6.2 Initial Problem. Section 6.2 Network Problems. 6.2 Initial Problem, cont d. Weighted Graphs. Weighted Graphs, cont d. Weighted Graphs, cont d

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1 Section 6.2 Network Problems Goals Study weighted graphs Study spanning trees Study minimal spanning trees Use Kruskal s algorithm 6.2 Initial Problem Walkways need to be built between the buildings on the campus pictured at right. 6.2 Initial Problem, cont d How can the walkways be constructed so that: Students can go from any building on campus to any other building without needing to walk outside. The total length of the walkways is minimized. The solution will be given at the end of the section. Weighted Graphs A graph in which each edge has a number associated with it is called a weighted graph. The number corresponding to an edge is called the weight of the edge. Weighted Graphs, cont d Typically the weights of the edges in a graph are assigned in a way that represents some physical situation or practical problem. The lengths of the edges in a graph do not have to be proportional to the stated weights. Weighted Graphs, cont d An example of a weighted graph is shown below. 1

2 Suppose Ed has: Example 1 A 35-minute drive from his home to his workplace A 15-minute drive from his workplace to his health club A 25 minute drive from his health club to his home Draw a weighted graph to represent this situation. Solution: Example 1, cont d The vertices of the graph will represent Ed s home (H), his workplace (W), and his health club (C). The weight of each edge will represent the driving time between the two places. For simplicity, we can draw all the edges as straight lines and with the same length. Example 1, cont d One possible weighted graph is shown below. Redundant Edges Redundant edges are edges that can be removed while still leaving a path between the two vertices. Example: If edge CH is removed from the previous graph, Ed can still get from home to his club by driving to work and then to the club. The edge CH is a redundant edge. Question: Are there any redundant edges in the graph? a. yes b. no A subgraph of a given graph is a set of vertices and edges chosen from among those of the original graph. Example: Removing edge CH creates the subgraph shown at right. Subgraphs 2

Subgraphs, cont d Example: Another subgraph could be created by removing vertex C and its two incident edges. Example 2 A bus line goes between the cities shown in the table.

3 Subgraphs, cont d Example: Another subgraph could be created by removing vertex C and its two incident edges. Example 2 A bus line goes between the cities shown in the table. a) Draw a weighted graph to represent the situation. b) Find two different routes from Selma to Anniston. Example 2, cont d a) Solution: A weighted graph is shown below. Example 2, cont d b) Solution: One possible route from Selma to Anniston is SMGA, with a total weight (length) of = 203 miles. Example 2, cont d b) Solution, cont d: Another possible route from Selma to Anniston is SBA, with a total weight (length) of = 222 miles. Trees One way to recognize that a graph contains redundant edges is to find a circuit in the graph. Recall that a circuit is a path that begins at a vertex and returns to that vertex without using any edges twice. A connected graph that has no circuits is called a tree. 3

4 Question: Choose the graph that is a tree. Trees, cont d The following figure gives examples of graphs that are trees and a graph that is not a tree. a. b. c. d. Example 3 This graph is not a tree because it contains several circuits. Find a subgraph that is a tree. Example 3, cont d Solution: Circuits are highlighted and edges are removed until the remaining subgraph has no circuits left. Spanning Trees A spanning tree is a subgraph that: Contains all the original vertices of the graph. Is connected. Contains no circuits. Spanning Trees, cont d Every connected graph has at least one spanning tree. An example of a spanning tree is shown below. 4

5 Minimal Spanning Trees A spanning tree with the smallest possible total weight is called a minimal spanning tree. Since a connected, weighted graph can have multiple spanning trees, there may be one or more than one of those spanning trees with the smallest total weight. Minimal Spanning Trees, cont d For the connected, weighted graph shown below: Subgraph (b) is not a minimal spanning tree. Subgraph (c) is a minimal spanning tree. Kruskal s Algorithm A method known as Kruskal s algorithm can be used to find a minimal spanning tree of a connected, weighted graph. Kruskal s algorithm begins with all the vertices of a graph and adds edges one by one, using the idea of acceptable and unacceptable edges. Kruskal s Algorithm, cont d Acceptable edges are Edges that do not share a vertex with any edges already chosen. Kruskal s Algorithm, cont d Acceptable edges are Edges that connect two components of the subgraph. Kruskal s Algorithm, cont d Acceptable edges are Edges that connect a component of the subgraph to a new vertex. 5

6 Kruskal s Algorithm, cont d Unacceptable edges are Edges that add to a component of the subgraph, but do not add a new vertex. Kruskal s Algorithm, cont d 1) Consider only the vertices of the graph. 2) Add the edge with the smallest weight. 3) Add the acceptable edge with the smallest weight. 4) Determine whether all vertices are connected by a path. a) If so, you have a minimal spanning tree. b) If not, repeat Step 3. Construct a minimal spanning tree for the weighted graph. Example 4 Example 4, cont d Solution: Start with the 4 vertices. Add the edge with the smallest weight. Example 4, cont d All unused edges are acceptable, so we simply choose the edge with the smallest weight. Select the edge with weight 2.6. Example 4, cont d The acceptable edges at this step were the ones with weights 4.8 and 3.7. Select the edge with weight

7 Example 4, cont d The subgraph is now connected, so this is a minimal spanning tree. The weight of the tree is = Initial Problem Solution How can a college build walkways of the minimum length possible so that students can go from any building on campus to any other building without needing to walk outside? Initial Problem Solution, cont d Create a weighted graph to represent the buildings and existing sidewalks on the campus. The weight of each edge is the distance between the two buildings. Initial Problem Solution, cont d Use Kruskal s algorithm to find a minimal spanning tree. The edges included in the minimal spanning tree represent the sidewalks that should be converted into covered walkways. Initial Problem Solution, cont d Under this plan, students can get from any building to any other building without going outside, although they may have to walk farther inside. The length of the walkways was minimized to a total of = 750 feet. 7

8 Section 6.3 Traveling-Salesperson Problem 6.3 Initial Problem Goals Study Hamiltonian circuits Study the Traveling-Salesperson Problem Use the Brute-Force Algorithm Use the Nearest-Neighbor Algorithm Use the Cheapest-Link Algorithm An author needs to find a route that will minimize the total distance she must drive between the cities. The solution will be given at the end of the section. Hamiltonian Circuits Example 1 A path that visits each vertex in a graph exactly once is called a Hamiltonian path. If a Hamiltonian path begins and ends at the same vertex, it is called a Hamiltonian circuit. The highlighted edges in the graph represent a Hamiltonian path. The path can be written AGFEDBC or CBDEFGA. Example 2 Hamiltonian Circuits, cont d The highlighted edges in the graph represent a Hamiltonian circuit. The circuit, starting at A, can be written AGFEDCBA Note: A Hamiltonian path or circuit must pass through every vertex of the graph, but it does not have to use every edge. A Hamiltonian path can be used, for example, to model a delivery truck stopping at every house in a neighborhood but not using every street. 8

9 Example 3 Find a Hamiltonian path in each graph or explain why you cannot. Example 3, cont d a) Solution: The path ABCD is a Hamiltonian path. Adding an edge to the path forms the Hamiltonian circuit ABCDA. Other Hamiltonian paths and circuits are possible. Example 3, cont d b) Solution: The path ABCD is a Hamiltonian path. There is no Hamiltonian circuit. Example 3, cont d c) There is no Hamiltonian path. Any Hamiltonian path must start at either A or F. Any path that starts at A will leave out either C, D, or F. Any path that starts at F will leave out either C, D, or A. Complete Graphs Example 4 A complete graph is a graph in which every pair of vertices is connected by exactly one edge. For each graph, determine whether or not it is complete. 9

Example 4, cont d Example 4, cont d a) Solution: Check each pair of vertices. There is exactly one edge between each pair of vertices, so the graph is complete.

10 Example 4, cont d Example 4, cont d a) Solution: Check each pair of vertices. There is exactly one edge between each pair of vertices, so the graph is complete. b) Solution: Check each pair of vertices. There is no edge from A to C, for example. The graph is not complete. Example 4, cont d Example 5 c) Solution: Check each pair of vertices. There are two edges from A to F, for example. The graph is not complete. Draw an example of a complete graph with five vertices. Count the number of edges in the graph. Example 5, cont d Solution: Two possible representations of such a graph are shown below. A complete graph with 5 vertices has 10 edges. Theorem on Complete Graphs The number of edges in a complete graph always follows a certain pattern. A complete graph with n vertices n always has ( n 1) edges. 2 10

Question: A graph has 20 vertices and 195 edges. Is the graph complete? a. yes b. no Complete Graphs, cont d In general, a graph may or may not have a Hamiltonian path or circuit.

11 Question: A graph has 20 vertices and 195 edges. Is the graph complete? a. yes b. no Complete Graphs, cont d In general, a graph may or may not have a Hamiltonian path or circuit. A complete graph, however, always has a Hamiltonian circuit. Any list of all the vertices in the graph will be a Hamiltonian path. Adding the starting vertex to the end of the list describing the path will create a Hamiltonian circuit. List all the Hamiltonian paths and all the Hamiltonian circuits in the graph. Example 6 Solution: Example 6, cont d Example 6, cont d Note that the complete graph in the previous example had 3 vertices and 6 different Hamiltonian circuits. This is the case because each circuit can begin at one of the 3 vertices, travel to one of 2 vertices second, and travel to the 1 remaining vertex last. There are 3(2)(1) = 6 Hamiltonian circuits. Theorem on Complete Graphs The number of Hamiltonian paths in a complete graph with n vertices is n! = n(n 1)(n 2)(n 3) (1). The symbol n! is read n factorial. 11

12 Question: How many possible Hamiltonian circuits exist in a complete graph with 9 vertices? a. 81 b. 40,320 c. 362,880 d. 3,265,920 Traveling-Salesperson Problem A salesperson who must visit various cities and then return home wants to minimize the cost (measured in distance, time, or money) of his or her trip. Because Hamiltonian circuits can be applied to find this type of least-cost trip, the problem of finding a least-cost Hamiltonian circuit is called the Traveling-Salesperson Problem. Complete Weighted Graphs The cost of a path in a weighted graph is the sum of the weights assigned to the edges in the path. When costs are assigned to each edge in a complete graph, the graph is called a complete weighted graph. Question: What is the cost of the path ABD in the graph below? a. 26 b. 14 c. 24 d. 41 Example 7 Example 7, cont d List all possible Hamiltonian paths, and their costs, in the complete weighted graph. Find the path of least cost. Solution: The graph has 4 vertices, so there are 4! = 24 different Hamiltonian paths. These paths and their costs are listed on the next slide. 12

Solution: Example 7, cont d Example 7, cont d Traveling-Salesperson, cont d There are 2 leastcost Hamiltonian circuits with a cost of 74.

13 Example 7, cont d Example 7, cont d There are 2 Hamiltonian paths with the lowest cost of 37. CDAB BADC Example 7, cont d List all possible Hamiltonian circuits that start and end at vertex A. Find the circuit at A of least cost. Solution: Example 7, cont d Example 7, cont d Traveling-Salesperson, cont d There are 2 leastcost Hamiltonian circuits with a cost of 74. ABCDEA AEDCBA These 2 circuits are mirror images of each other. The method used to solve the previous example is called the Brute-Force Algorithm. List all possible Hamiltonian circuits. Calculate the cost of each circuit. Find the 2 mirror image circuits of least cost. This method is very time-consuming. 13

14 Traveling-Salesperson, cont d Traveling-Salesperson, cont d Instead of the Brute-Force Algorithm, faster approximation algorithms are typically used. An approximation algorithm finds a Hamiltonian circuit that is either the leastcost circuit or is one that is not much more costly than the least-cost circuit. Two approximation algorithms will be studied: The Nearest-Neighbor Algorithm The Cheapest-Link Algorithm Nearest-Neighbor Algorithm 1) Specify a starting vertex. 2) If unvisited vertices remain, go from the current vertex to the unused vertex with the lowest-cost connecting edge. 3) If no unvisited vertices remain, return to the starting vertex to finish forming the circuit. Apply the algorithm to approximate the lowest-cost Hamiltonian circuit. Example 8 Solution: Example 8, cont d Let the starting vertex be A. The unvisited vertices are B, C, D, and E. The choice of edges are AB: 13, AC: 11, AD: 17, or AE: 15. Choose edge AC and move to vertex C, the new current vertex. Example 8, cont d The unvisited vertices are B, D, and E. The choice of edges are CB: 20, CD: 12, or CE: 14. Choose edge CD. 14

Example 8, cont d The current vertex is now D. The unvisited vertices are B and E. The choice of edges are DB: 19 or DE: 16. Choose edge DE. Example 8, cont d The current vertex is now E.

15 Example 8, cont d The current vertex is now D. The unvisited vertices are B and E. The choice of edges are DB: 19 or DE: 16. Choose edge DE. Example 8, cont d The current vertex is now E. The unvisited vertex is B. Choose edge EB. Example 8, cont d The current vertex is now B. There are no unvisited vertices. Return to vertex A using edge BA. Example 8, cont d The Hamiltonian circuit is ACDEBA. Its cost is 70. This is at or near the lowest possible cost. Cheapest-Link Algorithm 1) All edges are acceptable and no edges have been selected. 2) Choose an acceptable edge of smallest weight. Cheapest-Link Algorithm, cont d 3) If a Hamiltonian circuit is not formed, determine the set of acceptable edges and repeat Step 2. Unacceptable edges are those that either share a vertex with two selected edges or that will close a circuit that is not Hamiltonian. 4) If the selected edges form a Hamiltonian circuit, this is the low-cost Hamiltonian circuit. 15

Apply the algorithm to approximate the lowest-cost Hamiltonian circuit. Example 9 Solution: Example 9, cont d All edges are acceptable. Select the edge of lowest cost: AC.

16 Apply the algorithm to approximate the lowest-cost Hamiltonian circuit. Example 9 Solution: Example 9, cont d All edges are acceptable. Select the edge of lowest cost: AC. Example 9, cont d All edges are still acceptable. Select the edge of lowest cost: CD. Example 9, cont d Edges BC, CE, and AD are unacceptable. Select the acceptable edge of lowest cost: AB. Example 9, cont d Edges BC, CE, BD, and AD are unacceptable. Select the acceptable edge of lowest cost: DE. Example 9, cont d The only remaining acceptable edge is BE. The low-cost circuit is complete, with a weight of

Topics Covered. Introduction to Graphs Euler s Theorem Hamiltonian Circuits The Traveling Salesman Problem Trees and Kruskal s Algorithm

Topics Covered. Introduction to Graphs Euler s Theorem Hamiltonian Circuits The Traveling Salesman Problem Trees and Kruskal s Algorithm Graph Theory Topics Covered Introduction to Graphs Euler s Theorem Hamiltonian Circuits The Traveling Salesman Problem Trees and Kruskal s Algorithm What is a graph? A collection of points, called vertices