Generator-offset property: Difference between revisions

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Let S(a,b,c) be a scale word in three '''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 m-steps in the new words correspond 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 m-steps in the new words correspond 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

Σ1: P P P P ... P I

Σ1: P P P P ... P I

Σ2: [some mos]

Σ2: [some mos]

Σ3: [some mos]

Σ3: [some mos]

(Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)

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 pattern (up to knowing the order of the step sizes) and even the same mode:

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 pattern (up to knowing the order of the step sizes) and even the same mode:

S1: L ... L s

S1: L ... L s

S2: L' ... L' s'

Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting 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.

Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting 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.

We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). Only 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, then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.

We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). Only 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, then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.

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’. 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:

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 we denote T. Here the values in each component differ by at most 1, and α ≠ α’. 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

Σ = Q Q Q R Q Q R R T

Λ1 = α α α α α α α α α’

Λ1 = α α α α α α α α α’

Λ2 = β β β β’ β β β β β’

Λ2 = β β β β’ β β β β β’

Λ3 = γ γ γ γ’ γ γ γ γ γ

Λ3 = γ γ γ γ’ γ γ γ γ γ

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

We don’t know which P’s have preimage Q and which P’s have preimage 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).

Λ2 is λβ μβ’, 2 ≤ μ ≤ ceil(n/2).

Λ3 is a (λ±1)β (μ∓1)β’ mos.

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.

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

1 … n

Σ = Q … I

Σ = Q … T

Λ1 = α … α α’

Λ1 = α … α α’

Λ2 = β (pattern) β’

Λ2 = β (pattern) β’

Λ3 = γ ( ditto ) γ

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).

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:

Case 1: Λ2 is the mos (n − 2)β 2β’.

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.

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

Σ = Q … Q R Q … Q T

Λ1 = α … α α α … α α’

Λ1 = α … α α α … α α’

Λ2 = β … β β’ β … β β’

Λ2 = β … β β’ β … β β’

Λ3 = γ … γ γ’ γ … γ γ

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).

# the stack has one T and does not contain any R (since it’s more than f − 1 generators away).

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

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End Case 1.]

[Case 2: Λ2 has fewer β than β’.

[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.

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 number of k-steps stacked is odd, then there are two sizes that do not contain T and one size that contains T; if the number of k-steps stacked is even, then there is one size that does not contain T and two sizes that contain T. Hence S is SV3.

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

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[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, since Λ3 is a mos. We thus have the following cases:

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, since Λ3 is a mos. We thus have the following cases:

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

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.)

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.

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, i.e. the word w[i] w[i+1] ... w[j] where indices are taken to be elements of '''Z'''/n'''Z'''. 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:

Use “w[i:j]” to denote the slice of the cyclic word w that includes both endpoints, i.e. the word w[i] w[i+1] ... w[j] where indices are taken to be elements of '''Z'''/n'''Z'''. 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

Σ = Q R Q Q R Q T

Λ1 = α α α α α α α’

Λ1 = α α α α α α α’

Λ2 = β β’ β β β’ β β’

Λ2 = β β’ β β β’ β β’

Λ3 = γ γ’ γ γ γ’ γ γ

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 = 2: QR QQ RQ TQ RQ QR QT => S is equivalent to abacaba.

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

# t = 3: QRQ QRQ TQR QQR QTQ RQQ RQT => S is equivalent to abacaba.

(This also implies S is SV3.)

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[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.

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.

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]

@@ Line 116: / Line 116: @@
 Let S(a,b,c) be a scale word in three '''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 m-steps in the new words correspond 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 S<sub>1</sub> (via identifying b ~ c), S<sub>2</sub> (via identifying a ~ c) and S<sub>3</sub> (via identifying a ~ b). Suppose S<sub>1</sub>'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 S<sub>2</sub> and S<sub>3</sub>, respectively. These two-"step-size" words, which we call Σ<sub>2</sub> and Σ<sub>3</sub>, must be mosses, since m-steps in the new words correspond to mk-steps in the mos words S<sub>1</sub> and S<sub>2</sub>, which come in at most two sizes. Both Σ<sub>2</sub> and Σ<sub>3</sub> are single-period mosses, since (k, n) = 1.
   index: 1 2 3 4 ...   n
-  Σ1:    P P P P ... P I
+  Σ<sub>1</sub>:    P P P P ... P I
-  Σ2:    [some mos]
+  Σ<sub>2</sub>:    [some mos]
-  Σ3:    [some mos]
+  Σ<sub>3</sub>:    [some mos]
 (Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)
-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 pattern (up to knowing the order of the step sizes) and even the same mode:
+Assume that S<sub>2</sub>'s generator is also a k-step, and that Σ<sub>2</sub>'s I is located at index n. Then Σ<sub>1</sub> and Σ<sub>2</sub> are the same mos pattern (up to knowing the order of the step sizes) and even the same mode:
-  S1:    L  ... L  s
+  S<sub>1</sub>:    L  ... L  s
   S2:    L' ... L' s'
-Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting 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.
+Assume the L of S<sub>1</sub> (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S<sub>1</sub> come from a. Then the steps of S<sub>2</sub> corresponding to the L of S<sub>1</sub> must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S<sub>1</sub> and S<sub>2</sub> are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting 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.
-We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). Only 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, then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.
+We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). Only two k-steps of S can project to P in S<sub>1</sub>, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S, then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S<sub>3</sub>, contradicting the mos property of S<sub>3</sub>.
-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’. 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:
+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 we denote T. Here the values in each component differ by at most 1, and α ≠ α’. Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ. The cyclic words Λ<sub>1</sub> = the pattern of α and α’, Λ<sub>2</sub> = the pattern of β and β’, and Λ<sub>3</sub> = 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
+  Σ  = Q Q Q R  Q Q R R T
-  Λ1 = α α α α  α α α α α’
+  Λ<sub>1</sub> = α α α α  α α α α α’
-  Λ2 = β β β β’ β β β β β’
+  Λ<sub>2</sub> = β β β β’ β β β β β’
-  Λ3 = γ γ γ γ’ γ γ γ γ γ
+  Λ<sub>3</sub> = γ γ γ γ’ γ γ γ γ γ
-We don’t know which P’s have preimage Q and which P’s have preimage R, but we now do know that Σ2 has one more R than Σ3, and thus that:
+We don’t know which P’s have preimage Q and which P’s have preimage R, but we now do know that Σ<sub>2</sub> has one more R than Σ<sub>3</sub>, and thus that:
-Λ2 is λβ μβ’, 2 ≤ μ ≤ ceil(n/2).
+Λ<sub>2</sub> is λβ μβ’, 2 ≤ μ ≤ ceil(n/2).
-Λ3 is a (λ±1)β (μ∓1)β’ mos.
+Λ<sub>3</sub> 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.
@@ Line 151: / Line 151: @@
 We can assume that the first k-step in Σ is Q:
 …         n
-  Σ  = Q …         I
+  Σ  = Q …         T
-  Λ1 = α …       α α’
+  Λ<sub>1</sub> = α …       α α’
-  Λ2 = β (pattern) β’
+  Λ<sub>2</sub> = β (pattern) β’
-  Λ3 = γ ( ditto ) γ
+  Λ<sub>3</sub> = γ ( ditto ) γ
-Since, by our assumption, Λ3 has two γ’s in a row, Λ3 must have more γ than γ’, so μ &minus; 1 < n/2. Since Λ3 is a mos, μ &minus; 1 ≥ 1. So we have 2 ≤ μ ≤ ceil(n/2).
+Since, by our assumption, Λ<sub>3</sub> has two γ’s in a row, Λ<sub>3</sub> must have more γ than γ’, so μ &minus; 1 < n/2. Since Λ<sub>3</sub> is a mos, μ &minus; 1 ≥ 1. So we have 2 ≤ μ ≤ ceil(n/2).
 We have three cases to consider:
-Case 1: Λ2 is the mos (n &minus; 2)β 2β’.
+Case 1: Λ<sub>2</sub> is the mos (n &minus; 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.
+For Λ<sub>2</sub> 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
+  Σ  = Q … Q R  Q … Q  T
-  Λ1 = α … α α  α … α  α’
+  Λ<sub>1</sub> = α … α α  α … α  α’
-  Λ2 = β … β β’ β … β  β’
+  Λ<sub>2</sub> = β … β β’ β … β  β’
-  Λ3 = γ … γ γ’ γ … γ  γ
+  Λ<sub>3</sub> = 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 &minus; 1 generators away).
+# the stack has one T and does not contain any R (since it’s more than f &minus; 1 generators away).
 These give exactly three distinct sizes for every interval class. Hence S is SV3.
@@ Line 179: / Line 179: @@
 End Case 1.]
-[Case 2: Λ2 has fewer β than β’.
+[Case 2: Λ<sub>2</sub> 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.
+Since Λ<sub>3</sub> has more β than β’, Λ<sub>2</sub> is floor(n/2)β ceil(n/2)β’, and Λ<sub>3</sub> 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 Λ<sub>2</sub> is ββ’ββ’…ββ’β’. It is now easy to see that if the number of k-steps stacked is odd, then there are two sizes that do not contain T and one size that contains T; if the number of k-steps stacked is even, then there is one size that does not contain T and two sizes that contain T. Hence S is SV3.
 In this case we have Σ = QRQR…QRI, thus S satisfies the generator-offset property.
@@ Line 189: / Line 189: @@
 [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) &minus; 1) + 1 (Λ3 might have chunks of size floor(n/μ) &minus; 1 and floor(n/μ) instead) = 2*floor(n/μ) &minus; 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 &minus; x, which must be 0 or 1, since Λ3 is a mos. We thus have the following cases:
+Λ<sub>2</sub> 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 Λ<sub>3</sub> has a chunk of γs of that same size. Λ<sub>3</sub> also has a chunk that goes over n and loops back around, which must be of size y = at least 2*(floor(n/μ) &minus; 1) + 1 (Λ<sub>3</sub> might have chunks of size floor(n/μ) &minus; 1 and floor(n/μ) instead) = 2*floor(n/μ) &minus; 1, and at most 2*(floor(n/μ)+1) + 1 = 2*floor(n/μ) + 3 (if Λ<sub>3</sub> has chunks of size floor(n/μ) and ceil(n/μ)). The difference between the chunk sizes is y &minus; x, which must be 0 or 1, since Λ<sub>3</sub> is a mos. We thus have the following cases:
-BEGIN CASEBLOCK 3.x: [ Here chunk of Λ2 means chunk of β, and chunk of Λ3 means chunk of γ.
+BEGIN CASEBLOCK 3.x: [ Here chunk of Λ<sub>2</sub> means chunk of β, and chunk of Λ<sub>3</sub> means chunk of γ.
 [Case 3.1: (x, y) = (floor(n/μ), 2*floor(n/μ) &minus; 1)
-Since y &minus; x = floor(n/μ) &minus; 1 and floor(n/μ) ≥ 2, we have: x = floor(n/μ) = 2 and y &minus; x = 1; hence y = 2*floor(n/μ) &minus; 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.)
+Since y &minus; x = floor(n/μ) &minus; 1 and floor(n/μ) ≥ 2, we have: x = floor(n/μ) = 2 and y &minus; x = 1; hence y = 2*floor(n/μ) &minus; 1 = 3. The chunk in Λ<sub>3</sub> whose size is 3 is made from two chunks in Λ<sub>2</sub> of size 1. (So Λ<sub>2</sub> has chunks of size 1 and 2, and Λ<sub>3</sub> 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.
+Λ<sub>2</sub> has two consecutive chunks of size 1. Since chunk sizes form a mos, Λ<sub>2</sub> 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, i.e. the word w[i] w[i+1] ... w[j] where indices are taken to be elements of '''Z'''/n'''Z'''. Λ2 has only two chunks of size 1, Λ2[(n &minus; 1):(n &minus; 1)] and Λ2[1:1], since otherwise Λ3 would have a chunk of size 1 within Λ3[1:n &minus; 1]. Thus Λ2 has exactly one chunk of size 2. Thus Λ2 = ββ’βββ’ββ’ and Λ3 = γγ’γγγ’γγ. Thus we have:
+Use “w[i:j]” to denote the slice of the cyclic word w that includes both endpoints, i.e. the word w[i] w[i+1] ... w[j] where indices are taken to be elements of '''Z'''/n'''Z'''. Λ<sub>2</sub> has only two chunks of size 1, Λ<sub>2</sub>[(n &minus; 1):(n &minus; 1)] and Λ<sub>2</sub>[1:1], since otherwise Λ<sub>3</sub> would have a chunk of size 1 within Λ<sub>3</sub>[1:n &minus; 1]. Thus Λ<sub>2</sub> has exactly one chunk of size 2. Thus Λ<sub>2</sub> = ββ’βββ’ββ’ and Λ<sub>3</sub> = γγ’γγγ’γγ. Thus we have:
 2  3 4 5  6 7
-  Σ =  Q R  Q Q R  Q I
+  Σ =  Q R  Q Q R  Q T
-  Λ1 = α α  α α α  α α’
+  Λ<sub>1</sub> = α α  α α α  α α’
-  Λ2 = β β’ β β β’ β β’
+  Λ<sub>2</sub> = β β’ β β β’ β β’
-  Λ3 = γ γ’ γ γ γ’ γ γ
+  Λ<sub>3</sub> = 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 = 2: QR QQ RQ TQ RQ QR QT => S is equivalent to abacaba.
-# t = 3: QRQ QRQ IQR QQR QIQ RQQ RQI => S is equivalent to abacaba.
+# t = 3: QRQ QRQ TQR QQR QTQ RQQ RQT => S is equivalent to abacaba.
 (This also implies S is SV3.)
@@ Line 217: / Line 217: @@
 [Case 3.2: (x, y) = (floor(n/μ) + 1, 2*floor(n/μ) &minus; 1).
-Impossible: y = 2*floor(n/μ) &minus; 1 can only occur if Λ3 has chunks of size floor(n/μ) &minus; 1 and floor(n/μ), which contradicts the size of x.
+Impossible: y = 2*floor(n/μ) &minus; 1 can only occur if Λ<sub>3</sub> has chunks of size floor(n/μ) &minus; 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) &minus; 1 and floor(n/μ), which contradicts the size of x.
+Impossible: y = 2*floor(n/μ) can only occur if Λ<sub>3</sub> has chunks of size floor(n/μ) &minus; 1 and floor(n/μ), which contradicts the size of x.
 End Case 3.3]