Generator-offset property: Difference between revisions
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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 = | In this case we have Σ = QRQR…QRT, thus S satisfies the generator-offset property. | ||
End Case 2.] | End Case 2.] | ||