Generator-offset property

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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, ...,δm 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, with basis p (period), g, δ1, ...,δm (though the specific tuning used may be of lower rank).

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). All odd GO scales are SGA, and aside from odd GO scales, only xyxz satisfies this property. The Zarlino and diasem scales above are both SGA. Blackdye is GO but not SGA.

Theorems

Proposition 1 (Properties of SGA scales)

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. S = aX bY bZ is obtained from the (single-period) 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).
  5. The two alternants differ by replacing one Y with a Z.
  6. S is pairwise-mos (PMOS). That is, the result of identifying any two step sizes of S is always a mos.
  7. S is monotone-mos (MMOS). That is, each of the following operations results in a mos: setting L = M, setting M = s, and setting s = 0.

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

[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 |g3g2|). Thus we have 4 distinct sizes for k-steps:

  1. a1, a2 and a3 are clearly distinct.
  2. a4a3 = g1g2 != 0, since the scale is a non-degenerate AG scale.
  3. a4a1 = g3g2 = (g3 + g1) − (g2 + g1) != 0. This is exactly the chroma of the mos generated by g0.
  4. a4a2 = g1 − 2 g2 + g3 = (g3g2) + (g1g2) = (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. This proves (3).

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. This proves (2).

(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 preimage of) the 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 T be the subword consisting only of Y's and Z's. If T has a substring of length j 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".]

The generator of aX 2bW must have an odd number of W steps; if it had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction. This with (4) immediately gives (5).

For (6) and (7), we know SGA scales are MV3 and that MV3s project to mosses when one removes all instances of one step size, in particular s. Thus it suffices to prove that the result of equating any two step sizes is a mos. Equating Y and Z equates the swung alternants, resulting in the mos aX 2bW by (5). It remains to prove that equating X and Y equates the generator with the difference between the start of the 2nd chain with the end of the 1st one or vice versa; similarly for equating X and Z. Consider the swung-generator-alternant chain g1 g2 .... g1 g2 g3. Assume that the offset g1 = δ has one more Y (and one fewer Z) than g2 = g - δ. Then g3 (which becomes the imperfect generator of aX 2bW) either has one more X or one fewer X than both g1 and g2. If g3 has one more X, then g3 + g1 is equated to g1 + g2 = g, as desired. If g3 has one fewer X, then simply use the inverted generator class. Assume that the offset g1 = δ has one fewer Y (and one more Z) than g2. Then we use the inverted generator class iff g3 has one more X than both g1 and g2. [math]\square[/math]

Proposition 2 (Odd GO 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 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]\square[/math]

Proposition 3 (Properties of even GO scales)

GO scales of even size have the following properties:

  1. They are two parallel mosses of size n/2 generated by g, with offset a k-step with k odd
  2. They are not MV3
  3. They are not chiral

Proof

(1) and (2) were proved in the proof of Prop 1. (3) is easy to check using (1). [math]\square[/math]

Open conjectures

Conjecture 4 ("MV3 Structure Theorem")

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)

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.

Falsified conjectures