Lecture 23

CSE 331Oct 28, 2009

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Will not always do COMPLETE proofs

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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 (di > dj)

(b) O’ has one less inversion than O

(c) Max lateness(O’) ≤ Max lateness(O)

(a.5) Swap i and j to get O’Repeat O(n2) times

Repeat O(n2) times

Exercise: Prove by induction.

No consecutive inversion No inversions

Exercise: Prove by induction.

No consecutive inversion No inversions

Same latenessSame latenessSame lateness

Max lateness(O’) ≤ Max lateness(O)

ii jjO

iijjO’

di > dj

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’ =

t1 t3

t2

t3 - di < t3 - dj = Lateness of j in O

Rest of today

Shortest Path Problem

http://xkcd.com/85/

Reading AssignmentSec 2.5of [KT]