22
Chapter 5 Answers 1 Chapter 5 Answers Lesson 5.1 1. Planar 2. Nonplanar 3. Planar 4. Planar
Chapter 5 Answers 1
Chapter 5 Answers
Lesson 5.11. Planar
2. Nonplanar3. Planar
4. Planar
Chapter 5 Answers 2
5. Hint: move A and B around.
6. A K5 graph has five vertices of degree 4. The graph in Exercise 5 only has threevertices of degree 4.
7. Look for five vertices that have degree 4 or greater. Pick one of these verticesand mark four paths to the four other vertices; move to one of those and markpaths to the other three. Continue in the same manner until ten edges aredrawn. A K5 requires ten distinct edges.The graph does contain a K5 subgraph as shown in the graph below:
8.
9. Graphs will vary but complements should be nonplanar.10. When the circuits are drawn on the board, a single board can be used only if the
graph is planar. If it is not planar, the conductive paths cross and possibly shortthe circuit.
Chapter 5 Answers 3
11. a.
K2,3
b.
K2,4
12. a. {A,B,C,D,E,F} and {G}b. {P,N,R} and {Q,M,S}c. {T,Y,W,Z} and {X,V,U}d. {H, J, K, M} and {I, L}
13. a. Yes, the graph can be divided into two distinct sets.b. No, the graph cannot be divided into two distinct sets.c. Yes, the graph can be divided into two distinct sets.
14. Answers will vary. One possible algorithm is as follows:
1. Start at any vertex and label it with an *.2. Label all vertices adjacent to it with a circle.3. Label each vertex adjacent to a circle with an *.4. Continue in this manner until all vertices are labeled. If no two adjacent
vertices have the same labels, the graph is bipartite.
Chapter 5 Answers 4
15. 6, 12, mn16. When m and n are both even17. The chromatic number for any bipartite graph is 2.18. 30 handshakes, bipartite19. Answers will vary. The following is one example: Mary and her four friends
have five flavors of soft drinks from which to choose. Each friend can choosetwo of his/her favorite flavors. Draw a graph of the situation.
20. No. The graph is a K3,3 and is not planar. Therefore, it cannot be redrawn withoutedges crossing.
21. a and d; because all of the edges and vertices of the original graph are in a and d.22. The converse of the statement is: If a graph is not planar, then it contains a K5 or
a K3,3 subgraph. Most students will think the statement is true, but Exercises 23and 24 show it to be false.
23. No, No.24.
Lesson 5.21. a and b.
c. SDCBSd. SDCBSe. Yes.
Chapter 5 Answers 5
2. a and b.
c. SBDCSd. SDBCSe. No.
3. a and b.
c. SBDCSd. SBDCSe. Yes.
Chapter 5 Answers 6
4. a and b.
c. SDCEBSd. SBDCESe. No.
5. a. 0.36 seconds, about 24 hoursb. 1.008 × 10–8 seconds, 0.002 seconds, about 18.8 hours
6. Answers will vary. Possible examples: shortest distance for a bank courier tomake his/her route and return to the bank, a robot moving from a “home”location to other places in the warehouse and back “home,” and so on.
7. Shortest possible circuit: SACBS. Total distance: 18.75 mm.8. a.
b. GBDACGc. 86.5 minutes
Chapter 5 Answers 7
Lesson 5.31. 6, 12, 11, C, BC. The shortest path from A to E is AHGE.
2. ABECDF (11)3. If you are looking for a path from one vertex to a second and find the desired
path before all vertices are circled.4. S to M: 2 S to N: 5 S to K: 4 S to J: 4 S to L: 65. a. Albany, C, E, H, Ladue
b. Albany, B, D, Fenton, G, K, Ladue. This problem yields a different solutionthan part a, because you have to find two solutions and then add them: firstyou must find the shortest path from Albany to Fenton, then a path withFenton as Start.
6. By placing subtotals on the vertices.7. b. 7.58. Answers will vary. Possible modifications are as follows:
Step 1: Label the starting vertex S and circle it. Examine all directed edges from Sto the other vertices. Darken the directed edge with the shortest length and circlethe vertex at the other endpoint of the darkened directed edge.Step 2: Examine all directed edges from circled vertices to uncircled ones.Step 3: Using only circled vertices and darkened directed edges between thevertices that are circled, find the lengths of all paths from S to each vertex beingexamined. Choose the vertex and directed edge that yield the shortest path.Circle this vertex and darken this directed edge. Ties are broken arbitrarily.Step 4. Repeat Steps 2 and 3 until all vertices are circled. The darkened edges ofthe graph form the shortest routes from S to every other vertex in the graph.
Chapter 5 Answers 8
9.
Shortest path: DCAB. Least charge: 1 + 2 + 7 = 10.
Lesson 5.41. BCEFB, CDEC, BCDEFB, BCFB, CEFC, CDEFC2. a. Yes.
b. No, it is not connected.c. Yes.d. No, it contains a cycle.e. No, it is not connected.f. Yes.
3. a. 5 vertices
Chapter 5 Answers 9
b. 6 vertices
4.
Number ofVertices
Number ofEdges
1 0
2 1
3 2
4 3
n n – 1
a. 18 edgesb. 16 verticesc. The number of edges = the number of vertices – 1.
5. The graph will no longer be a tree because it will not be connected.6.
Chapter 5 Answers 10
7. a.Number of
VerticesSum of the Degrees of
the VerticesRecurrenceRelation
1 0 S1 = 02 2 S2 = S1 + 23 4 S3 = S2 + 24 6 S4 = S3 + 25 8 S5 = S4 + 26 10 S6 = S5 + 2
b. Sn = Sn–1 + 28.
If the tree is arranged in this order, each vertex has degree 2 except the twovertices on each end. They have degree 1. Hence Sn = 2n – 2.
9. Answers will vary. Possible forest:
10. 5 edges11.
The leaves of the tree indicate the people with no children.12. Rhombus
Chapter 5 Answers 11
13. Two possible subgraphs that are trees are shown below.
14. Answers will vary. One possible network is shown by the following graph:
15. It is not possible to have a tree with 8 vertices and 10 edges. If a graph with 8vertices has more than 7 edges, it has to have at least one cycle in the graph.
Lesson 5.51. One possible spanning tree:
2. One possible spanning tree:
Chapter 5 Answers 12
3. A spanning tree does not exist because the graph is not connected.4. One possible spanning tree:
5. A spanning tree does not exist because the graph is not connected.6. These are possible answers.
7. a. Yes.b. E and J.c. DE and DJ.d. F, K, I
8. This is one of many possibilities.
Chapter 5 Answers 13
9. Every connected graph with one edge has a spanning tree because if a graph hasonly one edge, it is its own spanning tree.Assume that any connected graph with k edges has a spanning tree. Let G be aconnected graph with k + 1 edges. If G has no cycles, then it is a spanning tree. IfG has a cycle, remove one edge from the cycle. The resulting graph G' is stillconnected and has k vertices. We know from our assumption that it has aspanning tree. And since the spanning tree for G' includes all of the vertices of G,it is also a spanning tree for G.
10.
The algorithm can be modified by labeling a vertex n + 1 if there is a directededge to it from a vertex labeled n.
11. The minimum weight is 10.12. The minimum weight is 18.13. The minimum weight is 55.14. The minimum weight is 46.15. $2,10016. $3,80017. Find the edge of longest length rather than shortest length.18. Sample graph:
Minimal weight = 10
Chapter 5 Answers 14
19.
Minimal weight = 1820. a. Weight 28
b. A to F is 10; A to B is 3; A to C is 11; A to E is 10; A to D is 14.
c. No. It is a spanning tree, but in this example it is not minimal. Its total weightis 32, which is greater than 28.
21. Answers will vary.TSP: Asks for a subgraph that includes all of the vertices. It is the shortest routefrom a specific vertex to each of the other vertices and back to the startingvertex. This subgraph is not a tree.Shortest Path: Asks for the shortest path from a specific vertex to another vertexin the graph or possibly to every other vertex in the graph.Minimum Spanning Tree: Asks for the subgraph of the graph that includes all ofthe vertices of the graph and has the least weight.
Chapter 5 Answers 15
Lesson 5.61.
2.
3. Binary Tree.a. V is level 2.b. C is the parent.c. G and H are children.
4. Not a binary tree5. Binary Tree
a. V is level 1.b. A is the parent.c. B is the child.
6. Binary Treea. V is level 3.b. E is the parent.c. No children.
Chapter 5 Answers 16
7. There are 17 questions in the book.
8.
9.
10.
Chapter 5 Answers 17
11.
12. 3 2 * 8 2 3 * – +13. 5 1 * 13 + 2 9 3 / * +14. 4 3 + 8 * 2 +15. 6 4 – 1 + 8 2 / 1 * *16. a. 33
b. 7c. 9d. 14e. 5f. 2
17. a. 2 3 6 * + 4 1 + –b. 5 3 – 2 * 7 6 2 / – +
18.
Chapter 5 Answers 18
19.
20. ABDEGHCFI21. 12) + * 3 2 – 8 * 2 3
13) + + * 5 1 13 * 2 / 9 314) + * + 4 3 8 215) * + – 6 4 1 * / 8 2 1
22. 1723. a. 8
b. 9c. 5d. 2
Review1. The following are important points made in Chapter 5.
• Graphs can be classified as planar and nonplanar.• Planar graphs have chromatic numbers less than or equal to four.• Certain situations can be modeled with bipartite graphs.• A special kind of graph called a tree can be used to organize information.• Three types of optimization problems (traveling salesperson problem,
shortest route problem, and minimum spanning tree) can be modeled withweighted graphs.
• Algorithms exist that help solve these three optimization problems.• A rooted tree can be used to model arithmetic expressions.• A tree traversal (organized procedure) can be used to obtain the value of an
expression tree.
Chapter 5 Answers 19
2. Planar
3.
4. a. 2b. 2c. 2d. 2
5. a. The vertices of the graph can be divided into two sets so that each edge of thegraph has one endpoint in each set.
b. Yes, all possible edges from one set of vertices to the other are drawn.c. Yes.
d. 2
Chapter 5 Answers 20
6. One possible solution:
7. a. O-SCM-Ob. 314 ftc. Hamiltonian Circuit
8. Total weight: 9 + 23 +31 + 35 + 28 = 1269.
10. 3, 4, 5, n11. The circuit does not include all the vertices of the graph.12. a. Home-T-P-G-H-BB
b. 14 milesc.
13. The total weight of a minimum spanning tree for the graph is 23 miles.14. a. Yes, it is a connected graph with no cycles.
b. Yes, it is a connected graph with no cycles.c. No, the graph contains a cycle.d. No, the graph is not connected.
Chapter 5 Answers 21
15. The graph will no longer be a tree because it will contain a cycle, multiple edges,or a loop.
16.
17. a.
b. Total cost = $23,000.18. One possible solution.
19. Problems similar to those in Lesson 5.5.20.
1 2 3 4 5 6
1H 1T 2H 2T 3H 3T 4H 4T 5H 5T 6H 6T
Chapter 5 Answers 22
21.
22. 4 6 3 – + 5 2 * +23. 1824. Any expression is possible. The following is just one example:
The expression: 3 * (2 + 6) – 5The expression tree:
The postorder listing: 3 2 6 + * 5 –
