Tuning-Math Digests messages 3275 - 3299

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Message: 3275

Date: Tue, 15 Jan 2002 21:40:32

Subject: Re: algorithm sought

From: clumma

>>Can't it always just be 60 degrees?  There is that point where
>>this is no longer the closest packing -- that's bad I presume...
>>Paul once posted something from Mathworld...
>
>That's exactly what I've been talking about and presenting for
>the last few months!

Wow, sorry I've been so dense.  There's a big hole in my algebra
where polynomials of degree >2 should be.  I was amazed in 10th
grade when my questions about cubic and higher polynomials were
explicitly dismissed, and I haven't had algebra since.

>For the 5-limit, it is
> 
> ||3^a 5^b|| = sqrt(a^2 + ab + b^2)
> 
> For the 7-limit
> 
> ||3^a 5^b 7^c|| = sqrt(a^2 + b^2 + c^2 + ab + ac + bc)

Is there a post where the derivation for this is given?

>Beyond that we need to decide if 3 stays the same size as 5, 7,
>and 11, or is half as long.

You can expect me to become completely confused when dealing
with prime limits.  I've spent all my time thinking in odd-limits,
and frankly I think all the stuff having to do with music supports
this.  Why bother with different lengths -- unless you're trying
to weight for consonance, which is a different matter?  It is true
that log(n) lengths unifies the taxicab metric for odd- and prime-
limits.  I understand that.  But I still want to visualize my
9-edges apart from my 3-edges.  Musically, I don't care if the
polynomials are easier to factor with prime coefficients.  I want
the option of keeping my lengths unweighted, and thinking in the
9-limit.  Can your distance stuff be adapted to odd limits?

>Forget the densest packing, we want the packing corresponding to
>octave classes of intervals.

What does this mean?

-Carl


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Message: 3276

Date: Tue, 15 Jan 2002 21:40:46

Subject: Re: more graph theory terminology

From: paulerlich

--- In tuning-math@y..., "genewardsmith" <genewardsmith@j...> wrote:
> --- In tuning-math@y..., "clumma" <carl@l...> wrote:
> 
> > So it seems a 1-diameter subgraph is also called a clique? 
> 
> Unless the author defines clique to be maximal. The cliques of a  
>graph give us consonant chords in the sense I've been using it on 
>the masses of asses thread.

Look up "saturated" in Monz' dictionary -- perhaps this is relevant?


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Message: 3277

Date: Wed, 16 Jan 2002 03:21:54

Subject: Re: algorithm sought

From: genewardsmith

--- In tuning-math@y..., "clumma" <carl@l...> wrote:

> >>>Hence both 9/5 and 9/7 are consonant with 1.
> >> 
> >>But they should have the same distance from 1 as 3/2,
> >>or any other tonality-diamond pitch, no?

In fact, ||3/2|| = 1, not 2.

> >Why? Certainly if we were picky about it, we'd want a precise
> >length to be associated with each interval, and whether these are
> >equal or un-equal, we'd almost certainly be in non-Euclidean space
> >pretty fast. 

No, everthing gets a length and it is all Euclidean.

> Okay, I guess I was just thinking too taxicab-ish.  But I'm a
> long way from understanding this stuff.  What can it do that
> a
taxicab metric can't?

It's not trying to be a consonance measure; what it does is the same
kind of stuff as the lattice diagrams with triangles and tetrahedra
and what not. These are called the "root lattices" A2 and A3, and have
corresponding bilinear forms or "Gram matrices"; it can be generalized
to An but then we need to decide what to do about the prime power
problem. My o-limit quadratic forms correspond to lattices which are
unions of translates of An.

By taking things a certain fixed distance from points, we get various
kinds of things such as diamonds, hexanies and so forth; I was
starting to explain this once but got the impression people understood
it in their own way. One nice thing about the lattice point of view is
that these things can be enumerated via theta functions, which may be
worth explaining. This is all also connected to the transformation
groups Robert and I were playing with.


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Message: 3278

Date: Wed, 16 Jan 2002 07:16:41

Subject: Re: algorithm sought

From: genewardsmith

--- In tuning-math@y..., "clumma" <carl@l...> wrote:

> I wouldn't think we'd need to break any of the the Euclidean
> axioms, either way.  If you define anything > 3-D non-Euclidean,
> then...

Please don't do that! "Euclidean" means, more or less, affine R^n with inner product--it has R^n as a vector space as a model.

> What's the problem?

Is 9 the same length as 7, or twice as long?

> For radius r, what determines what you get?  Wether you start on
> a vertex or not, etc.?  

Right--it's what the center is--a vertex, a face, and edge, a "deep hole" center such as the center of the octahedron, or what have you.


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Message: 3279

Date: Wed, 16 Jan 2002 03:26:43

Subject: Re: [tuning] Re: badly tuned remote overtones

From: monz

HOLD THE PRESSES!

I wrote:

> From: monz <joemonz@xxxxx.xxx>
> To: <tuning-math@xxxxxxxxxxx.xxx>
> Sent: Tuesday, January 15, 2002 11:10 PM
> Subject: Re: [tuning-math] [tuning] Re: badly tuned remote overtones
>
>  c = 12*4 = 48     c =  8*6 = 48
>                                      b =  5*9 =  45
>  b = 11*4 = 44
>                   (bb=  7*6 = 42)
>  a = 10*4 = 40
>
> ...
>
> These are all the unison-vectors implied by Schoenberg's diagram:
> 
> E  5*6=30 : Eb 4*7=28  =  15:14
> B 11*4=44 : Bb 7*6=42  =  22:21
> B  5*9=45 : B 11*4=44  =  45:44
> B  5*9=45 : Bb 7*6=42  =  15:14
> F 16*4=64 : F  7*9=63  =  64:63
> F 11*6=66 : F 16*4=64  =  33:32
> F 11*6=66 : F  7*9=63  =  22:21
> A  9*9=81 :(A 20*4=80) =  81:80
> C 11*9=99 :(C 24*4=96) =  33:32 



I've been writing again and again about the inconsistency
of Scheonberg's notation of the 11th harmonic in the diagram
on p 24 in the Carter translation.  Well, I was just stunned
to find out that in the original 1911 German edition, the
11th harmonic of F is written as "b", *which in German means
b-flat*! (what we call "b" is written as "h" in German). 
It has been silently changed in the Carter translation,
and so my writings on Schoenberg's theories must all be
revised.

So in the above diagram, 22:21 is both a unison-vector
(F 11*6=66 : F  7*9=63) and a chromatic-vector used as
a unison-vector (B 11*4=44 : Bb 7*6=42), and 45:44 is
a unison vector (B  5*9=45 : B 11*4=44) -- this is incorrect.

The note I've been giving as B 11*4=44 must be called
Bb 11*4=44.  This has the effect of changing 45:44 into
a chromatic-vector, and making 22:21 consistently a
unison-vector.



Since Schoenberg spends a good deal of space in
_Harmonielehre_ on ways to use the 12-EDO scale to
construct harmonies that lie outside traditional tonal
theory, we must assume that he intends to distinguish
pairs of notes that are separated by approximately a
semitone, so that the 15:14 chromatic-vector cannot be
considered a unison-vector.  Therefore, the total list
of unison-vectors implied by Schoenberg's 1911 diagram is:

Bb 11*4=44 : Bb 7*6=42  =  22:21
F  16*4=64 : F  7*9=63  =  64:63
F  11*6=66 : F 16*4=64  =  33:32
F  11*6=66 : F  7*9=63  =  22:21
A   9*9=81 :(A 20*4=80) =  81:80
C  11*9=99 :(C 24*4=96) =  33:32 

But because 22:21, 33:32, and 64:63 form a dependent triplet
(any one of them can be found by multiplying the other two),
this does not suffice to create a periodicity-block, which
needs another independent unison-vector.

So I will give Paul Erlich the benefit of the doubt
and assume that Schoenberg was following tradition as
closely as possible in his note naming, by thinking in
terms of meantone, and therefore the most likely candidate
for the other independent UV is the 5-limit diesis
128:125 = [ 7  0 -3 ] .

So ...

 matrix

   2  3  5  7 11   unison-vectors  ~cents   

[  1  0  0  0  0 ]  =    2:1     0   
[  7  0 -3  0  0 ]  =  128:125  41.05885841   
[ -5  1  0  0  1 ]  =   33:32   53.27294323   
[  6 -2  0 -1  0 ]  =   64:63   27.2640918   
[ -4  4 -1  0  0 ]  =   81:80   21.5062896   


fractional inverse      

[ 12  0   0   0   0 ]
[ 19 -1   0   0   3 ] 
[ 28 -4   0   0   0 ] 
[ 34  2   0 -12  -6 ] 
[ 41  1  12   0  -3 ]
 
determinant = | 12 |


The only cardinality given by this matrix is 12,
with the following mapping:

h12(2)
h12(
h12(
h12(
h12(



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Message: 3280

Date: Wed, 16 Jan 2002 05:03:35

Subject: Iterated CPS

From: genewardsmith

If S is a set of octave-equivalence classes, we can denote by CPS(S)
the set of ratios of these classes, and we then can iterate this
function, giving us CPS^2(S) = CPS(CPS(S)) and so forth. In octave
reduced form, starting from the diminished 7th tetrad of the 31-et,
the following would be an example (where "0" denotes the absence of an
11-limit consonance):


[0, 7, 15, 23]
edges   3   6   6   6   connectivity   1   3   3   3

[1, 7/6, 7/5, 5/3]
[7/6, 1, 6/5, 10/7]
[7/5, 6/5, 1, 6/5]
[5/3, 10/7, 6/5, 1]


[0, 7, 8, 15, 16, 23, 24]
edges   6   15   15   18   connectivity   1   3   3   5

[1, 7/6, 6/5, 7/5, 10/7, 5/3, 12/7]
[7/6, 1, 0, 6/5, 11/9, 10/7, 16/11]
[6/5, 0, 1, 7/6, 6/5, 7/5, 10/7]
[7/5, 6/5, 7/6, 1, 0, 6/5, 11/9]
[10/7, 11/9, 6/5, 0, 1, 7/6, 6/5]
[5/3, 10/7, 7/5, 6/5, 7/6, 1, 0]
[12/7, 16/11, 10/7, 11/9, 6/5, 0, 1]


[0, 1, 7, 8, 9, 14, 15, 16, 17, 22, 23, 24, 30]
edges   19   46   48   63   connectivity   2   6   6   9

[1, 0, 7/6, 6/5, 11/9, 11/8, 7/5, 10/7, 16/11, 18/11, 5/3, 12/7, 0]
[0, 1, 8/7, 7/6, 6/5, 4/3, 11/8, 7/5, 10/7, 8/5, 18/11, 5/3, 0]
[7/6, 8/7, 1, 0, 0, 7/6, 6/5, 11/9, 5/4, 7/5, 10/7, 16/11, 5/3]
[6/5, 7/6, 0, 1, 0, 8/7, 7/6, 6/5, 11/9, 11/8, 7/5, 10/7, 18/11]
[11/9, 6/5, 0, 0, 1, 9/8, 8/7, 7/6, 6/5, 4/3, 11/8, 7/5, 8/5]
[11/8, 4/3, 7/6, 8/7, 9/8, 1, 0, 0, 0, 6/5, 11/9, 5/4, 10/7]
[7/5, 11/8, 6/5, 7/6, 8/7, 0, 1, 0, 0, 7/6, 6/5, 11/9, 7/5]
[10/7, 7/5, 11/9, 6/5, 7/6, 0, 0, 1, 0, 8/7, 7/6, 6/5, 11/8]
[16/11, 10/7, 5/4, 11/9, 6/5, 0, 0, 0, 1, 9/8, 8/7, 7/6, 4/3]
[18/11, 8/5, 7/5, 11/8, 4/3, 6/5, 7/6, 8/7, 9/8, 1, 0, 0, 6/5]
[5/3, 18/11, 10/7, 7/5, 11/8, 11/9, 6/5, 7/6, 8/7, 0, 1, 0, 7/6]
[12/7, 5/3, 16/11, 10/7, 7/5, 5/4, 11/9, 6/5, 7/6, 0, 0, 1, 8/7]
[0, 0, 5/3, 18/11, 8/5, 10/7, 7/5, 11/8, 4/3, 6/5, 7/6, 8/7, 1]


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Message: 3281

Date: 16 Jan 2002 23:59:34 -080

Subject: optimal? (heuristic, kees stuff)

From: paul@xxxxxxxxxxxxx.xxx

I think the premise of calculating the optimal tuning for a given error
measure, for use with that error measure, may be flawed, in the
context of my heuristic. The tempering process may be a
non-optimizing, but still quite simple and reasonable, one . . .
hopefully the argument by which I derived the heuristic will spark
someone's mind . . .

Here it is again:

*******************************************************************

The thinking goes as follows. The "length" of a unison vector 
n/d, n~=d, in the Tenney lattice with taxicab metric, or van Prooijen 
lattice with triangular-taxicab metric, is proportional to log(n) + 
log(d) (hence approx. proportional to log(d)), and also to 
the "number" (in some weighted sense) of consonant intervals making 
up that unison vector. Thus, in order to temper this unison vector 
out (assuming that other UVs being tempered out, if any, are 
orthogonal to this one), one must temper each consonant interval 
involved by an "average" amount proportional to w/log(d), where w is 
the musical width of the unison vector.

w=log(n/d)
w~=n/d-1
w~=(n-d)/d

Hence the amount of tempering implied by the unison vector is approx. 
proportional to

(n-d)/(d*log(d))

Yes?

*******************************************************************


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Message: 3282

Date: Wed, 16 Jan 2002 03:43:28

Subject: ERROR IN CARTER'S SCHOENBERG (Re: badly tuned remote overtones)

From: monz

HOLD THE PRESSES!

I wrote:

> From: monz <joemonz@xxxxx.xxx>
> To: <tuning-math@xxxxxxxxxxx.xxx>
> Sent: Tuesday, January 15, 2002 11:10 PM
> Subject: Re: badly tuned remote overtones
>
>
>  c = 12*4 = 48     c =  8*6 = 48
>                                      b =  5*9 =  45
>  b = 11*4 = 44
>                   (bb=  7*6 = 42)
>  a = 10*4 = 40
>  g =  9*4 = 36     g =  6*6 = 36     g =  4*9 =  36
>
> ...
>
> These are all the unison-vectors implied by Schoenberg's diagram:
> 
> E  5*6=30 : Eb 4*7=28  =  15:14
> B 11*4=44 : Bb 7*6=42  =  22:21
> B  5*9=45 : B 11*4=44  =  45:44
> B  5*9=45 : Bb 7*6=42  =  15:14
> F 16*4=64 : F  7*9=63  =  64:63
> F 11*6=66 : F 16*4=64  =  33:32
> F 11*6=66 : F  7*9=63  =  22:21
> A  9*9=81 :(A 20*4=80) =  81:80
> C 11*9=99 :(C 24*4=96) =  33:32 



I've been writing again and again about the inconsistency
of Scheonberg's notation of the 11th harmonic in the diagram
on p 24 in the Carter translation.  Well, I was just stunned
to find out that in the original 1911 German edition, the
11th harmonic of F is written as "b", *which in German means
b-flat*! (what we call "b" is written as "h" in German). 

It has been changed *without comment* in the Carter translation
(or perhaps it's simply an easily-overlooked error), and
so my writings on Schoenberg's theories must all be revised.

The corrupt diagram appears on p 24 of Carter's 1978 English
translation, published as _Theory of Harmony_.  The original
diagram is on p 23 of the 1911 edition of _Harmonielehre_.


So in the above diagram, 22:21 is both a unison-vector
(F 11*6=66 : F  7*9=63) and a chromatic-vector used as
a unison-vector (B 11*4=44 : Bb 7*6=42), and 45:44 is
a unison vector (B  5*9=45 : B 11*4=44) -- this is incorrect.

The note I've been giving as B 11*4=44 must be called
Bb 11*4=44.  This has the effect of changing 45:44 into
a chromatic-vector, and making 22:21 consistently a
unison-vector.



Since Schoenberg spends a good deal of space in
_Harmonielehre_ on ways to use the 12-EDO scale to
construct harmonies that lie outside traditional tonal
theory, we must assume that he intends to distinguish
pairs of notes that are separated by approximately a
semitone, so that the 15:14 chromatic-vector cannot be
considered a unison-vector.  Therefore, the total list
of unison-vectors implied by Schoenberg's 1911 diagram is:

Bb 11*4=44 : Bb 7*6=42  =  22:21
F  16*4=64 : F  7*9=63  =  64:63
F  11*6=66 : F 16*4=64  =  33:32
F  11*6=66 : F  7*9=63  =  22:21
A   9*9=81 :(A 20*4=80) =  81:80
C  11*9=99 :(C 24*4=96) =  33:32 


But because 22:21, 33:32, and 64:63 form a dependent triplet
(any one of them can be found by multiplying the other two),
this does not suffice to create a periodicity-block, which
needs another independent unison-vector.


So I will give Paul Erlich the benefit of the doubt
and assume that Schoenberg was following tradition as
closely as possible in his note naming, by thinking in
terms of meantone, and therefore the most likely candidate
for the other independent UV is the 5-limit diesis
128:125 = [ 7  0 -3 ] .

So ...

 matrix

   2  3  5  7 11   unison-vectors  ~cents   

[  1  0  0  0  0 ]  =    2:1     0   
[  7  0 -3  0  0 ]  =  128:125  41.05885841   
[ -5  1  0  0  1 ]  =   33:32   53.27294323   
[  6 -2  0 -1  0 ]  =   64:63   27.2640918   
[ -4  4 -1  0  0 ]  =   81:80   21.5062896   


fractional inverse      

[ 12  0   0   0   0 ]
[ 19 -1   0   0   3 ] 
[ 28 -4   0   0   0 ] 
[ 34  2   0 -12  -6 ] 
[ 41  1  12   0  -3 ]
 
determinant = | 12 |


The only cardinality given by this matrix is 12,
with the following mapping:

h12(2)  =  12
h12(3)  =  19
h12(5)  =  28
h12(7)  =  34
h12(11) =  41


So with references of "F", "C", and "G",
our 12-EDO mapping is:

2  3  5  7  11

G  D  B  F   C  
C  G  E  Bb  F
F  C  A  Eb  Bb

and 11 is given consistently as the 12-EDO "4th" above
the fundamental, and never as the "augmented 4th", as
Carter's edition implies.



-monz



 




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Message: 3283

Date: Wed, 16 Jan 2002 05:16:42

Subject: Re: algorithm sought

From: clumma

>>But they should have the same distance from 1 as 3/2,
>>or any other tonality-diamond pitch, no?
>
> In fact, ||3/2|| = 1, not 2.

This was my objection.
 
>>Why? Certainly if we were picky about it, we'd want a precise
>>length to be associated with each interval, and whether these
>>are equal or un-equal, we'd almost certainly be in non-Euclidean
>>space pretty fast. 
> 
> No, everthing gets a length and it is all Euclidean.

I wouldn't think we'd need to break any of the the Euclidean
axioms, either way.  If you define anything > 3-D non-Euclidean,
then...

>>Okay, I guess I was just thinking too taxicab-ish.  But I'm a
>>long way from understanding this stuff.  What can it do that
>>a taxicab metric can't?
> 
>It's not trying to be a consonance measure; what it does is the
>same kind of stuff as the lattice diagrams with triangles and
>tetrahedra and what not. These are called the "root lattices" A2
>and A3, and have corresponding bilinear forms or "Gram matrices";
>it can be generalized to An but then we need to decide what to do
>about the prime power problem. My o-limit quadratic forms
>correspond to lattices which are unions of translates of An.

What's the problem?

>By taking things a certain fixed distance from points, we get
>various kinds of things such as diamonds, hexanies and so forth;
>I was starting to explain this once but got the impression people
>understood it in their own way. One nice thing about the lattice
>point of view is that these things can be enumerated via theta
>functions, which may be worth explaining. This is all also
>connected to the transformation groups Robert and I were playing
>with.

For radius r, what determines what you get?  Wether you start on
a vertex or not, etc.?  I believe your stuff works, but I'm not
clear on how to use it to turn out what I want.

-Carl


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Message: 3284

Date: Wed, 16 Jan 2002 12:54 +0

Subject: Re: algorithm sought

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <a21q6u+hlkj@xxxxxxx.xxx>
Me:
> >I did work out the general, multidimensional case, but I don't
> >have the result to hand.  Probably, something like
> >
> > x' = x + y*cos(theta) + z*cos(theta)
> > y' = y*sin(theta) + z*cos(theta)
> > z' = z*sin(phi)
> > 
> >will work.  I don't know how to get from theta, the angle between
> >axes, to phi, the angle between the z axis and the x-y plane.
> >Certainly not in general.

Carl:
> Can't it always just be 60 degrees?  There is that point where
> this is no longer the closest packing -- that's bad I presume...
> Paul once posted something from Mathworld...

That's what Gene seems to have done, so go with that.  I was thinking of 
the general case with a free parameter that can be set to give either 
triangular or rectangular lattice, or anything in between.

Me:
> >Oh, for the algorithm, trying all combinations of consonances
> >above the tonic should work.  That'll be O(n**m) where m is the
> >number of notes in the chord, but shouldn't be a problem for the
> >kind of numbers we're talking about.


Carl:
> What's the double-star?  Combinations of notes or intervals?
> Notes gives you CPSs, and intervals don't get all the chords
> because some chords contain more than one instance of an
> interval.  Plus, order matters, at least for lists of 2nds,
> so we wind up with the procedure I described when complaining
> about my lack of a scheme compiler.

** is Fortran for exponentiation.  Actually, that formula's wrong, but 
probably isn't important anyway.  The number of consonances is more 
significant than the number of notes in the chords.

I'm guessing notes (that is, intervals relative to the 1/1) will give a 
more manageable calculation.  That also means it is the usual combinations 
algorithm, which I posted before in Python, and should be easier in 
Scheme.

Example: 5-limit

Consonances are 1/1, 5/4, 6/5, 4/3, 3/2, 8/5, 5/3.  1/1 is redundant, so 
take all pairs of the others.

1/1 5/4 6/5 contains an interval of 25:24, so throw it out.

1/1 5/4 4/3 contains an interval of 16:15, so throw it out.

1/1 5/4 3/2 is consonant

1/1 5/4 8/5 contains, what, 32:25?  So throw that out.

1/1 5/4 5/3 is consonant

1/1 6/5 4/3 contains an interval of 10:9, so throw it out.

and so on.

The consonance checking is also a combinations problem: take each pair of 
notes.  You can also exclude the 1/1, because everything's consonant 
relative to that, so we're only choosing 1 from 1 in the 5-limit.

These calculations are laborious to do by hand, but shouldn't be a problem 
for a computer.


                Graham


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Message: 3285

Date: Wed, 16 Jan 2002 13:57 +0

Subject: Re: badly tuned remote overtones

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <004401c19e0a$eba97f20$af48620c@xxx.xxx.xxx>
monz wrote:

> inverse    
> 
> [ 12  0   0   0  0 ]
> [ 19 -1   0   0  2 ]      1
> [ 28 -4   0   0 -4 ]  *  --
> [ 34  2   0 -12 -4 ]     12
> [ 41  1  12   0 -2 ]
> 
> 
> OK, so I can see that the first column vector of the
> inverse gives the 12-EDO homomorphism:

...

> But now what do all those other column vectors mean?
> They all start with 0 ... what does that mean?

Starting with 0 means they'll do as octave-equivalent generator mappings.  
The first is some kind of meantone because it starts with -1 and -4, so a 
third is four fifths.  Then the 2 means that 7:4 approximates as a minor 
seventh, rather than an augmented sixth.  That sounds right for 
Schoenberg's assumptions.  Then the 1 means that 11:8 approximates the 
same as 4:3.  You could check Arnie's spelling against these, to see if he 
really did need enharmonic equivalence.

The other columns will be other temperaments.  Looks like two that have a 
pair of 12-equal scales a comma apart to get better 7:4 or 11:8 
approximations.  And than a temperament that divides the octave into two 
equal parts.  Could be diaschismic.  Yes, that's right, an 11-limit 
Paultone, or whatever that's calling itself these days.


                       Graham


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Message: 3286

Date: Wed, 16 Jan 2002 15:27:22

Subject: Re: Hi Dave K.

From: dkeenanuqnetau

--- In tuning-math@y..., "paulerlich" <paul@s...> wrote:
> Dave, perhaps you can help with this Kees lattice business. It's 
pure 
> math, but Gene appears to have no interest in it. It seems to me 
that 
> we should be able, based on Kees's lattice with a 'taxicab' metric, 
> be able to define an error function so that
...

Sorry. No time. But here's how I'd approach it. Set up a spreadsheet 
with the ten examples in 10 rows. Columns would be n and d of ratio to 
be tempered out, and for each (octave-equivalent) interval the error 
in cents and the number of generators required, then the errror and 
gens functions of n and d that you propose. Then I'd try a weighted 
rms error and a weighted rms gens where the weights can be changed for 
each interval (indep. wts for err and gens). I'd calculate the 
least-squares difference between the weighted rms results and your 
proposed functions of n and d. Then I'd fool around with the error 
weights to minimise the error in the errors and fool around with the 
gens weights to minimise the error in the gens. I hope this makes 
sense.


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Message: 3287

Date: Wed, 16 Jan 2002 15:37:18

Subject: Re: tuning dictionaries vs math dictionaries

From: dkeenanuqnetau

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:
> My whole intention is to understand, and help others
> to understand, the work that's gone on at tuning-math
> for the last several months.  Gene, Paul, Graham, and
> Dave are the only members posting who seem to follow it.

I'm not following it any more. They lost me somewhere back before 
"torsion".


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Message: 3288

Date: Wed, 16 Jan 2002 09:37:00

Subject: Re: badly tuned remote overtones

From: monz

> From: <graham@xxxxxxxxxx.xx.xx>
> To: <tuning-math@xxxxxxxxxxx.xxx>
> Sent: Wednesday, January 16, 2002 5:57 AM
> Subject: [tuning-math] Re: badly tuned remote overtones
>
>
> In-Reply-To: <004401c19e0a$eba97f20$af48620c@xxx.xxx.xxx>
> monz wrote:
> 
> > inverse    
> > 
> > [ 12  0   0   0  0 ]
> > [ 19 -1   0   0  2 ]      1
> > [ 28 -4   0   0 -4 ]  *  --
> > [ 34  2   0 -12 -4 ]     12
> > [ 41  1  12   0 -2 ]
> > 
> > 
> > OK, so I can see that the first column vector of the
> > inverse gives the 12-EDO homomorphism:
> 
> ...
> 
> > But now what do all those other column vectors mean?
> > They all start with 0 ... what does that mean?
> 
> Starting with 0 means they'll do as octave-equivalent
> generator mappings.  The first is some kind of meantone
> because it starts with -1 and -4, so a third is four fifths. ...


I noticed that myself ... thanks, Graham!


> ...  Then the 2 means that 7:4 approximates as a minor 
> seventh, rather than an augmented sixth.  That sounds right
> for Schoenberg's assumptions.  Then the 1 means that
> 11:8 approximates the same as 4:3.  You could check Arnie's
> spelling against these, to see if he really did need
> enharmonic equivalence.


OK, I noticed these also, since I figured out that this column
was some sort of meantone, but didn't really think about the
"minor 7th vs. augmented 6th" question ... I simply saw that
it gave the numbers of a typical linear mapping but with the
signs reversed.  Thanks for that!

If you see my last post (ERROR IN CARTER'S SCHOENBERG) you'll
see that Schoenberg did indeed consistently spell the note
which represents the 11th harmonic the same as that which
represents 4:3.  This was big news to me, because the English
translation has a misprint in the diagram, and I've based years
of research on that error, until today.


> The other columns will be other temperaments.  ...


OK, I assumed that too, after seeing that the first two
columns both gave temperaments.  Thanks for confirmation.


> ... Looks like two that have a pair of 12-equal scales a
> comma apart to get better 7:4 or 11:8 approximations.  And
> than a temperament that divides the octave into two equal
> parts.  Could be diaschismic.  Yes, that's right, an 11-limit 
> Paultone, or whatever that's calling itself these days.


Now *those* are things I'd never figure out.  I'd be
interested in knowing more.  But please base analysis on
what I now think is the most appropriate unison-vector matrix:


 matrix

   2  3  5  7 11   unison-vectors  ~cents   

[  1  0  0  0  0 ]  =    2:1     0   
[  7  0 -3  0  0 ]  =  128:125  41.05885841   
[ -5  1  0  0  1 ]  =   33:32   53.27294323   
[  6 -2  0 -1  0 ]  =   64:63   27.2640918   
[ -4  4 -1  0  0 ]  =   81:80   21.5062896   


fractional inverse      

[ 12  0   0   0   0 ]
[ 19 -1   0   0   3 ] 
[ 28 -4   0   0   0 ] 
[ 34  2   0 -12  -6 ] 
[ 41  1  12   0  -3 ]
 
determinant = | 12 |


And when I wrote after this:

> The only cardinality given by this matrix is 12,


I was wrong about that, correct?  As Graham shows,
the other columns with all those zeros do indeed
express temperament mappings.



-monz




 




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Message: 3289

Date: Wed, 16 Jan 2002 11:14:23

Subject: Fw: copy of tuning-math post I can't send now (Schoenberg)

From: monz

Graham couldn't access the list, so he sent me this reply.


> From: Graham Breed <graham.breed@xxxx.xx.xx>
> To: <joemonz@xxxxx.xxx>
> Sent: Wednesday, January 16, 2002 10:11 AM
> Subject: copy of tuning-math post I can't send now
>
>
> monz wrote:
>
> > If you see my last post (ERROR IN CARTER'S SCHOENBERG) you'll
> > see that Schoenberg did indeed consistently spell the note
> > which represents the 11th harmonic the same as that which
> > represents 4:3.  This was big news to me, because the English
> > translation has a misprint in the diagram, and I've based years
> > of research on that error, until today.
>
> That's consistent 12-equal spelling, though.  The matrix suggests a
> consistent meantone spelling, which is stricter.  That is, 7:4 is always a
> minor seventh and never an augmented sixth.
>
> > > The other columns will be other temperaments.  ...
> >
> >
> > OK, I assumed that too, after seeing that the first two
> > columns both gave temperaments.  Thanks for confirmation.
>
> Each column matches up with a row from the original matrix.  The
temperament
> shown by the column doesn't temper out the comma shown by the relevant
row.
> So the meantone column is what you get by not tempering out an enharmonic
> diesis.
>
> > > ... Looks like two that have a pair of 12-equal scales a
> > > comma apart to get better 7:4 or 11:8 approximations.  And
> > > than a temperament that divides the octave into two equal
> > > parts.  Could be diaschismic.  Yes, that's right, an 11-limit
> > > Paultone, or whatever that's calling itself these days.
> >
> >
> > Now *those* are things I'd never figure out.  I'd be
> > interested in knowing more.  But please base analysis on
> > what I now think is the most appropriate unison-vector matrix:
>
> ...
>
> > fractional inverse
> >
> > [ 12  0   0   0   0 ]
> > [ 19 -1   0   0   3 ]
> > [ 28 -4   0   0   0 ]
> > [ 34  2   0 -12  -6 ]
> > [ 41  1  12   0  -3 ]
> >
> > determinant = | 12 |
>
> See the third column has all zeros except for a 12 right at the bottom.
> That means, trivially, it has a greatest common divisor of 12.  (We don't
> count zeros in the gcd.)  As a rule of thumb, the GCD of a column tells
you
> how many *equal* steps the equivalence interval is being divided into.  In
> this case, we're dividing the octave into 12 equal steps.
>
> Dividing the octave this way is the same as defining a new period to be a
> fraction of the original equivalence interval.  Divide the whole column
> through by the GCD, and you get the mapping within the period.  In this
> case, that gives [0 0 0 0 1].  The first zero tells you that the octave
> takes the same value as it does in 12-equal.  Not much of a surprise.  The
> next four zeros tell you that 3:1, 5:1 and 7:1 are also taken from
12-equal.
> And the 1 in the last column tells you that 11:1 is one generator step
away
> from it's value in 12-equal.  For 11:1 to be just, you'd have a 51 cent
> generator.
>
> This is a fairly trivial example, and not much use as a temperament.  But
> you could realise it by having two keyboards tuned a quartertone apart.
To
> play an 11-limit otonality, you'd have C-E-G-Bb-D on one keyboard, and F
on
> the other.  There are some more respectable temperaments that work this
way.
> Like my multiple-29, where 1 and 3 (and 9) come from 1 keyboard, and 5, 7,
> 11 and 13 (and 15) from the other.
>
> The next column is the same idea, but it's 7:1 that's on the second
> keyboard.
>
> The last column is different now.  I don't recognize it, but it would mean
> dividing the octave into 3 equal parts, and taking 5:4 as an exact
> third-octave, same as it is in 12-equal.  It's the column that doesn't
> temper out the syntonic comma.
>
>
>                          Graham









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Message: 3290

Date: Wed, 16 Jan 2002 11:21:46

Subject: Re: copy of tuning-math post I can't send now

From: monz

Wow, thanks Graham!  This helps a lot!


-monz

----- Original Message -----
From: Graham Breed <graham.breed@xxxx.xx.xx>
To: <joemonz@xxxxx.xxx>
Sent: Wednesday, January 16, 2002 10:11 AM
Subject: copy of tuning-math post I can't send now


> I can't access my home e-mail now, so I can't send this properly.  If I
> don't manage to do so with in the next half hour, you can reply to copy to
> the list.
>
>
> monz wrote:
>
> > If you see my last post (ERROR IN CARTER'S SCHOENBERG) you'll
> > see that Schoenberg did indeed consistently spell the note
> > which represents the 11th harmonic the same as that which
> > represents 4:3.  This was big news to me, because the English
> > translation has a misprint in the diagram, and I've based years
> > of research on that error, until today.  <etc>












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Message: 3291

Date: Wed, 16 Jan 2002 19:39:50

Subject: Re: algorithm sought

From: clumma

>Carl:
>> Can't it always just be 60 degrees?  There is that point where
>> this is no longer the closest packing -- that's bad I presume...
>> Paul once posted something from Mathworld...
> 
>That's what Gene seems to have done, so go with that.  I was
>thinking of the general case with a free parameter that can be
>set to give either triangular or rectangular lattice, or anything
>in between.

Okay.  Thanks for the follow-up.

>I'm guessing notes (that is, intervals relative to the 1/1) will
>give a more manageable calculation.

...than intervals.  You're probably right.  It seems to get around
the problem of having more than one instance of the same thing,
turing n^2 into n!.  But for my ultimate purpose, I'll need to
consider all inversions if I use notes, which I think turns it
back into n^2.

>That also means it is the usual combinations algorithm, which I
>posted before in Python, and should be easier in Scheme.

I didn't see your python post, but here's the scheme:

(define combo
  (lambda (k ls)
     (if (zero? k)
        (list (list))
        (if (null? ls)
           (list)
           (append (combo k (cdr ls))
              (map (lambda (y) (cons (car ls) y))
                 (combo (sub1 k) (cdr ls))))))))

You can't count lines of code in scheme like you can in
a block-structured imperitive language.  And comparing
expressions -- () pairs -- to # of lines isn't fair.  Frankly,
I don't no how they measure algorithms in scheme, but I'll
bet egg's benedict you can't get more compact than this,
considering this exposes some of the actual operations on
the UTM tape (car and cdr).

>The consonance checking is also a combinations problem: take each
>pair of notes.  You can also exclude the 1/1, because everything's
>consonant relative to that, so we're only choosing 1 from 1 in
>the 5-limit.
> 
>These calculations are laborious to do by hand, but shouldn't be
>a problem for a computer.

Without a compiler, I think any of these brute-force methods
are all off limits, so to speak.  Scheme compilers are all mucho
bucks, industrial (though Chez hints at plans to release a
consumer version within the year!), except ones that "compile"
to C, which I've never been able to get to work.  Anyhow, it's
high time I learned another language.

In the mean time, I'm convinced there's a better way than
brute force.  Gene may have found one...

-Carl


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Message: 3292

Date: Wed, 16 Jan 2002 20:02:06

Subject: Re: algorithm sought

From: clumma

>>What's the problem?
> 
>Is 9 the same length as 7, or twice as long?

I suppose I should plug these choices in, and see how the results
differ... having a hard time seeing how there could be uncertainty.
Either a given r exists that encloses all consonances and excludes
everything else, or not.

>>For radius r, what determines what you get?  Wether you start on
>>a vertex or not, etc.?  
> 
>Right--it's what the center is--a vertex, a face, and edge, a
>"deep hole" center such as the center of the octahedron, or what
>have you.

Okay, so now all we need are rules that tell us what we get
for a given r and origin pair.

I don't suppose there's any hope of getting subsets directly
from this method?

-Carl


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Message: 3293

Date: Wed, 16 Jan 2002 20:12:28

Subject: yahoo spaced out (was re: algorithm sought)

From: clumma

What is supposed to be the point of the space-removal
thing?  I can't fathom it.  And "view unformatted message"
does nothing (my eyes, the goggles do nothing!).

Needless to say, my beautiful indenting was destroyed.

> I didn't see your python post, but here's the scheme:
> 
> (define combo
>   (lambda (k ls)
>      (if (zero? k)
>         (list (list))
>         (if (null? ls)
>            (list)
>            (append (combo k (cdr ls))
>               (map (lambda (y) (cons (car ls) y))
>                  (combo (sub1 k) (cdr ls))))))))

-C.


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Message: 3294

Date: Wed, 16 Jan 2002 21:30:35

Subject: Re: badly tuned remote overtones

From: genewardsmith

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> If you see my last post (ERROR IN CARTER'S SCHOENBERG) you'll
> see that Schoenberg did indeed consistently spell the note
> which represents the 11th harmonic the same as that which
> represents 4:3.  This was big news to me, because the English
> translation has a misprint in the diagram, and I've based years
> of research on that error, until today.

I told you that you were the man to sort Schoenberg out. :)

> I was wrong about that, correct?  As Graham shows,
> the other columns with all those zeros do indeed
> express temperament mappings.

They give what I call "vals", but these are ones which define your PB.


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Message: 3295

Date: Wed, 16 Jan 2002 00:11:26

Subject: Re: algorithm sought

From: paulerlich

--- In tuning-math@y..., "genewardsmith" <genewardsmith@j...> wrote:
> --- In tuning-math@y..., "paulerlich" <paul@s...> wrote:
> 
> > > ||3^a 5^b 7^c|| = sqrt(a^2 + 4b^2 + 4c^2 + 2ab + 2ac + 4bc)
> > > 
> > > would be the length of 3^a 5^b 7^c. Everything in a radius of 2 
of 
> > > anything will be consonant.
> 
> > What happens to 9:5 and 9:7?
> 
> ||9/5|| = ||3^2 5^(-)|| = sqrt(4+4+0-4+0+0) = 2
> 
> ||9/7|| = ||3^2 7^(-1)|| = sqrt(4+0+4+0-4+0) = 2
> 
> Hence both 9/5 and 9/7 are consonant with 1.

I'm very impressed, Gene! And there's a Euclidean model for this? 
Will it break down in, say, the 15-limit?


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Message: 3296

Date: Wed, 16 Jan 2002 21:36:31

Subject: Re: algorithm sought

From: genewardsmith

--- In tuning-math@y..., "clumma" <carl@l...> wrote:
> >>What's the problem?
> > 
> >Is 9 the same length as 7, or twice as long?
> 
> I suppose I should plug these choices in, and see how the results
> differ... having a hard time seeing how there could be uncertainty.
> Either a given r exists that encloses all consonances and excludes
> everything else, or not.

If you want the 9-limit, 9 is the same length as 7. If you want the 
7-limit, 9 is twice as long. 

> Okay, so now all we need are rules that tell us what we get
> for a given r and origin pair.
> 
> I don't suppose there's any hope of getting subsets directly
> from
this method?

I'll put it on my list of things to think about. I'm afraid I haven't
really given the problem any thought, partly because at the moment I'm
more interested in tempered versions of it.


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Message: 3297

Date: Wed, 16 Jan 2002 00:12:48

Subject: Re: algorithm sought

From: clumma

>>> ||3^a 5^b 7^c|| = sqrt(a^2 + 4b^2 + 4c^2 + 2ab + 2ac + 4bc)
>>> 
>>>would be the length of 3^a 5^b 7^c. Everything in a radius of 2
>>>of anything will be consonant.
>>
>>What happens to 9:5 and 9:7?
>
> ||9/5|| = ||3^2 5^(-)|| = sqrt(4+4+0-4+0+0) = 2
> 
> ||9/7|| = ||3^2 7^(-1)|| = sqrt(4+0+4+0-4+0) = 2
> 
>Hence both 9/5 and 9/7 are consonant with 1.

But they should have the same distance from 1 as 3/2,
or any other tonality-diamond pitch, no?

-Carl


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Message: 3298

Date: Wed, 16 Jan 2002 21:50:26

Subject: Re: algorithm sought

From: clumma

>>>Is 9 the same length as 7, or twice as long?
>> 
>>I suppose I should plug these choices in, and see how the results
>>differ... having a hard time seeing how there could be uncertainty.
>>Either a given r exists that encloses all consonances and excludes
>>everything else, or not.
> 
>If you want the 9-limit, 9 is the same length as 7. If you want the 
>7-limit, 9 is twice as long. 

So, problem solved.  n is a fair variable.

>>Okay, so now all we need are rules that tell us what we get
>>for a given r and origin pair.
>> 
>>I don't suppose there's any hope of getting subsets directly
>>from this method?
> 
>I'll put it on my list of things to think about. I'm afraid I
>haven't really given the problem any thought, partly because
>at the moment I'm more interested in tempered versions of it.

Cool!  One thing it will allow us to do is subtract out the
natural JI chords and leave only magic chords in your lists
on the main list.

-Carl


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Message: 3299

Date: Wed, 16 Jan 2002 22:07:40

Subject: Re: yahoo spaced out (was re: algorithm sought)

From: dkeenanuqnetau

--- In tuning-math@y..., "clumma" <carl@l...> wrote:
> What is supposed to be the point of the space-removal
> thing?  I can't fathom it.  And "view unformatted message"
> does nothing (my eyes, the goggles do nothing!).

I think Yahoo screwed up and forgot to put </PRE> around 
everything. The workaround is, when formatting matters, tell your 
readers, if they are viewing it on the web they need to click Message 
Index then Expand Messages.


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