Recursive structure of MOS scales: Difference between revisions

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== Finding the MOS pattern from nLms ==
== Finding the MOS pattern from nLms ==


Example: 7L5s
Example:  
 
7L5s
 
7 chunks
7 chunks
5 = 1 + 1 + 1 + 1 + 1 + 0 + 0
5 = 1 + 1 + 1 + 1 + 1 + 0 + 0
5L2s
5L2s
LLLsLLs -> Ls Ls Ls L Ls Ls L
LLLsLLs -> Ls Ls Ls L Ls Ls L


== Finding a generator ==
== Finding a generator ==

Revision as of 23:07, 14 April 2021

By looking at the "tetrachords" L..s of an MOS scale in word form and giving them the names "L" and "s", we get out another MOS scale. The MOS thus obtained preserves a number of important properties, such as which interval is the generator. To find properties of complex MOS word patterns, we can then just compare them to the simpler ones, whose properties we know.

Recursive structure

Let $w$ be an MOS scale word such that there are strictly fewer $s$ and $L$. (by MOS we mean that the word is left- and right-periodic and has max variety 2). We can separate the MOS into chunks of $L\dots s$, which will come in two varieties where one has one more $L$ than the other.

(Proof: Suppose we had three chunks L...s with $n$, $n+1$ and $n+2$ 'L's. Then we have a length n+2 subword that's 'L's, one that has one s at the end and one that has two 's's on either side.)

By naming the larger chunk $L'$ and the smaller chunk $s'$, we end up with another MOS.

Example: LLsLsLsLLsLs becomes L's's'L's', becomes Ls.

(Proof: by a similar argument to the previous proof, if the resulting word was not max variety 2 then the original was not either.)

It is clear that the operation is reversible.

Finding the MOS pattern from nLms

Example:

7L5s

7 chunks

5 = 1 + 1 + 1 + 1 + 1 + 0 + 0

5L2s

LLLsLLs -> Ls Ls Ls L Ls Ls L

Finding a generator