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Shortest Paths

Shortest Paths. Text Discrete Mathematics and Its Applications (5 th Edition) Kenneth H. Rosen Chapter 9.6 Based on slides from Chuck Allison, Michael T. Goodrich, and Roberto Tamassia By Longin Jan Latecki. BOS. NY. CHI. SF. DEN. ATL. LA. MIA. Weighted Graphs.

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Shortest Paths

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  1. Shortest Paths Text Discrete Mathematics and Its Applications(5th Edition) Kenneth H. Rosen Chapter 9.6 Based on slides from Chuck Allison, Michael T. Goodrich, and Roberto Tamassia By Longin Jan Latecki

  2. BOS NY CHI SF DEN ATL LA MIA Weighted Graphs Graphs that have a number assigned to each edge are called weighted graphs.

  3. BOS NY CHI SF DEN ATL LA MIA Weighted Graphs MILES 860 2534 191 1855 722 908 957 760 606 834 349 2451 1090 595

  4. BOS NY CHI SF DEN ATL LA MIA Weighted Graphs FARES $129 $79 $39 $99 $59 $69 $89 $79 $99 $89 $39 $129 $69

  5. BOS NY CHI SF DEN ATL LA MIA Weighted Graphs FLIGHT TIMES 4:05 2:10 0:50 2:55 1:50 2:10 2:20 1:55 1:40 2:45 2:00 3:50 1:15 1:30

  6. Weighted Graphs • A weighted graph is a graph in which each edge (u, v) has a weight w(u, v). Each weight is a real number. • Weights can represent distance, cost, time, capacity, etc. • The length of a path in a weighted graph is the sum of the weights on the edges. • Dijkstra’s Algorithm finds the shortest path between two vertices.

  7. Dijkstra's Algorithm

  8. Dijkstra Animation • Demo

  9. Problem: shortest path from a to z f b d 5 5 4 7 3 4 1 2 a z 3 4 c 5 e 5 g

  10. 1 20 15 5 35 75 10 2 3 7 50 40 35 15 4 6 10

  11. Theorems Dijkstra’s algorithm finds the length of a shortest path between two vertices in a connected simple undirected weighted graph G=(V,E). The time required by Dijkstra's algorithm is O(|V|2). It will be reduced to O(|E|log|V|) if heap is used to keep {vV\Si : L(v) < }, where Si is the set S after iteration i.

  12. The Traveling Salesman Problem • The traveling salesman problem is one of the classical problems in computer science. • A traveling salesman wants to visit a number of cities and then return to his starting point. Of course he wants to save time and energy, so he wants to determine the shortest cycle for his trip. • We can represent the cities and the distances between them by a weighted, complete, undirected graph. • The problem then is to find the shortest cycle (of minimum total weight that visits each vertex exactly one). • Finding the shortest cycle is different than Dijkstra’s shortest path. It is much harder too, no polynomial time algorithm exists!

  13. The Traveling Salesman Problem • Importance: • Variety of scheduling application can be solved as atraveling salesmen problem. • Examples: • Ordering drill position on a drill press. • School bus routing. • The problem has theoretical importance because it represents a class of difficult problems known as NP-hard problems.

  14. THE FEDERAL EMERGENCY MANAGEMENT AGENCY • A visit must be made to four local offices of FEMA, going out from and returning to the same main office in Northridge, Southern California.

  15. FEMA traveling salesmanNetwork representation

  16. 40 2 3 25 35 50 40 50 1 4 65 45 30 80 Home

  17. FEMA - Traveling Salesman • Solution approaches • Enumeration of all possible cycles. • This results in (m-1)! cycles to enumerate for a graph with m nodes. • Only small problems can be solved with this approach.

  18. FEMA – full enumeration Possible cycles Cycle Total Cost 1. H-O1-O2-O3-O4-H 210 2. H-O1-O2-O4-O3-H 195 3. H-O1-O3-O2-O3-H 240 4. H-O1-O3-O4-O2-H 200 5. H-O1-O4-O2-O3-H 225 6. H-O1-O4-O3-O2-H 200 7. H-O2-O3-O1-O4-H 265 8. H-O2-O1-O3-O4-H 235 9. H-O2-O4-O1-O3-H 250 10. H-O2-O1-O4-O3-H 220 11. H-O3-O1-O2-O4-H 260 12. H-O3-O1-O2-O4-H 260 Minimum For this problem we have (5-1)! / 2 = 12 cycles. Symmetrical problemsneed to enumerate only (m-1)! / 2 cycles.

  19. FEMA – optimal solution 40 2 3 25 35 50 40 1 50 4 65 45 30 80 Home

  20. The Traveling Salesman Problem • Unfortunately, no algorithm solving the traveling salesman problem with polynomial worst-case time complexity has been devised yet. • This means that for large numbers of vertices, solving the traveling salesman problem is impractical. • In these cases, we can use efficient approximation algorithms that determine a path whose length may be slightly larger than the traveling salesman’s path, but

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