1. Discrete Mathematics and Its Applications (5th Edition)
Shortest Paths
Text
Discrete Mathematics and Its Applications (5th Edition)
Kenneth H. Rosen
Chapter 10.6
Based on slides from Chuck Allison, Michael T. Goodrich, and Roberto Tamassia
By Longin Jan Latecki

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

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

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

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

15. Dijkstra Animation Youtube video:
Java animation:

16. Problem: shortest path from a to zd 5 5 4 7 3 4 1 2 a z 3 4 c 5 e 5 g a b c d e f g z S ∞ x
4(a)
3(a)

17. 1 2 3 4 5 6 7 S ∞ x
15(1)
35(1)
20(1) 1 5 7 2 3 4 6 20 40 15 35 10 50 75

18. 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.

19. 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).

20. The Traveling Salesman Problem
Importance:
Variety of scheduling application can be solved as a traveling salesmen problem.
Examples:
Ordering drill position on a drill press.
School bus routing.

21. 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.

22. FEMA traveling salesman Network representation

23. 40 2 3 25 35 50 40 50 1 4 45 65 30 80
Home

24. 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.

25. FEMA – full enumeration
Possible cycles
Cycle Total Cost
1. H-O1-O2-O3-O4-H 210
2. H-O1-O2-O4-O3-H
3. H-O1-O3-O2-O3-H
4. H-O1-O3-O4-O2-H
5. H-O1-O4-O2-O3-H
6. H-O1-O4-O3-O2-H
7. H-O2-O3-O1-O4-H
8. H-O2-O1-O3-O4-H
9. H-O2-O4-O1-O3-H
10. H-O2-O1-O4-O3-H
11. H-O3-O1-O2-O4-H
12. H-O3-O1-O2-O4-H 260
Minimum
For this problem we have (5-1)! / 2 = 12 cycles. Symmetrical problems need to enumerate only (m-1)! / 2 cycles.

26. FEMA – optimal solution40 2 3 25 35 50 40 1 50 4 45 65 30 80
Home

27. The Traveling Salesman Problem
Finding the shortest cycle is different than Dijkstra’s shortest path. It is much harder too, no algorithm exists with polynomial worst-case time complexity!
This means that for large numbers of vertices, solving the traveling salesman problem is impractical.
The problem has theoretical importance because it represents a class of difficult problems known as NP-hard problems.
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.