Generator-offset property: Difference between revisions
m →Proposition 1 (Properties of SGA scales): combining (4) and (5) (to add the assumption that n is odd) and renumbering |
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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. | 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…QRT, and ''S'' is well-formed with respect to the generator Q + R, thus ''S'' satisfies the generator-offset property. | In this case we have Σ = QRQR…QRT, and ''S'' is well-formed with respect to the generator Q + R, thus ''S'' satisfies the generator-offset property. By Proposition 1, ''S'' is SV3. | ||
'''Case 3:''' 3 ≤ μ ≤ floor(''n''/2). | '''Case 3:''' 3 ≤ μ ≤ floor(''n''/2). | ||