Interleaving: Difference between revisions

Line 26:

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 ~~zero interval~~), ''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>

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.

@@ Line 26: / Line 26: @@
 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 zero interval), ''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>
+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.