Generator-offset property

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

• the scale can be built by stacking two alternating generators (called alternants), which do not necessarily take up the same number of steps
• the scale is generated by two chains of generators separated by an 'offset, and the lengths of the chains differ by at most one.

The Zarlino (3L 2M 2S) JI scale is an example of an AG 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 a periodic scale) of size n is AG if it satisfies the following equivalent properties:

1. 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. S is generated by two chains of generators separated by a fixed interval; either both chains are of size n/2, or one chain has size (n + 1)/2 and the second has size (n − 1)/2.

These are equivalent, since the separating interval can be taken to be g₁ and the generator of each chain = g₁ + g₂. 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).

Related properties

• A strengthening of the generator-offset property, tentatively named split-generator-alternant (SGA), states that g₁ and g₂ can be taken to be 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 AG scales and xyxz can satisfy this property. The Zarlino and diasem scales above are both SGA. Blackdye is AG but not SGA.

Theorems

Theorem 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).

[Note: This is not true with SGA replaced with AG; 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.)

Now we 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. [math]\displaystyle{ \square }[/math]

Open conjectures

Conjecture 2

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

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

then it is AG. (a strong converse to Theorem 1)

Conjecture 3

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.

Falsified conjectures

Conjecture -1

An odd generator-offset scale is SGA.

Counterexamples

This is false for the 5-note scale 0 250 300 700 750 1200 with alternants 700 and 50: the 1-steps form the chain 250 50 400 50 450 and the 2-steps form the chain 300 450 700 250 500.

A 7-note counterexample: 0 210 220 700 710 920 930 1200