polya counting method (fwd)

Not long ago I sent a brief note about this to someone else by email. In
case it's useful here, I'm forwarding it. I don't think there's anything
in it that wasn't in Robert Walker's messages of December 2000.

---------- Forwarded message ----------
Date: Mon, 21 Apr 2003 23:59:22 -0400 (EDT)
From: jon wild <wild@fas.harvard.edu>
Subject: polya counting method

Here v briefly is Polya's algorithm, as applied to counting the set-class
population of any chromatic universe.

For each operation in the equivalence group (i.e. each transposition and
each inversion) you look at its cyclic structure. E.g. mod 12, the 12
transpositions are as follows:

T_0: 12 cycles of 1 note
T_1: 1 cycle of 12 notes
T_2: 2 cycles of 6 notes
T_3: 3 cycles of 4 notes
T_4: 4 cycles of 3 notes
T_5: 1 cycle of 12 notes
T_6: 6 cycles of 2 notes
T_7: 1 cycle of 12 notes
T_8: 4 cycles of 3 notes
T_9: 3 cycles of 4 notes
T_10: 2 cycles of 6 notes
T_11: 1 cycle of 12 notes

For the inversions, the odd ones are all

I_odd: 6 cycles of 2 notes.

The even ones are all:

I_even: 5 cycles of 2 notes and 2 cycles of 1 note.

Now, for each operation you write (x^n + 1)^b, where b is the number of
cycles of length n (e.g. for T8 you write (x^3 + 1)^4 )

For any operation that has cycles of different lengths, like an even
inversion, you do the above for each cycle length and multiply together,
so for e.g. I0 you write (x^2 + 1)^5 * (x+1)^2

Now you have an expression for each operation, which you add together and
divide by the total number of operations (24), obtaining:

((x+1)^12 + 7*(x^2+1)^6 + 2(x^3+1)^4 + 2(x^4+1)^3 + 6(x^2+1)^5*(x+1)^2
+2(x^6+1)^2 + 4(x^12+1))/24

If you expand this polynomial (I advise the use of a software package
like Maple to do this), you get

x^12 + x^11 + 6x^10 + 12x^9 + 29x^8 + 38x^7 + 50x^6 + 38x^5 + 29x^4 +
12x^3 + 6x^2 + x + 1

and the coefficient of x^n in this expression is the number of n-chords.

--- In tuning-math@yahoogroups.com, jon wild <wild@f...> wrote:
> If you expand this polynomial (I advise the use of a software
package
> like Maple to do this), you get
>
> x^12 + x^11 + 6x^10 + 12x^9 + 29x^8 + 38x^7 + 50x^6 + 38x^5 +
29x^4 +
> 12x^3 + 6x^2 + x + 1
>
> and the coefficient of x^n in this expression is the number of n-
chords.

Simply amazing. Will this work for finding all the Tn types (if you
leave out the terms for inversion?). And in regards to your files,
would it be possible to post them on the internet somewhere (I have
DSL) I'm especially interested in sets like C{24,12}.

I know how to reduce for and then count Tn and TnI types (using a
less advanced technique, and then only for each C{m,n} at a time). Of
course, to reduce for Z-relations all you can do is count the
interval vectors. But as you said that is pretty trivial.

I've been able to run certain sets (in IGOR) for up to 30 in
the "numerator" and 9 in the "denominator" for C{n,m}(Sorry for the
bad terminology), counting the interval vectors. Is there anyway you
could go higher than n=31? I'm looking for patterns in the factors
of interval vector counts etc. The fact that you can count all the
subsets for ETs up to 31 is very cool, I'm sure that would keep
me busy for hours. Thanks again

Paul

--- In tuning-math@yahoogroups.com, jon wild <wild@f...> wrote:

> If you expand this polynomial (I advise the use of a software
package
> like Maple to do this), you get
>
> x^12 + x^11 + 6x^10 + 12x^9 + 29x^8 + 38x^7 + 50x^6 + 38x^5 +
29x^4 +
> 12x^3 + 6x^2 + x + 1
>
> and the coefficient of x^n in this expression is the number of n-
chords.

Thanks; I was just looking at this polynomial and trying to reverse
engineer the process. This isn't hard at all, is it?

I would just love to see the cyclic structures for other symmetries.
Like S3XC4 or S3XS4 etc. Do you know if Robert Walker or anyone has
done this?

Paul

--- In tuning-math@yahoogroups.com, jon wild <wild@f...> wrote:
>
> Not long ago I sent a brief note about this to someone else by
email. In
> case it's useful here, I'm forwarding it. I don't think there's
anything
> in it that wasn't in Robert Walker's messages of December 2000.
>
>
> ---------- Forwarded message ----------
> Date: Mon, 21 Apr 2003 23:59:22 -0400 (EDT)
> From: jon wild <wild@f...>
> Subject: polya counting method
>
>
> Here v briefly is Polya's algorithm, as applied to counting the set-
class
> population of any chromatic universe.
>
> For each operation in the equivalence group (i.e. each
transposition and
> each inversion) you look at its cyclic structure. E.g. mod 12, the
12
> transpositions are as follows:
>
> T_0: 12 cycles of 1 note
> T_1: 1 cycle of 12 notes
> T_2: 2 cycles of 6 notes
> T_3: 3 cycles of 4 notes
> T_4: 4 cycles of 3 notes
> T_5: 1 cycle of 12 notes
> T_6: 6 cycles of 2 notes
> T_7: 1 cycle of 12 notes
> T_8: 4 cycles of 3 notes
> T_9: 3 cycles of 4 notes
> T_10: 2 cycles of 6 notes
> T_11: 1 cycle of 12 notes
>
> For the inversions, the odd ones are all
>
> I_odd: 6 cycles of 2 notes.
>
> The even ones are all:
>
> I_even: 5 cycles of 2 notes and 2 cycles of 1 note.
>
> Now, for each operation you write (x^n + 1)^b, where b is the
number of
> cycles of length n (e.g. for T8 you write (x^3 + 1)^4 )
>
> For any operation that has cycles of different lengths, like an even
> inversion, you do the above for each cycle length and multiply
together,
> so for e.g. I0 you write (x^2 + 1)^5 * (x+1)^2
>
> Now you have an expression for each operation, which you add
together and
> divide by the total number of operations (24), obtaining:
>
> ((x+1)^12 + 7*(x^2+1)^6 + 2(x^3+1)^4 + 2(x^4+1)^3 + 6(x^2+1)^5*
(x+1)^2
> +2(x^6+1)^2 + 4(x^12+1))/24
>
> If you expand this polynomial (I advise the use of a software
package
> like Maple to do this), you get
>
> x^12 + x^11 + 6x^10 + 12x^9 + 29x^8 + 38x^7 + 50x^6 + 38x^5 +
29x^4 +
> 12x^3 + 6x^2 + x + 1
>
> and the coefficient of x^n in this expression is the number of n-
chords.

Jon (or anyone),

Please see question below Thanks!

--- In tuning-math@yahoogroups.com, jon wild <wild@f...> wrote:
>
>
> Not long ago I sent a brief note about this to someone else by
email. In
> case it's useful here, I'm forwarding it. I don't think there's
anything
> in it that wasn't in Robert Walker's messages of December 2000.
>
>
> ---------- Forwarded message ----------
> Date: Mon, 21 Apr 2003 23:59:22 -0400 (EDT)
> From: jon wild <wild@f...>
> Subject: polya counting method
>
>
> Here v briefly is Polya's algorithm, as applied to counting the set-
class
> population of any chromatic universe.
>
> For each operation in the equivalence group (i.e. each
transposition and
> each inversion) you look at its cyclic structure. E.g. mod 12, the
12
> transpositions are as follows:
>
> T_0: 12 cycles of 1 note
> T_1: 1 cycle of 12 notes
> T_2: 2 cycles of 6 notes
> T_3: 3 cycles of 4 notes
> T_4: 4 cycles of 3 notes
> T_5: 1 cycle of 12 notes
> T_6: 6 cycles of 2 notes
> T_7: 1 cycle of 12 notes
> T_8: 4 cycles of 3 notes
> T_9: 3 cycles of 4 notes
> T_10: 2 cycles of 6 notes
> T_11: 1 cycle of 12 notes
>
> For the inversions, the odd ones are all
>
> I_odd: 6 cycles of 2 notes.
>
> The even ones are all:
>
> I_even: 5 cycles of 2 notes and 2 cycles of 1 note.

Could anyone tell me why this is the cycle structure for inversions?
(Even and odd). I understand the cycle structure of transpositions,
but not inversions. Thanks!

> Now, for each operation you write (x^n + 1)^b, where b is the
number of
> cycles of length n (e.g. for T8 you write (x^3 + 1)^4 )
>
> For any operation that has cycles of different lengths, like an even
> inversion, you do the above for each cycle length and multiply
together,
> so for e.g. I0 you write (x^2 + 1)^5 * (x+1)^2
>
> Now you have an expression for each operation, which you add
together and
> divide by the total number of operations (24), obtaining:
>
> ((x+1)^12 + 7*(x^2+1)^6 + 2(x^3+1)^4 + 2(x^4+1)^3 + 6(x^2+1)^5*
(x+1)^2
> +2(x^6+1)^2 + 4(x^12+1))/24
>
> If you expand this polynomial (I advise the use of a software
package
> like Maple to do this), you get
>
> x^12 + x^11 + 6x^10 + 12x^9 + 29x^8 + 38x^7 + 50x^6 + 38x^5 +
29x^4 +
> 12x^3 + 6x^2 + x + 1
>
> and the coefficient of x^n in this expression is the number of n-
chords.
>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> > For the inversions, the odd ones are all
> >
> > I_odd: 6 cycles of 2 notes.
> >
> > The even ones are all:
> >
> > I_even: 5 cycles of 2 notes and 2 cycles of 1 note.
>
> Could anyone tell me why this is the cycle structure for inversions?
> (Even and odd). I understand the cycle structure of transpositions,
> but not inversions. Thanks!

Clearly inversions have to be products of transpositions, and can fix
at most one note, depending on whether the inversion is set up to fix
a note or an interval. So these are the choices. Anyway, you can
simply look at them and see what you get.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > > For the inversions, the odd ones are all
> > >
> > > I_odd: 6 cycles of 2 notes.
> > >
> > > The even ones are all:
> > >
> > > I_even: 5 cycles of 2 notes and 2 cycles of 1 note.
> >
> > Could anyone tell me why this is the cycle structure for
inversions?
> > (Even and odd). I understand the cycle structure of
transpositions,
> > but not inversions. Thanks!
>
> Clearly inversions have to be products of transpositions, and can
fix
> at most one note, depending on whether the inversion is set up to
fix
> a note or an interval. So these are the choices. Anyway, you can
> simply look at them and see what you get.
>
Wouldn't the whole tone series, inverted by T2I fix all 6 notes?
But I have another question. How would I do an polynomial Expansion
in Python? (For example, (x+1)^3 -> 3x^3+3x^2+3x+3. Thanks.

On 11/18/05, Paul G Hjelmstad <paul_hjelmstad@allianzlife.com> wrote:

> Wouldn't the whole tone series, inverted by T2I fix all 6 notes?
> But I have another question. How would I do an polynomial Expansion
> in Python? (For example, (x+1)^3 -> 3x^3+3x^2+3x+3. Thanks.

It sounds like you want a symbolic algebra package, rather than a
scripting language.

Graham