lecture 39
DESCRIPTION
Lecture 39. CSE 331 Dec 9, 2009. Announcements. Please fill in the online feedback form. 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 c e (can beTRANSCRIPT
Lecture 39
CSE 331Dec 9, 2009
Announcements
Please fill in the online feedback form
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