14 Graph Theory. Exercise Set 14-1

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1 14 Graph Theory Exercise Set A graph in this chapter consists of vertices and edges. In previous chapters the term was used as a visual picture of a set of ordered pairs defined by a relation or function. 2. A loop is an edge that connects a vertex to itself. A circuit begins and ends at the same vertex, but can cross over several edges to do this. 3. A circuit begins and ends at the same vertex, but a path may end at a vertex different from where it started. 4. Answers may vary. 5. A graph is connected if there is at least one path between any pair of vertices. 6. Answers may vary. 7. Two vertices are adjacent if there is an edge connecting them. In the context of Exercise 6 it would mean there is a non-stop flight between two cities. 8. Not every graph has a bridge. 9. A coloring is a method of coloring all vertices on a graph so that any pair of vertices joined by an edge have different colors. 10. The chromatic number of a graph is the smallest number of colors that can be used to color a graph. 11. If 5 is the chromatic number, then 5 is the smallest number of colors that can be used. So, 6 would be more than necessary, but ok, and 4 wouldn t be enough. 12. Graph coloring for a map means that no two states that share a border have the same color. 13. The vertices are A, B, C, D, E, F, and G. 14. There are nine edges: AB, AC, CD, DB, BE, EF, EG, FG, and the loop at G. 15. B, G, and F are adjacent to E since there is an edge connecting them to E. 16. A, C, D, and F are even with 2 edges each, G is even with 4 edges. 17. B and E are odd with 3 edges each. 18. The edge from B to E is a bridge since removing it would make the graph disconnected. 19. Answers may vary; one such path is A, B, E, F 20. Answers may vary; one such circuit is A, B, D, C, A 21. Vertex G has a loop. 22. Edges AC, CD, DB, EF, and the loop at G are not included in the path. 23. A, B, E, D is not a path because there is no edge connecting E and D. 24. E, F, G is not a circuit since it doesn t end at the same vertex where it began. 25. Chromatic number is Chromatic number is Chromatic number is

2 28. Chromatic number is Chromatic number is Smallest number of colors is Smallest number of colors is Smallest number of colors is

3 46. Smallest number of colors is Smallest number of colors is Smallest number of colors is Smallest number of colors is Smallest number of colors Smallest number of colors is Smallest number of colors is Smallest number of colors is Answers may vary. 56. Let D = Daisy, R = Rose, L = Lilac, S = Sunflower, I = Iris, M = Marigold, and T = Tulip. 49. Smallest number of colors is No; a vertex is adjacent to itself if there is a loop, so it will be adjacent to a vertex with the same color. 58. The countries lined up on top need to have colors different from the long country but not different from each other: only one they share a border with. 59. Answers may vary. 60. New Jersey 707

4 61. Answers may vary. 62. Answers may vary. 63. a) It means that the house is split into distinct parts that are not connected. b) There s a room that connects one part of a room to another. c) There s a door that, if removed, would make it impossible to get from one part of the house to another. 64. a) It would mean that the graph was drawn wrong. A state can t have a border with itself. b) There s a state that has to be driven through in order to get from one group of states to another. c) A circuit represents a trip through several states that ends back in the original state. Exercise Set An Euler path passes through every edge exactly once; a path passes through a subset of the vertices and edges of the graph. 2. An Euler path must pass every edge once, but may not end where it started; an Euler circuit must pass every edge exactly once and end where it began. 3. A graph has an Euler circuit if all vertices are even. 4. If there are exactly two odd vertices the graph has an Euler path but not an Euler 5. A graph with more than two odd vertices has no Euler paths or circuits. 6. Fleury s Algorithm is used to find an Euler path or 7. The graph has an Euler 8. The graph has an Euler path but not an Euler 9. The graph has neither an Euler path nor an Euler 10. The graph has an Euler path but not an Euler 11. (a) Vertex A B C D E F G H I Degree Since there are exactly 2 odd vertices there is an Euler path, but not an Euler (b) One Euler path is: A, B, C, A, I, C, D, G, I, H, G, F, E, D, F 12. (a) Vertex A B C D E F G Degree Since all vertices are even there is an Euler (b) One Euler circuit is: A, E, B, F, C, G, D, C, B, A 13. (a) Vertex A B C D E F G H I Degree Since all vertices are even there is an Euler (b) One Euler circuit is: A, B, D, H, I, G, D, C, G, F, E, C, A 14. (a) Vertex A B C D E F G Degree Since exactly two vertices are odd there is an Euler path. (b) One Euler path is: B, A, E, G, F, E, B, C, F, D, C 15. (a) Vertex A B C D E F G H I Degree Since there are more than 2 odd vertices there is neither an Euler path nor an Euler 708

5 16. (a) Vertex A B C D E F G Degree Since there are more than 2 odd vertices there is neither an Euler path nor an Euler 17. (a) Vertex A B C Degree Since all vertices are even there is an Euler (b) One Euler circuit is: A, B, A, C, B, C, A 18. (a) Vertex A B C D E F Degree Since there are more than 2 odd vertices there is neither an Euler path nor an Euler 19. (a) Vertex A B C D E F Degree Since there are more than 2 odd vertices there is neither an Euler path nor an Euler 20. (a) 21. Vertex A B C D E F Degree Since there are no odd vertices the graph has an Euler (b) One Euler circuit is: A, B, C, D, B, F, D, E, F, A Vertex A B C D Degree Since there are no odd vertices the graph has an Euler One Euler circuit is: A, B, C, D, C, A Vertex A B C D E Degree Since there exactly two odd vertices the graph has an Euler path. One Euler path is: E, B, C, E, D, C, A, D Vertex A B C D Degree Since there are exactly two odd vertices the graph has an Euler path. One Euler path is: A, B, C, A, D, B Vertex A B C D E F Degree Since there exactly two odd vertices the graph has an Euler path. One Euler path is: A, B, E, D, A, C, F, E Vertex A B C D E Degree Since there exactly two odd vertices the graph has an Euler path. One Euler path is: A, B, E, D, A, C, D 709

6 Vertex A B C D E Degree Since there more than two odd vertices the graph has neither an Euler path nor an Euler 30. Vertex A B C D E F G Degree Since there exactly two odd vertices the graph has an Euler path. One Euler path is: A, B, E, D, C, F, A, G, E 28. Vertex A B C D E F G H I Degree Since there more than two odd vertices the graph has neither an Euler path nor an Euler 31. Vertex A B C D E F Degree Since there no odd vertices the graph has an Euler One Euler circuit is: A, F, D, C, E, F, B, C, A Vertex A B C D E F G Degree Since there exactly two odd vertices the graph has an Euler path. One Euler path is: B, A, D, E, B, C, G, F, D Vertex MT ND SD NE WY ID Degree Since there no odd vertices the graph has an Euler One Euler circuit is: WY, ID, MT, ND, SD, MT, WY, NE, SD, WY 710

7 Let the streets be the edges and the intersections be the vertices. Vertex WV VA NC KY TN Degree Since there exactly two odd vertices the graph has an Euler path. One Euler path is: KY, WV, VA, TN, KY, VA, NC, TN 33. Let the streets be the edges and the intersections be the vertices. Vtx A B C D E F G H I J K L Deg Since there no odd vertices the graph has an Euler One Euler circuit is: A, B, E, F, J, I, L, K, H, I, E, D, H, G, C, D, A Vertex A B C D E F G Degree Vertex H I J K L M N Degree Since there more than two odd vertices the graph does not have an Euler circuit nor an Euler path. 35. Neither can be done. Since all four vertices are odd, either removing one bridge or adding one more will make two of the vertices even and the other two odd, in which case there is still no Euler 36. Using just the bridges, the vertices corresponding to New Jersey and Brooklyn are odd, while the other three are even, so there is an Euler path but not an Euler If you include the tunnels, New Jersey and Manhattan are odd vertices while the other three are even, so again there s an Euler path but no Euler 37. Path: you can pass through every door in the house exactly once. Circuit: you can do so and end up in the room you started in. 38. Path: you can plan a trip that crosses each state border exactly once. Circuit: you can plan a trip that begins and ends in the same state and crosses every border exactly once. 39. Graphs will of course vary, but such a graph cannot have an Euler circuit because any graph with a bridge will always have at least two odd vertices. 711

8 40. Graphs will vary, but a graph with a loop can have an Euler A loop counts as two edges emanating from a vertex, so adding a loop at a vertex will not change it from even to odd or vice versa. In fact, that proves that if a graph has an Euler circuit without a loop, it will still have one with a loop added. If it has no Euler circuit without a loop, adding a loop won t change that. 41. Draw an edge connecting any two unconnected odd vertices. 42. No. It would need to have no odd vertices, but currently has four of them; no single edge can change more than two vertices. 43. There would be no way to traverse every edge if the graph were not connected. 44. It is not possible to have exactly one odd vertex in a graph. 45. Answers may vary. 46. n 1 1+ n , The results are getting close to e. n Exercise Set An Euler path passes through each edge exactly once, a Hamilton path passes through each vertex exactly once. 2. A Hamilton circuit ends at the same vertex where it began, but a Hamilton path does not. 3. Answers may vary. 4. Answers may vary. 5. In a complete graph every pair of vertices is connected by an edge. 6. A weighted graph is one where the edges are labeled with some quantity of interest, like distance or time. 7. The typical traveling sales person problem is the problem of trying to find the most efficient way to visit each city (vertex). The optimal solution is the one which is most efficient (least time or shortest distance, for example). 8. The brute force method to find an optimal solution would be to try every possible Hamilton path and see which is most efficient. 9. The optimal solution is a Hamilton circuit for a complete weighted graph for which the sum of the weights of the edges traversed is the smallest possible number. You can only find the optimal solution using the Brute force method, which can be really time-consuming. The approximate optimal solution will find one, but it might not be the optimal one. 10. The nearest neighbor method means you start at a designated vertex and then continue by choosing edges with the least weight that go to a vertex that you haven t already used. For the cheapest link algorithm, start with the cheapest edge. Then pick the next cheapest edge that doesn t result in 3 edges coming from one vertex or in a complete circuit that leaves out some of the vertices. Repeat until complete. 11. Answers may vary., two possibilities are: A, B, C, D, E, F and A, C, D, E, F, B. 12. Answers may vary., two possibilities are: A, B, C, E, D and A, D, E, C, B. 13. Answers may vary., two possibilities are: A, B, E, F, J, I, L, K, H, G, C, D and A, B, E, D, C, G, H, K, L, I, J, F 14. Answers may vary., two possibilities are: B, A, C, D, E, F, G and A, B, C, D, G, F, E. 15. Answers may vary., two possibilities are: A, B, E, C, D and A, B, E, D, C. 712

9 16. Answers may vary., two possibilities are: E, B, A, D, C and A, D, C, B, E. 17. Answers may vary., two possibilities are: A, B, D, E, C, F, G, H and A, B, D, E, C, F, H, G. 18. Answers may vary., two possibilities are: A, B, C, D, E, F and A, F, B, C, D, E. 19. Answers may vary., two possibilities are: A, B, C, E, D, A and C, B, A, D, E, C. 20. Answers may vary., two possibilities are: A, B, D, E, F, G, H, C, A and B, A, C, H, G, F, E, D, B. 21. Answers may vary., two possibilities are: A, B, D, G, F, E, H, I, C, A and D, B, A, C, I, H, E, F, G, D. 22. Answers may vary., two possibilities are: A, B, C, E, D, F, A and B, C, E, F, D, A, B. 23. Answers may vary., two possibilities are: A, B, C, D, E, A and A, D, B, E, C, A. 24. Answers may vary., two possibilities are: A, C, B, D, E, F, G, A and G, F, E, D, B, C, A, G. 25. The number of Hamilton circuits for a complete graph with 3 vertices is (3 1)! = 2! = 2 1 = 2. It is reasonable to use the brute force method. 26. The number of Hamilton circuits for a complete graph with 6 vertices is (6 1)! = 5! = = 120. Seriously, using the brute force method would be crazy. 27. The number of Hamilton circuits for a complete graph with 9 vertices is (9 1)! = 8! = = 40,320. Should this even be a question? That would be ridiculous. 28. The number of Hamilton circuits for a complete graph with 11 vertices is (11 1)! = 10! = = 3,628,800. Ok, we get it... if the number of vertices is big, it is ludicrous to use the brute force method. 29. There are (4 1)! = 6 possible Hamilton circuits, they are (with corresponding total weights): P, Q, S, R, P: = 235 P, R, S, Q, P: = 235 P, S, R, Q, P: = 200 P, Q, R, S, P: = 200 P, R, Q, S, P: = 235 P, S, Q, R, P: = 235 There are two optimal solutions P, Q, R, S, P and P, S, R, Q, P, each with a total weight of There are (4 1)! = 6 possible Hamilton circuits, they are (with corresponding total weights): W, X, Z, Y, W: = 1,017 W, Y, X, Z, W: = 1,338 W, Z, Y, X, W: = 1,101 W, X, Y, Z, W: = 1,101 W, Y, Z, X, W: = 1,017 W, Z, X, Y, W: = 1,338 There are two optimal solutions W, X, Z, Y, W and W, Y, Z, X, W, each with a total weight of 1, Starting at A the nearest neighbor is C with a weight of 6, then B with 7, then E with 12, then D with 10, and back to A with 7. The approximate optimal path is A, C, B, E, D, A with total weight = Starting at A the nearest neighbor is F with weight of 16, then B with 18, C with 18, E with 15, D with 23 then back to A with 19. The approximate optimal circuit is A, F, B, C, E, D, A with total weight of = Starting at A the nearest neighbor is B with weight of 122, then C with 89, then D with 96, then E with 173 then back to A with 171. The approximate optimal circuit is A, B, C, D, E, A with total weight of =

10 34. Starting at A the nearest neighbor is C with 8. At C, there are two possiblilities: B or E, each with 12. Let s try each. First, move to B with 12, then F with 13, then E with 13, then D with 22 and then back to A with 17. The is A, C, B, F, E, D, A with total weight of = 85. Now, try the second possible solution. Move A to C with 8, C to E with 12, E to B with 15, B to F with 13, F to D with 19, and D to A with 17. The total weight is = 84. So, the optimal circuit is A, C, E, B, F, D, A. 35. Pick edge from A to C with a weight of 6. Pick edge from C to B with a weight of 7. Pick edge from B to E with a weight of 12. Pick edge from E to D with a weight of 10 and pick edge from D to A with a weight of 7. The approximate optimal path is A, C, B, E, D, A with total weight = 42. It is the same as the nearest neighbor algorithm. 36. Pick edge from E to C with a weight of 15. Pick edge from F to A with a weight of 16. Pick edge from B to C with a weight of 18. Pick edge from B to F with a weight of 18. Pick edge from E to D with weight of 23. Pick edge from D to A with an edge of 19. The approximate optimal path is A, D, E, C, B, F, A with total weight of = 109. It is the same as the nearest neighbor algorithm. This is a different circuit than the nearest neighbor method, but it has the same total weight. 37. Pick edge from B to C with a weight of 89. Pick edge from B to E with a weight of 94. Pick edge from C to D with a weight of 96. Pick edge from A to D with a weight of 150. Pick edge from E to A with a weight of 171. The approximate optimal path is A, D, C, B, E, A with total weight = 600. This is a better result than the nearest neighbor algorithm. 38. Pick edge from A to C with a weight of 8. Pick edge from F to A with a weight of 9. Pick edge from C to B with a weight of 12. Pick edge from E to F with a weight of 13. Pick edge from B to D with a weight of 21. Pick edge from D to E with a weight of 22. The approximate optimal path is A, C, B, D, E, F, A with a total weight of = Let T = Pitt, L = Phil, B = Balt, and W = Wash. 40. Let T = Pitt, L = Phil, B = Balt, and W = Wash. The possibilities, with corresponding total distances, are: T, L, W, B, T: = 723 T, L, B, W, T: = 688 T, W, B, L, T: = 688 T, W, L, B, T: = 725 T, B, W, L, T: = 723 T, B, L, W, T: = 725 There are two optimal paths, T, L, B, W, T and T, W, B, L, T, each with a total distance of 688 miles. 41. Let T = Pitt, L = Phil, B = Balt, and W = Wash. Starting at T, the nearest neighbor is B with distance of 244, then W with 38, then L with 136 and back to T with 305. The approximate optimal solution is then, T, B, W, L, T with total distance of = 723 miles. This route is longer than the optimal solution. 42. Let T = Pitt, L = Phil, B = Balt, and W = Wash. Pick edge from W to B with a weight of 38. Pick edge from B to L with a weight of 100. Pick edge from T to W with a weight of 245. Pick edge from L to T with a weight of 305. The approximate optimal path is T, W, B, L, T with a total weight of = 688. This is the optimal solution, making it better than the nearest neighbor approximate solution. 43. Let N = New York, D = Cleveland, O = Chicago, and B = Baltimore. 714

11 44. Let N = New York, D = Cleveland, O = Chicago and B = Baltimore. The possible circuits with corresponding costs are: O, N, D, B, O: $450 + $375 + $300 + $325 = $1,450 O, B, N, D, O: $325 + $200 + $375 + $250 = $1,150 O, D, N, B, O: $250 + $375 + $200 + $325 = $1,150 O, N, B, D, O: $450 + $200 + $300 + $250 = $1,200 O, D, B, N, O: $250 + $300 + $200 + $450 = $1,200 O, B, D, N, O: $325 + $300 + $375 + $450 = $1,450 There are two optimal routes, O, B, N, D, O and O, D, N, B, O, each with a total cost of $1, Let N = New York, D = Cleveland, O = Chicago, and B = Baltimore. Starting at O, the nearest neighbor is D at a cost of $250, then B at $300, then N at $200, and back to O for $450. The approximate optimal solution is O, D, B, N, O with a total cost of $1,200. This route is $50 more expensive than the optimal solution. 46. Let N = New York, D = Cleveland, O = Chicago, and B = Baltimore. Pick the edge from B to N with a weight of 200. Pick the edge from O to D with a weight of 250. Pick the edge from D to B with a weight of 300. Pick the edge from N to O with a weight of 450. The approximate optimal solution is O, D, B, N, O with a total cost of = $1,200. This route is $50 more expensive than the optimal solution You would have to be crazy to do brute force. There are 5! = 120 possible routes. 49. Starting at H, the nearest neighbor is 3 with a weight of 1:05, then 1 with a weight of 1:40, then 4 with a weight of 2:15, then 5 with a weight of 1:15, then 2 with a weight of 1:15, then H with a weight of 4:30. The approximate optimal solution is H, 3, 1, 4, 5, 2, H with a total time of 1:05 + 1:40 + 2:15 + 1:15 + 1:15 + 2:15 = 12: Pick the edge from H to 3 with a weight of 1:05. Pick the edge from 4 to 5 with a weight of 1:15. Pick the edge from 5 to 2 with a weight of 1:15. Pick the edge from 3 to 1 with a weight of 1:40. Pick the edge 1 to 4 with a weight of 2:15. Pick the edge from 2 to H with a weight of 4:30. The approximate optimal solution is H, 3, 1, 4, 5, 2, H with a total time of 1:05 + 1:40 + 2:15 + 1:15 + 1:15 + 2:15 = 12:00. The answers for the nearest neighbor and cheapest link are the same

12 Brute Force Method: RB, DD, VB, HV, RB = 265 RB, DD, HV, VB, RB = 320 RB, VB, DD, HV, RB = 325 RB, VB, HV, DD, RB = 320 RB, HV, VB, DD, RB = 265 RB, HV, DD, VB, RB = 325 The optimal solution using the Brute Force Method is RB, DD, VB, HV, RB with a total weight of: = 265 or the reverse. Nearest Neighbor Method: Starting at RB, the nearest neighbor is HV with a weight of 45, then VB at 50, then DD at 90 and back to RB for 80. The approximate optimal solution is RB, DD, VB, HV, RB with a total weight of: = Brute Force Method: RB, DD, VB, HV, RB = RB, DD, HV, VB, RB = RB, VB, DD, HV, RB = RB, VB, HV, DD, RB = RB, HV, VB, DD, RB = RB, HV, DD, VB, RB = The optimal solution using the Brute Force Method is RB, DD, VB, HV, RB with a total weight of: = or the reverse. Nearest Neighbor Method: Starting at RB, the nearest neighbor is HV with a weight of 15.5, then VB at 15.7, then DD at 39.9 and back to RB for The approximate optimal solution is RB, DD, VB, HV, RB with a total weight of: = The same route as in Exercise 53 is most efficient, but even though that route is less than 8 miles shorter than the next-best route, it saves almost an hour. 55. Answers may vary. 56. Answers may vary. 57. Answers may vary. 58. Answers may vary. 59. Answers may vary. 60. Answers may vary. 61. Answers may vary. 716

13 62. (a) If there are n vertices, then you have n 1 to choose from for the next visit. The graph is complete therefore we are assured that we can visit any of the other n 1 vertices. (b) After visiting 2 vertices, you have n 2 choices for the next one. (c) Using the fundamental counting principle there are (n 1)(n 2)(n 3) (n (n 1)) = (n 1)(n 2)(n 3) 1 = (n 1)! ways to visit all n vertices. 63. No; in a complete graph, every vertex is connected to every other, and there s only one way to do that. 64. If a complete graph has n vertices, then every vertex has degree of n 1 since each vertex is connected to all of the others and there are n 1 others. There s an Euler circuit if all vertices are even; this happens if n is odd. There could be a Euler path but no circuit if exactly two vertices are odd. This can t happen for n > Answers may vary. 66. Answers may vary. Exercise Set A graph may have circuits but a tree has no circuits. 2. A tree cannot be a complete graph because a complete graph has circuits. 3. A spanning tree is a tree found by removing edges from a graph until it is a tree. 4. A minimum spanning tree for a weighted graph is the spanning tree whose edges add up to the least amount possible. 5. Starting with the lowest weighted edge, pick the next lowest connected to it, and then continue in this fashion until all vertices are connected. Delete all other edges and you will have a minimum spanning tree. 6. Answers may vary. 7. The graph is not a tree because it contains the circuit A, B, C, A. 8. The graph is a tree. 9. The graph is a tree. 10. The graph is not a tree because it contains the circuit A, D, E, A. 11. The graph is not a tree because it is disconnected. 12. The graph is a tree. 13. The graph is a tree. 14. The graph is a tree. 15. The graph is not a tree because it contains the circuit A, B, C, A. 16. The graph is a tree

14 The lowest weighted edge is AE, followed by ED and EB, then the edge BC. Keep these edges for the minimum spanning tree pictured. The total weight is = The lowest weighted edge is BD, next is DA, and the next that doesn t make a circuit is BC. Keep these three edges for the minimum spanning tree pictured. The total weight is = The lowest weighted edge is BC, followed by DF. The next lowest weighted that doesn t make a circuit is either CD or BF. Regardless of which of these edges is chosen, the next two lowest weighted edges that do not form a circuit are AB and DE. The minimum spanning tree containing CD is pictured. The total weight is = The lowest weighted edge is AB, followed by CD, and BC. The lowest weighted edges that do not form a circuit are EF, CE, and EG. Keep these edges for the minimum spanning tree pictured. The total weight is = The lowest weighted edge is GE, followed by EF. The lowest weighted edges that do not form a circuit are AI, IG, BI, IH, ID, and IC. Keep these edges for the minimum spanning tree pictured. The total weight is = The lowest weighted edge is FE, followed by DC, BC and FG, AB, and ED. Keep these edges for the minimum spanning tree pictured. The total weight is = The lowest weighted edge is BE followed by CF and BF. The lowest weighted edges that do not form a circuit are DF and AE. Keep these edges for the minimal spanning tree pictured. The total weight is =

15 30. The lowest weighted edges are CD and GI followed by EF. The next lowest weighted edge that does not form a circuit is either GF or IF. Regardless of which is chosen, the next two lowest weighted edges are DE and GH. The next lowest weighted edge that does not form a circuit is BC. The next lowest weighted edge that does not form a circuit is either AH or AB. The minimal spanning tree containing edges GI and AB is pictured. The total weight is = The minimum spanning tree is pictured below. 34. The minimum spanning tree is pictures below:: 1,110 2,370 The shortest distance is 1,450 2,200 1, , , ,450 = 7,130. At a cost of $290 per foot to install, the minimum cost is 7130(290) = $2,067, When the graph can be represented by a tree, every room can be reached from any other room using some path, but you can t start in one room and return to that room without retracing part of your route. 36. Answers can vary depending on how you find the distances. Distances shown here are from park to park, not city to city. Weighted graph and Minimum spanning tree: The least amount of wire needed is = 28 feet. 32. The minimum spanning tree is pictured below. The shortest distance is = 1,400 feet. 33. The minimum spanning tree is pictured below. Optimal route: Pit Cle Col Cin Det Pit. It follows the minimum spanning tree as long as possible without retracing an edge. (The reverse route works too, and finishes with three edges of the spanning tree.) 37. Answers may vary. 38. a) 10 b) Answers may vary. 39. Answers may vary. 40. Answers may vary. 41. Answers may vary. 42. Answers may vary. The shortest distance is = 162 feet. 719

16 Review Exercises 1. The vertices are A, B, C, D, E, and F. 2. There are four edges connected to E, therefore E is an even vertex. 3. There are three edges connected to C, therefore C is an odd vertex. 4. The vertex B has a loop since there is an edge from B to itself. 5. C and E are connected to F by an edge, therefore C and E are adjacent to F. 6. The edge EF is a bridge, since if it is removed the graph will be disconnected. 7. Since there is a path between every pair of vertices, the graph is connected. 8. The graph has 9 edges, the loop at B and BC, BA, BE, AE, ED, CF and EF. 9. Yes. Each has three vertices with degree 3, one with degree 4 and one with degree Chromatic number: Chromatic number for exercise 11: 3 Chromatic number for exercise 12: All vertices have even degree, so there is a Euler Circuit. 16. It doesn t have a path or circuit because there are three vertices (V, X, and Y) of odd degree. 17. There are exactly two odd vertices (E and G), so it has an Euler path but no Euler 18. The graph from Exercise 15 has an Euler circuit; one possible answer is: A, B, C, D, E, G, F, A, G, C, A. 19. The graph from Exercise 17; one possible answer: E, D, C, A, B, C, E, B, G, E, F, G. 20. There are 4 odd vertices, so there isn t a circuit or a path. 21. There is a Euler path. One possible path: C, E, W, F, V, C, W, V. 22. In the nearest neighbor method, you always choose the edge with the lowest weight connected to the edges you ve already chosen. In cheapest link, you choose the lowest-weight edges overall, as long as they don t form a circuit or revisit a vertex. 23. Answers may vary; one Hamilton path is A, E, B, D, C, H. 24. Answers may vary; one Hamilton circuit is A, B, C, D, E, H, F, A. 25. A complete graph with 12 vertices has (12 1)! = 11! = = 39,916,800 Hamilton circuits. 720

17 26. Starting at A the nearest neighbor is D with a weight of 22, then C with 23, then B with 23, then E with 26, and finally back to A with 31. So an optimal circuit is A, D, C, B, E, A for a total weight of = a) It is a tree because there s exactly one path connecting each pair of vertices. b) It isn t a tree because it s disconnected, so there s no path connecting certain pairs of vertices C, D, F, E, C = 1,415 C, D, E, F, C = 1,385 C, F, D, E, C = 1,839 C, F, E, D, C = 1,396 C, E, F, D, C =1,415 C, E, D, F, C = 1,839 The cheapest route is C, D, E, F, C or C, F, E, D, C; $1, Pick the edge from F to E with a weight of 230. Pick the edge from C to D with a weight of 256. Pick the edge from D to F with a weight of 560. Pick the edge from E to C with a weight of 732. The approximate optimal solution is C, D, F, E, C or C, E, F, D, C with a total weight of = $1,415. The cheapest link approximate solution is 1,415 1,396 = $19 more than the optimal solution. 32. The lowest weighted edge is DE followed by AC and AB. The next lowest weighted edges that do not form a circuit are FG, CH, HG, and HE. Keep these edges and the minimum spanning tree is pictured. The total weight is = A minimum spanning tree connects all vertices with exactly one path that has the lowest possible overall weight. In a traveling salesperson problem, we find a circuit that visits each vertex once and has minimum weight. 34. The minimum cost is = $

18 Chapter Test 1. a) There are four edges connected to vertex C, therefore the degree of vertex C is 4. b) There is a loop at vertex E since there is an edge from E to itself. c) Answers may vary; one possible path is A, B, C, E, D, F. d) The edge FD is a bridge since removing it would cause the graph to be disconnected. e) C is adjacent to A, B, D, and E since there is an edge connecting C to each of these vertices. f) C and E are even. A, B, D, and F are odd. 2. Answers may vary. One possibility: 7. To plow each street only once would mean finding an Euler path. One Euler path is A, D, E, A, B, E, F, B, C, F. 8. Let A = Adamsburg, T = Trafford, W = White Oak and C = Turtle Creek. 3 a) b) It would require 3 colors 5. a) Answers may vary; one Euler path is B, A, F, B, C, D, E, C, F. b) Answers may vary; one Hamilton path is A, B, F, C, E, D. The Hamilton path will typically vary from the Euler path. 9. The possible paths with their corresponding distances are: A, T, C, W, A: = 78 A, T, W, C, A: = 64 A, C, W, T, A: = 64 A, C, T, W, A: = 78 A, W, T, C, A: = 78 A, W, C, T, A: = 78 The shortest distance is 64 miles. 10. Nearest Neighbor Method: Starting at A the nearest neighbor is C at 18 miles, then W at 12 miles, then T at 14 miles, and back to A at 20 miles. So the approximate optimal path is A, C, W, T, A and the approximate optimal distance is = 64 miles. The result is the same as the one obtained by the brute force method. Cheapest Link Algorithm: Pick the edge from W to C with a weight of 12. Pick the edge from W to T with a weight of 14. Pick the edge from C to A with a weight of 18. Pick the edge from A to T with a weight of 20. So the approximate optimal path is A, C, W, T, A and the approximate optimal distance is = 64 miles. The result is the same as the one obtained by the brute force method. 722

19 a) If the graph is constructed so that the roads are edges then Euler paths would be used since he wants to cross every road only once. b) Letting the attractions be the vertices the friends would want to use an optimal Hamilton path so that they visit every attraction exactly once with the least amount of walking. c) Letting the delivery locations be the vertices the drive would want to find an optimal Hamilton path so that every location is visited exactly once in the least amount of time. d) Letting the colleges they visit be vertices, the friends will want to find an optimal Hamilton circuit so that they visit each college exactly once and travel the shortest possible distance. e) Letting the streets be edges the police officer will want to find an Euler path so that he travels down every street. f) Letting the classrooms be the vertices, the IT team will want to find a minimum spanning tree so that every classroom is wired and the least amount of wire is used. 723