Download e-book for kindle: Computational linguistic text processing: logical form, by Rodolfo Delmonte

By Rodolfo Delmonte

ISBN-10: 1600217001

ISBN-13: 9781600217005

Semantic Interpretation & Logical shape; Discourse version & Inferential techniques; reading Spatiotemporal destinations; Discourse relatives & Rhetorical constitution; Discourse kin & Causality; textual content new release; Linguistically-Based Semantic assessment for try out Entailment; Reasoning from Discourse version for query Answering & textual content Paraphrasing

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Extra info for Computational linguistic text processing: logical form, semantic interpretation, discourse relations and question answering

Example text

1 Let {b0 , . . , bn } ⊆ L be an LLL reduced Z-basis, with respect to the parameter 1 n γ = 43 . Hence we ||b0 || ≤ 2 4 · det(L) n+1 , and since det(L) = | det(B)| = c we conclude ||b0 || ≤ < 1. There are q, p1 , . . , pn ∈ Z such that b0 = b0 q + n i=1 bi pi = [qc, qβ1 − p1 , . . , qβn − pn ] ∈ L. We may assume that q ≥ 0. 2 n 2 Moreover, assume that q = 0, then ||b0 || = i=1 pi ≥ 1, a contradiction. n(n+1) Hence we have q ≥ 1. Then we have 2− 4 · n+1 q = qc ≤ ||b0 || ≤ , hence n(n+1) n+1 1 q ≤ 2 4 · −n , thus ≤ 2 4 · q − n .

We have Φi | gcd(Ψ, Θ − αi ) and Φj | gcd(Ψ, Θ − αi ), as well as Φj | gcd(Ψ, Θ − αj ) and Φi | gcd(Ψ, Θ − αj ). Hence given an Fq -basis {Θ0 , . . , Θr−1 } ⊆ UΨ , where we may assume Θ0 = 1, by Fq -linearity there is an element Θk , for some k ∈ {1, . . , r − 1}, having the same distinguishing property. Thus we have the following deterministic algorithm: We successively for k ∈ {1, . . , r − 1} and α ∈ Fq compute gcd(Ψ, Θk − α) ∈ Fq [X], where Ψ ∈ Fq [X] runs through all the factors of Ψ found so far; here we initially have Ψ = Ψ, and whenever 1 = gcd(Ψ, Θk ) = Ψ we replace Ψ by the non-trivial factors found; we terminate as soon as a total of r factors has been found.

Note that the typical choice is γ = 34 , yielding α = 2. 7) Proposition. 6), let B be LLL reduced. Then: i−1 a) For 1 ≤ j ≤ i ≤ n we have ||bj || ≤ α 2 · ||bi ||. n−1 1 b) We have ||b1 || ≤ α 4 · det(L) n . n−1 c) We have ||b1 || ≤ α 2 · min(L). n(n−1) n d) We have det(L) ≤ i=1 ||bi || ≤ α 4 · det(L). 2 2 2 Proof. For i ∈ {2, . . , n} we have ||bi || ≥ (γ− 14 )·||bi−1 || = α1 ·||bi−1 || . Hence for 2 2 2 2 i−1 2 1 ≤ j ≤ i ≤ n we get ||bj || ≤ αi−j · ||bi || . Thus ||bi || = ||bi || + j=1 µ2ij · ||bj || ≤ i−1 2 2 (1 + (α − 1) · j=1 αi−j−1 ) · ||bi || = αi−1 · ||bi || .

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Computational linguistic text processing: logical form, semantic interpretation, discourse relations and question answering by Rodolfo Delmonte


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