At the next level we store only islands at any given

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Unformatted text preview: n or equal to , we can stop dividing. We make k as small as possible in order to divide the length T segment into as few as possible chunks, thereby making the ﬁnal linear chunks as large as possible, while still meeting the constraint that the chunk is ≤ . We can solve for k exactly, as k = logb (T / ) . Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-40 (of 180) k-best without the penalty Island Summary Scratch More analysis at the top level, we need b − 1 islands (since we don’t store the right most one). Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-41 (of 180) k-best without the penalty Island Summary Scratch More analysis at the top level, we need b − 1 islands (since we don’t store the right most one). At the next level, we store only islands at any given time for one of the length T /b segments, and within that segment we store, again at most b − 1 islands. Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-41 (of...
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This document was uploaded on 04/05/2014.

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