Subsets in 12-tET and Meantone Keyboards?

First I list different subsets, minus their symmetrical sets, giving
assymetrical sets, first, for full subsets, and then reduced for the
dihedral group:

Triads (19)

5(3) + 2(2)
-5(1)
5(2) + 2(2) = 7(2)

Triads, Dihedral (12)

5(2) + 2(1)
-5
5 + 2 = 7

Tetrads (43)

5(7) + 2(4)
-5(3)
5(4) + 2(4) = 7(4)

Tetrads, Dihedral

5(5) + 2(2)
-5(3)
5(2) + 2(2) = 7(2)

Pentads

5(10) + 2(8) (Torsional)
-5(2)
5(8) + 2(8) = 7(8)

Pentads, Dihedral

5(6) + 2(4)
-5(2)
5(4) + 2(4) = 7(4)

Hexads ?

5(16)
-5(4)
5(12)

Hexads, Dihedral

5(10)
-5(4)
5(6)

All but hexads give assymmetrical sets that are 7y.

Now, let's look at corresponding meantone temperaments (All except 29
which is schismic)

Let's use the example of 43 tetrachords and 43 meantone:

So for 43-tET meantone, based on the Encyclopedia, is 5t + 2s.
Therefore, t=7 and s=4.

Now if I understand correctly, diatonic wholestep is 7, and diatonic
semitone is 4, therefore, the chromatic semitone (black key) is 3.

Therefore, black keys add up to 15 (5 *3) and white keys minus black
keys to 28:

I get

7 7 4 7 7 7 4 total
3 3 3 3 3 black
4 4 4 4 4 4 4 remainder

Here's the weird correspondence to sets:

43 tetrachords = 15 symmetrical and 28 asymmetrical

So one can map sets (in 12t-ET) to scales (in other ET's)

The Polya math for tetrachords:

Symmetrical: Based on (x^2+1)^6. The tetrachord part is just 15 based
on Binom(6,2))

Total:

1/12 [Binom(12,4) + Phi(2) Binom(6,2) Phi(4) Binom(3,1)]

This is (495 + 15 + 6)/12 = 43

So, as far as I can see, the symmetrical sets correspond to
black keys and the asymmetrical sets to what remains.

I guess a super simple correspondence would be mapping the
5 symmetrical sets of triads in 12-tET (Reduced) to the black keys in
12-tET:

How about

F# -> (0,1,2)
C# -> (0,2,4)
G# -> (0,3,6)
D# -> (0,4,8)
A# -> (0,5,10)->(0,2,7)

There might be other isometries, such as the groups of 3,2 black keys
and even or odd generator of the symmetrical set (1,2,3,4,5: three
are odd and two are even.)

If anyone has read so far, please answer the questionnaire:

1) I think you may be onto something
2) This is the stupidest thing I have ever seen you post
3) "Splunge"

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<phjelmstad@...> wrote:
>
> First I list different subsets, minus their symmetrical sets, giving
> assymetrical sets, first, for full subsets, and then reduced for
the
> dihedral group:
>
> Triads (19)
>
> 5(3) + 2(2)
> -5(1)
> 5(2) + 2(2) = 7(2)
>
> Triads, Dihedral (12)
>
> 5(2) + 2(1)
> -5
> 5 + 2 = 7
>
> Tetrads (43)
>
> 5(7) + 2(4)
> -5(3)
> 5(4) + 2(4) = 7(4)
>
> Tetrads, Dihedral
>
> 5(5) + 2(2)
> -5(3)
> 5(2) + 2(2) = 7(2)
>
> Pentads
>
> 5(10) + 2(8) (Torsional)
> -5(2)
> 5(8) + 2(8) = 7(8)
>
> Pentads, Dihedral
>
> 5(6) + 2(4)
> -5(2)
> 5(4) + 2(4) = 7(4)
>
> Hexads ?
>
> 5(16)
> -5(4)
> 5(12)
>
> Hexads, Dihedral
>
> 5(10)
> -5(4)
> 5(6)
>
> All but hexads give assymmetrical sets that are 7y.
>
> Now, let's look at corresponding meantone temperaments (All except
29
> which is schismic)
>
> Let's use the example of 43 tetrachords and 43 meantone:
>
> So for 43-tET meantone, based on the Encyclopedia, is 5t + 2s.
> Therefore, t=7 and s=4.
>
> Now if I understand correctly, diatonic wholestep is 7, and
diatonic
> semitone is 4, therefore, the chromatic semitone (black key) is 3.
>
> Therefore, black keys add up to 15 (5 *3) and white keys minus
black
> keys to 28:
>
> I get
>
> 7 7 4 7 7 7 4 total
> 3 3 3 3 3 black
> 4 4 4 4 4 4 4 remainder
>
> Here's the weird correspondence to sets:
>
> 43 tetrachords = 15 symmetrical and 28 asymmetrical
>
> So one can map sets (in 12t-ET) to scales (in other ET's)
>
> The Polya math for tetrachords:
>
> Symmetrical: Based on (x^2+1)^6. The tetrachord part is just 15
based
> on Binom(6,2))
>
> Total:
>
> 1/12 [Binom(12,4) + Phi(2) Binom(6,2) Phi(4) Binom(3,1)]
>
> This is (495 + 15 + 6)/12 = 43
>
> So, as far as I can see, the symmetrical sets correspond to
> black keys and the asymmetrical sets to what remains.
>
> I guess a super simple correspondence would be mapping the
> 5 symmetrical sets of triads in 12-tET (Reduced) to the black keys
in
> 12-tET:
>
> How about
>
> F# -> (0,1,2)
> C# -> (0,2,4)
> G# -> (0,3,6)
> D# -> (0,4,8)
> A# -> (0,5,10)->(0,2,7)
>
> There might be other isometries, such as the groups of 3,2 black
keys
> and even or odd generator of the symmetrical set (1,2,3,4,5: three
> are odd and two are even.)
>
> If anyone has read so far, please answer the questionnaire:
>
> 1) I think you may be onto something
> 2) This is the stupidest thing I have ever seen you post
> 3) "Splunge"
>
> Paul Hj

Here are the layouts:

19:

3 3 2 3 3 3 2
1 1 1 1 1
2 2 2 2 2 2 2

12:

2 2 1 2 2 2 1
1 1 1 1 1
1 1 1 1 1 1 1

43:

7 7 4 7 7 7 4
3 3 3 3 3
4 4 4 4 4 4 4

29: (schismic)

5 5 2 5 5 5 2
3 3 3 3 3
2 2 2 2 2 2 2

66: (doubled)

10 10 8 10 10 10 8
2 2 2 2 2
8 8 8 8 8 8 8

38: (doubled)

6 6 4 6 6 6 4
2 2 2 2 2

80: (very skeptical)

16 16 16 16 16
4 4 4 4 4
12 12 12 12 12

50: (maybe)

10 10 10 10 10
4 4 4 4 4
6 6 6 6 6

Now map each subset to "12tET-like" sets in their new temperament.

Each assymetrical set corresponds to a white note n-tuple
Each symmetrical set corresponds to a black note n-tuple

Where does the diminished 7 tetrad land in 43-tET?

In the middle of the 3-set black note, that is the middle of G#?

Generators: Even: 0,6 + C(5,1) flanges
Odd: no 0,6 + C(5,2) flanges

I need to work out the well-ordering, but I think (0,3,6,9) would
have to be in the middle of the middle of the blacks, thus, G#(0,1,0)

Paul Hj

If you all can excuse just one more post on this, if anything it is a
reference for myself:

Subsets represented: 12,19,29,43,38,66,50,80
Meantone temperaments: 12,19,43,38,66,50,80
Fake meantone: 29
Doubled: 38(19), 66(33), 80(40)

Meantone Keyboards <-> Subsets of 12-tET

Improved layouts and correspondences:

80 is (40 * 2) tET, it is faulty in that big white step minus small
white step doesn't equal a black step:

12 12 10 12 12 12 10
4 4 0 4 4 4 0
8 8 10 8 8 8 10

Also, the bottom row is not all the same number

Now 66 is perfect (66 tET = (33 * 2)

10 10 8 10 10 10 8
2 2 0 2 2 2 0
8 8 8 8 8 8 8

50 has the same problem as 80

8 8 5 8 8 8 5
4 4 0 4 4 4 0
4 4 5 4 4 4 5

38 works, (19 * 2)

6 6 4 6 6 6 4
2 2 0 2 2 2 0
4 4 4 4 4 4 4

43 works

7 7 4 7 7 7 4
3 3 0 3 3 3 0
4 4 4 4 4 4 4

Now 29 works as a "fake meantone" with a real bad major third:

5 5 2 5 5 5 2
3 3 0 3 3 3 0
2 2 2 2 2 2 2

12 always worked

2 2 1 2 2 2 1
1 1 0 1 1 1 0
1 1 1 1 1 1 1

Now the correspondence with sets:

80&50 hexachords
66&38 pentachords
43&29 tetrachords
19&12 triads

6 dyads - Nope, doesn't work

(The second numbers in each type have mirror image sets chopped off
(dihedral reduction)

In all cases, the first row is total sets, the second row corresponds
to symmetrical sets, and the last row to asymmetrical sets.

THESIS:

Each set in a subset group of 12-tET maps to a pitch in it's
corresponding temperament. So, 43 subsets each map one-to-one to a
pitch in 43t-ET.

Now we can see that symmetrical pentachords are based on {Reflection
+ 0 + Dyad}: (So 12-> 10,11,0,1,2)

12, 13, 14, 15, 23, 24, 25, 34, 35, 45 (These are 1,0,1,0,1,0,1,1,0,1
weight parity)

Let's map each one to a black key in 66 or 38-tET:

2 2 0 2 2 2 0

The six 1's go to the second cluster 2 2 2
The four 0's go tht first cluster 2 2

1010
101
10
1

How about (13,24,35 & 15 for the 0's) and
(12,23,34,45 & 14,25 for the 1's)

I dunno

I can map the 15 tetrachords to 3 3 0 3 3 3 based on mod 3 weighting
where 0,1,2,3,4,5,6,7,8,9,10,11->0,1,2,3,2,1,0,1,2,3,2,1,0 and set
3=0 for 3 weight classes but don't have as nice a mapping as with 5
symmetrical triads or 10 symmetrical pentads. Might be nothing.

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<phjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <phjelmstad@> wrote:
> >
> > First I list different subsets, minus their symmetrical sets,
giving
> > assymetrical sets, first, for full subsets, and then reduced for
> the
> > dihedral group:
> >
> > Triads (19)
> >
> > 5(3) + 2(2)
> > -5(1)
> > 5(2) + 2(2) = 7(2)
> >
> > Triads, Dihedral (12)
> >
> > 5(2) + 2(1)
> > -5
> > 5 + 2 = 7
> >
> > Tetrads (43)
> >
> > 5(7) + 2(4)
> > -5(3)
> > 5(4) + 2(4) = 7(4)
> >
> > Tetrads, Dihedral
> >
> > 5(5) + 2(2)
> > -5(3)
> > 5(2) + 2(2) = 7(2)
> >
> > Pentads
> >
> > 5(10) + 2(8) (Torsional)
> > -5(2)
> > 5(8) + 2(8) = 7(8)
> >
> > Pentads, Dihedral
> >
> > 5(6) + 2(4)
> > -5(2)
> > 5(4) + 2(4) = 7(4)
> >
> > Hexads ?
> >
> > 5(16)
> > -5(4)
> > 5(12)
> >
> > Hexads, Dihedral
> >
> > 5(10)
> > -5(4)
> > 5(6)
> >
> > All but hexads give assymmetrical sets that are 7y.
> >
> > Now, let's look at corresponding meantone temperaments (All
except
> 29
> > which is schismic)
> >
> > Let's use the example of 43 tetrachords and 43 meantone:
> >
> > So for 43-tET meantone, based on the Encyclopedia, is 5t + 2s.
> > Therefore, t=7 and s=4.
> >
> > Now if I understand correctly, diatonic wholestep is 7, and
> diatonic
> > semitone is 4, therefore, the chromatic semitone (black key) is
3.
> >
> > Therefore, black keys add up to 15 (5 *3) and white keys minus
> black
> > keys to 28:
> >
> > I get
> >
> > 7 7 4 7 7 7 4 total
> > 3 3 3 3 3 black
> > 4 4 4 4 4 4 4 remainder
> >
> > Here's the weird correspondence to sets:
> >
> > 43 tetrachords = 15 symmetrical and 28 asymmetrical
> >
> > So one can map sets (in 12t-ET) to scales (in other ET's)
> >
> > The Polya math for tetrachords:
> >
> > Symmetrical: Based on (x^2+1)^6. The tetrachord part is just 15
> based
> > on Binom(6,2))
> >
> > Total:
> >
> > 1/12 [Binom(12,4) + Phi(2) Binom(6,2) Phi(4) Binom(3,1)]
> >
> > This is (495 + 15 + 6)/12 = 43
> >
> > So, as far as I can see, the symmetrical sets correspond to
> > black keys and the asymmetrical sets to what remains.
> >
> > I guess a super simple correspondence would be mapping the
> > 5 symmetrical sets of triads in 12-tET (Reduced) to the black
keys
> in
> > 12-tET:
> >
> > How about
> >
> > F# -> (0,1,2)
> > C# -> (0,2,4)
> > G# -> (0,3,6)
> > D# -> (0,4,8)
> > A# -> (0,5,10)->(0,2,7)
> >
> > There might be other isometries, such as the groups of 3,2 black
> keys
> > and even or odd generator of the symmetrical set (1,2,3,4,5: three
> > are odd and two are even.)
> >
> > If anyone has read so far, please answer the questionnaire:
> >
> > 1) I think you may be onto something
> > 2) This is the stupidest thing I have ever seen you post
> > 3) "Splunge"
> >
> > Paul Hj
>
> Here are the layouts:
>
> 19:
>
> 3 3 2 3 3 3 2
> 1 1 1 1 1
> 2 2 2 2 2 2 2
>
> 12:
>
> 2 2 1 2 2 2 1
> 1 1 1 1 1
> 1 1 1 1 1 1 1
>
> 43:
>
> 7 7 4 7 7 7 4
> 3 3 3 3 3
> 4 4 4 4 4 4 4
>
> 29: (schismic)
>
> 5 5 2 5 5 5 2
> 3 3 3 3 3
> 2 2 2 2 2 2 2
>
> 66: (doubled)
>
> 10 10 8 10 10 10 8
> 2 2 2 2 2
> 8 8 8 8 8 8 8
>
> 38: (doubled)
>
> 6 6 4 6 6 6 4
> 2 2 2 2 2
>
> 80: (very skeptical)
>
> 16 16 16 16 16
> 4 4 4 4 4
> 12 12 12 12 12
>
> 50: (maybe)
>
> 10 10 10 10 10
> 4 4 4 4 4
> 6 6 6 6 6
>
> Now map each subset to "12tET-like" sets in their new temperament.
>
> Each assymetrical set corresponds to a white note n-tuple
> Each symmetrical set corresponds to a black note n-tuple
>
> Where does the diminished 7 tetrad land in 43-tET?
>
> In the middle of the 3-set black note, that is the middle of G#?
>
> Generators: Even: 0,6 + C(5,1) flanges
> Odd: no 0,6 + C(5,2) flanges
>
> I need to work out the well-ordering, but I think (0,3,6,9) would
> have to be in the middle of the middle of the blacks, thus, G#
(0,1,0)
>
> Paul Hj
>