Paul's lattice math and my diagrams (was: Hey Carl [old])

> From: paulerlich <paul@stretch-music.com>
> To: <tuning-math@yahoogroups.com>
> Sent: Wednesday, December 26, 2001 1:45 PM
> Subject: [tuning-math] Re: Microtemperament and scale structure
>
>
> --- In tuning-math@y..., "monz" <joemonz@y...> wrote:
>
> > OK, I'm only up to late August, so there's obviously some more
> > important stuff coming up.
>
> Monz, it means a lot to me that you are taking the time and trouble
> to go through this stuff, and hopefully incorporate as much of it as
> possible into your webpages. It all feels a little more "meaningful",
> somehow . . .

Thanks, Paul! Most people around here (any of the tuning lists,
but especially this one and HE) hold you and your work in pretty
high repute... so coming from you, this feels like some cachet
being bestowed upon me. :)

So now, if you don't mind backtracking a bit further, I've saved
this one for last and think you'll get something out of it...

> From: Paul Erlich <paul@stretch-music.com>
> To: <tuning-math@yahoogroups.com>
> Sent: Saturday, July 28, 2001 10:40 AM
> Subject: [tuning-math] Re: Hey Carl
>
>
> Here's why the hypothesis should work.
>
> Take an n-dimensional lattice, and pick n independent
> unison vectors. Use these to divide the lattice into
> parallelograms or parallelepipeds or hyperparallelepipeds,
> Fokker style. Each one contains an identical copy of a single
> scale (the PB) with N notes. Any vector in the lattice now
> corresponds to a single generic interval in this scale no
> matter where the vector is placed (if the PB is CS, which
> it normally should be). Now suppose all but one of the
> unison vectors are tempered out. The "wolves" now divide
> the lattice into parallel strips, or layers, or hyper-layers.
> The "width" of each of these, along the direction of the
> chromatic unison vector (the one that remains untempered),
> is equal to the length of exactly one of this chromatic
> unison vector.

Paul, this is *exactly* what's going on in my "meantone
acoustical rational implications lattices" here:
http://www.ixpres.com/interval/monzo/meantone/lattices/lattices.htm

The 5-limit periodicity blocks are bounded by 2 unison-vectors,
one of which is tempered out (the 81:80 syntonic comma) and
one of which isn't -- and that one is the one which appears
at the end of each meantone chain.

This is not really clear in the diagrams that exist on the
webpage now, because they all have arbitrary limits of +/- 27
generators. I'm preparing some new diagrams which *do* show
proper Fokker periodicity-blocks for these (and other)
meantones.

>
> Now let's go back to "any vector in the lattice". This vector,
> added to itself over and over, will land one back at a pitch
> in the same equivalence class as the pitch one started with,
> after N iterations (and more often if the vector represents
> a generic interval whose cardinality is not relatively prime
> with N). In general, the vector will have a length that is
> some fraction M/N of the width of one strip/layer/hyperlayer,
> measured in the direction of this vector (NOT in the direction
> of the chromatic unison vector). M must be an integer, since
> after N iterations, you're guaranteed to be in a point in the
> same equivalence class as where you started, hence you must be
> an exact integer M strips/layers/hyperlayers away. As a
> special example, the generator has length 1/N of the width
> of one strip/layer/hyperlayer, measured in the direction of
> the generator.

This is precisely what was in my mind when I came up with
these meantone lattices.

> Anyhow, each occurence of the vector will cross either
> floor(M/N) or ceiling(M/N) boundaries between
> strips/layers/hyperlayers. Now, each time one crosses
> one of these boundaries in a given direction, one shifts
> by a chromatic unison vector. Hence each specific occurence
> of the generic interval in question will be shifted by
> either floor(M/N) or ceiling(M/N) chromatic unison vectors.
> Thus there will be only two specific sizes of the interval
> in question, and their difference will be exactly 1 of the
> chromatic unison vector. And since the vectors in the chain
> are equally spaced and the boundaries are equally spaced,
> the pattern of these two sizes will be an MOS pattern.

Isn't this exactly how my pseudo-code works? (posted here:
</tuning-math/messages/2069?expand=1>).

> QED -- right?

Is that "Quod Erat Demonstrandum" or "Quite Easily Done"?
(or "Quickly Ends Dandruff"? ;-) )

-monz

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--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> > Here's why the hypothesis should work.
> >
> > Take an n-dimensional lattice, and pick n independent
> > unison vectors. Use these to divide the lattice into
> > parallelograms or parallelepipeds or hyperparallelepipeds,
> > Fokker style. Each one contains an identical copy of a single
> > scale (the PB) with N notes. Any vector in the lattice now
> > corresponds to a single generic interval in this scale no
> > matter where the vector is placed (if the PB is CS, which
> > it normally should be). Now suppose all but one of the
> > unison vectors are tempered out. The "wolves" now divide
> > the lattice into parallel strips, or layers, or hyper-layers.
> > The "width" of each of these, along the direction of the
> > chromatic unison vector (the one that remains untempered),
> > is equal to the length of exactly one of this chromatic
> > unison vector.
>
>
> Paul, this is *exactly* what's going on in my "meantone
> acoustical rational implications lattices" here:
> http://www.ixpres.com/interval/monzo/meantone/lattices/lattices.htm

But alas, you are not getting the infinite strips I refer to above.

> The 5-limit periodicity blocks are bounded by 2 unison-vectors,
> one of which is tempered out (the 81:80 syntonic comma) and
> one of which isn't -- and that one is the one which appears
> at the end of each meantone chain.

Right -- but since the 81:80 _is_ tempered out, your lattices should
be proceeding infinitely in the direction of the 81:80.
> >
> > Now let's go back to "any vector in the lattice". This vector,
> > added to itself over and over, will land one back at a pitch
> > in the same equivalence class as the pitch one started with,
> > after N iterations (and more often if the vector represents
> > a generic interval whose cardinality is not relatively prime
> > with N). In general, the vector will have a length that is
> > some fraction M/N of the width of one strip/layer/hyperlayer,
> > measured in the direction of this vector (NOT in the direction
> > of the chromatic unison vector). M must be an integer, since
> > after N iterations, you're guaranteed to be in a point in the
> > same equivalence class as where you started, hence you must be
> > an exact integer M strips/layers/hyperlayers away. As a
> > special example, the generator has length 1/N of the width
> > of one strip/layer/hyperlayer, measured in the direction of
> > the generator.
>
>
> This is precisely what was in my mind when I came up with
> these meantone lattices.

Really? I don't see the strips, and I don't see how the generator
could be said to have any property resembling this in your lattices.

> > Anyhow, each occurence of the vector will cross either
> > floor(M/N) or ceiling(M/N) boundaries between
> > strips/layers/hyperlayers. Now, each time one crosses
> > one of these boundaries in a given direction, one shifts
> > by a chromatic unison vector. Hence each specific occurence
> > of the generic interval in question will be shifted by
> > either floor(M/N) or ceiling(M/N) chromatic unison vectors.
> > Thus there will be only two specific sizes of the interval
> > in question, and their difference will be exactly 1 of the
> > chromatic unison vector. And since the vectors in the chain
> > are equally spaced and the boundaries are equally spaced,
> > the pattern of these two sizes will be an MOS pattern.
>
> Isn't this exactly how my pseudo-code works? (posted here:
> </tuning-math/messages/2069?expand=1>).

Monz, I don't see anything in your pseudocode that would give you any
of this -- have you actually managed to produce MOSs with it?