Interleaving: Difference between revisions

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== Ternary interleaved scales ==
== Ternary interleaved scales ==
Conjecture: Given a [[ternary]] [[step signature]] of the form {{nowrap|''a'''''X'''''b'''''Y'''(''a'' + ''b'')'''Z'''}} where gcd(''a'', ''b'') = 1, there exists a unique single-period 2-interleaved ternary [[word|scale word]] with that step signature imposing no nontrivial linear relations between step sizes. This ternary scale word consists of ''a''-many '''XZ''' subwords and ''b''-many '''YZ''' subwords arranged in a MOS pattern (like the steps of ''a'''''L'''''b'''''s''') and consists of an interleaved pair of two {{nowrap|''a''('''X''' + '''Z''')''b''('''Y''' + '''Z''')}} subsets, offset by '''Z'''. [[Blackdye]] ('''sLmLsLmLsL''', 5'''L'''2'''m'''3'''s''') and [[whitedye]] ('''LsLsLsmsLsLsms''', 5'''L'''2'''m'''7'''s''') are examples of this.
Given a [[ternary]] [[step signature]] of the form {{nowrap|''a'''''X'''''b'''''Y'''(''a'' + ''b'')'''Z'''}} where gcd(''a'', ''b'') = 1, there always exists a single-period 2-interleaved ternary [[word|scale word]] with that step signature imposing no nontrivial linear relations between step sizes. This ternary scale word consists of ''a''-many '''XZ''' subwords and ''b''-many '''YZ''' subwords arranged in a MOS pattern (like the steps of ''a'''''L'''''b'''''s''') and consists of an interleaved pair of two {{nowrap|''a''('''X''' + '''Z''')''b''('''Y''' + '''Z''')}} subsets, offset by '''Z'''. [[Blackdye]] ('''sLmLsLmLsL''', 5'''L'''2'''m'''3'''s''') and [[whitedye]] ('''LsLsLsmsLsLsms''', 5'''L'''2'''m'''7'''s''') are examples of this.


=== Lemma ===
Conjecture: It is the ''unique'' one with this property.
If ''S'' consists of the subwords '''XZ''' and '''YZ''' arranged in the pattern of a single-period binary circular word ''w''(''x'', ''y'') where {{pipe}}''w''{{pipe}} > 2, and ''k'' is an odd number greater than 1 and less than {{pipe}}''S''{{pipe}} - 1, then the class of ''k''-steps has more than 3 abstract intervals.
 
{{proof|contents=Denote by {{pipe}}''w''{{pipe}} the length of subword ''w'' in letters and by ‖''w''‖ the interval subtended by subword ''w'' in its circular word.
 
Let ''A'' be the set of all (''k'' - 1)/2-step intervals of ''w''. {{pipe}}''A''{{pipe}} = 1 implies that ''k'' = 1 mod {{pipe}}''w''{{pipe}}, so {{pipe}}''A''{{pipe}} &ge; 2 and contains at least two intervals '''w'''<sub>1</sub> and '''w'''<sub>2</sub>.
 
Say {{pipe}}''A''{{pipe}} = 2. (If {{pipe}}''A''{{pipe}} &ge; 3 then the proof is easy.) Say ''B'' is the set of all subwords of ''w'' (not intervals) subtending '''w'''<sub>1</sub>, and ''C'' is the set of all subwords of ''w'' subtending '''w'''<sub>2</sub>.
 
Choose ''b'' in ''B'' and ''c'' in ''C'' and consider the subwords ''b''('''XZ''', '''YZ''') and ''c''('''XZ''', '''YZ''') of ''S''.
 
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 {{pipe}}''w''{{pipe}} = 3, then these are the whole word ''w''; either way we have four abstract intervals.
 
==> If {{pipe}}''w''{{pipe}} > 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) '''Z'''''b''('''XZ''', '''YZ'''), '''Z'''''c''('''XZ''', '''YZ'''), ''b''('''XZ''', '''YZ''')'''X''', ''c''('''XZ''', '''YZ''')'''X''' => done
 
(d) is similar to (a).
 
(c) We have '''Z'''''b''('''XZ''', '''YZ'''), '''Z'''''c''('''XZ''', '''YZ'''), ''b''('''XZ''', '''YZ''')'''Y''' and ''c''('''XZ''', '''YZ''')'''X'''. If the sizes of ''b''('''XZ''', '''YZ''')'''Y''' and ''c''('''XZ''', '''YZ''')'''X''' are different we are done. If they are the same, we have that ''by'' and ''cx'' subtend the same interval in ''w'', hence ''b'' has one more ''x'' and one fewer ''y'' than ''c''.
 
By scooting ''bx'' one letter to the left in ''w'', we find either (i) ''xb'' or (ii) ''yb''. The {{pipe}}''b''{{pipe}}-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 ''xby'' = ''xb'xy'', and we have ‖''p''('''XZ''', '''YZ''')'''X'''‖ as a fourth interval size.
 
(ii), (1) => ''b'' = ''b'y'', ''p'' = ''yb' '', continue scooting to the left until you find an ''x'' to the left.
 
(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).
}}
 
=== Attempted proof of Conjecture ===
{{proof|contents=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'''<sup>''q''</sup> a maximal subword of consecutive '''Z'''s has ''q'' odd, and they all must be the same length and separated by one non-'''Z''' letter '''W''':
 
'''ZZ ZZ ... ZW''' (in strand ''S''<sub>1</sub>)
 
'''Z ZZ ... ZZ WZ''' (in strand ''S''<sub>2</sub>)
 
Moreover, the offset '''δ''' = '''Z''', i.e. ''S''<sub>2</sub> is separated by the interval '''Z''' to the right of ''S''<sub>1</sub>.
 
If any maximal subword of consecutive '''Z'''s has ''q'' > 1, then the scale can be split into two subwords of length ''a'' + ''b'', ''w''<sub>1</sub> with the maximal number of consecutive '''Z''''s and ''w''<sub>2</sub> with the minimal number of '''Z'''s. Scoot ''w''<sub>1</sub> 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 ''S''<sub>1</sub> and beginning in ''S''<sub>2</sub>. 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 '''W'''<sub>1</sub>, ..., '''W'''<sub>''r''</sub>. 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 nontrivial 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 2: gcd(2(''a'' + ''b''), ''k'') > 1. ''k'' being even contradicts the interleaving property, hence ''k'' = ''a'' + ''b'' which must be odd.
 
Scoot the (''a'' + ''b'')-letter subword across ''s''. On ''a'' + ''b'' consecutive notes, the interval subtended by this word is '''δ''', and on the other ''a'' + ''b'' notes, the interval subtended by this word is ‖''s''‖ - '''δ'''.
 
We have '''V'''w'''W''' where '''V''' != '''W''', ‖'''V'''''w''‖ = '''δ''', ‖''w'''''W'''‖ = ‖''s''‖ - '''δ'''
 
=> '''δ''' - '''V''' = ‖''w''‖ = ‖''s''‖ - '''δ''' - '''W'''
 
=> 2'''δ''' + '''W''' = ‖''s''‖ + '''V'''. If this linear relation is nontrivial, it's a contradiction. If this linear relation is trivial, the third letter '''U''' occurs an even number of times in ''s'', '''V''' occurs an odd number of times in ''s'', and '''W''' occurs an even number of times. Since gcd(''a'', ''b'') = 1, {'''U''', '''W'''} != {'''X''', '''Y'''}, and so either {'''U''', '''W'''} = {'''X''', '''Z'''} or {'''U''', '''W'''} = {'''Y''', '''Z'''}. But this is again a contradiction since gcd(''a'' + ''b'', ''b'') = gcd(''a'', ''a'' + ''b'') = gcd(''a'', ''b'') = 1.
}}


== Generalizations ==
== Generalizations ==

Revision as of 09:45, 6 January 2025

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)
  • 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 [mk, Mk], 1 ≤ kn − 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 [mk, Mk]. The reverse implication follows.

For the forward implication, we wish to show that the interleaving condition is violated if mk < Mk and δ ∈ [mk, Mk] for some k, 1 ≤ kn − 1. We first observe that if mk < Mk, 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 [mk, Mk]. Moreover, such closed intervals [t0, t1] or [t1, t0], taken over all non-edE subsets comprised of stacked k-steps in S, must cover [mk, Mk]. Indeed, if such a subset in S has the k-step Mk, 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 "Mk" and "smaller" are replaced with "mk" and "larger".

The covering of [mk, Mk] constructed above grants us a stacked pair t0, t1 of unequal k-steps in S such that δ ∈ [t0, t1] ⊆ [mk, Mk]. 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 always exists a single-period 2-interleaved ternary scale word with that step signature imposing no nontrivial linear relations between step sizes. 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.

Conjecture: It is the unique one with this property.

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.