Recursive structure of MOS scales

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

If you have aLbs, there are b chunks. In that case (a % b) of the chunks will have ceil(a/b) "L"s, the rest will have floor(a/b) "L"s.

Example:

We take 5L7s, which will have 7 chunks

The chunks must partition the 5 "L"s, and 5 = 1 + 1 + 1 + 1 + 1 + 0 + 0.

So we get 5L2s, which we know is LLLsLLs and turns back to sL sL sL s sL sL s

Finding a generator

The recursive structure also allows a recursive algorithm to find the generator:

To find the generator you reduce to a simple pattern you know the generator of, then plug everything back in.

If you do that to 5L 7s you get 5L 2s: LLLsLLs. You know that the generator of 5L 2s is LLLs, so using the above procedure for finding the pattern, the generator of 5L 7s is LsLsLss. If you don't know anything you just reduce all the way to 1L 1s and plug everything back in.

Tree of MOS patterns

MOS patterns have parents and children. If 1L 1s is the root of the tree, with any generator between 0\1 (for paucitonic 1L 1s) and 1\2 (for equalized 1L 1s) Every node aL bs has two children, aL (a+b)s and (a+b)L as (these MOS patterns are sister MOS patterns; they are called such because they have the same parent). The generator range of aL bs splits at the mediant of the endpoints of the parent interval (which is the generator size for basic aL bs), giving the generator ranges of the two child patterns.

The tree of MOS patterns starts with:

1L 1s
1L 2s        2L 1s
1L 3s 3L 1s  3L 2s 2L 3s

The corresponding tree of generator ranges looks like:

(0\1, 1\2)
(0\1, 1\3)            (1\3, 1\2)
(0\1, 1\4) (1\4, 1\3) (1\3, 2\5) (2\5, 1\2)