Shortest Paths Todays announcements: PA3 due 29 Nov 23:59 Final - PowerPoint PPT Presentation

Shortest Paths Today’s announcements: ◮ PA3 due 29 Nov 23:59 ◮ Final Exam, 10 Dec 12:00, OSBO A Today’s Plan: ◮ Dijkstra’s shortest path algorithm 22 7 3 10 s 4 6 8 10 12 4 5 8 7 16 12 8 4 0 Find the shortest path 2 4 8 16 from s to t . 45 4 7 23 6 1 2 3 5 3 8 1 8 9 3 1 4 1 t 1 / 6

Dijkstra’s Single-Source Shortest Path Algorithm (another breadth-first search with a priority queue) 3 10 A B C Assume edge weights are non-negative. -5 1 2 D E Dijkstra’s algorithm is a greedy algorithm (it makes the current best choice without considering future consequences). Intuition: Find shortest paths from source in order of length. ◮ Start at the source vertex (shortest path length = 0) ◮ The next shortest path extends some already discovered shortest path by one edge. Why? ◮ Find it (by considering all one-edge extensions) and repeat. 2 / 6

Dijkstra’s Algorithm Pseudocode 1. Unmark all vertices 2. Initialize the dist to each vertex to ∞ 3. Initialize the dist to the source to 0 4. While there are unmarked vertices left in the graph ◮ Select the unmarked vertex v with the lowest dist ◮ Mark v by setting SPlength=dist ◮ For each edge ( v , w ) ◮ dist( w ) = min(dist( w ), dist( v ) + weight of ( v , w )) 2 2 C B F 1 1 3 9 4 A 10 2 H 2 8 1 1 D E G 7 4 vertex A B C D E F G H dist SPlength 3 / 6

Correctness: The Cloud Proof u Q s P y cloud ◮ Assume Dijkstra’s algorithm finds the correct shortest path to the first k vertices it visits (the cloud ). ◮ But it fails on the ( k + 1)st vertex u . ◮ So there is some shorter path, P , from s to u . ◮ Path P must contain a first vertex y not in the cloud. ◮ But since the path, Q , to u is the shortest path out of the cloud, the path on P upto y must be at least as long as Q . ◮ Thus the whole path P is at least as long as Q . Contradiction (What do we use in that last step?) 4 / 6

Data Structures for Dijkstra’s Algorithm n times: Select the unmarked vertex with the lowest dist. removeMin m times: dist( w ) = min { dist( w ), dist( v ) + weight of ( v , w ) } decreasePriority Runtime: (adjacency matrix or adjacency list?) 5 / 6

Fibonacci Heaps ◮ Another implementation of a Priority Queue ◮ Amortized O (1) time for decreasePriority. ◮ O (log n ) time for removeMin Dijkstra’s uses n removeMins and m decreasePriorities Runtime with Fibonacci heaps: 6 / 6