Clusters of Steps in Moments of Symmetry

Take a look at the following scale, 21L5s
LLLLLsLLLLsLLLLsLLLLsLLLLs

We can group the clusters of large steps as follows:
(LLLLL)s(LLLL)s(LLLL)s(LLLL)s(LLLL)s

Notice how the clusters of large steps come in two different sizes:
-Narrow Cluster: 4 large steps (LLLL)
-Wide Cluster: 5 large steps (LLLLL)

Hmmm, so it appears cluster come in **two distinct sizes** and that they spread themselves out **evenly** - remind you of anything? More on this later…

So let's assume the more general case, the MOS ALBs (assuming GCD(A,B)=1, A>B). The sizes of the clusters of "A-steps" can be represented as so:
-Narrow Cluster: floor(A/B)
-Wide Cluster: ceil(A/B) = floor(A/B)+1

The number of times each cluster appears can be found with these expressions:
#Narrow Clusters = B*(floor(A/B)+1)-A
#Wide Clusters = A-B*floor(A/B)

The sum of the above two expressions is B (this is trivially true if you think about it).

In the case where B>A, simply replace all B's with A's and vice versa in the above formulas.

Ryan Avella

That's beautiful. I was aware of this pattern, and had intended to try to work it out, but never got round to it.

Andy Milne

--- In tuning-math@yahoogroups.com, "Ryan Avella" <domeofatonement@...> wrote:
>
> Take a look at the following scale, 21L5s
> LLLLLsLLLLsLLLLsLLLLsLLLLs
>
> We can group the clusters of large steps as follows:
> (LLLLL)s(LLLL)s(LLLL)s(LLLL)s(LLLL)s
>
> Notice how the clusters of large steps come in two different sizes:
> -Narrow Cluster: 4 large steps (LLLL)
> -Wide Cluster: 5 large steps (LLLLL)
>
> Hmmm, so it appears cluster come in **two distinct sizes** and that they spread themselves out **evenly** - remind you of anything? More on this later…
>
>
> So let's assume the more general case, the MOS ALBs (assuming GCD(A,B)=1, A>B). The sizes of the clusters of "A-steps" can be represented as so:
> -Narrow Cluster: floor(A/B)
> -Wide Cluster: ceil(A/B) = floor(A/B)+1
>
> The number of times each cluster appears can be found with these expressions:
> #Narrow Clusters = B*(floor(A/B)+1)-A
> #Wide Clusters = A-B*floor(A/B)
>
> The sum of the above two expressions is B (this is trivially true if you think about it).
>
> In the case where B>A, simply replace all B's with A's and vice versa in the above formulas.
>
>
> Ryan Avella
>

--- In tuning-math@yahoogroups.com, "Ryan Avella" <domeofatonement@...> wrote:
> So let's assume the more general case, the MOS ALBs (assuming GCD(A,B)=1, A>B). The sizes of the clusters of "A-steps" can be represented as so:

I just realized that my conditions for the general case aren't complete. It is true that GCD(A,B) must be 1, but this is not sufficient. We must also make sure that B does not divide A at all.

Here are the revised conditions for a MOS ALBs, which should be sufficient:
1.) A > B > 1
2.) GCD(A,B) = 1

As before, the theorem still holds true if all B's are switched with A's, and vice versa. Hence, the theorem does not discriminate between large and small steps.

Interestingly, this theorem can be used to craft a sort of "transform" from one MOS scale to a more primitive MOS. For example, 22L5s transforms into 2L3s, which then transforms into 1L1s. This transform essentially takes the clusters of steps of a MOS and turns them into the basic building blocks of a new MOS.

Ryan Avella

--- In tuning-math@yahoogroups.com, "Ryan Avella" <domeofatonement@...> wrote:

> Interestingly, this theorem can be used to craft a sort of "transform" from one MOS scale to a more primitive MOS. For example, 22L5s transforms into 2L3s, which then transforms into 1L1s. This transform essentially takes the clusters of steps of a MOS and turns them into the basic building blocks of a new MOS.

Makes me wonder about a context-free language for MOS.

On Thu, Nov 8, 2012 at 9:07 AM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> Makes me wonder about a context-free language for MOS.

Relevant is Erv Wilson's Rabbit Sequence:
http://www.anaphoria.com/RabbitSequence.pdf

His words are over the alphabet {A, b}, where ironically his A
corresponds to our s and his b corresponds to our L. Then for each
word there's a binary fork; the top route on each fork replaces all A
with Ab, and the bottom route on each fork does the same thing but
swaps A and b afterward.

I have this hunch that a similar context-free grammar approach will be
strongly related to "omnitetrachordal" scales, which I've started
calling "subperiodic" scales to reflect the cases where 4/3 isn't the
interval at which the tetrachords repeat (what I'm calling the
"subperiod").

-Mike