Interleaving: Difference between revisions

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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 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''~~. Assume~~ ''t''0 < ''t''1 ~~and δ ∈ [~~''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, 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, 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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 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 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''. Assume ''t''<sub>0</sub> < ''t''<sub>1</sub> and δ ∈ [''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>, 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>, 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 ==