Generator-offset property: Difference between revisions

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== Theorems ==

=== ~~Theorem~~ 1 ===

=== Proposition 1 ===

Let ''S'' be a 3-step-size scale word in L, M, and s, and suppose ''S'' is SGA. Then:

# ''S'' is unconditionally MV3 (i.e. MV3 regardless of tuning).

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Now we only need to see that SGA + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning. <math>\square</math>

=== Proposition 2 ===

Suppose that a periodic scale satisfies the following:

* is generator-offset (chain lengths off by 1)

* has odd length ''n''

* all instances of the generator are ''k''-steps for a fixed k.

Then the scale is SGA.

==== Proof ====

Assume that ''k'' is even. (If ''k'' is not even, invert the generator.) On some tonic p we have a chain of ceil(''n''/2) generators and on some other note ''p''' (not on the first chain) we'll have floor(''n''/2) generators. We have ''k'' ceil(''n''/2) notes on ''p'' and ''k'' floor(''n''/2) notes on ''p''' = ''p'' + offset.

We must have gcd(''k'', ''n'') = 1. If not, since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and the ''k''-steps must form more than 2 parallel chains.

By modular arithmetic we have rk mod n = k/2 iff r = ceil(n/2) mod n. (Since gcd(2, ''n'') = 1, 2 is multiplicatively invertible mod ''n'', and we can multiply both sides by 2 to check this claim: we get (''n'' + 1)k = ''nk + k = k'' (mod ''n''), which is true.) This proves that the offset, which must be reached after ceil(''n''/2) generator steps, is a k/2 step (If the offset wasn't reached in ceil(n/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) <math>\square</math>

== Open conjectures ==

@@ Line 15: / Line 15: @@
 == Theorems ==
-=== Theorem 1 ===
+=== Proposition 1 ===
 Let ''S'' be a 3-step-size scale word in L, M, and s, and suppose ''S'' is SGA. Then:
 # ''S'' is unconditionally MV3 (i.e. MV3 regardless of tuning).
@@ Line 59: / Line 59: @@
 Now we only need to see that SGA + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning. <math>\square</math>
+=== Proposition 2 ===
+Suppose that a periodic scale satisfies the following:
+* is generator-offset (chain lengths off by 1)
+* has odd length ''n''
+* all instances of the generator are ''k''-steps for a fixed k.
+Then the scale is SGA.
+==== Proof ====
+Assume that ''k'' is even. (If ''k'' is not even, invert the generator.) On some tonic p we have a chain of ceil(''n''/2) generators and on some other note ''p''' (not on the first chain) we'll have floor(''n''/2) generators. We have ''k'' ceil(''n''/2) notes on ''p'' and ''k'' floor(''n''/2) notes on ''p''' = ''p'' + offset.
+We must have gcd(''k'', ''n'') = 1. If not, since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and the ''k''-steps must form more than 2 parallel chains.
+By modular arithmetic we have rk mod n = k/2 iff r = ceil(n/2) mod n. (Since gcd(2, ''n'') = 1, 2 is multiplicatively invertible mod ''n'', and we can multiply both sides by 2 to check this claim: we get (''n'' + 1)k = ''nk + k = k'' (mod ''n''), which is true.) This proves that the offset, which must be reached after ceil(''n''/2) generator steps, is a k/2 step (If the offset wasn't reached in ceil(n/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) <math>\square</math>
 == Open conjectures ==