Yahoo Tuning Groups Ultimate Backup tuning-math Survey IV

<245/243, 64/63>

ets: 5, 17, 22, 27, 49

Map:

[-1 1]
[-1 2]
[ 3 6]
[-4 2]

Adjusted map:

[ 0 1]
[-1 2]
[-9 6]
[ 2 2]

Generators: a = .4080126504 (~4/3) = 19.99261987 / 49

Related systems: 27+22 and 22+5

Errors:

3: 8.43
5: 7.15
7: 10.40

Comparison to 27 and 49:

3: 9.16 8.25
5: 13.67 5.52
7: 8.95 10.77

This is essentially the 27+22 system of the 49 et

<126/125, 245/243>

Map:

[4 -1]
[3 1]
[5 1]
[5 2]

Generators a~9/7 = 17.005/46 and b~7/5 = 22.202/46

Adjusted map:

[0 1]
[7 -1]
[9 -1]
[13 -2]

Generators a = .3696836546 = 17.00544811 / 46; b = 1

Errors and 46-et errors:

3: 3.39 2.39
5: 6.27 4.99
7:-1.76 -3.61

Pretty close to the 27+19 system of 46-et

<225/224, 1728/1715>

ets: 9, 22, 31, 53, 84

Map (no adjustment needed)

[ 0 1]
[ 7 0]
[-3 3]
[ 8 1]

Generators: a = 18.99284544 / 84; b = 1

This is the Orwell, of course.

Errors compared to 53 and 84

3: -2.670 -0.068 -1.955
5: -0.293 -1.408 -0.599
7: 1.785 4.759 2.603

For the 7-limit, the Orwell may as well be taken as the classic
George 19/84; for the 11-limit the 53, which of course is also good,
looks a little better.

<126/125, 81/80>

Map:

[ 1 1]
[ 1 2]
[ 0 4]
[-3 7]

Adjusted map:

[0 1]
[-1 2]
[-4 4]
[-10 7]

Generators: a = 0.4194600621 = 13.00326081 / 31

This is, of course, meantone--essentially, 1/4-comma meantone

Errors and 31-et comparison

3: -5.31 -5.18
5: 0.28 0.78
7: -2.35 -1.08

🔗Paul Erlich <paul@stretch-music.com>

🔗graham@microtonal.co.uk

🔗Paul Erlich <paul@stretch-music.com>

--- In tuning-math@y..., graham@m... wrote:
> In-Reply-To: <9u33hf+h7bd@e...>
> Paul wrote:
>
> > Shouldn't this have made your top ten at
> > <http://x31eq.com/limit7.txt>?
>
> No.
>
> > Why didn't it?
>
> Because there are 35 temperaments that scored higher.

Isn't your score

complexity*max_error?

For this temperament get

13*(<8.03)=<104.39,

while for #10 on your list I get

6*17.848=107.088

So it looks like it scores better than your #10.

What am I doing wrong?

🔗graham@microtonal.co.uk

🔗Paul Erlich <paul@stretch-music.com>

🔗genewardsmith@juno.com

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Can we get Gene to compute the TM-reduced basis for each linear
> temperament for some distance down your 7-limit rankings? Are they
> listed anywhere beyond the top ten?

Here's his top ten. In each case, I checked for reduction, both
torsion and Minkowski, and found one unreduced set. This I reduced by
means of the commas of the wedge invartiant and LLL. In each case
below, I give the reduced basis, the wedge invariant, the length of
the wedge invariant (which is proportional to number of steps in a
typical scale for this temperament), and two badness measures--the
first is rms generator steps times rms 7-limit error, in units of
step-cents, and the second is rms generator steps squared times rms
7-limit error, which should correspond more directly to Graham's
measure.

(1) <3136/3125, 2401/2400>

[16,2,5,6,37,-34] length = 53.348

8.832 step-cents
89.117 step^2-cents

(2) <21/20, 27/25>

[2,3,1,-6,4,0] length = 8.124

58.751 sc
99.858 s2c

(3) <225/224, 1029/1024>

[6,-7,-2,15,20,-25] length = 36.592

11.321 sc
78.277 s2c

(4) <1728/1715, 225/224>

[7,-3,8,27,7,-21] length = 36.620

18.651 sc
134.353 s2c

Graham had <839808/823543, 2109375/2097152>, which has 7-torsion

(5) <50/49, 36/35>

[2,0,4,-2,5,1] length = 7.071

38.274 sc
76.548 s2c

(6) <81/80, 126/125>

[1,4,10,12,-13,4] length = 21.119

11.734 sc
37.567 s2c

(7) <126/125, 1728/1715>

[10,9,7,-9,17,-9] length = 26.096

18.050 sc
98.129 s2c

(8) <25/24, 49/48>

[4,2,2,-1,8,-6] length = 11.180

52.944 sc
117.066 s2c

(9) <50/49, 64/63>

[-2,4,4,-2,-12,11] length = 17.464

32.710 sc
98.129 s2c

(10) <49/48, 126/125>

[6,5,3,-7,12,-6] length = 17.292

42.910 sc
150.013 s2c

In terms of step-cents, I get from best to worst 1,3,6,7,4,9,10,8,2;
so the second-best on Graham's list is the worst here. Paultone is
promoted from 9th to 6th, and meantone from 6th to 3rd. Step-cents,
which seems to me easier to justify as units, is working better so
far as I can see.

In terms of step^2-cents, I get 6,5,3,1,7,9,2,8,4,10. This is closer
to Graham's list, but it isn't clear to me why these units should be
preferred. The 9th and 7th on his list tie by this measure, by the
way.

🔗graham@microtonal.co.uk

In-Reply-To: <9u4bn5+jj4l@eGroups.com>
Gene wrote:

> In terms of step-cents, I get from best to worst 1,3,6,7,4,9,10,8,2;
> so the second-best on Graham's list is the worst here. Paultone is
> promoted from 9th to 6th, and meantone from 6th to 3rd. Step-cents,
> which seems to me easier to justify as units, is working better so
> far as I can see.

Well, the current second is a curiosity, with a worst error of 44 cents.
What happened to 5? My ordering by step cents is 1 3 4 6 7 11 22 23 20
56 of the original ordering. The original 2 is now 48. That new number
10, old 56, is 29+31, which looks reasonable to me.

> In terms of step^2-cents, I get 6,5,3,1,7,9,2,8,4,10. This is closer
> to Graham's list, but it isn't clear to me why these units should be
> preferred. The 9th and 7th on his list tie by this measure, by the
> way.

It's still interesting how far it diverges from my list. The idea of
squaring the complexity was to bring some sanity to the lower limits,
where absurdly accurate temperaments were dominating the list. I've put a
cap on the complexity since, which solves that problem.

It looks like the 7-limit list might make more sense with step-cents, but
not the 5-limit one. Meantone drops from 3 to 7. And the 11-limit is
getting over-complex, with Miracle dropping to 3 behind 80+72 and 46+26.
It's 9 in the 9-limit list, Magic dropped out completely.

I suppose you'll be wanting those new lists

<http://x31eq.com/limit5.stepcents>
<http://x31eq.com/limit7.stepcents>
<http://x31eq.com/limit9.stepcents>
<http://x31eq.com/limit11.stepcents>
<http://x31eq.com/limit13.stepcents>
<http://x31eq.com/limit15.stepcents>
<http://x31eq.com/limit17.stepcents>
<http://x31eq.com/limit19.stepcents>

Graham

🔗Paul Erlich <paul@stretch-music.com>

🔗genewardsmith@juno.com

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> > the length of
> > the wedge invariant (which is proportional to number of steps in
a
> > typical scale for this temperament),

> Fascinating -- how does that work out?

It's simply a way of trying to interpret what the area defined by the
paralleogram of two generators or two vals--which is an invariant of
the temperament--means in some rough sense. Since area is assoicated
to counts of lattice points in the coordiante projections of this
paralleogram, it seemed a reasonable way to look at it.

> > and two badness measures--the
> > first is rms generator steps
>
> What does rms generator steps mean?

I count the number of generator steps of the non-octave generator to
{3,5,7,5/3,7/3,7/5}, take the mean and then the root-mean-square of
the deviations from the mean, and finally multiply by how many octave-
generators there are per octave--that is, by the number in the top
right hand entry of the "map" matrix.

I've been finding the wedge product an easier way of getting to the
map matrix than by previous method of LLL reduction, which is why I
haven't put any LLL-reduced maps in Survey IX. The right-hand column
can be read from the wedge invaiant immediately, and the top right
entry is simply the gcd of this. The other three entries can be
chosen on the basis of being integers and giving the correct wedge
invariant when the wedge product is taken with the other column; this
gives three linear equations in three unknowns, which has a
parametrized solution. This could be automated by Graham and
generalized to higher prime limits if he wanted to try it.

🔗Paul Erlich <paul@stretch-music.com>

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
>
> > > the length of
> > > the wedge invariant (which is proportional to number of steps
in
> a
> > > typical scale for this temperament),
>
> > Fascinating -- how does that work out?
>
> It's simply a way of trying to interpret what the area defined by
the
> paralleogram of two generators or two vals--which is an invariant
of
> the temperament--means in some rough sense. Since area is
assoicated
> to counts of lattice points in the coordiante projections of this
> paralleogram, it seemed a reasonable way to look at it.

Confused. Can you give a low-limit example?

> > > and two badness measures--the
> > > first is rms generator steps
> >
> > What does rms generator steps mean?
>
> I count the number of generator steps of the non-octave generator to
> {3,5,7,5/3,7/3,7/5}, take the mean and then the root-mean-square of
> the deviations from the mean, and finally multiply

Multiply what? The mean or the root-mean-square of the deviations
from the mean?

> by how many octave-
> generators there are per octave--that is, by the number in the top
> right hand entry of the "map" matrix.

🔗graham@microtonal.co.uk

genewardsmith@juno.com () wrote:

> I've been finding the wedge product an easier way of getting to the
> map matrix than by previous method of LLL reduction, which is why I
> haven't put any LLL-reduced maps in Survey IX. The right-hand column
> can be read from the wedge invaiant immediately, and the top right
> entry is simply the gcd of this. The other three entries can be
> chosen on the basis of being integers and giving the correct wedge
> invariant when the wedge product is taken with the other column; this
> gives three linear equations in three unknowns, which has a
> parametrized solution. This could be automated by Graham and
> generalized to higher prime limits if he wanted to try it.

Your right hand column is my left hand column. Well, so far I'm getting
the other column from the adjoint, which is equivalent to some kind of
wedge product, as you confirmed before. So you have a way to go straight
to the period mapping? Also, will this work with octave equivalent
matrices? If I were to rely on them now, I could get the generator
mapping by algebra (with caveats regarding torsion), but I'd need to use
the metric to get the period mapping.

It might be interesting to try other methods, but the problem I'd like you
to set your mind to is going from the map to the simplest set of unison
vectors. That's something that's not working currently. I can go from
unison vectors to maps no problem.

Graham

🔗genewardsmith@juno.com

--- In tuning-math@y..., graham@m... wrote:

> It might be interesting to try other methods, but the problem I'd
like you
> to set your mind to is going from the map to the simplest set of
unison
> vectors. That's something that's not working currently. I can go
from
> unison vectors to maps no problem.

One way to do this would be to treat the unison vector problem in the
exact same way as the generator map problem. We can find the wedge
invariant from the map by taking the wedge product of the column
vectors, and directly from the wedge invariant we can read off a
5-limit unison vector. We then can find the exponent of 7 for the
other unison vector via the gcd of the exponents of the first, and
the rest using the requirement that the wedge invariant is invariant.
Once we have this pair of vectors we should have something which we
can readily reduce further via LLL and Minkowski reduction.

How well would this work? Why don't you give a list of hard examples
and we'll see.

🔗genewardsmith@juno.com

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
> --- In tuning-math@y..., genewardsmith@j... wrote:

> > It's simply a way of trying to interpret what the area defined by
> the
> > paralleogram of two generators or two vals--which is an invariant
> of
> > the temperament--means in some rough sense. Since area is
> assoicated
> > to counts of lattice points in the coordiante projections of this
> > paralleogram, it seemed a reasonable way to look at it.

> Confused. Can you give a low-limit example?

Let's try 50/49^64/63 = [-2,4,4,-2,-12,11]

We have, in the above order:

5-7: -2

[2 -2]^(-1) [1/2 -1]
[0 -1] = [ 0 -1]

The determinant is -2, corresponding to the first entry, so we have
two classes in the block of 5-7 equivalence classes. Using the
inverse matrix columns, we can calculate a block based on 0<=u<1,
0<=v<1 where u = i/2 and v = -i-j. The block turns out to be
<1, 5/7>

7-3: 4

[ 0 2]^(-1) [ 1/2 0]
[-1 -2] = [-1/4 0]

Block: <1,7,1/3,7/3>

3-5: 4

Block <1,5,1/3,5/3>

2-3: -2

Block <1,3>

2-5: -12

Block: <1,2,4,8,16,32,10,20,40,80,160,320>

2-7: 11

Block: <1,2/7,4/7,8/7,16/7,32/7,4/49,8/49,16/49,32/49,64/49>

Fodder for some kind of product set here, perhaps.

> > I count the number of generator steps of the non-octave generator
to
> > {3,5,7,5/3,7/3,7/5}, take the mean and then the root-mean-square
of
> > the deviations from the mean, and finally multiply
>
> Multiply what? The mean or the root-mean-square of the deviations
> from the mean?

RMS of deviations from mean.

🔗Paul Erlich <paul@stretch-music.com>

🔗genewardsmith@juno.com

🔗Paul Erlich <paul@stretch-music.com>

🔗genewardsmith@juno.com

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Oh dear . . . how about a couple of 5-limit examples?

In the 5-limit, the wedge product becomes the cross-product. The
cross-product of two ets gives the intersection of the kernels, and
the cross-product of two commas gives the common et. So, it's not the
same.

In the 11-limit, the wedge product of two ets gives a 10-dimensional
vector which represents a linear temperament, and of two commas a 10-
dimensional vector which represents a planar temperament, so we don't
have something betwixt and between. Wedging three ets gives a 10-
dimensional planar temperament invariant, which can be compared to
two commas, and three commas gives a linear temperament invariant, as
with two ets. I could go on and describe the 13-limit, but the point
is the best way to get a handle on the 7-limit is to look at it, not
at the 5-limit!

🔗Paul Erlich <paul@stretch-music.com>

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
>
> > Oh dear . . . how about a couple of 5-limit examples?
>
> In the 5-limit, the wedge product becomes the cross-product. The
> cross-product of two ets gives the intersection of the kernels, and
> the cross-product of two commas gives the common et. So, it's not
the
> same.
>
> In the 11-limit, the wedge product of two ets gives a 10-
dimensional
> vector which represents a linear temperament, and of two commas a
10-
> dimensional vector which represents a planar temperament, so we
don't
> have something betwixt and between. Wedging three ets gives a 10-
> dimensional planar temperament invariant, which can be compared to
> two commas, and three commas gives a linear temperament invariant,
as
> with two ets. I could go on and describe the 13-limit, but the
point
> is the best way to get a handle on the 7-limit is to look at it,
not
> at the 5-limit!

I'm trying to get an _intuitive_ handle on it . . . sigh . . . better
get those math books you mentioned!

🔗genewardsmith@juno.com

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> I'm trying to get an _intuitive_ handle on it . . . sigh . . .
better
> get those math books you mentioned!

If you have library priviledges at a university library, the subject
area is multilinear algebra, where exterior algebra and bilinear
forms are important, and tensor products and differential forms can,
let us hope, be safely ignored. However you could simply try to get a
feel for it by doing some calculations. Since 64/63 = 2^6 3^-2 7^-1,
I might write it 6 [2] - 2 [3] - [7], where the brakets are intended
to signal the number is a vector. Then 50/49 = [2] + 2 [5] - 2 [7],
and

50/49^64/63 = ([2]-2[5]-2[7])^(6[2]-2[3]-[7]). Taking the wedge
product would be great practice...since a^b=b^a, a^a=0, so [2]^[2]=0,
[3]^[3]=0, [5]^[5]=0, [7]^[7]=0. On the other hand,
[2]^[7]= - [7]^[2], and so forth. Why not treat it as one of your
challenges and work it out?

🔗Paul Erlich <paul@stretch-music.com>

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
>
> > I'm trying to get an _intuitive_ handle on it . . . sigh . . .
> better
> > get those math books you mentioned!
>
> If you have library priviledges at a university library, the
subject
> area is multilinear algebra, where exterior algebra and bilinear
> forms are important, and tensor products and differential forms
can,
> let us hope, be safely ignored. However you could simply try to get
a
> feel for it by doing some calculations. Since 64/63 = 2^6 3^-2 7^-
1,
> I might write it 6 [2] - 2 [3] - [7], where the brakets are
intended
> to signal the number is a vector. Then 50/49 = [2] + 2 [5] - 2 [7],
> and
>
> 50/49^64/63 = ([2]-2[5]-2[7])^(6[2]-2[3]-[7]). Taking the wedge
> product would be great practice...since a^b=b^a, a^a=0, so [2]^[2]
=0,
> [3]^[3]=0, [5]^[5]=0, [7]^[7]=0. On the other hand,
> [2]^[7]= - [7]^[2], and so forth. Why not treat it as one of your
> challenges and work it out?

What will this tell me _intuitively_?

P.S. look over at harmonic_entropy if you haven't lately.