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Lecture 22. CSE 331 Oct 26, 2009. Blog posts. Please sign up if you have not. 32 of you have signed up: 4 lectures with no volunteers. If I have a pick a blogger I will only pick ONE/lecture. Will lose out on 5% of your grade. Nov 6 is now open. How to do better in this class.

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## Lecture 22

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**Lecture 22**CSE 331 Oct 26, 2009**Blog posts**Please sign up if you have not 32 of you have signed up: 4 lectures with no volunteers If I have a pick a blogger I will only pick ONE/lecture Will lose out on 5% of your grade Nov 6 is now open**How to do better in this class**Look up the blog post Helpful for job interviews!**Two definitions for schedules**Idle time Max “gap” between two consecutively scheduled tasks Idle time =1 Idle time =0 f=1 Inversion (i,j) is an inversion if i is scheduled before j but di > dj For every i in 1..n do i Schedule job i from si=f to fi=f+ti j f=f+ti j i 0 idle time and 0 inversions for greedy schedule**We proved last lecture**Any two schedules with 0 idle time and 0 inversions have the same max lateness**Proving greedy is optimal**Any two schedules with 0 idle time and 0 inversions have the same max lateness Greedy schedule has 0 idle time and 0 inversions**Will prove today**Any two schedules with 0 idle time and 0 inversions have the same max lateness Greedy schedule has 0 idle time and 0 inversions There is an optimal schedule with 0 idle time and 0 inversions**Optimal schedule with 0 idle time**≥ ≥ = = Lateness “Only” need to convert a 0 idle optimal ordering to one with 0 inversions (and 0 idle time)**Today’s agenda**“Exchange” argument to convert an optimal solution into a 0 inversion one Shortest Path problem

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