Generator-offset property: Difference between revisions

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=== Theorem 4 (PWF implies SV3 and either GO or abacaba) ===

Let S(a,b,c) be a scale word in three step sizes a, b, c. Suppose S is PWF. Then S is SV3 and has an odd number of notes. Moreover, S is either GO or equivalent to the scale word abacaba.

Let S(a,b,c) be a scale word in three abstract (can be thought as equuvalent to Z-linearly independent) step sizes a, b, c. Suppose S is PWF. Then S is SV3 and has an odd number of notes. Moreover, S is either GO or equivalent to the scale word abacaba.

==== Proof ====

Suppose S has n notes (after dealing with small cases, we may assume n >= 7) and S projects to single-period mosses S1 (via identifying b ~ c), S2 (via identifying a ~ c) and S3 (via identifying a ~ b). Suppose S1's generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S2 and S3, respectively. These two-"step-size" words, which we call Σ2 and Σ3, must be mosses, since every m-step in the new words coresponds to mk-steps in the mos words S1 and S2, which come in at most two sizes. Both Σ2 and Σ3 are single-period mosses, since (k, n) = 1.

index: 1 2 3 4 ... n

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These give exactly three distinct sizes for every interval class. Hence S is SV3.

In this case S has two chains of Qs, one with floor(n/2) notes and and one with ceil(n/2) notes. Thus S also satisfies the generator-offset property with generator Q.

In this case S has two chains of Qs, one with floor(n/2) notes and and one with ceil(n/2) notes. Every instance of Q must be a k-step, since Q = αa + βb + γc is the only way to write Q in the basis (a, b, c). Thus S also satisfies the generator-offset property with generator Q.

End Case 1.]

@@ Line 114: / Line 114: @@
 === Theorem 4 (PWF implies SV3 and either GO or abacaba) ===
-Let S(a,b,c) be a scale word in three step sizes a, b, c. Suppose S is PWF. Then S is SV3 and has an odd number of notes. Moreover, S is either GO or equivalent to the scale word abacaba.
+Let S(a,b,c) be a scale word in three abstract (can be thought as equuvalent to Z-linearly independent) step sizes a, b, c. Suppose S is PWF. Then S is SV3 and has an odd number of notes. Moreover, S is either GO or equivalent to the scale word abacaba.
 ==== Proof ====
-Suppose S has n notes (after dealing with small cases, we may assume n >= 7) and S projects to single-period mosses S1 (via identifying b ~ c), S2 (via identifying a ~ c) and S3 (via identifying a ~ b). Suppose S1's generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S2 and S3, respectively. These two-"step-size" words, which we call Σ2 and Σ3, must be mosses, since every m-step in the new words coresponds to mk-steps in the mos words S1 and S2, which come in at most two sizes.  Both Σ2 and Σ3 are single-period mosses, since (k, n) = 1.
+Suppose S has n notes (after dealing with small cases, we may assume n >= 7) and S projects to single-period mosses S1 (via identifying b ~ c), S2 (via identifying a ~ c) and S3 (via identifying a ~ b). Suppose S1's generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S2 and S3, respectively. These two-"step-size" words, which we call Σ2 and Σ3, must be mosses, since every m-step in the new words coresponds to mk-steps in the mos words S1 and S2, which come in at most two sizes. Both Σ2 and Σ3 are single-period mosses, since (k, n) = 1.
   index: 1 2 3 4 ...   n
@@ Line 175: / Line 175: @@
 These give exactly three distinct sizes for every interval class. Hence S is SV3.
-In this case S has two chains of Qs, one with floor(n/2) notes and and one with ceil(n/2) notes. Thus S also satisfies the generator-offset property with generator Q.
+In this case S has two chains of Qs, one with floor(n/2) notes and and one with ceil(n/2) notes. Every instance of Q must be a k-step, since Q = αa + βb + γc is the only way to write Q in the basis (a, b, c). Thus S also satisfies the generator-offset property with generator Q.
 End Case 1.]