Interleaving: Difference between revisions

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# Given an E-equivalent scale ''S'', offsets δ within (0, min({step sizes in ''S''})) are called ''small'' in the context of floughtening ''S''. Small offsets are significant because the resulting flought scale has a structure that closely mimics the underlying scale structure: if ''S'' is a circular word <math>w(a_1, a_2, ..., a_n)</math> then Fl(''s''; δ) uses the same circular word but with δ followed by the difference between δ and every step size in w, namely <math>w(\delta b_1, \delta b_2, ..., \delta b_n)</math> where <math>b_i = a_i - \delta</math>.

# A flought 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 Fl(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|title=Proof of the offset constraints|contents=

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 ''S''1, ''S''2 denote the two copies of ''S'' separated by δ, where ''S''1(0) = '''0''' (the unison), ''S''2(0) = δ. Assume that the scale ''F'' is the union of ''S''1 and ''S''2, and ''F''(0) = '''0'''. Let <math>m_k = \min \mathcal{D}_k(S)</math> and <math>M_k = \max \mathcal{D}_k(S).</math>

Suppose δ > 0 is not in any intervals [''m''''k'', ''M''''k''], 1 ≤ ''k'' ≤ ''n'' − 1, ''n'' = len(''S''). Then for any ''k'', ''S''1(''k'') falls between adjacent notes of ''S''2. The same holds when we reverse the roles of ''S''1 and ''S''2 and use the offset ''E'' − δ; since the union <math>\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''(''n''0), ''S''(''n''0 + ''k'')) (''S''(''n''0 + ''k''), ''S''(''n''0 + 2''k'')), whose sizes ''t''0, ''t''1 are unequal and both contained in [''m''''k'', ''M''''k'']. Moreover, such intervals [''t''0, ''t''1] or [''t''1, ''t''0], taken over all non-ed''E'' circles of ''k''-steps in ''S'', must cover [''m''''k'', ''M''k]. Indeed, if a circle of stacked ''k''-steps in ''S'' has the ''k''-step ''M''''k'', that circle must also have a ''k''-step smaller than ''k''/gcd(''n'', ''k'') steps of ''n''/gcd(''n'', ''k'')-ed''E'', and by symmetry, the previous clause also holds when "''M''''k''" and "smaller" are replaced with "''m''''k''" and "larger".

The cover constructed above grants us a stacked pair ''t''0, ''t''1 of unequal ''k''-steps in ''S'' such that δ ∈ [''t''0, ''t''1]. Assume ''t''0 < ''t''1. (If ''t''0 > ''t''1, take equave complements and use the offset ''E'' − δ.) Then the corresponding occurrence of the ''k''-step ''t''0 in ''S''2 is shifted into the closed interval ''I'' corresponding to the ''k''-step ''t''1 in ''S''1. But we then have ''k'' + 1 notes of ''S''2 within ''I''. Assuming none of these notes coincide with a note of ''S''1 (otherwise, interleaving would be violated), each of the ''k'' + 1 notes must fall within one of the ''k'' scale steps subtended by ''t''0 in ''S''1. By the pigeonhole principle, at least one of these steps in ''S''1 must contain two consecutive notes of ''S''2 in its interior, breaking the interleaving condition as desired.}}

== Some flought scales ==

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

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== Proofs ==

=== 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 ''S''1, ''S''2 denote the two copies of ''S'' separated by δ, where ''S''1(0) = '''0''' (the unison), ''S''2(0) = δ. Assume that the scale ''F'' is the union of ''S''1 and ''S''2, and ''F''(0) = '''0'''. Let <math>m_k = \min \mathcal{D}_k(S)</math> and <math>M_k = \max \mathcal{D}_k(S).</math>

Suppose δ > 0 is not in any intervals [''m''''k'', ''M''''k''], 1 ≤ ''k'' ≤ ''n'' − 1, ''n'' = len(''S''). Then for any ''k'', ''S''1(''k'') falls between adjacent notes of ''S''2. The same holds when we reverse the roles of ''S''1 and ''S''2 and use the offset ''E'' − δ; since the union <math>\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''(''n''0), ''S''(''n''0 + ''k'')) (''S''(''n''0 + ''k''), ''S''(''n''0 + 2''k'')), whose sizes ''t''0, ''t''1 are unequal and both contained in [''m''''k'', ''M''''k'']. Moreover, such intervals [''t''0, ''t''1] or [''t''1, ''t''0], taken over all non-ed''E'' circles of ''k''-steps in ''S'', must cover [''m''''k'', ''M''k]. Indeed, if a circle of stacked ''k''-steps in ''S'' has the ''k''-step ''M''''k'', that circle must also have a ''k''-step smaller than ''k''/gcd(''n'', ''k'') steps of ''n''/gcd(''n'', ''k'')-ed''E'', and by symmetry, the previous clause also holds when "''M''''k''" and "smaller" are replaced with "''m''''k''" and "larger".

The cover constructed above grants us a stacked pair ''t''0, ''t''1 of unequal ''k''-steps in ''S'' such that δ ∈ [''t''0, ''t''1]. Assume ''t''0 < ''t''1. (If ''t''0 > ''t''1, take equave complements and use the offset ''E'' − δ.) Then the corresponding occurrence of the ''k''-step ''t''0 in ''S''2 is shifted into the closed interval ''I'' corresponding to the ''k''-step ''t''1 in ''S''1. But we then have ''k'' + 1 notes of ''S''2 within ''I''. Assuming none of these notes coincide with a note of ''S''1 (otherwise, interleaving would be violated), each of the ''k'' + 1 notes must fall within one of the ''k'' scale steps subtended by ''t''0 in ''S''1. By the pigeonhole principle, at least one of these steps in ''S''1 must contain two consecutive notes of ''S''2 in its interior, breaking the interleaving condition as desired. {{qed}}

[[Category:Scale]]

[[Category:Terms]]

[[Category:Articles with proofs]]

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 # Given an E-equivalent scale ''S'', offsets δ within (0, min({step sizes in ''S''})) are called ''small'' in the context of floughtening ''S''. Small offsets are significant because the resulting flought scale has a structure that closely mimics the underlying scale structure: if ''S'' is a circular word <math>w(a_1, a_2, ..., a_n)</math> then Fl(''s''; δ) uses the same circular word but with δ followed by the difference between δ and every step size in w, namely <math>w(\delta b_1, \delta b_2, ..., \delta b_n)</math> where <math>b_i = a_i - \delta</math>.
 # A flought 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 Fl(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|title=Proof of the offset constraints|contents=
+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 ''S''<sub>1</sub>, ''S''<sub>2</sub> denote the two copies of ''S'' separated by δ, where ''S''<sub>1</sub>(0) = '''0''' (the unison), ''S''<sub>2</sub>(0) = δ. Assume that the scale ''F'' is the union of ''S''<sub>1</sub> and ''S''<sub>2</sub>, and ''F''(0) = '''0'''. Let <math>m_k = \min \mathcal{D}_k(S)</math> and <math>M_k = \max \mathcal{D}_k(S).</math>
+Suppose δ > 0 is not in any intervals [''m''<sub>''k''</sub>, ''M''<sub>''k''</sub>], 1 &le; ''k'' &le; ''n'' &minus; 1, ''n'' = len(''S''). Then for any ''k'', ''S''<sub>1</sub>(''k'') falls between adjacent notes of ''S''<sub>2</sub>. The same holds when we reverse the roles of ''S''<sub>1</sub> and ''S''<sub>2</sub> and use the offset ''E'' &minus; δ; since the union <math>\bigcup_{k=1}^{n-1} [m_k, M_k]</math> is invariant under taking equave complements, neither is ''E'' &minus; δ within any [''m''<sub>''k''</sub>, ''M''<sub>''k''</sub>]. The reverse implication follows.
+For the forward implication, we wish to show that the interleaving condition is violated if ''m''<sub>''k''</sub> < ''M''<sub>''k''</sub> and δ ∈ [''m''<sub>''k''</sub>, ''M''<sub>''k''</sub>] for some ''k'', 1 &le; ''k'' &le; ''n'' &minus; 1. We first observe that if ''m''<sub>''k''</sub> < ''M''<sub>''k''</sub>, then ''S'' has some pair of stacked ''k''-steps, say (''S''(''n''<sub>0</sub>),&nbsp;''S''(''n''<sub>0</sub>&nbsp;+&nbsp;''k''))&nbsp;(''S''(''n''<sub>0</sub>&nbsp;+&nbsp;''k''), ''S''(''n''<sub>0</sub>&nbsp;+&nbsp;2''k'')), whose sizes ''t''<sub>0</sub>, ''t''<sub>1</sub> are unequal and both contained in [''m''<sub>''k''</sub>, ''M''<sub>''k''</sub>]. Moreover, such intervals [''t''<sub>0</sub>, ''t''<sub>1</sub>] or [''t''<sub>1</sub>, ''t''<sub>0</sub>], taken over all non-ed''E'' circles of ''k''-steps in ''S'', must cover [''m''<sub>''k''</sub>, ''M''<sub>k</sub>]. Indeed, if a circle of stacked ''k''-steps in ''S'' has the ''k''-step ''M''<sub>''k''</sub>, that circle must also have a ''k''-step smaller than ''k''/gcd(''n'', ''k'') steps of ''n''/gcd(''n'', ''k'')-ed''E'', and by symmetry, the previous clause also holds when "''M''<sub>''k''</sub>" and "smaller" are replaced with "''m''<sub>''k''</sub>" and "larger".
+The cover constructed above grants us a stacked pair ''t''<sub>0</sub>, ''t''<sub>1</sub> of unequal ''k''-steps in ''S'' such that δ ∈ [''t''<sub>0</sub>, ''t''<sub>1</sub>]. Assume ''t''<sub>0</sub> < ''t''<sub>1</sub>. (If ''t''<sub>0</sub> > ''t''<sub>1</sub>, take equave complements and use the offset ''E'' &minus; δ.) Then the corresponding occurrence of the ''k''-step ''t''<sub>0</sub> in ''S''<sub>2</sub> is shifted into the closed interval ''I'' corresponding to the ''k''-step ''t''<sub>1</sub> in ''S''<sub>1</sub>. But we then have ''k'' + 1 notes of ''S''<sub>2</sub> within ''I''. Assuming none of these notes coincide with a note of ''S''<sub>1</sub> (otherwise, interleaving would be violated), each of the ''k'' + 1 notes must fall within one of the ''k'' scale steps subtended by ''t''<sub>0</sub> in ''S''<sub>1</sub>. By the pigeonhole principle, at least one of these steps in ''S''<sub>1</sub> must contain two consecutive notes of ''S''<sub>2</sub> in its interior, breaking the interleaving condition as desired.}}
 == Some flought scales ==
 Flought scales can easily be built from a harmonic series mode as the strand: for example, if ''n''::2''n'' is the strand, then (2''n'' + 1)/''2n'' always works as the offset (e.g. strand 5:6:7:8:9:10, offset 10:11). Here are some other examples:
@@ Line 24: / Line 33: @@
 == Proofs ==
 === 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 ''S''<sub>1</sub>, ''S''<sub>2</sub> denote the two copies of ''S'' separated by δ, where ''S''<sub>1</sub>(0) = '''0''' (the unison), ''S''<sub>2</sub>(0) = δ. Assume that the scale ''F'' is the union of ''S''<sub>1</sub> and ''S''<sub>2</sub>, and ''F''(0) = '''0'''. Let <math>m_k = \min \mathcal{D}_k(S)</math> and <math>M_k = \max \mathcal{D}_k(S).</math>
-Suppose δ > 0 is not in any intervals [''m''<sub>''k''</sub>, ''M''<sub>''k''</sub>], 1 &le; ''k'' &le; ''n'' &minus; 1, ''n'' = len(''S''). Then for any ''k'', ''S''<sub>1</sub>(''k'') falls between adjacent notes of ''S''<sub>2</sub>. The same holds when we reverse the roles of ''S''<sub>1</sub> and ''S''<sub>2</sub> and use the offset ''E'' &minus; δ; since the union <math>\bigcup_{k=1}^{n-1} [m_k, M_k]</math> is invariant under taking equave complements, neither is ''E'' &minus; δ within any [''m''<sub>''k''</sub>, ''M''<sub>''k''</sub>]. The reverse implication follows.
-For the forward implication, we wish to show that the interleaving condition is violated if ''m''<sub>''k''</sub> < ''M''<sub>''k''</sub> and δ ∈ [''m''<sub>''k''</sub>, ''M''<sub>''k''</sub>] for some ''k'', 1 &le; ''k'' &le; ''n'' &minus; 1. We first observe that if ''m''<sub>''k''</sub> < ''M''<sub>''k''</sub>, then ''S'' has some pair of stacked ''k''-steps, say (''S''(''n''<sub>0</sub>),&nbsp;''S''(''n''<sub>0</sub>&nbsp;+&nbsp;''k''))&nbsp;(''S''(''n''<sub>0</sub>&nbsp;+&nbsp;''k''), ''S''(''n''<sub>0</sub>&nbsp;+&nbsp;2''k'')), whose sizes ''t''<sub>0</sub>, ''t''<sub>1</sub> are unequal and both contained in [''m''<sub>''k''</sub>, ''M''<sub>''k''</sub>]. Moreover, such intervals [''t''<sub>0</sub>, ''t''<sub>1</sub>] or [''t''<sub>1</sub>, ''t''<sub>0</sub>], taken over all non-ed''E'' circles of ''k''-steps in ''S'', must cover [''m''<sub>''k''</sub>, ''M''<sub>k</sub>]. Indeed, if a circle of stacked ''k''-steps in ''S'' has the ''k''-step ''M''<sub>''k''</sub>, that circle must also have a ''k''-step smaller than ''k''/gcd(''n'', ''k'') steps of ''n''/gcd(''n'', ''k'')-ed''E'', and by symmetry, the previous clause also holds when "''M''<sub>''k''</sub>" and "smaller" are replaced with "''m''<sub>''k''</sub>" and "larger".
-The cover constructed above grants us a stacked pair ''t''<sub>0</sub>, ''t''<sub>1</sub> of unequal ''k''-steps in ''S'' such that δ ∈ [''t''<sub>0</sub>, ''t''<sub>1</sub>]. Assume ''t''<sub>0</sub> < ''t''<sub>1</sub>. (If ''t''<sub>0</sub> > ''t''<sub>1</sub>, take equave complements and use the offset ''E'' &minus; δ.) Then the corresponding occurrence of the ''k''-step ''t''<sub>0</sub> in ''S''<sub>2</sub> is shifted into the closed interval ''I'' corresponding to the ''k''-step ''t''<sub>1</sub> in ''S''<sub>1</sub>. But we then have ''k'' + 1 notes of ''S''<sub>2</sub> within ''I''. Assuming none of these notes coincide with a note of ''S''<sub>1</sub> (otherwise, interleaving would be violated), each of the ''k'' + 1 notes must fall within one of the ''k'' scale steps subtended by ''t''<sub>0</sub> in ''S''<sub>1</sub>. By the pigeonhole principle, at least one of these steps in ''S''<sub>1</sub> must contain two consecutive notes of ''S''<sub>2</sub> in its interior, breaking the interleaving condition as desired. {{qed}}
 [[Category:Scale]]
 [[Category:Terms]]
 [[Category:Articles with proofs]]