Chapter 6. The Traveling-Salesman Problem. Section 1. Hamilton circuits and Hamilton paths.

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1 Chapter 6. The Traveling-Salesman Problem Section 1. Hamilton circuits and Hamilton paths. Recall: an Euler path is a path that travels through every edge of a graph once and only once; an Euler circuit is a circuit that travels through every edge of a graph once and only once; A Hamilton path is a path that travels through every vertex of a graph once and only once; a Hamilton circuit is a circuit that travels through every vertex of a graph once and only once. Look at the examples on page 206. They show that Euler circuits and Hamilton circuits have almost nothing to do with each other. In the last chapter, we learned a simple rule for whether or not there exists an Euler circuit. No such rule has yet been discovered for the existence of a Hamilton circuit! However, we will mostly be interested in a special type of graph which has a Hamilton circuit. Section 2. Complete graphs. Look at the graphs on p. 207 (or the blackboard). They are called complete graphs. There is exactly one edge connecting each pair of vertices. If a complete graph has 2 vertices, then it has 1 edge. If a complete graph has 3 vertices, then it has 1+2=3 edges. If a complete graph has 4 vertices, then it has 1+2+3=6 edges. If a complete graph has N vertices, then it has (N-1)= (N-1)*N/2 edges. The complete graph with 4 vertices is written K4, etc. 1

2 Every complete graph has a Hamilton circuit. K3 has 6 of them: ABCA, BCAB, CABC and their mirror images ACBA, BACB, CBAC. The first three circuits are the same, except for what vertex we choose as starting point; the other three are the same as the first three, except for the direction in which we travel: A A B C B C We ll ignore starting points (but not direction of travel), and say that K3 has two Hamilton circuits. In Table 6-2, p.208, the book shows that K4 has 6=2*3 Hamilton circuits. Similarly, K5 has 24=2*3*4 Hamilton circuits. These numbers are called factorials: 3!=1*2*3, 4!=1*2*3*4, and k!=1*2*3*... *k. The complete graph with N vertices, KN, has (N-1)! Hamilton circuits. Look at Table 6-4 on p.209 for some of these numbers. 10!= != != != Notice how quickly these numbers become huge! 2

3 Section 3. Traveling-salesman problems. In these problems, you want to travel to several locations (these are the vertices). You can get from each location to every other location, and you pick out one best way to do that (this will be the edge connected those two vertices). Note that this is a complete graph. Our problem is to visit all of the locations exactly once, that is, to find a Hamilton circuit there are many of these we want to find the best one! Best might mean fastest, shortest, cheapest or something else. Look at the book s examples 4, 5 and 6. Attached to each edge is a number, called the weight of that edge, representing how good that route is ( smaller is better ). The total weight of a Hamilton circuit is the sum of the weights of all the edges in that circuit. So the traveling-salesman problem, TSP for short, is to find the Hamilton circuit with the smallest total weight. 3

4 Section 4. Simple strategies for solving TSPs. Section 5. The brute-force and nearest-neighbor algorithms. We can solve a TSP by listing all Hamilton circuits, calculating the total weight of each one, and then chosing the one which has least total weight. This is called the brute-force algorithm. Example: A 6 B C 8 D This is K4; there are 3!=6 Hamilton circuits (ignoring start point). Hamilton circuit total weight mirror-image circuit 1 A,B,C,D,A =26 A,D,C,B,A 2 A,B,D,C,A =23 A,C,D,B,A 3 A,C,B,D,A =21 A,D,B,C,A We see that the third family of circuits has the least total weight. This method is good if we only have a few vertices, but we ve seen that the number of Hamilton circuits grows very rapidly as we increase the number of vertices. It quickly becomes impossible to do the calculation by hand. Even with a computer, this method becomes impractical with fairly small numbers of vertices (around 20). 4

5 Here is another method which takes a lot less work: The nearest-neighbor algorithm: 1. Choose any starting vertex. 2. Whenever you reach any vertex, look at the weights of all the edges that lead to vertices you haven t visited yet. Choose the one of least weight. 3. Once you ve reached the last vertex, go back to the starting point. Example: A 6 B C 8 D If we start at A, the nearest neighbor is C, then B, then we have no choice but to go to D, then A. That gives us the optimal Hamiltonian circuit A,C,B,D,A of total weight 21. If we start at C, we go to A, B, D, C. This is a non-optimal Hamiltonian circuit of total weight 23. So the nearest-neighbor algorithm takes less work, but doesn t necessarily give the best answer. However, it usually produces an answer which is close to best. 5

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