For any b t 1d so that d logb t the algorithm has ot

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Unformatted text preview: re-root in T ! Prof. Jeff Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-48 (of 180) k-best without the penalty Island Summary Scratch Setting the parameters Alternatively, can take b = segment lengths). √ T (so log base changes for different we get memory complexity of √ √ O(b × logb (T )) = O( T × 2) = O( T ) memory compute becomes O(T logb (T )) = O(2 × T ) = O(T ) compute. √ so algorithm takes at most twice as long but uses 2/ T as much memory. Hence, asymptotically, this means that there is an algorithm that is still linear in T while memory usage is only square-root in T ! for any b = T 1/d (so that d = logb T ), the algorithm has O(T 1/d ) memory and O(dT ) compute. Prof. Jeff Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-48 (of 180) k-best without the penalty Island Summary Scratch Setting the parameters Alternatively, can take b = segment lengths). √ T (so log base changes for different we get memory complexity of √ √ O(b × l...
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This document was uploaded on 04/05/2014.

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