Lecture 39

CSE 331Dec 9, 2009

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Sample final has been posted

Graded HW 9 on Friday

Shortest Path Problem

Input: (Directed) Graph G=(V,E) and for every edge e has a cost ce (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

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 { cv,w + OPT(i-1, w)} }

Path uses ≤ i-1 edgesPath uses ≤ i-1 edges Best path through all neighbors

Best path through all neighbors

Some consequencesOPT(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 { cv,w + OPT(i-1, w)} }

OPT(n-1,v) is shortest path cost between v and tOPT(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

Today’s agenda

Finish Bellman-Ford algorithm

Look at a related problem: longest path problem