Chapter 4
Teaching Notes, Chapter 4: For those who wish to take a quick look at graph theory before
going on to other topics, we recommend Sections 4.1 and 4.2. If you wish to cover more topics, but
not the whole chapter, the following dependencies might be useful.
Section
Section 4.3
Section 4.4
Section 4.5
Section 4.6
Depends upon
Sections 4.1 and 4.2
Sections 4.1 and 4.2
Sections 4.1, 4.2, and 4.3
Sections 4.1, 4.2, 4.3, and 4.5
Teaching Notes, Section 4.1: Asking students to work Exploratory Exercise 2 in pairs will be
useful in helping them understand the proof of Theorem 4; asking them to complete Exploratory
Exercise 4 will assist in understanding Theorems 1 and 2.
Related Website: Visit this math forum site at Drexel University for an introduction to Euler’s
Königsburg bridge problem: http://mathforum.org/isaac/problems/bridges1.html
Exploratory Exercise Set 4.1
1.
Answers will vary.
2. Answers will vary.
3.
(1) Start at one of the towns with odd degree, call it S.
(2) Drive from town to town, making random choices about where to go, but not passing over
any section of road twice. Since there are a finite number of sections of road, we must
eventually enter a town from which there are no uninspected sections of road on which to
leave. Theorem 1 ensures that this town at which we must stop is the second town of odd
degree, call it D.
(3) Examine the map. If all sections of road have been inspected, we are done. Suppose there is
some section that has not been inspected. Since the map is connected, there is some town T
that has already been visited in a previous trip but is also the terminus of a section of road
that has not been inspected. Further, since we inspect roads connected to towns two at a
time (enter on one, leave on another), any town with roads that have not been inspected will
have an even number of such roads
(4) Begin a side trip at town T, again taking care to use only road sections that have not been
inspected. By Theorem 2, you will eventually stop at T. At this point you have inspected
all sections, or you can return to Step 3 to prepare for another side trip.
Because there are a finite number of road sections on the map, we will inspect all sections in a
finite number of side trips. Finally, we can design one single economical inspection tour for the
Inspector. We will start at S, and travel the original tour with the proviso that each time we
come to a town that is the beginning of a side trip (like T) we will take the side trip beginning
and ending at that town before continuing. Eventually we will end the trip at D.
4. Answers will vary.
57
58
Chapter 4
Exercise Set 4.1
1. Answers are not unique.
Salesman
Inspector
1
S
1
S
2
10 4
5
4
3
2
5
8
9
3
7
6
2. Answers are not unique:
Salesman
Inspector
6
3
4
1
2
3
4
S
7
5
5
8
6
9
10
2
S
1
3.
Map
(a)
(b)
(c)
(d)
(e)
(f)
Number of Towns of
Odd Degree
4
2
2
8
4
0
Number of Towns
of Even Degree
3
4
4
4
1
8
Route for Inspector
(Yes or No)
No
Yes
Yes
No
No
Yes
4.
Map
Sum of the Degrees
Number
of Towns
of Roads
(a)
18
9
(b)
16
8
(c)
12
6
(d)
32
16
(e)
24
12
(f)
20
10
All sums are even numbers as stated in Theorem 6. Observe that the sum of the degrees of all
towns is twice the number of roads.
Graph Theory
59
5. Since all destinations have odd degree, you cannot plan a walk that will take you over every
bridge once.
A A
D
B
C
6. You can walk around the city, crossing each bridge exactly once. You can do this because only
two destinations in the city have an odd number of bridges (an odd degree). One possible route is
shown below.
2
1
4
3
S
5
8
9
6
7
7. (a)
(b) In the roadmap in (a) each door is represented by a road. Since every door must be passed
through, this is equivalent to passing over every road.
(c) Yes. By Theorem 5, since only two of the rooms have odd degree, a path through the house
can be found that passes through each door exactly once.
(d)
2
3
6
1
4
8.
9.
5
There is no path through this house using each door exactly once because four rooms have an
odd number of doors (odd degree).
A salesman can visit each of the towns in this road system exactly once.
1
S
11
10
5
9
6
4
3
7
8
2
60
Chapter 4
10. If the citizens of Königsberg build a bridge from island B to shore D, then only two destinations
would have odd degree, and the problem can be solved.
11. Answers will vary
1
2
3
4
2
5
3
6
1
8
10
7
T1
9
2
3
1
S
2
1
13 14
5
11
6
9
3
T2
4
10
16
15
7
8
12
12. The sum of an even integer and an odd integer is an odd integer, the sum of two odd integers is an
even integer, and the sum of two even integers is even. It follows that if a town has an odd
number of towns of odd degree, the sum of the degrees would be odd, in contradiction of
Theorem 6. More specifically, suppose there is a single town of odd degree. The sum of the
degrees of all other towns would be some even integer, hence the sum of the degrees of all towns
would be odd.
13. When the Salesman starts at a town, he can travel to either of two towns. But at each
successive town there is a single road that will carry him to a town other than the one he just left.
If this one choice takes him back to a town he has already visited, before he gets to all towns, then
the road system is not connected. Any town not yet visited would have no roads to any of the
towns on the trip he just completed.
14. Answer is not unique.
15. By Theorem 1, we must begin and end our trip in a town of odd degree. With four towns, we can
only begin in one, and end in one, leaving two towns with sections of road that have not been
inspected.
Teaching Notes, Section 4.2: One of our learning objectives when teaching any mathematics
course involves convincing students that they can learn from reading the text in addition to learning
from our brilliant exposition. Two or three times each semester we begin the class by assigning new
reading to be done in class, and then assigning problems related to the reading to be done in class,
usually followed by group discussion. We have used Section 4.2 for this purpose on several
occasions. We generally ask the students to solve Exploratory Exercises 1 and 2 and Exercise 8
individually and then discuss answers in pairs or groups of four.
Related Websites: Visit this web site for an exhaustive discussion of the traveling salesman problem
including historical notes and pictures of Hamilton’s original “Voyage Around the World Game”:
www.math.princeton.edu/tsp/
Links to a variety of web pages related to graph theory can be found at this site:
http://archives.math.utk.edu/topics/discreteMath.html
Graph Theory
61
Exploratory Exercise Set 4.2
1.
Answers will vary.
2.
(a) e6
(b) x, y
(c) x, w
(d) e1, e7
(e) e5
(f) (x, e6, u, e4, w)
(g) (x, e1, v, e7, x, e1, v, e2, z, e3, w)
(h) (x, e7, v, e2, z, e3, w, e4, u, e6, x)
(i) (w, e4, u, e5, u, e4, w)
3.
(a) (1) The computers
(2) If the computers are connected by a cable.
(3)
A
B
E
F
C
(b)
D
(1) Species of animals
(2) Whether the animals compete for the same food resources
(3)
Hawk
Squirrel
Racoon
Woodpeckers
Mice
Owl
Opossum
4. (a)
(1) Start at any vertex on the graph, call it S.
(2) Form a path by proceeding from vertex to edge to vertex, at each vertex making a
random choice about which edge to use, but not choosing any edge twice. Since there
are a finite number of edges, you must eventually use a vertex for which there are no
unused edges that you can choose, so the path must end. Because all vertices are of
even degree, Theorem 8 ensures us that this vertex at which we must stop is S.
(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 connected, there is some vertex V that is
part of a previous path but is the endpoint of an edge that has not been used. Because
we used edges connected to vertices two at a time, we observe that V and any other
vertex that is the endpoint of unused edges is the endpoint of an even number of unused
edges.
(4) Start an auxiliary path at vertex V, again choosing only edges that have not previously
been chosen. Again, you must eventually stop, and your stopping point by Theorem 8
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 initial vertex in an
auxiliary path (like V), we will use the edges and vertices of that auxiliary path before
continuing.
(b) There is an Euler circuit since each vertex will has even degree.
(c) Answers will vary.
62
Chapter 4
(d) In step (1) the initial path should begin at a vertex of odd degree; Theorems 7 and 8 ensure
that it will end at the other vertex of odd degree. The reasoning in the remaining steps is very
much the same.
v1
v2
5.
(a) (1)
(2)
v1
v2
v3
v4
(b) (1)
b
1
0
1
0
1
c
0
1
0
1
1
d
0
0
1
0
1
e
1
1
1
1
0
(c) (1)
v3
v4
b
0
0
1
1
1
b
d
e
c
a
d
0
0
1
1
0
e
1
1
0
1
1
c
1
1
1
0
0
a
1
0
1
0
0
(2)
a
0
1
0
1
a
b
c
d
b
1
0
1
1
c
0
1
0
1
d
1
1
1
0
Exercise Set 4.2
1.
V = { P, Q }
2.
V = { a, b, c }
3.
V = { w, x, y, z }
4.
V = { p, q, r, s, t, u }
b
5. (a)
a
P
p
0
1
0
0
1
p
q
r
s
t
q
1
0
1
1
1
r
0
1
0
1
1
s
0
1
1
0
1
t
1
1
1
1
0
E = { {P,Q} }
E = { {a, b}, {a, c}, {b, c} }
E = { {w, x}, {w, y}, {w, z}, {x, y}, {x, z}, {y, z} }
E = { {p, q}, {q, r}, {s, t}, {s, u}, {t, u} }
(b)
d
z
x
e
c
y
f
Q
R
6.
7.
8.
2
(2)
a
0
1
0
0
1
a
b
c
d
e
(c)
v1
T
Loops: e5
Loops: e1, e3
(a) e1, e7
(c) e3, e4
(e) (v, e1, u, e3, x)
(g) (v, e1, u, e3, x, e4, u, e3, x)
(i) x
Parallel Edges: e1, e2, e3
Parallel Edges: e5, e6
(b) u, y
(d) e2
(f) (v, e7, w, e6, y, e5, x)
(h) (y, e6, w, e7, v, e1, u, e3, x, e5, y)
(j) u
Graph Theory
9.
63
Answers are not unique.
(a)
(b)
(c)
10.
11.
12.
13.
14.
15.
16.
17.
Answers will vary.
Since there are exactly two vertices of odd degree, there exists an Euler path.
Since there are exactly two vertices of odd degree, there exists an Euler path.
(c, a, e, f, b, c, e, d, f, a, b, d)
There does not exist a Euler path or circuit because there are four vertices of odd degree.
Because all vertices have even degree, there is an Euler circuit.
(a, c, e, d, b, a)
There is no Hamilton path or circuit as the far left and right vertices cannot be reached without
repeating a vertex.
Suppose there is a Hamilton path or circuit. Any such path would have to both start and end at
one of the two vertices of degree one, else a vertex adjacent to a vertex of degree one would be
visited twice. Assume that we have a Hamilton path that starts at one such vertex and ends at
the other. As that path moves from vertices b and f to vertices c, d, and e, it necessarily visits
vertices b and f a total of three times. Thus one of b and f is visited at least twice. Thus there is
no Hamilton path.
18. (a)
(b)
(c)
(d)
19. Label the vertices in the following order: a
b
d
c
i
k
m
p
s
g
n
q
e
h
l
f
j
o
r
t
64
Chapter 4
A Hamilton circuit is (a, f, t, s, c, i, m, p, q, r, o, j, e, h, l, n, k, g, d, b, a).
Teaching Notes, Section 4.3:
Related Website: Visit this web site for java applets that illustrate many of the fundamental ideas of
graph theory including the concept of isomorphic graphs: www.utc.edu/%7Ecpmawata/petersen/
Exploratory Exercise Set 4.3
1. (a)
Adjacent vertices in G1:
Adjacent vertices in G2
Vertices adjacent to p: s and q
Vertices adjacent to j: k, l, and m
Vertices adjacent to q: p, s, and r
Vertices adjacent to k: j, l
Vertices adjacent to s: p, q, and r
Vertices adjacent to l: m, k, and j
Vertices adjacent to r: q, s
Vertices adjacent to m: j, l
(b) No, as p is adjacent to two vertices, and j is adjacent to three vertices
(c) q  j
ql
sl
sj
pk
pm
rm
rk
2.
(a) a  x
ay
az
bx
by
bz
by
bz
bx
cz
cx
cy
cz
cx
cy
ay
az
ax
(b) Graphs in which b  z, because they both have degree 2.
(c) b and z are paired together because they have the same degree
(d) Answers will vary
aw
ax
ay
az
bx
bx
by
bz
bw
cw
cy
cz
cw
cx
dz
dz
dw
dx
dy
ay
(e) The first vertex can be assigned one of n vertices, the second vertex, n – 1 vertices, and so
forth. The multiplication principle ensures that there are n! isomorphisms.
3.
(a) (1) (a, b, c, d, a)
(2) (a, b, d, c, a)
(a, d, b, c, a)
(a, c, d, b, a)
(a, c, b, d, a)
(a, d, c, b, a)
(3) At the first vertex, we can choose one of 3 edges; at the second vertex, we can choose
one of 2 edges; and at the last vertex we have one edge to choose. 3  2  1 = 6.
(4) At each vertex we have 6 Hamiltonian circuits. 4  6 = 24.
(b) (1)
(a, b, c, a)
(b, c, a, b)
(c, a, b, c)
a
(a, c, b, a)
(b, a, c, b)
(c, b, a, c)
b
c
Graph Theory
65
(2)
4.
(a, b, c, d)
(b, a, c, d)
(c, a, b, d)
(d, a, b, c)
(a, b, d, c)
(b, a, d, c)
(c, a, d, b)
(d, a, c, b)
(a, c, b, d)
(b, c, a, d)
(c, b, a, d)
(d, b, a, c)
(a, c, d, b)
(b, c, d, a)
(c, b, d, a)
(d,
b,
c,
a)
(a,
d,
b,
c)
(b, d, a, c)
c
(c, d, a, b)
(d, c, a, b)
(a, d, c, b)
d
(b, d, c, a)
(c, d, b, a)
(d, c, b, a)
(3) Each of the n vertices has (n – 1)! different Hamiltonian circuits.
n  (n –1)! = n!
(c) (1) 20! = 2,432,902,008,176,640,000
(2) 2,432,902,008s
(3) Approximately 77 years
(a) (1) The function is clearly one-to-one and onto. Further, this table shows that the function
preserves adjacency in both directions.
Adjacent vertices in G1:
Adjacent vertices in G2
Vertices adjacent to a: b and d
Vertices adjacent to z = f(a): x = f(b), w =
f(d)
Vertices adjacent to b: a and d
Vertices adjacent to x = f(b): z = f(a), w =
f(d)
Vertices adjacent to c: d
Vertices adjacent to y = f(c): w = f(d)
Vertices adjacent to d: a, b and c
Vertices adjacent to w = f(d): z = f(a), x =
f(b), y = f(c),
a
b
(2)
a
b
c
d
a
0
1
0
1
b
1
0
0
1
c
0
0
0
1
d
1
1
1
0
z
x
y
w
z
0
1
0
1
(3) The adjacency matrices are the same.
(b) (1) p  x
qw
rz
sv
ty
(2)
p q r s
t
p 0 1 0 1
1
x
q 1 0 1 0
0
w
r 0 1 0 1
0
z
s 1 0 1 0
1
v
t 1 0 0 1
0
y
(3) The matrices are the same.
(4) Yes.
x
1
0
0
1
y
0
0
0
1
w
1
1
1
0
x
0
1
0
1
1
w
1
0
1
0
0
z
0
1
0
1
0
v
1
0
1
0
1
y
1
0
0
1
0
66
Chapter 4
Exercise Set 4.3
1. (a) 10 handshakes
(b) 3 bicycle rides
(c) 6 cables
(d) A complete graph
2.
3.
K7
K6
Answers will vary
4.
5. (a)
Ben
Luis
Lois
(c)
Dell
John
A
(b) K5
C
(d) K2,3
D
B
E
6. (a) The graphs do not have the same number of vertices.
(b) The first graph has no vertices of degree 1, and the second graph does.
7. (a) No. The fact that they have the same number of vertices ensures that there is a one-to-one
and onto function between the sets of vertices. However, there may not be such a function
that preserves adjacencies. For example, think about K3 and the graph with three vertices but
no edges.
Graph Theory
67
(b) No. They may not have the same number of vertices. For example, think about K3 and a
simple graph on four vertices with three edges.
8. (a) 1 (b) 3 (c) 6 (d) 10 (e) 1 + 2 + … + (n – 1) or
( n  1)( n)
2
9. In a complete graph on n vertices, all vertices have degree of n – 1. Therefore, if n is an odd
integer, the degree of each vertex is even and vice versa. By Theorem 10, if n is odd and greater
than 3, Kn has an Euler circuit. K2 has an Euler path but no Euler circuit. If n is even and greater
than 2, Kn has too many vertices of odd degree to have an Euler path.
10. p  l, q  m, r  n, s  o, t  p
11. a  z, b  x, c  y, d  w
12. Vertex d has degree 3, and no vertex in the second graph has degree 3.
13. Vertex a has degree 1, and no vertex in the second graph has degree 1.
14. a  v, b  x, c  z, d  w, e  y
15. Vertex u has degree 2, and no vertex in the first graph has degree 2.
16.
17.
.
18. (a) V1 = {a, b}, V2 ={c, d, e}
(b) V1 = {p, r}, V2 = {s, t}
19. (a) a and b are adjacent
(b) a and c are adjacent
(c) b and c are adjacent
(d) G is not bipartite, because it is impossible to place a, b, and c in two disjoint sets having the
property that vertices within sets are not adjacent.
20. (a) Yes (b) Yes (c) Yes (d) Yes
(e) 2,n will always have an Euler circuit for odd n as there will be exactly 2 vertices of odd
degree.
(f) 2,n will always have an Euler circuit for even n as all vertices will have even degree.
21. (a) No (b) Yes (c) No (d) No
(e) K3,n has n vertices of degree 3. By Theorem 9, if a graph has vertices of odd degree, there is
no Euler path unless there are exactly 2 vertices with odd degree. Thus, by Theorem 9, there
will be no Euler path.
22. If both m and n are even, then all vertices will have even degree, and there will be an Euler circuit.
23. If m is odd, then for all n except 2, the number of vertices of odd degree will be some number
other than 2. Thus by Theorem 9, there will be no Euler path.
24. (a){Q, R}
68
Chapter 4
(b)
Function between
edge sets
Edges of Edges of
G1
G2
{A, E}
{Q, R}
{E, D}
{R, S}
{D, A}
{S, Q}
{D, C}
{S, T}
{C, B}
{T, P}
{B, A}
{P, Q}
25. (a)
(b)
(c)
(d)
(e)
(f)
Because {a1, a2, …, ak}are all adjacent to u and f is an isomorphism.
Because the isomorphism function is one-to-one.
Then f(u) has the same number of neighbors as u, and hence the same degree.
Because the isomorphism function is onto.
By definition of isomorphism, f(a) is adjacent to f(u) if and only if a is adjacent to u.
{a1, a2, …, ak} was defined to be the set of neighbors of u.
Teaching Notes, Section 4.4: Much of the utility of graphs in real world applications turns
around organizing complex information in a graph and then using some kind of searching algorithm
to find a “best” solution of all solutions represented in the graph. In this section we explore several
algorithms for finding the “shortest path” through a graph. Though the algorithms are somewhat
technical, this study gives a clear indication of this utility of graphs and provides an excellent
example of the power of iteration.
The material can be handled at several different levels. You can focus primary attention on the
geometric shortest path algorithm which is the most intuitive of the algorithms presented. At a
second level, you can next discuss Dijkstra’s shortest-path algorithm. [Although the geometric
shortest-path algorithm is not identical to the classic Dijkstra algorithm, the geometric shortest-path
algorithm does serve as a useful stepping stone toward understanding Dijkstra’s algorithm.] Finally,
at a third level you can discuss how the Dijkstra algorithm can be implemented using just tables and
lists (and hence could be implemented on a computer).
Whatever your choice in algorithms to present, you need to take care to ensure students
understand their power. With the small graphs that we are able to analyze by hand, it is easy to feel
like trial and error is as efficient as the algorithms. Making sure that students spend some time on
Exploratory Exercise 4 and Exercise 7 will help them to understand the power that these algorithms
provide when applied to large weighted graphs.
Related Website: Visit this web site for description of many of the algorithms that are critical in
graph theory including many that are discussed in this text: http://www.csee.usf.edu/~maurer/graphs/
Exploratory Exercise Set 4.4
1. (a) Vertices represent subway stations. Edges represent a subway line running between two
stations, and weights represent the travel time between stations.
(b) Vertices and edges as above, with weights representing distance between stations.
(c) Vertices and edges as above, with weights representing the fare to ride between two stations.
2. (a) The next vertex circled is B, and the first edge shaded is {A, B}, because the weight of edge
{A , B} is less than the weight of {A, D}.
Graph Theory
69
(b) (1)
3
B
A
0
(2)

C
2
3
A
0
3
5
D
4
E
3
4
0
3
(a)
Z
5
2
C
3
4
Z
2
8
E
5
9
4
4.
2
E

E
2
D
5
D
4
Z

3
5
B
A
4
2
3
(b)
3
C
2
D
2
3

3. (a) (1) (A, B, C, Z) and (A, D, E)
(2)
B
2
A
2
5
C
2
Z

2
4
3
B
(1) and (2)
Path
(A, B, Z)
(A, B, C, A, G, Z )
(A, B, C, D, F, Z)
(A, B, C, D, F, E,
C, A, G, Z)
(A, B, C, E, F, Z)
(A, B, C, E, F, D,
C, A, G, Z)
(A, C, B, A, G, Z)
Length
130
280
120
390
Path
(A, C, B, Z)
(A, C, D, F, Z)
(A, C, D, F, E, C, B, Z)
(A, C, D, F, E, C, B, A,
G, Z)
(A, C, E, F, Z)
(A, C, E, F, D, C, B, Z)
130
360
280
(A, C, E, F, D, C, B, A,
G, Z)
(A, G, Z)
(3) Path (A, B, C, D, F, Z) is the shortest with length 120.
B
(b) Step 1
30
100
30
40
A
D
Z
10
90
C
F 10
50
10
E
70
60
G
Length
220
150
330
360
160
330
390
130
70
Chapter 4
Step 2
B
100
30
30
40 D
A
Z
10
90
C
F 10
50
10
E
70
60
G
Step 3 (First Iteration)
B
100
30
30
D
40
A
Z
10
90
C
F 10
50
10
E
70
60
G
B
Step 3 (Second Iteration)
100
30
30
40
A
D
Z
10
90
C
F 10
50
10
E
70
60
G
Step 3 (Third Iteration)
B
30
100
30
A
D
40
90
Z
10
C
F 10
50
10
E
70
60
G
Graph Theory
71
Step 3 (Fourth Iteration)
B
30
100
30
40
A
D
Z
10
90
F 10
C
50
10
E
70
60
Step 3 (Fifth Iteration)
G
B
30
100
30
40
A
D
Z
10
90
C
F 10
50
10
E
70
60
G
Thus the shortest path is (A, B, C, D, F, Z) and has length 120.
72
Chapter 4
(c)
Initial Labels
Vertex
L(u)
P(u)
A
0
B
C
D
E
F
G
Z
30
90
∞
∞
∞
60
∞
First Iteration
Vertex
L(u)
P(u)
A
0
A
A
B
C
D
E
F
G
Z
A
Third Iteration
Vertex
L(u)
P(u)
A
0
B
C
D
E
F
G
Z
30
90
100
110
∞
60
130
30
60
∞
∞
∞
60
130
Second Iteration
Vertex
L(u)
P(u)
A
0
A
B
B
C
D
E
F
G
Z
A
B
Fourth Iteration
Vertex
L(u)
P(u)
A
0
A
B
C
C
B
C
D
E
F
G
Z
A
B
Sixth Iteration
Vertex
L(u)
P(u)
A
0
30
90
100
110
110
60
130
B
C
D
E
F
G
Z
A
B
C
C
D
A
B
30
90
100
110
110
60
120
B
C
D
E
F
G
Z
30
90
100
110
110
60
120
3
a
3
A
B
C
C
D
A
F
1
3
d
b
3
1
1
c
5
e
3
z
a
1
3
d
3
3
1
1
c
A
B
A
B
C
C
D
A
F
Thus the shortest path from
A to Z is (A, B, C, D, F, Z),
with length 120.
Exercise Set 4.4
1.
b
A
B
C
C
Fifth Iteration
Vertex
L(u)
P(u)
A
0
Seventh Iteration
Vertex
L(u)
P(u)
A
0
B
30
A
C
90
B
D
100
C
E
110
C
F
110
D
G
60
A
Z
120
F
(d) Answers will vary.
5.
Answers will vary.
6.
Answers will vary.
30
60
100
110
∞
60
130
5
e
z
Graph Theory
73
b
3
a
3
1
1
3
z
1
b
3
b
3
3
c
a
d
5
1
3
3
1
3
3
z
1
c
d
d
1
e
3
1
a
3
5
e
Thus a shortest path is (a, c, e, z) with length 7.
z
1
c
5
e
e
2.
a
7
e
2
3
a
7
2
c
3
c
2
5
z
2
5
6
z
6
9
b
9
3
b
3
d
d
e
a
7
e
2
3
a
7
2
c
3
c
2
5
z
2
5
6
z
6
9
b
9
3
b
3
d
d
e
a
7
e
2
3
a
7
2
c
3
c
2
5
z
2
5
6
z
6
9
b
9
3
b
3
d
d
Thus the shortest path is (a, b, d, c, e, z) and has weight 11.
3.
2
∞
3
0
3
3
b
1
a
d
3
3
z
1
1
c
1
∞
5
e
∞
0
3
a
1
b
3
1
3
4
d
3
z
1
1
c
1
∞
5
e
6
74
Chapter 4
2
2
4
b
3
3
0
1
a
5
1 2
0
z
a
1
3
c
1
1
3
7
z
1
5
1
4
3
1
a
1
c
6
3
0
e
5
2
b
3
d
c
6
1
e
5
6
Thus a shortest path is (a, c, d, z) with length 7.
4.
0
a
e
a
7
2
3
2
5
6
z
b
0
∞7
a
e
7
2
d
5
14
z
5
6
9
3
2
3
2
z
7
d
0
8
a
e
7
2
3
14
2
5
z
1
c
9
6
6
1
c
9
6
b
3
2
0
d
b
5
2
3
0
e
7
2
d
5
7
a
8
a
3
e
7
2
3
11
2
5
6
c
b
2
3
10
2
z
1
7
6
c
9
6
d
∞
1
c
9
3
3
2
6
7
b
2
2
2
1
c
e
7
∞
9
6
b
2
3
5
Thus the shortest path is (a, b, d, c, e, z) of length 10.
z
1
d
5
7
z
1
e
∞
3
3
1
0
7
1
6
d
3
z
1
e
3
d
3
4
b
a
∞
3
3
1
c
0
3
3
1
4
b
d
Graph Theory
75
5. Length 530
∞
∞
200
300
100
200
140
30
80
0
60
120
∞
Omega
350
Alpha
∞
50
300
150
450
300
∞
∞
∞
200
300
100
200
140
30
80
0
60
120
650
Omega
350
Alpha
∞
50
300
150
450
300
750
330
∞
200
300
100
200
140
30
80
0
60
120
380
Omega
350
Alpha
50
300
∞
150
450
300
750
330
200
470
300
30
100
200
80
0
140
380
60
120
Omega
350
Alpha
50
300
150
300
450
750
∞
76
Chapter 4
330
200
470
300
100
200
140
30
120
380
80
0
60
Omega
350
Alpha
50
300
∞
150
450
300
530
330
200
470
300
100
200
140
30
120
380
80
0
60
Omega
350
Alpha
50
300
530
150
450
300
530
330
200
470
300
100
200
140
30
120
380
80
0
60
Omega
350
Alpha
50
300
150
450
300
6. Length 3:30
530
530
2:20
1:10
Beta
2:00
1:00
:40
:50
1:00
3:00
1:30
Gamma
1:10
Graph Theory
7.
77
(a) and (b)
Path
(A, B, Z)
(A, B, C, E, Z)
(A, B, C, D, E, Z)
(A, D, E, Z)
(A, D, C, E, Z)
(A, D, C, B, Z)
(A, D, E, C, B, Z)
Length
7
8
11
7
6
9
14
(c) A shortest path is (A, D, C, E, Z) and has length 6.
8.
A path of shortest length is (a, c, d, e, g, z) and has length 16.
∞
4
d
b
6
∞
4
f
b
∞
d
f
0
∞
0
∞
a
z
a
z
c
e
g
c
e
g
3
∞
∞
3
9
∞
6
4
b
d
∞
4
f
b
6
d
11
f
0
∞
0
∞
a
z
a
z
c
e
g
c
e
g
3
9
∞
3
7
∞
6
4
b
d
11
4
f
b
6
d
11
f
0
∞
0
18
a
z
a
z
c
e
g
c
e
g
3
7
12
3
7
12
6
4
b
d
6
11
4
f
b
0
16
0
a
z
a
d
11
f
16
z
c
e
g
c
e
g
3
7
12
3
7
12
78
Chapter 4
9.
Third Iteration
Fourth Iteration
Vertex
L(u)
P(u)
Vertex
L(u)
P(u)
A
0
A
0
B
5
C
B
4
C
C
2
A
C
2
A
D
6
C
D
6
C
E
8
C
E
8
C
F
10
D
F
10
D
A shortest path is (A, C, D, F) and has length 10.
10. A shortest path is (a, c, d, z) of length 7.
Fifth Iteration
Vertex
L(u)
A
0
B
4
C
2
D
6
E
8
F
10
Initial Labels
Vertex
L(u)
P(u)
a
0
b
4
a
c
1
a
d
∞
e
∞
z
∞
Third Iteration
Vertex
L(u)
P(u)
a
0
b
2
c
c
1
a
d
4
c
e
6
c
z
7
d
Second Iteration
Vertex
L(u)
P(u)
a
0
b
2
c
c
1
a
d
4
c
e
6
c
z
∞
Fifth Iteration
Vertex
L(u)
P(u)
a
0
b
2
c
c
1
a
d
4
c
e
6
c
z
7
d
α
11.
β
120
250
50
First Iteration
Vertex
L(u)
P(u)
a
0
b
2
c
c
1
a
d
4
c
e
6
c
z
∞
Fourth Iteration
Vertex
L(u)
P(u)
a
0
b
2
c
c
1
a
d
4
c
e
6
c
z
7
d
ω
240
100
γ
90
270
150
σ
130
κ
290
P(u)
C
A
C
C
D
Graph Theory
79












-120
250
200
210
50
120
-320
220
90
170
250
320
-100
230
250
200
220
100
-130
150
210
90
230
130
-260
50
170
250
150
260
--
12. (a) Find the shortest path from A to Q, then the shortest path from Q to Z.
(b) Path (A, C, E, Q, G, Z) of length 11.
13. (a)
(b)
Vertex
Vertex
Length
From a
0
2
6
5
7
10
a
b
c
d
e
z
a
b
c
d
e
f
g
z
Length
From a
0
4
3
6
7
11
12
16
Exploratory Exercise Set 4.5
1.
We can connect house to utilities in each case as illustrated below:
(a)
H
H
1
2
Gas
Water
(b)
(c)
H1
Gas
H2
Water
H3
H1
Gas
H2
Water
H3
H4
80
Chapter 4
(d)
H4
H1
Gas
H2
H5
H3
Water
Gas
(e)
Houses
Water
2.
(a) Yes, as seen in this drawing.
H1
Gas
3.
(b)
(c)
(d)
(a)
H2
Water
Electric
No, K3,3 is nonplanar. See Example 4.
No, after experiments with several drawings we see that K3,3 is a subgraph of K3,4.
No, as K3,3 will always be a subgraph of K3,n
E=9
(b) R = 3
(c) V = 4
(d) R = 2
Graph Theory
81
Exercise Set 4.5
1.
Graph
G1
G2
G3
G4
2.
a
Vertices (v)
4
4
4
4
Edges (e)
5
3
8
6
3.
c
b
a
c
4.
c
e
b
5.
f
d
e
b
d
d
a
v–e+r
2
2
2
2
Regions (r)
3
1
6
4
c
e
b
a
6.
7.
10.
11.
12.
(a)
(c)
6
(a)
(a)
(a)
(f)
d
e = 6, v = 4; r = 4; 4 – 6 + 4 = 2
(b) e = 6, v = 5; r = 3; 5 – 6 + 3 = 2
e = 6, v = 5; r = 3; 5 – 6 + 3 = 2
(d) e = 7, v = 6; r = 3; 6 – 7 + 3 = 2
8. 8
9. 12
(b, c, e, b)
(b) (a, d, f, e, b, a)
(c) (a, b, c, d, f, e, c, a)
(a, b, c, a)
(b) 3
(c) (a, b, c, d, a)
(d) 4
e = 10
(b) v = 5
(c) r = e – v + 2 = 7
(d) 21
(e) 20
N cannot be greater than or equal to 21 and less than or equal to 20. Thus there is no planar
representation of K5
13.
14.
Polyhedron
P1
P2
P3
P4
(a) Case 1:
Case 2:
15.
Vertices (v)
Edges (e)
8
12
4
6
12
18
9
14
(b) Case 1: v = 2, e = 1, r = 1
v–e+r=2
Case 2: v = 1, e = 1, r = 2
v–e+r=2
Faces (f)
6
4
8
7
v–e+f
2
2
2
2
Edge {a, f} was removed, and edges {a, x} and {x, f} were added. Edge {c, d} was removed,
and edges {c, y} and {y, d} were added.
16. Edge {t, b} was removed, and edges {t, x} and {x, b} were added. Edge {b, d} was removed,
and {b, y} and {y, d} were added.
82
Chapter 4
17.
18.
19. Each graph is formed by elementary subdivisions on K3,3 or K5, or it contains a subgraph
isomorphic to one of K3,3 or K5.
20. When n ≥ 5, the graph Kn will contain a subgraph isomorphic to K5.
Teaching Notes, Section 4.6:
Related Website: Visit this web site for a discussion of a new proof of the four color theorem that is
less computationally intensive than the original Appel/Haken proof. On a more elementary level, a
thorough discussion of the history of the four color problem is found at this site:
http://www.math.gatech.edu/~thomas/FC/fourcolor.html
Exploratory Exercise Set 4.6
1.
(a) The minimum number of aquaria that can be used is three.
A
Red
B
C
Blue
Green
E
F
G
Red
Green
D
Red
Green
(b) The tasks can be done with four teachers.
C
H
Blue
F
Green
Red
L
Blue
B
Red
R
M
Green
Yellow
Graph Theory
2.
3.
4.
83
Answers will vary.
(a) (1) 2
(2) 2
(3) 2
(4) The chromatic number of a cycle with an even number of vertices is 2.
(b) (1) 3
(2) 3
(3) 3
(4) The chromatic number of a cycle with an odd number of vertices is 3.
(5)The chromatic number of a simple graph containing a cycle with an odd number of
vertices is greater than or equal to 3.
(a)
C = {a, b, j, c, i, d, h, f, e}
Blue
b
a
Red
c
Red
j
Blue
C = {a, b, j, c, i, d, h, f, e, g}
i
Red
h
Blue
f
Blue
j
Blue
i
Red
b
h
Blue
c
d
f
e
g
i
h
Step 2, C = {a, b, g}
Blue
b
c
d
f
Blue
g
e
h
g
Red
d
c
Red
(b) Initially, C = { }
Step 1, C = {a}
Red
a
Blue
e
Blue
b
a
Red
Red
a
Red
d
i
Blue
e
f
Blue
g
Red
84
Chapter 4
Step 3, C = {a, b, g, c, h}
Red
a
Blue
b
Red
c
d
f
Blue
g
e
i
h
Red
Step 4, C = {a, b, g, c, h, d, e, i}
Blue
b
Red
a
Red
c
e
Blue
Blue
g
Blue
d
f
i
Blue
h
Red
Step 5, C = {a, b, g, c, h, d, e, i, f}
Blue
b
Red
a
e
Blue
Blue
g
h
Red
5.
Red
c
Blue
d
f Red
i
Blue
(c) (1) If u and v were adjacent, then the graph would contain a cycle with an odd number of
edges.
(2) If w and u were adjacent, then there would be a cycle with an odd number of edges.
(a)
G
G´
(1) 4
4
(2) 4
3
(3) 4 is less than 5
3 is less than 5
(b) (1) The vertex has degree 0, thus D + 1 = 1. The vertex only needs one color since it is not
adjacent to any other vertices.
(2) The graph could be connected, in which case D = 1, and D + 1 = 2. Since the
vertices are adjacent, they must have different colors, hence 2 colors.
Or, the vertices may not be adjacent, in which case, D + 1 = 1, and each vertex can be
the same color.
Graph Theory
85
(3) None of the vertices may be adjacent, D + 1 = 1, and each vertex can be the same
color. The chromatic number is 1.
In this case, with two vertices adjacent, D = 1, and D + 1 = 2. The two vertices that are
adjacent must have different colors, and the third vertex can be either color. The
chromatic number is 2.
In the case where the graph is K2,1 D = 2, and D + 1 = 3. The chromatic number is 2.
In the case where the graph is K3, D = 2, and D + 1 = 3. This graph consists of a cycle
with an odd number of edges, and requires 3 colors.
(4)





G´ has k vertices.
The maximum of the degree of vertices in G´ is less than or equal to D.
By the inductive hypothesis, we know that G´ can be colored with D + 1 colors.
D since D is the maximum of the degrees of vertices of G.
Since V is adjacent to at most D vertices in G´, we can certainly color V using
one of D + 1 vertices.
Answers will vary.
6.
Exercise Set 4.6
1.
R
B
2.
R
R
G
B
R
Y
3.
4.
R
Y
Y
R
Y
R
G
B
B
G
5.
6.
R
R
G
B
Y
B
B
Y
G
G
B
86
7.
8.
9.
Chapter 4
(1) 3
(4) 4
(a)
(a)
(2) 2
(3) 3
(5) 2
(6) 3
(b) All graphs in which the degree of all vertices is 0.
R
G
B
B
R
G
(b)
R
G
B
R
B
(c)
G
10. (a)
R
G
B
G
B
R
(b)
(c)
R
B
G
B
G
R
11. (a)
(b)
(c)
R
R
G
B
G
R
B
G
R
B
B
G
Graph Theory
87
12. (a)
R
B
R
B
G
G
B
R
R
Y
R
G
B
G
R
B
G
(b)
(c)
R
B
R
G
G
B
B
R
Y
R
R
G
B
R
G
B
G
13. Three displays are necessary.
cardinals
horned owls
R
rabbits
blue jays
B
G
foxes
R
R
squirrels
B
bobcats
B
C1
14. Five meeting times are necessary.
R
B
C2
C6
W
C3
C5
G
Y
C4 Y
88
Chapter 4
15. Three freight cars are necessary.
Blue
A
Green
F
Green
E
Red
B
Blue
G
Blue
D
Green
C
16. Four different channels are necessary.
7
1
6
2
5
3
4
Green
Blue
Red
White
Red
Blue
17.
18.
(a)
(b)
(a)
(b)
(c)
(d)
19.
(a)
(b)
(c)
Red
3
(b) 5
(c) 6
K5 and K6 are not planar graphs.
If either V1 or V2 were empty, the chromatic number would be 1.
If there were any edges incident with pairs of vertices in V1, then the pair of vertices could
not both be colored Red.
If there were any edges incident with pairs of vertices in V2, then the pair of vertices could
not both be colored Blue.
V1 and V2 are disjoint, non-empty sets so that each edge is incident with a vertex in V1 and a
vertex in V2.
Because no pair of vertices in V1 is adjacent nor is any pair of vertices in V2 adjacent.
Since the graph is connected and has more than one vertex, each vertex of the graph must
be adjacent to at least one other vertex. Thus at least two colors are required to color the
graph.
In (a) we showed that two colors will color the graph. In (b) we showed that one color is
not sufficient. Thus the chromatic number is 2.