Hendecachords in 22-tET

(I checked Wikipedia...Hendecagons have 11 sides, so Hendecachords
have 11 notes)

Some fun facts about 11-note chords in 22-tET:

32,066 chords after reducing for transposition (easily done, one 1
has non-primitive order: (0,2,4,6,10,12,14,16,18,20)

16,159 chords after reducing for asymmetry (252 are symmetrical)

16,080 chord-partitions (after reducing for 94 chords complementible
to themselves)

8,359 chords after reducing for complementability and assymetry
(7,800 are complementible Z-related: That is, Z-related to their
complement)

7,944 chords after reducing for noncomplementable Z-relations (I call
strangely-Z-related)

94 chords in the "core of partitions"
31,972 chords (out of 32,066) which aren't

638 chords in the "core of symmetry of partitions"
7,721 chords (out of 8,359) which aren't
638 chords which are

Still have to figure out more specific values (if I find the
total of symmetrical sets which have themselves as complements, then
I can fill out the entire grid...)

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> (I checked Wikipedia...Hendecagons have 11 sides, so Hendecachords
> have 11 notes)
>
> Some fun facts about 11-note chords in 22-tET:
>
> 32,066 chords after reducing for transposition (easily done, one 1
> has non-primitive order: (0,2,4,6,10,12,14,16,18,20)
>
> 16,159 chords after reducing for asymmetry (252 are symmetrical)
>
> 16,080 chord-partitions (after reducing for 94 chords
complementible
> to themselves)
>
> 8,359 chords after reducing for complementability and assymetry
> (7,800 are complementible Z-related: That is, Z-related to their
> complement)
>
> 7,944 chords after reducing for noncomplementable Z-relations (I
call
> strangely-Z-related)
>
> 94 chords in the "core of partitions"
> 31,972 chords (out of 32,066) which aren't
>
> 638 chords in the "core of symmetry of partitions"
> 7,721 chords (out of 8,359) which aren't
> 638 chords which are
>
> Still have to figure out more specific values (if I find the
> total of symmetrical sets which have themselves as complements,
then
> I can fill out the entire grid...)
>
> Paul Hj

With the help of the Online Encyclopedia of Integer Sequences,
here is the full grid for hendachords in 22-et:

110 110 32
7690 7690 496+31=527
7690 7690 496+31=527

32 11-chords have themselves exactly as a complement, and are
symmetrical. The reason for breaking out 496+31 in the right hand
squares is because 31 11-chords have themselves as a complement,
but are asymmetrical. The top row is for symmetrical 11-chords
which total to 252. 110 are complementary-Z-related. 7690 are
asymmetrical and complementary-Z-related. 527, once again, are
complementary and either have themselves as a complement (31) or
have their mirror image as a complement (496). So the second and
third rows are the asymmetrical sets. (7690+7690+527) and their
mirror images.

The right hand column is regular non-Z related sets. The first and
second columns are sets which are Z-related (complements of each
other, but with different structure. You actually find these in
a roundabout way: Finding the partitions (16,080 with 94 as the core
of partitions; and then reducing for asymmetry: 638 chords are the
core of symmetry of partitions, which is 110+32+496)

Adding 7690+31 gives a total which is the total number of partitions,
mentioned above (16,080 chords). I also discovered something else
this week: 32+496+496 is chords for which the complement can be found
by mirror image (and is not Z-related) This is 1024 or 2^n-1. This is
true for any value. The reason it is always a power of two is because
the complement can be found by "swapping out" along a line of
symmetry.

Any thoughts? Any thing that needs clarifying?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> > (I checked Wikipedia...Hendecagons have 11 sides, so
Hendecachords
> > have 11 notes)
> >
> > Some fun facts about 11-note chords in 22-tET:
> >
> > 32,066 chords after reducing for transposition (easily done, one
1
> > has non-primitive order: (0,2,4,6,10,12,14,16,18,20)
> >
> > 16,159 chords after reducing for asymmetry (252 are symmetrical)
> >
> > 16,080 chord-partitions (after reducing for 94 chords
> complementible
> > to themselves)
> >
> > 8,359 chords after reducing for complementability and assymetry
> > (7,800 are complementible Z-related: That is, Z-related to their
> > complement)
> >
> > 7,944 chords after reducing for noncomplementable Z-relations (I
> call
> > strangely-Z-related)
> >
> > 94 chords in the "core of partitions"
> > 31,972 chords (out of 32,066) which aren't
> >
> > 638 chords in the "core of symmetry of partitions"
> > 7,721 chords (out of 8,359) which aren't
> > 638 chords which are
> >
> > Still have to figure out more specific values (if I find the
> > total of symmetrical sets which have themselves as complements,
> then
> > I can fill out the entire grid...)
> >
> > Paul Hj
>
> With the help of the Online Encyclopedia of Integer Sequences,
> here is the full grid for hendachords in 22-et:
>
> 110 110 32
> 7690 7690 496+31=527
> 7690 7690 496+31=527
>
> 32 11-chords have themselves exactly as a complement, and are
> symmetrical. The reason for breaking out 496+31 in the right hand
> squares is because 31 11-chords have themselves as a complement,
> but are asymmetrical. The top row is for symmetrical 11-chords
> which total to 252. 110 are complementary-Z-related. 7690 are
> asymmetrical and complementary-Z-related. 527, once again, are
> complementary and either have themselves as a complement (31) or
> have their mirror image as a complement (496). So the second and
> third rows are the asymmetrical sets. (7690+7690+527) and their
> mirror images.
>
> The right hand column is regular non-Z related sets. The first and
> second columns are sets which are Z-related (complements of each
> other, but with different structure. You actually find these in
> a roundabout way: Finding the partitions (16,080 with 94 as the
core
> of partitions; and then reducing for asymmetry: 638 chords are the
> core of symmetry of partitions, which is 110+32+496)
>
> Adding 7690+31 gives a total which is the total number of
partitions,
> mentioned above (16,080 chords). I also discovered something else
> this week: 32+496+496 is chords for which the complement can be
found
> by mirror image (and is not Z-related) This is 1024 or 2^n-1. This
is
> true for any value. The reason it is always a power of two is
because
> the complement can be found by "swapping out" along a line of
> symmetry.
>
> Any thoughts? Any thing that needs clarifying?

I realized that I am using the term "partition" in a non-orthodox way.
What I mean is counting sets and their complement as one set, or
you might say the number of ways 22 can be divided into 11+11
different sets, (so that A+B or A+A is counted only once. Hmm. I'm
still not satisfied with that description...)

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> Any thoughts? Any thing that needs clarifying?

Just as 12-et has the Mathieu group M12 to go with it, 22-et has the
(3-transitive) Mathieu group M22 to go with it. That might be worth
exploring.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Any thoughts? Any thing that needs clarifying?
>
> Just as 12-et has the Mathieu group M12 to go with it, 22-et has the
> (3-transitive) Mathieu group M22 to go with it. That might be worth
> exploring.

Cool! Thanks. Does the Mathieu group correspond to a Steiner system?
(I know, Paul: Look it up!)

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@m...> wrote:
> >
> > > Any thoughts? Any thing that needs clarifying?
> >
> > Just as 12-et has the Mathieu group M12 to go with it, 22-et has the
> > (3-transitive) Mathieu group M22 to go with it. That might be worth
> > exploring.
>
> Cool! Thanks. Does the Mathieu group correspond to a Steiner system?
> (I know, Paul: Look it up!)

Okay I looked it up. It does correspond to the Steiner system
S(3,6,22): (Summarizing) Out of 22 notes, each triple (triad) will
appear in exactly one of the hexachord blocks...

It's kind of cool: 21 blocks of 6 (hexachords) contain any given point
(note) and there are 77 total hexachord blocks in the Steiner system.

This is based on b=vr/k and r=(C(v-1,t-1)/C(k-1,t-1) where S(t,k,v)=S
(3,6,22)

It's 3-transitive, which corresponds to the first variable in S(3,6,22)

Although it doesn't seem to apply to music directly, it could be
employed as a compositional tool in a form of stochastic music...

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> Although it doesn't seem to apply to music directly, it could be
> employed as a compositional tool in a form of stochastic music...

It seems to me there are lots of possibilites. For instance, given any
triad, there is a unique hexachord, and hence a unique complementary
triad. That ought to be worth something.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Although it doesn't seem to apply to music directly, it could be
> > employed as a compositional tool in a form of stochastic music...
>
> It seems to me there are lots of possibilites. For instance, given any
> triad, there is a unique hexachord, and hence a unique complementary
> triad. That ought to be worth something.

Yes...say does anyone know where I could get a listing of the 132
Steiner system blocks for S(5,6,12) and the 77 blocks for S(3,6,22)?
(Both blocks of hexachords in their respective systems) I might try
sci.math...

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@m...> wrote:
> >
> > > Although it doesn't seem to apply to music directly, it could
be
> > > employed as a compositional tool in a form of stochastic
music...
> >
> > It seems to me there are lots of possibilites. For instance,
given any
> > triad, there is a unique hexachord, and hence a unique
complementary
> > triad. That ought to be worth something.
>
> Yes...say does anyone know where I could get a listing of the 132
> Steiner system blocks for S(5,6,12) and the 77 blocks for S(3,6,22)?
> (Both blocks of hexachords in their respective systems) I might try
> sci.math...
>
> Paul Hj

Also, the order of each Mathieu group is a multiple of C(v,t)

The remaining factors:

M11 is 24 2^3 * 3
M12 is 120 2^3 * 3 * 5
M22 is 288 2^5 * 3^2
M23 is 1152 2^7 * 3^2
M24 is 5760 2^7 * 3^2 * 5

I understand the C(v,t) portion (C(22,3) in the case of M22 and S
(3,6,22)) Where do these other factors come in? Why is the number
of triads,(1540) for example, multiplied by 288 to give the order of
M22 (443,520). I guess I need some help understanding the
automorphism groups of Steiner systems --- Thanks

> Yes...say does anyone know where I could get a listing of the 132
> Steiner system blocks for S(5,6,12) and the 77 blocks for S(3,6,22)?
> (Both blocks of hexachords in their respective systems) I might try
> sci.math...
>
> Paul Hj

Someone on sci.math recommended that I use Golay codes to generate
Steiner Systems (blocks) . Anyone familiar with algebraic coding and
Golay codes?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> Someone on sci.math recommended that I use Golay codes to generate
> Steiner Systems (blocks) . Anyone familiar with algebraic coding and
> Golay codes?

Chapter 11 of "Sphere Packings, Lattices and Groups" is "The Golay
Codes and The Mathieu Groups", which concentrates mostly on M24 (since
M22 and M23 can be dervied from it.) I don't have Huppert and
Blackburn, but if I recall correctly they started with the group of
the projective plane over the field GF(4) of four elements, which is a
permutation group on the 21 points of the plane, and then worked their
way up.

Unless you are very ambitious the easiest way to get these would not
be to construct them yourself, but to find someone who has them; I
used to be that was a long time ago. I think there may be a Lecture
Notes in Math volume which actually lists them.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Someone on sci.math recommended that I use Golay codes to
generate
> > Steiner Systems (blocks) . Anyone familiar with algebraic coding
and
> > Golay codes?
>
> Chapter 11 of "Sphere Packings, Lattices and Groups" is "The Golay
> Codes and The Mathieu Groups", which concentrates mostly on M24
(since
> M22 and M23 can be dervied from it.) I don't have Huppert and
> Blackburn, but if I recall correctly they started with the group of
> the projective plane over the field GF(4) of four elements, which
is a
> permutation group on the 21 points of the plane, and then worked
their
> way up.
>
> Unless you are very ambitious the easiest way to get these would
not
> be to construct them yourself, but to find someone who has them; I
> used to be that was a long time ago. I think there may be a Lecture
> Notes in Math volume which actually lists them.

Thanks. I'll keep looking. I printed out a list that I found on the
Internet, and then found that it is missing two sets (S(5,6,12)
should have 132 hexads, and they listed only 130.) Shows you can't
trust the internet. Which was too bad because it was listed by
weight, which I found useful. The funny thing is the list is still
perfectly symmetrical, so I don't know where the missing sets would
be. I will try Lecture Notes in Math Thanks again