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2.4. A block splitting pattern 41Proof.P t,k = P[ ∑dγ ′ < ∑ dδ ′ ]======B∑∆−1∑∆=1 Γ=0B∑∆−1∑∆=1 Γ=0B∑∆=1 Γ=0B∑∆−1∑∆=1 Γ=0B∑∆−1∑∆=1 Γ=0B∑∆−1∑∆−1∑∆=1 Γ=0∑∑∑∑∑∑d γ′ i =Γ0≤γ ′ i∑d γ′ i =Γ0≤γ ′ i∑d γ′ i =Γ0≤γ ′ i∑d γ′ i =Γ0≤γ ′ i∑d γ′ i =Γ0≤γ ′ i∑P (γ ′ , δ ′ )∑d δ′ i =∆0≤δ ′ i ≤βi∑∑d δ′ i =∆0≤δ ′ i ≤βi∑∑d δ′ i =∆0≤δ ′ i ≤βi∑∑d δ′ i =∆0≤δ ′ i ≤βi∑∑2 −∆−Γ−2d ∑∑∏P (γ i, ′ δ i)′d∏P (δ i)P ′ (γ i ′ | δ i)′d2 −∆−d+#{i|δi=βi} ∏ dP (γ ′ i | δ ′ i)2 −∆−Γ−2d+2#{i|δi=βi} 1 δ ′i =β i,γ i ′ =0d δ′ i =∆0≤δ ′ i ≤βid γ′ i =Γ0≤γ ′ i∑2 2#{i|δi=βi} 1 δ ′∑i =β i,γ i ′ =0 .d δ′ i =∆0≤δ ′ i ≤βiThe above discussion also shows that a ∈ S t,k is equivalent to ∑ d ɛ′ i < k − w H(t). Moreover,as was already mentioned above, if that inequality is verified, then each block overflows into thenext one. That is exactly the situation that was studied in the previous subsection when t wasmade of one block, so that the computations we did there are still valid (only to compute #C t,k,iwith i < 0 where only a’s in S t,k are enumerated, but not with i ≥ 0) and we get the followingproposition.Proposition 2.4.10. Let k ≥ 2 and t ∈ ( Z/(2 k − 1)Z ) ∗verifyingThenP t,k =min(α i ) ≥iB−1∑E=0∑∑d∑β i − 1 = B − 1 = k − w H (t) − 1 .i=1d ei=E0≤e iB−1∑= 2 −B 3 −dE=0∏P (e i )d2 −E ∑∑d ei=E0≤e i∏d()4 max(1,min(ei,βi)) − 1.