A scale satisfies the generator-offset property (also GO, alternating generator or AG) if it satisfies the following properties:
- The scale is generated by two chains of stacked copies of a generator.
- The two chains are separated by an offset (the difference between the first note of the second chain and the first note of the first chain).
- 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.)
- 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:
- 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.
- 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:
- 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.
- 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).
- 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.
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:
- S is abstractly SV3 (i.e. SV3 for almost all tunings).
- S is of the form ax by bz for some permutation (x, y, z) of (L, M, s).
- The length of S is either odd, or 4 (and S is of the form xyxz).
- S = aX bY bZ is obtained from some mode of 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, fixing the mode of aX 2bW).
- The two alternants differ by replacing one Y with a Z.
- S is pairwise-mos. That is, the following operations each result in a mos: setting L = M, setting L = s, and setting M = s.
- S is elimination-mos. That is, "tempering out" any one step size results in a mos.
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.]
Assuming SGA, we have two chains of generator g0 (going right). The two cases are:
CASE 1: EVEN LENGTH O-O-...-O (n/2 notes) O-O-...-O (n/2 notes)
CASE 2: ODD LENGTH 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:
- from g1 ... g1, we get a1 = (r − 1)/2*g0 + g1 = (r + 1)/2 g1 + (r − 1)/2 g2
- from g2 ... g2, we get a2 = (r − 1)/2*g0 + g2 = (r − 1)/2 g1 + (r + 1)/2 g2
- from g2 (...even # of gens...) g1 g3 g1 (...even # of gens...) g2, we get a3 = (r − 1)/2 g1 + (r − 1)/2 g2 + g3
- 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:
- a1, a2 and a3 are clearly distinct.
- a4 − a3 = g1 − g2 != 0, since the scale is a non-degenerate GO scale.
- a4 − a1 = g3 − g2 = (g3 + g1) − (g2 + g1) != 0. This is exactly the chroma of the mos generated by g0.
- 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-length, unconditionally SV3, GO scale must be of the form xy...xyxz. But this pattern is not abstractly SV3 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 come in exactly 3 sizes:
- kg1 + (k − 1)g2
- (k − 1)g1 + kg2
- (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 length => abstractly SV3. But the argument in case 2 above works for any interval class (qbstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 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 the perfect generator is 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 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, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [Whenever the boundaries of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the boundary and a minimum with another choice, guaranteeing that intermediate values are reached.]
Any generator of aX 2bW must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). This with (4) immediately gives (5).
For (6), odd-numbered SGA scales are Fokker blocks (in the 2-dimensional lattice generated by the generator and the offset). To see this, consider the following lattice depiction of such a scale:
x x x ... x x x x ... x x
and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A 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).
For (7), consider the mos aX 2bW as chunks of X separated by W (tempering Y and Z together into W). Eliminating every other W turns it into a mos, because the sum of sizes of consecutive chunks of X (1st chunk with 2nd chunk, 3rd with 4th, ...) must form a mos. This is because the chunk sizes of X form a mos, and taking every kth note of an n-note mos where k divides n yields a mos. Since the result of setting X = 0 is the mos bY bZ, S is elimination-mos.
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.
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.)
Proposition 3 (Properties of even GO scales)
GO scales of even size have the following properties:
- They are two parallel mosses of size n/2 generated by g, with offset a k-step with k odd
- They are not SV3
- They are not chiral
(1) and (2) were proved in the proof of Prop 1. (3) is easy to check using (1).
Theorem 4 (PWF implies SV3 and either GO or abacaba)
Let S(a,b,c) be a scale word in three Z-linearly independent step sizes a, b, c. Suppose S is PWF. Then S is SV3 and has an odd number of notes. Moreover, S is either GO 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 single-period mosses S1 (via identifying b ~ c), S2 (via identifying a ~ c) and S3 (via identifying a ~ b). Suppose S1's generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S2 and S3, respectively. These two-"step-size" 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 S1 and S2, which come in at most two sizes. Both Σ2 and Σ3 are single-period mosses, since (k, n) = 1.
index: 1 2 3 4 ... n Σ1: P P P P ... P I Σ2: [some mos] Σ3: [some mos]
(Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)
Assume that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos pattern (up to knowing the order of the step sizes) and even the same mode:
S1: L ... L s S2: L' ... L' s'
Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 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 S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.
We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). Only two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S, then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.
So assume Q = (α, β, γ) != R = (α, β’, γ’) are the two preimages of P in S. Then I has preimage (α’, β’’, γ’’), which by abuse we also denote I. Here the values in each component differ by at most 1, and α != α’. Recall that I has preimage (α’, β’’, γ’’). Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ. The cyclic words Λ1 = the pattern of α and α’, Λ2 = the pattern of β and β’, and Λ3 = the pattern of γ and γ must form mosses. By way of illustration, the chain of k-steps might look like this in S:
1 2 3 4 5 6 7 8 9 Σ = Q Q Q R Q Q R R I Λ1 = α α α α α α α α α’ Λ2 = β β β β’ β β β β β’ Λ3 = γ γ γ γ’ γ γ γ γ γ
We don’t know which P’s have preimage Q and which P’s have preimage R, but we now do know that Σ2 has one more R than Σ3, and thus that:
Λ2 is λβ μβ’, 2 <= μ <= ceil(n/2). Λ3 is a (λ±1)β (μ∓1)β’ mos.
Since neither are multimosses, and at least one of μ and (μ∓1) are even, it is now immediate that n is odd.
We can assume that the first k-step in Σ is Q:
1 … n Σ = Q … I Λ1 = α … α α’ Λ2 = β (pattern) β’ Λ3 = γ ( ditto ) γ
Since, by our assumption, Λ3 has two γ’s in a row, Λ3 must have more γ than γ’, so μ - 1 < n/2. Since Λ3 is a mos, μ - 1 >= 1. So we have 2 <= μ <= ceil(n/2).
We have three cases to consider:
Case 1: Λ2 is the mos (n-2)β 2β’.
For Λ2 to be a mos, the first occurrence of R must be at either f = floor(n/2) or ceil(n/2). We may assume that it is at f; otherwise flip the chain and reindex the words to start at 2f.
1 … f … 2f n Σ = Q … Q R Q … Q I Λ1 = α … α α α … α α’ Λ2 = β … β β’ β … β β’ Λ3 = γ … γ γ’ γ … γ γ
We need only consider stacks up to f-many k-steps. Either:
- the stack has only preimages of P’s and it either contains the preimage of the fth P or not; or
- the stack has one I and does not contain any R (since it’s more than f-1 generators away).
These give exactly three distinct sizes for every interval class. Hence S is SV3.
In this case S has two chains of Qs, one with floor(n/2) notes and and one with ceil(n/2) notes. Every instance of Q must be a k-step, since 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.
End Case 1.]
[Case 2: Λ2 has fewer β than β’.
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 size of the stack of k-steps is odd, then there are two sizes that do not contain I and one size that contains I; if the stack size is even, then there is one size that does not contain I and two sizes that contain I. Hence S is SV3.
In this case we have Σ = QRQR…QRI, thus S satisfies the generator-offset property.
End Case 2.]
[Case 3: μ = anything else from 3 to floor(n/2).
Λ2 has a chunk of βs (after the first β’) of size x = either floor(n/μ) (>= floor(n/floor(n/2)) = 2) or ceil(n/μ) (= floor(n/μ) + 1). Hence Λ3 has a chunk of γs of that same size. Λ3 also has a chunk that goes over n and loops back around, which must be of size y = at least 2*(floor(n/μ)-1) + 1 (Λ3 might have chunks of size floor(n/μ)-1 and floor(n/μ) instead) = 2*floor(n/μ) - 1, and at most 2*(floor(n/μ)+1) + 1 = 2*floor(n/μ) + 3 (if Λ3 has chunks of size floor(n/μ) and ceil(n/μ)). The difference between the chunk sizes is y - x, which must be 0 or 1, since Λ3 is a mos. We thus have the following cases:
BEGIN CASEBLOCK 3.x: [ Here chunk of Λ2 means chunk of β, and chunk of Λ3 means chunk of γ.
[Case 3.1: (x, y) = (floor(n/μ), 2*floor(n/μ) - 1) Since y-x = floor(n/μ) - 1 and floor(n/μ) >= 2, we have: x = floor(n/μ) = 2 and y-x = 1; hence y = 2*floor(n/μ) - 1 = 3. The chunk in Λ3 whose size is 3 is made from two chunks in Λ2 of size 1. (So Λ2 has chunks of size 1 and 2, and Λ3 has chunks of size 2 and 3.)
Λ2 has two consecutive chunks of size 1. Since chunk sizes form a mos, Λ2 has more chunks of size 1 than it has chunks of size 2.
Use “w[i:j]” to denote the slice of the cyclic word w that includes both endpoints. Λ2 has only two chunks of size 1, Λ2[n-1:n-1] and Λ2[1:1], since otherwise Λ3 would have a chunk of size 1 within Λ3[1:n-1]. Thus Λ2 has exactly one chunk of size 2. Thus Λ2 = ββ’βββ’ββ’ and Λ3 = γγ’γγγ’γγ. Thus we have:
1 2 3 4 5 6 7 Σ = Q R Q Q R Q I Λ1 = α α α α α α α’ Λ2 = β β’ β β β’ β β’ Λ3 = γ γ’ γ γ γ’ γ γ
Suppose a step of S is reached by stacking t-many k-steps. We have three cases after accounting for equave complements:
- t = 1: S is equivalent to abacaba.
- t = 2: QR QQ RQ IQ RQ QR QI => S is equivalent to abacaba.
- t = 3: QRQ QRQ IQR QQR QIQ RQQ RQI => S is equivalent to abacaba.
(This also implies S is SV3.) End Case 3.1]
[Case 3.2: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) - 1).
Impossible: y = 2*floor(n/μ) - 1 can only occur if Λ3 has chunks of size floor(n/μ) - 1 and floor(n/μ), which contradicts the size of x. End Case 3.2]
[Case 3.3: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ))
Impossible: y = 2*floor(n/μ) can only occur if Λ3 has chunks of size floor(n/μ) - 1 and floor(n/μ), which contradicts the size of x. End Case 3.3]
The remaining cases are all impossible because they imply y - x >= 2:
[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)]
END CASEBLOCK 3.x]
End Case 3.]
Conjecture ("SV3 Structure Theorem")
If a 3-step size scale word ax by cz is
- abstractly SV3 (strict variety 3, i.e. every k-step except multiples of the equave comes in exactly 3 sizes, for almost all tunings),
- has odd length,
- has gcd(a, b, c) = 1,
- is not of the form xyzyx or xyxzxyx,
then it is GO (and therefore SGA). (a converse to Proposition 1)
By Theorem 4, this conjecture would follow from proving that such scales are PWF.
If a 3-step size scale word that is ax by cz is abstractly SV3, then two of a, b, c are equal.
Conjecture ("MV3 Sequences")
Given any two generators, we can iterate them to any number of notes and see what the maximum-variety of the resulting scale is. In particular, we can look at those scale sizes which are MV3, and thus compute the MV3 sequence for the pair of generators (similar to the "MOS sequence" one can compute for one generator). Thus, for any pair of generators, we can form the associated sequence of increasingly large MV3 scales.
Surprisingly, for almost all pairs of generators, this sequence seems to terminate after some (usually relatively small) scale. That is, if we simply take all possible pairs of generators between 0 and 1200 cents, and for each pair we compute the MV3 sequence for all generator pairs up to some maximum N, such as 1000, we can easily see that most points will have only a few entries in it, after which no MV3 scales are apparently generated. It would seem to be true that as the two generators get closer and closer in size, the MV3 sequence gets longer and longer, until when the two generators are equal you have an infinite-length sequence (corresponding to MOS).
It is pretty easy to see this behavior is true if we simply compute the MV3 sequences up to any very large N, far beyond the scale sizes we typically use in music theory, but it would be good to have a proof.
This heading has those open questions for which no conjecture has yet been formed either way. (These can be updated as necessary)
- Given any arbitrary MOS (or DE, etc) scale with at least three notes per period, is there *always* a MV3 GO scale which can be derived as a "detempering" of that scale? Or is this only true for some MOS's? For instance, the MOS LLsLLLs has the MV3 GO scale LmsLmLs as a detempering. Does a similar MV3 detempering exist for every possible DE scale with at least three notes per period, or at least for strict MOS's with one period per octave (e.g. well-formed scales)?
- The scale tree is a great way to analyze MOS scales. For any generator, we can compute the various MOS's it forms if we simply look at the scale tree, and indeed MOS "words" like LLsLLLs can be identified with regions on the scale tree (in this situation the interval between 4/7 and 3/5). A similar "scale plane" should exist for GO-MV3 scales, where given some word representing a GO-MV3 scale, we can look at the set of points on the generator plane which generates it; these seem to often be triangles, with the lines corresponding to MOS's and the vertices corresponding to EDOs (though is this always true?). What is the big picture of this scale plane? Can we use Viggo Brun's algorithm for this, generalizing the theory of continued fractions? Is there some simple formula we can use to predict, given some GO-MV3 scale, which region on the scale plane it corresponds to? Can we plot simple generator-size-proportions as points in this space? And so on.
- In the theory of MOS, there is a second scale tree that is less frequently talked about, which Erv Wilson calls the "Rabbit Sequence" (Erv Wilson's original version, interactive version 1, interactive version 2). This is a tree for which each MOS word has two children, depending on if the MOS is "soft" (with L/s < 2) or "hard" (with L/s > 2). For instance, LsLss has the two children LLsLLLs and ssLsssL. Does a similar scale plane exist for these GO-MV3 scales?
All mosses have an MV3 detempering (Counterexample: LsLsLsLsLs)
LsLsLsLsLs does not have an MV3 detempering: Wolog we “detemper” L to L and M.
LsMsLsMsLs is not MV3: LsM LsL sMs sLs
LsMsLsMsMs can be rotated to MsLsMsLsMs which is not MV3 by symmetry with LsMsLsMsLs.
LsMsMsMsMs is not MV3: LsM MLM sMs sLs
MsLsLsLsLs not MV3 for the same reason as LsMsMsMsMs