Re: [tuning] Re: The CPS blues...[combination product sets]

Joseph wrote:
>Is that because it is the 13th scale in the Hexany group that was
>developed in Scala. The factor 13, obviously, has nothing to do with
>the factoring of this scale, yes (??)

Yes, the numbering of the hexany scales in the archive is arbitrary.

Carl wrote:
>There doesn't seem to be a way to create stellated CPS's in Scala, yet
>the scale archive contains files like steleiko.scl... Were those made
>by hand?

Yes, I think by John Chalmers. As far as I know, Erv Wilson never gave
a complete definition how to stellate the general n-out-of-m CPS.
The only thing I did once is derive a possible formula for the number
of tones in a stellated CPS.

Manuel Op de Coul coul@ezh.nl

MANUEL!
his definition (word of month) would be, complete each of the 8 triads in the hexany with
the missing tetrad member. i believe in his letter to Fokker he includes a written definition
which i can't get to at the moment!

MANUEL.OP.DE.COUL@EZH.NL wrote:

> Yes, I think by John Chalmers. As far as I know, Erv Wilson never gave
> a complete definition how to stellate the general n-out-of-m CPS.
> The only thing I did once is derive a possible formula for the number
> of tones in a stellated CPS.
>

-- Kraig Grady
North American Embassy of Anaphoria island
www.anaphoria.com

Erv Wilson has indeed defined and calculated all CPSs through the 70 tone,
4-out-of-8 factor hebdomakontany.

Here's a beginner's guide to making stellate n-anies.

Example: a stellate hexany.

A hexany consists of six tones, each generated by taking two out of out a
set of four factors. So, given the factors A,B,C & D, the hexany has the
tones AB, AC, AD, BC, BD, CD. (Any numbers can serve as factors, a good set
for beginners is the set of factors 1,3,5,7).

A stellate hexany will add the tones necessary to make complete tetrads
(A:B:C:D or 1/A:B:C:D) out of each triad found in the basic hexany. The
eight triads are:

AB : AC : AD (all tones with A as a factor)
AB : BC : BD (all tones with B as a factor)
AC : BC : CD (all tones with C as a factor)
AD : BD : CD (all tones with D as a factor)
BC : BD : CD (all tones without A as a factor)
AC : AD : CD (all tones without B as a factor)
AB : AD : BD (all tones without C as a factor)
AB : AC : BC (all tones without D as a factor)

Adding the following eight tones: AA, BB, CC, DD, BCD, CD/B, BD/C, BC/D,
completes the tetrads:

AA : AB : AC : AD
AB : BB : BC : BD
AC : BC : CC : CD
AD : BD : CD : DD
BCD : CD : BD : BC
CD : CD/B : AD : AC
BD : AD : BD/C : AB
BC : AC : AB : BC/D

An eikosany and a hebdomakontany will be made stellate by likewise
completing, respectively, all hexads and ogdoads.

So, Daniel, stellation is only defined for an m-out-of-n CPS if n = m*2,
right?

From: Paul H. Erlich <PERLICH@ACADIAN-ASSET.COM>

> So, Daniel, stellation is only defined for an m-out-of-n CPS if n = m*2,
> right?
>

I believe so -- the CPSs where n does not equal 2m are not symmetrical when
put on a triangular graph (so they don't look like stars or mandalas); but
the point is a bit trivial as these sets are all subsets of the next-higher
n=2m CPS.

>I believe so -- the CPSs where n does not equal 2m are not symmetrical when
>put on a triangular graph

Well, they are symmetrical, but less so (i.e., with a smaller symmetry
group).

>but
>the point is a bit trivial as these sets are all subsets of the next-higher
>n=2m CPS.

Can you elaborate on what you see as "trivial" here?

And next-higher? Meaning with more factors? The choice of whice additional
factor(s) to include seems completely divorced from the concept of
stellation.

I think the best single illustration of the CPS concept is

http://www.anaphoria.com/dal16.html

One should be able to grasp the CPS concept, as well as its particular
manifestations, by examining this diagram.

From: Paul H. Erlich <PERLICH@ACADIAN-ASSET.COM>
> >I believe so -- the CPSs where n does not equal 2m are not symmetrical
when
> >put on a triangular graph
>
> Well, they are symmetrical, but less so (i.e., with a smaller symmetry
> group).
>

No -- sorry, I'm wrong about this. On page 3 of the "Letter to John
Chalmers", Wilson indicates five possible solutions for a "stellated
dekateserany". (All five are present in the stellated Eikosany.) It
appears that the n=2m CPSs will each have single stellations while the
non-2=2m CPSs will have m possible stellations.

> >but
> >the point is a bit trivial as these sets are all subsets of the
next-higher
> >n=2m CPS.
>
> Can you elaborate on what you see as "trivial" here?
>

The 2)4 tone hexany will contain all the 1)1, 1)2, 2)2, 1)3, 2)3, and 3)3
subsets; the same goes for the stellate form.

> And next-higher? Meaning with more factors? The choice of whice additional
> factor(s) to include seems completely divorced from the concept of
> stellation.
>

Yes, with more factors. The structure of the CPS system (which Wilson -- in
the _Letter to A.D. Fokker_ associates with Pascal's triangle) is
independent of any particular factors, and the assignment of factors to the
CPS graphs is arbitrary.

Reading through the Fokker and Chalmers letters, I notice that Wilson does
not indicate how Pascal's triangle might be used to determine the number of
tones in a stellated CPS -- anyone have a shortcut?

Daniel Wolf wrote,

>On page 3 of the "Letter to John
>Chalmers", Wilson indicates five possible solutions for a "stellated
>dekateserany". (All five are present in the stellated Eikosany.)

Any chance this letter might be made available for perusal?

Daniel & Paul!
Have a request in to put both "letter to Chalmers' and "letter to fokker" up and would
expect the latter to go up sooner (i have a copy here). On pages 2 of the latter and the page
where the stellate hexany is illustrated he refers to this figure as a mandala. I would guess
he chose this term out of its property of being self mirroring "union of four elements",
thus a symbol of wholeness. He also states "Analogs to this figure exist in hexadic and
ogdoadic, in all even-numbered orders of Tone Space."
As much as Pascal triangle predicts many aspect of the CPS, much appears to be left out.
which lead me to compile many of it properties independent of the triangle as a reference, it
might be possible, but i fail to see an easy way. The easiest short cut seems to be to count
the number of factors involved and double it to include the reciprocals. Whereas one might
not want a stellate of every triad found in (for instance) the 4)6 set, as this set contains 6
subharmonic pentads, I have suggested and discussed with Erv the possibility of completing
these pentads into hexads by 6 subharmonic flutes. such a thing he found agreeable. It could
be mirrored with 6 harmonic flutes for that matter based on the 2)6 cps. Such factors could
play an important role in finding the the best constant structure for these structures.

Daniel Wolf wrote:

> From: Paul H. Erlich <PERLICH@ACADIAN-ASSET.COM>
> > >I believe so -- the CPSs where n does not equal 2m are not symmetrical
> when
> > >put on a triangular graph
> >
> > Well, they are symmetrical, but less so (i.e., with a smaller symmetry
> > group).
> >
>
> No -- sorry, I'm wrong about this. On page 3 of the "Letter to John
> Chalmers", Wilson indicates five possible solutions for a "stellated
> dekateserany". (All five are present in the stellated Eikosany.) It
> appears that the n=2m CPSs will each have single stellations while the
> non-2=2m CPSs will have m possible stellations.
>
> > >but
> > >the point is a bit trivial as these sets are all subsets of the
> next-higher
> > >n=2m CPS.
> >
> > Can you elaborate on what you see as "trivial" here?
> >
>
> The 2)4 tone hexany will contain all the 1)1, 1)2, 2)2, 1)3, 2)3, and 3)3
> subsets; the same goes for the stellate form.
>
> > And next-higher? Meaning with more factors? The choice of whice additional
> > factor(s) to include seems completely divorced from the concept of
> > stellation.
> >
>
> Yes, with more factors. The structure of the CPS system (which Wilson -- in
> the _Letter to A.D. Fokker_ associates with Pascal's triangle) is
> independent of any particular factors, and the assignment of factors to the
> CPS graphs is arbitrary.
>
> Reading through the Fokker and Chalmers letters, I notice that Wilson does
> not indicate how Pascal's triangle might be used to determine the number of
> tones in a stellated CPS -- anyone have a shortcut?
>
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-- Kraig Grady
North American Embassy of Anaphoria island
www.anaphoria.com

Daniel wrote,

>It
>appears that the n=2m CPSs will each have single stellations while the
>non-2=2m CPSs will have m possible stellations.

Hmm. I would think the 1-out-of-3 and 2-out-of-3 CPSs would each have one
stellation:

ab/c------a-----ac/b a^2
\ / \ / / \
\ / \ / / \
\ / \ / / \
b-------c and ab-------ac
\ / / \ / \

\ / / \ / \
\ / / \ / \
bc/a b^2------bc-----c^2

while for the 1-out-of-4 (and 3-out-of-4) CPSs, one could at best complete
the tetrad at each incomplete dyad:

cd/b------cd/a
\`. ,'/
ac/b-----\-`c'-/----bc/a
\`. ,\/|\/. ,'/
\ac/d /\|/\bc/d /
ad/b--------d-------bd/a
\`.\ /,'/ \`.\|/,'/
\ `a'-/---\-`b' /
\ |\/. , \/| /
\|/\ab/d /\|/
ad/c \ | /bd/c
\|/
ab/c

(which is equivalent to completing the hexany at each triad).

What other possible stellations are there, and in what sense are they
trivial?

Daniel!
What is the "stellated dekateserany". It escapes me for the moment and what are the
variations possible. It sounds like in the Eikosany, the hexany can be stellated by tone
forming tetrads using the two factors not found in the hexany. For instance in the 1-3-7-9
hexany found in the 1-3-7-9-11-15 eikosany, 11 and 15 and their reciprocals are added to make
tetrads.

> Daniel Wolf wrote,
>
> >On page 3 of the "Letter to John
> >Chalmers", Wilson indicates five possible solutions for a "stellated
> >dekateserany". (All five are present in the stellated Eikosany.)
>
>

-- Kraig Grady
North American Embassy of Anaphoria island
www.anaphoria.com

Daniel Wolf wrote,

>On page 3 of the "Letter to John
>Chalmers", Wilson indicates five possible solutions for a "stellated
>dekateserany".

Wait a minute -- I thought the dekatesserany was the already stellated
hexany (see http://www.ixpres.com/interval/dict/dekates.htm). Is there a
bare CPS that is a dekatesserany? It seems unlikely, since the number 14
doesn't appear until the 14th row of Pascal's triangle.

Still curious about the 3- and 4-factor cases I asked about yesterday . . .

Manuel Op de Coul wrote,

>If I apply this to the 1)4 tetrany, the outcome is 4+12=16.
>This is when all 6 dyads are completed to tetrads.

That is the case I illustrated in my post of Thu 6/8/00 7:39 PM.

>But I think another way to stellate it is to complete the 4 triads
>to tetrads. Then the number of tones is 4+4=8.

Nice idea, but you can't do that. The only complete tetrad those triads are
a part of is the original tetrad -- no other complete tetrads contain them.
Remember, the 3-d space is filled with tetrahedra and octahedra; the
tetrahedra meet at edges but not at faces. If you stellate the faces of the
tetrahedron with octahedra (hexanies), you get the exact same 16-tone
structure mentiones above.

>Your formula gives 4+2=6 tones for the 3)4 tetrany, so it's not
>symmetrical between M and N-M, is that correct?

ooh -- not a good sign.

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: N! N! N!
: ------- + -------------- + ------------
: M!(N-M)! (M-2)!(1+N-M)! (N-M-2)!(1+M)!

: ...but it looks like we're facing some negative factorials. I also tried
: expressing it as a single fraction, but got some nasty polynomials.

The first negative factorial results from dividing by (M-1), which
is dividing by zero if M=1, so you mustn't do that.
We can rewrite the last two terms to a single fraction and with a little
bit of work we get

3 3
N! (M + (N-M) - N) N!
------- + ------------------
M!(N-M)! (M+1)!(N-M+1)!

And this form nicely shows the symmetry between N and N-M.

Manuel Op de Coul coul@ezh.nl

PLEASE! DON'T SEND ANY MESSAGES TO FOLLOWING ADRESS!!
BARBARAD@CYBERHIGHWAY.NET

I AM SICK RECEVING THIS MASSAGES I DO NOT NOW WHERE THEY COM FROM

On June 8, I demonstrated the (partially) stellated 1)4 CPS:

>complete the tetrad at each incomplete dyad:

> cd/b------cd/a
> \`. ,'/
>ac/b-----\-`c'-/----bc/a
> \`. ,\/|\/. ,'/
> \ac/d /\|/\bc/d /
>ad/b--------d-------bd/a
> \`.\ /,'/ \`.\|/,'/
> \ `a'-/---\-`b' /
> \ |\/. , \/| /
> \|/\ab/d /\|/
> ad/c \ | /bd/c
> \|/
> ab/c

>(which is equivalent to completing the hexany at each triad).

So, if I understand Carl and Manuel correcly, their proposal for the
"superstellation," which corresponds with Wilson's original definition of
stellation at least for (n-2))n CPSs, would involve completing all the
triads to tetrads. Let's first redraw the diagram above with all consonant
intervals drawn in so that these triads (as well as the hexanies) are more
clearly seen:

cd/b------cd/a
,'/ \`. ,'/ \`.
ac/b-/---\-`c'-/---\bc/a
|\/. ,\/|\/. ,\/|
|/\ac/d-/\|/\bc/d /\|
ad/b--------d-------bd/a
\`.\|/,'/ \`.\|/,'/
\ `a'-/---\-`b' /
\ |\/.\ /,\/| /
\|/\ab/d /\|/
ad/c-----/bd/c
`.\|/,'
ab/c

(some of the consonant intervals are almost completely hidden).

Now, each of the four hexanies has four of its triads as part of complete
tetrads, and the other four incomplete. Completing all of these would mean
adding 16 notes to this set, bringing the total to 32. However, Manuel's
"superstellated 1)4)" only has 20 notes.

However, looking back at Manuel's post, it seems we only want to complete
those triads formed entirely from tones added by the partial stellation, and
not the triads that include a tone from the original CPS (but why???). So
we'd only add 4 tones to the diagram above, giving:

acd/bb-----cd/b------cd/a-----bcd/aa
\`. ,'/ \`. ,'/ \`. ,'/
\ac/b-/---\-`c'-/---\bc/a /
\ |\/. ,\/|\/. ,\/| /
\|/\ac/d-/\|/\bc/d /\|/
ad/b--------d-------bd/a
\`.\|/,'/@\`.\|/,'/
\ `a'-/---\-`b' /
\ |\/.\|/,\/| /
\|/\ab/d /\|/
ad/c-----/bd/c
\`.\|/,'/
\ab/c /
\ | /
\|/
abd/cc

where the "@" represents abc/dd, whose connections with ac/d and bc/d are
completely obscured (it would probably have been better to diagram the
superstellated 3)4 CPS, instead).

It is curious how the (super)stellation of an otonal tetrad adds 10 utonal
tetrads, but no more otonal tetrads.

By the way, this diagram is also a depiction of the Rubik's Cube-like puzzle
"Pyraminx". It's one big tetrahedron, and it appears to be made up of a
bunch of smaller tetrahedra, but if you take it apart (I actually did this
as a child), you'll be suprised to see, in addition to 10 little tetrahedra,
the four octahedra (the hexanies) lurking within, each providing three of
the externally visible triangles. The innermost tetrahedron, opposite in
orientation to all the others as well as the big tetrahedon, is made up in
the puzzle of mostly empty space, with a tetrahedral joint in the center,
connecting the four octahedra and allowing each to rotate freely about an
axis passing through the nearest vertex of the big tetrahedron and through
the center. . .

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Please contact us at:
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All phone, fax and email remain the same.

Manuel, now that I think about it, your construction appears eminently
logical. Once again, I'd like to state my conjecture (which I mentioned in a
previous response to Carl) as to the "why" of Wilson stellation, or what
we're calling "superstellation". It seems to get you the biggest fully
connected region in the lattice that includes the initial CPS but no
additional copies of that CPS. So in the case I just discussed, the
(super)stellated otonal tetrad, trying to complete any more dyads or triads
will result in the appearance of another otonal tetrad. As it stands, it
contains only the original otonal tetrad, as well as 10 utonal tetrads and 4
hexanies.

_________________________________________________

We have moved!

As of June 26, 2000, Acadian Asset Management will be at a
new location in Boston's financial district.

Please contact us at:
Acadian Asset Management
Ten Post Office Square, 8th Floor
Boston, MA 02109.

All phone, fax and email remain the same.