Ternary scale theorems

This is an expert page. It is written to allow experienced readers to learn more about the advanced elements of the topic.

A ternary scale is a scale with three (positive) step sizes, with no other constraints such as maximum variety. This page documents known properties of subtypes of ternary scales and their proofs.

Conventions

• Bolded Latin variables refer to step vectors (linear combinations of step sizes).
• Indices for all words are 0-indexed.
  • If s is a circular word and i < 0 or i ≥ len(s), we first replace i with i % len(s) before using it as an argument in s[-].
• The notation s(X₁, ..., X_r) is used for an r-ary scale word with variables X₁, ..., X_r possibly standing in for any sizes. If s(X, Y) = XXY then s(A, B) = AAB.
• We leave the distinction between linear words (words in the ordinary sense) and circular words up to context. We usually also elide the distinction between subwords and the interval sizes that subtend them.
• For a word w and letter x, |w|_x denotes the number of occurrences of the letter x in w. For a step vector size v, |v|_x is similar.

Definitions

• A circular word s (representing the steps of a periodic scale) of size n is generator-offset if it satisfies the following properties. The following conditions do not 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).
  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.
• 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 if its period is the same as its equave. A multiMOS or multiperiod MOS is a non-primitive MOS. A MOS aL bs is primitive iff gcd(a, b) = 1. This corresponds to the term single-period in common xen parlance. Any multiMOS can be constructed from a primitive MOS by repeating the MOS pattern multiple times, e.g. if 3L 2s is LLsLs, then 9L 6s is LLsLsLLsLsLLsLs.
• An n-ary scale is a scale with n different step sizes. Binary and ternary are used when n = 2 and 3, respectively.
• A well-formed generator sequence (WFGS) is a generator sequence GS(x₁, ..., x_r) with the following properties:
  • There exists a positive integer k such that for every generator x_i in the GS recipe GS(x₁, ..., x_r), every occurrence of x_i in the scale subtends k steps. This implies that the gap between the next higher equave and the result of stacking len(scale) − 1 of the generators in the recipe, called the closing generator, or the imperfect generator since it is analogous to the imperfect generator in MOS scales, also subtends this number of steps.
  • The closing generator must be distinct from all of the generators used in the generator sequence and occur only once in the scale.
• The property of having a WFGS of period 2, denoted AGS (alternating generator sequence) in this article, is important as it is equivalent to being an odd-regular MV3 scale; see below.
• An odd-step is a k-step where k is odd; an even-step is defined similarly.
• Given a linear or circular word s with a step size X, define E_X(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. The length of a scale s is denoted len(s).
• 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. primitive MOSes).

Theorem 1 (Properties of AGS scales)

Let s be a ternary scale word in L, M, and s of length n, and suppose s is AGS. 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, then the result of identifying the two equinumerous step sizes is a primitive MOS.
5. If n is odd, s = aX bY bZ is obtained from some mode of the (primitive) MOS aX 2bW by replacing all the Ws successively with alternating Ys and Zs (or alternating Zs and Ys for the other chirality, fixing the mode of aX 2bW). The two alternants differ by replacing one Y with a Z. In other words, s is odd-regular in our classification of MV3 scales.

Proof

Let e be the equave of s.

Assuming AGS, we have two chains of the aggregate generator g (going right). In the diagrams below, O represents a note and - represents a generator g. 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 ≤ N where N is the 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 − 1)*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 g₁ and g₂ subtend the same number of steps by the AGS assumption, each is an odd-step. All multiples of the aggregate generator g must be even-steps, and those intervals that are "offset" by g₁ must be odd-steps. Letting M be the subset consisting 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 of k-steps reached by stacking r generators:

1. from g₁ g₂ ... g₁, we get a₁ = ⁠r − 1/2⁠ * g + g₁ = ⌈r⁄2⌉ g₁ + ⌊r⁄2⌋ g₂
2. from g₂ g₁ ... g₂, we get a₂ = ⁠r − 1/2⁠ * g + g₂ = ⌊r⁄2⌋ g₁ + ⌈r⁄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 − ⁠3/2⁠g₁ + ⁠r − n/2⁠ − ⁠1/2⁠g₂ (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 − ⁠1/2⁠g₁ + ⁠r − n/2⁠ − ⁠3/2⁠g₂ (mod e).

Since n > 0, these are all distinct by ℤ-linear independence; hence there are at least 4 sizes for k-steps. A 1-step must be reached by stacking an odd number of generators, thus by applying this argument to 1-steps, we see that there must be at least 4 step sizes in some tuning, a contradiction. Thus g₁ and g₂ must themselves be step sizes. Thus we see that an even-length AGS ternary 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)

Let n ≥ 3 and let g₁, g₂ be the two alternants. Let g₃ be the closing generator after stacking alternating g₁ and g₂. Then the generator circle is (g₁ g₂)^⌊n/2⌋ g₃. If a step is formed by stacking k generators, we may assume that k is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:

1. ⌈k⁄2⌉g₁ + ⌊k⁄2⌋g₂
2. ⌊k⁄2⌋g₁ + ⌈k⁄2⌉g₂
3. ⌊k⁄2⌋g₁ + ⌊k⁄2⌋ g₂ + g₃

(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-k subwords of the (linear) word (g₁ g₂)^⌊n⁄2⌋, we see that the first two sizes must both occur ⁠n − k/2⁠ times. This proves (2).

Statement (3)

We only need to see that if len(s) is odd and s is AGS, s is abstractly SV3. But the argument in case 2 above works when you substitute any odd-step interval classes in s instead of a 1-step (abstract SV3 wasn't used). To get even-step interval classes, we can take octave complements. Hence any interval class in such a scale comes in (abstractly) exactly 3 sizes.

Statement (4)

The n − 1 stacked AGS terms are identified when the equinumerous step sizes are equated. Thus we have a binary scale with a generator (occurring at n − 1 positions), hence being a primitive MOS.

Statement (5)

By part (2), we have that s has step signature aX bY bZ, a odd. By part (4), we have that T(X, W) = s(X, W, W) is a MOS scale aX2bW. If b = 1, there's nothing to prove, so assume b > 1.

Consider the two generators in the GS of s, which are lifts of the generator iX + jW of T(X, W), where gcd(j, 2b) = 1. Assume, possibly after inverting the generator, that the imperfect generator of T has j − 1 Ws and the perfect generator has j Ws.

Claim 1: Deleting Xs from the generator subwords of s gives every j-step subword in the scale E_X(s)(Y, Z), the scale word obtained by deleting all X's from s. These j-step subwords are adjacent and alternating under the ordering induced by the AGS stack.

Proof:

• A.1. Say that the generator of T has k steps.
• A.2.i. The imperfect generator of T occurs only at one position. Call the unique imperfect position p.
• A.2.ii. Say that the number of X steps in a perfect generator is i, and the number of W steps in a perfect generator is j, we have that k = i + j.
• A.2.iii. We know from MOS theory that letter counts in k-steps (for any fixed k) differ by at most 1. Assume, possibly after taking the equave complement, that the imperfect generator has one more X: the imperfect generator has (i + 1)-many X's, and (j − 1)-many W's.
• A.3.i. Recall that p is the unique position such that the k-letter slice I = T[p : p + k] abelianizes to the imperfect generator.
• A.3.ii. Scooting the slice I to the right yields I_R := T[p + 1 : p + 1 + k]. Since its abelianization is a perfect generator, I_R has i-many X's and j-many W's.
• A.3.iii. Since I_R gains a W and loses an X relative to I, the lost letter X is at the leftmost position of I's window, which is p.
• A.3.iv. Conclusion: T[p], the leftmost letter of I = T[p : p + k], is X.
• B.1. Now we go back to the original necklace s. Lift each perfect generator window (we have n − 1 perfect windows) of T to s.
• B.2. By the hypothesis that s has an AGS, and since the AGS descends to stacking a single generator in the template MOS T, the lifted generators g₁ and g₂ alternate in their counts of Y and also alternate in their counts of Z.
• B.3. For a MOS binary word, the count of a given letter in a generator is coprime to the total count of that letter in one period of the MOS. By this fact applied to T, gcd(j, 2b) = 1.
• B.4. Hence, since every instance of the generator in T has j-many W letters, every instance of g₁ and every instance of g₂ has j-many non-X letters.
• C.1. Importantly, deleting X's gives windows of length j, such that when you project adjacent lifted generators (by deleting X's) to the binary necklace U := E_X(s)(Y, Z), the resulting j-step windows in U are adjacent and do not overlap.
• C.2. Moreover, for every j-step window U[q : q + j], there exists an (i + j)-step window s[r : r + i + j] such that s[r] is the non-X that corresponds to U[q] under step deletion. Since by subclaim A, the unique imperfect (i + j)-step window in s begins in an X, we know that s[r : r + i + j] is perfect.
• C.3. We need only stack 2b ≤ n − 1 generators (to get 2b-many j-step windows downstairs) to witness the alternation. Under the ordering induced by this stacking, the 1st j-step subword of U and the last (2b-th) j-step window differ due to parity. Since gcd(j, 2b) = 1, this visits every note of U.

Claim 2: If a binary necklace U has b Ys and b Zs, 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 − 1/2⁠}u (mod e) and (vu)^{⁠m − 1/2⁠}v (mod e), and the step sizes alternate because u and v do.

These two claims prove that E_X(S) = (YZ)^b and that the two GS generators' sizes differ by replacing one Y for a Z. [math]\displaystyle{ \square }[/math]

Theorem 2 (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 primitive MOSes). Then s is SV3 and has an odd number of notes. Moreover, s is either generator-offset 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 primitive MOSes 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 MOSes, 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 primitive 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 primitive MOSes.

 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 s steps 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 circular word Λ₁ formed by the a-components of the k-steps in P is α...αα′. Since Σ₂ is a primitive MOS pattern of βb + (n − β)(a~c) and β′a + (n − β′)(a~c), the circular word Λ₂ = the pattern of β and β′ must be a primitive MOS. Similarly, Λ₃ = the pattern of γ and γ′ is a primitive MOS.

Suppose Λ₂ is the MOS λβ μβ′. Then Λ₃ is the MOS (λ ± 1)γ (μ ∓ 1)γ′. Since both Λ₂ and Λ₃ are primitive, 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 = (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 ≤ ⌈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 = ⌊n/2⌋ or ⌈n/2⌉. We may assume that it is at f; otherwise reverse 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 a window stacking argument shows that the second type of fk-step ((f − 1)Q + T) alternates with the first type ((f − 1)Q + R), and fQ occurs only once, so s has generator sequence GS((f − 1)Q + T, (f − 1)Q + R). Since n is odd, s is odd-regular.

Case 2: μ ≥ ⌈n/2⌉, i.e. Λ₂ has fewer β than β′.

Since Λ₃ has more β than β′, Λ₂ is ⌊n/2⌋β ⌈n/2⌉β′, and Λ₃ is ⌈n/2⌉γ ⌊n/2⌋γ′. There is a unique mode of ⌈n/2⌉γ ⌊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 has generator sequence GS(Q, R). We thus have that s is odd-regular.

Case 3: 3 ≤ μ ≤ ⌊n/2⌋.

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

[math]\displaystyle{ 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 ⌊n/μ⌋ − 1 and ⌊n/μ⌋, and the upper bound is reached if Λ₃ has chunks of sizes ⌊n/μ⌋ and ⌈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) = (⌊n/μ⌋, 2*⌊n/μ⌋ − 1).

Since y − x = ⌊n/μ⌋ − 1, we have x = ⌊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 − 2] and Λ₂[0], since otherwise Λ₃ would have a chunk of size 1 within Λ₃[0 : n − 1]. 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) = (⌊n/μ⌋ + 1, 2*⌊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) = (⌊n/μ⌋ + 1, 2*⌊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) = (⌊n/μ⌋ + 1, 2*⌊n/μ⌋ + 1)
• Case 3.5: (x, y) = (⌊n/μ⌋ + 1, 2*⌊n/μ⌋ + 2)
• Case 3.6: (x, y) = (⌊n/μ⌋ + 1, 2*⌊n/μ⌋ + 3)
• Case 3.7: (x, y) = (⌊n/μ⌋, 2*⌊n/μ⌋)
• Case 3.8: (x, y) = (⌊n/μ⌋, 2*⌊n/μ⌋ + 1)
• Case 3.9: (x, y) = (⌊n/μ⌋, 2*⌊n/μ⌋ + 2)
• Case 3.10: (x, y) = (⌊n/μ⌋, 2*⌊n/μ⌋ + 3)

[math]\displaystyle{ \square }[/math]

Theorem 3 (PMOS scales are balanced)

All pairwise-MOS scales are balanced.

Proof

For any individual letter X, identify letters other than it to get a MOS. Since MOS words are balanced, the block balance for any letter X is at most 1, as required by the balance property. [math]\displaystyle{ \square }[/math]

Theorem 4 (Generator-offset structure of even-regular scales)

Definition (Even-regular scale)

A primitive ternary scale s is even-regular if len(s) is even and s is equivalent to a word constructed from taking the MOS 2aX 2cZ with a odd and gcd(a, c) = 1, and replacing every other X with Y. In particular, s has step signature equivalent to aX aY bZ with a odd and b even. For example, LsLsLmsLsLsm (achiral diachrome, 5L 2m 5s) is an even-regular scale.

Theorem

If s = s(X, Y, Z) is even-regular, then:

1. s consists of two generator chains, each with len(s)/2 notes;
2. the generator has the same interval class as some generator of the MOS 2aW 2cZ;
3. the two generator chains are offset by a len(s)/2-step interval;
4. s is balanced.

Proof

The result of substituting Y with X (let us call this map p) is the MOS M = 2aX 2cZ, which has exactly 2 periods since gcd(a, c) = 1. M thus consists of two generator chains separated by the period of M, which has a + c = len(s) steps. It thus suffices for there to exist k, 0 < k < a + c, such that every perfect k-step generator has the same preimage in s, which will be our desired generator. Suppose that the perfect k-step of M is iW + jZ where 0 < i < a. Since a is odd, possibly after taking the period-complement we may assume that i is even. Hence each subword w of s such that its projection p(w) subtends a perfect k-step satisfies |w|_X = |w|_Y = i/2. It plainly follows that every such w satisfies |w|_X = |w|_Y = ⁠i/2⁠ and |w|_Z = j.

It remains to show that s is balanced. Any k-step subword has either j or j + 1 Zs for some j since the result of conflating X and Y is a MOS, and k-step subwords for both possibilities exist when 0 < k < len(s)/2. If the number of non-Z letters in a k-step subword is even, then there is only one possibility for the number of X and the number of Y. If the number of non-Z letters in a k-step subword is odd, then both the number of Xs and the number of Ys differ by at most 1. [math]\displaystyle{ \square }[/math]

Theorem 5 (Classification of MV3 scales)

In the following, equivalent means "is the same circular word after permuting X, Y, and Z." This means that XYXZXYX is equivalent to YZYXYZY, or XZXYXZX, and so on.

Theorem 5.1 (Classification of ternary balanced scales)

1. A primitive balanced ternary scale s is pairwise-MOS; conversely, pairwise-MOS scales are balanced. Such a scale satisfies one of the following:
   1. sporadic balanced: s is equivalent to XYXZXYX, the ternary Fraenkel word, with step signature 4X2Y1Z.
   2. odd-regular: len(s) is odd, and s is equivalent to a word constructed from taking the brightest mode of the MOS cXbZ with c even and gcd(c, b) = 1, and replacing every other X with Y. We assume X > Z when constructing the MOS. In particular, s has step signature aXaYbZ where b is odd (with a = c/2).
   3. even-regular: len(s) is even, and s is equivalent to a word constructed from taking the brightest mode of the MOS 2aX2cZ with a odd and gcd(a, c) = 1, and replacing every other X with Y. In particular, s has step signature aXaYbZ with a odd and b even.
2. All primitive balanced ternary scales are MV3.
3. A balanced primitive ternary scale is SV3 if and only if it is not even-regular.
4. Odd-regular balanced primitive ternary scales have a generator sequence of period 2.

(Condensed: All single-period balanced ternary scales that are not the Fraenkel word are aX aY bZ. In this case, if b is odd, then the scale is odd-regular. If b is even, then the scale is even-regular.)

Proof

For 5.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced.

We will first prove that a balanced circular word is primitive iff the gcd of the step signature is 1. Proof sketch: let d be the gcd of the step signature. (n/d)-step multisets come in 1 size, namely the equave divided by d, because if some letter count differs, then we get 3 values for this letter count for (n/d)-step multisets by the discrete IVT.

It remains to show that (a) ternary balanced words are pairwise-MOS (b) if a > b > c, then s is equivalent to the Fraenkel word (c) assuming a != b = c any s that is not odd-regular or even-regular is not balanced.

(a) Let s be a ternary balanced word; then for any given letter y the number of ys in a subword of any given length L varies by at most 1. Thus the same is true when we count all non-y letters in any subword of length L; thus when we equate x and z, the count of the resulting letter in any subword of length L differs by 1. Being a binary balanced word is one characterization of the MOS property.

(b) The following proof is taken from "Balanced Sequences and Optimal Routing", by Altman, Gaujal, and Hordijk (2000).

Let W be the (balanced) right-infinite word made by concatenating infinitely many copies of s. We use the following steps, using the balance property:

(i) The sequence XZX must appear in W.

There are two consecutive Xs with no Y in between since a > b. This means either XX or XZX appears. If XX appears, then a Z is necessarily surrounded by two Xs.

(ii) The sequence YXXY and XYXXYX must appear in W.

There exists a pair of consecutive Ys with no Z in between. Thus we have a subword of the form YXⁿY. Now, n ≤ 1 is not possible because of the presence of XZX and Y-balance. n ≥ 3 implies the existence of X^n-1ZX^n-1 by X-balance which is incompatible with YXⁿY because of Y-balance. Therefore, n = 2. Note that this also implies the presence of subwords XX and XYXXYX.

(iii) The sequence XYXZXYX appears in W.

The sequence W must contain a Z. This Z is necessarily surrounded by two Xs since XX exists by Step (ii). This group is necessarily surrounded by two Ys since YXXY exists, and consequently, necessarily surrounded by two Xs because XYXXYX exists. We get the sequence XYXZXYX.

(iv) W = (XYXZXYX)^ω.

No letter around this word can be a Z because YXXY exists. None can be a Y since XZX exists. Therefore, they have to be two Xs. Then note that the two surrounding letters cannot be Z (because of the existence of XYXXYX) nor X (because of the existence of YXZ) so they are Y, then followed by X (because XX exists). At this point, we have the sequence “_XYXXYXZXYXXYX_”. Both _s are necessarily Zs. To end the proof, note that we have obtained the configuration around every Z and this determines the whole sequence. Thus W = (XYXZXYX)^ω.

(c) The scale made by taking s and conflating Y and Z into the letter W must be a MOS. To this scale we may imagine substituting a scale made of an equal amount of Y and Z letters into the "slot letters" W letter by letter. Let t₁ be a length-k subword of the form YX^k-2Y under the projection. We may assume that the chunk sizes of the MOS are k - 2 and k - 1, or k - 2 and k - 3. Either way, there exists some subword with (k - i)-many Xs, i = 1 or 2, and two Zs. This violates balance because t₁ contains zero Zs.

For 5.1.2: Suppose s is balanced and has at least three sizes for k-steps, a_iX + b_iY + c_iZ = (a_i, b_i, c_i) for i ∈ {1, 2, 3}. We may assume (a₂, b₂, c₂) = (a₁, b₁ + 1, c₁ − 1). Then either (a₃, b₃, c₃) = (a₁ + 1, b₁, c₁ − 1) or (a₃, b₃, c₃) = (a₁ − 1, b₁ + 1, c₁). In both cases, by balancedness applied to subwords of length k, the three vectors represent the only possible interval sizes.

For 5.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are not SV3, observe that (a + c)-steps come in only 2 sizes in such a scale s: ⌊a/2⌋X + ⌈a/2⌉Y + cZ and ⌈a/2⌉X + ⌊a/2⌋Y + cZ, since the underlying MOS 2aX2cY only has the (a + c)-step aX + cZ. The construction replaces the Xs in these subwords with alternating Xs and Ys; either of X or Y may occur first, corresponding to the two possible sizes, since a is odd and thus the (a + c)-step subword s[k − 1 : k + a + c − 1] becomes the subword s[k + a + c − 1 : k + 2a + 2c − 1] via interchanging X and Y.

Claim 5.1.4 can be verified by noting that such scales are PWF and using Theorem 4. [math]\displaystyle{ \square }[/math]

Theorem 5.2 (Classification of MV3 scales)

A primitive MV3 scale is either

1. balanced (classified by the previous theorem),
2. sporadic non-balanced: equivalent to XYZYX, or
3. twisted: equivalent to a word constructed as follows:
   • Start with the brightest multiMOS word kcXkbZ with c being an even number.
   • Interchange a Z and an X at some (possibly more than one) of the boundaries of these copies of the MOS word w. Here, the boundary of two consecutive copies of w is the last letter of the first word and the first letter of the second word. (At the ends of the whole multiMOS word, the boundaries are just the first and last letters of the word.) For example, let w be the multiMOS word 8X6Z, XXZXZXZXXZXZXZ. Then the border between the copies of the MOS subword XXZXZXZ are w[6]w[7] and w[13]w[0] (using 0-based numbering).
   • Replace every other X with Y in w. (Thus in particular, twisted MV3 scales have step signature kaXkaYkbZ)

Proof

Most of this has been proved by Bulgakova, Buzhinsky and Goncharov (2023), "On balanced and abelian properties of circular words over a ternary alphabet"; however, the designations sporadic, odd-regular, and even-regular for the classes are ours.

Note: The xen term "brightest MOS word" is equivalent to "Christoffel word" in the paper, and similarly "brightest multiMOS word" is equivalent to "powers of a Christoffel word". Also see Glossary for combinatorics on words for more equivalents between xen community terms and standard academic terminology.

Theorem 6 (Even-regular scales as (contra)interleavings)

Let s be a primitive even-regular scale of MOS substitution type ax(ky kz) where a is even and gcd(a, k) = 1. Let n = |s| = a + 2k.

1. If n is singly even, then s is a contrainterleaving of the two opposite chiralities of an odd-regular scale.
2. If n is doubly even and > 4, then s is an interleaving of two copies of a smaller even-regular scale.
3. If n = 4, then s = xyxz is an interleaving of a 2-note MOS.

Proof

Statement 3 is trivial and is included for completeness. We assume n > 4. The a = 2k case means that k = gcd(a, k) = 1, and a = 2. This is the trivial n = 4 case. Thus a ≠ 2k.

The 2-step intervals of s must be:

1. if a > 2k: y + z, otherwise: 2x
2. x + y
3. x + z

We also know that s is of the form w(x, y, z)w(x, z, y). Hence the number of occurrences of x + y = the number of occurrences of x + z, counting all 2-steps in all of s.

Write s₁ for the scale word made from stacked 2-steps from the 0-degree, and let s₂ be as follows:

• In the singly even case, let s₂ be the circular word of 2-steps starting at the (n/2)-degree. We know that they differ only by interchanging y and z, hence that they have the same period. Hence both s₁ and s₂ are primitive.
• In the doubly even case, start from the mode of s whose template MOS is the brightest mode. Let s₂ be offset at a generator of the even-regular scale, which we choose to have the same interval class as a bright generator of the MOS ax 2kX. This is what induces the equality of s₁ and s₂ (in particular, the two scales have the same period, thus they are both primitive): Let s_t be the period of the brightest mode of the template MOS, and let g be its bright generator class. Then the slice s_t[-g +1 : 1] is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of s₁ and s₂, we turn that slice into the bright generator, hence swapping s_t[-g] and s_t[-g + 1]. Note that g must be odd since it generates a 2-period MOS. So (under 0-indexing) the first letter's index is even and the second letter's index is odd, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (n/2)-step, that doesn't affect the parity since n/2 is even.

We prove that s₁ and s₂ are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both s₁ and s₂. The word of stacked 2-steps of the template MOS (which is of the form w(x, X, X)w(x, X, X)), which is itself a MOS word, consists of letters (1) x + X and (2) 2X if more X's than x's, 2x if more x's than X's. The word of stacked 2-steps from our chosen offset is also this same MOS word. Thus it remains to handle the cases (1) and (2) above. Whenever the letter x + X is encountered, the number of the last letters that are equated to X that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence (since n > 4) the letter 2X resp. 2x serves as the non-slot letter, and the letters (x + X) serve as the slot letters where a 2-period filling MOS word (a repetition of (x+y)(x+z)) is substituted.

Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for s₁ and one in the template MOS s₂, due to index parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps of the step size that occurs less frequently in the template MOS ax 2kX, which is true by assumption (a and 2k are both even).

• In the singly even case, since there are evenly many slot letters in both s₁ and s₂, there are oddly many non-slot letters in both. Since s₁ and s₂ differ by interchanging y and z, they have "opposite" filling letters, x + y being the opposite of x + z. This makes s₁ and s₂ opposite chiralities of an odd-regular MV3 scale.
• In the doubly even case, the number of non-slot letters in s₁ and s₂ is even, and we have a filling MOS of period 2. Since s₁ and s₂ are both primitive, they are both even-regular scales. [math]\displaystyle{ \square }[/math]

Open problems

1. Classify all twisted SV3 scales, thereby completing the classification of all abstractly SV3 scales.
2. Conjecture: If a twisted MV3 is not SV3, then it is constructed from kaXkbZ where k is composite.

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 scale with at least three notes per period, is there *always* a MV3 generator-offset 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 generator-offset 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)?
   • Yes. For an axby MOS with gcd(a, b) = 1, if one of a and b is even, detemper x resp. y into two step sizes. The result is a 1-period odd-regular MV3. If neither is even, assume a > b. Then use (a - b)xbybz, which is a 1-period even-regular MV3 since gcd(a - b, b) = gcd(a, b) = 1.
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 generator-offset-MV3 scales, where given some word representing a generator-offset-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 generator-offset-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 generator-offset-MV3 scales?

External links

• Scale word theorems formalized in Lean 4 (WIP)