# To identify it can partition dq1t 1 t2 into separate

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Unformatted text preview: {q1:T (2)}. In latter case, must exist diﬀerence from both 1st and 2nd best. To identify it, can partition DQ1:T (1, t(2) ) into separate sets of paths based on where this diﬀerence happens. Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-20 (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 second best path. Third best is either in one of original blocks {DQ1:T (1, t)}t=t(2) that did not contain the second best . . . . . . or otherwise is in the same original block that contained the ∗ ∗ second best q1:T (3) ∈ DQ1:T (1, t(2) ) \ {q1:T (2)}. In latter case, must exist diﬀerence from both 1st and 2nd best. To identify it, can partition DQ1:T (1, t(2) ) into separate sets of paths based on where this diﬀerence happens. Since DQ1:T (1, t(2) ) is the same as ﬁrst best upto but not including time t(2) , there is no partition before time t(2) . Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-20 (of 180) k-best without the penalty Island Summary Scratch Finding the 3rd best - partitioning DQ1:T (1, t(2) ) The ﬁrst time (meaning the frame closest to t = t(2) ) at which there ∗ is a diﬀerence between the second best path q1:T (2) and third best ∗ (2) may be either: path q1:T Prof....
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