The pigeonhole principle and strict propriety/coherence

It occurs to me that by the pigeonhole principle, in n-et a scale of
m>n/2 steps cannot be strictly proper. Hence, 12's "unique badness"
for the diatonic scale is simply due to the fact that it doesn't have
enough notes to do a proper job analyzing it. The same with 19 and
Meantone[12]; you really want 31 for looking at it and all the variants.

On 5/27/06, Gene Ward Smith <genewardsmith@coolgoose.com> wrote:
> It occurs to me that by the pigeonhole principle, in n-et a scale of
> m>n/2 steps cannot be strictly proper. Hence, 12's "unique badness"
> for the diatonic scale is simply due to the fact that it doesn't have
> enough notes to do a proper job analyzing it. The same with 19 and
> Meantone[12]; you really want 31 for looking at it and all the variants.

Isn't the 3-note scale in 5-equal [1,3,5] a counterexample to what you
just said?

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...>
wrote:

> Isn't the 3-note scale in 5-equal [1,3,5] a counterexample to what you
> just said?

Yes, and another counterexample is the 10-note proper scale in 19-et
which I had already posted about! Both of these are scales of all odd
integers up to an odd integer, which is always going to work. If you
have an odd number n, then you can put the n-1 numbers less than n
into (n-1)/2 pairs, [i,i+1] etc which was the kind of pigeonholing I
was thinking about; these would be the differences between interval
classes.

If n is even, you can go [1,4,5,8 ... 4i, 4i+1 ... n-3, n], which also
gives a proper scale, this time exactly n/2 in size. So you can get a
proper scale of size (n+1)/2 consisting of odd integers when n is odd,
and a scale of size n/2 consisting of integers congruent to 0 or 1 mod
4 when n is even. Proving these were maximal would do it.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> If n is even,

Divisible by 4. If it's only divisible by 2, it you can take a period
of n/2, and treat that like an odd scale, ending up with a scale of
size 2n+2.

On 5/27/06, Gene Ward Smith <genewardsmith@coolgoose.com> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <genewardsmith@...> wrote:
> >
> > If n is even,
>
> Divisible by 4. If it's only divisible by 2, it you can take a period
> of n/2, and treat that like an odd scale, ending up with a scale of
> size 2n+2.

But wait... Even in 12-equal, the octatonic scale [1, 3, 4, 6, 7, 9,
10, 12] is strictly proper, right? I'd say you're making some
unwarrented assumptions. =P

Keenan

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...>
wrote:

> But wait... Even in 12-equal, the octatonic scale [1, 3, 4, 6, 7, 9,
> 10, 12] is strictly proper, right? I'd say you're making some
> unwarrented assumptions. =P

I was leaping to conclusions without thinking enough, a bad habit of
mine. But excluding scales with a period smaller than an octave seems
like a reasonable way to reformulate the question. Actually, from
Balzano's point of view these fail to exhibit "uniqueness", so he'd
want to exclude them.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:

> I was leaping to conclusions without thinking enough, a bad habit of
> mine. But excluding scales with a period smaller than an octave seems
> like a reasonable way to reformulate the question. Actually, from
> Balzano's point of view these fail to exhibit "uniqueness", so he'd
> want to exclude them.

The basic thing which is screwing my conclusion up are classes of k
scale steps with only one interval in them. If that occurs, it means
that s[i+k]-s[k] = c, which means s is quasiperiodic with period k, so
k divides n, the et number. If we exclude those, or consider them
separately, then s is quasiperiodic with period n. There is then only
a single class with one interval in it, which is the octave/unison class.
s[i+n]-s[i] = m, where m is the number of scale elements. The other
classes must have at least two members since we are assuming no
periods smaller than an octave. So 2*(m-1)+1 <= n, which entails m <=
(n+1)/2, which is the pigeonhole business I was raving about.