Generator-offset property: Difference between revisions

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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''.

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 ~~generators~~ corresponding to a step are:

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:

# ''kg''1 + (''k'' − 1)''g''2

# (''k'' − 1)''g''1 + ''kg''2

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 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''.
-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 generators corresponding to a step are:
+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:
 # ''kg''<sub>1</sub> + (''k'' &minus; 1)''g''<sub>2</sub>
 # (''k'' &minus; 1)''g''<sub>1</sub> + ''kg''<sub>2</sub>