Generator-offset property: Difference between revisions

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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-degenerate AG scale.

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

# ''a''4 − ''a''1 = ''g''3 − ''g''2 = (''g''3 + ''g''1) − (''g''2 + ''g''1) != 0. This is exactly the chroma of the mos generated by ''g''0.

# ''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-length, 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''. This proves (3).

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-length, unconditionally SV3, GO scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally SV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''. This proves (3).

In case 2, let (2, 1) − (1, 1) = ''g''1, (1, 2) − (2, 1) = ''g''2 be the two alternants. 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 combinations of alternants corresponding to a step are:

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 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-degenerate AG scale.
+# ''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-degenerate GO scale.
 # ''a''<sub>4</sub> &minus; ''a''<sub>1</sub> = ''g''<sub>3</sub> &minus; ''g''<sub>2</sub> = (''g''<sub>3</sub> + ''g''<sub>1</sub>) &minus; (''g''<sub>2</sub> + ''g''<sub>1</sub>) != 0. This is exactly the chroma of the mos generated by ''g''<sub>0</sub>.
 # ''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-length, 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''. This proves (3).
+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-length, unconditionally SV3, GO scale must be of the form ''xy...xyxz''. But this pattern is not unconditionally SV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''. This proves (3).
 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 alternants. 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 combinations of alternants corresponding to a step are: