Generator-offset property

A scale satisfies the generator-offset property (also GO, alternating generator or AG) if it satisfies the following properties:

1. The scale is generated by two chains of stacked copies of a generator.
2. The two chains are separated by an offset (the difference between the first note of the second chain minus the first note of the first chain).
3. The lengths of the chains differ by at most one. (1-3 can be restated as: The scale can be built by stacking two alternating generators (called alternants), which do not necessarily take up the same number of steps.)
4. The generator always occurs as the same number of steps. For example, the generator is never both a 2-step and a 3-step.

Plot of at-most-decatonic MV3 GO scale regions in the plane.

The Zarlino (3L 2M 2S) JI scale is an example of a GO scale, because it is built by stacking alternating 5/4 and 6/5 generators. 7-limit diasem (5L 2M 2S) is another example, with generators 7/6 and 8/7.

More formally, a cyclic word S (representing the steps of a periodic scale) of size n is GO if it satisfies the following properties:

1. S is generated by two chains of stacked generators g separated by a fixed offset δ; either both chains are of size n/2, or one chain has size (n + 1)/2 and the second has size (n − 1)/2. Equivalently, S can be built by stacking a single chain of alternants g₁ and g₂, resulting in a circle of the form either g₁ g₂ ... g₁ g₂ g₁ g₃ or g₁ g₂ ... g₁ g₂ g₃.
2. The scale is well-formed with respect to g, i.e. all occurrences of the generator g are k-steps for a fixed k.

This doesn't imply that g₁ and g₂ are the same number of scale steps. For example, 5-limit blackdye has g₁ = 9/5 (a 9-step) and g₂ = 5/3 (a 7-step).

More generally, we say that a scale is m-GO if both of the following hold:

1. the scale consists of m + 1 chains of stacked generator g (implying m offsets δ₁, ...,δ_m from a fixed chain), each chain having either j or j + 1 notes.
2. The scale is well-formed with respect to g.

(Thus 1-GO is the same thing as GO.)

An m-GO scale can be interpreted as a scale in a rank-(m + 2) regular temperament, with basis p (period), g, δ₁, ...,δ_m (though the specific tuning used may be of lower rank).

Other definitions

• A strengthening of the generator-offset property, tentatively named the swung-generator-alternant property (SGA), states that the alternants g₁ and g₂ can be taken to always subtend the same number of scale steps, thus both representing "detemperings" of a generator of a single-period mos scale (otherwise known as a well-formed scale). All odd GO scales are SGA, and aside from odd GO scales, only xyxz satisfies this property. The Zarlino and diasem scales above are both SGA. Blackdye is GO but not SGA.

Theorems

Proposition 1 (Properties of SGA scales)

Let S be a 3-step-size scale word in L, M, and s, and suppose S is SGA. Then:

1. S is unconditionally MV3 (i.e. MV3 regardless of tuning).
2. S is of the form ax by bz for some permutation (x, y, z) of (L, M, s).
3. The length of S is either odd, or 4 (and S is of the form xyxz).
4. S = aX bY bZ is obtained from some mode of the (single-period) mos aX 2bW by replacing all the W's successively with alternating Y's and Z's (or alternating Z's and Y's for the other chirality, fixing the mode of aX 2bW).
5. The two alternants differ by replacing one Y with a Z.
6. S is pairwise-mos. That is, the following operations result in a mos: setting L = M, setting L = s, and setting M = s.
7. S is elimination-mos. That is, "tempering out" any one step size results in a mos.

In particular, odd GO scales always satisfy these properties (see Proposition 2 below).

[Note: This is not true with SGA replaced with GO; blackdye is a counterexample that is MV4.]

Proof

Assuming SGA, we have two chains of generator g₀ (going right). The two cases are:

CASE 1: EVEN LENGTH
O-O-...-O (n/2 notes)
O-O-...-O (n/2 notes)

and

CASE 2: ODD LENGTH
O-O-O-...-O ((n+1)/2 notes)
O-O-...-O ((n-1)/2 notes).

Label the notes (1, j) and (2, j), 1 ≤ j ≤ (chain length), for notes in the upper and lower chain respectively.

In case 1, let g₁ = (2, 1) − (1, 1) and g₂ = (1, 2) − (2, 1). We have the chain g₁ g₂ g₁ g₂ ... g₁ g₃.

Let r be odd and r ≥ 3. Consider the following abstract sizes for the interval class (k-steps) reached by stacking r generators:

1. from g₁ ... g₁, we get a₁ = (r − 1)/2*g₀ + g₁ = (r + 1)/2 g₁ + (r − 1)/2 g₂
2. from g₂ ... g₂, we get a₂ = (r − 1)/2*g₀ + g₂ = (r − 1)/2 g₁ + (r + 1)/2 g₂
3. from g₂ (...even # of gens...) g₁ g₃ g₁ (...even # of gens...) g₂, we get a₃ = (r − 1)/2 g₁ + (r − 1)/2 g₂ + g₃
4. from g₁ (...odd # of gens...) g₁ g₃ g₁ (...odd # of gens...) g₁, we get a₄ = (r + 1)/2 g₁ + (r − 3)/2 g₂ + g₃.

Choose a tuning where g₁ and g₂ are both very close to but not exactly 1/2*g₀, resulting in a scale very close to the mos generated by 1/2 g₀. (i.e. g₁ and g₂ differ from 1/2*g₀ by ε, a quantity much smaller than the chroma of the n/2-note mos generated by g₀, which is |g₃ − g₂|). Thus we have 4 distinct sizes for k-steps:

1. a₁, a₂ and a₃ are clearly distinct.
2. a₄ − a₃ = g₁ − g₂ != 0, since the scale is a non-degenerate GO scale.
3. a₄ − a₁ = g₃ − g₂ = (g₃ + g₁) − (g₂ + g₁) != 0. This is exactly the chroma of the mos generated by g₀.
4. a₄ − a₂ = g₁ − 2 g₂ + g₃ = (g₃ − g₂) + (g₁ − g₂) = (chroma ± ε) != 0 by choice of tuning.

By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g₁ and g₂ must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form xy...xyxz. But this pattern is not unconditionally SV3 if n ≥ 6, since 3-steps come in 4 sizes: xyx, yxy, yxz and xzx. Thus n = 4 and the scale is xyxz. This proves (3).

In case 2, let (2, 1) − (1, 1) = g₁, (1, 2) − (2, 1) = g₂ be the two alternants. Let g₃ be the leftover generator after stacking alternating g₁ and g₂. Then the generator circle looks like g₁ g₂ g₁ g₂ ... g₁ g₂ g₃. Then the combinations of alternants corresponding to a step come in exactly 3 sizes:

1. kg₁ + (k − 1)g₂
2. (k − 1)g₁ + kg₂
3. (k − 1)g₁ + (k − 1) g₂ + g₃,

if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times. This proves (2).

(The above holds for any odd n ≥ 3.)

For (1), we now only need to see that SGA + odd length => unconditionally SV3. But the argument in case 2 above works for any interval class (unconditional SV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.

For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. If Y’s and Z’s don't alternate perfectly, then (ignoring X's) you have two consecutive Y's somewhere and two consecutive Z's somewhere else. Assume that g_pf = iX + jW with j >=2. (If this is not true, invert the generator, since b > 1.)

In aX bY bZ, consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) (the preimage of) the imperfect generator. Since a + 2b >= 5, are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to AV3. [Let T be the subword consisting only of Y's and Z's. If T has a substring of length j that's not contained in a perfect generator, you can go somewhere else to find it, since the numbers of Y's and Z's change one at a time and reach a maximum and a minimum somewhere, guaranteeing that intermediate values are reached "on the other side".]

Any generator of aX 2bW must have an odd number of W steps; if a generator had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction since the generator of aX bW' can be generated by a generator of aX 2bW. This with (4) immediately gives (5).

For (6), odd-numbered SGA scales are Fokker blocks (in the 2-dimensional lattice generated by the generator and the offset). To see this, consider the following lattice depiction of such a scale:

x x x ... x 
x x x ... x x

and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A Fokker block has the property that tempering out by each of the chromas gives two mosses. These correspond to two of the temperings X = Y, Y = Z and X = Z. The third tempering follows by symmetry (by taking the other chirality).

For (7), consider the mos aX 2bW as chunks of X separated by W (tempering Y and Z together into W). Eliminating every other W turns it into a mos, because the sum of sizes of consecutive chunks of X (1st chunk with 2nd chunk, 3rd with 4th, ...) must form a mos. This is because the chunk sizes of X form a mos, and taking every kth note of an n-note mos where k divides n yields a mos. Since the result of setting X = 0 is the mos bY bZ, S is elimination-mos.

Proposition 2 (Odd GO scales are SGA)

Suppose that a periodic scale satisfies the following:

• is generator-offset
• has odd size n.

Then the scale is SGA.

Proof

Assume that the generator is a k-step and k is even. (If k is not even, invert the generator.) On some tonic p we have a chain of ceil(n/2) notes and on some other note p' = p + offset (not on the first chain) we'll have floor(n/2) notes.

We must have gcd(k, n) = 1. If not, since n is odd, gcd(k, n) is an odd number at least 3, and the k-steps must form more than 2 parallel chains.

By modular arithmetic we have rk mod n = k/2 iff r = ceil(n/2) mod n. (Since gcd(2, n) = 1, 2 is multiplicatively invertible mod n, and we can multiply both sides by 2 to check this.) This proves that the offset, which must be reached after ceil(n/2) generator steps, is a k/2-step, as desired. (If the offset wasn't reached in ceil(n/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.)

Proposition 3 (Properties of even GO scales)

GO scales of even size have the following properties:

1. They are two parallel mosses of size n/2 generated by g, with offset a k-step with k odd
2. They are not SV3
3. They are not chiral

Proof

(1) and (2) were proved in the proof of Prop 1. (3) is easy to check using (1).

Open conjectures

Conjecture ("SV3 Structure Theorem")

Conjecture 1

If a 3-step size scale word ax by cz is

1. unconditionally SV3 (strict variety 3, i.e. every k-step except multiples of the equave comes in exactly 3 sizes),
2. has odd length,
3. has gcd(a, b, c) = 1,
4. is not of the form xyzyx or xyxzxyx,

then it is GO (and therefore SGA). (a converse to Theorem 1)

Conjecture ("MV3 Sequences")

Given any two generators, we can iterate them to any number of notes and see what the maximum-variety of the resulting scale is. In particular, we can look at those scale sizes which are MV3, and thus compute the MV3 sequence for the pair of generators (similar to the "MOS sequence" one can compute for one generator). Thus, for any pair of generators, we can form the associated sequence of increasingly large MV3 scales.

Surprisingly, for almost all pairs of generators, this sequence seems to terminate after some (usually relatively small) scale. That is, if we simply take all possible pairs of generators between 0 and 1200 cents, and for each pair we compute the MV3 sequence for all generator pairs up to some maximum N, such as 1000, we can easily see that most points will have only a few entries in it, after which no MV3 scales are apparently generated. It would seem to be true that as the two generators get closer and closer in size, the MV3 sequence gets longer and longer, until when the two generators are equal you have an infinite-length sequence (corresponding to MOS).

It is pretty easy to see this behavior is true if we simply compute the MV3 sequences up to any very large N, far beyond the scale sizes we typically use in music theory, but it would be good to have a proof.

Open questions

This heading has those open questions for which no conjecture has yet been formed either way. (These can be updated as necessary)

1. Given any arbitrary MOS (or DE, etc) scale with at least three notes per period, is there *always* a MV3 GO scale which can be derived as a "detempering" of that scale? Or is this only true for some MOS's? For instance, the MOS LLsLLLs has the MV3 GO scale LmsLmLs as a detempering. Does a similar MV3 detempering exist for every possible DE scale with at least three notes per period, or at least for strict MOS's with one period per octave (e.g. well-formed scales)?
2. The scale tree is a great way to analyze MOS scales. For any generator, we can compute the various MOS's it forms if we simply look at the scale tree, and indeed MOS "words" like LLsLLLs can be identified with regions on the scale tree (in this situation the interval between 4/7 and 3/5). A similar "scale plane" should exist for GO-MV3 scales, where given some word representing a GO-MV3 scale, we can look at the set of points on the generator plane which generates it; these seem to often be triangles, with the lines corresponding to MOS's and the vertices corresponding to EDOs (though is this always true?). What is the big picture of this scale plane? Can we use Viggo Brun's algorithm for this, generalizing the theory of continued fractions? Is there some simple formula we can use to predict, given some GO-MV3 scale, which region on the scale plane it corresponds to? Can we plot simple generator-size-proportions as points in this space? And so on.
3. In the theory of MOS, there is a second scale tree that is less frequently talked about, which Erv Wilson calls the "Rabbit Sequence" (Erv Wilson's original version, interactive version 1, interactive version 2). This is a tree for which each MOS word has two children, depending on if the MOS is "soft" (with L/s < 2) or "hard" (with L/s > 2). For instance, LsLss has the two children LLsLLLs and ssLsssL. Does a similar scale plane exist for these GO-MV3 scales?

Falsified conjectures

All mosses have an MV3 detempering (Counterexample: LsLsLsLsLs)

LsLsLsLsLs does not have an MV3 detempering: Wolog we “detemper” L to L and M.

LsMsLsMsLs is not MV3: LsM LsL sMs sLs

LsMsLsMsMs can be rotated to MsLsMsLsMs which is not MV3 by symmetry with LsMsLsMsLs.

LsMsMsMsMs is not MV3: LsM MLM sMs sLs

MsLsLsLsLs not MV3 for the same reason as LsMsMsMsMs