Ternary scale theorems

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 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, ..., 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.

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.
• For the generator-offset property, see the article.
• A strengthening of the generator-offset property, here 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 primitive mos scale (otherwise known as a well-formed scale). This is simply the property of having a generator sequence of period 2. All odd generator-offset scales are SGA, and aside from odd generator-offset scales, the only ternary scales to satisfy SGA are (xy)^rxz, r ≥ 1. The Zarlino and diasem scales above are both SGA. Blackdye is generator-offset but not SGA.
• 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 mosses).

Theorem 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 (primitive) 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.

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

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

Proof

Let e be the equave of S.

Assuming SGA, 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 ≤ (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 S is generator-offset it is well-formed with respect to the aggregate generator g = (g₂ + g₁). Since g₁ and g₂ subtend the same number of steps, 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 = ceil(r/2) g₁ + floor(r/2) g₂
2. from g₂ g₁ ... g₂, we get a₂ = (r − 1)/2 * g + g₂ = floor(r/2) g₁ + ceil(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 − 3/2)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.

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, abstractly SV3, generator-offset 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 n ≥ 3 and 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 is (g₁ g₂)^floor(n/2) g₃. If a step is formed by stacking k generators, k must be odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:

1. ceil(k/2)g₁ + floor(k/2)g₂
2. floor(k/2)g₁ + ceil(k/2)g₂
3. floor(k/2)g₁ + floor(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)^floor(n/2), we see that the first two sizes must both occur (n − k) times. This proves (2).

Statement (3)

We 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 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)

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

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. 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 E_X(S)(Y, Z), the scale word obtained by deleting all X's from S.

Proof: Consider the "imperfect alternant" subword of S on index p, which is I = S[p : p + i + j], and suppose the result of deleting all X's from I has a j-step subword w 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 non-X step than I by our assumption. Thus the word I must both begin and end in a non-X letter. Removing all the X's from I results in a word that is j + 1 letters long and is the j-step word w 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 of S 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 necklace 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 alternate because u and v do.

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

Statement (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). [math]\displaystyle{ \square }[/math]

Theorem 2 (Odd generator-offset 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 g 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.

Assume 1 < gcd(k, n) > n and n ≥ 5. Since n is odd, d = gcd(k, n) is an odd number at least 3, and by well-formedness with respect to the generator, there must be a circle of n/d < floor(n/2) notes formed by g, contrary to the assumption of GO. Thus, gcd(k, n) = 1.

By modular arithmetic, if 0 ≤ r < n, 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. (As [k] is a generator of ℤ/n, stacking (n − 1)-many k-steps must visit every note exactly once. Thus if the offset wasn't reached in (n + 1)/2 steps, the two generator chains wouldn't have the assumed lengths.) [math]\displaystyle{ \square }[/math]

Theorem 3 (Properties of even generator-offset ternary scales)

A primitive generator-offset ternary scale s of even size 6 or greater, where the generator g is an even-step, has the following properties:

1. s is a union of two copies of a primitive mos M of size n/2 generated by g; thus it is a flought scale obtained by taking two offset copies of said primitive mos.
2. s is not SV3.
3. s is not chiral.
4. If M = M(y, z) is the primitive MOS necklace above, then s = M(XY, XZ) for some assignment of variable names X, Y, and Z to the three letters of s.

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

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 primitive mosses). 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 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 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 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 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 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 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 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]\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 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)

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

Theorem 5 (PWF scales are balanced)

All pairwise-well-formed scales are balanced.

Proof

Let S be a PWF scale. The case where S is equivalent to XYXZXYX can be manually verified, so by Theorem 4, the only remaining case is when S can be constructed by stacking two alternating sizes, g₁ and g₂, of k-steps. We assume that S has scale signature aX bY bZ where a is odd. This k corresponds to the class of generators of the MOS aX 2bW. This MOS is obtained from S by applying the function π such that π(X) = X and π(Y) = π(Z) = W. Naturally, π applies to abstract interval sizes, or more properly, Parikh vectors. Additionally, we can choose k so that the two sizes of k-steps in π(S) are:

• the perfect generator g = tX + (k − t)W (note (k − t) is odd by a previous proof), and
• the imperfect generator i = (t + 1)X + (k − t − 1)W.

We have π(g₁) = π(g₂) = g. Let h be the size in S such that π(h) = i. Hence only one k-step subword h has this size in S. By Theorem 1, we also may assume |g₁|_Y = |g₂|_Y + 1 and |g₁|_Z = |g₂|_Z − 1.

As S is periodic and gcd(len(S), k) = 1, it suffices to count letters in stacks of k-steps in S. The j-step on any note of S can be computed by reducing a stack of k-step subwords on that note, which alternate in size between g₁ and g₂. Note that:

1. Such a stack has a unique number m = m(j) of k-steps for a given j.
2. Either such a stack has i as one of its k-steps, or it does not.

Depending on point (2), m may be even or odd. If m is even, then in any stack in S that has i, the number of stacked g generators in the projection is odd, hence the number of non-X letters in the corresponding km-step word in S is odd. Each incremental shift of the boundary that does not result in including i results in one Y being swapped for a Z, or vice versa, while the number of X steps remains t. In summary:

• |g₁|_X = |g₂|_X = t,
• |g₁|_Y = |g₂|_Z = ceil((k − t)/2),
• |g₁|_Z = |g₂|_Y = floor((k − t)/2),
• |i|_X = (t + 1),
• |i|_Y = |i|_Z = (k − t − 1)/2 = floor((k − t)/2).

This proves that the set of j-steps is balanced. The argument for when m is odd is similar. [math]\displaystyle{ \square }[/math]

Theorem 6 (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 6.1 (Classification of ternary balanced scales)

1. A primitive balanced MV3 scale S is one of the following:
   1. If len(S) is odd, then S is equivalent to XYXZXYX, or to a word constructed from taking the brightest mode of the MOS aX bZ with a even and gcd(a, b) = 1, and replacing every other X with Y. We assume X > Z when constructing the MOS.
   2. If len(S) is even, then S is equivalent to a word constructed from taking the brightest mode of the MOS 2aX 2bZ with a odd and gcd(a, b) = 1, and replacing every other X with Y.
2. All primitive balanced ternary scales are MV3.
3. Balanced primitive ternary scales not of type (3) are always SV3.

Theorem 6.2 (Classification of MV3 scales)

1. A primitive MV3 scale is either (1) balanced, (2) equivalent to XYZYX, or (3) equivalent to a "twisted" word constructed as follows:
   1. Start with the brightest multimos word kaX kbZ with a being an even number.
   2. 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[7]w[8] and w[14]w[1] (using one-based numbering).
   3. Replace every other X with Y in w.

Proof

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

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.