything remaining of that exponent.

[-1 0 0 0]
T=1

>3. If any nonzero exponents remain, go back to step one, otherwise
>stop.

T=... whoops, I forgot an absolute value here. The correct
value is 2.

-Carl

🔗Carl Lumma <ekin@lumma.org>

>>>>So what? You still get an infinite number representing each
>>>>interval, since you can multiply by arbitary powers of the dummy
>>>>comma 9/3^2.
>>>
>>>An infinite number from where? If you look at the algorithm, that
>>>dummy comma has zero length.
>>
>>If it has length zero then we are not talking about a lattice at all,
>>though a quotient of it (modding out the dummy comma) might be. In a
>>symmetrical lattice it necessarily has the same length as, for
>>example, 11/3^2, which is of length sqrt(1^2+2^2-1*2)=sqrt(3).

I don't know where sqrt would be coming from. I thought everything
would have to have whole number lengths.

>>It
>>does *not* have the same length as 11/9, which is of length one, of
>>course.
>
>Yes, you're onto something here. In the unweighted lattice there is
>a point for 9/3^2, which lies on the diameter-1 hull.

That was based on the rather simplistic idea that one steps out and
two back leaves you one away from where you started.

-Carl

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Graham Breed <graham@microtonal.co.uk>

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> >> My recollection is that Paul H.'s algorithm assigns a unique
> >> >> lattice route (and therefore hanzo) to each 9-limit interval.
> >> >
> >> >So what? You still get an infinite number representing each
> >> >interval, since you can multiply by arbitary powers of the dummy
> >> >comma 9/3^2.
> >>
> >> An infinite number from where? If you look at the algorithm,
that
> >> dummy comma has zero length.
> >
> >How do you get that????
>
> >Given a Fokker-style interval vector (I1, I2, . . . In):
>
> [-2 0 0 1]
>
> >1. Go to the rightmost nonzero exponent; add its absolute value
> >to the total.
>
> T=1
>
> >2. Use that exponent to cancel out as many exponents of the
opposite
> >sign as possible, starting to its immediate left and working right;
> >discard anything remaining of that exponent.
>
> [-1 0 0 0]
> T=1
>
> >3. If any nonzero exponents remain, go back to step one, otherwise
> >stop.
>
> T=... whoops, I forgot an absolute value here. The correct
> value is 2.
>
> -Carl

You mean 1, right? It's certainly not zero, though. So you have an
infinite chain of duplicates for each pitch, spaced at increments of
1 unit . . .

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:
> Carl:
> >>The usual point of odd-limit is to get octave equivalence, and
> >>therefore I'd say the 2s should be dropped from the basis.
>
> Paul E:
> > Bad move -- for one thing, you can't detect torsion.
>
> Can't you? I'm not sure it's even relevant here, but I thought I
went
> through this in excruciating detail some time ago and showed that
you
> can detect torsion with octave-equivalent vectors. In fact, I seem
to
> remember running the calculation in parallel using both methods,
and
> showing that I always got the same results.

Right, but that was a specific set of calculations, not a general
proof. You were just looking at 'linear' temperaments, if I recall
correctly, but torsion can afflict all types of temperaments. Plus it
seemed your method was far less elegant.

> No, the outstanding problem is that we can't go from the
> octave-equivalent mapping to an optimized generator size. But I'm
sure
> it can be done if anybody cared. The situation as I left it was
that
> nobody did.
>
>
> Graham

I care, and I hope Gene does too. I'd like to see this revisited.

🔗Carl Lumma <ekin@lumma.org>

>> >> >So what? You still get an infinite number representing each
>> >> >interval, since you can multiply by arbitary powers of the dummy
>> >> >comma 9/3^2.
>> >>
>> >> An infinite number from where? If you look at the algorithm,
>> >> that dummy comma has zero length.
>> >
>> >How do you get that????
>>
>> >Given a Fokker-style interval vector (I1, I2, . . . In):
>>
>> [-2 0 0 1]
>>
>> >1. Go to the rightmost nonzero exponent; add its absolute value
>> >to the total.
>>
>> T=1
>>
>> >2. Use that exponent to cancel out as many exponents of the
>> >opposite sign as possible, starting to its immediate left and
>> >working right; discard anything remaining of that exponent.
>>
>> [-1 0 0 0]
>> T=1
>>
>> >3. If any nonzero exponents remain, go back to step one, otherwise
>> >stop.
>>
>> T=... whoops, I forgot an absolute value here. The correct
>> value is 2.
>
>You mean 1, right?

No, T=1 and you've got to add what's left of the 3 exponent, abs(-1),
to it.

>It's certainly not zero, though. So you have an
>infinite chain of duplicates for each pitch, spaced at increments of
>1 unit . . .

...

-C.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> >> >So what? You still get an infinite number representing each
> >> >> >interval, since you can multiply by arbitary powers of the
dummy
> >> >> >comma 9/3^2.
> >> >>
> >> >> An infinite number from where? If you look at the algorithm,
> >> >> that dummy comma has zero length.
> >> >
> >> >How do you get that????
> >>
> >> >Given a Fokker-style interval vector (I1, I2, . . . In):
> >>
> >> [-2 0 0 1]
> >>
> >> >1. Go to the rightmost nonzero exponent; add its absolute value
> >> >to the total.
> >>
> >> T=1
> >>
> >> >2. Use that exponent to cancel out as many exponents of the
> >> >opposite sign as possible, starting to its immediate left and
> >> >working right; discard anything remaining of that exponent.
> >>
> >> [-1 0 0 0]
> >> T=1
> >>
> >> >3. If any nonzero exponents remain, go back to step one,
otherwise
> >> >stop.
> >>
> >> T=... whoops, I forgot an absolute value here. The correct
> >> value is 2.
> >
> >You mean 1, right?
>
> No, T=1 and you've got to add what's left of the 3 exponent, abs(-
1),
> to it.

Oh yeah, you're right.

> >It's certainly not zero, though. So you have an
> >infinite chain of duplicates for each pitch, spaced at increments
of
> >1 unit . . .
>
> ...

2 units.

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> > >> I don't know where sqrt would be coming from. I thought
everything
> > >> would have to have whole number lengths.
> > >
> > >I'm using Euclidean distances.
> >
> > I wish you had said that.
>
> If I'm talking about symmetric lattices, that would be assumed.

You didn't assume it the other week, when it seemed (even upon
clarification) that you referred to *two* possible metrics in the
symmetric lattice -- the euclidean one, and the 'taxicab' one. The
latter is what Paul Hahn's algorithm gives you.

But now you say this. So now I wonder what you were *really* talking
about.

> All
> these diagrams of hexagons and tetrahedrons and octahedrons that
> people draw are Euclidean;

That's a bogus assumption. We live in a nearly euclidean universe so
people have no choice in the matter when it comes to diagrams. If I
draw a Tenney lattice, which assumes a taxicab metric, how am I
supposed to avoid drawing it in a way that you'd interpret as
euclidean??

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > I care, and I hope Gene does too. I'd like to see this revisited.
>
> You can do anything with octave-equivalent vectors that you can do
> with monzos if you can recover the monzos. This means you can't do
> some things, but since commas are small, you can recover the commas,
> and hence can eg take two octave-equivalent 7-limit vectors and get
> the corresponding temperament.

Sure. That's how I generated most of those graphs of commas -- a
large search in n-1 dimensional space, filling in the 2 component
later by assuming each interval was between -600 and 600 cents. But I
don't think this is what Graham was doing.

> What I don't understand is why anyone would want to. You can also
> throw away the 7, and keep the 2,3 and 5. Would anyone propose doing
> that? It strikes me as an absurd proceedure. Where octave-equivalent
> vectors are useful is in octave-equivalent contexts.

Which would seem to be most musical contexts of interest on these
lists.

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> > > >> I don't know where sqrt would be coming from. I thought
> everything
> > > >> would have to have whole number lengths.
> > > >
> > > >I'm using Euclidean distances.
> > >
> > > I wish you had said that.
> >
> > If I'm talking about symmetric lattices, that would be assumed.
>
> You didn't assume it the other week, when it seemed (even upon
> clarification) that you referred to *two* possible metrics in the
> symmetric lattice -- the euclidean one, and the 'taxicab' one.

Those are two different lattices, and my point is that if someone says
"symmetric lattice" without qualification they presumably mean the
stuff you guys already knew about when I got here.

> That's a bogus assumption. We live in a nearly euclidean universe so
> people have no choice in the matter when it comes to diagrams. If I
> draw a Tenney lattice, which assumes a taxicab metric, how am I
> supposed to avoid drawing it in a way that you'd interpret as
> euclidean??

You tell us it's a Tenney lattice. This is really the only possible
method.

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul Erlich <perlich@aya.yale.edu>

But they both can only be drawn with the usual tetrahedra and
octahedra, right? So they *look* like the same lattice, and the
choice of metric has to be specified *verbally*.

> and my point is that if someone says
> "symmetric lattice" without qualification they presumably mean the
> stuff you guys already knew about when I got here.

Which pretty much used a 'taxicab' metric (as per Paul Hahn's
algorithm) when a metric was used at all.

> > That's a bogus assumption. We live in a nearly euclidean universe
so
> > people have no choice in the matter when it comes to diagrams. If
I
> > draw a Tenney lattice, which assumes a taxicab metric, how am I
> > supposed to avoid drawing it in a way that you'd interpret as
> > euclidean??
>
> You tell us it's a Tenney lattice. This is really the only possible
> method.

So you see why your assumption was bogus, yes?

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> > > You didn't assume it the other week, when it seemed (even upon
> > > clarification) that you referred to *two* possible metrics in the
> > > symmetric lattice -- the euclidean one, and the 'taxicab' one.
> >
> > Those are two different lattices,
>
> But they both can only be drawn with the usual tetrahedra and
> octahedra, right? So they *look* like the same lattice, and the
> choice of metric has to be specified *verbally*.

Strictly speaking, only one of them can be drawn at all.

> > and my point is that if someone says
> > "symmetric lattice" without qualification they presumably mean the
> > stuff you guys already knew about when I got here.
>
> Which pretty much used a 'taxicab' metric (as per Paul Hahn's
> algorithm) when a metric was used at all.

Are you saying you actually were clueless about this until I arrived?
Did you not know a hexany was a literal octahedron, and tetrads
literal tetrahedra?

> So you see why your assumption was bogus, yes?

Most of the world means "Euclidean" when they say "lattice" at all,
and don't bother to say so; and assumes it when they speak of
octahedra or other geometrical figures. If you want to do things
differently, it is up to you to specify but please don't claim I am
being ambiguous. I'm not; I'm doing things how pretty well everyone
does them.

I'd be interested to know what people meant by "lattice" when I
arrived, though, and whether we have been miscommunicating from the start.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > > > You didn't assume it the other week, when it seemed (even
upon
> > > > clarification) that you referred to *two* possible metrics in
the
> > > > symmetric lattice -- the euclidean one, and the 'taxicab'
one.
> > >
> > > Those are two different lattices,
> >
> > But they both can only be drawn with the usual tetrahedra and
> > octahedra, right? So they *look* like the same lattice, and the
> > choice of metric has to be specified *verbally*.
>
>
> Strictly speaking, only one of them can be drawn at all.

So you can't draw a graph?

> > > and my point is that if someone says
> > > "symmetric lattice" without qualification they presumably mean
the
> > > stuff you guys already knew about when I got here.
> >
> > Which pretty much used a 'taxicab' metric (as per Paul Hahn's
> > algorithm) when a metric was used at all.
>
> Are you saying you actually were clueless about this until I
arrived?
> Did you not know a hexany was a literal octahedron, and tetrads
> literal tetrahedra?

That's the way we were drawing them, so I don't know what you could
mean by 'clueless'.

> I'd be interested to know what people meant by "lattice" when I
> arrived, though, and whether we have been miscommunicating from the
>start.

In any case, hopefully it's clear to you that Paul Hahn's algorithm
(at least if we stick to 7-limit and don't mess with 9-limit) gives
the taxicab distance in the symmetric oct-tet lattice. Since it
measures the number of 'consonances' needed to get from one note to
another, it's certainly a meaningful measure, while the euclidean
distance doesn't seem to signify anything musically meaningful.

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> Most of the world means "Euclidean" when they say "lattice" at all,
> and don't bother to say so...

Here's the Wikipedia article:

http://en.wikipedia.org/wiki/Lattice_(group)

The definition is assuming a Euclidean metric, since R^n without
qualification means R^n under a Euclidean norm. It then goes on to
reference usage in materials science (assumed Euclidean) and
computational physics, where you are generally thinking Euclidean, but
where the lattice is really a kind of graph, and so may be close to
what people are thinking here. Then it finishes talking about Lie
groups and Haar measure, pretty much off on a tangent.

Here is World of Mathematics:

http://mathworld.wolfram.com/PointLattice.html

"Formally, a lattice is a discrete subgroup of Euclidean space"

This entry is explicitly ruling out other possibilities! I prefer to
say something like it is a discrete subgroup of a finite dimensional
normed real vector space with compact quotient, but norms other than
L2 norms come up seldom; you can read long and sophisticated books
stuffed full of lattices which never bother to mention them.

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> > Strictly speaking, only one of them can be drawn at all.
>
> So you can't draw a graph?

A figure in Tenney space is not a mere graph; it has far more structure.

> > Are you saying you actually were clueless about this until I
> arrived?
> > Did you not know a hexany was a literal octahedron, and tetrads
> > literal tetrahedra?
>
> That's the way we were drawing them, so I don't know what you could
> mean by 'clueless'.

Did you or did you not know you could put 7-limit note classes into a
symmetric lattice pattern in three dimensional Euclidean space?

> In any case, hopefully it's clear to you that Paul Hahn's algorithm
> (at least if we stick to 7-limit and don't mess with 9-limit) gives
> the taxicab distance in the symmetric oct-tet lattice. Since it
> measures the number of 'consonances' needed to get from one note to
> another, it's certainly a meaningful measure, while the euclidean
> distance doesn't seem to signify anything musically meaningful.

I agree it is meaningful and quite interesting; I don't agree
Euclidean is meaningless. You've been sleeping through my postings if
you think that; Euclidean draws finer distinctions than Hahn's measure
does, and those distinctions do relate to things musical.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > > Strictly speaking, only one of them can be drawn at all.
> >
> > So you can't draw a graph?
>
> A figure in Tenney space is not a mere graph; it has far more
structure.

OK . . . but can't you think of it as existing in euclidean space and
define a new kind of distance in which one is restricted to traveling
along the edges?

> > > Are you saying you actually were clueless about this until I
> > arrived?
> > > Did you not know a hexany was a literal octahedron, and tetrads
> > > literal tetrahedra?
> >
> > That's the way we were drawing them, so I don't know what you
could
> > mean by 'clueless'.
>
> Did you or did you not know you could put 7-limit note classes into
a
> symmetric lattice pattern in three dimensional Euclidean space?

Since we were actually doing so, it seems absurd to question whether
we knew that we could do so. We simply rarely or never measured
euclidean distances in this structure, since they don't carry a
straightforward musical meaning.

> > In any case, hopefully it's clear to you that Paul Hahn's
algorithm
> > (at least if we stick to 7-limit and don't mess with 9-limit)
gives
> > the taxicab distance in the symmetric oct-tet lattice. Since it
> > measures the number of 'consonances' needed to get from one note
to
> > another, it's certainly a meaningful measure, while the euclidean
> > distance doesn't seem to signify anything musically meaningful.
>
> I agree it is meaningful and quite interesting; I don't agree
> Euclidean is meaningless. You've been sleeping through my postings
if
> you think that;

I don't appreciate that comment.

> Euclidean draws finer distinctions than Hahn's measure
> does,

Clearly -- but are they meaningful distinctions?

> and those distinctions do relate to things musical.

Namely . . . ?

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
> >
> > > > Strictly speaking, only one of them can be drawn at all.
> > >
> > > So you can't draw a graph?
> >
> > A figure in Tenney space is not a mere graph; it has far more
> structure.
>
> OK . . . but can't you think of it as existing in euclidean space and
> define a new kind of distance in which one is restricted to traveling
> along the edges?

Sure, and that may be a good approach. You are now using a Euclidean
lattice, and defining a corridor-travel distance function on top of
it. It is certainly familiar to us that as-the-crow-flies and as the
taxicab drives are both valid and can exist simultaneously.

> > Did you or did you not know you could put 7-limit note classes into
> a
> > symmetric lattice pattern in three dimensional Euclidean space?
>
> Since we were actually doing so, it seems absurd to question whether
> we knew that we could do so. We simply rarely or never measured
> euclidean distances in this structure, since they don't carry a
> straightforward musical meaning.

It never occurred to you to do orthogonal transformations? In any
case, you should be aware that there *are* differences between
intervals like 15/14 and 21/20 as compared to 25/24, 36/35 or 49/48,
and that these again are different from 50/49. They all take two
7-limit consonaces to get to, which is why they all turn up in the
stepwise problem I've recently been posting on, and Hahn's measure
captures that. They behave differently in terms of how the chord
relationships work, and the finer distinctions of the Euclidean metric
capture that.

> I don't appreciate that comment.

Nevertheless, it seems you've been missing a lot.

> > Euclidean draws finer distinctions than Hahn's measure
> > does,
>
> Clearly -- but are they meaningful distinctions?

See above. Yes, of course they are meaningful. If you want to try to
linearly transform 50/49 to 49/48, be my guest, but don't be surprised
when it works very differently than a linear transformation sending
15/14 to 21/20.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> > wrote:
> > > --- In tuning-math@yahoogroups.com, "Paul Erlich"
<perlich@a...>
> > wrote:
> > >
> > > > > Strictly speaking, only one of them can be drawn at all.
> > > >
> > > > So you can't draw a graph?
> > >
> > > A figure in Tenney space is not a mere graph; it has far more
> > structure.
> >
> > OK . . . but can't you think of it as existing in euclidean space
and
> > define a new kind of distance in which one is restricted to
traveling
> > along the edges?
>
> Sure, and that may be a good approach. You are now using a Euclidean
> lattice, and defining a corridor-travel distance function on top of
> it. It is certainly familiar to us that as-the-crow-flies and as the
> taxicab drives are both valid and can exist simultaneously.
>
> > > Did you or did you not know you could put 7-limit note classes
into
> > a
> > > symmetric lattice pattern in three dimensional Euclidean space?
> >
> > Since we were actually doing so, it seems absurd to question
whether
> > we knew that we could do so. We simply rarely or never measured
> > euclidean distances in this structure, since they don't carry a
> > straightforward musical meaning.
>
> It never occurred to you to do orthogonal transformations?

We've transformed between the 48 elements of the symmetry group of
the 7-limit lattice. Are "orthogonal transformations" the ones that
involve a 90-degree rotation, or something else?

> In any
> case, you should be aware that there *are* differences between
> intervals like 15/14 and 21/20 as compared to 25/24, 36/35 or 49/48,
> and that these again are different from 50/49.

We could refer to these as "meta", "ortho", and "para", respectively.

> They all take two
> 7-limit consonaces to get to, which is why they all turn up in the
> stepwise problem I've recently been posting on, and Hahn's measure
> captures that. They behave differently in terms of how the chord
> relationships work, and the finer distinctions of the Euclidean
metric
> capture that.

But there's nothing *special* about the euclidean metric here. The
chord relationships work differently, but what is it about them that
implies the specific values that the euclidean metric gives?

> > I don't appreciate that comment.
>
> Nevertheless, it seems you've been missing a lot.

I don't think so.

> > > Euclidean draws finer distinctions than Hahn's measure
> > > does,
> >
> > Clearly -- but are they meaningful distinctions?
>
> See above. Yes, of course they are meaningful. If you want to try to
> linearly transform 50/49 to 49/48, be my guest, but don't be
surprised
> when it works very differently than a linear transformation sending
> 15/14 to 21/20.

You misunderstood my question, and in fact I find this response
highly condescending.

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> > It never occurred to you to do orthogonal transformations?
>
> We've transformed between the 48 elements of the symmetry group of
> the 7-limit lattice. Are "orthogonal transformations" the ones that
> involve a 90-degree rotation, or something else?

I meant isometries--Euclidean distance preserving mappings--fixing the
origin, and sending the lattice to itself. This defines the
automorphism group aut(L) of the lattice L. Below is a definition of
othogonal transformation.

http://mathworld.wolfram.com/OrthogonalTransformation.html

> > In any
> > case, you should be aware that there *are* differences between
> > intervals like 15/14 and 21/20 as compared to 25/24, 36/35 or 49/48,
> > and that these again are different from 50/49.
>
> We could refer to these as "meta", "ortho", and "para", respectively.

I called them shell 2, shell 3 and shell 4 before, which is lattice
terminology.

> > They all take two
> > 7-limit consonaces to get to, which is why they all turn up in the
> > stepwise problem I've recently been posting on, and Hahn's measure
> > captures that. They behave differently in terms of how the chord
> > relationships work, and the finer distinctions of the Euclidean
> metric
> > capture that.
>
> But there's nothing *special* about the euclidean metric here. The
> chord relationships work differently, but what is it about them that
> implies the specific values that the euclidean metric gives?

The Euclidean metric gives the most refined distinctions. The Hahn
metric replaces spheres with rhombic dodecahedra. This means to figure
out what it means in general, and not just between lattice points,
you'd need to work that out, which sounds like a bit of a pain but is
probably worth doing; but it also means you end up conflating a lot of
things the Euclidean metric classifies as distinct. If you want to
really understand things, Euclidean is often the way to go.

Here's an example: from a Euclidean point of view, 81/80 and 1029/1024
are symmetrically located with respect to 1, and this is interesting;
we can immediately conclude that 81/80 planar and 1029/1024 planar can
be transformed between. From a Hahn point of view, they are still
symmetrically located, and this is less interesting, because we would
need further analysis to tell us that we can, in fact, transform 81/80
planar to 1029/1024 planar.

> > See above. Yes, of course they are meaningful. If you want to try to
> > linearly transform 50/49 to 49/48, be my guest, but don't be
> surprised
> > when it works very differently than a linear transformation sending
> > 15/14 to 21/20.
>
> You misunderstood my question, and in fact I find this response
> highly condescending.

Touchy touchy. Why should I be the only one on this thread to be
condescended to, if that really is what the above involves? Is it in
fact so obvious that 49/48<-->50/49 is a very different propostion
than 15/14<-->21/20, and if so, why? It's tough if I am going to be
dumped on half the time for obscurity, and the other half for
condescendingly explaining the blindingly obvious.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > > It never occurred to you to do orthogonal transformations?
> >
> > We've transformed between the 48 elements of the symmetry group
of
> > the 7-limit lattice. Are "orthogonal transformations" the ones
that
> > involve a 90-degree rotation, or something else?
>
> I meant isometries--Euclidean distance preserving mappings--fixing
the
> origin, and sending the lattice to itself.

So are there 48 of these for the 7-limit symmetrical (oct-tet)
lattice? Did I forget to count mirror inversions?

> > > They all take two
> > > 7-limit consonaces to get to, which is why they all turn up in
the
> > > stepwise problem I've recently been posting on, and Hahn's
measure
> > > captures that. They behave differently in terms of how the chord
> > > relationships work, and the finer distinctions of the Euclidean
> > metric
> > > capture that.
> >
> > But there's nothing *special* about the euclidean metric here.
The
> > chord relationships work differently, but what is it about them
that
> > implies the specific values that the euclidean metric gives?
>
> The Euclidean metric gives the most refined distinctions. The Hahn
> metric replaces spheres with rhombic dodecahedra. This means to
figure
> out what it means in general, and not just between lattice points,
> you'd need to work that out, which sounds like a bit of a pain but
is
> probably worth doing; but it also means you end up conflating a lot
of
> things the Euclidean metric classifies as distinct. If you want to
> really understand things, Euclidean is often the way to go.

It's perfectly clear that Hahn's metric 'conflates' things. But
Euclidean seems arbitrary among an infinite class of possibilities
here.

> Here's an example: from a Euclidean point of view, 81/80 and
1029/1024
> are symmetrically located with respect to 1, and this is
interesting;
> we can immediately conclude that 81/80 planar and 1029/1024 planar
can
> be transformed between. From a Hahn point of view, they are still
> symmetrically located, and this is less interesting, because we
would
> need further analysis to tell us that we can, in fact, transform
81/80
> planar to 1029/1024 planar.

Hahn's metric isn't intended to tell us anything about configuration,
so I don't see this as a flaw.

> > > See above. Yes, of course they are meaningful. If you want to
try to
> > > linearly transform 50/49 to 49/48, be my guest, but don't be
> > surprised
> > > when it works very differently than a linear transformation
sending
> > > 15/14 to 21/20.
> >
> > You misunderstood my question, and in fact I find this response
> > highly condescending.
>
> Touchy touchy. Why should I be the only one on this thread to be
> condescended to, if that really is what the above involves? Is it in
> fact so obvious that 49/48<-->50/49 is a very different propostion
> than 15/14<-->21/20, and if so, why?

It's immediately evident from looking at the lattice.

> It's tough if I am going to be
> dumped on half the time for obscurity, and the other half for
> condescendingly explaining the blindingly obvious.

It's possible to explain the blindingly obvious without being
condescending. Don't take it personally, but do try to improve.

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> So are there 48 of these for the 7-limit symmetrical (oct-tet)
> lattice? Did I forget to count mirror inversions?

No; there are 48. In fcc coordinates, there are 8 sign changes and six
permutation, and 8*6 = 48.

> It's perfectly clear that Hahn's metric 'conflates' things. But
> Euclidean seems arbitrary among an infinite class of possibilities
> here.

Hardly arbitrary if you accept the above 48 transformations as
interesting, since they are orthogonal. The Euclidean metric falls out
immediately in terms of the invariants of the group, actually.

> > Here's an example: from a Euclidean point of view, 81/80 and
> 1029/1024
> > are symmetrically located with respect to 1, and this is
> interesting;
> > we can immediately conclude that 81/80 planar and 1029/1024 planar
> can
> > be transformed between. From a Hahn point of view, they are still
> > symmetrically located, and this is less interesting, because we
> would
> > need further analysis to tell us that we can, in fact, transform
> 81/80
> > planar to 1029/1024 planar.
>
> Hahn's metric isn't intended to tell us anything about configuration,
> so I don't see this as a flaw.

I'm pointing out that Euclidean gives us more information.

> > Is it in
> > fact so obvious that 49/48<-->50/49 is a very different propostion
> > than 15/14<-->21/20, and if so, why?
>
> It's immediately evident from looking at the lattice.

It's immediately obvious from looking at the *Euclidean* lattice.
Aren't you proving my point for me?

> > It's tough if I am going to be
> > dumped on half the time for obscurity, and the other half for
> > condescendingly explaining the blindingly obvious.
>
> It's possible to explain the blindingly obvious without being
> condescending. Don't take it personally, but do try to improve.

First I need to learn what is blindingly obvious and what isn't. It
seems I am often wrong in those assessments; in other words, to me
"blindingly obvious" is not itself blindingly obvious. And if I am
supposed to stop being condescending, will other people undertake to
do so also?

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > So are there 48 of these for the 7-limit symmetrical (oct-tet)
> > lattice? Did I forget to count mirror inversions?
>
> No; there are 48. In fcc coordinates, there are 8 sign changes and
six
> permutation, and 8*6 = 48.
>
>
> > It's perfectly clear that Hahn's metric 'conflates' things. But
> > Euclidean seems arbitrary among an infinite class of
possibilities
> > here.
>
> Hardly arbitrary if you accept the above 48 transformations as
> interesting, since they are orthogonal.

I don't get it. How does this list of transformations depend on your
metric?

> The Euclidean metric falls out
> immediately in terms of the invariants of the group, actually.

Can you explain this?

> > > Here's an example: from a Euclidean point of view, 81/80 and
> > 1029/1024
> > > are symmetrically located with respect to 1, and this is
> > interesting;
> > > we can immediately conclude that 81/80 planar and 1029/1024
planar
> > can
> > > be transformed between. From a Hahn point of view, they are
still
> > > symmetrically located, and this is less interesting, because we
> > would
> > > need further analysis to tell us that we can, in fact,
transform
> > 81/80
> > > planar to 1029/1024 planar.
> >
> > Hahn's metric isn't intended to tell us anything about
configuration,
> > so I don't see this as a flaw.
>
> I'm pointing out that Euclidean gives us more information.

But all kinds of other shapes, besides a sphere, would also imply
these same distinct shells.

> > > Is it in
> > > fact so obvious that 49/48<-->50/49 is a very different
propostion
> > > than 15/14<-->21/20, and if so, why?
> >
> > It's immediately evident from looking at the lattice.
>
> It's immediately obvious from looking at the *Euclidean* lattice.
> Aren't you proving my point for me?

Absolutely not. For instance, for 50:49 or any other "para" dyad with
Hahn distance 2, there's only one note consonant with both pitches in
the dyad. So it's straightforward to take the "lattice" (or whatever
you want to call it) and transform 50:49 to any dyad which has this
property (such as 9:8, 25:18, 25:16, etc.) while it would take some
severe gymnastics to transform it to one that doesn't have this
property,

> > > It's tough if I am going to be
> > > dumped on half the time for obscurity, and the other half for
> > > condescendingly explaining the blindingly obvious.
> >
> > It's possible to explain the blindingly obvious without being
> > condescending. Don't take it personally, but do try to improve.
>
> First I need to learn what is blindingly obvious and what isn't.

No you don't -- just explain it without being condescending, and even
if people already know it, they won't be offended.

> And if I am
> supposed to stop being condescending, will other people undertake to
> do so also?

I'm doing my best, and still improving, I think.

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> > Hardly arbitrary if you accept the above 48 transformations as
> > interesting, since they are orthogonal.
>
> I don't get it. How does this list of transformations depend on your
> metric?

It defines a metric.

> > The Euclidean metric falls out
> > immediately in terms of the invariants of the group, actually.
>
> Can you explain this?

I checked Wikipedia, and it directed me to the Molien series article,
which doesn't exist. Maybe I should write one, which could serve as an
explanation.

> > I'm pointing out that Euclidean gives us more information.
>
> But all kinds of other shapes, besides a sphere, would also imply
> these same distinct shells.

They imply distict shells, but not the *same* distinct shells.

> > > > Is it in
> > > > fact so obvious that 49/48<-->50/49 is a very different
> propostion
> > > > than 15/14<-->21/20, and if so, why?
> > >
> > > It's immediately evident from looking at the lattice.
> >
> > It's immediately obvious from looking at the *Euclidean* lattice.
> > Aren't you proving my point for me?
>
> Absolutely not. For instance, for 50:49 or any other "para" dyad with
> Hahn distance 2, there's only one note consonant with both pitches in
> the dyad. So it's straightforward to take the "lattice" (or whatever
> you want to call it) and transform 50:49 to any dyad which has this
> property (such as 9:8, 25:18, 25:16, etc.) while it would take some
> severe gymnastics to transform it to one that doesn't have this
> property,

Which is clear from the Euclidean lattice, which says that 9/8, 25/18,
25/16, 50/49 are all located at a distance of 2 from the unison, and
(this is harder; you can use the invariant stuff I mentioned as one
method) that they are all in the same geometric relationship to the
lattice. It is hardly clear using Hahn distance, which says all of the
above are at a distance of 2, but also says 10/9, 16/15, 25/24, 36/35
etc for which the Euclidean distance is sqrt(3) are at a distance of
2, and even 15/14, 21/20 etc for which the Euclidean distance is
sqrt(2) are at a distance of 2. The Hahn distance is making far less
refined distictions, and is not providing the help in sorting these
questions out that the Euclidean distance immediately gives.

> > > > It's tough if I am going to be
> > > > dumped on half the time for obscurity, and the other half for
> > > > condescendingly explaining the blindingly obvious.
> > >
> > > It's possible to explain the blindingly obvious without being
> > > condescending. Don't take it personally, but do try to improve.
> >
> > First I need to learn what is blindingly obvious and what isn't.
>
> No you don't -- just explain it without being condescending, and even
> if people already know it, they won't be offended.

Your claim is that I was being condescending. On what basis are you
making it? What, exactly, was condescending about my remark?

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > > Hardly arbitrary if you accept the above 48 transformations as
> > > interesting, since they are orthogonal.
> >
> > I don't get it. How does this list of transformations depend on
your
> > metric?
>
> It defines a metric.

That seems impossible to me. Please convince me.

> > > I'm pointing out that Euclidean gives us more information.
> >
> > But all kinds of other shapes, besides a sphere, would also imply
> > these same distinct shells.
>
> They imply distict shells, but not the *same* distinct shells.

But what's special about the shells you get from the Euclidean metric
as opposed to the shells you get from using, instead of a sphere,
some other rounded shape with all the symmetries of the rhombic
dodecahedron?

> > > > > Is it in
> > > > > fact so obvious that 49/48<-->50/49 is a very different
> > propostion
> > > > > than 15/14<-->21/20, and if so, why?
> > > >
> > > > It's immediately evident from looking at the lattice.
> > >
> > > It's immediately obvious from looking at the *Euclidean*
lattice.
> > > Aren't you proving my point for me?
> >
> > Absolutely not. For instance, for 50:49 or any other "para" dyad
with
> > Hahn distance 2, there's only one note consonant with both
pitches in
> > the dyad. So it's straightforward to take the "lattice" (or
whatever
> > you want to call it) and transform 50:49 to any dyad which has
this
> > property (such as 9:8, 25:18, 25:16, etc.) while it would take
some
> > severe gymnastics to transform it to one that doesn't have this
> > property,
>
> Which is clear from the Euclidean lattice, which says that 9/8,
25/18,
> 25/16, 50/49 are all located at a distance of 2 from the unison, and
> (this is harder; you can use the invariant stuff I mentioned as one
> method) that they are all in the same geometric relationship to the
> lattice.

>It is hardly clear using Hahn distance,

I repeat: Hahn distance is not intended to give configuration
information anyway. But configuration, not metrics, is what I was
referring to above.

> which says all of the
> above are at a distance of 2, but also says 10/9, 16/15, 25/24,
36/35
> etc for which the Euclidean distance is sqrt(3) are at a distance of
> 2, and even 15/14, 21/20 etc for which the Euclidean distance is
> sqrt(2) are at a distance of 2. The Hahn distance is making far less
> refined distictions, and is not providing the help in sorting these
> questions out that the Euclidean distance immediately gives.

We are going around in circles with this conversation.

> > > > > It's tough if I am going to be
> > > > > dumped on half the time for obscurity, and the other half
for
> > > > > condescendingly explaining the blindingly obvious.
> > > >
> > > > It's possible to explain the blindingly obvious without being
> > > > condescending. Don't take it personally, but do try to
improve.
> > >
> > > First I need to learn what is blindingly obvious and what isn't.
> >
> > No you don't -- just explain it without being condescending, and
even
> > if people already know it, they won't be offended.
>
> Your claim is that I was being condescending. On what basis are you
> making it? What, exactly, was condescending about my remark?

I don't care to dwell on this any longer, but for the record, here it
was:

"Yes, of course they are meaningful. If you want to
try to linearly transform 50/49 to 49/48, be my guest, but don't be
surprised . . ."

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> > Scales which have centers in holes--the octahedra and
> > tetahedra--instead of lattice points, for starters.
>
> OK. But why would a *sphere* be the most useful delimiting shape for
> constructing such scales?

I didn't claim it would--in fact, you were the one expressing
skepticism about the value of the Hahn metric for non-lattice points.

I feel it wouldn't, since the resulting
> scales can actually have 'concavities' of a mild sort, indicating
> suboptimality in terms of consonances-per-note.

Concavities? You are claiming it is possible for the convex hull to
contain lattice points not inside the sphere? I think for any norm,
including Euclidean and Hahn, that is ruled out by the triangle
inequality.

In any case, with spheres the shells are less populous, so you get
more scale possibilities--for good or ill, I don't know.

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
> >
> > > > Hardly arbitrary if you accept the above 48 transformations as
> > > > interesting, since they are orthogonal.
> > >
> > > I don't get it. How does this list of transformations depend on
> your
> > > metric?
> >
> > It defines a metric.
>
> That seems impossible to me. Please convince me.

It's easiest to see in fcc coordinates, where the group consists of
sign changes and permutations of [x, y, z]. What homogenous polynomial
of degree two in x, y and z is invariant under the operations of the
group? It is easy to check that x^2+y^2+z^2 fits the bill, and not
difficult to show it is the *only* degree two polynomial, up to a
constant factor, which does. That, however tells us that Euclidean
distance from the unison is preserved by the operations of the group,
or in other words, the group is a group of orthogonal transformations.
The geometry comes right out of the group. One way to look at it is to
put the origin at the center of a hexany instead, and what you are
looking at are rotations and reflections--Euclidean
self-congruences--of an octahedron.

> But what's special about the shells you get from the Euclidean metric
> as opposed to the shells you get from using, instead of a sphere,
> some other rounded shape with all the symmetries of the rhombic
> dodecahedron?

Obviously the sphere is far easier to manage, just for starters. You
might try to get a polynomial of even degree in x, y, z which is not a
power of x^2+y^2+z^2 but is invariant under the group operations and
which has the same value on the twelve points [+-1, +1, 0] and its
permutations, and use this to define a norm. There are independent
invariants of degree four and six; the simplest possibility would be
to use

[(x^2+y^2+z^2)^2 + K (x+y+z)(-x+y+z)(x-y+z)(x+y-z)]^(1/4)

as the norm, where K is a constant. What values of K make this a norm?
Hell if I know. :)

> I repeat: Hahn distance is not intended to give configuration
> information anyway. But configuration, not metrics, is what I was
> referring to above.

Then you should *love* Euclidean metrics, which have angles and all
the rest of it, and are ideal for analyzing configurations.

> I don't care to dwell on this any longer, but for the record, here it
> was:
>
> "Yes, of course they are meaningful. If you want to
> try to linearly transform 50/49 to 49/48, be my guest, but don't be
> surprised . . ."

That is offensive mostly because you choose to be offended by it; and
I would guess you choose to be offended by it because of
ongoing--perhaps chronic--irritation with me. If you are upset because
Dave left, he gave his reasons, and I wasn't in them, by the way. I
was irritated when Carl seemed to suggest I didn't know beans about
symmetric lattices, but he meant nothing offensive by what he said, so
my reaction was really my problem.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > > Scales which have centers in holes--the octahedra and
> > > tetahedra--instead of lattice points, for starters.
> >
> > OK. But why would a *sphere* be the most useful delimiting shape
for
> > constructing such scales?
>
> I didn't claim it would--in fact, you were the one expressing
> skepticism about the value of the Hahn metric for non-lattice
points.

I retract that skepticism. But all your "shell" posts used the
euclidean metric, and this question was aimed at that.

> I feel it wouldn't, since the resulting
> > scales can actually have 'concavities' of a mild sort, indicating
> > suboptimality in terms of consonances-per-note.
>
> Concavities? You are claiming it is possible for the convex hull to
> contain lattice points not inside the sphere?

No -- that's why I put 'concavities' in quotes. But the smallest
polyhedron large enough to contain all the edges in the scale will
have concavities in general if you use a sphere (i.e., euclidean
metric), which is indicative of suboptimality in terms of consonances-
per-note.

> In any case, with spheres the shells are less populous, so you get
> more scale possibilities--for good or ill, I don't know.

I think, beyond a certain shell, you begin to miss out on the most
desirable scale possibilities entirely: those where the smallest
polyhedron large enough to contain all the edges in the scale is
convex.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> > wrote:
> > > --- In tuning-math@yahoogroups.com, "Paul Erlich"
<perlich@a...>
> > wrote:
> > >
> > > > > Hardly arbitrary if you accept the above 48 transformations
as
> > > > > interesting, since they are orthogonal.
> > > >
> > > > I don't get it. How does this list of transformations depend
on
> > your
> > > > metric?
> > >
> > > It defines a metric.
> >
> > That seems impossible to me. Please convince me.
>
> It's easiest to see in fcc coordinates, where the group consists of
> sign changes and permutations of [x, y, z]. What homogenous
polynomial
> of degree two in x, y and z is invariant under the operations of the
> group? It is easy to check that x^2+y^2+z^2 fits the bill, and not
> difficult to show it is the *only* degree two polynomial, up to a
> constant factor, which does.

But not all metrics are degree two polynomials.

> That, however tells us that Euclidean
> distance from the unison is preserved by the operations of the
group,
> or in other words, the group is a group of orthogonal
transformations.
> The geometry comes right out of the group. One way to look at it is
to
> put the origin at the center of a hexany instead, and what you are
> looking at are rotations and reflections--Euclidean
> self-congruences--of an octahedron.

Convince me that Hahn's metric doesn't yield the exact same list of
48 orthogonal transformations.

>
> > But what's special about the shells you get from the Euclidean
metric
> > as opposed to the shells you get from using, instead of a sphere,
> > some other rounded shape with all the symmetries of the rhombic
> > dodecahedron?
>
> Obviously the sphere is far easier to manage, just for starters. You
> might try to get a polynomial of even degree in x, y, z which is
not a
> power of x^2+y^2+z^2 but is invariant under the group operations and
> which has the same value on the twelve points [+-1, +1, 0] and its
> permutations, and use this to define a norm. There are independent
> invariants of degree four and six; the simplest possibility would be
> to use
>
> [(x^2+y^2+z^2)^2 + K (x+y+z)(-x+y+z)(x-y+z)(x+y-z)]^(1/4)
>
> as the norm, where K is a constant. What values of K make this a
norm?
> Hell if I know. :)

But you seem to be retracting your claim about that the list of
orthogonal transformations defines a metric -- yes?

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
> >
> > > No -- that's why I put 'concavities' in quotes. But the
smallest
> > > polyhedron large enough to contain all the edges in the scale
> will
> > > have concavities in general if you use a sphere (i.e.,
euclidean
> > > metric), which is indicative of suboptimality in terms of
> consonances-
> > > per-note.
> >
> > Do you have an example of this?
>
> I'm sure you've posted some already. Basically, if the convex hull
is
> not a rhombic dodecahedron but has the same symmetry as one, you're
> suboptimal in terms of consonances-per-note -- you can improve the
> ratio by either adding or taking away notes until you do have a
> rhombic dodecahedron.

Or cuboctahedron (there may be other possibilities).

If you can create a depiction of the scales you obtained from
spherical shells, you'll see the 'mild concavities' I'm talking about.

This would be a lot easier to depict in ASCII if we were talking
about 5-limit instead of 7-limit . . .

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> But not all metrics are degree two polynomials.

The point is that not all finite groups of matricies gives you such a
thing, and that in this case we are dealing with a group which has the
orthogonality property, which means the group reprsentation itself has
a Eucildean character.

> Convince me that Hahn's metric doesn't yield the exact same list of
> 48 orthogonal transformations.

When I talked about 49/48->50/49, you became upset with me, and now
you ask me this! Let's see--a Hahn isometry would require that you
send 3,5,7 to three independent 7-limit consonances, which gives you
12 choose 3 = 220 possibilities, clearly a lot more than 48. If we
have something in Hahn shell n, does it necessarily stay in Hahn shell
n? Either way, we are counting the number consonance steps it takes to
get to the note-class, so it seems it should. I suppose the thing to
do is to work out an example, but it certainly isn't clear how Hahn
distinquishes the 48 orthogonal transformations among the rest.

> But you seem to be retracting your claim about that the list of
> orthogonal transformations defines a metric -- yes?

No, it clearly defines a metric--singling out the Euclidean metric
because of the uniqueness of the degree two invariant; you seem to
take that to mean a unique metric. I have invariants of degrees four
and six to play with, and any homogenous polynomial I cook up from
them which leads to convex unit balls (from setting the polynomial
equal to one) will define a metric which is designed to have the
specific property that the 48 transformations are isometries of this
metric.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > But not all metrics are degree two polynomials.
>
> The point is that not all finite groups of matricies gives you such
a
> thing, and that in this case we are dealing with a group which has
the
> orthogonality property, which means the group reprsentation itself
has
> a Eucildean character.
>
> > Convince me that Hahn's metric doesn't yield the exact same list
of
> > 48 orthogonal transformations.
>
> When I talked about 49/48->50/49, you became upset with me, and now
> you ask me this! Let's see--a Hahn isometry would require that you
> send 3,5,7 to three independent 7-limit consonances, which gives you
> 12 choose 3 = 220 possibilities, clearly a lot more than 48.

Gene, you are wrong. A Hahn isometry would require that you send
3,5,7 to three independent 7-limit consonances *which are consonant
with one another*! Only 48 possibilities.

> but it certainly isn't clear how Hahn
> distinquishes the 48 orthogonal transformations among the rest.

As long as we're following the definition of orthogonal
transformation that you gave, you're transforming the entire lattice,
so this seems perfectly clear to me.

> > But you seem to be retracting your claim about that the list of
> > orthogonal transformations defines a metric -- yes?
>
> No, it clearly defines a metric--singling out the Euclidean metric
> because of the uniqueness of the degree two invariant; you seem to
> take that to mean a unique metric.

What else could "defines a metric" be taken to mean, if not "yields a
unique choice of metric"?

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Graham Breed <graham@microtonal.co.uk>

Paul Erlich wrote (2nd March):

> Right, but that was a specific set of calculations, not a general > proof. You were just looking at 'linear' temperaments, if I recall > correctly, but torsion can afflict all types of temperaments. Plus it > seemed your method was far less elegant.

What was inelegant about the torsion finder? I don't remember anything. I thought it was easier than for the octave-specific case because you don't have to check for a common divisor in the left hand column because the left hand column isn't there. So torsion should show up iff the adjoint matrix has a common divisor.

That certainly should work, because if it gives different results to the octave-specific case it means the left hand (octave-specific) column must miss a common divisor that the rest of the adjoint has. For the simple case where the common divisor is 2, that means a power of each odd prime is equal to an odd number of octaves plus an even number of some comma. Divide it through, and you get a non-square rational number to be the square of another rational number, and so there's a formula for the exact solution of the square root of a rational number! This runs into problems with the fundamental theorem of arithmetic, so it can't be possible.

Anyway, I'm less likely to miss something in the tests than that proof. And the general case doesn't matter, only the specific cases that refer to interesting temperaments. I don't even think all of those matter -- if the overwhelming majority of cases of torsion are discovered and dealt with, that's good enough. There are lots of sets of unison vectors that don't give sensible results, and there's always a set of unison vectors to give a specific temperament without torsion.

The linear temperament issue doesn't matter. Torsion is a property of periodicity blocks, and it shouldn't matter how many chromatic unison vectors you choose (or which ones). I think I checked the whole adjoint matrix anyway. I thought we could prove that (for sensibly small unison vectors) the octave-equivalent adjoint is a subset of the octave-specific adjoint.

> I care, and I hope Gene does too. I'd like to see this revisited.

The problem is that you have to consider modulo arithmetic in the optimization problem. Let's take quarter comma meantone as the example. The rule is that four fifths have to add up to a 5:4. In other words, the generator is a quarter of a 5:4. Well, 5:4 is 386.3 cents so the generator must be 96.6 cents. You can see this is wrong because 96.6 is absurdly inaccurate for a 3:2 of 702.0 cents.

There are four different results of division by 4 in modulo arithmetic. So a quarter of 5:4 could be 96.6, 396.6, 696.6 or 996.6 cents. Obviously you choose the best one.

The old way of finding the minimax is to set the intervals between every consonant interval and every other consonant interval to be just. Then take whichever of these tunings gives the best result. The same should work for the octave equivalent case, so long as you take all the different generators that make each interval just.

The newer way is to trace the worst error function downhill until you get to the bottom. I didn't expect this to work, but it does because the function only has one local minimum. I think it's that bit less likely to work in the octave-equivalent case, but you could give it a try.

For the RMS, it gets more complicated because the total error can't be represented as a single quadratic function anymore. Maybe your numerical packages can solve this. The only way I can think is to find the minimax using the method above. The problem is that it would take a lot longer, whereas currently this is a simple calculation. Perhaps only solve for one consonance, and hope you get in the right basin of attraction. Still, it can certainly be done.

Graham

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Graham Breed <graham@microtonal.co.uk>

Gene Ward Smith wrote:

> I think this would be clearer with some examples. Let's say you have
> 21 and 41. How do you get miracle out of the pair of them by your
> method? Then the same question for 1029/1024 and 16875/16807.

Even with the octave specific method, 21 and 41 don't give miracle, but:

3/62, 58.4 cent generator

basis:
(1.0, 0.048647720621243257)

mapping by period and generator:
[(1, 0), (1, 12), (1, 27), (3, -4), (2, 30)]

mapping by steps:
[(41, 21), (65, 33), (95, 48), (115, 59), (142, 72)]

highest interval width: 34
complexity measure: 34 (41 for smallest MOS)
highest error: 0.008440 (10.128 cents)

1029/1024 and 16875/16807 confuse the program, which suggests that whatever I was doing before won't work here. Hmm. Anyway, using 3:1 as the chroma gives
[[ 1, 0, 0, 0],
[ 0, 1, 0, 0],
[-10, 1, 0, 3],
[ 0, 3, 4, -5]]

and an adjoint of

[[ 12, 0, 0, 0],
[ 24, 12, 0, 0],
[ 22, -14, 5, 3],
[ 32, -4, 4, 0]]

So there is some spurious torsion in the second column. I wonder how I got rid of it before. The way the algorithm actually works, it discovers that a period of 6 will work before it tries the period of 3 that it thinks will work.

Well, if you need to find torsion with octave-equivalent vectors, you can always work out the periodicity block I suppose.

Graham

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:
> Gene Ward Smith wrote:
>
> > I think this would be clearer with some examples. Let's say you have
> > 21 and 41. How do you get miracle out of the pair of them by your
> > method? Then the same question for 1029/1024 and 16875/16807.
>
> Even with the octave specific method, 21 and 41 don't give miracle, but:

The two vals <21 33 49 59| and <41 65 95 115| should lead to miracle;
if they don't something is wrong. The same goes in the 11-limit for
<21 33 49 59 72| and <41 65 95 115 142|.

> 3/62, 58.4 cent generator
>
> basis:
> (1.0, 0.048647720621243257)
>
> mapping by period and generator:
> [(1, 0), (1, 12), (1, 27), (3, -4), (2, 30)]
>
> mapping by steps:
> [(41, 21), (65, 33), (95, 48), (115, 59), (142, 72)]

This is the two vals <41 65 95 115 142| and <21 33 48 59 72|, which
leads to the question why the second val for 21-equal; it doesn't seem
like first choice. The temperament in question, with TM basis {100/99,
245/243, 1029/1024}, is associated firmly to 41 at any rate.

> So there is some spurious torsion in the second column. I wonder how I
> got rid of it before. The way the algorithm actually works, it
> discovers that a period of 6 will work before it tries the period of 3
> that it thinks will work.
>
> Well, if you need to find torsion with octave-equivalent vectors, you
> can always work out the periodicity block I suppose.

I think you are making a good case for the claim the best plan is to
use wedgies.

🔗Graham Breed <graham@microtonal.co.uk>

Gene Ward Smith wrote:

> The two vals <21 33 49 59| and <41 65 95 115| should lead to miracle;
> if they don't something is wrong. The same goes in the 11-limit for
> <21 33 49 59 72| and <41 65 95 115 142|.

Why should they? They never have before.

> This is the two vals <41 65 95 115 142| and <21 33 48 59 72|, which
> leads to the question why the second val for 21-equal; it doesn't seem
> like first choice. The temperament in question, with TM basis {100/99,
> 245/243, 1029/1024}, is associated firmly to 41 at any rate.

That val is for the best equal temperament, isn't it? Can you find a better one?

> I think you are making a good case for the claim the best plan is to
> use wedgies.

Wedgies give identical results to matrices, so what difference does it make?

Graham

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:
> Gene Ward Smith wrote:
>
> > The two vals <21 33 49 59| and <41 65 95 115| should lead to miracle;
> > if they don't something is wrong. The same goes in the 11-limit for
> > <21 33 49 59 72| and <41 65 95 115 142|.
>
> Why should they? They never have before.

If you get a different answer, I think you must be using an incorrect
algorithm.

> > This is the two vals <41 65 95 115 142| and <21 33 48 59 72|, which
> > leads to the question why the second val for 21-equal; it doesn't seem
> > like first choice. The temperament in question, with TM basis {100/99,
> > 245/243, 1029/1024}, is associated firmly to 41 at any rate.
>
> That val is for the best equal temperament, isn't it? Can you find a
> better one?

That would depend on your definition of better; but the one with the
TOP tuning closest to JI is the "standard" val, <21 33 48 59 73|. If
we put that together with 41, we get an 11-limit temperament which
adds 245/242 rather than 385/384 to 225/224 and 1029/1024, leading to
a temperament with a half-secor generator. Incidentally, if we add
243/242 to 225/224 and 1029/1024, what does your program give?

> > I think you are making a good case for the claim the best plan is to
> > use wedgies.
>
> Wedgies give identical results to matrices, so what difference does
it make?

At this point it is by no means clear your matrix methods are giving
correct answers, and in any case you seem to be working a lot harder
for them.

🔗Graham Breed <graham@microtonal.co.uk>

Gene Ward Smith wrote:

> That would depend on your definition of better; but the one with the
> TOP tuning closest to JI is the "standard" val, <21 33 48 59 73|. If
> we put that together with 41, we get an 11-limit temperament which
> adds 245/242 rather than 385/384 to 225/224 and 1029/1024, leading to
> a temperament with a half-secor generator. Incidentally, if we add > 243/242 to 225/224 and 1029/1024, what does your program give?

My definition is the closest to JI. <21 33 48 59 72| has a worst 11-limit error of 0.81 scale steps, but for <21 33 49 59 73| it's 0.92 steps. There isn't a standard val because the temperament is inconsistent.

I get this temperament from the nearest prime approximations of 21 and 41 and also the unison vectors you gave:

3/62, 58.4 cent generator

basis:
(1.0, 0.048631497540919798)

mapping by period and generator:
[(1, 0), (1, 12), (3, -14), (3, -4), (4, -11)]

mapping by steps:
[(41, 21), (65, 33), (95, 49), (115, 59), (142, 73)]

highest interval width: 38
complexity measure: 38 (41 for smallest MOS)
highest error: 0.008391 (10.069 cents)
unique

Which looks like what you said, but not miracle. You get miracle from that other set.

> At this point it is by no means clear your matrix methods are giving
> correct answers, and in any case you seem to be working a lot harder
> for them.

They're giving exactly the same results as the wedgies.

Graham

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Graham Breed <graham@microtonal.co.uk>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Graham Breed <graham@microtonal.co.uk>

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

🔗Gene Ward Smith <gwsmith@svpal.org>

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

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?

As
> there's no way of knowing where a wedge product comes from I have to
> treat contorsion the same way as torsion.

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

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 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.

>
> 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.

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.

🔗Graham Breed <graham@microtonal.co.uk>

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

🔗Gene Ward Smith <gwsmith@svpal.org>

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

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.

> > 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.

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.

🔗Graham Breed <graham@microtonal.co.uk>

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:

http://66.98.148.43/~xenharmo/regular.html

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

🔗Gene Ward Smith <gwsmith@svpal.org>

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

> 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:
>
> http://66.98.148.43/~xenharmo/regular.html

No, in fact it doesn't. Half of the notes are not being mapped to. If
there was something to correspond to 2/sqrt(3), half of a fourth, you
would be OK but the whole point of contorsion is that there isn't.

> That's somewhat vague as to whether contorsion is allowed. First it
> says that an "icon" has to be epimorphic.

Nothing about it is vague that I can see. The epimorphic mapping, as
stated, is from the p-limit for some prime p to the icon.

But then "given an icon" we
> have to "find a homomorphic mapping".

And the homomorphic mapping is from the icon to cents or something of
that sort. This is the tuning stage and it would not make sense to ask
that this be epimorphic.

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.

How do you propose to map the 5-limit to Vicentino's system? You get a
map to only half of it.

> 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.

Contemperaments?

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

And why is producing contemperaments from wedge products, which can
easily be done, any more pointless than producing them any other way?

> 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.

Good!

🔗Carl Lumma <ekin@lumma.org>

>>there *are* differences between
>>intervals like 15/14 and 21/20 as compared to 25/24, 36/35 or 49/48,
>>and that these again are different from 50/49. They all take two
>>7-limit consonaces to get to, which is why they all turn up in the
>>stepwise problem I've recently been posting on, and Hahn's measure
>>captures that. They behave differently in terms of how the chord
>>relationships work, and the finer distinctions of the Euclidean metric
>>capture that.
>
>What kind of chord relationships? If they all take 2 consonances
>to get to, none of these are common-tone-preserving. Can you
>illustrate with an example?

Here's an example you did give

""from a Euclidean point of view, 81/80 and 1029/1024
are symmetrically located with respect to 1, and this is interesting;
we can immediately conclude that 81/80 planar and 1029/1024 planar can
be transformed between""

but I don't know what you mean by "transformed between" here.

-C.