Generator-offset property: Difference between revisions

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O-O-...-O ((n-1)/2 notes).

Label the notes (1, ''k'') and (2, ''k''), 1 ≤ ''k'' ≤ (chain length), for notes in the upper and lower chain respectively.

Label the notes (1, ''j'') and (2, ''j''), 1 ≤ ''j'' ≤ (chain length), for notes in the upper and lower chain respectively.

In case 1, let ''g''1 = (2, 1) − (1, 1) and ''g''2 = (1, 2) − (2, 1). We have the chain ''g''1 ''g''2 ''g''1 ''g''2 ... ''g''1 ''g''3.

Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class reached by stacking ''r'' generators:

Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class (''k''-steps) reached by stacking ''r'' generators:

# from ''g''1 ... ''g''1, we get ''a''1 = (''r'' − 1)/2*''g''0 + ''g''1 = (''r'' + 1)/2 ''g''1 + (''r'' − 1)/2 ''g''2

# from ''g''2 ... ''g''2, we get ''a''2 = (''r'' − 1)/2*''g''0 + ''g''2 = (''r'' − 1)/2 ''g''1 + (''r'' + 1)/2 ''g''2

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# from ''g''1 (...odd # of gens...) ''g''1 ''g''3 ''g''1 (...odd # of gens...) ''g''1, we get ''a''4 = (''r'' + 1)/2 ''g''1 + (''r'' − 3)/2 ''g''2 + ''g''3.

Choose a tuning where ''g''1 and ''g''2 are both very close to but not exactly 1/2*''g''0, resulting in a scale very close to the mos generated by 1/2 ''g''0. (i.e. ''g''1 and ''g''2 differ from 1/2*''g''0 by ε, a quantity much smaller than the chroma of the ''n''/2-note mos generated by ''g''0, which is |''g''3 − ''g''2|). Thus we have 4 distinct sizes for k-steps:

Choose a tuning where ''g''1 and ''g''2 are both very close to but not exactly 1/2*''g''0, resulting in a scale very close to the mos generated by 1/2 ''g''0. (i.e. ''g''1 and ''g''2 differ from 1/2*''g''0 by ε, a quantity much smaller than the chroma of the ''n''/2-note mos generated by ''g''0, which is |''g''3 − ''g''2|). Thus we have 4 distinct sizes for ''k''-steps:

# ''a''1, ''a''2 and ''a''3 are clearly distinct.

# ''a''4 − ''a''3 = ''g''1 − ''g''2 != 0, since the scale is a non-trivial AG.

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# ''a''4 − ''a''2 = ''g''1 − 2 ''g''2 + ''g''3 = (''g''3 − ''g''2) + (''g''1 − ''g''2) = (chroma ± ε) != 0 by choice of tuning.

By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''1 and ''g''2 must themselves be step sizes. Thus we see that an even-cardinality, unconditionally MV3, AG scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally MV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and

By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''1 and ''g''2 must themselves be step sizes. Thus we see that an even-cardinality, unconditionally MV3, AG scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally MV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''.

''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''.

In case 2, let (2, 1) − (1, 1) = ''g''1, (1, 2) − (2, 1) = ''g''2 be the two alternating generators. Let ''g''3 be the leftover generator after stacking alternating ''g''1 and ''g''2. Then the generator circle looks like ''g''1 ''g''2 ''g''1 ''g''2 ... ''g''1 ''g''2 ''g''3. Then the generators corresponding to a step are:

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Now we only need to see that AG + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.

== Conjectures ==

=== Conjecture 2 ===

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   O-O-...-O ((n-1)/2 notes).
-Label the notes (1, ''k'') and (2, ''k''), 1 ≤ ''k'' ≤ (chain length), for notes in the upper and lower chain respectively.
+Label the notes (1, ''j'') and (2, ''j''), 1 ≤ ''j'' ≤ (chain length), for notes in the upper and lower chain respectively.
 In case 1, let ''g''<sub>1</sub> = (2, 1) &minus; (1, 1) and ''g''<sub>2</sub> = (1, 2) &minus; (2, 1). We have the chain ''g''<sub>1</sub> ''g''<sub>2</sub> ''g''<sub>1</sub> ''g''<sub>2</sub> ... ''g''<sub>1</sub> ''g''<sub>3</sub>.
-Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class reached by stacking ''r'' generators:
+Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class (''k''-steps) reached by stacking ''r'' generators:
 # from ''g''<sub>1</sub> ... ''g''<sub>1</sub>, we get ''a''<sub>1</sub> = (''r'' &minus; 1)/2*''g''<sub>0</sub> + ''g''<sub>1</sub> = (''r'' + 1)/2 ''g''<sub>1</sub> + (''r'' &minus; 1)/2 ''g''<sub>2</sub>
 # from ''g''<sub>2</sub> ... ''g''<sub>2</sub>, we get ''a''<sub>2</sub> = (''r'' &minus; 1)/2*''g''<sub>0</sub> + ''g''<sub>2</sub> = (''r'' &minus; 1)/2 ''g''<sub>1</sub> + (''r'' + 1)/2 ''g''<sub>2</sub>
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 # from ''g''<sub>1</sub> (...odd # of gens...) ''g''<sub>1</sub> ''g''<sub>3</sub> ''g''<sub>1</sub> (...odd # of gens...) ''g''<sub>1</sub>, we get ''a''<sub>4</sub> = (''r'' + 1)/2 ''g''<sub>1</sub> + (''r'' &minus; 3)/2 ''g''<sub>2</sub> + ''g''<sub>3</sub>.
-Choose a tuning where ''g''<sub>1</sub> and ''g''<sub>2</sub> are both very close to but not exactly 1/2*''g''<sub>0</sub>, resulting in a scale very close to the mos generated by 1/2 ''g''<sub>0</sub>. (i.e. ''g''<sub>1</sub> and ''g''<sub>2</sub> differ from 1/2*''g''<sub>0</sub> by ε, a quantity much smaller than the chroma of the ''n''/2-note mos generated by ''g''<sub>0</sub>, which is |''g''<sub>3</sub> &minus; ''g''<sub>2</sub>|). Thus we have 4 distinct sizes for k-steps:
+Choose a tuning where ''g''<sub>1</sub> and ''g''<sub>2</sub> are both very close to but not exactly 1/2*''g''<sub>0</sub>, resulting in a scale very close to the mos generated by 1/2 ''g''<sub>0</sub>. (i.e. ''g''<sub>1</sub> and ''g''<sub>2</sub> differ from 1/2*''g''<sub>0</sub> by ε, a quantity much smaller than the chroma of the ''n''/2-note mos generated by ''g''<sub>0</sub>, which is |''g''<sub>3</sub> &minus; ''g''<sub>2</sub>|). Thus we have 4 distinct sizes for ''k''-steps:
 # ''a''<sub>1</sub>, ''a''<sub>2</sub> and ''a''<sub>3</sub> are clearly distinct.
 # ''a''<sub>4</sub> &minus; ''a''<sub>3</sub> = ''g''<sub>1</sub> &minus; ''g''<sub>2</sub> != 0, since the scale is a non-trivial AG.
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 # ''a''<sub>4</sub> &minus; ''a''<sub>2</sub> = ''g''<sub>1</sub> &minus; 2 ''g''<sub>2</sub> + ''g''<sub>3</sub> = (''g''<sub>3</sub> &minus; ''g''<sub>2</sub>) + (''g''<sub>1</sub> &minus; ''g''<sub>2</sub>) = (chroma ± ε) != 0 by choice of tuning.
-By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''<sub>1</sub> and ''g''<sub>2</sub> must themselves be step sizes. Thus we see that an even-cardinality, unconditionally MV3, AG scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally MV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and
+By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''<sub>1</sub> and ''g''<sub>2</sub> must themselves be step sizes. Thus we see that an even-cardinality, unconditionally MV3, AG scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally MV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''.
-''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''.
 In case 2, let (2, 1) &minus; (1, 1) = ''g''<sub>1</sub>, (1, 2) &minus; (2, 1) = ''g''<sub>2</sub> be the two alternating generators. Let ''g''<sub>3</sub> be the leftover generator after stacking alternating ''g''<sub>1</sub> and ''g''<sub>2</sub>. Then the generator circle looks like ''g''<sub>1</sub> ''g''<sub>2</sub> ''g''<sub>1</sub> ''g''<sub>2</sub> ... ''g''<sub>1</sub> ''g''<sub>2</sub> ''g''<sub>3</sub>. Then the generators corresponding to a step are:
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 Now we only need to see that AG + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.
 == Conjectures ==
 === Conjecture 2 ===