Generator-offset property: Difference between revisions

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A scale satisfies the '''alternating generator property''', or the '''AG''' property for short, if it satisfies the following equivalent properties:

* the scale can be built by stacking alternating generators, for example 7/6 and 8/7.

* the scale is generated by two chains of generators separated by a fixed interval; either both chains are of size m, or one chain has size m and the second has size m-1.

More formally, a cyclic word S (representing a [[periodic scale]]) is AG if it satisfies the following equivalent properties:

# S can be built by stacking a single chain of alternating generators g1 and g2, resulting in a circle of the form either g1 g2 ... g1 g2 g1 g3 or g1 g2 ... g1 g2 g3.

# S is generated by two chains of generators separated by a fixed interval; either both chains are of size m, or one chain has size m and the second has size m-1.

These are equivalent, since the separating interval can be taken to be g1 and the generator of each chain = g1 + g2.

== Theorems ==

'''Theorem 1''' If a 3-step-size scale word ''S'' in L, M, and s is both AG and unconditionally MV3, then the scale is of the form ax by bz for (x,y,z) some permutation of (L, M, s); and the scale's cardinality is either odd, or 4 (and is of the form xyxz). Moreover, any odd-cardinality AG scale is MV3.

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

=== Proof of Thm 1 ===

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

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.

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)~~:

In case 1, let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). We have the chain g1 g2 g1 g2... g1 g3.

# 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

Let r be odd and r >= 3. Consider the following abstract sizes for the interval class reached by stacking r generators:

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

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

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

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

# from g2 (...even # of gens...) g1 g3 g1 (...even # of gens...) g2, we get a3 = (r-1)/2 g1 + (r-1)/2 g2 + g3

# from g1 (...odd # of gens...) g1 g3 g1 (...odd # of gens...) g1, we 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:

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# 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~~.)

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-cardinality, MV3, AG scale must be of the form xy...xyxz. But this pattern is not MV3 if n >=6, 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) = 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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(The above holds for any odd n >= 3.)

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

Now 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 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) ===~~

[[Category:Theory]]

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

[[Category:AG scales| ]]

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+A scale satisfies the '''alternating generator property''', or the '''AG''' property for short, if it satisfies the following equivalent properties:
+* the scale can be built by stacking alternating generators, for example 7/6 and 8/7.
+* the scale is generated by two chains of generators separated by a fixed interval; either both chains are of size m, or one chain has size m and the second has size m-1.
+More formally, a cyclic word S (representing a [[periodic scale]]) is AG if it satisfies the following equivalent properties:
+# S can be built by stacking a single chain of alternating generators g1 and g2, resulting in a circle of the form either g1 g2 ... g1 g2 g1 g3 or g1 g2 ... g1 g2 g3.
+# S is generated by two chains of generators separated by a fixed interval; either both chains are of size m, or one chain has size m and the second has size m-1.
+These are equivalent, since the separating interval can be taken to be g1 and the generator of each chain = g1 + g2.
 == Theorems ==
+'''Theorem 1''' If a 3-step-size scale word ''S'' in L, M, and s is both AG and unconditionally MV3, then the scale is of the form ax by bz for (x,y,z) some permutation of (L, M, s); and the scale's cardinality is either odd, or 4 (and is of the form xyxz). Moreover, any odd-cardinality AG scale is MV3.
-=== AG + unconditionally MV3 implies "ax by bz" and that the scale's cardinality is odd or 4 ===
+=== Proof of Thm 1 ===
 '''Assuming both AG and unconditional MV3''', we have two chains of generator g0 (going right). The two cases are:
   O-O-...-O (m notes)
@@ Line 11: / Line 21: @@
 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):
+In case 1, let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). We have the chain g1 g2 g1 g2... g1 g3.
-# 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
+Let r be odd and r >= 3. Consider the following abstract sizes for the interval class reached by stacking r generators:
-# from g2 (even) g1 g3 g1 (even) g2, get a3 = (r-1)/2 g1 + (r-1)/2 g2 + g3
+# from g1 ... g1, we get a1 = (r-1)/2*g0 + g1 = (r+1)/2 g1 + (r-1)/2 g2
-# from g1 (odd) g1 g3 g1 (odd) g1, get a4 = (r+1)/2 g1 + (r-3)/2 g2 + g3.
+# from g2 ... g2, we get a2 = (r-1)/2*g0 + g2 = (r-1)/2 g1 + (r+1)/2 g2
+# from g2 (...even # of gens...) g1 g3 g1 (...even # of gens...) g2, we get a3 = (r-1)/2 g1 + (r-1)/2 g2 + g3
+# from g1 (...odd # of gens...) g1 g3 g1 (...odd # of gens...) g1, we 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:
@@ Line 23: / Line 35: @@
 # 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.)
+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-cardinality, MV3, AG scale must be of the form xy...xyxz. But this pattern is not MV3 if n >=6, 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) = 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:
@@ Line 33: / Line 45: @@
 (The above holds for any odd n >= 3.)
-This proof shows that AG and unconditionally-MV3 scales must have cardinality odd or 4.
+Now 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 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) ===
+[[Category:Theory]]
-We prove that 3-DE + not abcba implies PMOS, which is known to imply MV3.
+[[Category:AG scales| ]]<!--Main article-->