Generator-offset property: Difference between revisions

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These are all distinct by '''Z'''-linear independence. By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g1 and g2 must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form (xy)''r''xz. (Note that (xy)''r''xz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).

In case 2, let (2, 1) − (1, 1) = g1, (1, 2) − (2, 1) = g2 be the two alternants. Let g3 be the leftover generator after stacking alternating g1 and g2. Then the generator circle looks like g1 g2 g1 g2 ... g1 g2 g3. ~~Then~~ the combinations of alternants corresponding to a step come in exactly 3 sizes:

In case 2, let (2, 1) − (1, 1) = g1, (1, 2) − (2, 1) = g2 be the two alternants. Let g3 be the leftover generator after stacking alternating g1 and g2. Then the generator circle looks like g1 g2 g1 g2 ... g1 g2 g3. Assuming that a step is an odd number of generators, the combinations of alternants corresponding to a step come in exactly 3 sizes:

# ''k''g1 + (''k'' − 1)g2

# (''k'' − 1)g1 + ''k''g2

# (''k'' − 1)g1 + (''k'' − 1) g2 + g3,

# (''k'' − 1)g1 + (''k'' − 1) g2 + g3

~~if a step is an odd number of generators~~ (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times. This proves (2).

(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times. This proves (2).

(The above holds for any odd ''n'' ≥ 3.)

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 These are all distinct by '''Z'''-linear independence. 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)<sup>''r''</sup>xz. (Note that (xy)<sup>''r''</sup>xz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).
-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 come in exactly 3 sizes:
+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>. Assuming that a step is an odd number of generators, the combinations of alternants corresponding to a step come in exactly 3 sizes:
 # ''k''g<sub>1</sub> + (''k'' &minus; 1)g<sub>2</sub>
 # (''k'' &minus; 1)g<sub>1</sub> + ''k''g<sub>2</sub>
-# (''k'' &minus; 1)g<sub>1</sub> + (''k'' &minus; 1) g<sub>2</sub> + g<sub>3</sub>,
+# (''k'' &minus; 1)g<sub>1</sub> + (''k'' &minus; 1) g<sub>2</sub> + g<sub>3</sub>
-if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times. This proves (2).
+(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times. This proves (2).
 (The above holds for any odd ''n'' ≥ 3.)