Ternary scale theorems: Difference between revisions

** 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''('''X'''₁, ..., '''X'''_''r'') is used for an ''r''-ary scale word with variables '''X'''₁, ..., '''X'''_''r'' possibly standing in for any sizes. If ''s''('''X''', '''Y''') = '''XXY''' then ''s''('''A''', '''B''') = '''AAB'''.

* 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.

* 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 '''g'''₁ and '''g'''₂ are the same number of scale steps. For example, 5-limit [[blackdye]] has '''g'''₁ = 9/5 (a 9-step) and '''g'''₂ = 5/3 (a 7-step).

*# ''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 '''g'''₁ and '''g'''₂, resulting in a circle of the form either '''g'''₁ '''g'''₂ ... '''g'''₁ '''g'''₂ '''g'''₁ '''g'''₃ or '''g'''₁ '''g'''₂ ... '''g'''₁ '''g'''₂ '''g'''₃.

* 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 ''a'''''L''' ''b'''''s''' 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 3'''L''' 2'''s''' is '''LLsLs''', then 9'''L''' 6'''s''' 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 strengthening of the generator-offset property, here the ''swung-generator-alternant property'' (SGA), states that the alternants '''g'''₁ and '''g'''₂ 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''')^''r'''''XZ''', ''r'' ≥ 1. The Zarlino and diasem scales above are both SGA. [[Blackdye]] is generator-offset but not SGA.

* 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].

# The length of ''s'' is odd, or ''s'' is equivalent to ('''xy''')^''r'''''xz''' for some integer ''r'' ≥ 1.

# If ''n'' is odd, ''s'' is of the form ''a'''''x''' ''b'''''y''' ''b'''''z''' for some permutation ('''x''', '''y''', '''z''') of ('''L''', '''M''', '''s''').

# 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'''.

# If ''n'' is odd, ''s'' = ''a'''''X''' ''b'''''Y''' ''b'''''Z''' is obtained from some mode of the (primitive) MOS ''a'''''X''' 2''b'''''W''' 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 ''a'''''X''' 2''b'''''W'''). The two alternants differ by replacing one '''Y''' with a '''Z'''.

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.

In case 1, let '''g'''₁ = (2, 1) − (1, 1), '''g'''₂ = (1, 2) − (2, 1), and '''g'''₃ = (1, 1) − (''n''/2, 2) = ((−''n''/2 − 1)*'''g'''₁ − ''n''/2*'''g'''₂) mod '''e'''. We assume that '''g'''₁, '''g'''₂ and '''e''' are ℤ-linearly independent. We have the chain '''g'''₁ '''g'''₂ '''g'''₁ '''g'''₂ ... '''g'''₁ '''g'''₃ which visits every note in ''s''.  

Since ''s'' is generator-offset it is well-formed with respect to the aggregate generator '''g''' = ('''g'''₂ + '''g'''₁). Since '''g'''₁ and '''g'''₂ subtend the same number of steps by the SGA assumption, each is an odd-step. All multiples of the aggregate generator '''g''' must be even-steps, and those dyads that are "offset" by '''g'''₁ 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 ('''g'''₃ + '''g'''₁), the imperfect generator of the MOS generated by '''g''', subtends the same number of steps as '''g'''. Thus '''g'''₂ and '''g'''₃ subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''₂ with '''g'''₃ in the next part.

# from '''g'''₁ '''g'''₂ ... '''g'''₁, we get a₁ = (''r'' − 1)/2 * '''g''' + '''g'''₁ = {{ceil|''r''/2}} '''g'''₁ + {{floor|''r''/2}} '''g'''₂  

# from '''g'''₂ '''g'''₁ ... '''g'''₂, we get a₂ = (''r'' − 1)/2 * '''g''' + '''g'''₂ = {{floor|''r''/2}} '''g'''₁ + {{ceil|''r''/2}} '''g'''₂

# from '''g'''₂ (...even # of gens...) '''g'''₁ '''g'''₃ '''g'''₁ (...even # of gens...) '''g'''₂, we get a₃ = (''r'' − 1)/2 '''g'''₁ + (''r'' − 1)/2 '''g'''₂ + '''g'''₃ ≡ (''r'' − ''n''/2 − 3/2)'''g'''₁ + (''r'' − ''n''/2 − 1/2)'''g'''₂ mod e.

# from '''g'''₁ (...odd # of gens...) '''g'''₁ '''g'''₃ '''g'''₁ (...odd # of gens...) '''g'''₁, we get a₄ = (''r'' + 1)/2 '''g'''₁ + (''r'' − 3)/2 '''g'''₂ + '''g'''₃ ≡ (''r'' − ''n''/2 − 1/2)'''g'''₁ + (''r'' − ''n''/2 − 3/2)'''g'''₂ 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 '''g'''₁ and '''g'''₂ must themselves be step sizes. Thus we see that an even-length SGA ternary scale must be of the form (xy)^''r''xz. (Note that (xy)^''r''xz is not SV3, since it has only two kinds of 2-steps, '''xy''' and '''xz'''.) This proves (1).

In case 2, let ''n'' ≥ 3 and let (2, 1) − (1, 1) = '''g'''₁, (1, 2) − (2, 1) = '''g'''₂ be the two alternants. Let '''g'''₃ be the closing generator after stacking alternating '''g'''₁ and '''g'''₂. Then the generator circle is ('''g'''₁ '''g'''₂)^{{{floor|''n''/2}}} '''g'''₃. 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:

# {{ceil|''k''/2}}'''g'''₁ + {{floor|''k''/2}}'''g'''₂

# {{floor|''k''/2}}'''g'''₁ + {{ceil|''k''/2}}'''g'''₂

# {{floor|''k''/2}}'''g'''₁ + {{floor|''k''/2}} '''g'''₂ + '''g'''₃

(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 ('''g'''₁ '''g'''₂)^{{{floor|''n''/2}}}, we see that the first two sizes must both occur (''n'' − ''k'')/2 times. This proves (2).

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 MOSes. 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).

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

Consider the two generators in the GS of ''s'', which are detemperings of the generator ''i'''''X''' + ''j'''''W''' of ''T''('''X''', '''W'''), where gcd(''j'', 2''k'') = 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.

Proof: Consider the subword for the closing generator 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''. Shifting ''I'' one step to the left and one step to the right, ''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'' 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'' can all be obtained by deleting '''X'''s from detempered perfect generators; 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'', 2''b'') = 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'', 2''b'') = 1, there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd because gcd(''m'', 2''b'') = 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'''. {{Qed}}

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.  

# 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''.

Suppose ''s'' has ''n'' notes (after dealing with small cases, we may assume ''n'' ≥ 7) and ''s'' projects to primitive MOSes ''s''₁ (via identifying '''b''' with '''c'''), ''s''₂ (via identifying '''a''' with '''c'''), and ''s''₃ (via identifying '''a''' with '''b'''). Suppose ''s''₁'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 ''s''₂ and ''s''₃, 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 ''s''₁ and ''s''₂, which come in at most two sizes. Since ''s''₁ 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.

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 β'''b''' + (''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.

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:

  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 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 2''f''.

  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

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''' = α'''a''' + β'''b''' + γ'''c''' 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'''.

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.

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>

'''Case 3.1:''' (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} &minus; 1).

Since ''y'' &minus; ''x'' = {{floor|''n''/μ}} &minus; 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'' &minus; 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:

  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

# ''t'' = 1: ''s'' is equivalent to '''abacaba'''.

# ''t'' = 2: ''s'' is '''QR QQ RQ TQ RQ QR QT''' => ''s'' is equivalent to '''abacaba'''.

# ''t'' = 3: ''s'' is '''QRQ QRQ TQR QQR QTQ RQQ RQT''' => ''s'' is equivalent to '''abacaba'''.

'''Case 3.2''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} &minus; 1) is impossible: Here (''x'', ''y'') = (4, 5). But then Λ₂ has a chunk of size &lt; 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.

* '''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)

Let ''s'' be a PWF (thus primitive) 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, '''g'''₁ and '''g'''₂, of ''k''-steps. We assume that ''s'' has [[step signature]] ''a'''''X''' ''b'''''Y''' ''b'''''Z''' where ''a'' is odd. This ''k'' corresponds to a class of generators of the primitive MOS ''a'''''X''' 2''b'''''W'''. This MOS is obtained from ''s'' by applying the letterwise substitution function π such that π('''X''') = '''X''' and π('''Y''') = π('''Z''') = '''W'''. Naturally, π applies to linear words, circular words, and step vector sizes. Additionally, we can choose ''k'' so that the two sizes of ''k''-steps in π(''s'') are:

* the perfect generator '''g''' = ''t'''''X''' + (''k'' &minus; ''t'')'''W''' (note (''k'' &minus; ''t'') is odd by a previous proof), and

* the imperfect generator '''i''' = (''t'' + 1)'''X''' + (''k'' &minus; ''t'' &minus; 1)'''W'''.

We have π('''g'''₁) = π('''g'''₂) = '''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 |'''g'''₁|_'''Y''' = |'''g'''₂|_'''Y''' + 1 and |'''g'''₁|_'''Z''' = |'''g'''₂|_'''Z''' &minus; 1 (the other case corresponds to the opposite chirality).

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 '''g'''₁ and '''g'''₂. Note that:

# Such a stack has a unique number ''m'' = ''m''(''j'') of ''k''-steps for a given ''j''.

Depending on point (2), ''m'' may be even or odd. If ''m'' is odd, then in any stack in ''s'' that does not have '''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''. On the other hand, '''i''' has an even number of non-'''X''' steps, thus the numbers of '''Y''' and '''Z''' are the same in '''v''' and equal to min(|'''u'''₁|_'''Y''', |'''u'''₂|_'''Y''').

* |'''u'''₁|_'''X''' = |'''u'''₂|_'''X''' = |v|_'''X''' &minus; 1

* |'''u'''₁|_'''Y''' = |'''u'''₂|_'''Y''' + 1

* |'''u'''₁|_'''Z''' = |'''u'''₂|_'''Z''' &minus; 1

* |'''v'''|_'''Y''' = |'''v'''|_'''Z''' = |'''u'''₁|_'''Y''' = min(|'''u'''₁|_'''Y''', |'''u'''₂|_'''Y''').

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 2''a'''''X'''2''c'''''Z''' with ''a'' odd and gcd(''a'', ''c'') = 1, and replacing every other '''X''' with '''Y'''. In particular,  ''s'' has [[step signature]] equivalent to ''a'''''X'''''a'''''Y'''''b'''''Z''' with ''a'' odd and ''b'' even. For example, '''LsLsLmsLsLsm''' (achiral [[diachrome]], 5'''L'''2'''m'''5'''s''') is an even-regular scale.

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

The result of substituting '''Y''' with '''X''' (let us call this map ''p'') is the MOS ''M'' = 2''a'''''X'''2''c'''''Z''', 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 &lt; ''k'' &lt; ''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 ''i'''''W''' + ''j'''''Z''' where 0 &lt; ''i'' &lt; ''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''.

## '''odd-regular''': len(''s'') is odd, and ''s'' is equivalent to a word constructed from taking the brightest mode of the MOS ''c'''''X'''''b'''''Z''' with ''c'' even and gcd(''c'', ''b'') = 1, and replacing every other '''X''' with '''Y'''. We assume '''X''' &gt; '''Z''' when constructing the MOS. In particular, ''s'' has [[step signature]] ''a'''''X'''''a'''''Y'''''b'''''Z''' where ''b'' is odd (with ''a'' = ''c''/2).

## '''even-regular''': len(''s'') is even, and ''s'' is equivalent to a word constructed from taking the brightest mode of the MOS 2''a'''''X'''2''c'''''Z''' with ''a'' odd and gcd(''a'', ''c'') = 1, and replacing every other '''X''' with '''Y'''. In particular,  ''s'' has [[step signature]] ''a'''''X'''''a'''''Y'''''b'''''Z''' with ''a'' odd and ''b'' even.

(b) Assume that the projection ''p''_'''YZ'''(''s'') identifying '''Y''' and '''Z''' of a primitive PMOS scale ''s'' with signature ''ar'''''X''' ''b'''''Y''' ''c'''''Z'''  is an ''r''-period MOS, ''r'' &gt; 1, with step signature ''ar'''''X''' ''dr'''''W'''. We claim that neither ''b'' nor ''c'' is divisible by ''r''. Since ''p''_'''XY'''(''s'') is the MOS (''ar'' + ''b'')'''W''' ''c'''''Z''' and  ''p''_'''XZ'''(''s'') is the MOS (''ar'' + ''c'')'''W''' ''b'''''Y''', if either ''b'' and ''c'' is divisible by ''r'', then the distributions of two of the letters have ''r'' periods. Then the distribution of the third letter also has ''r'' periods, meaning that ''s'' itself has ''r'' periods, a contradiction. It suffices to show that ''r'' = 2. If ''r'' &gt; 2, then...

For 7.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, ''a''_''i'''''X''' + ''b''_''i'''''Y''' + ''c''_''i'''''Z''' = (''a''_''i'', ''b''_''i'', ''c''_''i'') for ''i'' = 1, 2, 3. We may assume (''a''₂, ''b''₂, ''c''₂) = (''a''₁, ''b''₁ + 1, ''c''₁ - 1). Then either (''a''₃, ''b''₃, ''c''₃) = (''a''₁ + 1, ''b''₁, ''c''₁ - 1) or (''a''₃, ''b''₃, ''c''₃) = (''a''₁ - 1, ''b''₁ + 1, ''c''₁). In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible interval sizes.