Generator-offset property: Difference between revisions

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Λ2 = β W(β, β’) β’

Λ3 = γ W(γ, γ’) γ

where W is a word in two letters of length n-2.

where W is a word in two letters of length 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).

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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.

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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:

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@@ Line 155: / Line 155: @@
   Λ<sub>2</sub> = β W(β, β’) β’
   Λ<sub>3</sub> = γ W(γ, γ’) γ
-where W is a word in two letters of length n-2.
+where W is a word in two letters of length n &minus; 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).
@@ Line 190: / Line 190: @@
 [Case 3: μ = anything else from 3 to floor(n/2).
-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:
+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 Λ<sub>2</sub> means chunk of β, and chunk of Λ<sub>3</sub> means chunk of γ.
@@ Line 199: / Line 199: @@
 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'''. 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:
+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:
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