Interleaving: Difference between revisions

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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 offsets. 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 ~~''F''(0) =~~ ''S''1(0) = 1/1, ''S''2(0) = δ ~~and~~ the scale ''F'' is the union of ''S''1 and ''S''2. Let <math>m_k = \min \mathcal{D}_k(S)</math> and <math>M_k = \max \mathcal{D}_k(S).</math>

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

Suppose δ 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 of the [''m''''k'', ''M''''k''] is invariant under taking equave complements, neither is ''E'' − δ within any [''m''''k'', ''M''''k'']. The reverse implication follows.

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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 assert that if this holds, 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'']. This is because such intervals [''t''0, ''t''1] or [''t''1, ''t''0] 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".

Now assume 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 (~~which~~ would ~~violate interleaving~~), 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.

Now assume 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 ==

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 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 offsets. 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 ''F''(0) = ''S''<sub>1</sub>(0) = 1/1, ''S''<sub>2</sub>(0) = δ and the scale ''F'' is the union of ''S''<sub>1</sub> and ''S''<sub>2</sub>. Let <math>m_k = \min \mathcal{D}_k(S)</math> and <math>M_k = \max \mathcal{D}_k(S).</math>
+Let ''S''<sub>1</sub>, ''S''<sub>2</sub> denote the two copies of ''S'' separated by δ, where ''S''<sub>1</sub>(0) = 1/1, ''S''<sub>2</sub>(0) = δ, the scale ''F'' is the union of ''S''<sub>1</sub> and ''S''<sub>2</sub>, and ''F''(0) = 1/1. Let <math>m_k = \min \mathcal{D}_k(S)</math> and <math>M_k = \max \mathcal{D}_k(S).</math>
 Suppose δ 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 of the [''m''<sub>''k''</sub>, ''M''<sub>''k''</sub>] 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.
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 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 assert that if this holds, 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>]. This is because such intervals [''t''<sub>0</sub>, ''t''<sub>1</sub>] or [''t''<sub>1</sub>, ''t''<sub>0</sub>] 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".
-Now assume 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> (which would violate interleaving), 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.
+Now assume 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 ==