Yahoo Tuning Groups Ultimate Backup tuning Re: Towards a Hyper MOS (was Re: two-dimensional three-term BP scales)

Hi Dan,

Thinking again about your conjecture:
Ex. scale, your 6 note mode
0 2 4 3 3 4 4
in 20-tet
from post
http://www.egroups.com/message/tuning/17032
is
SLMMLL
so intervals are
LL LM MM LS = 8 7 6 6
LLM LLS LMM LMS = 11 10 10 9
etc.

So it works by having two distinct (in LMS notation) intervals which are the same size in
each class.

So one can see why one can't adjust the intervals of this scale by just nudging some of
the notes up or down in pitch a little to make it into a scale that uses ratios, because
it would disturb the equivalence.

Your conjecture implies that all the trivalent scales with an even number of notes will
have two distinct (in LMS) intervals the same size in at least one of the classes,
otherwise one could make all the numbers rational by using small adjustments in pitch to
nudge them to nearby rationals.

The other way however, a rational scale could easily have two distinct intervals (in LMS)
which are the same size, e.g.
S = 5/4, L = 25/16, then L = SS, etc. etc.

So your conjecture so far would leave open the possibility of a rational trivalent scale
of this type with an odd number of notes. Do you think that might be possible? Or might it
be that all rational trivalent scales have to have all three interval sizes in each class
distinct (in LMS) notation. That would be a very strong result, seems to me, if true.

0 2 4 3 3 4 4 seems to be a counter example to your conjecture that all such scales have
to include the midpoint - in degrees from start of scale, the notes are:

0 2 6 9 12 16 20

- if I understood it correctly.

I'd be interested if you have any worked out reasoning for your conjecture that a
trivalent scale with an even number of notes always has to be a subset of n-tet.

I haven't yet got on to reading the rest of your post slowly, but certainly will do so,
and look forward to it.

Robert

🔗ligonj@northstate.net

🔗Robert Walker <robert_walker@rcwalker.freeserve.co.uk>

Hi Dan,

Just realised what you must mean by including the midpoint

> 0 2 4 3 3 4 4 seems to be a counter example to your conjecture that all such scales have
> to include the midpoint - in degrees from start of scale, the notes are:

- if every mode had to include the midpoint, it would be a scale that consisted of identical
copies in first and second half of the octave (is this what is called a periodicity block?).

But you must mean just that one of the modes of the scale includes the mid-point. In fact,
if it is trivalent, and also strictly proper, then it is impossible for all the modes to include the mid-point.

I plan to add an option to FTS to search the entire SCALA scale archive for trivalent scales
(and Myhill ones). It will be easy to do, prob. a couple of dozen extra lines of code or so,
as it already has option to do searches, and sort by properties of the steps of the scale,
so just a matter of adding some extra options. We will also be able to search the SCALA modes archive
in the same way.

Then we can sort the trivalent scales found according to number of notes, number and size of generators
in cents, periodicity of the generator construction (= 2 for alternating ones), and number
of duplicate intervals for each interval class, if any.

Should make it easy to do a first test of conjectures, or search for possible counter
examples.

Prob. do that tomorrow.

On your 6 note trivalent scale:
2 4 3 3 4 4
S L M M L L

with intervals
LL LM MM LS = 8 7 6 6
LLM LLS LMM LMS = 11 10 10 9

- it works because LS = MM

Could do any other scale with S = 2M-L

E.g. try M = 4, L = 5, then S = 3
3 5 4 4 5 5
=
0 3 8 12 16 21 26
in 26-tet.
which is indeed trivalent.

Since LS = MM, then SLMMLL is the same size as MMMMLL.

So the midpoint will be LMM, and so, one of the intervals, for this particular 6 note trivalent scale construction,
whatever the choice of L, M, S with LS = MM.

So that accords with your conjecture so far anyway.

Robert

🔗D.Stearns <STEARNS@CAPECOD.NET>

🔗Robert Walker <robert_walker@rcwalker.freeserve.co.uk>

Hi Jacky:

> I think if you keep working on this, your posts will help me to grasp
> it better - which has already been hugely helpful. Thanks for
> the "breakdown for dummies"! This has been an extrememly cryptic one
> for me to follow (perhaps this was intended).

Great, thanks!

Sorry, this post is probably cryptic because part of it is a mistake.

This is probably what made ot cryptic:

http://www.egroups.com/message/tuning/17578
> Your conjecture implies that all the trivalent scales with an even number of notes will
> have two distinct (in LMS) intervals the same size in at least one of the classes,
> otherwise one could make all the numbers rational by using small adjustments in pitch to
> nudge them to nearby rationals.

All those double negatives confuse it.

What I was saying is that if a scale is trivalent, and there are no duplicated
intervals like LS = MM the same size, you are then free to make tiny adjustments to
the L and M as you like, and so, move both to nearby rationals, by adding some very small
amount to each, some tiny fraction of a cent.

Then the thought was that S would become rational too. But of course it won't in
general. It will if there is only one S. But if there are k of them, all you end up
showing is that S^k is rational. S itself could easily turn out to be irrational.

Of course, you have infinitely many choices for the micro-adjustments to L and M.
But since there are also infinitely many possible irrational values for S, it
could be that all of these yield irrational values for S.

Nor can I see any way to rescue this idea at present - can't see any way to show
that any of these infinitely many possible values of S has to be rational.
Maybe there is some relevant number theoretical result?

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

I was hoping to show that if Dan's conjecture about irrational intervals occurring in any
trivalent scale with an even number of notes is true, then the only scales with an even number of notes
that are trivalent would be those ones with duplicated intervals, because (I thought), the notes
of the other type of trivalent scales can be nudged by tiny increments to nearby rationals.

Does that make some sense? It is rather tricky to express clearly. Certainly no intention
to be cryptic; I'm doing the best I can.

Robert

🔗Robert Walker <robert_walker@rcwalker.freeserve.co.uk>

Hi Dan,

Just looked at your recent posts, and they are making a lot more sense than they did - not
quite there yet, but if I work slowly through the earlier ones, I may well get it without having
to ask many more questions. We'll see.

This is how far I've got:

First, I'm trying to understand your [a,b] index, where they come from originally, and how you turn them into [a,b,c] indices.

I'll try re-writing your post in my own words, so see if this agrees with what you have in mind.

The notation itself is clear - [a,b,c] is the number of S, M and L step sizes,
and [a,b] is the number of S and L step sizes, respectiviely.

Next, you have a way of converting any arbitrary [a,b] index into a Myhill scale
with the desired number of small and large step sizes, right?

To do that, you need to make adjacent fractions, i.e. fractions r/s and t/u where
r*u and t*s differ by 1.

In preparation for that, you need to use a weighting.

You use as your weights..

golden ratio phi to make generalised "golden" scales

2, to make generalised "equal" scales

or 1+sqrt(2), to make generalised "silver" scales.

1/phi is the limiting value of the series

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

I know this one - 3 =2+1, 5 =3+2, 8 = 5+3,...

1/2 is limiting value of
1/1.5, 1.5/3.25, 3.25/6.375, 6.375/12.8125, 12.8125/25.59375, ...

Where does this sequence come from??
(
= 2/3 6/13 26/51 102/205 410/819
= 2*(1/3, 3/13, 13/51, 51/205, 205/819,...)
= 2*(1/3 3/13 13/(3*17) 3*17/(5*41) 5*41/(3^2*7*13)
)

1/(sqrt(2)+1) is the limiting value of the series

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

Let W be one of these three weights (or any other weight one likes)

Here's the generalized formula for deriving a weighted generator for the scale

To make a weighted generator X for scale with steps [a,b], with P as the
periodicity (formal octave)

use

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

where A and B are chosen such that A/a and B/b are adjacent fractions.

Take [5,4] as an example, i.e. we want to make a Myhill scale
with 5 large steps and 4 small ones. Let's make a Bohlen-Pierce scale,
so P = 3.

First, you need A, B such that A/a and B/b are adjacent fractions.

Substituting for a and b, the adjacent fractions are A/5 and B/4.

I.e. we want A*4 and B*5 to differ by 1 (that's the definition of an adjacent fraction).

You can use A=B=1, or alternatively, A=4 and B=3.
Let's use A=4 and B=3, i.e. A/a B/b = 4/5 3/4.

So formula gives
X = (P/(a+W*b))*(A+W*B) = (3/(5+W*4))*(4+W*3)

The first ex in the BP post is phi weighted. Writing g = phi (golden ratio):
X = (3/(5+g*4))*(4+g*3)

(reworking your calc:)
With g = 1.6180339887
I make this 2.3153816, or 1453.5 ¢

But you got 1468¢

Let's use that value instead, to follow what happens next -

We are making a 9 note scale. I.e 9 uses of this generator.

I get the same numbers you do:

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

For the reworked value of X it would be:
0 cents 1453.5 cents 1005 cents 556.59 cents 108.13 cents 1561.6 cents 1113.2 cents 664.72 cents 216.27 cents 1902 cents

Either way, it's Myhill, and 9 notes, as desired.

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

Right, so now I know how to make a Myhill scale by your construction, but have no idea why it works.

To check I understand this so far, let's try for a 17 note scale with 10 small and 7 large intervals.
[a,b] = [10,7]

Want A/10 and B/7 as adjacent fractions. For instance, A=3 and B =2 would do.

A/a B/b = 3/10 2/7.

So formula gives (for W=g)
X = (P/(a+W*b))*(A+W*B) = (3/(10+g*7))*(3+g*2) = 0.87723883

Since it is less than 1, use it's reciprocal as the generator.

This indeed makes a 17 note Myhill scale!

0 cents 138.8 cents 226.8 cents 365.5 cents 453.5 cents 592.3 cents 680.3 cents 819 cents 907 cents 1046 cents 1134 cents 1273 cents 1361 cents 1499 cents 1587 cents 1726 cents 1814 cents 1902 cents

How do you do it? Some kind of wizardry I think...

Well, maybe we'll find out later.

Let's keep going.

To go back to the example of [5,4], and using generator of X=1468¢ to keep in line with your expo:

You use this to make a generalised Fibonacci sequence starting with the 5 and 4:
5, 4, 9, 13, 21, ...

You want the fourth term of this sequence. Repeat X that many times, and find difference
between it and the periodicity P, and result will be the generalised Pythagorean comma.

13*1468¢ = 19084¢ = 1848.4512¢ (on reducing into range 1/1 to 3/1), comma of
1901.955-1848.4512 =~ 53.5 cents

Again I get a slightly different value from your 63¢, so to follow the rest, let's use your value.

Now you take your original Myhill scale generated using

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

Keep the first half of it unchanged, while moving all the intervals in the second half back
by the generalised pythagorean comma (with middle note 166 staying with the first half)

0---1468---1034---600---166 (unchanged
1634---1200---766---332 -> 1570----1136-----702-----268

where 1634 -> 1570 etc. - now you are using 64¢ for the comma.

And here we are:

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

Well, that part is easy to follow.

It gives a scale which can also be generated by my alternating generators method:

0 cents 1570 cents 1468 cents 1136 cents 1034 cents 702.1 cents 600.1 cents 268.1 cents 166.1 cents 1902 cents
with the generators
1570 cents and -102 cents.

Also I can see why I got different results when trying to reproduce your construction.

I was removing the comma from alternate intervals like this:

0---1468---1034---600---166---1634---1200---766---332
->
0---1404---1034---536---166---1570---1200---702---332

1404-----536-----1570----702
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
0----1034-----166-----1200-----332

My remarks about making the generators almost equal, and getting the
Wilson 17-note scale generated by 1/1 5/4 3/2 from the one
generated by 1/1 (3/2)^2 3/2 can't have made that much sense
in terms of your construction - hopefully they will now.

Well that explains a lot, and has clarified a lot.

I'll have a look at the rest of your post next.

But how _do_ you know that your weighting formula will work
for the Myhill scales in the first place? Or are you using a
published mathematical result from somewhere - if so, maybe
you could give the ref, so that I can look it up.

If published, maybe it is somewhere in the photocopies I lost in the coach station
on the way up to Skye over Christmas - I haven't yet gone back to the
library to redo them all.

Robert

🔗D.Stearns <STEARNS@CAPECOD.NET>

🔗Robert Walker <robert_walker@rcwalker.freeserve.co.uk>

Hi Dan,

> 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.

Thanks!

For some people it comes naturally perhaps, but I have to work very hard
at it.

Maths is good for developing clarity of ideas, but not so good for helping one
to express them clearly in ordinary english sentences.

You think something is so clear, such as the help files for a computer program, then
someone comes along and reads them, and they can't figure out all sorts of things,
and one can see why too, one sees that certain things just aren't explained, or are
explained in an extremely obscure fashion.

Then they all have to be re-worked again!

Luckily I've had spells of studying philosophy, especially philosophy of mathematics,
which involved a fair amount of essay writing for my tutors, which helps a lot. Also
studying books on English composition. Then once one sees what is needed, one
finds examples of good clear style here and there, and tries to learn what one can
from them.

After a while it takes off and begins to come more naturally.

Maths is particularly tricky to explain, and mathematicians often have a lot of trouble
trying to explain their ideas, especially to someone in another field, or new to their
field.

I think it is also especially hard to explain a new idea when you are doing it for
the first time - much easier if it has been explained before, as you can look at someone
else's explanation of it, and use the best points of that, and improve it, so with time,
the idea gets easier and easier to express.

Robert