Re: [tuning-math] Digest Number 858

Paul Hj wrote:

> --- 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?).

Yes - it's like magic. And if you want to do it reduced by M5 (the
operator that maps the chromatic scale to the cycle of fifths), you just
include those transforms too.

By the way I noticed that Julian Hook is giving a paper on Polya counting
methods in music theory at the national meeting of the American
Mathematical Society - they're having a special session on music and
maths, in Phoenix, in January 2004.

> 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'm emailing you a file of C(12,24), gzipped. Ask me if you've got any
questions. The only place I've got those files at the moment is a unix
account, with no public_html folder to share things in... if you have a
unix account and would like to email me a username/password, I could scp
them to you. Otherwise it'll be a bit trickier and you'll have to ask me
again in a couple of weeks as I'm a bit bogged down now!

> 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).

My files are really the result of a brute-force search, whose numbers I
checked by Polya afterwards.

> 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?

For that I'd need a library of bitwise operations that simulates 64-bit
architecture in C. And a faster processor! 31-tet took about 6 hours to
generate, I think... Each tet takes twice as long as the last one.

See you --Jon

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

> Yes - it's like magic. And if you want to do it reduced by M5 (the
> operator that maps the chromatic scale to the cycle of fifths), you
>just include those transforms too.

Cool - I've always called this "1-5" symmetry. After reducing for the
Z-relation, the count for hexachords is 35. Reducing for 1-5 symmetry
brings this down to 26. I use the letters of the alphabet to arrange
the 35 hexachords in a nice 7 X 5 grid, which I could upload to the
database. Each column is based on a particular interval, (1 through
5, excluding the tritone) in terms of high frequency of interval
count. Hexachords that have 1-5 symmetry are labeled, for example, B1
and B5. Oh, do you suppose you could give me the transforms for M5-
reduction?
>
> I'm emailing you a file of C(12,24), gzipped. Ask me if you've got
>any questions.

Thanx! I'll look at it.

Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, jon wild <wild@f...> wrote:
>

> > I'm emailing you a file of C(12,24), gzipped. Ask me if you've
> > got any questions.

Hmm! You get a count of 56822, and I get a count of 61822, for
C{12,24}. Off by 5000. I base my result on 924 symmetrical sets.
Perhaps I am wrong...Now to reduce for Z-relations...

Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
> > --- In tuning-math@yahoogroups.com, jon wild <wild@f...> wrote:
> >
>
> > > I'm emailing you a file of C(12,24), gzipped. Ask me if you've
> > > got any questions.
>
> Hmm! You get a count of 56822, and I get a count of 61822, for
> C{12,24}. Off by 5000. I base my result on 924 symmetrical sets.
> Perhaps I am wrong...Now to reduce for Z-relations...

Okay, here it is. Fed the file into Excel and used Dcount.

1 1020
2 41676
3 138
4 12512
5 0
6 876
7 0
8 600

So 1020 + 41676/2 + 138/3 + 12512/4 + 876/6 + 600/8 = 25253

Paul Hj. wrote: (we need the second letter in case Paul Hahn comes back!)

> > > I'm emailing you a file of C(12,24), gzipped. Ask me if you've
> > > got any questions.
>
>> Hmm! You get a count of 56822, and I get a count of 61822, for
>> C{12,24}. Off by 5000. I base my result on 924 symmetrical sets.
>> Perhaps I am wrong...Now to reduce for Z-relations...
>
> Okay, here it is. Fed the file into Excel and used Dcount.
>
> 1 1020
> 2 41676
> 3 138
> 4 12512
> 5 0
> 6 876
> 7 0
> 8 600
>
> So 1020 + 41676/2 + 138/3 + 12512/4 + 876/6 + 600/8 = 25253

I'm elsewhere now and the file is unavailable to me, but I'm positive
there were some Z-related groups of size 12 in there. Not sure why
our tallies don't match--do you want to apply your newfound knowledge and
see what Mr. Polya has to say about the matter, or should I?

--- In tuning-math@yahoogroups.com, jon wild <wild@f...> wrote:
>
> Paul Hj. wrote: (we need the second letter in case Paul Hahn comes
back!)
>
> > > > I'm emailing you a file of C(12,24), gzipped. Ask me if you've
> > > > got any questions.
> >
> >> Hmm! You get a count of 56822, and I get a count of 61822, for
> >> C{12,24}. Off by 5000. I base my result on 924 symmetrical sets.
> >> Perhaps I am wrong...Now to reduce for Z-relations...
> >
> > Okay, here it is. Fed the file into Excel and used Dcount.
> >
> > 1 1020
> > 2 41676
> > 3 138
> > 4 12512
> > 5 0
> > 6 876
> > 7 0
> > 8 600
> >
> > So 1020 + 41676/2 + 138/3 + 12512/4 + 876/6 + 600/8 = 25253
>
> I'm elsewhere now and the file is unavailable to me, but I'm
positive
> there were some Z-related groups of size 12 in there. Not sure why
> our tallies don't match--do you want to apply your newfound
knowledge and
> see what Mr. Polya has to say about the matter, or should I?

Can you? A couple things - my error - 56822 is correct. Also, the
above tally adds up to 56822. Definitely did not see 12 in your file
(I scrolled all the way through it). 25253 is prime, which is a
disappointment, oh well. I'm betting 25253 is right.
(How would Polya calculate that?)