925 qt2 1 qt2 1 1 qt2 qt2 1

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Unformatted text preview: , T ) = q1:T ∈ DQ1:T (1, t(2) ) : qt(2) = qt(2) (2), . . . , ∗ ∗ qT −1 = qT −1 (2), qT = qT (2) Prof. Jeﬀ Bilmes (9.23) EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 (9.24) page 9-22 (of 180) k-best without the penalty Island Summary Scratch Finding the 3rd best - partitioning DQ1:T (1, t(2) ) Also, since the 2nd and 1st best is the same up to time t(2) − 1, we have another characterization of DQ1:T (1, t(2) , t(2) ) as: ∗ ∗ DQ1:T (1, t(2) , t(2) ) = q1:T : q1 = q1 (1), q2 = q2 (1), . . . , (9.25) ∗ ∗ ∗ qt(2) −1 = qt(2) −1 (1), qt(2) = qt(2) (1), qt(2) = qt(2) (2) (9.26) Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-23 (of 180) k-best without the penalty Island Summary Scratch Finding the 3rd best - partitioning DQ1:T (1, t(2) ) Also, since the 2nd and 1st best is the same up to time t(2) − 1, we have another characterization of DQ1:T (1, t(2) , t(2) ) as: ∗ ∗ DQ1:T (1, t(2) , t(2) ) = q1:T : q1 = q1 (1), q2 = q2 (1), . . . , (9.25) ∗ ∗ ∗ qt(2) −1 = qt(2) −1 (1), qt(2) = qt(2) (1), qt(2) = qt(2) (2) (...
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This document was uploaded on 04/05/2014.

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