Tuning-Math Digests messages 2125 - 2149

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Message: 2125

Date: Tue, 27 Nov 2001 07:52:53

Subject: Standardizing our wedge product

From: genewardsmith@xxxx.xxx

To get into the standization fun over on the other list, I decided we 
could standardize a basis for the 7-limit wedge product, getting the 
product of vals and intervals to correspond. I propose:

Let [u1, u2, u3, u4] and [v1, v2, v3, v4] be two intervals in the 
prime-power basis. Then

[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u1*v2-v1*u2, u1*v3-v1*u3, u1*v4-v1*u4, 
u2*v3-v2*u3, u2*v4-v2*u4, u3*v4-v3*u4]

Let [u1,u2,u3,u4] and [v1,v2,v3,v4] be two vals in the prime-
valuation basis (that is, h12 would be [12,19,28,34] and so forth.) 
Then

[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u3*v4-v3*u4, v2*u4-u2*v4, u2*v3-v2*u3, 
u1*v4-v1*u4, v1*u3-u1*v3, u1*v2-v1*u2]

This now means that a given linear temperament gives the same wedge 
product, up to sign, whether defined in terms of vals or intervals. 
For instance

2401/2400^2375/4374 = h270^h441 = [34, -22, -1, -18, 27, -18]

Since this is an invariant of the temperament, it would be a good 
thing to use to refer to it, but for the fact that it is opaque and 
does not immediately tell us how to define the temperament. However, 
it does allow us to put vals and intervals to the test, and see right 
away if they are giving us the same system, or any torsion.


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Message: 2126

Date: Tue, 27 Nov 2001 08:39:00

Subject: Torsion problems on my list

From: genewardsmith@xxxx.xxx

I checked my list, and the only two torsion problems were
<50/49, 2048/2025> (paultone) and <81/80,3136/3125> (meantone). 
Apparently having a name with "tone" in it is dangerous. :)


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Message: 2127

Date: Tue, 27 Nov 2001 09:16:22

Subject: Re: Survey

From: genewardsmith@xxxx.xxx

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

> Rather than "Minkowski reduced", why don't we call this particular 
> definition the definition of "Tenney-Minkowski" reduced; or, if 
> Minkowski himself isn't really involved in any big way, "Tenney-
> Smith" reduced?

It's bad form to name something after yourself, and the hypenated 
names are too long, but I could go for "TM-reduced".


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Message: 2128

Date: Tue, 27 Nov 2001 10:16:16

Subject: Re: Standardizing our wedge product

From: genewardsmith@xxxx.xxx

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

Since no one is likely to complain about my proposed standard, I'd 
better do it myself. I think it should be changed--the standard I 
gave would be a usual one in most applications on the interval side, 
and simply made the vals correspond. However, this isn't any old 
wedge product, it is supposed to be useful for music theory. From 
that point of view, it makes sense to have the first entry 3-limit, 
the next two 5-limit, the next three 7-limit, and so forth. This 
allows a more easy comparison from one prime limit to another, which 
should help for extending temperaments and classifying them. Hence I 
now will propose instead:

> Let [u1, u2, u3, u4] and [v1, v2, v3, v4] be two intervals in the 
> prime-power basis. Then
 
[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u1*v2-v1*u2, u1*v3-v1*u3, u2*v3-v2*u3, 
u1*v4-v1*u4, u2*v4-v2*u4, u3*v4-v3*u4]

> Let [u1,u2,u3,u4] and [v1,v2,v3,v4] be two vals in the prime-
> valuation basis (that is, h12 would be [12,19,28,34] and so forth.) 
> Then

[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u3*v4-v3*u4, v2*u4-u2*v4, u1*v4-v1*u4, 
u2*v3-v2*u3, v1*u3-u1*v3, u1*v2-v1*u2]

Thus for example for paultone, we get

64/63^50/49 = [2,12,-4,-11,4,2]

h10^h12 = [-2,-12,4,11,-4,-2]

We can also use this on the maps of generators to primes, since they 
are also vals; from

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

we get [0,-1,2,2]^[2,4,3,4] = [2,12,-4,-11,4,2]. We see that three 
seemingly very different defintions, in terms of two intervals, two 
equal temperaments, or two maps of generators to primes, give us the 
same wedge invariant. This strikes me as useful. It is also not too 
difficult to extract the temperament from the wedge invariant, using 
the fact that u^u=0 and testing for vals or intervals which give 0 in 
a wedge product with the invariant, so it does work, though not 
transparently, as a means of classifying temperaments.


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Message: 2129

Date: Tue, 27 Nov 2001 10:54:29

Subject: Re: Standardizing our wedge product

From: genewardsmith@xxxx.xxx

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

Since no one is likely to complain about my proposed standard, I'd 
better do it myself. I think it should be changed to make it more 
useful from the point of view of music theory, where it makes sense 
to have the first entry 3-limit, the next two 5-limit, the next three 
7-limit, and so forth. This allows a more easy comparison from one 
prime limit to another, which should help for extending temperaments 
and classifying them. Hence I now will propose instead:

> Let [u1, u2, u3, u4] and [v1, v2, v3, v4] be two intervals in the 
> prime-power basis. Then

[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u3*v4-v3*u4, v2*u4-u2*v4, u2*v3-v2*u3, 
u1*v4-v1*u4, v1*u3-u1*v3, u1*v2-v1*u2] 

> Let [u1,u2,u3,u4] and [v1,v2,v3,v4] be two vals in the prime-
> valuation basis (that is, h12 would be [12,19,28,34] and so forth.) 
> Then

[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u1*v2-v1*u2, u1*v3-v1*u3, u1*v4-v1*u4, 
u2*v3-v2*u3, u2*v4-v2*u4, u3*v4-v3*u4]

Thus for example for paultone, we get

64/63^50/49 = [2,-4,-4,-11,-12,2]

h10^h12 = [-2,4,4,11,12,-2]

We can also use this on the maps of generators to primes, since they 
are also vals; from

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

we get [0,-1,2,2]^[2,4,3,4] = [2,-4,-4,-11,-12,2]. We see that three 
seemingly very different defintions, in terms of two intervals, two 
equal temperaments, or two maps of generators to primes, give us the 
same wedge invariant. This strikes me as useful. It is also not too 
difficult to extract the temperament from the wedge invariant, using 
the fact that u^u=0 and testing for vals or intervals which give 0 in 
a wedge product with the invariant, so it does work, though not 
transparently, as a means of classifying temperaments.

We can see how the change in the definition works to our advantage 
with the septimal version of meantone, which of course extends the 5-
limit version. We have

126/125^81/80 = h19^h31 = [1,4,10,4,13,12]

Here the [1,4,10] at the beginning is the 5-limit part. 

I gave the following as "map" and "adjusted map"

[ 1  1]  [ 0  1]
[ 1  2]  [-1  2]
[ 0  4]  [-4  4]
[-3  7]  [-10 7]

If we take the wedge product here, we get

[1,1,0,-3]^[1,2,4,7] = [0,-1,-4,-10]^[1,2,4,7] = [1,4,10,4,13,12]


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Message: 2130

Date: Tue, 27 Nov 2001 13:32 +0

Subject: Re: More fun with wedge products

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9tvaik+ehul@xxxxxxx.xxx>
In article <9tvaik+ehul@xxxxxxx.xxx>, genewardsmith@xxxx.xxx () wrote:

> The triple wedge product of 7-limit intervals is basically a val. We 
> can also take wedge products of vals, and the triple product can be 
> identified with an interval. Taking a wedge product of two intervals 
> or two vals gives us something in the Twilight Zone halfway between 
> an interval and a val, and we can compare them directly.

What's a val?

These wedge products seem to behave like the vector products we use in 
physics.  Are they similar?  The same thing?

> The {64/63, 49/48} system has 5, 10, and 15 for vals--in other words, 
> these are common kernel elements. If we write them as vectors, we get
> 
> h5: [ 5]  h10: [10] h15: [15]
>     [ 8]       [16]      [35]
>     [12]       [23]      [35]
>     [14]       [28]      [42]

So these are the normal ET mappings.  How did you get them?  I notice that 
h15 isn't the same as h5+h10.  Is that right?

> Taking wedge products gives us:
> 
> h5^h10  = [0, -5, 0, -8, 0, 14]
> h15^h10 = [0, -5, 0, -8, 0, 14]
> h5^h15  = [0, -5, 0, -8, 0, 14]
> 
> On the other hand, we also have
> 
> 64/63^49/48 = [-14, 0, 8, 0, -5, 0]
> 
> This is the the same as the previous wedge products read backwards, 
> up to a change of sign, and this is no accident. The basis element
> v2^v3 for vals corresponds to e5^e7 for intervals, v2^v7 corresponds 
> to e3^e5, and so forth. These wedge products tell us that neither the 
> vals nor the intervals are giving us torsion, and that they are all 
> giving us the same linear temperament. The wedge product, in fact, 
> could be used to identify linear temperaments, rather than Minkowski 
> reduced pairs of intervals.

Could the wedge product be used to remove the torsion, or only recognize 
it?  I'd still like a routing for working backwards from the mapping to 
the unison vectors.

What algorithm are you using to find the temperaments?  I don't remember 
seeing it.  It might be useful to see your wedge product routine as well, 
if they turn out to be important.  My unison vectors to temperament 
routine has been able to handle all the torsion I've thrown at it so far, 
so it doesn't really matter if I don't spot it in time.


                          Graham


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Message: 2131

Date: Tue, 27 Nov 2001 17:11 +0

Subject: Re: Wedge products

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9tv600+a9o8@xxxxxxx.xxx>
Gene wrote:

> four dimensions. If we take a quadruple wedge product, we get a 4 
> choose 4 = one-demensional space, with basis element {e2^e3^e5^e7}; 
> this can therefore be identified with a scalar, which makes the 
> quadruple wedge product the determinant.

Okay, if you were to do that in octave-equivalent space, this would be the 
Fokker determinant.  In octave-equivalent 7-limit space, it's the scalar 
triple product, which Fokker would have recognised as being a kind of 
volume.


> The triple wedge product, 
> being in effect a thing which takes an interval and sends it to an 
> integer, can be identified with a val. The triple wedge product of 
> three 7-limit intervals is the val we get by putting the three 
> intervals in as rows of a 4x4 matrix and making the top row the four 
> basis vectors.

That's the same as what I'm getting by taking the left hand column of the 
adjoint, with an arbitrary row to make up the numbers.  So with octave 
specific unison vectors, it describes the related equal temperament.  With 
octave equivalent vectors it gives you the mapping.  In three dimensions, 
it's the same as the vector product.


And all this has something to do with wedge products?



                 Graham


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Message: 2132

Date: Wed, 28 Nov 2001 20:00:12

Subject: Re: Hey Graham

From: genewardsmith@xxxx.xxx

--- 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.


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Message: 2133

Date: Wed, 28 Nov 2001 16:31:43

Subject: Hey Graham

From: Paul Erlich

--- 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
<404 Not Found * Search for http://www.microtonal.co.uk/limit7.txt in Wayback Machine>?
Why didn't it?


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Message: 2134

Date: Wed, 28 Nov 2001 16:46:00

Subject: Re: Survey VII

From: Paul Erlich

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

> <875/864, 50/49> Minkowski reduced
> 
> Ets: 22, 26, 48
> 
> Map:
> 
> [2  2]
> [1 -3]
> [3  0]
> [4  1]
> 
> Adjusted map:
> 
> [0  2]
> [4  1]
> [3  3]
> [3  4]
> 
> Generators: a = .2713428065 (~6/5) = 13.0244547 / 48; b = 1/2
> 
> Errors and the 48-et:
> 
> 3:  0.49   -1.96
> 5: -9.48  -11.31
> 7:  8.01    6.17
> 
> Here's another system with very flat major thirds, and Paul, I 
> suppose, would prefer it in the 18+4 version of the 22-et, or 22+4 
of 
> the 26 et, over 26+22 in the 48-et.

Why would you say that? I only liked 26-tET in contexts where 81:80 
was vanishing (probably since my brain could find something familiar 
in such music) -- otherwise, I might not. But none of these comments 
should be construed to mean that I'd be opposed to something *in 
principle* -- only that the result didn't appeal to me when I tried 
to get short-term musical satisfaction out of it.


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Message: 2135

Date: Wed, 28 Nov 2001 16:51 +0

Subject: Re: Hey Graham

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9u33hf+h7bd@xxxxxxx.xxx>
Paul wrote:

> Shouldn't this have made your top ten at
> <404 Not Found * Search for http://www.microtonal.co.uk/limit7.txt in Wayback Machine>?

No.

> Why didn't it?

Because there are 35 temperaments that scored higher.


             Graham


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Message: 2136

Date: Wed, 28 Nov 2001 17:01:02

Subject: Re: Hey Graham

From: Paul Erlich

--- In tuning-math@y..., graham@m... wrote:
> In-Reply-To: <9u33hf+h7bd@e...>
> Paul wrote:
> 
> > Shouldn't this have made your top ten at
> > <404 Not Found * Search for http://www.microtonal.co.uk/limit7.txt in Wayback Machine>?
> 
> 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?


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Message: 2137

Date: Wed, 28 Nov 2001 17:09 +0

Subject: Re: Hey Graham

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9u358e+nleq@xxxxxxx.xxx>
> Isn't your score
> 
> complexity*max_error?

No.  It's complexity^2*max_error.


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Message: 2138

Date: Wed, 28 Nov 2001 19:00:43

Subject: Re: Hey Graham

From: Paul Erlich

--- 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?


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Message: 2139

Date: Thu, 29 Nov 2001 20:05:27

Subject: Re: Graham's Top Ten

From: Paul Erlich

--- 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?


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Message: 2140

Date: Thu, 29 Nov 2001 03:57:25

Subject: Graham's Top Ten

From: genewardsmith@xxxx.xxx

--- 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.


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Message: 2141

Date: Thu, 29 Nov 2001 20:12:20

Subject: Re: Survey IX

From: Paul Erlich

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

> <2401/2400, 6144/6125>
> 
> Minkowski reduction: <2401/2400, 3136/3125>
> 
> Wedge invariant: [16,2,5,6,37,-34] length = 53.3479
> 
> Ets: 31,68,99,130
> 
> Map:
> 
> [ 0  1]
> [16 -1]
> [ 2  2]
> [ 5  2]
> 
> Generators: a = .1615916143 = 15.99756982 / 99; b = 1
> 
> Errors and 99 et:
> 
> 3: 0.604 1.075
> 5: 1.506 1.565
> 7: 0.724 0.871
> 
> Measures: 8.8323 sc; 89.1168 s2c
> 
> This beats Miracle in terms of step-cents, but not with step^2 
cents. 
> The fact that so much 5 and 7 is available is the reason why. I was 
a 
> little disturbed to find that not only was this not Minkowski 
> reduced, it was also not even LLL-reduced. How did it get on my 
list?
> Have I done this one already? :(

Yes, you did do this one already. I asked/pointed out that it was the 
same as Graham's #1 on 404 Not Found * Search for http://www.microtonal.co.uk/limit7.txt in Wayback Machine -- now 
here it is again. Somehow, it made it into your survey _twice_.

> <2401/2400,5120/5103> reduced
> 
> Wedge invariant: [2,25,13,-40,-15,35] length = 62.0322
> 
> Ets: 41,58,99,140,239
> 
> Map:
> 
> [ 0  1]
> [ 2  1]
> [15 -1]
> [13 -5]
> 
> Generators: a = .2928926789 (~49/40) = 70.00135026 / 239; b = 1
> 
> Errors and 239 et
> 
> 3: .987 .974
> 5: .467 .297
> 7: .300 .212
> 
> Measures: 7.3682 sc; 92.7798 s2c
> 
> Another valuable system, particularly if you are fond of 7/5.

Graham -- shouldn't this be on your list? Its complexity is only 15, 
yet it has lower errors than your #1. Perhaps this shows a limitation 
of your method?

> 
> 
> <1029/1024,126/125>
> 
> Wedge invariant: [9,5,-3,-21,30,-13] length = 40.3113
> 
> Ets: 15,16,31,46,77
> 
> Map:
> 
> [ 0  1]
> [ 9  1]
> [ 5  2]
> [-3  3]
> 
> Generators: a = .06475590616 = 2.007433091 / 31; b = 1
> 
> Errors and 31 et
> 
> 3: -2.59  -5.19
> 5:  2.22   0.78
> 7: -1.95  -1.08
> 
> Measures: 22.068 sc; 158.834 s2c

I've played with this system -- it extends pretty well into the 11-
limit, right? How would you describe this in terms of an ET+ET 
genesis?


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Message: 2142

Date: Thu, 29 Nov 2001 07:26:27

Subject: Survey IX

From: genewardsmith@xxxx.xxx

<225/224, 1029/1024> reduced

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

Ets: 31,41,72

Map:

[ 0  1]
[ 6  1]
[-7  3]
[-2  3]

Generators: a = .09714412267 = 6.994376832 / 72 (secor); b = 1

Errors and 72-et

3: -2.517  -1.955
5: -2.324  -2.980
7: -1.972  -2.159

Measures: 11.3213 sc; 78.2771 s2c

This is, of course, the 7-limit version of Miracle.


<2401/2400, 6144/6125>

Minkowski reduction: <2401/2400, 3136/3125>

Wedge invariant: [16,2,5,6,37,-34] length = 53.3479

Ets: 31,68,99,130

Map:

[ 0  1]
[16 -1]
[ 2  2]
[ 5  2]

Generators: a = .1615916143 = 15.99756982 / 99; b = 1

Errors and 99 et:

3: 0.604 1.075
5: 1.506 1.565
7: 0.724 0.871

Measures: 8.8323 sc; 89.1168 s2c

This beats Miracle in terms of step-cents, but not with step^2 cents. 
The fact that so much 5 and 7 is available is the reason why. I was a 
little disturbed to find that not only was this not Minkowski 
reduced, it was also not even LLL-reduced. How did it get on my list?
Have I done this one already? :(

In any case, this is a very interesting system.


<2401/2400,5120/5103> reduced

Wedge invariant: [2,25,13,-40,-15,35] length = 62.0322

Ets: 41,58,99,140,239

Map:

[ 0  1]
[ 2  1]
[15 -1]
[13 -5]

Generators: a = .2928926789 (~49/40) = 70.00135026 / 239; b = 1

Errors and 239 et

3: .987 .974
5: .467 .297
7: .300 .212

Measures: 7.3682 sc; 92.7798 s2c

Another valuable system, particularly if you are fond of 7/5.


<1029/1024,126/125>

Wedge invariant: [9,5,-3,-21,30,-13] length = 40.3113

Ets: 15,16,31,46,77

Map:

[ 0  1]
[ 9  1]
[ 5  2]
[-3  3]

Generators: a = .06475590616 = 2.007433091 / 31; b = 1

Errors and 31 et

3: -2.59  -5.19
5:  2.22   0.78
7: -1.95  -1.08

Measures: 22.068 sc; 158.834 s2c


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Message: 2143

Date: Thu, 29 Nov 2001 21:04:19

Subject: Re: Survey IX

From: genewardsmith@xxxx.xxx

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

> > <2401/2400,5120/5103> reduced
> > 
> > Wedge invariant: [2,25,13,-40,-15,35] length = 62.0322
> > 
> > Ets: 41,58,99,140,239
> > 
> > Map:
> > 
> > [ 0  1]
> > [ 2  1]
> > [15 -1]
> > [13 -5]

> Graham -- shouldn't this be on your list? Its complexity is only 
15, 
> yet it has lower errors than your #1. Perhaps this shows a 
limitation 
> of your method?

It shows a limitation of my brain which you've probably already 
noticed. If you look at the wedge invatiant, you'll see the "15" 
should be "25", so it isn't nearly as good as I thought by looking at 
my typo.

> > 
> > 
> > <1029/1024,126/125>
> > 
> > Wedge invariant: [9,5,-3,-21,30,-13] length = 40.3113
> > 
> > Ets: 15,16,31,46,77
> > 
> > Map:
> > 
> > [ 0  1]
> > [ 9  1]
> > [ 5  2]
> > [-3  3]

> I've played with this system -- it extends pretty well into the 11-
> limit, right? How would you describe this in terms of an ET+ET 
> genesis?

It extends nicely to the 11-limit, with [9,5,-3,7] being the 
generator map. For et plus et, all you need to do is see if two ets 
on the et list add to a third. It's done pretty well by 16+15 in the 
31-et, and also by 31+15=46, 46+31=77.


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Message: 2144

Date: Thu, 29 Nov 2001 10:45 +0

Subject: Re: Graham's Top Ten

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9u4bn5+jj4l@xxxxxxx.xxx>
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

<3 4 5 7 8 9 10 12 15 16 18 19 22 23 25 26 27 28 29 31 *>
<4 5 6 9 10 12 15 16 18 19 22 26 27 29 31 35 36 37 41 42 *>
<5 12 19 22 26 27 29 31 41 46 50 53 58 60 68 70 72 77 80 84 *>
<22 26 29 31 41 46 58 72 80 87 89 94 111 113 118 121 130 145 149 152 *>
<26 29 41 46 58 72 80 87 94 111 113 121 130 149 159 166 171 183 190 198 *>
<29 41 58 72 80 87 94 111 121 130 149 159 183 190 198 212 217 224 241 253 *>
<58 72 80 94 111 121 149 159 183 217 253 282 301 311 320 364 388 400 414 422 *>
<80 94 111 121 217 282 311 320 364 388 400 422 436 460 525 581 597 624 643 653 *>


                        Graham


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Message: 2145

Date: Thu, 29 Nov 2001 21:17:42

Subject: Re: Graham's Top Ten

From: genewardsmith@xxxx.xxx

--- 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.


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Message: 2146

Date: Thu, 29 Nov 2001 21:22:12

Subject: Re: Graham's Top Ten

From: Paul Erlich

--- 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.


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Message: 2147

Date: Thu, 29 Nov 2001 21:18:47

Subject: Re: Survey IX

From: Paul Erlich

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

> > > <1029/1024,126/125>
> > > 
> > > Wedge invariant: [9,5,-3,-21,30,-13] length = 40.3113
> > > 
> > > Ets: 15,16,31,46,77
> > > 
> > > Map:
> > > 
> > > [ 0  1]
> > > [ 9  1]
> > > [ 5  2]
> > > [-3  3]
> 
> > I've played with this system -- it extends pretty well into the 
11-
> > limit, right? How would you describe this in terms of an ET+ET 
> > genesis?
> 
> It extends nicely to the 11-limit, with [9,5,-3,7] being the 
> generator map.

Yes -- I thought those 11s would fall right in there!

> For et plus et, all you need to do is see if two ets 
> on the et list add to a third. It's done pretty well by 16+15 in 
the 
> 31-et,

That's where I found it, on my guitar.

Graham, where does this one come in on your rankings?


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Message: 2148

Date: Thu, 29 Nov 2001 22:03 +0

Subject: Re: Graham's Top Ten

From: graham@xxxxxxxxxx.xx.xx

genewardsmith@xxxx.xxx () 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


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Message: 2149

Date: Thu, 29 Nov 2001 22:03 +0

Subject: Re: Survey IX

From: graham@xxxxxxxxxx.xx.xx

> Graham, where does this one come in on your rankings?

11 in the 7-limit, 17 in the 11-limit.


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