Eureka! -- part 1

The kind of two-dimensional generalization of the Farey series we're
interested for calculating the harmonic entropy of triads is
discussed in

http://www.lim.univ-mrs.fr/~thiel/print/remy-thiel-iwcia00.pdf

See the bottom of page 5 and the following.

The generalization is called a "Farey set".

A "regular triangle" is the 2-d analogue of consecutive Farey
fractions. In the 1-d case, it's well known that two fractions p/q
and r/s in a Farey series are consecutive if and only if the
determinant

|p q|
|r s|

equals ±1. Similarly, three triads a:b:c, d:e:f, and g:h:i in the
Farey plane form a "regular triangle" if and only if the determinant

|a b c|
|d e f|
|g h i|

equals ±1. Apparently, Minkowski proved that there are no triads
of
the Farey set inside, or on the borders of, a regular triangle.

Mediants in the Farey plane are defined, though these are still
points rather than line segments. In the 1-D case, if the next-higher-
order Farey series has a point between p/q and r/s, it's the mediant
(p+r)/(q+s).

If, in the 2-D case, the next-higher-order Farey series provides a
point in the regular triangle, that point will be the mediant of two
of the vertices of the triangle: either
(a+d):(b+e):(c+f)
or
(a+g):(b+h):(c+i)
or
(d+g):(e+h):(f+i).

Now the partition of the Farey series in the 1-D case into a set of
exhaustive and mutually exclusive regions is trivial. But in the 2-D
case, a partition into regular triangles (called a "triangulation")
is almost never unique. However, one can define compatible sequences
of triangulations in which each triangulation is a refinement of the
triangulation of the Farey set of next-lowest order...

>http://www.lim.univ-mrs.fr/~thiel/print/remy-thiel-iwcia00.pdf

Great! How'd you find it?

So triads will be represented as points on the coordinate plane? Assuming
yes...

>(y/x, z/x)

Both coordinates for each point in ^F_n (Farey set order n) will
themselves belong to F_n (Farey series order n) as fractions, right?

Nothing in the ^F_n definition seems to care what order the coordinates
are in, so the plot of triads on the coordinate plane will be symmetrical
with respect to otonality and utonality if we plot triads using their
two interior intervals?

>If, in the 2-D case, the next-higher-order Farey series provides a point
>in the regular triangle, that point will be the mediant of two of the
>vertices of the triangle:

"The" mediant; no ^F_(n+1) will add two points to a Farey triangle of
F_n?

"If"; not every (n+1) gives every triangle a mediant? That seems bad.

I'm not really clear on what makes a given progression of triangulations
"compatible".

-Carl

Carl wrote,

>Great! How'd you find it?

Google search for "Farey web".

>So triads will be represented as points on the coordinate plane? Assuming
>yes...

How could it be otherwise?

I presume you're up to speed on my triangular triad plots?

>Both coordinates for each point in ^F_n (Farey set order n) will
>themselves belong to F_n (Farey series order n) as fractions, right?

Yes, but the converse is not true.

>Nothing in the ^F_n definition seems to care what order the coordinates
>are in, so the plot of triads on the coordinate plane will be symmetrical
>with respect to otonality and utonality if we plot triads using their
>two interior intervals?

A major triad will be there if n>5, while a minor triad will only be there
if n>14. Their positions are symmetrical with respect to one another, but
any given ^F_n plot will favor otonalities over utonalities.

>"The" mediant; no ^F_(n+1) will add two points to a Farey triangle of
>F_n?

No.

>"If"; not every (n+1) gives every triangle a mediant? That seems bad.

Not so bad.

>I'm not really clear on what makes a given progression of triangulations
>"compatible".

If every line segment appearing in one triangulation also appears in the
next-higher-order triangulation.

>>Great! How'd you find it?
>
> Google search for "Farey web".

Google strikes again.

> I presume you're up to speed on my triangular triad plots?

I was vaguely up to speed as of December 2000.

>>Both coordinates for each point in ^F_n (Farey set order n) will
>>themselves belong to F_n (Farey series order n) as fractions, right?
>
> Yes, but the converse is not true.

You mean that taking k=2 combinations of F_n gives a superset of
^F_n because both fractions in each ^F_n ordered pair must have
the same denominator?

>>Nothing in the ^F_n definition seems to care what order the
>>coordinates are in, so the plot of triads on the coordinate plane
>>will be symmetrical with respect to otonality and utonality if we
>>plot triads using their two interior intervals?
>
>A major triad will be there if n>5, while a minor triad will only
>be there if n>14.

How so? What are their ordered pairs?

>>"The" mediant; no ^F_(n+1) will add two points to a Farey triangle of
>>F_n?
>
>No.

So "a mediant" would be more accurate? That's really bad.

>>"If"; not every (n+1) gives every triangle a mediant? That seems
>>bad.
>
>Not so bad.

Wait- not every consecutive pair of fractions in a Farey series gets
a mediant with (n+1), right? Otherwise, the number of new Farey
fractions as n goes up couldn't fluctuate as much as Schroeder says
(he uses phi for this function, IIRC). I _could_ just check this,
even have a scheme proggie to do it, but I'm at work, and really, I
should be... working. :(

>>I'm not really clear on what makes a given progression of
>>triangulations "compatible".
>
> If every line segment appearing in one triangulation also appears
> in the next-higher-order triangulation.

It was my understanding that triangulation series starts with an
arbitrary choice at some small n (n=2 only has 2 points, but n=3
allows many possible triagulations), and works it's way up. How
would a line segment disappear?

-Carl

Carl wrote,

>>>Both coordinates for each point in ^F_n (Farey set order n) will
>>>themselves belong to F_n (Farey series order n) as fractions, right?
>
>> Yes, but the converse is not true.

>You mean that taking k=2 combinations of F_n gives a superset of
>^F_n because both fractions in each ^F_n ordered pair must have
>the same denominator?

That sounds right.

>>A major triad will be there if n>5, while a minor triad will only
>>be there if n>14.

>How so? What are their ordered pairs?

Major: (6,4) (5,4)

Minor: (12,10) (15,10)

Although the coordinates simply reduce to (3/2),(5/4) or vice verse, the
major will appear in ^F_6 and higher, while the minor will only appear in
^F_15 and higher.

>>>"The" mediant; no ^F_(n+1) will add two points to a Farey triangle of
>>>F_n?
>>
>>No.

>So "a mediant" would be more accurate?

I don't see why you'd say that.

>That's really bad.

:(

>>>"If"; not every (n+1) gives every triangle a mediant? That seems
>>>bad.
>
>>Not so bad.

>Wait- not every consecutive pair of fractions in a Farey series gets
>a mediant with (n+1), right?

Right.

>>>I'm not really clear on what makes a given progression of
>>>triangulations "compatible".
>
>> If every line segment appearing in one triangulation also appears
>> in the next-higher-order triangulation.

>It was my understanding that triangulation series starts with an
>arbitrary choice at some small n (n=2 only has 2 points, but n=3
>allows many possible triagulations), and works it's way up. How
>would a line segment disappear?

One could find a triangulation for N=100 without looking at any lower
triangulations. But the way you're doing it, you'll get a compatible
progression, by definition.

>>>A major triad will be there if n>5, while a minor triad will only
>>>be there if n>14.
>
>>How so? What are their ordered pairs?
>
>Major: (6,4) (5,4)
>
>Minor: (12,10) (15,10)

Whoa. Each chord is a line segment? I thought each chord
was a point, ie Maj=(5/4, 3/2) Min=(6/5, 3/2).

>>>>"The" mediant; no ^F_(n+1) will add two points to a Farey triangle of
>>>>F_n?
>>>
>>>No.
>
>>So "a mediant" would be more accurate?
>
>I don't see why you'd say that.

It says,

"Theorem 2 (Monkemeyer) Let (^Q,^R,^S) be a regular triangle of
^F_n. Let ^P be a point of ^F_(n+1), included in the triangle
and distinct from the vertices; then ^P is the mediant of
2 of the vertices."

I guess my question is, how do you know which two vertices the
point is a mediant with?

>>Wait- not every consecutive pair of fractions in a Farey series gets
>>a mediant with (n+1), right?
>
> Right.

Not so bad then.

>>>>I'm not really clear on what makes a given progression of
>>>>triangulations "compatible".
>>
>>> If every line segment appearing in one triangulation also appears
>>> in the next-higher-order triangulation.
>
>>It was my understanding that triangulation series starts with an
>>arbitrary choice at some small n (n=2 only has 2 points, but n=3
>>allows many possible triagulations), and works it's way up. How
>>would a line segment disappear?
>
>One could find a triangulation for N=100 without looking at any lower
>triangulations. But the way you're doing it, you'll get a compatible
>progression, by definition.

Okay, then I just need to figure out how to do N=100 from scratch.
Glad this was last in this message, as I've clearly got to iron out
some of the above before this.

-Carl

>>>>A major triad will be there if n>5, while a minor triad will only
>>>>be there if n>14.
>
>>>How so? What are their ordered pairs?
>
>>Major: (6,4) (5,4)
>
>>Minor: (12,10) (15,10)

>Whoa. Each chord is a line segment?

Nope.

>I thought each chord
>was a point, ie Maj=(5/4, 3/2) Min=(6/5, 3/2).

Correct. My ordered pairs were (y,x) (z,x), and the point corresponding to
them is (y/x, z/x).

>I guess my question is, how do you know which two vertices the
>point is a mediant with?

You have to calculate it!

>>I thought each chord
>>was a point, ie Maj=(5/4, 3/2) Min=(6/5, 3/2).
>
>Correct. My ordered pairs were (y,x) (z,x), and the point corresponding to
>them is (y/x, z/x).

Okay- I guess I'm just not familiar with this usage of "ordered pairs".
Anyway, according to the paper, yx and zx must be in lowest terms, so how
are you going to keep the minor triad out of ^F_6?

>>I guess my question is, how do you know which two vertices the
>>point is a mediant with?
>
>You have to calculate it!

Damn!

-Carl

>Okay- I guess I'm just not familiar with this usage of "ordered pairs".
>Anyway, according to the paper, yx and zx must be in lowest terms,

Where do you see that? No, x:y:z must be in lowest terms.

>so how
>are you going to keep the minor triad out of ^F_6?

Because x=10.

>>Okay- I guess I'm just not familiar with this usage of "ordered pairs".
>>Anyway, according to the paper, yx and zx must be in lowest terms,
>
>Where do you see that? No, x:y:z must be in lowest terms.

Ah, right.

>>so how are you going to keep the minor triad out of ^F_6?
>
>Because x=10.

Check.

So, the fact that triangulations aren't unique means that a given pair
of points may be consecutive with more than one third point?

-Carl

>So, the fact that triangulations aren't unique means that a given pair
>of points may be consecutive with more than one third point?

No, it doesn't mean that. Any pair of points is simply a member of an
ordinary Farey series. It's with triples of points that the complexities
come in.

>>So, the fact that triangulations aren't unique means that a given pair
>>of points may be consecutive with more than one third point?
>
>No, it doesn't mean that. Any pair of points is simply a member of an
>ordinary Farey series. It's with triples of points that the complexities
>come in.

I guess I don't understand how to construct a triangulation, then.

For a given ^F_n, the points are fixed. Every pair of points must
participate in ____ regular triangles. By your above statement, I take
it that the answer can only be zero or one.

-Carl

Every pair of neighboring points must form an edge for two regular
triangles. I guess I misunderstood what you meant by "consecutive".

-----Original Message-----
From: Carl Lumma [mailto:carl@...]
Sent: Friday, May 11, 2001 4:05 PM
To: harmonic_entropy@yahoogroups.com
Subject: [harmonic_entropy] re: Eureka! -- part 1

>>So, the fact that triangulations aren't unique means that a given pair
>>of points may be consecutive with more than one third point?
>
>No, it doesn't mean that. Any pair of points is simply a member of an
>ordinary Farey series. It's with triples of points that the complexities
>come in.

I guess I don't understand how to construct a triangulation, then.

For a given ^F_n, the points are fixed. Every pair of points must
participate in ____ regular triangles. By your above statement, I take
it that the answer can only be zero or one.

-Carl

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>Every pair of neighboring points must form an edge for two regular
>triangles. I guess I misunderstood what you meant by "consecutive".

Cool. I guess I'm waiting for part 2, then. Thanks, Paul!

-C.