Generator-offset property

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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 an interval called generator.
  2. The two chains are separated by a different interval called the offset (the difference between the first note of the second chain and 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) a1 and a2. Note that a1 does not need to subtend, i.e. occur as, the same number of steps as a2.)
  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. The 7-limit version of diasem (5L 2M 2S) is another example, with generators 7/6 and 8/7.

For a generalization, see alternating generator sequence.

Mathematical definition

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).

Conventions used on this page

  • Non-italicized Latin variables refer to interval sizes, for example step sizes.
  • Indices for all words are 1-indexed.
    • If S is a circular word and i < 1 or i > len(S), we first replace i with i % len(S) + 1 before using it as an argument in S[-].
  • The notation S(X1, ..., Xr) is used for an r-ary scale word with variables X1, ..., Xr possibly standing in for any sizes. If S(X, Y) = XXY then S(A, B) = AAB.
  • We leave the distinction between linear and cyclic words up to context. We usually also elide the distinction between subwords and the interval sizes that subtend them.

Other definitions

  • A scale or scale word is a circular word with a chosen size for its equave. As we're not working with scales with distinct equaves simultaneously, all three terms are effectively synonymous for our purposes.
  • A scale is primitive, or single-period, if its period is the same as its equave. A multimos or multiperiod mos is a non-primitive mos. A mos aLbs is primitive iff gcd(a, b) = 1.
  • An n-ary scale is a scale with n different step sizes. Binary and ternary are used when n = 2 and 3 respectively.
  • 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, the only ternary scales to satisfy SGA are (xy)rxz, r ≥ 1. The Zarlino and diasem scales above are both SGA. Blackdye is GO but not SGA.
  • An odd-step is a k-step where k is odd; an even-step is defined similarly.
  • Given a linear or cyclic word S with a step size X, define EX(S) as the scale word resulting from deleting all instances of X from S.
  • By a subword, substring, or slice of a word S, denoted S[i : j] (j > i), we mean S[i] S[i + 1] ... S[j − 1].
  • Given a mos aX bY, a chunk of X's is a maximal (possibly length 0) substring made of X's, bounded by Y's. We do not include the boundary Y's.
  • Length is another term for a scale's size.
  • A projection of a ternary scale is the operation of equating two of its step sizes.
  • A ternary scale is pairwise-well-formed if all its projections are well-formed (i.e. single-period mosses).

Theorems

Proposition 1 (Properties of SGA scales)

Let S be a ternary scale word in L, M, and s of length n, and suppose S is SGA. Then:

  1. The length of S is odd, or S is equivalent to (xy)rxz for some integer r ≥ 1.
  2. If n is odd, S is of the form ax by bz for some permutation (x, y, z) of (L, M, s).
  3. If n is odd, S is abstractly SV3 (i.e. SV3 for almost all tunings).
  4. If n is odd, 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). The two alternants differ by replacing one Y with a Z.
  5. If n is odd, S is pairwise-mos. That is, the following operations each result in a mos: setting L = M, setting L = s, and setting M = s.
  6. If n is odd, S is elimination-mos. That is, EX(S), EY(S), EZ(S) are all mosses.

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

Let e be the equave of S.

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 ≤ (number of notes in the chain), for notes in the upper and lower chain respectively.

Statement (1)

In case 1, let g1 = (2, 1) − (1, 1), g2 = (1, 2) − (2, 1), and g3 = (1, 1) − (n/2, 2) = (−n/2*g1 − g1 − n/2*g2) mod e. We assume that g1, g2 and e are Z-linearly independent. We have the chain g1 g2 g1 g2 ... g1 g3 which visits every note in S.

Since S is GO it is well-formed with respect to g = (g2 + g1). Since g1 and g2 subtend the same number of steps, all multiples of the generator g must be even-steps, and those intervals that are "offset" by g1 must be odd-steps. Letting M be the subset of all even-numbered notes (which are generated by g) and considering M as a scale by dividing degree indices in M by two, M is well-formed with respect to g, thus M (and its offset) must be a mos subset. Hence (g3 + g1), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus g2 and g3 subtend the same number of steps, a fact we need in order to be able to substitute one instance of g2 with g3 in the next part.

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 ≡ (r − n/2 − 3/2)g1 + (r − n/2 − 1/2)g2 mod e.
  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 ≡ (r − n/2 − 1/2)g1 + (r − n/2 − 3/2)g2 mod e.

These are all distinct by Z-linear independence. 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, abstractly SV3, GO scale must be of the form (xy)rxz. (Note that (xy)rxz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).

Statement (2)

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. Assuming that a step is an odd number of generators, 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

(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.)

Statement (3)

For (3), we now only need to see that if len(S) is odd and S is SGA, S is abstractly SV3. But the argument in case 2 above works when you substitute any interval class in S instead of a 1-step (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes.

Statement (4)

For (4), assume S is aX bY bZ, a odd. If b = 1, there's nothing to prove, so assume b > 1.

Consider the two alternants, detemperings of the generator iX + jW of the aX 2bW mos T(X, W) = S(X, W, W) where gcd(j, 2k) = 1.

Claim 1: Deleting X's from the alternant subwords of S gives every j-step subword in the scale EX(S)(Y, Z), the scale word obtained by deleting all X's from S.

Proof: Assume, possibly after inverting the generator, that the imperfect generator of T has j + 1 W's and the perfect generator has j W's. Suppose that one j-step word R on index p of EX(S) is "contained in" the corresponding "imperfect alternant" of S, which is I = S[p : p + i + j]. By this we mean that EX(I) has R as a substring. Then S[p − 1: p − 1 + i + j] and S[p + 1 : p + 1 + i + j] are both detemperings of perfect generators of T, and have one fewer step that is Y or Z. Thus the word I must both begin and end in a letter that is either Y or Z. Removing all the X's from I results in a word that is j + 1 letters long and is the j-step we started with, with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this j-step. The rest of the j-step subwords are all contained in "perfect" alternants; take q ≠ p to be the index of the first letter of one such j-step subword (as contained in S) and use S[q : q + i + j].

Claim 2: If a binary scale U has b Y's and b Z's, gcd(j, 2b) = 1, and consecutively stacked j-steps in U occur in 2 alternating sizes, then U = (YZ)b.

Proof: Write u and v for the two sizes of j-steps. Since gcd(j, 2b) = 1, there exists m such that stacking m-many j-steps yields scale steps of U, and m is odd because gcd(m, 2b) = 1. Hence the scale steps of U are (uv)(m−1)/2u mod e and (vu)(m−1)/2v mod e, and the step sizes clearly alternate.

These two claims prove that EX(S) = (YZ)b and that the two alternants' sizes differ by replacing one Y for a Z.

Statement (5)

For (5), 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 rank-3 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).

Statement (6)

For (6), consider the mos aX 2bW as chunks of X separated by W (tempering Y and Z together into W). Eliminating every other W (corresponding to either Y or Z) turns it into a mos:

  1. The chunk sizes of X form a mos, and taking every kth note of an n-note mos yields a mos.
  2. The sum of sizes of consecutive chunks of X (1st chunk with 2nd chunk, 3rd with 4th, ...) must form a mos.
  3. If chunk sizes of a binary scale form a mos, the scale itself must be a mos; see Recursive structure of MOS scales, "Reflection of generators".

Lastly, EX(S) is the mos bY bZ; hence 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 note p we have a chain of (n + 1)/2 notes and on p′ = p + offset we'll have (n − 1)/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 by well-formedness with respect to the generator, the generators must form more than 2 parallel chains.

By modular arithmetic we have rk ≡ k/2 mod n iff r ≡ (n + 1)/2 mod n. (Note that both 2 and k are coprime with n, hence multiplicatively invertible mod n.) This proves that the offset, which must be reached after (n + 1)/2 k-steps, is a k/2-step, as desired. (If the offset wasn't reached in (n + 1)/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)

A primitive GO scale of even size where the generator g is an even-step (i.e. g subtends an even number of steps) has the following properties:

  1. It is a union of two primitive mosses of size n/2 generated by g
  2. It is not SV3
  3. It is not chiral
  4. It is a primitive mos word of two of the letters interleaved with the third letter, for example: XYXZXYXZXY, made by interleaving X's in the mos word YZYZY.

Proof

(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator g must be even-steps ..." to "These are all distinct by Z-linear independence", does not rely on S having the SGA property). (3) and (4) are easy to check using (1).

Theorem 4 (Classification of pairwise well-formed scales)

Let S(a, b, c) be a scale word in three Z-linearly independent step sizes a, b, c. Suppose S is pairwise well-formed (equivalently, all its projections are single-period mosses). Then S is SV3 and has an odd number of notes. Moreover, S is either GO or equivalent to the scale word abacaba.

Proof

If the generator of a projection of S is a k-step, the word of stacked k-steps in S is pairwise well-formed

Suppose S has n notes (after dealing with small cases, we may assume n ≥ 7) and S projects to single-period mosses S1 (via identifying b with c), S2 (via identifying a with c) and S3 (via identifying a with b). Suppose S1's generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking n-many k-steps, we get two words of length n of k-steps of S2 and S3, respectively. These binary words, which we call Σ2 and Σ3, must be mosses, since m-steps in the new words correspond to mk-steps in the mos words S1 and S2, which come in at most two sizes. Since S1 is a single-period mos, gcd(k, n) = 1. Hence when 0 < m < n, mk is not divisible by n and mk-steps come in exactly two sizes; hence both Σ2 and Σ3 are single-period mosses. 

index: 1 2 3 4 ...   n
Σ1:    P P P P ... P I
Σ2:    [some mos]
Σ3:    [some mos]

Below we write step sizes resulting from identification as a~b, b~c, and a~c.

Two sizes of k-steps in S project to S1's perfect generator

We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c).

Suppose for sake of contradiction that only one size of k-step (α, β, γ) in S projects to P in S1. Then projecting to S2 shows that S2's generator is the k-step (α + γ)*(a~c) + βb, and Σ2's imperfect generator is located at index n, like Σ1's imperfect generator is. Then S1 and S2 are the same mode of the same mos pattern (up to knowing which step size is the bigger one). Assume the L of S1 (it could be s, but it doesn't matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b's or all a~c's, thus these steps are all b's in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S is ternary.

Only two sizes of k-steps of S can project to P in S1, for if there are three sizes of k-steps (α, β, γ), (α, β′, γ′), (α, β′′, γ′′) in S that project to P, then β, β′ and β′′ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.

n is odd, etc.

Suppose Q = (α, β, γ) ≠ R = (α, β′, γ′) are the two k-steps in S that project to P. Then T = (α′, β′′, γ′′) projects to I. Here the values in each component differ by at most 1, and α ≠ α′. Then the cyclic word Λ1 formed by the a-components of the k-steps in P is α...αα′. Since Σ2 is a single-period mos pattern of βb + (n − β)(a~c) and β′a + (n − β′)(a~c), the cyclic word Λ2 = the pattern of β and β′ must be a single-period mos. Similarly, Λ3 = the pattern of γ and γ′ is a single-period mos.

Suppose Λ2 is the mos λβ μβ′. Then Λ3 is the mos (λ ± 1)γ (μ ∓ 1)γ′. Since neither Λ2 nor Λ3 are multimosses, and at least one of μ and (μ ∓ 1) are even, it is now immediate that n is odd.

Either β′′ = β or β′′ = β′. Assume β′′ = β′. Then γ′′ = γ, and Λ3 is (λ + 1)γ (μ − 1)γ′. Also assume that the first k-step in Σ is Q. Then we have: 

     1 …        n
Σ  = Q W(Q, R)  T
Λ1 = α …      α α′
Λ2 = β W(β, β′) β′
Λ3 = γ W(γ, γ′) γ

where W = W(x, y) is a word in two variables x and y, of length n − 2.

Case analysis

Since, by our assumption, Λ3 has two γ in a row, Λ3 must have more γ than γ′, so μ − 1 < n/2. Since Λ3 is a mos, μ − 1 ≥ 1. So we have 2 ≤ μ ≤ ceil(n/2).

We have three cases to consider:

Case 1: μ = 2, i.e. Λ2 is the mos (n − 2)β 2β′.

For Λ2 to be a mos, the first, and only, occurrence of R must be at either f = floor(n/2) or ceil(n/2). We may assume that it is at f; otherwise flip the chain and reindex the words to start at 2f. 

     1 …   f    … 2f n
Σ  = Q … Q R  Q … Q  T
Λ1 = α … α α  α … α  α′
Λ2 = β … β β′ β … β  β′
Λ3 = γ … γ γ′ γ … γ  γ

We need only consider stacks up to f-many k-steps. Either:

  1. the stack has only copies of Q and R; or
  2. the stack has one T and does not contain any R (since it's more than f − 1 generators away).

These give exactly three distinct sizes for every interval class. Hence S is SV3.

In this case S has two chains of Q, one with floor(n/2) notes and one offset by Q(f−1)R with ceil(n/2) notes. Every instance of Q must be a k-step, since by Z-linear independence Q = αa + βb + γc is the only way to write Q in the basis (a, b, c); so S is well-formed with respect to Q. Thus S also satisfies the generator-offset property with generator Q.

Case 2: μ ≥ ceil(n/2), i.e. Λ2 has fewer β than β′.

Since Λ3 has more β than β′, Λ2 is floor(n/2)β ceil(n/2)β′, and Λ3 is ceil(n/2)γ floor(n/2)γ′. There is a unique mode of ceil(n/2)γ floor(n/2)γ′ that both begins and ends with γ, namely γγ′γγ′…γγ′γ. Thus Λ2 is ββ′ββ′…ββ′β′. It is now easy to see that if the number of k-steps stacked is odd, then there are two sizes that do not contain T and one size that contains T; if the number of k-steps stacked is even, then there is one size that does not contain T and two sizes that contain T. Hence S is SV3.

In this case we have Σ = QRQR…QRT, and S is well-formed with respect to the generator Q + R, thus S satisfies the generator-offset property. By Proposition 1, S is SV3.

Case 3: 3 ≤ μ ≤ floor(n/2).

Λ2 has a chunk of β (after the first β′) of size x where x = floor(n/μ) ≥ floor(n/floor(n/2)) = 2 or x = ceil(n/μ) = floor(n/μ) + 1. Hence Λ3 has a chunk of γ of size x. Λ3 also has a chunk that contains Λ3[n : 2] as a subword. This chunk must be of size y, where

[math]2 \lfloor\frac{n}{\mu}\rfloor - 1 = 2 \big(\lfloor \frac{n}{\mu} \rfloor - 1\big) + 1 \leq y \leq 2 \big(\lfloor\frac{n}{\mu}\rfloor + 1 \big) + 1 = 2\lfloor\frac{n}{\mu}\rfloor + 3.[/math]

(The lower bound is reached if Λ3 has chunks of sizes floor(n/μ) − 1 and floor(n/μ), and the upper bound is reached if Λ3 has chunks of sizes floor(n/μ) and ceil(n/μ).)

The difference between the chunk sizes of Λ3 is y − x, which must be 1 since Λ3 is pairwise well-formed. We thus have the following subcases: (In the following, chunk of Λ2 means chunk of β, and chunk of Λ3 means chunk of γ.)

Case 3.1: (x, y) = (floor(n/μ), 2*floor(n/μ) − 1).

Since y − x = floor(n/μ) − 1, we have x = floor(n/μ) = 2 and y = 3. The chunk in Λ3 whose size was defined to be y is made from two consecutive chunks in Λ2 of size 1. (So Λ2 has chunks of size 1 and 2, and Λ3 has chunks of size 2 and 3.) Since chunk sizes of a mos themselves form a mos, Λ2 has more chunks of size 1 than it has chunks of size 2.

Λ2 has only two chunks of size 1, Λ2[n − 1] and Λ2[1], since otherwise Λ3 would have a chunk of size 1 within Λ3[1 : n]. Thus Λ2 has exactly one chunk of size 2. Thus Λ2 = ββ′βββ′ββ′ and Λ3 = γγ′γγγ′γγ. Thus we have: 

     1 2  3 4 5  6 7
Σ =  Q R  Q Q R  Q T
Λ1 = α α  α α α  α α′
Λ2 = β β′ β β β′ β β′
Λ3 = γ γ′ γ γ γ′ γ γ

Suppose a step of S is reached by stacking t-many k-steps. We have three cases after accounting for equave complements:

  1. t = 1: S is equivalent to abacaba.
  2. t = 2: S is QR QQ RQ TQ RQ QR QT => S is equivalent to abacaba.
  3. t = 3: S is QRQ QRQ TQR QQR QTQ RQQ RQT => S is equivalent to abacaba.

(This also implies S is SV3.)

Case 3.2: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) − 1) is impossible: Here (x, y) = (4, 5). But then Λ2 has a chunk of size < 3 because of the β' at index n, contradicting that x is one of the chunk sizes of Λ2.

Case 3.3: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ)) is impossible: Here (x, y) = (3, 4). But then Λ2 has a chunk of size 1 because of the β' at index n, and another chunk of size 0 or 2, contradicting that x is one of the chunk sizes of Λ2.

The remaining cases are all impossible because they imply y − x ≥ 2:

  • Case 3.4: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 1)
  • Case 3.5: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 2)
  • Case 3.6: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 3)
  • Case 3.7: (x, y) = (floor(n/μ), 2*floor(n/μ))
  • Case 3.8: (x, y) = (floor(n/μ), 2*floor(n/μ) + 1)
  • Case 3.9: (x, y) = (floor(n/μ), 2*floor(n/μ) + 2)
  • Case 3.10: (x, y) = (floor(n/μ), 2*floor(n/μ) + 3)

Theorem 5 (Classification of MV3 scales)

  1. A single-period MV3 is either (1) pairwise well-formed, (2) equivalent to XYZYX, or (3) a "twisted" word constructed as follows:
    1. Start with a power of a multimos word w(X, Z) = kaX kbZ such that a is even and each aX bZ subword of w is of the form XP(X, Z)Z where P(X, Z) is a palindrome.
    2. Interchange some of the Z's and X's at some (possibly none) of the borders of these copies of the mos word w.
    3. Replace every other X with Y in w.
  2. Single-period MV3 scales not of type (3) are always SV3.

Proof

Proven by Bulgakova, Buzhinsky and Goncharov (2023), "On balanced and abelian properties of circular words over a ternary alphabet" (and Theorem 4).

Open conjectures

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

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