<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

--- In tuning-math@y..., genewardsmith@j... wrote:

> <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

Hey Graham,

Shouldn't this have made your top ten at

<http://x31eq.com/limit7.txt>?

Why didn't it?

In-Reply-To: <9u33hf+h7bd@eGroups.com>

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.

Graham

--- 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?

In-Reply-To: <9u358e+nleq@eGroups.com>

> Isn't your score

>

> complexity*max_error?

No. It's complexity^2*max_error.

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

> In-Reply-To: <9u358e+nleq@e...>

> > Isn't your score

> >

> > complexity*max_error?

>

> No. It's complexity^2*max_error.

Oh . . .

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?

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

> Because there are 35 temperaments that scored higher.

I think it would be interesting to compare RMS error / standard

deviation of generators in cents per generator step to your scoring.

I'll work that out.

--- 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.

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

--- In tuning-math@y..., genewardsmith@j... 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?

> and two badness measures--the

> first is rms generator steps

What does rms generator steps mean?

--- 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.

--- 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.

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

--- 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.

--- 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.

--- In tuning-math@y..., genewardsmith@j... wrote:

> > > 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.

Why not the root-mean-square of the deviations from zero?

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

> Why not the root-mean-square of the deviations from zero?

Good point--the list was supposed to cover the reciprocals also, so

that we got the entire vertex figure, and the mean would be zero. I'd

better go change things.

I wrote,

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

Gene wrote,

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

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

--- 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!

--- 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!

--- 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?

--- 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.

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

> What will this tell me _intuitively_?

You need to start from knowing what something is before developing

the intuition.