What is the mapping from generators to primes in the 46-out-of-72

temperament?

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

> What is the mapping from generators to primes in the 46-out-of-72

> temperament?

I get that r = 2^(11/72) is the generator within a sqrt(2); we have

2 r^(-6) = 2^(42/72) ~ 3/2

2 r^(-11) = 2^(23/72) ~ 5/4

2^(1/2) r^2 = 2^(58/72) ~ 7/4

r^3 = 2^(33/72) ~ 11/8.

The complexity is 2*(3-(-11)) = 28.

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

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

>

> > What is the mapping from generators to primes in the 46-out-of-72

> > temperament?

>

> I get that r = 2^(11/72) is the generator within a sqrt(2);

Saw that on Graham's 11-limit and 13-limit results . . .

> we have

>

> 2 r^(-6) = 2^(42/72) ~ 3/2

>

> 2 r^(-11) = 2^(23/72) ~ 5/4

>

> 2^(1/2) r^2 = 2^(58/72) ~ 7/4

>

> r^3 = 2^(33/72) ~ 11/8.

>

Thanks a lot Bob.

So the 20-tone-per-octave MOS scale will have three 1:3:7:11 and

three 1/(1:3:7:11) chords (right?). Not enough to warrant too much

interest in this 20-tone-per-octave MOS at this point . . .

BUT:

merely within the 7-limit, the following 20-tone MOS scales (unless

some of them are double image) suggest themselves from PB

considerations:

2401 : 2400 commatic

64 : 63 commatic

25 : 24 chromatic

2401 : 2400 commatic

225 : 224 commatic

25 : 24 chromatic

2401 : 2400 commatic

225 : 224 commatic

28 : 27 chromatic

1029 : 1024 commatic

50 : 49 commatic

28 : 27 chromatic

225 : 224 commatic

1029 : 1024 commatic

25 : 24 chromatic

225 : 224 commatic

1029 : 1024 commatic

28 : 27 chromatic

2401 : 2400 commatic

1029 : 1024 commatic

25 : 24 chromatic

2401 : 2400 commatic

1029 : 1024 commatic

28 : 27 chromatic

How many distinct MOS scales are represented here? What are the

generators, and mapping from generators to primes, in each?

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

> Thanks a lot Bob.

You're welcome. Who's Bob?

> So the 20-tone-per-octave MOS scale will have three 1:3:7:11 and

> three 1/(1:3:7:11) chords (right?). Not enough to warrant too much

> interest in this 20-tone-per-octave MOS at this point . . .

I think you are asking for trouble in the form of torsion with this

20 business. Why 20? I could check all 7 choose 3 subsets of the

commas you give below, but is there a reason to think this will work?

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

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

>

> > Thanks a lot Bob.

>

> You're welcome. Who's Bob?

Oops, sorry Gene. It's late!

>

> > So the 20-tone-per-octave MOS scale will have three 1:3:7:11 and

> > three 1/(1:3:7:11) chords (right?). Not enough to warrant too

much

> > interest in this 20-tone-per-octave MOS at this point . . .

>

> I think you are asking for trouble in the form of torsion with this

> 20 business. Why 20? I could check all 7 choose 3 subsets of the

> commas you give below,

8 choose 3?

> but is there a reason to think this will work?

If there is one without torsion, then we have something to look at

that is potentially as interesting as Blackjack.

Do you have a reason do think there won't be one?

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

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

> > I could check all 7 choose 3 subsets of the

> > commas you give below,

>

> 8 choose 3?

>

Wait a minute . . . there are only 8 cases to check . . . I gave 8

triplets with a Fokker determinant of 20.

Paul wrote:

> How many distinct MOS scales are represented here? What are the

> generators, and mapping from generators to primes, in each?

I've written a test script. See

<http://x31eq.com/vectors.paul.py> and the results at

<http://x31eq.com/vectors.paul.out>. In summary

1/4, 356.0 cent generator

[(1, 0), (1, 2), (5, -9), (4, -4)]

1/20, 116.6 cent generator

[(1, 0), (1, 6), (3, -7), (3, -2)]

1/10, 116.6 cent generator

[(1, 0), (1, 6), (3, -7), (3, -2)]

1/10, 55.2 cent generator

[(1, 0), (2, -6), (2, 7), (3, 2)]

?

[(?, 0), (?, -3), (?, 1), (?, 1)]

1/20, 55.2 cent generator

[(1, 0), (2, -6), (2, 7), (3, 2)]

9/20, 578.1 cent generator

[(1, 0), (-1, 6), (6, -7), (4, -2)]

9/20, 578.1 cent generator

[(1, 0), (-1, 6), (6, -7), (4, -2)]

25:24, 1029:1024 and 225:224 fail, apparently because it wants a

half-octave generator, but doesn't give the usual clue. I think it should

come out like this:

1/9, 66.7 cent generator

basis:

(0.5, 0.0555981053341)

mapping by period and generator:

[[2, 0], [4, -3], [5, 1], [6, 1]]

mapping by steps:

[(10, 8), (17, 13), (26, 21), (31, 25)]

unison vectors:

[[1, 0, 2, -2], [10, -1, -2, -1]]

highest interval width: 4

complexity measure: 8 (10 for smallest MOS)

highest error: 0.248243 (297.892 cents)

Graham

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

> Paul wrote:

>

> > How many distinct MOS scales are represented here? What are the

> > generators, and mapping from generators to primes, in each?

>

> I've written a test script. See

> <http://x31eq.com/vectors.paul.py> and the results at

> <http://x31eq.com/vectors.paul.out>. In summary

>

>

>

> 1/4, 356.0 cent generator

> [(1, 0), (1, 2), (5, -9), (4, -4)]

>

> 1/20, 116.6 cent generator

> [(1, 0), (1, 6), (3, -7), (3, -2)]

>

> 1/10, 116.6 cent generator

> [(1, 0), (1, 6), (3, -7), (3, -2)]

>

>

> 1/10, 55.2 cent generator

> [(1, 0), (2, -6), (2, 7), (3, 2)]

>

>

> ?

> [(?, 0), (?, -3), (?, 1), (?, 1)]

>

>

>

> 1/20, 55.2 cent generator

> [(1, 0), (2, -6), (2, 7), (3, 2)]

>

> 9/20, 578.1 cent generator

> [(1, 0), (-1, 6), (6, -7), (4, -2)]

>

> 9/20, 578.1 cent generator

> [(1, 0), (-1, 6), (6, -7), (4, -2)]

>

So some of them _don't_ have torsion?

>

>

> 25:24, 1029:1024 and 225:224 fail, apparently because it wants a

> half-octave generator, but doesn't give the usual clue.

Gene?

Paul wrote:

> So some of them _don't_ have torsion?

The last figures were wrong (some of my unison vectors were wrong) so

here's what they should be:

[(-1, 2, 0), (-2, 0, -1), (-1, -2, 4)]

3/10, 356.0 cent generator

complexity measure: 11 (17 for smallest MOS)

highest error: 0.008316 (9.980 cents)

unique

[(-1, 2, 0), (2, 2, -1), (-1, -2, 4)]

1/10, 116.6 cent generator

complexity measure: 13 (21 for smallest MOS)

highest error: 0.002024 (2.428 cents)

unique

[(-3, 0, 1), (2, 2, -1), (-1, -2, 4)]

1/10, 116.6 cent generator

complexity measure: 13 (21 for smallest MOS)

highest error: 0.002024 (2.428 cents)

unique

[(-3, 0, 1), (0, 2, -2), (1, 0, 3)]

2/5, 231.2 cent generator

complexity measure: 8 (10 for smallest MOS)

highest error: 0.014573 (17.488 cents)

[(-1, 2, 0), (1, 0, 3), (2, 2, -1)]

1/10, 116.6 cent generator

complexity measure: 13 (21 for smallest MOS)

highest error: 0.002024 (2.428 cents)

unique

[(-3, 0, 1), (1, 0, 3), (2, 2, -1)]

1/10, 116.6 cent generator

complexity measure: 13 (21 for smallest MOS)

highest error: 0.002024 (2.428 cents)

unique

[(-1, 2, 0), (1, 0, 3), (-1, -2, 4)]

1/10, 116.6 cent generator

complexity measure: 13 (21 for smallest MOS)

highest error: 0.002024 (2.428 cents)

unique

[(-3, 0, 1), (1, 0, 3), (-1, -2, 4)]

1/10, 116.6 cent generator

complexity measure: 13 (21 for smallest MOS)

highest error: 0.002024 (2.428 cents)

unique

This one doesn't have torsion:

[(-3, 0, 1), (0, 2, -2), (1, 0, 3)]

conversion

[[10 0]

[16 -3]

[23 1]

[28 1]]

2/5, 231.2 cent generator

basis:

(0.5, 0.19264507794239583)

mapping by period and generator:

[(2, 0), (2, 3), (5, -1), (6, -1)]

mapping by steps:

[[6, 4], [9, 5], [14, 9], [17, 11]]

unison vectors:

[[1, 0, 2, -2], [-9, 1, 2, 1]]

highest interval width: 4

complexity measure: 8 (10 for smallest MOS)

highest error: 0.014573 (17.488 cents)

No, wait, it does give a 20 note periodicity block, but the whole adjoint

matrix divides through by 2. So this was falling victim to the trap I set

for the bad multiple-29 unison vectors. I've worked out some better ones

for that now, that I can try out next time I'm in Linux.

> > 25:24, 1029:1024 and 225:224 fail, apparently because it wants a

> > half-octave generator, but doesn't give the usual clue.

>

> Gene?

The first column of the adjoint does have a common factor of 2. But so do

all the others. Still, I changed the program to find the period a

different way. If we want to use only octave-equivalent matrices, I think

we'll still have to make sure we start with a good set of unison vectors,

and trust that left hand column.

Graham

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

> > > 25:24, 1029:1024 and 225:224 fail, apparently because it wants

a

> > > half-octave generator, but doesn't give the usual clue.

> >

> > Gene?

> The first column of the adjoint does have a common factor of 2.

But so do

> all the others.

I don't know what your usual clue is, but mine is common factors. We

have

[ a b c d]

det [ -5 2 2 -1] = -20a-32b-46c-56d = -2 h10

[-10 1 0 3]

[ -3 -1 2 0]

and therefore torsion.

In-Reply-To: <9pqocr+rc5u@eGroups.com>

Gene wrote:

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

>

> > > > 25:24, 1029:1024 and 225:224 fail, apparently because it wants

> a

> > > > half-octave generator, but doesn't give the usual clue.

> > >

> > > Gene?

>

> > The first column of the adjoint does have a common factor of 2.

> But so do

> > all the others.

>

> I don't know what your usual clue is, but mine is common factors. We

> have

>

> [ a b c d]

> det [ -5 2 2 -1] = -20a-32b-46c-56d = -2 h10

> [-10 1 0 3]

> [ -3 -1 2 0]

>

> and therefore torsion.

Oh yes, common factors in the left hand column are the definition of

torsion. But I meant the clue for the period not being the octave.

That's usually common factors in the second column, but not always. You

can set 7:5 to be a unison vector with octave equivalent matrices, and the

right mapping comes out, but no suggestion that the period is a

half-octave.

In the example above, all columns have a common factor of 2, so I divided

through by it before checking for divisions of the octave. This is

because of some inefficient unison vectors I have for the multiple-29

temperament that produce common factors of a power of 29 in all columns.

Divide through by those common factors, and you get the right answer.

It all depends on whether we take the adjoint matrix, or the inverse

multiplied by the lowest common denominator. It now looks like the

former, although I assumed the latter was simpler.

I've tightened up the program so that it doesn't need this clue about the

octave divisions. I also have better unison vectors for the multiple-29

case. I haven't studied the results in detail, but it's mostly working.

<http://x31eq.com/vectors.py>

<http://x31eq.com/vectors.out>

<http://x31eq.com/vectors.paul.py>

<http://x31eq.com/vectors.paul.out>

I'm doing additional calculations without needing the chromatic unison

vector. The problem is with octave equivalent matrices. It would be nice

to do all the calculations with them. The rules are:

1) Form a matrix with the chromatic unison vector in the top row, and

commatic unison vectors in the others.

2) The gcd of the left hand column of the adjoint is the number of equal

divisions of the octave.

3) Divide through by this and you have the mapping by generators modulo

the period.

These work fine as long as the unison vectors are well behaved. It'd be

nice if you could come up with a mathematical definition of this. But it

looks like they really have to be approximate unisons, rather than

approximate equal divisions of the octave (which don't work in the

octave-specific case anyway). And it has to be possible to specify all

consonant intervals as combinations of (commatic?) unison vectors.

Once the mapping's chosen, it should be easy to find the ET for which the

chromatic unison vector approximates to 0 steps, although I haven't worked

out the algorithm for that yet.

So, Paul's hypothesis can be made into a conjecture by saying "the above

method always works". I'm fairly confident that it will this time. All

we need is to convert it into mathematical language and prove it. Then we

can try submitting a paper to Perspectives of New Music.

Graham

In-Reply-To: <memo.305965@cix.compulink.co.uk>

Corrections already

> These work fine as long as the unison vectors are well behaved. It'd

> be nice if you could come up with a mathematical definition of this.

> But it looks like they really have to be approximate unisons, rather

> than approximate equal divisions of the octave (which don't work in the

> octave-specific case anyway). And it has to be possible to specify all

> consonant intervals as combinations of (commatic?) unison vectors.

The second criterion isn't right. Firstly, the unison vectors need to be

linearly independent. Secondly, taking each column independently, it must

be possible to combine the unison vectors to get any number you want in

there. Getting any consonant interval isn't possible unless you have a 1

note periodicity block. But this condition should remove torsion.

Graham

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

> Oh yes, common factors in the left hand column are the definition

of

> torsion. But I meant the clue for the period not being the

octave.

> That's usually common factors in the second column, but not

always. You

> can set 7:5 to be a unison vector with octave equivalent matrices,

and the

> right mapping comes out, but no suggestion that the period is a

> half-octave.

> In the example above, all columns have a common factor of 2, so I

divided

> through by it before checking for divisions of the octave.

All columns of what have a common factor of 2? I get

[ 1 0 0 0] [-20 0 0 0]

adj [ -5 2 2 -1] = [-32 -6 -2 -6]

[-10 1 0 3] [-46 -3 -1 -7]

[ -3 -1 2 0] [-56 2 -6 -2]

This has one column in the first matrix and one column in the second

divisible by 2. I don't see how a half-octave period even comes into

it.

> 1) Form a matrix with the chromatic unison vector in the top row,

and

> commatic unison vectors in the others.

> 2) The gcd of the left hand column of the adjoint is the number of

equal

> divisions of the octave.

You've lost me completely here. Left hand column of the adjoint of

what matrix is supposed to do this? It can hardly be the matrix of

unison vectors, with or without 2 included in the picture. Do you

maybe mean you divide through by the gcd to get the number of

divisions?

I don't see why you don't simply do what I do, and go for the whole

val, not just h(2). The number of divisions can be deceptive.

> These work fine as long as the unison vectors are well behaved.

It'd be

> nice if you could come up with a mathematical definition of this.

I thought I did.

In-Reply-To: <9ptp2v+3ilo@eGroups.com>

Gene wrote:

> > In the example above, all columns have a common factor of 2, so I

> divided

> > through by it before checking for divisions of the octave.

>

> All columns of what have a common factor of 2? I get

>

> [ 1 0 0 0] [-20 0 0 0]

> adj [ -5 2 2 -1] = [-32 -6 -2 -6]

> [-10 1 0 3] [-46 -3 -1 -7]

> [ -3 -1 2 0] [-56 2 -6 -2]

No, not that one.

> This has one column in the first matrix and one column in the second

> divisible by 2. I don't see how a half-octave period even comes into

> it.

It has torsion, but not a half-octave period.

I meant this one:

[(-3, 0, 1),

(0, 2, -2),

(1, 0, 3)]

Octave specifically,

[ 1 0 0 0]

[ 2 -3 0 1]

[ 1 0 2 -2]

[-10 1 0 3]

adjoint

[-20 0 0 0]

[-32 6 0 -2]

[-46 -2 -10 -6]

[-56 -2 0 -6]

> > 1) Form a matrix with the chromatic unison vector in the top row,

> and

> > commatic unison vectors in the others.

>

> > 2) The gcd of the left hand column of the adjoint is the number of

> equal

> > divisions of the octave.

>

> You've lost me completely here. Left hand column of the adjoint of

> what matrix is supposed to do this? It can hardly be the matrix of

> unison vectors, with or without 2 included in the picture. Do you

> maybe mean you divide through by the gcd to get the number of

> divisions?

Without 2 (this method is all for octave equivalent matrices). It's the

same as the gcd you take here:

"""

(4) 14/12 = 7/6 = 1+1/6, and the convergent to 7/6 is 1. We therefore

want A + B = 93.651 cents as our generator; this happens to be very

close to 1200/13 = 92.308 cents; not much of a surprise since

g+h12=h26 and gcd(12,14)=2. We therefore can use 1200/13 as our

generator, with a period of half an octave.

"""

GCD of 2, half octave period.

> I don't see why you don't simply do what I do, and go for the whole

> val, not just h(2). The number of divisions can be deceptive.

Why don't I do what? I'm using octave equivalent matrices for the

conjecture because one step using octave specific matrices involves

solving what I think could be called a (very simple) system of Diophantine

equations with only 1 unknown. Proving that that always works sounds

harder than proving the octave-specific case.

In fact, the number of notes in the periodicity block is the generator

mapping of the chromatic unison vector. From the definition of the

adjoint matrix, this will always be the same as the determinant. So we

can ignore the determinant if we really want to. But the result we get is

:

The number of periods to the octave

The mapping of generators to prime intervals

The number of notes in the periodicity block

Hopefully, we've defined away torsion. It'd be nice if we could get the

number of steps to a generator in the equivalent ET. Did we have an

algorithm for generating the periodicity block in order of pitch for

octave equivalent vectors?

> > These work fine as long as the unison vectors are well behaved.

> It'd be

> > nice if you could come up with a mathematical definition of this.

>

> I thought I did.

Did you? Where?

Graham