staring at stars [CPS sets]

Paul Erlich wrote, TD 668:

> 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.
>
Well, this is very "cool" to look at, but I'm not sure I'm learning
anything by staring at these stars... What determines the vector
positions of these factors in space?? I fear I'm going to need more
help...
__________ _______ ____ __ _
Joseph Pehrson

--- In tuning@egroups.com, Joseph Pehrson <josephpehrson@c...> wrote:
> Paul Erlich wrote, TD 668:
>
> > 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.
> >
> Well, this is very "cool" to look at, but I'm not sure I'm learning
> anything by staring at these stars... What determines the vector
> positions of these factors in space?? I fear I'm going to need more
> help...
> __________ _______ ____ __ _
> Joseph Pehrson

First of all, the entire figure is (in case you haven't caught this
already) an illustration of Pascal's triangle and its application to
combinatorics. Pascal's triangle, you may know, begins with a 1 and
each entry below that is the sum of the two entries below it
(assuming a sea of zeros where there are no numbers). Hence:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

The kth number (counting the first as zero) in the nth row (again
counting the first as zero) is known mathematically as "n choose k",
or the number of ways of choosing k objects out of a total of n. This
mathematical function is known as the "combination" (hence the first
word in "Combination Product Set). It is symbolized by

(n)
( )
(k),

except that there should be one big left parenthesis instead of
three, and one big right parenthesis instead of three. How does
Pascal's triangle work to give you this information? Well, let's say
you already know the number of ways of choosing k-1 objects out of n-
1 objects is

(n-1)
( )
(k-1),

and that the number of ways of choosing k objects out of n-1 objects
is

(n-1)
( )
( k ).

Now let's say you add one new object to your universe, so now you
have a total of n. How many ways are there to choose k objects now?
Well, you can either choose k objects out of the n-1 you had before,
and there are

(n-1)
( )
( k )

ways of doing that; or you can choose the new object along with k-1
objects out of the n-1 you had before, and there are

(n-1)
( )
(k-1)

ways of doing that. So the total number of possibilities is the sum
of those two numbers; i.e.,

(n) (n-1) (n-1)
( ) = ( ) + ( )
(k) (k-1) ( k ).

So given the facts that there is only 1 way to choose no objects and
only 1 way to choose all the objects; i.e.,

(n) (n)
( ) = 1, ( ) = 1,
(0) (n)

we have the outer "shell" of ones of Pascal's triangle; the previous
formula fills out the interior.

Now look at http://www.anaphoria.com/dal16.html again. Note that the
figures are arranged in a triangle. Count the number of "dots" in
each figure. It's Pascal's triangle!

Okay, so what about the shapes? Let's take the fourth row (n=3) as an
example, since it's simple but not so simple as to be "degenerate".
The three "objects" are the multiplicative factors A, B, and C. The
first "dot" on the left represents the set obtained by multiplying no
factors, which Wilson represents with the symbol for the empty set,
an "O" with a slash through it. He also writes the symbol

(0)
( )
(3)

to mean the mathematical operation 3-choose-0 (he's got it upside-
down relative to me): you're choosing zero elements out of a set of
3. Since 3-choose-0 equals 1 (as you can read off Pascal's triangle),
Erv writes a "1" near the dot as well. The result of choosing zero
factors out of three is a 1-note "scale", which Erv calls a "monany".

The next shape (representing 3-choose-1, which equals 3) after the
dot is a triangle:

B
/ \
/ \
A-----C

Choosing 1 element at a time out of 3 leads to three possible
products, namely the factors themselves: A, B, and C. This three-note
set Erv calls a "Triany" but it's really just a triad. Here Erv has
implicitly chosen to represent each of the "consonant" ratios A/C,
A/B, and B/C, by horizontal, upward-sloping, and downward-sloping
lines, respectively. This choice will hold through the rest of the
row.

The next shape is the upside-down triangle for 3-choose-2 = 3:

A*B---B*C
\ /
\ /
A*C

Clearly, each note in this triad comes from choosing two elements out
of the set and multiplying them together (thus the second word
of "Combination Product Set"). How does the configuration of these
notes come about? Well, note that the interval (remember, intervals
are _ratios_) between A*B and B*C is

A*B A
--- = -
B*C B,

and we have already chosen the horizontal line to represent the
interval A/B, with the numerator on the left end of the line and the
denominator on the right end. Similarly, the other two, diagonal
lines, represent the same interval, in the same orientation, as they
did in the previous triange. Check for yourself!

Finally, we come to 3-choose-3, where there is only one note,
obtained by multiplying all the elements: A*B*C. Since it is only one
note, it is really the same as the 3-choose-0 CPS: there is only one
kind of monany, but we've seen two kinds of trianies.

Now let's make this a little more concrete by assigning values to the
factors (you can assign any values you like, but I will deal with the
most familiar and useful case). A, B, and C will be 1, 3, and 5
(we'll allow factors of 2 for free due to octave equivalence). The
monanies are not very interesting, but the trianies are none other
than our familiar major and minor triad. The first one is 1:3:5, or
4:6:5, or the JI major triad; the second one is 1*3:1*5:3*5, or
3:5:15, or 12:10:15, or the JI minor triad.

The same logic applies to each row of the Figure 19
(http://www.anaphoria.com/dal16.html). The next row contains the
major tetrad and minor tetrad, and the hexany, which I've already
discussed quite a bit in the context of the four factors being 1, 3,
5, and 7. The following row shows pentads and dekanies; if we choose
1, 3, 5, 7, and 9 as our factors, the pentads are the complete otonal
and utonal chords of 9-limit harmony, and the dekanies are
interesting scales that you should work out for yourself as an
exercise. Note that in these diagrams and those in the next (and
last) row, not all the connecting lines (i.e., consonant intervals)
are drawn in by Erv; for example, in the 5-factor case he shows A/E
with a line but not B/D. One potentially confusing thing is that the
direction of A/E is parallel to that of B/D; hence sometimes these
intervals may overlap on the lattice diagram. So by only drawing some
of the lines, Erv staves off potentially confusing overlapping lines
by omitting lines for certain consonant intervals (fully half of them
in the 5-factor case, and two-thirds of them in the 6-factor case --
you will see the omitted lines as dotted lines in certain other
figures in the D'Alessandro paper).

Other than that, the orientations of the factors are simply chosen to
produce pretty, symmetrical shapes in which at least some of the
important intervallic and chordal relationships can be readily seen.
Here's a test: in the central diagram on the bottom, the Eikosany,
can you find the 30 consonant tetrads that Carl was referring to?
Each consonant tetrad will be configured exactly like either four out
of the six notes of the major hexad (the 1-out-of-6 CPS), or four out
of the six notes of the minor hexad (the 5-out-of-6 CPS). Recopying
the figure bigger and drawing in the dotted lines (as suggested by
the hexads in http://www.anaphoria.com/dal14.html) may help if you
have trouble. (Hint: of all the possible ways of taking four of the
six notes of the major hexad, and of all the possible ways of taking
four of the six notes of the minor hexad, the Eikosany will have
exactly one example of each -- 6-choose-4 is 15, and 15+15=30). If
this is too hard, go back to the hexany and show yourself how its 8
consonant triads (the ones that look like triangles with a note at
each vertex) are one each of the 4 possible ways of taking a triad
out of the major tetrad, and one each of the 4 possible ways of
taking a triad out of the minor tetrad. Then come back to the
Eikosany. If you want to "cheat", look at the factors that make up
the notes of the Eikosany (3 factors for each note) -- each consonant
otonal tetrad will have two factors in common to all notes (e.g.,
A*B*C, A*B*D, A*B*E, A*B*F), and each utonal tetrad will have two
factors absent from all notes (e.g., A*B*C, A*B*D, A*C*D, B*C*D).

Carl -- does that last hint for Joseph help you simplify your formula
any further?

I wrote:

> Pascal's triangle, you may know, begins with a 1 and
> each entry below that is the sum of the two entries below it

I meant the sum of the two entries _above_ it. Sorry!