Lecture 23 CSE 331 Oct 28, 2009

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Warning More work for YOU I am running behind on my topics schedule I wanted to show some of the cooler stuff Will not always do COMPLETE proofs More work for YOU

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 (di > dj) Repeat O(n2) times (a.5) Swap i and j to get O’ Exercise: Prove by induction. No consecutive inversion  No inversions (b) O’ has one less inversion than O (c) Max lateness(O’) ≤ Max lateness(O)

Max lateness(O’) ≤ Max lateness(O) di > dj Same lateness Same lateness O i j t1 t3 O’ j i t2 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’ = t3 - di < t3 - dj = Lateness of j in O

Rest of today Shortest Path Problem http://xkcd.com/85/

Reading Assignment Sec 2.5of [KT]