29 ee596awinter 2013dgms lecture 9 feb 6th 2013 page

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Unformatted text preview: n that one block. Prof. Jeff Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-24 (of 180) k-best without the penalty Island Summary Scratch Partitioning Thus, the blocks {DQ1:T (1, t)}t=t(2) and DQ1:T (1, t(2) , t) constitute the partitioning of DQ1:T \ {q1:T (1), q1:T (2)}. t≥t(2) Use same strategy as before: score each block (based on max path within), find the (a) max, and then compute it within that one block. We have max scores of previous blocks, we only need to find max scores of new blocks. Prof. Jeff Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-24 (of 180) k-best without the penalty Island Summary Scratch Scores of new blocks For a bit of notational simplicity, lets set t = t(2) in the following. Then for t = t = t(2) we have: pm DQ1:T (1, t , t ) = pm (DQ1:T (1, t )) ∗ maxq∈{q∗ (1),q∗ (2)} mt −1,t (qt −1 (2), q ) t t ∗ ∗ mt −1,t (qt −1 (2), qt (2)) (9.28) and for t > t(2) = t pm DQ1:T (1, t , t) = pm (DQ1:T (1, t )) Prof. Jeff Bilmes ∗ maxq=qt (2)...
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This document was uploaded on 04/05/2014.

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