CSE331 Lecture 22

CSE331 Lecture 22 - Greedy schedule has 0 idle time and 0...

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Lecture 22 CSE 331 Oct 21, 2011
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HW 5 due today Q1, Q2 and Q3 in separate piles I will not take any HW after 1:15pm
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Other HW related stuff HW 4 is available for pickup Solutions to HW 5 at the end of the lecture HW 6 has been posted (link on the blog)
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Some clarifications ANY correct algorithm/solution for a problem is OK Will mostly deal in proof ideas from now on
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Need another scribe for today Any volunteer?
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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 =1 Idle 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 0 idle time and 0 inversions for greedy schedule
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We proved last lecture Any two schedules with 0 idle time and 0 inversions have the same max lateness
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Proving greedy is optimal Any two schedules with 0 idle time and 0 inversions have the same max lateness
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Unformatted text preview: 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) Only need to convert a idle optimal ordering to one with inversions (and idle time) HW 5 due today Q1, Q2 and Q3 in separate piles I will not take any HW after 1:15pm Todays agenda Exchange argument to convert an optimal solution into a 0 inversion one Shortest Path problem Shortest Path problem http://xkcd.com/85/...
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This note was uploaded on 12/11/2011 for the course CSE 331 taught by Professor Rudra during the Fall '11 term at SUNY Buffalo.

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CSE331 Lecture 22 - Greedy schedule has 0 idle time and 0...

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