Generator-offset property: Difference between revisions
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We have three cases to consider: | We have three cases to consider: | ||
Case 1: Λ2 is the mos (n-2)β 2β’. | |||
For Λ2 to be a mos, the first occurrence of R must be at either f = floor(n/2) or ceil(n/2). We may assume that it is at f; otherwise flip the chain and reindex the words to start at 2f. | For Λ2 to be a mos, the first occurrence of R must be at either f = floor(n/2) or ceil(n/2). We may assume that it is at f; otherwise flip the chain and reindex the words to start at 2f. | ||
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We need only consider stacks up to f-many k-steps. Either: | We need only consider stacks up to f-many k-steps. Either: | ||
# the stack has only preimages of P’s and it either contains the preimage of the fth P or not; or | |||
# the stack has one I and does not contain any R (since it’s more than f-1 generators away). | |||
These give exactly three distinct sizes for k-steps. Hence S is SV3. | These give exactly three distinct sizes for k-steps. Hence S is SV3. | ||
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[Case 2: Λ2 has fewer β than β’. | [Case 2: Λ2 has fewer β than β’. | ||
Since Λ3 has more β than β’, Λ2 is floor(n/2)β ceil(n/2)β’, and Λ3 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 Λ2 is ββ’ββ’…ββ’β’. It is now easy to see that if the size of the stack of k-steps is odd, then there are two sizes that do not contain I and one size that contains I; if the stack size is even, then there is one size that does not contain I and two sizes that contain I. Hence S is SV3. | Since Λ3 has more β than β’, Λ2 is floor(n/2)β ceil(n/2)β’, and Λ3 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 Λ2 is ββ’ββ’…ββ’β’. It is now easy to see that if the size of the stack of k-steps is odd, then there are two sizes that do not contain I and one size that contains I; if the stack size is even, then there is one size that does not contain I and two sizes that contain I. Hence S is SV3. | ||