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. Moreover, S is either GO or equivalent to the scale word abacaba.

==== Proof ====

Suppose S has n notes (by excluding small cases, n >= 7) and S projects to single-period mosses S1 (a ~ b), S2 (b ~ c) and S3 (c ~ a). 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 words, which we call Σ2 and Σ3, must be mosses. Both Σ2 and Σ3 are single period, since (k, n) = 1.

index: 1 2 3 4 ... n

Σ1: P P P P ... P I

Σ2: [some mos]

Σ3: [some mos]

Assume that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos and even the same mode. Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying a and b, and all instances of s in S1 come from c. Then the corresponding steps of S2 must be either all a’s or all b~c’s. Thus these steps are all a’s or all b’s. This contradicts the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.

At most two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S (written in the basis a, b, c), then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S2, contradicting the mos property of S2.

So assume Q = (α, β, γ) != R = (α, β’, γ’) are the two preimages of P in S. Then I has preimage (α’, β’’, γ’’), which by abuse we also denote I. Here the values in each component differ by at most 1, and α != α’. Recall that I has preimage (α’, β’’, γ’’). Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ. The cyclic words Λ1 = the pattern of α and α’, Λ2 = the pattern of β and β’, and Λ3 = the pattern of γ and γ must form mosses. By way of illustration, the chain of k-steps might look like this in s:

1 2 3 4 5 6 7 8 9

Σ = Q Q Q R Q Q R R I

Λ1 = α α α α α α α α α’

Λ2 = β β β β’ β β β β β’

Λ3 = γ γ γ γ’ γ γ γ γ γ

We don’t know which P’s pull back to Q and which P’s pull back to R, but we now do know that Σ2 has one more R than Σ3, and thus that:

Λ2 is λβ μβ’, 2 <= μ <= ceil(n/2).

Λ3 is a (λ±1)β (μ∓1)β’ mos.

Since neither are multimosses, and at least one of μ and (μ∓1) are even, it is now immediate that n is odd.

We can assume that the first k-step in Σ is Q:

1 … n

Σ = Q … I

Λ1 = α … α α’

Λ2 = β (pattern) β’

Λ3 = γ ( ditto ) γ

Since, by our assumption, Λ3 has two γ’s in a row, Λ3 must have more γ than γ’, so μ - 1 < n/2. Since Λ3 is a mos, μ - 1 >= 1. So we have 2 <= μ <= ceil(n/2).

We have three cases to consider:

BEGIN CASEBLOCK x: [

[[Case 1: Λ2 is the mos (n-2)β 2β’.

For Λ2 to be a mos, the first occurrence of R must be at either f = floor(n/2) or ceil(n/2). We may assume that it is at f; otherwise flip the chain and reindex the words to start at 2f.

1 … f … 2f n

Σ = Q … Q R Q … Q I

Λ1 = α … α α α … α α’

Λ2 = β … β β’ β … β β’

Λ3 = γ … γ γ’ γ … γ γ

We need only consider stacks up to f-many k-steps. Either:

1. the stack has only preimages of P’s and it either contains the preimage of the fth P or not; or

2. the stack has one I and does not contain any R (since it’s more than f-1 generators away).

These give exactly three distinct sizes for k-steps. 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.

End Case 1.]

[Case 2: Λ2 has fewer β than β’.

Since Λ3 has more β than β’, Λ2 is floor(n/2)β ceil(n/2)β’, and Λ3 is ceil(n/2)γ floor(n/2)γ’. There is a unique mode of ceil(n/2)γ floor(n/2)γ’ that both begins and ends with γ, namely γγ’γγ’…γγ’γ. Thus Λ2 is ββ’ββ’…ββ’β’. It is now easy to see that if the size of the stack of k-steps is odd, then there are two sizes that do not contain I and one size that contains I; if the stack size is even, then there is one size that does not contain I and two sizes that contain I. Hence S is SV3.

In this case we have Σ = QRQR…QRI, thus S satisfies the generator-offset property.

End Case 2.]

[Case 3: μ = anything else from 3 to floor(n/2).

Λ2 has a chunk of βs (after the first β’) of size x = either floor(n/μ) (>= floor(n/floor(n/2)) = 2) or ceil(n/μ) (= floor(n/μ) + 1). Hence Λ3 has a chunk of γs of that same size. Λ3 also has a chunk that goes over n and loops back around, which must be of size y = at least 2*(floor(n/μ)-1) + 1 (Λ3 might have chunks of size floor(n/μ)-1 and floor(n/μ) instead) = 2*floor(n/μ) - 1, and at most 2*(floor(n/μ)+1) + 1 = 2*floor(n/μ) + 3 (if Λ3 has chunks of size floor(n/μ) and ceil(n/μ)). The difference between the chunk sizes is y - x, which must be 0 or 1.

The size difference within Λ3 (since it’s a mos) must be 0 or 1, so we have the following cases:

BEGIN CASEBLOCK 3.x: [ Here chunk of Λ2 means chunk of β, and chunk of Λ3 means chunk of γ.

[[Case 3.1: (x, y) = (floor(n/μ), 2*floor(n/μ) - 1)

Since y-x = floor(n/μ) - 1 and floor(n/μ) >= 2, we have: x = floor(n/μ) = 2 and y-x = 1; hence y = 2*floor(n/μ) - 1 = 3. The chunk in Λ3 whose size is 3 is made from two chunks in Λ2 of size 1. (So Λ2 has chunks of size 1 and 2, and Λ3 has chunks of size 2 and 3.)

Λ2 has two consecutive chunks of size 1. Since chunk sizes form a mos, Λ2 has more chunks of size 1 than it has chunks of size 2.

Use “w[i:j]” to denote the slice of the cyclic word w that includes both endpoints. Λ2 has only two chunks of size 1, Λ2[n-1:n-1] and Λ2[1:1], since otherwise Λ3 would have a chunk of size 1 within Λ3[1:n-1]. Thus Λ2 has exactly one chunk of size 2. Thus Λ2 = ββ’βββ’ββ’ and Λ3 = γγ’γγγ’γγ. Thus we have:

1 2 3 4 5 6 7

Σ = Q R Q Q R Q I

Λ1 = α α α α α α α’

Λ2 = β β’ β β β’ β β’

Λ3 = γ γ’ γ γ γ’ γ γ

Suppose a step of S is reached by stacking t-many k-steps. We have three cases after accounting for equave complements:

t = 1: S is equivalent to abacaba.

t = 2: QR QQ RQ IQ RQ QR QI => S is equivalent to abacaba.

t = 3: QRQ QRQ IQR QQR QIQ RQQ RQI => S is equivalent to abacaba.

(This also implies S is SV3.)

End Case 3.1]

[Case 3.2: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) - 1).

Impossible: y = 2*floor(n/μ) - 1 can only occur if Λ3 has chunks of size floor(n/μ) - 1 and floor(n/μ), which contradicts the size of x.

End Case 3.2]]

[Case 3.3: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ))

Impossible: y = 2*floor(n/μ) can only occur if Λ3 has chunks of size floor(n/μ) - 1 and floor(n/μ), which contradicts the size of x.

End Case 3.3]

The remaining cases are all impossible because they imply y - x >= 2:

[Case 3.4:(x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 1)]

[Case 3.5: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 2)]

[Case 3.6: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 3)]

[Case 3.7: (x, y) = (floor(n/μ), 2*floor(n/μ))]

[Case 3.8: (x, y) = (floor(n/μ), 2*floor(n/μ) + 1)]

[Case 3.9: (x, y) = (floor(n/μ), 2*floor(n/μ) + 2)]

[Case 3.10: (x, y) = (floor(n/μ), 2*floor(n/μ) + 3)]

]

END CASEBLOCK 3.x]

End Case 3.]

END CASEBLOCK x]

== Open conjectures ==

@@ 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. Moreover, S is either GO or equivalent to the scale word abacaba.
+==== Proof ====
+Suppose S has n notes (by excluding small cases, n >= 7) and S projects to single-period mosses S1 (a ~ b), S2 (b ~ c) and S3 (c ~ a). 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 words, which we call Σ2 and Σ3, must be mosses. Both Σ2 and Σ3 are single period, since (k, n) = 1.
+ index: 1 2 3 4 ...   n
+ Σ1:    P P P P ... P I
+ Σ2:    [some mos]
+ Σ3:    [some mos]
+Assume that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos and even the same mode. Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying a and b, and all instances of s in S1 come from c. Then the corresponding steps of S2 must be either all a’s or all b~c’s. Thus these steps are all a’s or all b’s. This contradicts the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.
+At most two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S (written in the basis a, b, c), then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S2, contradicting the mos property of S2.
+So assume Q = (α, β, γ) != R = (α, β’, γ’) are the two preimages of P in S. Then I has preimage (α’, β’’, γ’’), which by abuse we also denote I. Here the values in each component differ by at most 1, and α != α’. Recall that I has preimage (α’, β’’, γ’’). Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ. The cyclic words Λ1 = the pattern of α and α’, Λ2 = the pattern of β and β’, and Λ3 = the pattern of γ and γ must form mosses. By way of illustration, the chain of k-steps might look like this in s:
+2 3 4  5 6 7 8 9
+ Σ  = Q Q Q R  Q Q R R I
+ Λ1 = α α α α  α α α α α’
+ Λ2 = β β β β’ β β β β β’
+ Λ3 = γ γ γ γ’ γ γ γ γ γ
+We don’t know which P’s pull back to Q and which P’s pull back to R, but we now do know that Σ2 has one more R than Σ3, and thus that:
+Λ2 is λβ μβ’, 2 <= μ <= ceil(n/2).
+Λ3 is a (λ±1)β (μ∓1)β’ mos.
+Since neither are multimosses, and at least one of μ and (μ∓1) are even, it is now immediate that n is odd.
+We can assume that the first k-step in Σ is Q:
+…         n
+ Σ  = Q …         I
+ Λ1 = α …       α α’
+ Λ2 = β (pattern) β’
+ Λ3 = γ ( ditto ) γ
+Since, by our assumption, Λ3 has two γ’s in a row, Λ3 must have more γ than γ’, so μ - 1 < n/2. Since Λ3 is a mos, μ - 1 >= 1. So we have 2 <= μ <= ceil(n/2).
+We have three cases to consider:
+BEGIN CASEBLOCK x: [
+[[Case 1: Λ2 is the mos (n-2)β 2β’.
+For Λ2 to be a mos, the first occurrence of R must be at either f = floor(n/2) or ceil(n/2). We may assume that it is at f; otherwise flip the chain and reindex the words to start at 2f.
+…   f    … 2f n
+ Σ  = Q … Q R  Q … Q  I
+ Λ1 = α … α α  α … α  α’
+ Λ2 = β … β β’ β … β  β’
+ Λ3 = γ … γ γ’ γ … γ  γ
+We need only consider stacks up to f-many k-steps. Either:
+. the stack has only preimages of P’s and it either contains the preimage of the fth P or not; or
+. the stack has one I and does not contain any R (since it’s more than f-1 generators away).
+These give exactly three distinct sizes for k-steps. 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.
+End Case 1.]
+[Case 2: Λ2 has fewer β than β’.
+Since Λ3 has more β than β’, Λ2 is floor(n/2)β ceil(n/2)β’, and Λ3 is ceil(n/2)γ floor(n/2)γ’. There is a unique mode of ceil(n/2)γ floor(n/2)γ’ that both begins and ends with γ, namely γγ’γγ’…γγ’γ. Thus Λ2 is ββ’ββ’…ββ’β’. It is now easy to see that if the size of the stack of k-steps is odd, then there are two sizes that do not contain I and one size that contains I; if the stack size is even, then there is one size that does not contain I and two sizes that contain I. Hence S is SV3.
+In this case we have Σ = QRQR…QRI, thus S satisfies the generator-offset property.
+End Case 2.]
+[Case 3: μ = anything else from 3 to floor(n/2).
+Λ2 has a chunk of βs (after the first β’) of size x = either floor(n/μ) (>= floor(n/floor(n/2)) = 2) or ceil(n/μ) (= floor(n/μ) + 1). Hence Λ3 has a chunk of γs of that same size. Λ3 also has a chunk that goes over n and loops back around, which must be of size y = at least 2*(floor(n/μ)-1) + 1 (Λ3 might have chunks of size floor(n/μ)-1 and floor(n/μ) instead) = 2*floor(n/μ) - 1, and at most 2*(floor(n/μ)+1) + 1 = 2*floor(n/μ) + 3 (if Λ3 has chunks of size floor(n/μ) and ceil(n/μ)). The difference between the chunk sizes is y - x, which must be 0 or 1.
+The size difference within Λ3 (since it’s a mos) must be 0 or 1, so we have the following cases:
+BEGIN CASEBLOCK 3.x: [ Here chunk of Λ2 means chunk of β, and chunk of Λ3 means chunk of γ.
+[[Case 3.1: (x, y) = (floor(n/μ), 2*floor(n/μ) - 1)
+Since y-x = floor(n/μ) - 1 and floor(n/μ) >= 2, we have: x = floor(n/μ) = 2 and y-x = 1; hence y = 2*floor(n/μ) - 1 = 3. The chunk in Λ3 whose size is 3 is made from two chunks in Λ2 of size 1. (So Λ2 has chunks of size 1 and 2, and Λ3 has chunks of size 2 and 3.)
+Λ2 has two consecutive chunks of size 1. Since chunk sizes form a mos, Λ2 has more chunks of size 1 than it has chunks of size 2.
+Use “w[i:j]” to denote the slice of the cyclic word w that includes both endpoints. Λ2 has only two chunks of size 1, Λ2[n-1:n-1] and Λ2[1:1], since otherwise Λ3 would have a chunk of size 1 within Λ3[1:n-1]. Thus Λ2 has exactly one chunk of size 2. Thus Λ2 = ββ’βββ’ββ’ and Λ3 = γγ’γγγ’γγ. Thus we have:
+2  3 4 5  6 7
+ Σ =  Q R  Q Q R  Q I
+ Λ1 = α α  α α α  α α’
+ Λ2 = β β’ β β β’ β β’
+ Λ3 = γ γ’ γ γ γ’ γ γ
+Suppose a step of S is reached by stacking t-many k-steps. We have three cases after accounting for equave complements:
+t = 1: S is equivalent to abacaba.
+t = 2: QR QQ RQ IQ RQ QR QI => S is equivalent to abacaba.
+t = 3: QRQ QRQ IQR QQR QIQ RQQ RQI => S is equivalent to abacaba.
+(This also implies S is SV3.)
+End Case 3.1]
+[Case 3.2: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) - 1).
+Impossible: y = 2*floor(n/μ) - 1 can only occur if Λ3 has chunks of size floor(n/μ) - 1 and floor(n/μ), which contradicts the size of x.
+End Case 3.2]]
+[Case 3.3: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ))
+Impossible: y = 2*floor(n/μ) can only occur if Λ3 has chunks of size floor(n/μ) - 1 and floor(n/μ), which contradicts the size of x.
+End Case 3.3]
+The remaining cases are all impossible because they imply y - x >= 2:
+[Case 3.4:(x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 1)]
+[Case 3.5: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 2)]
+[Case 3.6: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) + 3)]
+[Case 3.7: (x, y) = (floor(n/μ), 2*floor(n/μ))]
+[Case 3.8: (x, y) = (floor(n/μ), 2*floor(n/μ) + 1)]
+[Case 3.9: (x, y) = (floor(n/μ), 2*floor(n/μ) + 2)]
+[Case 3.10: (x, y) = (floor(n/μ), 2*floor(n/μ) + 3)]
+]
+END CASEBLOCK 3.x]
+End Case 3.]
+END CASEBLOCK x]
 == Open conjectures ==