Recursive structure of MOS scales: Difference between revisions

Let w be an arbitrary moment-of-symmetry scale in either its brightest mode (where L's come before s's) or darkest mode (where s's come before L's). Let x be the number of L's in the scale and let y be the number of s's in the scale. Such a scale will fall under one of three general cases or one of three special cases described below.

Note that in this reduction process, the order of L's and s's may reverse. This is normal.

Just as reducing a moment-of-symmetry scale using the aforementioned process produces another moment-of-symmetry scale, reversing the reduction rules into production rules will produce another scale w' that is also a moment-of-symmetry scale. One of the following sets of production rules can be used:

Repeating these production rules will produce a scale that is more chromatic than the last, and every scale produced this way is another moment-of-symmetry scale.

The recursive nature of moment-of-symmetry scales allows for two algorithms to be made: one that can determine whether an arbitrary scale of large and small steps is a valid moment-of-symmetry scale and another that can construct a moment-of-symmetry scale from any number of large and small steps, without any prior knowledge of the scale's structure.  

This is the pseudocode for the chunking operation described above.

   

  integer x = 0;

  } else {

  string subscale = set.beginning_of_set

  }

  }

  subscale += "L"

   

  }

  }

  }

6L 1s, or LLLLLLs. This falls under one of the special cases and is a moment-of-symmetry scale.  

3L 6s, or LssLssLss. Since 3 and 6 share a common factor of 3, this scale consists of Lss repeated three times, which falls under the special cases and is a multi-period moment-of-symmetry scale.

=== General algorithm ===

This is the general algorithm for generating a moment of symmetry scale of the form xL ys represented as a string of L's and s's in its brightest mode. This algorithm will produce that scale without any prior knowledge of how the scale's steps are ordered. This algorithm is recursive. Let x be the number of L's and y the number of s's.  

## If both x and y are equal to 1, then the final scale is "Ls".

## If only x is equal to 1, then the final scale is L followed by y s's.

## If only y is equal to 1, then the final scale is x L's followed by s.

#### If x > y, every instance of an L in the prescale is replaced with one L and u s's, and every s replaced with one L and v s's. This produces the final scale in its brightest mode.

The code shown here is a Python function that implements the algorithm described above.<syntaxhighlight lang="python3" line="1">

def CalculateMos(x, y):

   

To find the generators for 5L 7s, we can start at step 2.3.1, as the mos is neither multi-period nor 1L 1s. Using these steps, we can recursively find the mos 2L 3s, which means we recursively find the mos 1L 1s.

To make sure that we have the correct mos, we can relate it back to 5L 2s, whose mos LLLsLLs. By replacing "Ls" with L, we can reduce our mos into that for 5L 2s.

The recursive structure also allows a recursive algorithm to find the generator, as a substring of the mos pattern. The reason why this works is that the reduction both ''preserves'' and ''reflects'' the generator; see the ''proofs'' section.

This is the general algorithm for finding a moment of symmetry scale's generator and its [[octave complement]] for a mos xL ys as two, not-necessarily-equal halves of the mos pattern xL ys. If the mos is multi-period, its generator and period complement are returned as two halves of the period. This algorithm will produce these intervals without any prior knowledge of how the scale's steps are ordered. This algorithm is recursive. Let x be the number of L's and y the number of s's.  

#### If y > x, every instance of an L in both intervals is replaced with u L's and one s, and every s replaced with v L's and one s. This produces the final scale's generator and complement.

The algorithm described here returns what is known as the chroma-positive generator, or bright generator, and is usually what is meant by ''the'' generator of a scale. The octave-complement returned this way is the scale's chroma-negative generator, or dark generator. More specifically, the bright generator is in its perfect form, and the dark generator is in its augmented form. If the perfect dark generator is needed instead, a quick way to produce it is to replace one L in the dark generator with an s.

If the number of steps in the generators is needed, simply count the total number of L's and s's in each generator.

The code shown here is a Python function that implements the algorithm described above.<syntaxhighlight lang="python3" line="1">

def CalculateMosGenerators(x, y):

</syntaxhighlight>

To find the generators for 5L 7s, we can start at step 2.3.1, as the mos is neither multi-period nor 1L 1s. Using these steps, we can recursively find the generators for the mos 2L 3s, which means we recursively find the generators for 1L 1s.

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

Assume there are more L's and s's, and that there is more than one s, and thus more than one chunk. Assume there is a generator. Assume the imperfect generator is bigger than the perfect generator. (If this isn't true, just use the inverted generator.) Because there are at least two chunk boundaries, and only one imperfect generator, there must be a chunk boundary with a perfect generator on top (the chunk boundary is on the left of the generator, e.g. <code>...|Ls|LLs|Ls...</code>). Because there's a chunk boundary to the left, there's an s just to the left of the left endpoint of the generator. If the rightmost step of this generator were an L, scooting the generator one step to the left would make it smaller, which contradicts the assumption that the imperfect generator is larger than perfect. Thus, the rightmost step of this generator is an s. That means the right endpoint of this generator falls on a chunk boundary. We already know the left endpoint also falls on a chunk boundary, so this perfect generator is still present as an interval in the reduced MOS.

This interval is, in fact, the generator of the reduced scale. Indeed, you can build this generator on all of the chunk boundaries of the non-reduced scale. All but one of these will be perfect. The perfect ones will have their right endpoint be a chunk boundary, as we showed before, so this is an interval that's on all but one tone of the reduced scale. This is a generator.  

Suppose there is a generator after chunking. Assume also that there are more L's than s's, and that the imperfect generator is larger than the perfect generator. This means that this interval, built on the chunk boundaries after expanding, will still be "a generator" on the chunk boundaries: it will be the same size on all but one of the chunk boundaries. Take one of the "perfect" (smaller) intervals. We showed previously that, before expansion, the step to its right was an L, just like its first step; therefore we can scoot it over for the entirety of the first chunk and keep it perfect. We now come to another chunk. If a "perfect" interval can be built on it then we can repeat. If the interval is imperfect, then we know that the step to its right is an s, which will make it smaller and thus perfect, and we can keep doing this for the remainder of the chunk. Therefore, the interval is a generator after expansion as well, since it is only imperfect in one position.

Every MOS can be reduced to nL 1s (or 1L ns). Each step of the reduction decreases either the number of L's or the number of s's (or both), so one of them must reach 1 at some point. ''(note that reducing further gets us to 1L 0s, which has a period, but no generator per se)''

It is clear that the MOS nL 1s has a unique generator, L (or its inversion). However, the previous proof showed that reduction reflects generators, and so by induction all MOS scales have a single generator.

[[Category:MOS scale]]