Re: two-dimensional three-term BP series

Hi Dan,

Thanks, glad that I understand now, at least some of it!

> And if limited to scales with an odd number of tones this works like a
> charm. Different weightings of the adjacent fractions can be used to

I've made a start at trying to prove that it will work.

Here is how it is going so far:

Let's take the Pythagorean diatonic scale as an example, as this is Myhill.

1/1 9/8 81/64 729/512 3/2 27/16 243/128 2/1
0 cents 203.9 cents 407.8 cents 611.7 cents 702 cents 905.9 cents 1110 cents 1200 cents

steps
9/8 9/8 9/8 256/243 9/8 9/8 256/243
0 cents 203.9 cents 203.9 cents 203.9 cents 90.22 cents 203.9 cents 203.9 cents 90.22
cents

L L L S L L S

Using 3/2 as the generator, let's number the notes according to the
order in which they are made:
0 2 4 6 1 3 5 0
where 0 means 1/1 is the starting note, the 1 means that 3/2 is the
first new one made, the 2 means 9/8 is the second, and so on.

Here 2/1 is labelled 0 by octave equivalence, as one is reducing all the notes
into the interval 1/1 2/1.

Now, when any Myhill scale is made from a generator, there must be a
point in the construction when the first S interval is made.

Suppose it is between note n and note n+k.

Then k uses of the generator make the S interval, so it also occurs
between 0 and k, and indeed, between any pair of notes with numbers
k apart.

In the Pythagorean diatonic scale, there is an S between 6 and 1.

So k = 5, so there must be another S between 5 and 0, as indeed there is.

The first S is made by subdividing a larger interval, and the other interval made
in this way is an L.

The interval subdivided is from 3 to 0, dividing it into 3 to 5 and 5 to 0.

The other S is between 4 and 6.

For the Pythagorean diatonic, the other interval is between 5 and 3, or between
4 and 6. So Ls are made using k = 2, and first L is between 0 and 2.

Now, lets try adding a small increment h to alternate generators. (h measured
in cents say, or any logarithmic notation).

This moves the odd numbered notes all up in pitch by h.

This will make the S intervals into two sizes
5 0 gets smaller as the 5 is moved up.
6 1 gets larger as the 1 is moved up.

The Ls all remain the same size as before. Why is this?

Since Ls are made using k =2, then any L is either formed from
two even numbered notes (say, 2 and 4), in which case both remain the same, or
it is formed from two odd numbered notes (say, 3 and 5), in which case, both move
up in pitch by the same amount. Either way, the L remains the same
size as before.

So doing this gives a scale with intervals of three sizes - the L, and two sizes for the
S.

Lets write kS for the k that generates the S, and kL for the one that generates the L.

(S and L here would normally be written as subscripts, but can't do that in plain text, so
I hope this is reasonably readable)

Then for this scale, kS = 5, and kL = 2

You can also get three sizes of step if the Ls are made using k odd, and the Ss using k
even.

If so, it would be the other way round - the Ss would remain the same as before,
and the Ls would be of two sizes.

Trying several scales, it looks as though kS + kL = number of notes in the scale. If this
is so, then the method will produce three step sizes for all scales with odd numbers of
notes.

Let's try an example. What if we make note 1 into 14/9. This, and any other scales used
here are just meant as an examples to help explain method, not particularly as ones to play in.

Instead of 1/1 3/2 9/4 (0 cents 702 cents 1404 cents), let's try the generator

1/1 14/9 9/4
0 cents 764.9 cents 1404 cents

so h = (14/9)/(3/2) = 28/27 = 62.96 cents.

Scale is

1/1 9/8 81/64 729/512 14/9 7/4 63/32 2/1
0 cents 203.9 cents 407.8 cents 611.7 cents 764.9 cents 968.8 cents 1173 cents 1200 cents

Steps
9/8 9/8 9/8 7168/6561 9/8 9/8 64/63
0 cents 203.9 cents 203.9 cents 203.9 cents 153.2 cents 203.9 cents 203.9 cents 27.26
cents

As expected, the Ls are the same as before, and the Ss are now of
two sizes, 7168/6561 and 64/63. (The first of which is now larger than
the original L, which can happen - that's okay as you still have three
step sizes)

As h increases, the smaller S gets smaller (e.g. with 11/7 it's 896/891)
and then at a certain point, e.g. 8/5, the 5 goes to the other side of the 0:
0 5 2 4 6 1 3 0

Increment here is 16/15 (111.7 cents) which is larger than the S at 256/243 (90.22 cents).

Result no longer has three step sizes.

For 1/1 8/5 9/4, steps are
81/80 10/9 9/8 9/8 4096/3645 9/8 10/9

However, if the increment is smaller than the original S, then resulting scale has three
step sizes.

I think your method keeps to increments smaller than the original S (or the original L if
it's the L that one is varying) - is that so?

Now, the conjecture is, that such a scale is always trivalent. One also wonders if it
might always be pairwise well formed.

I've had a little go at proving it, but can't see an easy way immediately. I wonder if
David Clampitt, or Norman Carey, or someone familiar with all the results in the field
would like to have a go as they could probably prove it more readily.

One can also work backwards.

For instance, starting with the Wilson 17 note scale generated by 1/1 5/4 3/2,
(generators 5/4 and 6/5 alternating)
try replacing by two equal intervals of size (3/2)^0.5 :
1/1 (3/2)^0.5 3/2

You get this 17 note Myhill scale:
1/1 56.8 cents 2187/2048 9/8 261 cents 351 cents 81/64 465 cents 555 cents 729/512 3/2 759
cents 6561/4096 27/16 963 cents 1052 cents 243/128 2/1

So Wilson's scale could be got by applying your tempering process to alternate generators
of this scale.

------------------------------------------------------------------------------------------

I wonder if you could show the two generators for your scales, either ratios
or in mathematical notation, as it will make it easier to follow the construction?

When the generators are in cents, one needs to know exactly what is going
on before one can figure out how the scale is constructed.

For instance, what are the generators for your two scales in the "two-dimensional
three-term BP scales" post?

I also have a few more questions, to do with some of your other remarks.

> give the generator within a given periodicity (P). And when a chain of
> these generators that is (a+b)-1 long results in a built-in ordering
> rule. This is why I've felt that some three-term analogue to adjacent
> fractions is what's really needed to completely generalize Trihill
> scales. This seems to be a very difficult problem though, and as no

Can you explain a bit more what this is about? What is the built in ordering
rule, and what do you mean by a three term analogue to adjacent fractions?

> Now if you look at the last example I gave in the "two-dimensional
> three-term BP scales" post, the three-term index of [1,4,4] gives a
> Tribonacci series of
>
> 1, 4, 4, 9, 17, 30, ...

Where do the later terms 9, 17, 30 come into the construction of the scale?
Do you mean that one can also make a 4 4 9 scale, and so on? If so, how?

>
> So the single generator model is converted into the generalized take
> on your initial two generator hunch by collapsing a single generator
> chain on itself, and
>
> 0---23---16---9---2---25---18---11---4
>
> becomes
>
> 26----19----12-----5
> / \ / \ / \ / \
> / \ / \ / \ / \
> 0----23----16-----9-----2
>
> (BTW, this is also the three-term Fibonacci to Tribonacci Yasser
> parallel that Paul Erlich initially posted about way back when.)

What are these numbers 0, 23, 16 etc?

I've often had trouble explaining ideas that I can see really clearly
as many mathematicians do at one stage or another. Nearly always,
it is because one uses concepts that one has got so used to that
one doesn't think to explain what they are. One just knows what they mean.

However, others are seeing them probably for the first time.

Usually if one explains enough of your terms and concepts, it then
becomes understandable to anyone - assuming the original idea is clear
as I get the impression your ideas are.

Robert Walker

Robert Walker wrote,

<< I've made a start at trying to prove that it will >>

Yes a simple proof would be neat, though I can't see any reason why it
wouldn't always work. (I'd also assume that a scale with an even
number of notes can only be trivalent if equally tempered.)

<< I think your method keeps to increments smaller than the original S
(or the original L if it's the L that one is varying) - is that so? >>

Well if the amount of S steps is less than the amount of L steps, the
amount of S steps remain the same but increases a comma (and by comma
here I mean a generalized comma in the sense that I explained in the
previous post) in size and the L steps that become M steps decrease a
comma in size, and L remains it's original size. So as you might
expect, if the amount of S steps is greater than the amount of L
steps, the amount of L steps remain the same but decreases a comma in
size and the S steps that become M steps increase a comma in size, and
S remains it's original size.

<< I wonder if you could show the two generators for your scales,
either ratios or in mathematical notation, as it will make it easier
to follow the construction? >>

If you think of a one-dimensional chain truncated so that it always
has one more interval (or note) than the second dimension (or plane)
then you should be able to visualize the generalization of this pretty
easily. So the simplest model for this is a

1/1---3/2---9/8

chain as

10/9
/ \
/ \
1/1---3/2

where a two-term scale of 1/1 9/8 3/2 2/1 becomes a three-term scale
of 1/1 10/9 3/2 2/1.

<< When the generators are in cents, one needs to know exactly what is
going on before one can figure out how the scale is constructed. >>

I give generators in cents when I weight the generator. This is a bit
more of a convoluted process. When I say [a,b], or [a,b,c], or [2,5],
or [2,2,3] I'm referring to a scale's index where the alphabetized
variables are small to large stepsizes. So [a,b] and [a,b,c] are the
generic generalized two and three stepsize indexes, and [2,5] and
[2,2,3] are just a familiar example of each. So [2,5] means a scale
consisting of 2 "small" steps and 5 large steps. [2,2,3] means a scale
consisting of 2 small, 2 medium, and 3 large steps.

To convert an [a,b] index into a single generator chain, you must
first convert the two terms into adjacent fractions -- fractions that
when cross-multiplied will differ by 1.

The process (or algorithm) I use to convert an [a,b] index into a
single generator chain usually uses three constants. I refer to these
as "apical" constants -- apical roughly meaning a bud which terminates
a stem; by analogy this would be the constant terminating the series.
Anyway, the constants I use here are (1+sqrt(5))/2, 2, and sqrt(2)+1,
but any weighting could conceivably be used depending on what one's
aims are.

Phi is derived from the series

1/1, 1/2, 2/3, 3/5, 5/8, ...

I call an [a,b] index weighted by this constant a generalized "golden"
scale.

2 is derived from series

1/1.5, 1.5/3.25, 3.25/6.375, 6.375/12.8125, 12.8125/25.59375, ...

I call an [a,b] index weighted by this constant a generalized "equal"
scale.

sqrt(2)+1 is derived from the following series

1/2, 2/5, 5/12, 12/29, 29/70, ...

I call an [a,b] index weighted by this constant a generalized "silver"
scale.

Here's the generalized formula for deriving a weighted generator for
any given [a,b] index within a given periodicity.

X = P/((a+W*b))*(A+W*B)

where:

"P" = any given periodicity

"W" = any given weight

"A"/"a", "B"/"b" = the two adjacent fractions of a given [a,b] index

and "X" = the resulting weighted generator

<< what are the generators for your two scales in the "two-dimensional
three-term BP scales" post? >>

It'll probably be clearer if I first go through the single generator
process. Okay, [a,b] = [5,4] here, so first I need to convert this
into adjacent fractions. You could use 1/5 and 1/4 or 4/5 and 3/4, I
used A/a B/b = 4/5 3/4.

The first example I gave was Phi weighted, so W = (1+sqrt(5))/2. And
as this is the Bohlen-Pierce scale P = 1:3. So now all that's left
undefined is X.

P/((a+W*b))*(A+W*B) = ~1468�

So now a chain of (a+b)-1 Xs gives

0---1468---1034---600---166---1634---1200---766---332

Now to convert this into a three-term scale you need to establish a
generalized Pythagorean comma. The difference between multiply X by
the forth term of the two-term Fibonacci series and P will result in
what I call the generalized Pythagorean comma. So we know X = ~1468�,
and that our two-term index is [5,4] which would result in a fourth
term of 13 as the series would be

5, 4, 9, 13, ...

So X*4T = ~63� (periodicity reduction is always assumed here).

Now truncating a chain of Xs so that it has one more interval than the
second dimension which is now decreased by the comma results in a Phi
weighted two-dimensional three-term BP of

1570----1136-----702-----268
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
0----1468----1034-----600-----166

<< Can you explain a bit more what this is about? What is the built in
ordering rule, and what do you mean by a three term analogue to
adjacent fractions? >>

The "built-in ordering rule" would be the arrangement of stepsizes, Ls
and Ss, that result from a single generator chain. The sudden omission
of this one seemingly simple detail when moving from a two-term to
n-term generalization has proved to be quite the royal pain the ass!
If one is interested as I am in a one size fits all generalization
here, one cannot arbitrarily order these stepsizes.

When I say "a three-term analogue to adjacent fractions" I mean
essentially the same thing as Stern-Brocot Tree

<http://206.4.57.253/editorial/knot/SB_tree.html>

if it were seeded with three terms instead of two.

<< Where do the later terms 9, 17, 30 come into the construction of
the scale? Do you mean that one can also make a 4 4 9 scale, and so
on? If so, how? >>

Yes, in the [1,4,4] you'd have a 9-out-of-30 where 30 is an equal
division of 1:3 and the two dimensions are 1548 and 1648, or
1:3^(23/30) and 1:3^(26/30)

26
/
/
0----23

And in the [4,4,9] you'd have a 17-out-of-56 where 56 is an equal
division of 1:3 and the two dimensions are 1460 and 1664, or
1:3^(43/56) and 1:3^(49/56)

49
/
/
0----43

<< Usually if one explains enough of your terms and concepts, it then
becomes understandable to anyone - assuming the original idea is clear
as I get the impression your ideas are. >>

Some folks here, like Margo Schulter for instance (and yourself for
that matter), really excel at "charitable narratives"... for whatever
reason this does not come easy for me, if at all. But hopefully some
of these explanations and such will help out some here.

let me know!,

--Dan Stearns