Wedgies from ets and generators

It's easy to compute a wedgie from a pair (n,g) where n is an equal
temperament and g is a generator, and this could be added to the list
of ways of cooking up candidate wedgies.

We first reduce n to m, where gcd(m,g)=1, by dividing out common
factors. We then form one val vector, the generator map, by taking
[0, hn(3)/g mod m, hn(5)/g mod m, hn(7)/g mod m], where the modulo is
taken in a range -m/2 < i <= m/2. The other val vector is simply the
n et map: [n, hn(3), hn(5), hn(7)]. We take the wedge product of
these two, and reduce it to the standard form wedgie, and there it is.
Of course this is a little arbitary in borderline cases, but those
are the least interesting anyway, and if desired a modulo in the range
-m/2 <= i < m/2 could also be tried.

In-Reply-To: <9v15pt+7bem@eGroups.com>
Gene wrote:

> It's easy to compute a wedgie from a pair (n,g) where n is an equal
> temperament and g is a generator, and this could be added to the list
> of ways of cooking up candidate wedgies.

This is a pair of numbers?

> We first reduce n to m, where gcd(m,g)=1, by dividing out common
> factors. We then form one val vector, the generator map, by taking
> [0, hn(3)/g mod m, hn(5)/g mod m, hn(7)/g mod m], where the modulo is
> taken in a range -m/2 < i <= m/2. The other val vector is simply the
> n et map: [n, hn(3), hn(5), hn(7)].

Those two vals are what I get by going from unison vectors (or a wedgie)
to linear temperaments. So hn(i) is the approximation to the ratio i/1 in
the linear temperament. What you're showing is the inverse process.

hn(3)/g mod m must be equivalent to Carey and Clampitt's inverse modulo.
We need integers p(i) such that (g*p(i)) mod m = hn(i) mod m. That's the
same as (g*p(i) - hn(i)) mod m = 0. That's how I understand it. If you
prefer to write it as p(i) = hn(i)/g mod m, all well and good.

> We take the wedge product of
> these two, and reduce it to the standard form wedgie, and there it is.

In my case "reduce it to the standard form wedgie" means getting the
complement() and simplify()ing it.

> Of course this is a little arbitary in borderline cases, but those
> are the least interesting anyway, and if desired a modulo in the range
> -m/2 <= i < m/2 could also be tried.

Yes, there are additional pieces of information you need, beyond the ET
and the generator. You need the list of prime intervals. You need a rule
for approximating them by integers. And you need a rule for choosing the
best of the mod n mappings. You won't get the simplest result for
11-limit Meantone31 by your rule. The 11-direction will come out as 13
fourths rather than 18 fifths. By some measures, this will be the
simplest 11-limit mapping of 31, so it's a big one to miss.

It is indeed easy when you ignore all the complications.

Graham

--- In tuning-math@y..., graham@m... wrote:

> > It's easy to compute a wedgie from a pair (n,g) where n is an
equal
> > temperament and g is a generator, and this could be added to the
list
> > of ways of cooking up candidate wedgies.
>
> This is a pair of numbers?

Correct, as for instance (31,13)

> Those two vals are what I get by going from unison vectors (or a
wedgie)
> to linear temperaments. So hn(i) is the approximation to the ratio
i/1 in
> the linear temperament.

If "i" is a rational number, hn(i) is how many steps in the n-et are
used to represent it.

> hn(3)/g mod m must be equivalent to Carey and Clampitt's inverse
modulo.

I don't know about Carey and Clampitt, but this is standard
mathematical notation. If hn(3)/g is not divisable by m then we may
reduce mod m. If you want to get fancy it's a matter
of "localization"; if hn(3)/g is in the localization of the integers
Zm at m, meaning reduced fractions whose numerator may be any integer
but whose denomimator is relatively prime to m, then we may reduce Zm
mod m by taking Zm/mZm just as we do Z/mZ.

> Yes, there are additional pieces of information you need, beyond
the ET
> and the generator. You need the list of prime intervals. You need
a rule
> for approximating them by integers. And you need a rule for
choosing the
> best of the mod n mappings. You won't get the simplest result for
> 11-limit Meantone31 by your rule. The 11-direction will come out
as 13
> fourths rather than 18 fifths.

It comes out as 18 when I use an alternative program which centers
around the mean of the 7-limit generators.

By some measures, this will be the
> simplest 11-limit mapping of 31, so it's a big one to miss.

If you really don't want to miss anything you could take all the
different equivalence possibilities up to your generator limit, and
test all of them. If I use the g^(5/3) cents measure, the two come
out almost the same: the 18 version as 313.6, and the -13 version as
311.4; presumably we want both.