Generator-offset property: Difference between revisions
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In this case we have Σ = QRQR…QRT, thus S satisfies the generator-offset property. | In this case we have Σ = QRQR…QRT, thus S satisfies the generator-offset property. | ||
===== Case 3 ===== | ===== Case 3 ===== | ||
μ | 3 ≤ μ ≤ 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) − 1) + 1 (Λ<sub>3</sub> 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 Λ<sub>3</sub> 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 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) − 1) + 1 (Λ<sub>3</sub> 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 Λ<sub>3</sub> 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 Λ<sub>3</sub> is a mos. We thus have the following cases: | ||