Interleaving: Difference between revisions
| Line 17: | Line 17: | ||
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'' − 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'' − 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. | ||
For a polyoffset, 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 completes the proof. | |||
== Some flought scales == | == Some flought scales == | ||