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set-theory quirk in 29-tet

🔗Jon Wild <wild@music.mcgill.ca>

10/5/2005 9:28:33 AM

Here's an odd bit of trivia I just found in my old files, though maybe Paul Hjelmstad will be the only interested party:

There are exactly twice as many decachordal set-classes as nonachordal set-classes in 29-tet.

173173 nonachords
346346 decachords

("set-class": equivalence class of pc collections, under transposition and inversion)

My guess is that's it's more of a coincidence, rather than something you can explain with a quirk of number theory in Z_29. Anyone know any better?

🔗hstraub64 <hstraub64@telesonique.net>

10/6/2005 6:11:02 AM

--- In tuning-math@yahoogroups.com, Jon Wild <wild@m...> wrote:
>
> Here's an odd bit of trivia I just found in my old files, though maybe
> Paul Hjelmstad will be the only interested party:
>

Oh, I am, too...

> There are exactly twice as many decachordal set-classes as nonachordal
> set-classes in 29-tet.
>
> 173173 nonachords
> 346346 decachords
>
> ("set-class": equivalence class of pc collections, under transposition
> and inversion)
>
> My guess is that's it's more of a coincidence, rather than something
> you can explain with a quirk of number theory in Z_29. Anyone know
> any better?

Well, according to basic combinatorics, the number of subsets of
cardinality k in a set of n elements is

[ n * (n-1) * (n-2) * ... * (n-k+1) ] / [ 1 * 2 * 3 * ... * k ]

From which we can derive that the number of subsets with cardinality
k+1 is the number of k-subsets multiplied with (n-k)/(k+1)

And in the case of n=29 and k=9, this yields exactly 2.

So it can be explained - but still a coincidence, in a way...
--
Hans Straub

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

10/6/2005 2:21:20 PM

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
wrote:
> --- In tuning-math@yahoogroups.com, Jon Wild <wild@m...> wrote:
> >
> > Here's an odd bit of trivia I just found in my old files, though
maybe
> > Paul Hjelmstad will be the only interested party:
> >
>
> Oh, I am, too...
>
> > There are exactly twice as many decachordal set-classes as
nonachordal
> > set-classes in 29-tet.
> >
> > 173173 nonachords
> > 346346 decachords
> >
> > ("set-class": equivalence class of pc collections, under
transposition
> > and inversion)
> >
> > My guess is that's it's more of a coincidence, rather than
something
> > you can explain with a quirk of number theory in Z_29. Anyone
know
> > any better?
>
> Well, according to basic combinatorics, the number of subsets of
> cardinality k in a set of n elements is
>
> [ n * (n-1) * (n-2) * ... * (n-k+1) ] / [ 1 * 2 * 3 * ... * k ]
>
> From which we can derive that the number of subsets with cardinality
> k+1 is the number of k-subsets multiplied with (n-k)/(k+1)
>
> And in the case of n=29 and k=9, this yields exactly 2.
>
> So it can be explained - but still a coincidence, in a way...
> --
> Hans Straub

Yes one runs across quite a bit of this sort of thing. It took me a
while to figure out the formulas for symmetric sets, (It varies,
depending on the evenness and oddness of n and k, among other things)
So a prime ET like 29 with k=10 is Combin(29,10)/29 and with k=9
is Combin(29,9)/29, which is exactly half, the number of symmetric
sets is 2002 and 1001 respectively, obviously one is half the other,
so the number of set types is just ((Combin(29,10)/29)+2002)/2 and
((Combin(29,9)/29)+1001)/2. Just to give an example of how weird
the symmetry sets are, in something as easy as 12-et you have
(1,1,6,5,15,10,20,10,15,5,6,1,1) but it can all be formulated.

🔗Jon Wild <wild@music.mcgill.ca>

10/7/2005 8:19:09 AM

Hans Straub wrote, quoting me:

>> There are exactly twice as many decachordal set-classes as nonachordal
>> set-classes in 29-tet.
>>
>> 173173 nonachords
>> 346346 decachords
>>
>> ("set-class": equivalence class of pc collections, under transposition
>> and inversion)
>>
>> My guess is that's it's more of a coincidence, rather than something
>> you can explain with a quirk of number theory in Z_29. Anyone know
>> any better?
>
> Well, according to basic combinatorics, the number of subsets of
> cardinality k in a set of n elements is
>
> [ n * (n-1) * (n-2) * ... * (n-k+1) ] / [ 1 * 2 * 3 * ... * k ]
>
> From which we can derive that the number of subsets with cardinality
> k+1 is the number of k-subsets multiplied with (n-k)/(k+1)
>
> And in the case of n=29 and k=9, this yields exactly 2.

But it's not this simple! You can't calculate the number of set-classes like this, because not all set-classes have the same degree of symmetry. (In other words, when you reduce the number of subsets into equivalence classes, not all equivalence classes have the same number of members.) Your approach would mean that there were twice as many octachords as heptachords in 23-tet (use k=7,n=23 in your formula), for example--which is false.

The only calculation I know of that gives this result directly (i.e. without enumerating all the set-classes explicitly) uses Polya's theorem, discussed here previously.

--Jon