CSE331 Lecture 38

# CSE331 Lecture 38 - Lecture 38 CSE 331 Dec 2, 2011 Review...

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Lecture 38 CSE 331 Dec 2, 2011

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Review Sessions etc. Atri: (at least ½ of) class next Friday Jiun-Jie: Extra office hour next Friday Jesse: Review Session on Mon, Dec 12 (pick time slot on blog) Temp grades hopefully sometime next week
Homework 10 Posted on the blog

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Shortest Path Problem Input: (Directed) Graph G=(V,E) and for every edge e has a cost c e (can be <0 ) t in V Output: Shortest path from every s to t 1 1 100 -1000 899 s t Shortest path has cost negative infinity Shortest path has cost negative infinity Assume that G has no negative cycle Assume that G has no negative cycle
When to use Dynamic Programming There are polynomially many sub-problems Optimal solution can be computed from solutions to sub-problems There is an ordering among sub-problem that allows for iterative solution Richard Bellman

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OPT(i,u) = cost of shortest path from u to t with at most i edges OPT(i,u) = min { OPT(i-1,u) , min (u,w) in E { c
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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 38 - Lecture 38 CSE 331 Dec 2, 2011 Review...

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