Excursions in Modern Mathematics Sixth Edition

1. Excursions in Modern MathematicsSixth Edition Peter Tannenbaum

2. Chapter 6The Traveling Salesman Problem Hamilton Joins the Circuit

3. The Traveling Salesman ProblemOutline/learning Objectives • To identify and model Hamilton circuit and Hamilton path problems. • To recognize complete graphs and state the number of Hamilton circuits that they have. • To identify traveling-salesman problems and the difficulties faced in solving them.

4. The Traveling Salesman ProblemOutline/learning Objectives • To implement brute-force, nearest-neighbor, repeated nearest-neighbor, and cheapest-link algorithms to find approximate solutions to traveling –salesman problems. • To recognize the difference between efficient and inefficient algorithms. • To recognize the difference between optimal and approximate algorithms.

5. The Traveling Salesman Problem 6.1 Hamilton Circuits and Hamilton Paths

6. The Traveling Salesman Problem • Hamilton path A path that visits each vertex of the graph once and only once. • Hamilton circuit A circuit that visits each vertex of the graph once and only once (at the end, of course, the circuit must return to the starting vertex).

7. The Traveling Salesman Problem Hamilton Circuit (Paths vs Euler Circuit Path Figure (a) shows a graph that has Euler circuits and has Hamilton circuits. One such Hamilton circuit is A, F, B, C, G, D, E, A. Note that once a graph has a Hamilton circuit, it automatically has a Hamilton path-- The Hamilton circuit can always be truncated into a Hamilton path by dropping the last vertex of the circuit.

8. The Traveling Salesman Problem Hamilton Circuit (Paths vs Euler Circuit Path Figure (b) shows a graph that has no Euler circuits but does have Euler paths (for example C, D, E, B, A, D), has no Hamilton circuits (sooner or later you have to go to C, and then you are stuck) but does have Hamilton paths (for example, A, B, E, D, C). Aha, a graph can have a Hamilton path but no Hamilton Circuit!

9. The Traveling Salesman Problem Hamilton Circuit (Paths vs Euler Circuit Path Figure (c) shows a graph that has neither Euler circuits nor paths (it has four odd vertices), has Hamilton circuits (for example A, B, C, D, E, A– there are plenty more), and consequently has Hamilton paths (for example, A, B, C, D, E).

10. The Traveling Salesman Problem Hamilton Circuit (Paths vs Euler Circuit Path Figure (d) shows a graph that has Euler circuits (the vertices are all even), has no Hamilton circuits (no matter what, your are going to have to go through E more than once!) but has Hamilton paths (for example, A, B, E, D, C).

11. The Traveling Salesman Problem Hamilton Circuit (Paths vs Euler Circuit Path Figure (e) shows a graph that has no Euler circuits but has Euler paths (F and G are the two odd vertices), had neither Hamilton circuits nor Hamilton paths.

12. The Traveling Salesman Problem Hamilton Circuit (Paths vs Euler Circuit Path Figure (f) shows a graph that has neither Euler circuits nor Euler paths (too many odd vertices), has neither Hamilton circuits nor Hamilton paths.

13. The Traveling Salesman Problem The lesson in the previous Example is that the existence of an Euler path or circuit in a graph tells us nothing about the existence of a Hamilton path or circuit in that graph. This is important because it implies that Euler’s circuit and path theorems from Chapter 5 are useless when it comes to Hamilton circuits and paths.

14. The Traveling Salesman Problem There are, however, nice theorems that identify special situations where a graph must have a Hamilton circuit. This best known of these theorems is Dirac’s theorem: If a connected graph has N vertices (N > 2) and all of them have degree bigger or equal to N / 2, then the graph has a Hamilton circuit.

15. The Traveling Salesman Problem 6.2 Complete Graphs

16. The Traveling Salesman Problem If a graph has a Hamilton circuit, then how many different Hamilton circuits does a it have? A graph with N vertices in which every pair of distinct vertices is joined by an edge is called a complete graph on N vertices and denoted by the symbol KN.

17. The Traveling Salesman Problem Number of Edges in KN • KN has N(N – 1)/2 edges. • Of all graphs with N vertices and no multiple edges or loops, KN has the most edges.

18. The Traveling Salesman Problem Hamilton Circuits in K4 If we travel the four vertices of K4 in an arbitrary order, we get a Hamilton path. For example, C, A, D, B is a Hamilton path.

19. The Traveling Salesman Problem Hamilton Circuits in K4 D, C, A, B is another Hamilton Path.

20. The Traveling Salesman Problem Hamilton Circuits in K4 Each of these Hamilton paths can be closed into a Hamilton circuit-- the path C, A, D, B begets the circuit D, A, D, B, C.

21. The Traveling Salesman Problem Hamilton Circuits in K4 The path D, C, A, B begets the circuit D, C, A, B, D.

22. The Traveling Salesman Problem Hamilton Circuits in K4 It is important to remember that the same Hamilton circuit can be written in many ways.

23. The Traveling Salesman Problem Hamilton Circuits in K4 For example, C, A, D, B, C is the same circuit as A, D, B, C, A– the only difference is that in the first case we used C as the reference point in the second case we used A.

24. The Traveling Salesman Problem Number of Hamilton Circuits in KN There are (N – 1)! Distinct Hamilton circuits in KN.

25. The Traveling Salesman Problem 6.3 Traveling Salesman Problems

26. The Traveling Salesman Problem The “ traveling salesman” is a convenient metaphor for many different important real-life applications, all involving Hamilton circuits in complete graphs but only occasionally involving salespeople.

27. The Traveling Salesman Problem Any graph whose edges have numbers attached to them is called a weighted graph, and the numbers are called the weights of the edges. The graph is called a complete weighted graph.

28. The Traveling Salesman Problem The problem we want to solve is fundamentally the same– find an optimal Hamilton circuit (a Hamilton circuit with least total weight) for the given weighted graph.

29. The Traveling Salesman Problem 6.4 Simple Strategies for Solving TSPs

30. The Traveling Salesman Problem • Strategy 1 (Exhaustive Search) • Make a list of all possible Hamilton circuits. For each circuit in the list, calculate the total weight of the circuit. From all the circuits, choose the circuit with smallest total weight.

31. The Traveling Salesman Problem • Strategy 2 (Go Cheap) • Start from the home city. From there go to the city that is the cheapest to get to. From each new city go to the next city that is cheapest to get to. When there are no more new cities to go to, go back home.

32. The Traveling Salesman Problem 6.5 The Brute-Force and Nearest Neighbor Algorithms

33. The Traveling Salesman Problem The Exhaustive Search strategy can be formalized into an algorithm generally known as the brute-force algorithm; the Go Cheap strategy can be formalized into an algorithm known as the nearest-neighbor algorithm. In both cases, the objective of the algorithm is to find an optimal (cheapest, shortest, fastest) Hamilton circuit in a complete weighted graph.

34. The Traveling Salesman Problem Algorithm 1: The Brute-Force Algorithm • Step 1. Make a list of all the possible Hamilton circuits of the graph.

35. The Traveling Salesman Problem Algorithm 1: The Brute-Force Algorithm • Step 2. For each Hamilton circuit calculate its total weight (add the weights of all the edges in the circuit).

36. The Traveling Salesman Problem Algorithm 1: The Brute-Force Algorithm • Step 3. Choose an optimal circuit (there is always more than one optimal circuit to choose from!).

37. The Traveling Salesman Problem Algorithm 2: The Nearest-Neighbor Algorithm • Start. Start at the designated starting vertex. If there is no designated starting vertex, pick any vertex.

38. The Traveling Salesman Problem Algorithm 2: The Nearest-Neighbor Algorithm • First step. From the starting vertex go to its nearest neighbor (the vertex for which the corresponding edge has the smallest weight.

39. The Traveling Salesman Problem Algorithm 2: The Nearest-Neighbor Algorithm • Middle steps. From each vertex go to its nearest neighbor, choosing only among the vertices that haven’t been yet visited. (If there is more than one, choose at random). Keep doing this until all the vertices have been visited.

40. The Traveling Salesman Problem • The brute-force algorithm is a classic example of what is formally known as an inefficient algorithm– an algorithm for which the number of steps needed to carry it out grows disproportionately with the size of the problem.

41. The Traveling Salesman Problem • The nearest-neighbor algorithm is an efficient algorithm. Roughly speaking, an efficient algorithm is an algorithm for which the amount of computational effort required to implement the algorithm grows in some reasonable proportion with the size of the input to the problem.

42. The Traveling Salesman Problem 6.6 Approximate Algorithm

43. The Traveling Salesman Problem A really good algorithm for solving TSP’s in general would have to be both efficient (like the nearest-neighbor) and optimal (like the brute-force). Unfortunately, nobody knows of such an algorithm. We will use the term approximate algorithm to describe any algorithm that produces solutions that are, most of the time, reasonably close to the optimal solution.

44. The Traveling Salesman Problem 6.7 The Repetitive Nearest-Neighbor Algorithm

45. The Traveling Salesman Problem • Let X be any vertex. Find the nearest-neighbor circuit using X as the starting vertex and calculate the total cost of the circuit. Algorithm 3: The Repetitive Nearest-Neighbor Algorithm

46. The Traveling Salesman Problem • We compute the nearest-neighbor circuit with A as the starting vertex, and we got A, C, E, D, B, A with a total cost of $773. Algorithm 3: The Repetitive Nearest-Neighbor Algorithm

47. The Traveling Salesman Problem • Repeat the process with each of the other vertices of the graph as the starting vertex. Algorithm 3: The Repetitive Nearest-Neighbor Algorithm

48. The Traveling Salesman Problem • If we use B as the starting vertex, the nearest-neighbor circuit takes us from B to C, then to A, E, D, and back to B, with a total cost of $722. Algorithm 3: The Repetitive Nearest-Neighbor Algorithm

49. The Traveling Salesman Problem • Remember we must start and end the trip at A– this very same circuit would take the form A, E, D, B, C, A. Algorithm 3: The Repetitive Nearest-Neighbor Algorithm

50. The Traveling Salesman Problem • The process is once again repeated using C, D, and E as the starting vertices with respective costs of $722, $722, and $741. Algorithm 3: The Repetitive Nearest-Neighbor Algorithm