1. Lecture 39 CSE 331 Dec 9, 2009

2. Announcements Please fill in the online feedback form Sample final has been posted Graded HW 9 on Friday

3. 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 Assume that G has no negative cycle

4. Recurrence Relation OPT(i,v) = cost of shortest path from v to t with at most i edges OPT(i,v) = min { OPT(i-1,v), min (v,w) in E { c v,w + OPT(i-1, w) } } Path uses ≤ i-1 edges Best path through all neighbors

5. Some consequences OPT(i,v) = shortest path from v to t with at most i edges OPT(i,v) = min { OPT(i-1,v), min (v,w) in E { c v,w + OPT(i-1, w) } } OPT(n-1,v) is shortest path cost between v and t Group talk time: How to compute the shortest path between s and t given all OPT(i,v) values Group talk time: How to compute the shortest path between s and t given all OPT(i,v) values

6. Today’s agenda Finish Bellman-Ford algorithm Look at a related problem: longest path problem