Generator-offset property: Difference between revisions

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'''Case 3:''' 3 ≤ μ ≤ floor(''n''/2).
'''Case 3:''' 3 ≤ μ ≤ floor(''n''/2).


Λ<sub>2</sub> has a chunk of β (after the first β′) of size ''x'' where ''x'' = floor(''n''/μ) ≥ floor(''n''/floor(''n''/2)) = 2 or ''x'' = ceil(''n''/μ) = floor(''n''/μ) + 1. Hence Λ<sub>3</sub> has a chunk of γ of size ''x''. Λ<sub>3</sub> also has a chunk that crosses index ''n'', which must be of size ''y'', where  
Λ<sub>2</sub> has a chunk of β (after the first β′) of size ''x'' where ''x'' = floor(''n''/μ) ≥ floor(''n''/floor(''n''/2)) = 2 or ''x'' = ceil(''n''/μ) = floor(''n''/μ) + 1. Hence Λ<sub>3</sub> has a chunk of γ of size ''x''. Λ<sub>3</sub> also has a chunk that contains Λ<sub>3</sub>[''n'' : 2]. This chunk must be of size ''y'', where  


<math>2 \lfloor\frac{n}{\mu}\rfloor - 1 = 2 \big(\lfloor \frac{n}{\mu} \rfloor - 1\big) + 1 \leq y \leq 2 \big(\lfloor\frac{n}{\mu}\rfloor + 1 \big) + 1 = 2\lfloor\frac{n}{\mu}\rfloor + 3.</math>
<math>2 \lfloor\frac{n}{\mu}\rfloor - 1 = 2 \big(\lfloor \frac{n}{\mu} \rfloor - 1\big) + 1 \leq y \leq 2 \big(\lfloor\frac{n}{\mu}\rfloor + 1 \big) + 1 = 2\lfloor\frac{n}{\mu}\rfloor + 3.</math>