Interleaving

A scale is (k-)interleaved or (k-)flought(ened) (/flɔːt/, rhymes with bought; this is a neo-Anglish term cognate with German geflochten 'braided') if it is made of k > 1 copies (called strands) of an n-note periodic scale s, and any two copies of s are interleaved so that any note of the first copy falls strictly between two notes of the other copy. The set of offsets that separate the strands from a fixed strand is a chord called the polyoffset, which is determined up to inversion and equave-equivalence for a given interleaved scale. An interleaved scale is thus a cross-set with a little additional structure. One can interleave or floughten a scale s by a certain polyoffset Δ (or: "Δ interleaves s" or "s is interleavable by Δ") if s is the strand scale of an interleaved scale with polyoffset Δ. Such a scale is denoted Interleave(s; Δ). The concept of interleaved scales is a generalization of bipentatonic scales.

Blackdye, Zil[14], and bicycle are examples of interleaved scales, because they each have two interleaved strands, respectively Pyth[5], Zarlino, and 8:9:10:11:13:14. The terminology, however, is intended to cover any number of strands and any choice of strand scale.

Some interleaved scales

Interleaved scales can easily be built from a harmonic series mode as the strand: for example, if n::2n is the strand, then (2n + 1)/2n always works as the offset (e.g. strand 5:6:7:8:9:10, offset 10:11). Here are some other examples:

• Interleave(12:14:16:18:21:24; 11:12)
• Interleave(12:14:16:18:21:24; 12:13:22)
• Interleave(12:14:16:18:21:24; 8:10:11)
  • User:Userminusone/Userminusone's_11_limit_15_tone_scale
• Interleave(12:14:16:18:21:24; 9:10:11)
  • Note: detempered 11-limit Porcupine[15]; well-formed generator sequence GS(10/9, 11/10, 12/11, 10/9, 11/10, 12/11, 10/9, 11/10, 189/176)
• Interleave(Pyth[5]; 8:10:11)
• Interleave(Pyth[5]; 9:10:11)
  • Note: detempered 2.3.5.11 Porcupine[15]; well-formed generator sequence GS(10/9, 11/10, 12/11)
• Interleave(9/8-14/11-4/3-3/2-56/33-21/11-2/1; 9/7)

Properties

1. The following is a necessary and sufficient condition for interleavability. Let S be a scale with equave E, [math]\displaystyle{ \mathcal{D}_k(S) }[/math] be the set of all k-step intervals of S, and Δ be a chord such that every interval of Δ falls within the open interval (0, E). Then the polyoffset chord Δ interleaves S if and only if no nonunison (positive) interval in Δ falls within [math]\displaystyle{ [\min \mathcal{D}_k(S), \max \mathcal{D}_k(S)] }[/math] for any k ∈ {0, ... len(S) - 1}.
2. For any periodic scale S with equave E, if δ is an offset and Interleave(S; δ) exists, then Interleave(S; δ) = Interleave(S; E - δ) = Interleave(S; δ + E). Thus, taking the equave complement of an offset in a polyoffset does not change the interleaved scale, nor does shifting any individual offset by equaves.
3. Given an E-equivalent scale S, offsets δ within the open interval (0, min({step sizes in S})) are called small in the context of interleaving S. Small offsets are significant because the resulting interleaved scale has a structure that closely mimics the underlying scale structure: if S is a circular word [math]\displaystyle{ w(a_1, a_2, ..., a_n) }[/math] then Interleave(s; δ) uses the same circular word but with δ followed by the difference between δ and every step size in w, namely [math]\displaystyle{ w(\delta b_1, \delta b_2, ..., \delta b_n) }[/math] where [math]\displaystyle{ b_i = a_i - \delta }[/math].
4. An interleaved scale is not always CS, even when the strand is CS and the scale has a generator sequence where every generator subtends the same number of steps. One such scale is Interleave(Zarlino; 32/25) = 25/24 9/8 75/64 5/4 125/96 4/3 375/256 3/2 25/16 5/3 225/128 15/8 125/64 2/1 which has GS(32/25 125/96 32/25 5/4).

Proof of the offset constraints

If the polyoffset has more than two notes, the interleaving condition only needs to hold for pairs of distinct strands, and hence the above property only needs to hold for pairs of notes in the polyoffset. This reduces the proof to the case of one offset δ.

Let S₁, S₂ denote the two copies of S separated by δ, where S₁(0) = 0 (the unison), S₂(0) = δ. Assume that the scale F is the union of S₁ and S₂, and F(0) = 0. Let [math]\displaystyle{ m_k = \min \mathcal{D}_k(S) }[/math] and [math]\displaystyle{ M_k = \max \mathcal{D}_k(S). }[/math]

Suppose δ > 0 is not in any closed intervals [m_k, M_k], 1 ≤ k ≤ n − 1, n = len(S). Then for any k, S₁(k) falls between adjacent notes of S₂. The same holds when we reverse the roles of S₁ and S₂ and use the offset E − δ; since the union [math]\displaystyle{ \bigcup_{k=1}^{n-1} [m_k, M_k] }[/math] is invariant under taking equave complements, neither is E − δ within any [m_k, M_k]. The reverse implication follows.

For the forward implication, we wish to show that the interleaving condition is violated if m_k < M_k and δ ∈ [m_k, M_k] for some k, 1 ≤ k ≤ n − 1. We first observe that if m_k < M_k, then S has some pair of stacked k-steps, say (S(n₀), S(n₀ + k)) (S(n₀ + k), S(n₀ + 2k)), whose sizes t₀, t₁ are unequal and both contained in [m_k, M_k]. Moreover, such closed intervals [t₀, t₁] or [t₁, t₀], taken over all non-edE subsets comprised of stacked k-steps in S, must cover [m_k, M_k]. Indeed, if such a subset in S has the k-step M_k, that subset must also have a k-step smaller than k/gcd(n, k) steps of n/gcd(n, k)-edE, and by symmetry, the previous clause also holds when "M_k" and "smaller" are replaced with "m_k" and "larger".

The covering of [m_k, M_k] constructed above grants us a stacked pair t₀, t₁ of unequal k-steps in S such that δ ∈ [t₀, t₁] ⊆ [m_k, M_k]. Assume t₀ < t₁. (If t₀ > t₁, take equave complements and use the offset E − δ.) Then the corresponding occurrence of the k-step t₀ in S₂ is shifted into the closed interval I corresponding to the k-step t₁ in S₁. But we then have k + 1 notes of S₂ within I. Assuming none of these notes coincide with a note of S₁ (otherwise, interleaving would be violated), each of the k + 1 notes must fall within one of the k scale steps subtended by t₀ in S₁. By the pigeonhole principle, at least one of these steps in S₁ must contain two consecutive notes of S₂ in its interior, breaking the interleaving condition as desired. [math]\displaystyle{ \square }[/math]

Ternary interleaved scales

Given a ternary step signature of the form aXbY(a + b)Z where gcd(a, b) = 1, there exists a unique single-period (abstractly) 2-interleaved ternary scale word with that step signature. This ternary scale word consists of a-many XZ subwords and b-many YZ subwords arranged in a MOS pattern (like the steps of aLbs) and consists of an interleaved pair of two a(X + Z)b(Y + Z) subsets, offset by Z. Blackdye (sLmLsLmLsL, 5L2m3s) and whitedye (LsLsLsmsLsLsms, 5L2m7s) are examples of this.

Lemma

If S consists of the subwords XZ and YZ arranged in the pattern of a single-period binary circular word w(x, y) where |w| > 2, and k is an odd number greater than 1 and less than |S| - 1, then the class of k-steps has more than 3 abstract intervals.

Proof: Denote by |w| the length of w in letters and by ||w|| the interval subtended by w in its scale word.

Let A be the set of all (k - 1)/2-step intervals of w. |A| = 1 implies that k = 1 mod |w|, so |A| ≥ 2 and contains at least two intervals w₁ and w₂.

Say |A| = 2. (If |A| ≥ 3 then the proof is easy.) Say B is the set of all subwords (not intervals) subtending w₁, and C is the set of all subwords ending in Z subtending w₂.

Choose b in B and c in C.

b(XZ, YZ)

c(XZ, YZ)

If k = 3, then we can just say b = x and c = y. Suppose YZX does not occur in S; then yx does not occur in w which is a contradiction, so YZX must occur.

=> ZYZXZ occurs

=> either XZYZXZ or YZYZXZ occurs (with suffix ZYZXZ)

==> If |w| = 3, then these are the whole word w; either way we have four abstract intervals.

==> If |w| > 3, then either ZYZXZXZ or ZYZXZYZ occurs (with prefix ZYZXZ).

Case XZYZYZXZ => XZY, YZY, ZYZ, ZXZ

Case YZYZYZXZ => YZY, YZX, ZYZ, ZXZ

So the case k = 3 is done.

For any other k, we similarly go through the cases (a) bx and cx, (b) bx and cy, (c) by and cx, (d) by and cy.

(a) Zb(XZ, YZ), Zc(XZ, YZ), b(XZ, YZ)X, c(XZ, YZ)X => done

(d) is similar to (a).

(b) We have Zb(XZ, YZ), Zc(XZ, YZ), b(XZ, YZ)X and c(XZ, YZ)Y. If the sizes of b(XZ, YZ)X and c(XZ, YZ)Y are different we are done. If they are the same, we have that bx and cy subtend the same interval in s, hence b has one more x and one fewer y than c.

By scooting bx one letter to the left in s, we find either (i) xb or (ii) yb. The |b|-letter prefix p of this subword subtends either the same interval as (1) b or (2) c.

(i), (1) => b = b'x, p = xb' , have xbx = xb'xx, continue scooting to the left until you find a y to the left, then same case as (ii), (2).

(ii), (1) => b = b'y, p = yb' , we have ||p(XZ, YZ)Y|| as a fourth interval size.

(i), (2) => b = b'y, p = xb' , we have ||p(XZ, YZ)X|| as a fourth interval size.

(ii), (2) => b =b'x, p = yb' , we have ||p(XZ, YZ)X|| as a fourth interval size.

(c) is similar to (b).

Proof of Theorem

Proof of Conjecture: Suppose the scale is made of two interleaved subsets offset by the abstract interval δ.

Case 1: gcd(2(a + b), k) = 1. If k = 1, then any Z^q a maximal subword of consecutive Zs has q odd, and they all must be the same length and separated by one non-Z letter W:

ZZ ZZ ... ZW (in strand S₁)

Z ZZ ... ZZ WZ (in strand S₂)

Moreover, the offset δ = Z, i.e. S₂ is separated by the interval Z to the right of S₁.

If any maximal subword of consecutive Zs has q > 1, then the scale can be split into two subwords of length a + b, w₁ with the maximal number of consecutive Z's and w₂ with the minimal number of Zs. We can also choose w₁ to begin in S₁ (because?)

Scoot w₁ to the right one step at a time until it loses one Z. This proves either that a non-Z letter is equal to Z or that the offset goes in the wrong direction. Hence q = 1, as desired.

If k > 1, stack the word of k-steps in the scale, yielding a circular word T, which traverses all notes of S since gcd(k, 2(a + b)) = 1. Since k is odd, the letters of this word alternate between beginning in S₁ and beginning in S₂. By a reasoning similar to the above, T has a letter δ between its two mutually interleaved strands. By the lemma this scale has step variety r > 3, say with letters W₁, ..., W_r. Let k' be the inverse of k mod 2(a + b). By stacking k' -step subwords of T, we end up with at least 4 different linear equations with 3 unknowns X, Y, and Z, implying a linear relation between X, Y, and Z (?). This is a contradiction as X, Y, and Z are assumed to not have a linear relation.

Case 3: gcd(2(a + b), k) > 1. k being even contradicts the interleaving property, hence k = a + b which must be odd. (To be continued)

Generalizations

Mutual interleavability

Two periodic scales [math]\displaystyle{ S, T : \mathbb{Z} \to \mathbb{R} }[/math] of the same length and equave are mutually interleavable if there exists [math]\displaystyle{ \delta\in\mathbb{R} }[/math] such that S and T + δ are interleaved. Note that though a given 2n-note scale being a mutually interleaved result of some pair of scales may be trivial, a given pair of scales being mutually interleavable is less so: for example, MMMM and Lsss are not mutually interleavable when s is too small.

A contrainterleaved scale is a mutually interleaved pair of the two chiralities of a chiral scale. Conjecture: All even-regular MV3 scales are contrainterleaved.