CSE331 Lecture 23

CSE331 Lecture 23 - Lecture 23 CSE 331 Oct 24, 2011 Warning...

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Lecture 23 CSE 331 Oct 24, 2011
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Warning Will not always do COMPLETE proofs More work for YOU
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Last lecture Convert optimal schedule O to Ô such that Ô has no inversions (a) Exists an inversion (i,j) such that i is scheduled right before j ( d i > d j ) (b) O’ has one less inversion than O (c) Max lateness( O’ ) Max lateness( O ) (a.5) Swap i and j to get O’ Repea t O(n 2 ) times Repea t O(n 2 ) times
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Same lateness Same lateness Same lateness Max lateness( O’ ) ≤ Max lateness( O ) i i j j O i i j j O’ d i > d j Lateness of j in O’ ≤ Lateness of j in O Lateness of j in O’ ≤ Lateness of j in O Lateness of i in O’ ≤ Lateness of j in O Lateness of i in O’ ≤ Lateness of j in O Lateness of i in O’ = t 1 t 3 t 2 t 3 - d i < t 3 - d j = Lateness of j in O
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Rest of today Shortest Path Problem http://xkcd.com/85/
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Reading Assignment Sec 2.5 of [KT]
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Shortest Path problem Input: Directed graph G=(V,E) Edge lengths, l e for e in E “start” vertex s in V Output:
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CSE331 Lecture 23 - Lecture 23 CSE 331 Oct 24, 2011 Warning...

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