Rank-3 scale theorems: Difference between revisions
→AG implies "ax by bz": Blackdye (LSLMLSLMLS) is a counterexample; it's an example of how the tacit assumption that the generator's size in scale steps is always coprime to the scale size can fail. |
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====== PMOS implies AG (except in the case xyxzxyx) (WIP) ====== | ====== PMOS implies AG (except in the case xyxzxyx) (WIP) ====== | ||
====== AG implies "ax by bz" ====== | ====== AG implies "ax by bz" under certain conditions ====== | ||
AG does ''not'' imply "ax by bz"; blackdye (LSLMLSLMLS) is a counterexample. | |||
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'''Assuming the alternating generator property''', we have two chains of generator g0 (going right). The two cases are: | '''Assuming the alternating generator property''', we have two chains of generator g0 (going right). The two cases are: | ||
O-O-...-O (m notes) | O-O-...-O (m notes) | ||
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Label the notes (1,k) and (2,k), 1 ≤ k ≤ m or m-1, for notes in the upper and lower chain respectively. | Label the notes (1,k) and (2,k), 1 ≤ k ≤ m or m-1, for notes in the upper and lower chain respectively. | ||
In case 1, let g1 = (2,1)-(1,m) and g2 = (1,1)-(2,m). The circle of stacked g0 generators is (starting from top left): (m-1 g0's) g1 (m-1 g0's) g2. A scale step is always a same number k (which must be odd) of such generators gi. Assume (after taking octave complement) that a single step takes less than half of the generators. So a word corresponding to the scale step is formed by one of: | <s>In case 1, let g1 = (2,1)-(1,m) and g2 = (1,1)-(2,m). The circle of stacked g0 generators is (starting from top left): (m-1 g0's) g1 (m-1 g0's) g2. A scale step is always a same number k (which must be odd) of such generators gi. Assume (after taking octave complement) that a single step takes less than half of the generators. So a word corresponding to the scale step is formed by one of: | ||
* k g0 | * k g0 | ||
* (k-1) g0 + g1 | * (k-1) g0 + g1 | ||
* (k-1) g0 + g2. | * (k-1) g0 + g2. | ||
It is clear that the last two sizes must occur the same number of times. | It is clear that the last two sizes must occur the same number of times.</s> | ||
In case 2, let (2,1)-(1,1) = g1, (1,2)-(2,1) = g2 be the two alternating generators. 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 generators corresponding to a step are: | In case 2, let (2,1)-(1,1) = g1, (1,2)-(2,1) = g2 be the two alternating generators. 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 generators corresponding to a step are: | ||
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* (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. QED. | 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. QED. | ||
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==== 3-DE implies MV3 (WIP) ==== | ==== 3-DE implies MV3 (WIP) ==== | ||