Generator-offset property

A scale satisfies the generator-offset property (also GO, alternating generator or AG) if it satisfies the following properties:

1. The scale is generated by two chains of stacked copies of a generator.
2. The two chains are separated by an offset (the difference between the first note of the second chain minus the first note of the first chain).
3. The lengths of the chains differ by at most one. (1-3 can be restated as: The scale can be built by stacking two alternating generators (called alternants), which do not necessarily take up the same number of steps.)
4. The generator always occurs as the same number of steps. For example, the generator is never both a 2-step and a 3-step.

The Zarlino (3L 2M 2S) JI scale is an example of a GO scale, because it is built by stacking alternating 5/4 and 6/5 generators. 7-limit diasem (5L 2M 2S) is another example, with generators 7/6 and 8/7.

More formally, a cyclic word S (representing the steps of a periodic scale) of size n is GO if it satisfies the following properties:

1. S is generated by two chains of stacked generators g separated by a fixed offset δ; either both chains are of size n/2, or one chain has size (n + 1)/2 and the second has size (n − 1)/2. Equivalently, S can be built by stacking a single chain of alternants g₁ and g₂, resulting in a circle of the form either g₁ g₂ ... g₁ g₂ g₁ g₃ or g₁ g₂ ... g₁ g₂ g₃.
2. The scale is well-formed with respect to g, i.e. all occurrences of the generator g are k-steps for a fixed k.

This doesn't imply that g₁ and g₂ are the same number of scale steps. For example, 5-limit blackdye has g₁ = 9/5 (a 9-step) and g₂ = 5/3 (a 7-step).

More generally, we say that a scale is m-GO if both of the following hold:

1. the scale consists of m + 1 chains of stacked generator g (implying m offsets δ₁, ...,δ_m from a fixed chain), each chain having either j or j + 1 notes.
2. The scale is well-formed with respect to g.

(Thus 1-GO is the same thing as GO.)

An m-GO scale can be interpreted as a scale in a rank-(m + 2) regular temperament (though the specific tuning used may be of lower rank), with basis p (period), g, δ₁, ...,δ_m.

Other definitions

• A strengthening of the generator-offset property, tentatively named the swung-generator-alternant property (SGA), states that the alternants g₁ and g₂ can be taken to always subtend the same number of scale steps, thus both representing "detemperings" of a generator of a single-period mos scale (otherwise known as a well-formed scale). Only odd GO scales and xyxz can satisfy this property. The Zarlino and diasem scales above are both SGA. Blackdye is GO but not SGA.

Theorems

Proposition 1

Let S be a 3-step-size scale word in L, M, and s, and suppose S is SGA. Then:

1. S is unconditionally MV3 (i.e. MV3 regardless of tuning).
2. S is of the form ax by bz for some permutation (x, y, z) of (L, M, s).
3. The cardinality (size) of S is either odd, or 4 (and S is of the form xyxz).
4. If |S| is odd, then S = aX bY bZ is obtained from the 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).

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

Proof

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

CASE 1: EVEN CARDINALITY
O-O-...-O (n/2 notes)
O-O-...-O (n/2 notes)

and

CASE 2: ODD CARDINALITY
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 ≤ (chain length), for notes in the upper and lower chain respectively.

In case 1, let g₁ = (2, 1) − (1, 1) and g₂ = (1, 2) − (2, 1). We have the chain g₁ g₂ g₁ g₂ ... g₁ g₃.

Let r be odd and r ≥ 3. Consider the following abstract sizes for the interval class (k-steps) reached by stacking r generators:

1. from g₁ ... g₁, we get a₁ = (r − 1)/2*g₀ + g₁ = (r + 1)/2 g₁ + (r − 1)/2 g₂
2. from g₂ ... g₂, we get a₂ = (r − 1)/2*g₀ + g₂ = (r − 1)/2 g₁ + (r + 1)/2 g₂
3. from g₂ (...even # of gens...) g₁ g₃ g₁ (...even # of gens...) g₂, we get a₃ = (r − 1)/2 g₁ + (r − 1)/2 g₂ + g₃
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.

Choose a tuning where g₁ and g₂ are both very close to but not exactly 1/2*g₀, resulting in a scale very close to the mos generated by 1/2 g₀. (i.e. g₁ and g₂ differ from 1/2*g₀ by ε, a quantity much smaller than the chroma of the n/2-note mos generated by g₀, which is |g₃ − g₂|). Thus we have 4 distinct sizes for k-steps:

1. a₁, a₂ and a₃ are clearly distinct.
2. a₄ − a₃ = g₁ − g₂ != 0, since the scale is a non-degenerate AG scale.
3. a₄ − a₁ = g₃ − g₂ = (g₃ + g₁) − (g₂ + g₁) != 0. This is exactly the chroma of the mos generated by g₀.
4. a₄ − a₂ = g₁ − 2 g₂ + g₃ = (g₃ − g₂) + (g₁ − g₂) = (chroma ± ε) != 0 by choice of tuning.

By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g₁ and g₂ must themselves be step sizes. Thus we see that an even-cardinality, unconditionally MV3, AG scale must be of the form xy...xyxz. But this pattern is not unconditionally MV3 if n ≥ 6, since 3-steps come in 4 sizes: xyx, yxy, yxz and xzx. Thus n = 4 and the scale is xyxz.

In case 2, let (2, 1) − (1, 1) = g₁, (1, 2) − (2, 1) = g₂ be the two alternants. Let g₃ be the leftover generator after stacking alternating g₁ and g₂. Then the generator circle looks like g₁ g₂ g₁ g₂ ... g₁ g₂ g₃. Then the combinations of alternants corresponding to a step are:

1. kg₁ + (k − 1)g₂
2. (k − 1)g₁ + kg₂
3. (k − 1)g₁ + (k − 1) g₂ + g₃,

if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times.

(The above holds for any odd n ≥ 3.)

For (1), we now only need to see that SGA + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.

For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. If Y’s and Z’s don't alternate perfectly, then (ignoring X's) you have two consecutive Y's somewhere and two consecutive Z's somewhere else. Assume that g_pf = iX + jW with j >=2. (If this is not true, invert the generator, since b > 1.)

In aX bY bZ, consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) the (“detempered”) imperfect generator. Since a + 2b >= 5, are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to MV3. [Let S be the subword consisting only of Y's and Z's If there is a substring of length j in the string that's not contained in a perfect generator, you can go somewhere else to find it, since the numbers of Y's and Z's change one at a time and reach a maximum and a minimum somewhere, guaranteeing that intermediate values are reached "on the other side".] [math]\displaystyle{ \square }[/math]

Proposition 2

Suppose that a periodic scale satisfies the following:

• is generator-offset (chain lengths off by 1)
• has odd length n

Then the scale is SGA.

Proof

Assume that the generator is a k-step and k is even. (If k is not even, invert the generator.) On some tonic p we have a chain of ceil(n/2) notes and on some other note p' = p + offset (not on the first chain) we'll have floor(n/2) notes.

We must have gcd(k, n) = 1. If not, since n is odd, gcd(k, n) is an odd number at least 3, and the k-steps must form more than 2 parallel chains.

By modular arithmetic we have rk mod n = k/2 iff r = ceil(n/2) mod n. (Since gcd(2, n) = 1, 2 is multiplicatively invertible mod n, and we can multiply both sides by 2 to check this.) This proves that the offset, which must be reached after ceil(n/2) generator steps, is a k/2-step, as desired. (If the offset wasn't reached in ceil(n/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) [math]\displaystyle{ \square }[/math]

Open conjectures

Conjecture 3

If a non-multiperiod 3-step size scale word is

1. unconditionally MV3,
2. has odd cardinality, and
3. is not of the form mx my mz, xyzyx or xyxzxyx,

then it is SGA. (a converse to Theorem 1)

Conjecture 4

If a non-multiperiod 3-step size scale word is

1. unconditionally MV3,
2. has even cardinality, and
3. is not of the form mx my mz,

then it is of the form W(x, y, z)W(y, x, z) for some word W in 3 variables.

Conjecture 5

An SGA scale is always pairwise-well-formed. That is, the result of identifying any two step sizes of an SGA scale is always a non-multiperiod mos.