Understanding hobbits

A long time ago, back before I had a halfway-decent understanding of
the actual math of regular temperament theory, I posted the following
algorithm for a hobbit. This was shot down by everyone:

Naive Mike's Algorithm:
1) Pick an n-limit regular temperament
2) For every ET that supports that temperament, a hobbit scale exists
3) To find the hobbit scale, start with that ET
4) Regress the scale to JI: for each scale step in that ET, pick the
simplest possible JI ratio that corresponds to it.
5) Now re-temper the resultant JI scale according to your original
regular temperament.

Everyone shot this down as being somehow wrong, including Gene, but I
don't think that it is. Now that I'm a bit more knowledgeable, here's
how I'd rephrase the above:

Enlightened Mike's Algorithm:
1) Pick a regular temperament
2) Find a val supporting the temperament, hence adding some additional
commas as "chromatic" unison vectors
3) For each integer i < v[1], find the simplest JI ratio j, as
measured by the OETES, such that v(j) = i
4) This is a JI transversal of the scale. Now tune the intervals
according to the temperament you picked in #1 and watch those wolf
intervals disappear

This is equivalent to Naive Mike's algorithm, with the single change
that I've specified that you use the octave-equivalent L2 seminorm to
judge which interval is "simplest," rather than leaving it open. What
exactly is wrong with this? It seems to be exactly what the definition
is here: http://xenharmonic.wikispaces.com/Hobbits

Also, is it true that if you use an unweighted OETES that you'll end
up getting hexagonal periodicity blocks when you do this? Methinks
this is the case.

-Mike

Mike Battaglia <battaglia01@gmail.com> wrote:
> A long time ago, back before I had a halfway-decent
> understanding of the actual math of regular temperament
> theory, I posted the following algorithm for a hobbit.
> This was shot down by everyone:
<snip>
> Everyone shot this down as being somehow wrong, including
> Gene, but I don't think that it is. Now that I'm a bit
> more knowledgeable, here's how I'd rephrase the above:

It isn't wrong. It just doesn't define a hobbit. (It may
define a just hobbit.)

> Enlightened Mike's Algorithm:
> 1) Pick a regular temperament
> 2) Find a val supporting the temperament, hence adding
> some additional commas as "chromatic" unison vectors
> 3) For each integer i < v[1], find the simplest JI ratio
> j, as measured by the OETES, such that v(j) = i
> 4) This is a JI transversal of the scale. Now tune the
> intervals according to the temperament you picked in #1
> and watch those wolf intervals disappear
>
> This is equivalent to Naive Mike's algorithm, with the
> single change that I've specified that you use the
> octave-equivalent L2 seminorm to judge which interval is
> "simplest," rather than leaving it open. What exactly is
> wrong with this? It seems to be exactly what the
> definition is here:
> http://xenharmonic.wikispaces.com/Hobbits

Hobbits use temperamental complexity, which is a
different kind of norm. (And I think it really is a norm,
given the right space.)

Is the L2 seminorm the thing that gives 15:8 and 5:3 equal
complexity?

> Also, is it true that if you use an unweighted OETES that
> you'll end up getting hexagonal periodicity blocks when
> you do this? Methinks this is the case.

Octave Equivalent Tenney-Euclidian Space? No, if it's
Tenney it isn't unweighted.

Hobbits will tend towards hexagons in whatever space you
define them. That's how you know they're different to
fokker blocks.

Graham

On Tue, Sep 27, 2011 at 8:27 AM, Graham Breed <gbreed@gmail.com> wrote:
>
> Hobbits use temperamental complexity, which is a
> different kind of norm. (And I think it really is a norm,
> given the right space.)

OK, I see. I now realize that my problem was that I didn't thoroughly
enough understand the OETES. After reading about it more, I now
realize that it also gives commas a length of zero, so it's a form of
temperamental complexity.

So am I correct in stating that the OETES isn't basis-invariant then?

> Is the L2 seminorm the thing that gives 15:8 and 5:3 equal
> complexity?

In JI, yes. If we're talking about the OETES specifically, I guess it
would depend on how it's mapped within the temperament. For example,
in 5-limit sensi, 5/3 is much less complex than 15/8, so I assume 5/3
would have lower OETES complexity.

> > Also, is it true that if you use an unweighted OETES that
> > you'll end up getting hexagonal periodicity blocks when
> > you do this? Methinks this is the case.
>
> Octave Equivalent Tenney-Euclidian Space? No, if it's
> Tenney it isn't unweighted.

I guess I just meant an octave-equivalent L2 seminorm then.

-Mike

Mike Battaglia <battaglia01@gmail.com> wrote:

> OK, I see. I now realize that my problem was that I
> didn't thoroughly enough understand the OETES. After
> reading about it more, I now realize that it also gives
> commas a length of zero, so it's a form of temperamental
> complexity.
>
> So am I correct in stating that the OETES isn't
> basis-invariant then?

I don't know. What's OETES?

The octave-equivalent temperamental complexity *should* be
basis-invariant, but I haven't checked that any formulation
is. What I know is that simplifying to a naive octave
equivalent temperamental complexity and then trying to
balance large and small intervals does end up basis
specific.

You don't need to do octave equivalent calculations. You
can work out octave specific intervals within one octave
and that'll probably be good enough.

> > Is the L2 seminorm the thing that gives 15:8 and 5:3
> > equal complexity?
>
> In JI, yes. If we're talking about the OETES
> specifically, I guess it would depend on how it's mapped
> within the temperament. For example, in 5-limit sensi,
> 5/3 is much less complex than 15/8, so I assume 5/3 would
> have lower OETES complexity.

It simplifies to counting generator steps for the rank 2
case.

> > > Also, is it true that if you use an unweighted OETES
> > > that you'll end up getting hexagonal periodicity
> > > blocks when you do this? Methinks this is the case.
> >
> > Octave Equivalent Tenney-Euclidian Space? No, if it's
> > Tenney it isn't unweighted.
>
> I guess I just meant an octave-equivalent L2 seminorm
> then.

Temperamental complexity is defined by (I think) the gram
matrix of the weighted mapping. Then you need to get rid
of the octaves. If you don't push towards a triangular
lattice it'll favor intervals that would have had lots of
2s in them. But that shouldn't matter much. You're always
comparing intervals of like size. You may find that you
get something like 9:8 instead of 10:9, but depending on
the geometry of the temperament you're looking at.

What does make a difference is to measure complexity
relative to the lattice midpoint of the scale instead of the
tonic. That's my version of compactness, which I think
approximates Honingh's.

Graham

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> So am I correct in stating that the OETES isn't basis-invariant then?

It's basis invariant if I understand what you mean by that.