Ternary scale theorems: Difference between revisions

* Non-italicized Latin variables refer to interval sizes, for example step sizes.

* 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.

* 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).

*# ''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.

* 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 primitive [[mos]] scale (otherwise known as a well-formed scale). 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.

* 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''.

* 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.

== 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:

# 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'' = ''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.

# 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'' is elimination-mos. That is, ''E''_X(''S''), ''E''_Y(''S''), ''E''_Z(''S'') are all mosses.-->

Let e be the equave of ''S''.

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

In case 1, let g₁ = (2, 1) &minus; (1, 1), g₂ = (1, 2) &minus; (2, 1), and g₃ = (1, 1) &minus; (''n''/2, 2) = ((&minus;''n''/2 &minus; 1)*g₁ &minus; ''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 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 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.

# from g₁ g₂ ... g₁, we get a₁ = (''r'' &minus; 1)/2*g + g₁ = ceil(''r''/2) g₁ + floor(''r''/2) g₂  

# from g₂ g₁ ... g₂, we get a₂ = (''r'' &minus; 1)/2*g + g₂ = floor(''r''/2) g₁ + ceil(''r''/2) g₂

# from g₂ (...even # of gens...) g₁ g₃ g₁ (...even # of gens...) g₂, we get a₃ = (''r'' &minus; 1)/2 g₁ + (''r'' &minus; 1)/2 g₂ + g₃ ≡ (''r'' &minus; ''n''/2 &minus; 3/2)g₁ + (''r'' &minus; ''n''/2 &minus; 1/2)g₂ mod e.

# from g₁ (...odd # of gens...) g₁ g₃ g₁ (...odd # of gens...) g₁, we get a₄ = (''r'' + 1)/2 g₁ + (''r'' &minus; 3)/2 g₂ + g₃ ≡ (''r'' &minus; ''n''/2 &minus; 1/2)g₁ + (''r'' &minus; ''n''/2 &minus; 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)^''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) &minus; (1, 1) = g₁, (1, 2) &minus; (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:

# ceil(''k''/2)g₁ + floor(''k''/2)g₂

# floor(''k''/2)g₁ + ceil(''k''/2)g₂

# 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'' &minus; ''k'') times. This proves (2).

Assume ''S'' is ''a''X ''b''Y ''b''Z, ''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 ''i''X + ''j''W of the ''a''X 2''b''W mos ''T''(X, W) = ''S''(X, W, W) where gcd(''j'', 2''k'') = 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" 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'' &minus; 1: ''p'' &minus; 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 are all contained in "perfect" alternants; take ''q'' &ne; ''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''&minus;1)/2}u mod e and (vu)^{(''m''&minus;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 alternants' sizes differ by replacing one Y for a Z.

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). {{qed}}<!--

==== Statement (6) ====

 

== Proposition 2 (Odd generator-offset scales are SGA) ==

If gcd(''k'', ''n'') > 1. 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. Thus, gcd(''k'', ''n'') = 1. By modular arithmetic, if 0 &le; ''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'' &minus; 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 either wouldn't be disjoint or wouldn't have the assumed lengths.) {{qed}}

== Proposition 3 (Properties of even generator-offset ternary scales) ==

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

# 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''.

(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). {{qed}}

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.

Suppose ''S'' has ''n'' notes (after dealing with small cases, we may assume ''n'' ≥ 7) and ''S'' projects to primitive mosses ''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 mosses, 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 mosses.

  Σ₁:    P P P P ... P I

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

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 ''S''₁. Then projecting to ''S''₂ shows that ''S''₂'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 ''S''₁ and ''S''₂ are the same mode of the same mos pattern (up to knowing which step size is the bigger one). Assume the L of ''S''₁ (it could be s, but it doesn't matter) is the result of identifying b and c, and all instances of s in ''S''₁ come from a. Then the steps of ''S''₂ corresponding to the L of ''S''₁ 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 ''S''₁ and ''S''₂ are the same mos pattern and mode). So ''S'' has only two step sizes (a and b), contradicting the assumption that ''S'' is ternary.

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'' &minus; β)(a~c) and β′a + (''n'' &minus; 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 &minus; 1)γ′. Also assume that the first ''k''-step in Σ is Q. Then we have:

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

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

# the stack has only copies of Q and R; or

# the stack has one T and does not contain any R (since it's more than ''f'' &minus; 1 generators away).

In this case S has two chains of Q, one with floor(''n''/2) notes and one offset by Q^{(''f''&minus;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.

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.

# ''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 < 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.

# A primitive MV3 is either (1) equivalent to XYZYX, (2) equivalent to XYXZXYX, (3) constructed from ''a''X ''b''Z with ''a'' even and gcd(''a'', ''b'') = 1 by replacing every other X with Y, (4) constructed from 2''a''X 2''b''Z with ''a'' odd and gcd(''a'', ''b'') = 1 by replacing every other X with Y, or (5) a "twisted" word constructed as follows:

## Start with a multimos word ''w''(X, Z) = ''ka''X ''kb''Z such that ''a'' is even and each ''a''X ''b''Z subword of ''w'' is of the form X''P''(X, Z)Z where ''P''(X, Z) is a palindrome.

## Interchange some of the Z's and X's at some of the boundaries of these copies of the mos word ''w''. Here, if ''w'' is a word and ''w'' = <i>u'uvv'</i>, the ''boundary'' between ''u'' and ''v'' is ''u''[len(''u'')]''v''[1] within w. (If w is a circular word and w = [w'] where w' is a linear word, replace w with w' in the previous sentence.) 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].

## Replace every other X with Y in ''w''.