Rank-3 scale theorems: Difference between revisions

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====== PMOS implies AG (except in the case xyxzxyx) (WIP) ======

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~~==== AG + unconditionally MV3 implies "ax by bz" and that the scale's cardinality is odd or 4 ====~~

~~'''Assuming both AG and unconditional MV3''', we have two chains of generator g0 (going right). The two cases are:~~

~~O-O-...-O (m notes)~~

~~and~~

~~O-O-O-...-O (m notes)~~

~~O-O-...-O (m-1 notes).~~

~~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 (even scale size n = 2^t r where r is odd), let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). We have the chain g1 g2 g1 g2... g1 g3. Suppose the k-step is the class generated by r generators (which is an odd number of generator steps):

~~# from g1 ... g1, get a1 = (r-1)/2*g0 + g1 = (r+1)/2 g1 + (r-1)/2 g2~~

~~# from g2 ... g2, get a2 = (r-1)/2*g0 + g2 = (r-1)/2 g1 + (r+1)/2 g2~~

~~# from g2 (even) g1 g3 g1 (even) g2, get a3 = (r-1)/2 g1 + (r-1)/2 g2 + g3~~

~~# from g1 (odd) g1 g3 g1 (odd) g1, get a4 = (r+1)/2 g1 + (r-3)/2 g2 + g3.~~

Choose a tuning where g1 and g2 are both very close to but not exactly 1/2*g0, resulting in a scale very close to the mos generated by 1/2 g0. (i.e. g1 and g2 differ from 1/2*g0 by ε, a quantity much smaller than the chroma of the n/2-note mos generated by g0, which is |g3 - g2|). Assuming n > 4, we have 4 distinct sizes for k-steps, a contradiction to unconditional-MV3:

~~# a1, a2 and a3 are clearly distinct.~~

~~# a4 - a3 = g1 - g2 != 0, since the scale is a non-trivial AG.~~

~~# a4 - a1 = g3 - g2 = (g3 + g1) - (g2 + g1) != 0. This is exactly the chroma of the mos generated by g0.~~

~~# a4 - a2 = g1 - 2 g2 + g3 = (g3 - g2) + (g1 - g2) = (chroma ± ε) != 0 by choice of tuning.~~

~~(For n = 4, the above argument doesn't work because a3 = a4, and xyxz is a counterexample.)~~

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:

~~# k g1 + (k-1) g2~~

~~# (k-1) g1 + k g2~~

~~# (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.

~~(The above holds for any odd n >= 3.)~~

~~This proof shows that AG and unconditionally-MV3 scales must have cardinality odd or 4.~~

~~==== An AG scale is unconditionally MV3 iff its cardinality is odd or 4 ====~~

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

~~==== An even-cardinality unconditional MV3 is of the form W(x,y,z)W(y,x,z) (WIP) ====~~

~~==== 3-DE implies MV3 (WIP) ====~~

~~We prove that 3-DE + not abcba implies PMOS, which is known to imply MV3.~~

[[Category:Fokker block]]

@@ Line 101: / Line 101: @@
 ====== PMOS implies AG (except in the case xyxzxyx) (WIP) ======
 -->
-==== AG + unconditionally MV3 implies "ax by bz" and that the scale's cardinality is odd or 4 ====
-'''Assuming both AG and unconditional MV3''', we have two chains of generator g0 (going right). The two cases are:
- O-O-...-O (m notes)
- O-O-...-O (m notes)
-and
- O-O-O-...-O (m notes)
- O-O-...-O (m-1 notes).
-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 (even scale size n = 2^t r where r is odd), let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). We have the chain g1 g2 g1 g2... g1 g3. Suppose the k-step is the class generated by r generators (which is an odd number of generator steps):
-# from g1 ... g1, get a1 = (r-1)/2*g0 + g1 = (r+1)/2 g1 + (r-1)/2 g2
-# from g2 ... g2, get a2 = (r-1)/2*g0 + g2 = (r-1)/2 g1 + (r+1)/2 g2
-# from g2 (even) g1 g3 g1 (even) g2, get a3 = (r-1)/2 g1 + (r-1)/2 g2 + g3
-# from g1 (odd) g1 g3 g1 (odd) g1, get a4 = (r+1)/2 g1 + (r-3)/2 g2 + g3.
-Choose a tuning where g1 and g2 are both very close to but not exactly 1/2*g0, resulting in a scale very close to the mos generated by 1/2 g0. (i.e. g1 and g2 differ from 1/2*g0 by ε, a quantity much smaller than the chroma of the n/2-note mos generated by g0, which is |g3 - g2|). Assuming n > 4, we have 4 distinct sizes for k-steps, a contradiction to unconditional-MV3:
-# a1, a2 and a3 are clearly distinct.
-# a4 - a3 = g1 - g2 != 0, since the scale is a non-trivial AG.
-# a4 - a1 = g3 - g2 = (g3 + g1) - (g2 + g1) != 0. This is exactly the chroma of the mos generated by g0.
-# a4 - a2 = g1 - 2 g2 + g3 = (g3 - g2) + (g1 - g2) = (chroma ± ε) != 0 by choice of tuning.
-(For n = 4, the above argument doesn't work because a3 = a4, and xyxz is a counterexample.)
-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:
-# k g1 + (k-1) g2
-# (k-1) g1 + k g2
-# (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.
-(The above holds for any odd n >= 3.)
-This proof shows that AG and unconditionally-MV3 scales must have cardinality odd or 4.
-==== An AG scale is unconditionally MV3 iff its cardinality is odd or 4 ====
-We only need to see that AG + odd cardinality => MV3. But the argument in case 2 above works for any interval class (MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.
-==== An even-cardinality unconditional MV3 is of the form W(x,y,z)W(y,x,z) (WIP) ====
-==== 3-DE implies MV3 (WIP) ====
-We prove that 3-DE + not abcba implies PMOS, which is known to imply MV3.
 [[Category:Fokker block]]