1. Lecture 22 CSE 331 Oct 26, 2009

2. Blog posts Please sign up if you have not If I have a pick a blogger I will only pick ONE/lecture Will lose out on 5% of your grade 32 of you have signed up: 4 lectures with no volunteers Nov 6 is now open

3. How to do better in this class Look up the blog post Helpful for job interviews!

4. Two definitions for schedules f=1 For every i in 1..n do Schedule job i from s i =f to f i =f+t i f=f+t i Idle time Inversion Max “gap” between two consecutively scheduled tasks Idle time =1Idle time =0 (i,j) is an inversion if i is scheduled before j but d i > d j i i j j i i j j 0 idle time and 0 inversions for greedy schedule

5. We proved last lecture Any two schedules with 0 idle time and 0 inversions have the same max lateness

6. 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

7. 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

8. 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)

9. Today’s agenda “Exchange” argument to convert an optimal solution into a 0 inversion one Shortest Path problem