Ultimate MOS Generalization Thread

(...of Ultimate Destiny)

Let's finally nail down what the best MOS analogues are in rank r > 2. I made a list of MOS properties we would like to generalize. Can anyone think of important MOS properties I've left out?

1. Melodic properties

1.1 Two steps: A MOS scale has 2 melodic steps when arranged in ascending order.
Possible generalizations: (a) r steps, or (b) 2^(r-1) steps

1.2 Evenness of distribution: Each kind of step is distributed as evenly as possible among all steps. More specifically, the pattern of steps forms a disjoint covering system of rational Beatty sequences (DCSRBS).

2. Properties related to interval inventory

2.1 Maximum variety (a.k.a. DE, pseudo-Myhill): A MOS scale has at most 2 different intervals in each class.
Possible generalizations: (a) r intervals, or (b) 2^(r-1) intervals

2.2 Mean variety: A MOS scale of n notes has the minimum possible mean variety over all n-note scales of the rank-2 temperament.

2.3 Chromata: A MOS scale has exactly 1 chroma, defined as a (positive) difference between intervals of the same class.
Possible generalizations: (a) r chromata, (b) 2^(r-1) chromata, (c) (3^(r-1)-1)/2 chromata, (d) some other number of chromata

2.4 Epimorphicity / constant structure: A MOS scale is associated with a val v such that every interval subtending n scale steps maps to n under the val (it's epimorphic). Assuming that v can be a factor in the wedge product of the temperament, so that all commas have v(c) = 0, this implies that no interval can appear in more than one class (constant structure).

3. Lattice properties

3.1 Convexity: The notes of a MOS scale form a convex set in period-equivalent lattice space.

3.2 (Free) Fokker block: A MOS satisfies Gene's definition of Fokker block at http://xenharmonic.wikispaces.com/Fokker+blocks . Equivalently, it consists of all lattice points inside an arbitrarily translated parallelotope with chromata as edges.

3.3 Lumma block / fixed Fokker block: A MOS scale satisfies the definition given by Carl Lumma for a Fokker block with parallelotope vertices fixed to lattice points.

4. Properties relating different MOSes of the same temperament.

4.1 Infinite sequence: The MOSes of a rank-2 temperament form an infinite sequence of increasing notes/period, where each MOS contains all smaller MOSes as subsets.
Possible generalization: (a) Instead of a linear sequence, rank-3 MOSes could form a 2-dimensional structure like a tree, planar directed acyclic graph (DAG), etc.

4.2 Universality: For any possible chord or pitch set of a rank-2 temperament, there is some MOS that contains it. This expresses the fact that MOSes are unlimited in size (or in rank > 2, shape).

4.3 Tempering down: If you temper out the chroma of an n-note MOS, you get an equal division of the period into n steps.
Possible generalizations: The straightforward generalization is that if you temper out *all* the chromata you get an equal temperament, which is equivalent to 2.4 (epimorphicity).
On the other hand, the existence of (r-1) chromata such that tempering out any (r-2) of them leads to a MOS is equivalent to property 3.2 (free Fokker block).

Now, it's impossible to generalize the strictest versions of all these properties to rank > 2 simultaneously, because some of them contradict each other. Here are the contradictions I know of:

1.1(a) + 1.2 + 4.2 => contradiction

This contradiction comes from a conjecture of Aviezri Fraenkel that for any covering system of n > 2 rational Beatty sequences, either two of the moduli (a.k.a. "numbers of steps") must be identical, or the moduli must be of the particular form 1,2,4...2^(n-1). The conjecture is actually still unproved in the general case, but
http://citeseer.ist.psu.edu/viewdoc/summary?doi=10.1.1.24.230
proves it for rank <= 6, which I dare to say is more than sufficient for any conceivable *musical* application.

If 1.1(a) and 1.2 both hold, then the width of the scale in at least one lattice direction is strictly limited, so 4.2 becomes impossible.

Since property 4.2 seems absolutely essential to me, this means that rank-3 MOS analogues must have 4 steps, or 3 steps arranged in an uneven (or less-than-perfectly-even) pattern.

2.1(a) + 4.2 => contradiction

Similar to the first contradiction, this assumes my (unproven but likely) conjecture about max-variety-3 scales. In rank >= 4 it's already firmly established because of http://dx.doi.org/10.1023/A:1006513923148

If 4.2 is to be preserved, max-variety-3 will not be a general property of rank-3 MOSes. They will always be a special subset.

2.2 + 3.3 => contradiction

I recently gave a counterexample that showed that the Lumma block does not always minimize the mean variety. Starting at, I believe, a minimum of 10 notes per period, there are situations where you can either minimize the mean variety, or stick to a Lumma block, but not both.

OPEN QUESTIONS:

Is it possible to construct arbitrarily large (free) Fokker blocks with only r different steps (e.g. 3 steps in rank 3)?

Is it possible to construct arbitrarily large (free) Fokker blocks that are DCSRBS?

Is there a simpler way to find (free) Fokker blocks that minimize mean variety than going through all possible Fokker blocks with the same chromata, one by one?

Is there a simple way to characterize all convex epimorphic scales, regardless of whether they are Fokker blocks?

Keenan

On Sat, Sep 24, 2011 at 4:52 PM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> (...of Ultimate Destiny)

Haha! Yes!

> 2.3 Chromata: A MOS scale has exactly 1 chroma, defined as a (positive) difference between intervals of the same class.
> Possible generalizations: (a) r chromata, (b) 2^(r-1) chromata, (c) (3^(r-1)-1)/2 chromata, (d) some other number of chromata

Where did you get (3^(r-1)-1)/2 from?

> 4.1 Infinite sequence: The MOSes of a rank-2 temperament form an infinite sequence of increasing notes/period, where each MOS contains all smaller MOSes as subsets.
> Possible generalization: (a) Instead of a linear sequence, rank-3 MOSes could form a 2-dimensional structure like a tree, planar directed acyclic graph (DAG), etc.

This is the one that's the most important to me. We looked before at
how rank-3 MOS's form triangles in Tenney tuning space, although I'm
still not sure what the rank-3 MOS "scale tree" will look like (a DAG
may very well be it).

If we're in rank-2 (let's say the generators are 2/1 and 3/1), an MOS
is what results if you add a chromatic unison vector. This isn't a
very useful definition by itself, because it means that MOS can be
formed by a chain of 3 3/2's, 4 3/2's, 5 3/2's, etc, rather than the
more familiar 2, 3, 5, 7, 12, 17, etc Pythagorean MOS series.

For any specific tuning of the generators, an MOS is what results if
you add a chromatic unison vector and the pitches remain in monotonic
ascending order with respect to how they'd map under the val defined
by your new vector. The MOS series is what you get if you look at the
sizes of the blocks that respect this definition and place them in
ascending order.

It should be noted that this might be tricky to generalize, because in
rank-2 octave-equivalent space, all periodicity blocks have torsion
except for the trivial block that only incorporates 1/1. I'm not sure
exactly how that's going to play out, but for now, let's just take the
definition at face value.

So if we're in rank-3, let's say that this is our periodicity block
(view fixed width)

#####
#####
#####
#####

This block has 20 notes. Let's say that this satisfies the definition
of an MOS above for some tuning of the generators. If we fix the
length along the y-axis and increase the length of the x-axis, hence
extending the block by uv1, we end up with this:

##########
##########
##########
##########

This block has 37 notes; let's assume that this is the next stopping
point that satisfies the definition of MOS above given our arbitrary
choice of generator tuning. However, we could also have fixed the
x-axis length and increased the y-axis length, hence extending the
block by uv2, we end up with this:

#####
#####
#####
#####
#####
#####
#####

This block has 35 notes; let's assume that this is the next stopping
point in the y-direction that satisfies the definition of MOS above
given our arbitrary choice of generator tuning. But what if we want to
increase the length of the block along the x and y axes? This extends
the block by uv1+uv2. We might now end up with this:

######
######
######
######
######

This block has 30 notes. There's nothing stopping this from being a
stopping point as well. So that's three directions that we can extend
it.

But what if our blocks are actually hexagons? You now have three sets
of parallel lines that you can extend things in, plus combinations of
those aswell. I can't check it out right now, but I conjecture it'll
give us 3!/3! + 3!/2! + 3!/1! = 1 + 3 + 6 = 10 different ways to
extend everything.

And then you have parallelograms that are also hexagons, which are the
triple wakalixes - not sure how that's going to work.

As a final thought, both parallelograms and hexagons can be broken
down into triangles (in interval space). Also, MOS's looked like
triangles in tuning space as well. Likewise, line segments in rank-2
interval space are MOS's, and they're also MOS's in rank-2 tuning
space. I'm not sure what I'm conjecturing, but rest assured I'm
conjecturing something right now.

> Is there a simple way to characterize all convex epimorphic scales, regardless of whether they are Fokker blocks?

Maybe what I'm conjecturing has something to do this. I dunno.

-Mike

An addendum to the last message - you can also extend the parallelogram in another way as well, which is by the uv1-uv2 direction. So that's 4 different directions, which can be reduced to a basis of two - but you might end up at unique stopping points along each direction.

There's something more important though...

On Sun, Sep 25, 2011 at 9:29 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> As a final thought, both parallelograms and hexagons can be broken
> down into triangles (in interval space). Also, MOS's looked like
> triangles in tuning space as well. Likewise, line segments in rank-2
> interval space are MOS's, and they're also MOS's in rank-2 tuning
> space. I'm not sure what I'm conjecturing, but rest assured I'm
> conjecturing something right now.
>
>> Is there a simple way to characterize all convex epimorphic scales, regardless of whether they are Fokker blocks?
>
> Maybe what I'm conjecturing has something to do this. I dunno.

I'm still not exactly sure what I'm saying, but here's something else related. Try this visualization out: imagine an MOS as a finite segment of some generator chain. Let's visualize it like this (fixed width)

#####################

or as an actual line, like this:

_____________________

Naively, you'd expect that when you get to rank-3, you'd just imagine two generator chains - however, you don't get a convex set if you do this. You'll get something like this (view fixed width)

# #
# #
# #
# #
#

or as a shape on the lattice, like this (fixed width)

\ /
\ /
\ /
\ /
\/

However, you can think of the two chains above as defining the below convex set, given by "filling out" the chain:

#########
#######
#####
###
#

Or, if viewed as a shape, like this

__________
\ /
\ /
\ /
\ /
\/

I find it very interesting that if you take two independent generator chains like that, that the minimal convex set containing your two chains will be a triangle.

One problem is that these triangles aren't going to be epimorphic wrt the lattice. If you tile the lattice with these triangles, arranging them so that each vertex is shared by three triangles, you'll find that there are complementary "holes" in the lattice. You can come up with a second set of triangles to fill these holes, which will look like this (fixed width goddamnit!)

# #
# ###
# => #####
# #######
# # # # # #########

/ /\
/ / \
/ => / \
/ / \
/_________ /________\

You can mix these with the other triangles in three different ways, generating three possible types of parallelogram:

__________
\ ~\
\ ~ \
\ ~ \
\ ~ \
\~________\

########@
######@##
####@####
##@######
@########

OR

__________
/~ /
/ ~ /
/ ~ /
/ ~ /
/________~/

@########
##@######
####@####
######@##
########@

OR

/\
/ \
/ \
/ \
/ ______ \
\ b>b>b>b>b>b> /
\ /
\ /
\ /
\/

#
###
#####
#######
@@@@@@@@@
#######
#####
###
#

A hexagon consists of these three blocks smashed together. These are the only ways to tile the lattice. Scalene triangles are also possible.

Just like you can extend a single generator chain to deeper MOS's along the MOS series, you can also extend a triangular generator chain as well, except you can extend it 6 different ways (you can extend either side in either direction).

I'm still not sure what this means, but it's something to think about. Maybe it means that triangles are in some sense going to form rank-3 proto-MOS's, and that sets of triangles can be combined in different ways to give you actual MOS's, whatever they may be.

Tuning-math destroyed my ASCII diagrams. I even posted it on Yahoo to try and avoid Gmail doing it. You'll have to copy and paste into notepad to figure out what I meant.

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

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> Where did you get (3^(r-1)-1)/2 from?

Just a little educated hunch, but now I realize I made a mistake and I should have guessed 4^(r-2). Also, the other choices were off by 1, obviously (for rank 2 we know there has to be only 1 chroma).

> For any specific tuning of the generators, an MOS is what results if
> you add a chromatic unison vector and the pitches remain in monotonic
> ascending order with respect to how they'd map under the val defined
> by your new vector. The MOS series is what you get if you look at the
> sizes of the blocks that respect this definition and place them in
> ascending order.

Yes.

> So if we're in rank-3, let's say that this is our periodicity block
> (view fixed width)
>
> #####
> #####
> #####
> #####
>
> This block has 20 notes. Let's say that this satisfies the definition
> of an MOS above for some tuning of the generators. If we fix the
> length along the y-axis and increase the length of the x-axis, hence
> extending the block by uv1, we end up with this:
>
> ##########
> ##########
> ##########
> ##########
>
> This block has 37 notes;

Uh, it looks like 40 to me.

[snip rest of post]

How about this way of thinking about it:

A MOS has two kinds of steps, L, and s. Specifically, let's say there are x L steps and y s steps, so we have the equation

xL + ys = P

where P is the period. When you go from one MOS to the next larger MOS, there's only one possible thing you can do: split every L into s and L-s (which is positive because L>s). Then your new MOS has the equation

x(L-s) + (x+y)s = P

Now, assume a rank-3 hyper-MOS ought to have 3 kinds of steps. (There's no good reason to assume this, because allowing 4 kinds of steps might make the other properties better overall. But let's assume it.) If my original hyper-MOS is

xL + ym + zs = P

then there are three basic things I could do to divide up the steps and produce more notes. I could cut an s out of each L,

x(L-s) + ym + (x+z)s = P

I could cut an m out of each L,

x(L-m) + (x+y)m + zs = P

or I could cut an s out of each m,

xL + y(m-s) + (y+z)s = P

There are also two more "composite" things I could do by applying two of the above in some order:

x(L-s) + y(m-s) + (x+y+z)s = P
x(L-m) + (x+y)(m-s) + (x+y+z)s = P

So if you represent a hyper-MOS as a triangle with x, y, and z as the specific vals (<x...|, <y...|, <z...|), the above gives you 5 possible ways to shrink the triangle by moving one or two vertices. The new triangles that contain the point representing the specific tuning you're using will be possible larger hyper-MOSes.

Keenan

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
> 1.2 Evenness of distribution: Each kind of step is distributed as evenly as possible among all steps. More specifically, the pattern of steps forms a disjoint covering system of rational Beatty sequences (DCSRBS).

Idea: What if DCSRBS is not the appropriate generalization?

DCSRBS definitely coincides with MOS imprints in the case of 2 kinds of steps, but for more kinds of steps it might be too restrictive. The problem is that DSCRBS demands that the positions of each kind of step independently must be perfectly evenly distributed. Thus, for example, no DCSRBS exists for the case 3a+2b+1c. If you demand that the 'a's are evenly distributed you need

a_a_a_

but then there's no way to fill in the blanks so that the 'b's are also evenly distributed, that is

b__b__

So, what about a different kind of generalization?

MOS imprints can also be characterized this way: If a MOS has x large steps and y small steps, draw a straight line between points (0,0) and (x,y) on a grid. Now approximate the line as closely as possible by the vertical and horizontal length-1 segments of the grid. Each segment in the x direction represents a large step and each segment in the y direction represents a small step.

In fact, if you don't ignore octaves, this is exactly what the infinite scale looks like. You simply pick the lattice points that are closest to a straight line through the octaves.

So perhaps the appropriate generalization of this "evenly distributed steps" property is a 3D discrete approximation of a line, rather than DCSRBS.

Highly relevant paper: http://hal.inria.fr/docs/00/10/28/72/PDF/D690.PDF

Keenan

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
> OPEN QUESTIONS:
>
> Is it possible to construct arbitrarily large (free) Fokker blocks with only r different steps (e.g. 3 steps in rank 3)?

A related question is this: Given a set of chromata, is there always a Fokker block with those chromata and only r steps?

The answer is no, because in 5-limit JI, there are 19 different 19-note Fokker blocks with the chromata 81/80 and 78732/78125 (sensipent comma), but all of them have 4 different steps.

> Is there a simple way to characterize all convex epimorphic scales, regardless of whether they are Fokker blocks?

A convex epimorphic scale is a convex lattice subset that tiles the lattice.

The following paragraph just argues that we're allowed to treat the convex lattice subsets as continuous convex polytopes that tile space, so feel free to skip.

The convex hull of a convex lattice subset is a convex polytope that does not contain any other lattice points. Any two convex sets have a hyperplane that separates them, so there is a hyperplane separating the scale from each of its translated copies. The translated copies come in pairs; for each copy translated by v there is another translated by -v. For one copy of each pair choose any separating hyperplane, but for the other, use that same hyperplane translated by -v. Now we have a bunch of separating hyperplanes that come in parallel planes separated by a lattice vector. Take the intersection of all the half-spaces bounded by these hyperplanes and containing the original scale. This is a convex polytope, and it tiles space in a face-to-face way.

http://dx.doi.org/10.1112/S0025579300010007 characterizes all convex polytopes that tile space. Convex epimorphic scales can be characterized as the intersections of such polytopes with the lattice.

For example, in 2D (such as the octave-equivalent lattice of a rank-3 temperament), the only convex polygons that tile space are parallelograms and hexagons. Therefore every rank-3 convex epimorphic scale consists of all the lattice points inside a hexagonal tile. (You don't even need to include parallelograms in this characterization, because every lattice subset defined by a parallelogram is equally well defined by a hexagon with two very short edges.)

Keenan