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Chapter 4 Answers 1

Chapter 4 Answers

Lesson 4.11.

2.

3.

4.

Chapter 4 Answers 2

5. Answers vary. Sample answer:a. A: 2, none; B: 4, A; C: 3, B; D: 7, C; E: 5, A; F: 10, E; G: 8, A; H: 3, A; I: 4, D, F, G,

and Hb.

c. In this case, the least amount of time would be 21 minutes.6. Answers vary.

Sample answer:Task Time (minutes) Prerequisites

Start 0 —

A. Do laundry 60 None

B. Purchase toiletries 45 None

C. Buy new shoes 50 None

D. Tell friends good-bye 30 None

E. Pack 70 A, B, C

F. Load car 45 E

G. Drive to college 90 D, F

b.

Chapter 4 Answers 3

c. In this case, the least amount of time is 265 minutes for path Start-AEFG-Finish.

7. a. A: 2, none; B: 4, none; C: 3, A; D: 3, C and B; E: 2, B; F: 1, E; G: 4, D and F.b. Task D can’t begin before A and C are complete and 5 minutes have elapsed.

G can’t start before 8 minutes have elapsed. Therefore, the project can’t becompleted in a time faster than 12 minutes. Also note that the path from theStart to A, C, D, G and the Finish is the longest path through the graph.

Lesson 4.21. EST for C through G: 7, 10, 11, 16, 23.

Minimum Project Time: 26.Critical Path: Start–ACEFG–Finish.

2. EST for C through H: 6, 5, 16, 16, 25, 22.Minimum Project Time: 32.Critical Paths: Start–ACEG–Finish, Start–ACFH–Finish.

3. EST for D through L: 4, 5, 6, 6, 6, 6, 9, 9, 15.Minimum Project Time: 20.Critical Path: Start–ADHKL–Finish.

4.

Minimum Project Time: 20.Critical Path: Start–BCFG–Finish.

5. a.

Chapter 4 Answers 4

b. Minimum Project Time: 22.c. Critical Path: Start–ADG–Finish.d. The minimum time is reduced to 21 days, to 20 days.e. No, below 8 days A is no longer on the critical path.

6. Answers will vary. Sample answer:

7. a. Minimum Project Time: 36Critical Path: Start–ADF–Finish.

8. a & b.

c. Minimum Project Time: 33d. Critical Path: Start–BDFH–Finish

9. a. Day 16, Day 17, Day 18. Both task G and the project will be delayed.b. Day 11.c. Day 5, Day 6, Day 5.

10. The graph below shows the ESTs and LSTs for each vertex:

Chapter 4 Answers 5

11. Sample algorithm. Answers will vary.

1. Begin with Finish and use the minimum project time.2. Subtract from the minimum project time the time that it takes to complete

the task(s) preceding it. This is the LST for the preceding tasks. Label thevertex with its LST.

3. If the time of a preceding task is dependent on more than one edge, choosethe earliest time.

4. Continue until all vertices are labeled.

Lesson 4.31.

2. a.

Chapter 4 Answers 6

b.

c.

d.

3. Graphs a and d are connected. Graph d is complete.4. a. Not adjacent: A&F, B&F, B&C, A&C, A&D, A&E, B&E, B&D, F&C, F&D

b. F, E, D, Cc. No, there is no path from A or B to the vertices C, D, E, or F.d. No, not every pair of vertices is adjacent. For example, B and C are not

adjacent.5.

6.

Chapter 4 Answers 7

7. a.

b.

c.

A B

C

D

E

8. a.

0 1 0 1 11 0 1 1 00 1 0 1 01 1 1 0 11 0 0 1 0

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b.

0 1 1 1 1 11 0 1 0 0 01 1 0 0 0 01 0 0 0 0 01 0 0 0 0 11 0 0 0 1 0

⎡

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⎤

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9.

0 1 1 0 01 0 0 0 01 0 0 1 10 0 1 0 10 0 1 1 0

⎡

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⎤

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⎥⎥⎥⎥⎥⎥

Chapter 4 Answers 8

a. All zerosb. Yes, along the main diagonal.c. Vertex adjacent to itself, there are two edges between two vertices.

10. The number of vertices that are adjacent to a given vertex.11. deg(V) = 3; deg(X) = 2; deg(Y) = 2; deg(Z) = 112. a. 3

b. n – 113. a. deg(B) = 2; deg(C) = 6, deg(D) = 3, deg(E) = 2

b.

1 0 1 0 00 0 2 0 01 2 0 1 20 0 1 1 00 0 2 0 0

⎡

⎣

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⎤

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c. Yes, you can tell the difference. A matrix for a graph will have 0s on the maindiagonal and 1s and 0s elsewhere. A matrix for a multigraph is characterizedby numbers greater than 1 in the matrix (indicating multiple edges) and/ornumbers other than 0s on the main diagonal (indicating loops).

14.K4 4 12 T4 = T3 + 6K5 5 20 T5 = T4 + 8K6 6 30 T6 = T5 + 10

Recurrence relation: Tn = Tn–1 + 2(n – 1)15. The sum of the degrees of all the vertices for a complete graph is an even

number.This is true for all graphs.The sum of the degrees of all the vertices is even because each edge contributes 2(one degree at each end of the edge) to the total.

16. a.

b. Impossiblec. In any graph, there always has to be an even number of vertices with odd

degree.17. Zeros on the main diagonal, ones everywhere else.

Chapter 4 Answers 9

18.K4 4 6 S4 = S3 + 3K5 5 10 S5 = S4 + 4K6 6 15 S6 = S5 + 5

Recurrence relation: Sn = Sn–1 + (n – 1)19. Take any two teams who play each other and call them A and B. If they play a

common third team, C, then the remaining two teams do not have two differentteams to play. Thus, A and B must play different teams for their second game.Suppose A plays C and B plays D.

If C plays D, then the fifth team has no one to play. Hence, C and D cannot playeach other. All that remains is the graph.

Lesson 4.41. a. Both. The degrees of all vertices are even.

b. Neither. Four vertices have odd degree.c. Euler Path. There are exactly two vertices of odd degree and the rest are

even.d. Both. The degrees of all vertices are even.

2. Sample graph:

3. New circuit: S, b, e, a, g, f, SFinal Circuit: S, e, f, a, b, c, S, b, e, a, g, f, S

Chapter 4 Answers 10

4. Answers may vary. Sample circuit: e, d, f, h, d, c, h, b, c, g, a, h, g, f, e5. a. If it were not connected there would be at least two vertices between which

no path would exist.b. Sample graph:

c. Sample graph:

6. No; yes; no; yes. No, if n is even—yes, if n is odd.7. No. If a graph is drawn of the present-day situation, there are still two vertices, B

and C, with odd degree.8. a. No, No.

b. It is possible to find a route repeating only one edge. For example, G, i, j, G, h,f, d, c, a, b, d, e, b, j, e, f, j, f, G where edge jf is inspected twice.

9. a. Sample digraph:

b. Sample digraph

10. a. V = {K, L, M, N, P} E = {KL, LP, LN, NM, MP, PK, PN}b. No, the indegree is not equal to the outdegree for vertices L and N.


11. a. Yes.b. No.c. Yes.d. No.

12. a. No, the outdegree doesn’t equal the indegree of each vertex.b. Yes, b or f.c. A digraph has a directed Euler path when the following three conditions are

met: 1) The outdegree equals the indegree at all vertices but two. 2) At one ofthose two vertices, the indegree is one greater than the outdegree. 3) At theother vertex, the outdegree is one greater than the indegree.

13. a.

a b c d eabcde

0 0 1 0 11 0 0 1 00 1 0 0 00 0 1 0 00 1 0 1 0

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b.

A B C DABCD

0 1 0 00 0 1 10 0 0 11 1 0 0

⎡

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⎤

⎦

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c.

s t v w xstvwx

0 0 0 1 00 0 0 0 00 1 0 0 00 0 0 0 01 1 1 1 0

⎡

⎣

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d.

W X Y ZWXYZ

0 1 0 10 0 1 01 0 0 00 1 0 0

⎡

⎣

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⎤

⎦

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14.

A B

CD

15. a.

b. There is no symmetry along this main diagonal. Symmetry could exist if foreach pair of adjacent vertices there would be directed edges in pairs with onegoing in one direction and one going the other. For example, A and B wouldbe connected with a directed edge AB and a directed edge BA.

c. 3, B’s outdegreed. 2, B’s indegree

16. Following is a sample Euler Circuit Program (TI-83 version):

Disp "HOW MANY ROWS?"Input R{R, R}→dim ([A])Disp "ENTER ELEMENTS"Disp "BY ROWS"For (I, 1, R)For (J, 1, R)Input XX→[A](I, J)EndEndFor (I, 1, R)0→SFor (J, 1, R)S + [A](I, J)→SEndS/2→YIf fPart (Y) ≠ 0ThenClrHome


Disp [A] "NO EULER"Disp "CIRCUIT EXISTS"StopElseEndEndClrHomeDisp [A]Disp "YOUR GRAPH HAS"Disp "AN EULER CIRCUIT"

Lesson 4.51. Graphs a and d have Hamiltonian circuits. For graphs b and c, the theorem does

not apply.2. Answers vary. Sample answers: Mail carriers visiting mailboxes that are located

at street corners and returning to the post office, traveling from home to severalcities and returning home, etc.

3. a.

b.

4. a. Many possible circuits can be found.b. Noc. A circuit can be found beginning at any vertex.

5. Hamiltonian circuit a, f, b, d, c, e, a6.

7. a. BCADb. DBCA, BCDA, CDBA


8. a.

b.

c. Impossible9.

10.When ties exist between two or more players (represented by vertices), thosevertices can be named in more than one order when finding a path through thegraph.

11. a. Sample algorithm:

1. Rank the vertices by descending order of outdegree.2. Begin the path with the vertex of greatest outdegree and continue in

order to the vertex of least outdegree.

b. The greatest difficulty arises when two vertices have equal outdegree and achoice must be made as to which one to visit first. Sometimes you move toone of the vertices and it’s not possible to get to the other one with the samelevel of outdegree. In this case, you must visit those vertices in reverse order.


12.

4 65 106 15

Sn = Sn–1 + (n – 1)13. If A and B were both transmitters (or both receivers) there would be no directed

edge between them.14. a.&b. Every tournament with 1 vertex (or 2 vertices) has a Hamiltonian path.

Assume a tournament with k vertices has a Hamiltonian path. Call thistournament of k vertices S. Since it has k vertices, we know from ourassumption that it has a Hamiltonian path. Call this path x1, x2, x3, …, xk. Nowintroduce a (k + 1)-st vertex, x, and directed edges that connect it to the othervertices in S. Call the new tournament T. If x has an indegree of 0, then x, x1,x2, x3, …, xk is a Hamiltonian path for T. Otherwise, follow the Hamiltonianpath for S until you get to the last directed edge pointing from S to vertex x.Follow it to the x. If all of the vertices of S have been visited, you have founda Hamiltonian path, x1, x2, x3, …, xk, x for T. If all of the vertices have not beenvisited, then choose an edge from x back to the next vertex in the path for Sand continue until all vertices of S have been visited.

15. a.

b.

c. Yes, Bd. The path BCDA indicates the ranking of the four candidates from winner to

loser.


16. a.

M =

0 0 1 1 11 0 1 1 00 0 0 1 10 0 0 0 10 1 0 0 0

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b.

M2 =

0 1 0 1 20 0 1 2 30 1 0 0 10 1 0 0 01 0 1 1 0

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c. The winner would be B.d. Entries in M3 would indicate the number of three-stage dominances (or paths

of length 3) of one player over another.

M3 =

1 2 1 1 10 3 0 1 31 1 1 1 01 0 1 1 00 0 1 2 3

⎡

⎣

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Lesson 4.61. a. 4

b. 2c. 2d. 3

2. a. Sample graph:

b. Sample graph:


3. a. List the vertices in order from the ones with the greatest degree to the oneswith the least.

b. Those not adjacent to it or adjacent to one with that color.c. The vertex with the next highest degree of those not already colored. Any

vertex not adjacent to that one or to one receiving the second color.d. When all of the vertices are colored.e. Answers vary. Sample coloring algorithm:

1. Make a list of the vertices in the graph, ordering them from highestdegree to lowest. Ties are broken arbitrarily.

2. Chose a color.3. Color the first uncolored vertex in the list and all uncolored vertices not

adjacent to it or another vertex of that color.4. If some vertices are uncolored, choose a new color, and go to Step 3;

otherwise stop. Your graph is colored.

4. 3.5. The graph has no edges.6. a. 2, 3, 4, 5

b. A complete graph KN would need N colors since every vertex in the graph isadjacent to every other vertex.

c. Good coloring algorithms should have no problem coloring a completegraph.

7. a. 2b. 3c. Difficulties with coloring algorithms may arise here. Instead of giving a

chromatic number of two, algorithms may indicate that three colors areneeded.

8. Many different graphs may need to be tried here. Three possible graphs to useare shown below:

9. 310. 411. 3


12. Sample coloring:

13. a. Sample coloring:

b. Sample coloring:

Review1. The following are important points made in Chapter 4.

• Graphs are useful tools in modeling real-world situations.• Graphs are made up of vertices and edges. Edges are drawn between vertices

to show particular relationships.• Graphs can be represented by diagrams, adjacency matrices, and vertex-edge

lists.


• Graphs are used to model the tasks that make up a project. In turn, this helpsorganize the work so the project can be completed in a minimal amount oftime.

• Euler circuits and Euler paths are paths through a graph that traverse eachedge of a graph exactly once. Algorithms have been developed for findingthe Euler circuit or path efficiently.

• Hamiltonian circuits and Hamiltonian paths are paths through the graph thatvisit each vertex of the graph exactly once. There is no easy way of findingout whether or not a graph has a Hamiltonian circuit or path.

• Graph coloring helps you to solve problems that deal with conflict such asscheduling problems.

2.

3. a. A,(0); B,(4); C,(4); D,(7); E,(6); F,(11); G,(10); H,(15); I,(16); J,(18)b. Minimum Project Time: 23.

4.Task Time Prerequisites

Start 0 —A 2 NoneB 3 NoneC 4 AD 4 A, BE 2 BF 3 CG 5 D, EH 7 F, GFinish


5. a. A,(0); B,(2); C,(2); D,(6); E,(6); F,(6); G,(9); H,(9); I,(14)b. Critical Path: Start–ABDHI–Finish.

Minimum Project Time: 17.6. a.

Start Finish

B

C

D

E

G

F

A

0

0

0

H

I

6

9

8

8

5

5

4

10

5

7

8

b. A,(0); B,(0); C,(0); D,(6); E,(9); F,(8); G,(13); H,(11); I,(18)c. Minimum Project Time: 26.d. Critical Path: Start–CFI–Finish.

7. a. Yes, a path exists from each vertex to every other vertex.b. No, not every pair of vertices is adjacent.c. A, D, or Cd. BCDE or BCAEe. Deg(C) = 4

f.

A B C D EABCDE

0 0 1 0 10 0 1 0 01 1 0 1 10 0 1 0 11 0 1 1 0

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⎣

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8. a. Sample graph:

E F

G

H

I

b. 4c. Yes, the degrees of all vertices are even.


9. Sample graph:

10. a. Euler path. Two vertices have odd degrees, and the remaining vertices haveeven degrees.

b. Euler circuit. All vertices have even degrees.11. a.

b.

12. a. Nob. Yesc.

13. a. Yesb. No, the graph has exactly two odd vertices. You would have to begin at one

of the vertices with an odd degree and end at the other.


14.

A B C D E FABCDEF

0 1 0 0 0 00 0 1 0 0 10 0 0 1 0 00 1 0 0 1 00 0 0 0 0 11 0 0 0 0 0

⎡

⎣

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⎤

⎦

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15. a.

b. There is no Condorcet winner because none of the candidates can beat all ofthe other candidates in a one-on-one race.

c. D, C, A, B; C, A, B, D; C, B, D, A; B, D, C, A; A, B, D, Cd. If the Hamiltonian path, B, D, C, A is chosen for a pairwise voting scheme, B

wins. The path shows that for B to “survive,” you need first to pair A and C.C wins. Next pair the winner, C, against D, and D wins. Finally, pair thewinner, D, against B, and B is the final winner of the election.

16. Sample graph:

17. a.

b. Three time slots. Sample schedule:Time 1—Math & Art, Time 2—Reading & History, Time 3—Science & French


18. a.

b. 4 frequencies19. a. No, the outdegrees are not equal to the indegrees at each vertex.

b. Yes, the outdegree equals the indegree at all vertices but two. At one of thosetwo vertices, the indegree is one greater than the outdegree and at the othervertex, the outdegree is one greater than the indegree.Answers will vary, but the paths must begin at B and end at D.

20. a&b. Sample graph and coloring:

c. Three colors

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