M12 as a compositional tool

I've found a method to use 11 Steiner hexad sets of M12 (Mathieu Group) (a double Steiner system, which has transpositional equivalence, so 11 x 12 = 132 total Steiners, and their inverses = 132 x 2) as a compositional tool.

One can create an 11 x 11 matrix with the rows Steiner sets and the columns interval vector values 1 through 11 of these sets.

The matrix

[[3,2,3,4,3,0,0,0,0,0,0],
[0,0,0,0,0,1,2,4,3,2,3],
[4,4,3,2,1,1,0,0,0,0,0],
[0,0,0,0,0,1,4,2,3,3,2],
[2,3,3,2,4,1,0,0,0,0,0],
[0,0,0,0,0,1,3,2,2,4,3],
[3,2,2,2,3,2,0,0,0,0,0],
[0,0,0,0,0,2,2,4,2,4,1],
[3,2,2,3,3,2,0,0,0,0,0],
[0,0,0,0,0,2,3,3,2,2,3],
[2,2,5,2,2,2,0,0,0,0,0]];

is then inverted so that other symmetries of hexachords (the sets
produced from these symmetries) can be found with this matrix as
a linear basis. I believe this could be used as a compositional
tool.

If anyone is interested I could provide more details. You need
to clear denominators (17 for "notched" sets for example). I also
work with a method that expands the "Outerzone" by a factor of 3,
(these are hexachords with 0 or 3 tritones) to clear denominators.

These has been effective so far for C4 X C3, D12, D12 with complementation (D12 X S2) and "Notched", which makes 60 virtual
hexachords sets. I found I can collapse sets of limited transposition
and create virtual sets with no degradation of meaning... and so forth. The "Interzone" doesn't present a problem with this.

The basic theory is that Steiner sets produce all the pentads exactly
once. I found I could find "Sister Steiner hexads" and analyze these
things also. The main increase is exactly sevenfold, with a
Steiner on each pentad line and six sisters. (Doubly so).

There are many interesting formulas to relate Steiner sets to the pentads, and methods to reconstruct based on partial information,
Also which could be used compositionally --- such as making sense
of the Z-relation based on interval vectors, and reconstructing
sets based on this.

PGH

--- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@...> wrote:
>
> I've found a method to use 11 Steiner hexad sets of M12 (Mathieu Group) (a double Steiner system, which has transpositional equivalence, so 11 x 12 = 132 total Steiners, and their inverses = 132 x 2) as a compositional tool.

Have you tried it out in practice?

> If anyone is interested I could provide more details. You need
> to clear denominators (17 for "notched" sets for example).

You should probably be using the adjoint rather than the inverse.

Yes, it works in practice. For example, I can back off sets to
go from 77 virtual set types (from 77 x 12 hexachords), down
to my 60 virtual sets, and by multiplying the inverse matrix
with my 60 sets totals (the interval vector), I get another
vector, which multiplied by a scaling factor (17 here), gives
a new vector, which multiplied by my original 11 x 11 matrix
gives the 60-set vector (scaled up by 17).

My 60 virtual sets produce my "notched" vector, which is 8 set types notched out in the Interzone (1 and 2 tritones) (Z-related assymmetricals) and 1 in the "Attic" (a zero tritone set)

I am sure the adjoint has value here, I will play with it, but
here I merely end up having to scale back by the determinant again.

PGH

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>
>
> --- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@> wrote:
> >
> > I've found a method to use 11 Steiner hexad sets of M12 (Mathieu Group) (a double Steiner system, which has transpositional equivalence, so 11 x 12 = 132 total Steiners, and their inverses = 132 x 2) as a compositional tool.
>
> Have you tried it out in practice?
>
> > If anyone is interested I could provide more details. You need
> > to clear denominators (17 for "notched" sets for example).
>
> You should probably be using the adjoint rather than the inverse.
>

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>
>
> --- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@> wrote:
> >
> > I've found a method to use 11 Steiner hexad sets of M12 (Mathieu Group) (a double Steiner system, which has transpositional equivalence, so 11 x 12 = 132 total Steiners, and their inverses = 132 x 2) as a compositional tool.
>
> Have you tried it out in practice?
>
> > If anyone is interested I could provide more details. You need
> > to clear denominators (17 for "notched" sets for example).
>
> You should probably be using the adjoint rather than the inverse.

Do you mean the adjugate? (Inverse * Determinant)?

--- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@...> wrote:

> > You should probably be using the adjoint rather than the inverse.
>
> Do you mean the adjugate? (Inverse * Determinant)?
>

Means the same. It's defined even if the determinant is zero, BTW.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>
>
> --- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@> wrote:
>
> > > You should probably be using the adjoint rather than the inverse.
> >
> > Do you mean the adjugate? (Inverse * Determinant)?
> >
>
> Means the same. It's defined even if the determinant is zero, BTW.

Right, thanks. I noticed that working with GAP, I could have a zero
determinant.

I found a better matrix for my purposes, it's smoother (can move
things around between, say 1st and 11th columns, 2 and 10 etc.)

[[2,1,2,2,1,0,2,2,1,1,1,]
[2,1,2,2,1,1,1,2,1,1,1,]
[2,2,1,1,1,1,0,1,2,2,2,]
[1,2,0,1,2,1,2,1,3,1,1,]
[1,1,2,1,2,1,2,1,1,2,1,]
[2,2,1,1,2,1,1,1,1,2,1,]
[2,1,1,1,2,2,1,1,1,1,2,]
[0,2,1,2,1,2,1,2,1,2,1,]
[1,1,1,1,1,2,2,2,1,1,2,]
[1,1,1,2,1,2,2,1,1,1,2,]
[1,1,3,1,1,2,1,1,2,1,1,]];

These are my sets E, B1, N1, R5, R5(Z), X, O1, V5, F, F(Z), H
but are based on a double Steiner system.

This one is nice because I can use my reduced vector (60 virtual
sets instead of 77) which eliminates mirrored Z-relations and
also the mirror of the forwards-complementable set (with no
tritones) but anyway, taking

(81,83,82,82,81,85,80,82,81,83,80) and multiplying by the inverse
of the above matrix

I get a vector which multiplied by 3 can be multiplied with the
11 x 11 matrix

(2,30,14,16,16,25,6,14,20,20,17) and works with a cross check.

I notice with the inverse matrix and here also i only need to scale
by 3, not the whole determinant of 225. Previously I only needed
425 of 31875 for scaling (highest ever needed) and the inverse
only by 1275 of 31875 with a common factor of 17 between them

I'd like to think interval vectors could be manipulated in the manner
of monzos and vals, I understand you use eigenmonzos, the problem
with interval vectors of Steiner sets is they have 15 intervals
to deal with and each note touches the others 5 times and the
remaining pentad has 10 intervals...but I think it's a rich
environment to study musical set theory in....the comma structure
is way to complicated though, and a value in an interval vector
slot isn't always a contiguous chain and so forth....

PGH

I don't normally top post, but it's mine anyway so ----

I found a better 11 x 11 matrix to use (with column and row totals):

11 x 11 MATRIX

E,2,1,2,2,1,0,2,2,1,1,1,15
B1,2,1,2,2,1,1,1,2,1,1,1,15
N1,2,2,1,1,1,1,0,1,2,2,2,15
R5,1,2,0,1,2,1,2,1,3,1,1,15
R5,1,1,2,1,2,1,2,1,1,2,1,15
X,2,2,1,1,2,1,1,1,1,2,1,15
O1,2,1,1,1,2,2,1,1,1,1,2,15
V5,0,2,1,2,1,2,1,2,1,2,1,15
F,1,1,1,1,1,2,2,2,1,1,2,15
F,1,1,1,2,1,2,2,1,1,1,2,15
H,1,1,3,1,1,2,1,1,2,1,1,15
15,15,15,15,15,15,15,15,15,15,15,165

Determinant: 225 (15^2)

MINVERSE (Inverse Matrix) (Not shown)

Different Source Vectors, and
New Vector = Source Vector * MINVERSE

IZ 95,95,95,95,95,100,95,95,95,95,95,1050,70
Vector *3,1,36,16,20,20,26,9,13,25,25,19,210,

OZ 10,10,10,10,10,5,10,10,10,10,10,105,7
Vector *3,20,-15,5,1,1,-5,12,8,-4,-4,2,21,

Full 105,105,105,105,105,105,105,105,105,105,105,1155,77
Vector 7,7,7,7,7,7,7,7,7,7,7,77,

Full*3 21,21,21,21,21,21,21,21,21,21,21,231,
Full*15 105,105,105,105,105,105,105,105,105,105,105,1155,

D12 IZ 69,67,69,66,68,72,69,67,69,66,68,750,50
Vector*3 3,123,48,75,60,108,27,24,105,90,87,750,

Notched IZ,72,74,73,74,72,80,72,74,73,74,72,810,54
Vector*3 -13,42,11,16,16,25,-3,8,23,23,14,162,

D12XS2 IZ,36,37,37,37,36,40,36,37,36,37,36,405,27
Vector*3 -4,18,5,7,7,13,0,5,11,11,8,81,

S2 IZ 48,47,48,47,48,50,47,48,47,48,47,525,35
Vector*3 -4,24,8,10,13,13,3,5,14,11,8,105,

D12 77,76,76,74,76,75.5,77,74,77,74,76,832.5,55.5
Vector*15 104,31.5,66.5,77.5,47.5,101.5,81,62,80,80,101,832.5,

Notched 80,83,81,82,81,85,80,82,82,83,81,900,60
Vector*3 4,27,16,17,17,20,9,16,19,19,16,180,

D12XS2 44,46,44,45,44,43.5,44,44,44,45,44,487.5,32.5
Vector*15 81,-1.5,43.5,37.5,22.5,58.5,54,63,30,45,54,487.5,

S2 57,57,57,56,58,53.5,56,58,56,57,57,622.5,41.5
Vector*15 146,-31.5,63.5,47.5,47.5,23.5,129,98,35,5,59,622.5,

The Full Vector is easy to prove, why it is Sevenfold. Since
each Steiner appears six times (once for each of its Pentads)
the hexads in these 6 Pentad rows can be found from the Steiner
Set. The other symmetries are a little more difficult to prove
but there are interesting patterns.

PGH

--- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@...> wrote:
>
> I've found a method to use 11 Steiner hexad sets of M12 (Mathieu Group) (a double Steiner system, which has transpositional equivalence, so 11 x 12 = 132 total Steiners, and their inverses = 132 x 2) as a compositional tool.
>
> One can create an 11 x 11 matrix with the rows Steiner sets and the columns interval vector values 1 through 11 of these sets.
>
> The matrix
>
> [[3,2,3,4,3,0,0,0,0,0,0],
> [0,0,0,0,0,1,2,4,3,2,3],
> [4,4,3,2,1,1,0,0,0,0,0],
> [0,0,0,0,0,1,4,2,3,3,2],
> [2,3,3,2,4,1,0,0,0,0,0],
> [0,0,0,0,0,1,3,2,2,4,3],
> [3,2,2,2,3,2,0,0,0,0,0],
> [0,0,0,0,0,2,2,4,2,4,1],
> [3,2,2,3,3,2,0,0,0,0,0],
> [0,0,0,0,0,2,3,3,2,2,3],
> [2,2,5,2,2,2,0,0,0,0,0]];
>
> is then inverted so that other symmetries of hexachords (the sets
> produced from these symmetries) can be found with this matrix as
> a linear basis. I believe this could be used as a compositional
> tool.
>
> If anyone is interested I could provide more details. You need
> to clear denominators (17 for "notched" sets for example). I also
> work with a method that expands the "Outerzone" by a factor of 3,
> (these are hexachords with 0 or 3 tritones) to clear denominators.
>
> These has been effective so far for C4 X C3, D12, D12 with complementation (D12 X S2) and "Notched", which makes 60 virtual
> hexachords sets. I found I can collapse sets of limited transposition
> and create virtual sets with no degradation of meaning... and so forth. The "Interzone" doesn't present a problem with this.
>
> The basic theory is that Steiner sets produce all the pentads exactly
> once. I found I could find "Sister Steiner hexads" and analyze these
> things also. The main increase is exactly sevenfold, with a
> Steiner on each pentad line and six sisters. (Doubly so).
>
> There are many interesting formulas to relate Steiner sets to the pentads, and methods to reconstruct based on partial information,
> Also which could be used compositionally --- such as making sense
> of the Z-relation based on interval vectors, and reconstructing
> sets based on this.
>
> PGH
>

In conclusion, (I know this is not the main discussion now -- but)

Every Pentad has an interval vector which adds up to 10. Every
Hexad has an interval vector which adds up to 15. Each Pentad
has one and only one "Steiner Pencil" which is the 5 interval
difference between its vector and its Steiner hexad vector (that
it belongs to).

These Pencils can be slid back and forth on the 11-vector, and
the results can be studied in terms of permutations, when
finding other symmetries. For example, for finding the 60 virtual
sets of D12 x S2. The Vectors based on Steiners equal, in total
exactly what I need to map to. And this equivalence is a matter
of permuting "Sliding Steiner Pencils" on Pentad rows (The rows
of a Pentad which each have seven Hexad parents).

PGH

--- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@...> wrote:
>
> I don't normally top post, but it's mine anyway so ----
>
> I found a better 11 x 11 matrix to use (with column and row totals):
>
> 11 x 11 MATRIX
>
> E,2,1,2,2,1,0,2,2,1,1,1,15
> B1,2,1,2,2,1,1,1,2,1,1,1,15
> N1,2,2,1,1,1,1,0,1,2,2,2,15
> R5,1,2,0,1,2,1,2,1,3,1,1,15
> R5,1,1,2,1,2,1,2,1,1,2,1,15
> X,2,2,1,1,2,1,1,1,1,2,1,15
> O1,2,1,1,1,2,2,1,1,1,1,2,15
> V5,0,2,1,2,1,2,1,2,1,2,1,15
> F,1,1,1,1,1,2,2,2,1,1,2,15
> F,1,1,1,2,1,2,2,1,1,1,2,15
> H,1,1,3,1,1,2,1,1,2,1,1,15
> 15,15,15,15,15,15,15,15,15,15,15,165
>
> Determinant: 225 (15^2)
>
>
> MINVERSE (Inverse Matrix) (Not shown)
>
>
> Different Source Vectors, and
> New Vector = Source Vector * MINVERSE
>
> IZ 95,95,95,95,95,100,95,95,95,95,95,1050,70
> Vector *3,1,36,16,20,20,26,9,13,25,25,19,210,
>
> OZ 10,10,10,10,10,5,10,10,10,10,10,105,7
> Vector *3,20,-15,5,1,1,-5,12,8,-4,-4,2,21,
>
> Full 105,105,105,105,105,105,105,105,105,105,105,1155,77
> Vector 7,7,7,7,7,7,7,7,7,7,7,77,
>
> Full*3 21,21,21,21,21,21,21,21,21,21,21,231,
> Full*15 105,105,105,105,105,105,105,105,105,105,105,1155,
>
> D12 IZ 69,67,69,66,68,72,69,67,69,66,68,750,50
> Vector*3 3,123,48,75,60,108,27,24,105,90,87,750,
>
> Notched IZ,72,74,73,74,72,80,72,74,73,74,72,810,54
> Vector*3 -13,42,11,16,16,25,-3,8,23,23,14,162,
>
> D12XS2 IZ,36,37,37,37,36,40,36,37,36,37,36,405,27
> Vector*3 -4,18,5,7,7,13,0,5,11,11,8,81,
>
> S2 IZ 48,47,48,47,48,50,47,48,47,48,47,525,35
> Vector*3 -4,24,8,10,13,13,3,5,14,11,8,105,
>
> D12 77,76,76,74,76,75.5,77,74,77,74,76,832.5,55.5
> Vector*15 104,31.5,66.5,77.5,47.5,101.5,81,62,80,80,101,832.5,
>
> Notched 80,83,81,82,81,85,80,82,82,83,81,900,60
> Vector*3 4,27,16,17,17,20,9,16,19,19,16,180,
>
> D12XS2 44,46,44,45,44,43.5,44,44,44,45,44,487.5,32.5
> Vector*15 81,-1.5,43.5,37.5,22.5,58.5,54,63,30,45,54,487.5,
>
> S2 57,57,57,56,58,53.5,56,58,56,57,57,622.5,41.5
> Vector*15 146,-31.5,63.5,47.5,47.5,23.5,129,98,35,5,59,622.5,
>
> The Full Vector is easy to prove, why it is Sevenfold. Since
> each Steiner appears six times (once for each of its Pentads)
> the hexads in these 6 Pentad rows can be found from the Steiner
> Set. The other symmetries are a little more difficult to prove
> but there are interesting patterns.
>
> PGH
>
> --- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@> wrote:
> >
> > I've found a method to use 11 Steiner hexad sets of M12 (Mathieu Group) (a double Steiner system, which has transpositional equivalence, so 11 x 12 = 132 total Steiners, and their inverses = 132 x 2) as a compositional tool.
> >
> > One can create an 11 x 11 matrix with the rows Steiner sets and the columns interval vector values 1 through 11 of these sets.
> >
> > The matrix
> >
> > [[3,2,3,4,3,0,0,0,0,0,0],
> > [0,0,0,0,0,1,2,4,3,2,3],
> > [4,4,3,2,1,1,0,0,0,0,0],
> > [0,0,0,0,0,1,4,2,3,3,2],
> > [2,3,3,2,4,1,0,0,0,0,0],
> > [0,0,0,0,0,1,3,2,2,4,3],
> > [3,2,2,2,3,2,0,0,0,0,0],
> > [0,0,0,0,0,2,2,4,2,4,1],
> > [3,2,2,3,3,2,0,0,0,0,0],
> > [0,0,0,0,0,2,3,3,2,2,3],
> > [2,2,5,2,2,2,0,0,0,0,0]];
> >
> > is then inverted so that other symmetries of hexachords (the sets
> > produced from these symmetries) can be found with this matrix as
> > a linear basis. I believe this could be used as a compositional
> > tool.
> >
> > If anyone is interested I could provide more details. You need
> > to clear denominators (17 for "notched" sets for example). I also
> > work with a method that expands the "Outerzone" by a factor of 3,
> > (these are hexachords with 0 or 3 tritones) to clear denominators.
> >
> > These has been effective so far for C4 X C3, D12, D12 with complementation (D12 X S2) and "Notched", which makes 60 virtual
> > hexachords sets. I found I can collapse sets of limited transposition
> > and create virtual sets with no degradation of meaning... and so forth. The "Interzone" doesn't present a problem with this.
> >
> > The basic theory is that Steiner sets produce all the pentads exactly
> > once. I found I could find "Sister Steiner hexads" and analyze these
> > things also. The main increase is exactly sevenfold, with a
> > Steiner on each pentad line and six sisters. (Doubly so).
> >
> > There are many interesting formulas to relate Steiner sets to the pentads, and methods to reconstruct based on partial information,
> > Also which could be used compositionally --- such as making sense
> > of the Z-relation based on interval vectors, and reconstructing
> > sets based on this.
> >
> > PGH
> >
>

CODA:

Now, there are 17 sets which get chopped off when one goes from
C4 X C3 -> D12 X S2. One merely eliminates itself in terms of
the 11 x 11 matrix. The other 16 can be divided in two, inverse
Z-related sets, and their complements) making eight. The vector is

(12,11,11,11,12,10,11,10,11,10,11) * (MINVERSE(11 x 11 MATRIX) =

(0,2,1,2,1,0,1,0,0,1,0), in terms of the original MATRIX.

In terms of this, two more sets map to themselves, leaving six.

This creates two cycles of three which permute pairs of interval
vector values +1,-1. Luckily, the swapping of the values corresponds
exactly to the free note (element) that moves, except that 1 and 5
act differently

B1->T1 moves 6, (map 4->10(2)), B1->D moves 5, (map 2->7 (5)), R5->T5 moves 5 (map 5->4); and

R5->L moves 6, (map 5->11 (1)), O1->Z moves 1 (map 1->2) and N1->R1
moves 1 (map 2->7 (5); so in each cycle, two are different. I get
6,5,5* and 6,1,1* where vector values swap 4->2, 2->5, 5->4 and
5->1, 1->2, 2->5 respectively so it is the third in each cycle
that does match. but the twists cancel producing 0 as a total
change.

Since the 77 virtual sets are a perfect sevenfold mapping of the MATRIX, eliminating this "notch" of vectors will leave 60 virtual
sets that also do, (because the sums are the same). This is due
to the fact that these 8 Steiners (B1,B1,R5,R5,R5,O1,F,F) sum to the
same as what they are eliminating (8 Z-related, asymmetrical hexachords)

PGH

--- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@...> wrote:
>
> In conclusion, (I know this is not the main discussion now -- but)
>
> Every Pentad has an interval vector which adds up to 10. Every
> Hexad has an interval vector which adds up to 15. Each Pentad
> has one and only one "Steiner Pencil" which is the 5 interval
> difference between its vector and its Steiner hexad vector (that
> it belongs to).
>
> These Pencils can be slid back and forth on the 11-vector, and
> the results can be studied in terms of permutations, when
> finding other symmetries. For example, for finding the 60 virtual
> sets of D12 x S2. The Vectors based on Steiners equal, in total
> exactly what I need to map to. And this equivalence is a matter
> of permuting "Sliding Steiner Pencils" on Pentad rows (The rows
> of a Pentad which each have seven Hexad parents).
>
> PGH
>
> --- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@> wrote:
> >
> > I don't normally top post, but it's mine anyway so ----
> >
> > I found a better 11 x 11 matrix to use (with column and row totals):
> >
> > 11 x 11 MATRIX
> >
> > E,2,1,2,2,1,0,2,2,1,1,1,15
> > B1,2,1,2,2,1,1,1,2,1,1,1,15
> > N1,2,2,1,1,1,1,0,1,2,2,2,15
> > R5,1,2,0,1,2,1,2,1,3,1,1,15
> > R5,1,1,2,1,2,1,2,1,1,2,1,15
> > X,2,2,1,1,2,1,1,1,1,2,1,15
> > O1,2,1,1,1,2,2,1,1,1,1,2,15
> > V5,0,2,1,2,1,2,1,2,1,2,1,15
> > F,1,1,1,1,1,2,2,2,1,1,2,15
> > F,1,1,1,2,1,2,2,1,1,1,2,15
> > H,1,1,3,1,1,2,1,1,2,1,1,15
> > 15,15,15,15,15,15,15,15,15,15,15,165
> >
> > Determinant: 225 (15^2)
> >
> >
> > MINVERSE (Inverse Matrix) (Not shown)
> >
> >
> > Different Source Vectors, and
> > New Vector = Source Vector * MINVERSE
> >
> > IZ 95,95,95,95,95,100,95,95,95,95,95,1050,70
> > Vector *3,1,36,16,20,20,26,9,13,25,25,19,210,
> >
> > OZ 10,10,10,10,10,5,10,10,10,10,10,105,7
> > Vector *3,20,-15,5,1,1,-5,12,8,-4,-4,2,21,
> >
> > Full 105,105,105,105,105,105,105,105,105,105,105,1155,77
> > Vector 7,7,7,7,7,7,7,7,7,7,7,77,
> >
> > Full*3 21,21,21,21,21,21,21,21,21,21,21,231,
> > Full*15 105,105,105,105,105,105,105,105,105,105,105,1155,
> >
> > D12 IZ 69,67,69,66,68,72,69,67,69,66,68,750,50
> > Vector*3 3,123,48,75,60,108,27,24,105,90,87,750,
> >
> > Notched IZ,72,74,73,74,72,80,72,74,73,74,72,810,54
> > Vector*3 -13,42,11,16,16,25,-3,8,23,23,14,162,
> >
> > D12XS2 IZ,36,37,37,37,36,40,36,37,36,37,36,405,27
> > Vector*3 -4,18,5,7,7,13,0,5,11,11,8,81,
> >
> > S2 IZ 48,47,48,47,48,50,47,48,47,48,47,525,35
> > Vector*3 -4,24,8,10,13,13,3,5,14,11,8,105,
> >
> > D12 77,76,76,74,76,75.5,77,74,77,74,76,832.5,55.5
> > Vector*15 104,31.5,66.5,77.5,47.5,101.5,81,62,80,80,101,832.5,
> >
> > Notched 80,83,81,82,81,85,80,82,82,83,81,900,60
> > Vector*3 4,27,16,17,17,20,9,16,19,19,16,180,
> >
> > D12XS2 44,46,44,45,44,43.5,44,44,44,45,44,487.5,32.5
> > Vector*15 81,-1.5,43.5,37.5,22.5,58.5,54,63,30,45,54,487.5,
> >
> > S2 57,57,57,56,58,53.5,56,58,56,57,57,622.5,41.5
> > Vector*15 146,-31.5,63.5,47.5,47.5,23.5,129,98,35,5,59,622.5,
> >
> > The Full Vector is easy to prove, why it is Sevenfold. Since
> > each Steiner appears six times (once for each of its Pentads)
> > the hexads in these 6 Pentad rows can be found from the Steiner
> > Set. The other symmetries are a little more difficult to prove
> > but there are interesting patterns.
> >
> > PGH
> >
> > --- In tuning-math@yahoogroups.com, "paulhjelmstad" <phjelmstad@> wrote:
> > >
> > > I've found a method to use 11 Steiner hexad sets of M12 (Mathieu Group) (a double Steiner system, which has transpositional equivalence, so 11 x 12 = 132 total Steiners, and their inverses = 132 x 2) as a compositional tool.
> > >
> > > One can create an 11 x 11 matrix with the rows Steiner sets and the columns interval vector values 1 through 11 of these sets.
> > >
> > > The matrix
> > >
> > > [[3,2,3,4,3,0,0,0,0,0,0],
> > > [0,0,0,0,0,1,2,4,3,2,3],
> > > [4,4,3,2,1,1,0,0,0,0,0],
> > > [0,0,0,0,0,1,4,2,3,3,2],
> > > [2,3,3,2,4,1,0,0,0,0,0],
> > > [0,0,0,0,0,1,3,2,2,4,3],
> > > [3,2,2,2,3,2,0,0,0,0,0],
> > > [0,0,0,0,0,2,2,4,2,4,1],
> > > [3,2,2,3,3,2,0,0,0,0,0],
> > > [0,0,0,0,0,2,3,3,2,2,3],
> > > [2,2,5,2,2,2,0,0,0,0,0]];
> > >
> > > is then inverted so that other symmetries of hexachords (the sets
> > > produced from these symmetries) can be found with this matrix as
> > > a linear basis. I believe this could be used as a compositional
> > > tool.
> > >
> > > If anyone is interested I could provide more details. You need
> > > to clear denominators (17 for "notched" sets for example). I also
> > > work with a method that expands the "Outerzone" by a factor of 3,
> > > (these are hexachords with 0 or 3 tritones) to clear denominators.
> > >
> > > These has been effective so far for C4 X C3, D12, D12 with complementation (D12 X S2) and "Notched", which makes 60 virtual
> > > hexachords sets. I found I can collapse sets of limited transposition
> > > and create virtual sets with no degradation of meaning... and so forth. The "Interzone" doesn't present a problem with this.
> > >
> > > The basic theory is that Steiner sets produce all the pentads exactly
> > > once. I found I could find "Sister Steiner hexads" and analyze these
> > > things also. The main increase is exactly sevenfold, with a
> > > Steiner on each pentad line and six sisters. (Doubly so).
> > >
> > > There are many interesting formulas to relate Steiner sets to the pentads, and methods to reconstruct based on partial information,
> > > Also which could be used compositionally --- such as making sense
> > > of the Z-relation based on interval vectors, and reconstructing
> > > sets based on this.
> > >
> > > PGH
> > >
> >
>