# 18 argmax mt1t qt1 2 q for t t2 1 t

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Unformatted text preview: −1 (2), q ) (9.18) ∗ ∈ argmax mt−1,t (qt−1 (2), q ) for t = t(2) + 1, . . . , T (9.19) q EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-19 (of 180) k-best without the penalty Island Summary Scratch The maximum score block Suppose t(2) ∈ argmaxt pm (DQ1:T (1, t)) is time that achieved the maximum. Then ∗ p(q1:T (2), x1:T ) = pm (DQ1:T (1, t(2) )). ¯ (9.16) ∗ and q1:T (2) ∈ DQ1:T (1, t(2) ). ∗ We can identify q1:T (2) using the following procedure, part of which is a copy, and part of which is dynamic programming: ∗ ∗ qt (2) ← qt (1) for t = 1, . . . , t(2) − 1 ∗ qt(2) (2) ∈ argmax q =q ∗(2) (1) t ∗ qt (2) (9.17) ∗ mt−1,t (qt(2) −1 (2), q ) (9.18) ∗ ∈ argmax mt−1,t (qt−1 (2), q ) for t = t(2) + 1, . . . , T (9.19) q Time cost: only an additional O(T N ). Additional memory cost: O(T ). Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-19 (of 180) k-best without the penalty Island Summary Scratch Finding the 3rd best ∗ the 2nd best is q1:T (2) ∈ DQ1:T (1, t(2) ) where t(2) is the index where there lies a diﬀerence between the ﬁrst and s...
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## This document was uploaded on 04/05/2014.

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