Chapter 5
Teaching Notes, Chapter 5: A quick stop in Section 5.1 will expand student knowledge of graph
theory by including a discussion of directed graphs. This discussion allows classes to revisit the
Traveling Salesman and Highway Inspector problems in one more context as well as discuss
tournament graphs.
Covering the introduction to trees in Section 5.2 allows your class to also cover Section 5.3 or
Section 5.4 or both. In Sections 5.3 and 5.4 we provide an intuitive discussion of some of the
applications (and associated algorithms) that have made the use of trees so critical in computer
science.
Exploratory Exercise Set 5.1
1.
2.
(a) (1)
(2)
(b) (1)
(2)
(3)
(4)
(5)
Answers will vary. Example: (a, e, d, b)
Answers will vary. Example: (a, e, d, b, c, d, c, e, b, a)
A path in a directed graph from u to u.
A circuit in which no edge is repeated.
A simple circuit that includes each edge of the directed graph.
A path that includes each vertex of the directed graph exactly once.
A circuit that includes each vertex of the directed graph exactly once except for the fact
that the initial vertex and the final vertex which are the same.
(a) (1)
(2)
a
c
Fred
b
d
Max
Louise
Tom
June
(b) (1)
(2)
(3)
(c) (1)
S = {a, b, c, d, e} R ={(a, b), (a, d), (c, a), (c, d), (e, d), (e, e)}
S = {x, y, z} R = {(x, y), (y, x), (y, z), (z, y), (z, x), (x, z), (x, x), (y, y), (z, z)}
S = {p, q} R = {(p, q), (q, p), (q, q), (p, p)}
Answers will vary. The relation R = {(a, a), (b, b), (c, c)} on S = {a, b, c} is an example
of a reflexive relation.
(2) There is a loop at each vertex.
(3) Answers will vary. The relation R = {(a, b), (b, a), (c, b), (b, c)} on S = {a, b, c} is an
example of a symmetric relation.
(4) If two vertices are connected by one edge, they are connected by a second edge in the
opposite direction.
(5) Answers will vary. The relation R = {(a, b), (b, c), (a, c)} on S = {a, b, c} is an example
of a transitive relation.
(6) If there is an edge directed from vertex x to vertex y and an edge directed from vertex y to
vertex z, there is an edge directed from vertex x to vertex z.
3. (a) (1) Start at any vertex on the directed graph, call it S.
(2) Form a path by moving from vertex to directed edge to vertex. At each vertex, randomly
choose a directed edge, but do not choose any directed edge that has been previously
chosen. Since there are a finite number of edges, you must eventually use a vertex that is
not the initial vertex of an unused edge, so the path must end. The fact that the indegree
of each vertex is equal to the outdegree assures that, with the exception of S, each time
we choose a vertex, it is the initial vertex of an unused edge. Hence, the vertex at which
we must stop is S.
89
90
Chapter 5
(3) Examine the graph. If all edges are listed in a path, we are done. Suppose there is some
edge not listed in a path. Since the graph is weakly connected, there is some vertex V that
is part of a previous path but is the endpoint of a directed edge that has not been used.
Because we used edges connected to vertices two at a time and because initially the
indegree of each vertex equaled the outdegree, we observe that V and any other vertex
that is the endpoint of unused edges has the property that the number of unused edges for
which it is the initial vertex equals the number of unused edges for which it is the
terminal vertex.
(4) Start an auxiliary path at vertex V, again choosing only directed edges that have not
previously been chosen. Again, you must eventually stop, and your stopping point will
be vertex V. At this point, you will have used all edges in the graph exactly one time, or
you can return to Step 3 to prepare to form another auxiliary path.
Because there are a finite number of edges in the graph, we will use all edges by forming a
finite number of paths. Finally, we can design one single path that will use all edges exactly
once. We will start at S, use vertices and edges in the order found in the original path with
the proviso that each time we come to a vertex that is the first vertex of an auxiliary path (like
V), we will use the edges and vertices of that auxiliary path before continuing.
(b) Since the indegree of each vertex equals the outdegree of that vertex, there is an Euler circuit
for this directed graph.
(c) If we randomly form our primary path we might follow path (a, b), (b, h), (h, j), (j, c), (c, h),
(h, a). At this point we would have to stop, but we would observe that there are unused edges
at vertex j. Starting a secondary path at vertex j, we might proceed (j, g), (g, f), (f, e), (e, j),
(j, i), (i, e), (e, i), (i, j). At this point we would again have to stop, but we would observe that
there are unused edges at vertex c. Starting a secondary path at vertex c, we might proceed
(c, d), (d, c), at which point we would again have to stop, but we would have used each of the
edges in the graph. We could then create an Euler path by following our primary path until
we reach vertex j, then following the first secondary path until we returned to j, continuing on
the primary path until we reach c, follow the second secondary path, then return to a on the
primary path. (a, b), (b, h), (h, j), (j, g), (g, f), (f, e), (e, j), (j, i), (i, e), (e, i), (i, j), (j, c), (c, d),
(d, c), (c, h), (h, a)
(d) Start the primary path at the vertex with outdegree one greater than indegree. Proceed as in
(a), but observe that vertex at which you have to stop will be the vertex with indegree one
greater than outdegree. At this point all vertices with unused edges will have the property
that the number of unused outward-bound edges is equal to the number of unused inwardbound edges. The rest of the discussion will follow as in (a).
4. (a)
(1)
(2)
v1
v2
v1
v2
v3
v4
v4
v3
Trees and Directed Graphs
(b)
91
(1)
a
b
c
d
a
0
0
0
1
b
1
0
1
0
c
1
1
0
1
d
0
1
0
0
v1
v2
v3
v4
v5
v6
(2)
v1
0
0
0
0
0
1
v2
1
0
0
0
0
0
v3
0
1
0
0
0
0
v4
0
0
1
0
0
0
v5
0
0
1
1
0
0
v6
1
0
1
0
1
0
5. Answers will vary.
Exercise Set 5.1
1. Vertices: {p, q, r, s} Edges: {(p, q), (p, r), (p, s), (q, r), (r, q), (r, s)}
2. Vertices: {a, b, c, d, e, f, g} Edges: {(b, d), (b, e), (a, b), (a, c), (c, g), (c, f)}
3. Answers will vary.
(a)
(b)
(c)
4.
Vertex
a
b
c
d
e
Indegree
0
2
1
1
3
Outdegree
2
2
1
2
0
5.
Vertex
Indegree
Outdegree
p
2
2
q
2
2
r
3
1
s
2
1
t
1
4
u
1
1
6. Since the directed graphs in Exercises 4 and 5 have three vertices in which the indegree is not
equal to the outdegree, there is not an Euler path/circuit for either graph.
7. (a) (O, H, I, J, G)
(b) (G, K, L, O, A, B, D, C)
(c) (G, K, L, O, H, I, J, G)
(d) No, for vertices {B, F, H, J, K, L} the indegree does not equal the outdegree.
8. There is no Euler path or circuit since for five vertices the indegree and the outdegree are not
equal.
9. There is no Euler path or circuit since for five vertices the indegree and the outdegree are not
equal.
10. There is an Euler circuit since for each vertex, the indegree is equal to the outdegree. Example:
(a, e, d, e, b, c, d, a, d, c, b, d, b, a)
92
Chapter 5
11.
Pork
Chicken
Mutton
Fish
Beef
12. Gnats beat Robins who beat Ants who beat Bats who beat Sloths.
Sloths beat Gnats who beat Robins who beat Ants who beat Bats.
13. (a)
(b) All teams may claim to be the winner. For
example, Chumps beat Lumps who beat Bumps
Chumps
Champs
who beat Champs.
Lumps
Bumps
Jets
Colts
14. (a)
Bills
Dolphins
Patriots
(b) Colts
(c)Yes. (d) Yes, since there are no circuits, there is a single Hamilton path
(Theorem 4).
15. (A, B, C, D), (A, D, B, C), (A, C, D, B)
16. (P, Q, T, R, S), (Q, T, P, R, S), (T, P, Q, R, S)
17. (a)
(b) (A, B, C, E, D)
B
(B, C, A, E, D)
(C, A, B, E, D)
A
C
Note: If the edge connecting E and D is
directed in the opposite direction, the
paths would read (A, B, C, D, E) and
so on.
(c) A, B, and C are winners.
E
D
B
18. (a)
A
C
E
D
Trees and Directed Graphs
93
(b) Answers will vary depending on directions given to the edge connecting A and C and the edge
connecting B and D. If directions given are (A, C) and (B, D), there are five Hamilton paths:
(A, B, C, D, E), (B, C, D, A, E), (B, D, A, C, E), (C, D, A, B, E), (D, A, B, C, E).
(c) {A, B, C, D}
19. The winners are the teams in set S.
20. (a)
B
A
C
E
D
(b) B follows A in the first Hamilton path while A follows B in the second Hamilton path. Hence
if we use the edges from A to B from the first Hamilton path [the single edge (A, B)]
followed by the edges from B to A in the second Hamilton path [(B, C), (C, A)], we create a
circuit.
Teaching Notes, Section 5.2: Don’t miss Exploratory Exercise 3 in which students are given a
chance to explore Cayley’s ground-breaking work with trees in his investigation of hydrocarbons.
Exploratory Exercise Set 5.2
1. (a) (1) Answers will vary. (2) Theorems 5 and 6.
(b) Answers will vary. (c) Answers will vary.
2. (a)
A
(d) Answers will vary.
(d)
A
A
A
B
B
A
(b)
A
A
C
G
B
C
D
C
D
E
F
(c)
A
A
F
B
G
A
G
C
D
C
A
E
F
E
94
Chapter 5
(e)
A
B
B
C
C
D
C
F
E
F
F
F
G
H
H
(f)
Number of
Competitors
2
3
4
5
6
7
8
Number of
Games
1
2
3
4
5
6
7
(g) If there are n players in a single elimination game, then there will be n – 1 games played.
(h)
A
B
B
C
C
C
E
D
E
E
I
F
E
G
H
H
I
Trees and Directed Graphs
95
(i) In a single elimination tournament with 500 entries, there must be 499 games played.
3. (a) (1) There are n carbon atoms, and 2n + 2 hydrogen atoms.
(2) There are a total of n + (2n + 2) = 3n + 2 vertices.
(3) 4(n) + (1)(2n + 2) = 6n + 2
(4) One-half of 6n + 2 or 3n + 1
(5) There are 3n + 2 vertices, and 3n + 1 edges. By Theorem 7, CnH2n+2 must be a tree.
(b) (1)
(2)
4. Answers will vary.
5. Answers will vary.
Exercise Set 5.2
1. Tree
3. Not a tree, this graph is not connected.
5. Tree
7. This confirms Theorem 6.
2. Not a tree, this graph contains a cycle.
4. Tree
6. Not a tree, this graph is not connected.
96
Chapter 5
8. (a)
(b)
(c)
(d)
(e)
(f)
Vertex A
Vertex A
Vertices W and R
Vertex Q
Vertices X and A
Vertex S
9. (a)
(b)
(c)
(d)
(e)
(f)
Vertex Y
Vertex W
Vertex R
Vertex W
Vertices W, R, and Y
Vertices Q and X
10. (a)
(b)
(c)
(d)
11. (a)
Assistant Superintendent: Instruction, Superintendent
Assistant Superintendent: Instruction
Director: Budget, Director: Transportation
Superintendent
(b)
Y
12. (a)
R
T
U
R
(g)
Z
W
Q
X
Z
R
W
S
(g)
T
U
(b)
R
T
R
S
U
Q
T
S
U
Q
P
P
13. (a) 14 (Theorem 6)
(b) 10 (Theorem 6)
14. Because this graph is not connected and because it has a cycle, it is not a tree. However, the
graph has 4 vertices and 3 edges.
15. Kn is a tree for n = 1 or n = 2. For larger values of n, Kn has a cycle.
16. Let X be a vertex in a tree and suppose X has two distinct parents, A and B. If R is the root of the
tree, there is a path C1 from R to A and a path C2 from R to B. Let Y be the last vertex that C1 and
C2 have in common. Then follow C1 from Y to A, then A to X, then X to B and back along C2 to Y.
This is a cycle. This is a contradiction.
Trees and Directed Graphs
97
17.
M FM
F
F
M
F
F
F
M
F
M
18. (a)
(b)
(c) There are ten such trees. We will seek a solution by focusing on the vertex with maximum
degree:
There is one tree in our set whose vertex of maximum degree has degree five.
There is one tree in our set whose vertex of maximum degree has degree four. Alter the tree
above to find it.
There are three trees in our set whose vertex of maximum degree has degree three. Here is one of
them. Modify the tree with maximum vertex of degree four to find the other two.
98
Chapter 5
There are three trees in our set whose vertex of maximum degree has degree two and that that
have two such vertices. Here are two of them.
There are two trees in our set whose vertex of maximum degree has degree two with the property
that there is one such vertex. Here is one of them.
19. (a) The simple circuit C is a cycle in the graph.
(b) Ĉ is a cycle in the graph.
(c) There are only a finite number of vertices in C.
(d) There are no repeated vertices in the simple circuit produced by the process other than the
initial and terminal vertices.
20. (a) In a tree there are no parallel edges since parallel edges between vertex u and v would create a
cycle.
(b) By definition, wk = ur = y and w1= u1 = x.
(c) Because path C1 is different from path C2, there must be an integer t with wt  ut. t must be
greater than 1 because w1= u1 = x.
(e) We know that wk = ur = y so k would be such an integer.
(f) This is a circuit because the initial vertex and the final vertex are the same. Since C1 is a
simple path, no edges are repeated in moving from wt – 1, along C1 to ws. Similarly, no edges
are repeated moving from ws to wt – 1, along C2. Since none of the vertices wt to ws from C1
are vertices of C2 there are no edges repeated. This is a simple circuit.
Teaching Notes, Section 5.3: In solving the problem of finding a minimum spanning tree for a
weighted graph, we have presented Prim’s algorithm because it is intuitively clear that as we proceed
through the steps of Prim’s algorithm, we indeed are creating a tree. However, some instructors using
this text have reported that they prefer to present Kruskal’s algorithm. Notice that Kruskal’s
algorithm is available to you in Exploratory Exercise 3.
Trees and Directed Graphs
99
Exploratory Exercise Set 5.3
1. (a) Answers will vary.
(b) Answers will vary.
0
0
1
1
0
0
1
1
1
1
2
2
2
2
2
2
2
2
2
2
3
2. (a) (3)
Tree 1
Tree 2
Tree 3
Tree 4
Tree 5
Tree 6
Tree 7
Tree 8
100
Chapter 5
(b) (1)
Tree
Tree 1
Tree 3
Tree 5
Tree 7
Weight
18
16
15
16
Tree
Tree 2
Tree 4
Tree 6
Tree 8
Weight
17
14
16
17
(2) The minimum spanning tree is Tree 4 with weight of 14.
(3)
6
6
b
3
d
5
3
4
a
b
3
e
c
4
d
5
a
e
c
4
6
6
b
d
3
5
a
b
3
4
3
e
c
4
3
4
d
5
a
4
c
4
3
e
3. (a) (1)
Edge
{a, b} {b, c} {a, c} {e, f} {b, d} {c, d} {d, e} {c, d} {d, f} {c, e}
Weight
1
1
2
2
3
3
3
3
4
4
(2) Choose {b, c} and shade it.
(3) {a, c} cannot be chosen because this would create a cycle. Choose {e, f}, an edge of
weight 2, and shade it.
(4) Choices will vary. We have chosen {b, d} and {d, e}. The resulting spanning tree is
shaded below.
3
4
b
1
d
1
f
3
3
a
2
c
4
2
e
(5) This is a spanning tree with weight 10. There is no spanning tree with smaller weight.
4. Answers will vary.
Exercise Set 5.3
1. Answers will vary.
2. Answers will vary.
Trees and Directed Graphs
101
3. Answers will vary.
(0)
(0)
(1)
(1)
(0)
(1)
(0)
(1)
(2)
(2)
(1)
(3)
(1)
(3)
(0)
(1)
(0)
(2)
(3)
(4)
(1)
(1)
(2)
(1)
(3)
(3)
(5)
(0)
(1)
(2)
(3)
(4)
(1)
(5)
(3)
(5)
(6)
(6)
(3)
(4)
(5)
102
Chapter 5
4. Answers will vary.
A
B
C
E
F
I
J
D
G
H
K
L
5. A spanning tree can be used to solve this problem. If the edges are labeled with weights
representing the cost to upgrade the computer link, then a minimum spanning tree will minimize
the cost to upgrade.
6. Answers will vary.
7. (a)
(b)
(c)
(d) The spanning tree resulting from a breadth-first search of Kn is a tree with one vertex of
degree n – 1 and n – 1 vertices of degree 1.
8. (a)
(b)
(c)
Answers will vary.
9.
P
4
4
R
6
3
S
3
P
2
4
5
Q
P
4
R
6
3
3
S
T
S
6
3
1
4
4
R
3
Q
P
2
4
2
4
3
Q
R
2
3
1
T
S
Q
2
S
T
P
6
3
1
Q
4
R
1
T
1
T
Trees and Directed Graphs
103
10. Answers will vary.
3
a
2
b 4
c
e2
f
2
d
3
g
4
h
3
i
11. Answers will vary.
a
2
b
5
13
6
g
a
g
a
11
i
2
b
6
c
5
7
3
h
11
i
2
b
6
c
5
7
3
11
i
2
b
6
c
5
7
5
f
2
3
3
h
h
11
i
2
b
6
c
5
7
8
11
i
13
3
8
3
h
11
i
2
b
6
c
5
7
5
e
f
2
13
3
8
3
h
11
i
2
b
6
c
5
7
5
e
d
g
f
2
13
9
5
e
6
e
13
a
8
3
d
g
f
3
13
9
h
6
g
8
3
d
13
6
f
2
13
9
a
5
e
13
e
d
g
5
2
13
9
3
6
a
8
c
7
6
f
2
d
g
a
5
e
13
g
6
5
d
9
h
13
9
8
3
6
b
13
f
3
d
2
6
e
13
a
5
2
13
9
c
7
d
9
6
f
2
13
3
3
h
8
11
i
104
Chapter 5
2
a
b
6
5
13
c
7
6
13
g
3
5
6
3
e
d
3
i
11
f
2
8
h
c
5
f
2
9
b
5
e
d
2
a
g
3
h
8
i
12. The minimum time is 20 hours.
4 hr
4 hr
3 hr
2 hr
4hr
3 hr
13. A minimum spanning tree would solve the problem. The minimum number of kilometers that
need to be paved would be 197 kilometers.
24
33
19
23
22
15
11
9
41
14. (a) In Steps 1 and 2, instead of choosing an edge with the least weight, an edge with the greatest
weight should be chosen.
(b) Below is found a maximum spanning tree with maximum weight 18.
Q
P
6
5
4
R
3
T
S
15. (a)
(b)
(c)
(d)
u and v are vertices in G, and G is connected.
C is a path from u to v in G'.
Modify C by replacing (vr, vr + 1) by (vr, vr  1, …, v2, v1, vk, vk  1, …, vr + 1) in C.
The new path created in (c) is a path from u to v in G'. Since any pair of vertices G' is
connected by a path, G' is connected.
Trees and Directed Graphs
105
Exploratory Exercise Set 5.4
1. (a) (1)
First
Second
Third
Fourth
(2)
First
Second
Third
Fourth
Fifth
Operation
(4 / 2)
22
3+4
7+5
Resulting Expression
3+22+5
3+4+5
7+5
12
Operation
(2+3)
52
4*25
100+3
1034
Resulting Expression
4*52+34
4*25+34
100+34
1034
99
Operation
(1+1)
22
16/4
Resulting Expression
16/22
16/4
4
Operation
(4 + 2)
(4 * 6)
(6 + 24)
(3 * 2)
(4 + 6)
30 / 10
Resulting Expression
(6+4*6))/(4+3*2)
(6+24)/(4+3*2)
30/(4+3*2)
30/(4+6)
30/10
3
(3)
First
Second
Third
(4)
First
Second
Third
Fourth
Fifth
Sixth
(b) (1)
+
5
+
↑
3
/
4
2
2
106
Chapter 5
(2)
(3)
_
/
+
4
3
*
2
(4)
1
2
+
2
+
↑
4
↑
16
1
3
/
+
+
6
4
*
*
+
4
4
3
2
2
2. Answers will vary.
3. (a)
First
Second
Third
Fourth
Fifth
Operation
2+4
6–3
3*5
15 + 3
3 * 18
Resulting Expression
663–5*3+*
635*3+
3 15 3 + *
3 18 *
54
First
Second
Third
Fourth
Operation
2/2
3+1
7*4
4 – 28
Resulting Expression
4731+*–
474*–
4 28 –
–24
First
Second
Third
Fourth
Fifth
Sixth
Operation
2+4
1–3
6 * (–2)
4/2
2–3
–12 + –1
Resulting Expression
613–*42/3–+
6 (–2) * 4 2 / 3 – +
(–12) 4 2 / 3 – +
(–12) 2 3 – +
(–12) (–1) +
–13
(b)
(c)
Trees and Directed Graphs
107
(d)
Operation
3–2
8/2
4+4
1+8
6*9
First
Second
Third
Fourth
Fifth
Resulting Expression
61482/++*
6144++*
618+*
69*
54
4. (a) (1) 2 + 3 * 5 – 4 – 6 / 2
This expression is not equivalent to the original. When we compute the original
expression, we get 16; when we compute the expression resulting from an inorder
traversal of the tree, we get 10.
(2) ((2 + (3 * 5)) – (4 – (6 / 2)))
This expression is equivalent to the original. When we compute this expression we get
16.
(b) (1) 2 3 5 * + 4 6 2 / – –
(2) When you evaluate using postorder traversal, the result is 16, the same answer you get
from the original expression.
(c) (1) 3 4 2 / 2  + 5 +
(2) 4 2 3 + 2  * 3 + 4 –
(3) 16 1 1 + 2  /
(4) 6 4 4 2 + * + 4 3 2 * + /
Exercise Set 5.4
1. (a) The left subtree of B consists of the vertices D and F and the connecting edge.
(b) The right child of E is the vertex G.
(c) F does not have a left child.
(d) The right subtree of D is the vertex F.
2. (a) The left subtree of B consists of the vertices D, H, and I, and the connecting
edges.
(b) The right child of E is the vertex J.
(c) F does not have a left child.
(d) The right subtree of D is the vertex I.
3. (a)
(b)
*
*
_
/
3
(c)
5
12
_
+
3
+
(d)
2
3
+
/
2
*
8
4
_
*
4
/
c
4
3
x
d
e
y
108
Chapter 5
/
(e)
+
+

*
_
3
6
5
1
3
*
(f)
/
_
+
/
c
a
4.
5.
6.
7.
8.
9.
8
7
(a)
(a)
(a)
(a)
(a)
(a)
(c)
(e)
10. (a)
(b)
11. (a)
d
b
e
f
Q, P, R
(b) Q, R, P
M, O, N, Q, P
(b) O, Q, P, N, M
H, D, B, I, E, A, J, F, K, C, G, L
(b) H, D, I, E, B, J, K, F, L, G, C, A
W, T, X, R, Y, U, Z, Q, V, S
(b) W, X, T, Y, Z, U, R, V, S, Q
18
(b) –21
(c) 31
(d) – 9
3 5 – 12 3 / *
(b) 2 3 + 4 *
243*+84/
(d) c d e /  x y * +
6 3 – 3 * 7 + 5 1 – 8 + / (f) a b – c / d e f / + *
bat, cat, dog, fish, frog, goat, hare, horse, llama, pig, toad
circuit, cycle, edge, graph, path, search, tree, vertex
Manuel
Bob
Tran
Rita
Ida
Sida
Ric
Pat
(b) {Bob, Ida, Manuel, Pat, Ric, Rita, Sida, Tran}
12. (a)
(b)
+
_
d
a
d
c
13.
A
C
B
E
D
F
+
_
d
*
*
+
_
b
e
a
f
c
Trees and Directed Graphs
109
14. Consider a binary tree with three vertices in which the root a has only a left child b, and b has
only a left child c. The postorder and inorder listing of the vertices gives the same result.
15. (a) M, N, P, O, Q, R
(b) A, B, D, H, E, I, C, F, J, G, K, L
16. (a) P, Q, R
(b) M, N, O, P, Q
(c) A, B, D, H, E, I, C, F, J, K, G, L
(d) Q, R, T, W, X, U, Y, Z, S, V
17. (a) 23
(b) 28
18. (a) * – 3 5 / 12 3 This evaluates to –8.
(b) * + 2 3 4 This evaluates to 20.
(c) – + 2 * 4 3 / 8 4 This evaluates to 12.
In all three cases, the prefix (Polish) expression yields the same result as following the order of
precedence on the original expressions.