## Your Turn

12.1

1.

On a graph the objects are represented with dots and their connects are represented with lines. The dots are the vertices. The vertices are

*p*,*q*,*r*,*s*, and*t*. The line segments joining the vertices are called edges. The edges are*pq*,*pr*,*pt*,*qr*,*qs*,*qt*,*rs*, and*st*.
The vertices are

*p*,*q*,*r*,*s*, and*t*. The edges are*pq*,*pr*,*pt*,*qr*,*qs*,*qt*,*rs*, and*st*.
12.2

12.3

12.4

12.5

1.

The objects that are represented with vertices are poker players. An edge between a pair of vertices would indicate the two players competed against each other at table. This is best represented as a graph because players from the same table never meet a second time. Also, a player cannot compete against themself; so, there is no need for a loop.

12.7

1.

The number of edges is half the number of the sum of degrees. If the sum of degrees is 6, then there are 3 edges.

3

Answers may vary. Two possible graphs are shown.

Graph 1:

Graph 2:

Graph 1:

Graph 2:

12.8

12.9

12.10

1.

Triangle: (

Quadrilateral: (

Pentagon: (

*a*,*b*,*e*or*a*,*d*,*e*)Quadrilateral: (

*b*,*c*,*d*,*e*or*a*,*b*,*e*,*d*)Pentagon: (

*a*,*b*,*c*,*d*,*e*)
12.11

12.12

12.13

1.

Graph

Graph

Graph

They do not have the same cycles. For example, Graph

*C*_{1}has 6 edges, but Graph*C*_{2}has 8.Graph

*C*_{1}has 4 vertices and Graph*C*_{2}has 5.Graph

*C*_{1}has no vertex of degree 4, but Graph*C*_{2}has one vertex of degree 4.They do not have the same cycles. For example, Graph

*C*_{2}has a pentagon cycle, but Graph*C*_{1}does not.
12.14

1.

Answers may vary.

- The total number of vertices in each graph is different. The Diamonds graph has 17 vertices while Dots graph has only 16.
- The degrees of vertices differ. The Diamonds graph has vertices of degrees 4 while the Dots graph does not.
- The graphs have different sizes of cyclic subgraphs. The Diamonds graph has 4 squares (4-cycles), while the Dots graph has 3 squares. Also, the Dots graph has 8-cycles while the Diamonds graph does not.

12.15

1.

Answers may vary. There are four possible isomorphisms:

*a − q*,*d − s*,*c − p*, and*b − r*.*a − p*,*d − s*,*c − q*, and*b − r*.*a − q*,*d − r*,*c − p*, and*b − s*.*a − p*,*d − r*,*c − q*, and*b − s*.

12.16

1.

Caden is correct because Graph

*E*could be untangled so that*d*is to the left of*c*and the edges*bc*and*ad*are no longer crossed. Maubi is incorrect because (*a*,*b*,*c*,*d*) is a quadrilateral in Graph*E*. Javier is incorrect because the portion of the graph he highlighted does not have 3 vertices and is not a triangle.
12.17

12.18

1.

The degrees are 0. There are no adjacent vertices in Graph

*N*since all vertices are adjacent in Graph*M*.
12.19

12.21

1.

To fly from Palm Beach to any other city, you must take the flight from PBI to TPA. To return, you must take the flight from TPA to PBI. This means that the graph of any itinerary that begins and ends at PBI covers the same edge twice and is not a circuit.

2.

The degree of PBI is 1 meaning that there is only one edge connecting PBI to other parts of the graph.

12.22

12.23

12.24

1.

Graphs

*G*,*H*, and*I*are connected; so, they each have a single component with the vertices {*a*,*b*,*c*,*d*}. Graph*F*is disconnected with two components, {*a*,*c*,*d*} and {*b*}. Graph*J*is disconnected with two components, {*a*,*d*} and {*b*,*c*}. Graph*K*is disconnected with three components, {*a*}, {*b*,*d*}, and {*c*}.
12.25

1.

Each pair of vertices would be connected by a path that was at most 6 edges long. So, the graph would be connected.

12.26

12.27

12.28

12.29

12.30

1.

In a complete graph with three or more vertices, any three vertices form a triangle which is a cycle. Any edge in a graph that is part of a cycle cannot be a bridge so, there are no bridges in a complete graph. Also, any edge that is part of a triangle cannot be a local bridge because the removal of any edge of a triangle will leave a path of only two edges between the vertices of that edge. So, there are no bridges or local bridges in a complete graph.

12.31

12.32

1.

The graph has Euler trails. They must begin and end at vertices

*u*and*v*. An example is*v*→*w*→*x*→*u*→*z*→*y*→*w*→*u*
12.33

12.34

1.

$n!=6!=6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=720$ and $\left(n-1\right)!=(6-1)!=5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$

12.35

12.36

12.37

12.39

12.42

12.43

1.

*V*→*W*→*X*→*Y*→*Z*→*V**V*→*W*→*X*→*Z*→*Y*→*V**V*→*W*→*Y*→*X*→*Z*→*V**V*→*W*→*Y*→*Z*→*X*→*V**V*→*W*→*Z*→*X*→*Y*→*V**V*→*W*→*Z*→*Y*→*X*→*V**V*→*X*→*W*→*Y*→*Z*→*V**V*→*X*→*W*→*Z*→*Y*→*V**V*→*X*→*Y*→*W*→*Z*→*V**V*→*X*→*Y*→*Z*→*W*→*V*(reverse of 6)*V*→*X*→*Z*→*W*→*Y*→*V**V*→*X*→*Z*→*Y*→*W*→*V*(reverse of 4)*V*→*Y*→*X*→*W*→*Z*→*V**V*→*Y*→*X*→*Z*→*W*→*V*(reverse of 5)*V*→*Y*→*W*→*X*→*Z*→*V**V*→*Y*→*W*→*Z*→*X*→*V*(reverse of 11)*V*→*Y*→*Z*→*X*→*W*→*V*(reverse of 2)*V*→*Y*→*Z*→*W*→*X*→*V*(reverse of 8)*V*→*Z*→*X*→*Y*→*W*→*V*(reverse of 3)*V*→*Z*→*X*→*W*→*Y*→*V*(reverse of 15)*V*→*Z*→*Y*→*X*→*W*→*V*(reverse of 1)*V*→*Z*→*Y*→*W*→*X*→*V*(reverse of 7)*V*→*Z*→*W*→*X*→*Y*→*V*(reverse of 13)*V*→*Z*→*W*→*Y*→*X*→*V*(reverse of 9)

12.44

1.

The officer should travel from Travis Air Force Base to Beal Air Force Base, to Edwards Air Force Base, to Los Angeles Air Force Base, and return to Travis.

12.45

12.46

1.

The star topology is a tree, and the tree topology is a tree. The ring topology and the mesh topology are not trees because they contain cycles.

12.48

2.

There are 6 vertices and 5 edges, which confirms Graph

*I*is a tree, because the number of edges is one less than the number of vertices.
12.50

12.51

1.

The three edges must include one edge from list A and one pair of edges from list B.

List A:

List B:

List A:

*be*,*eh*,*hi*,*gi*,*bg*List B:

*ac*and*ad*,*ac*and*af*,*ac*and*cd*,*ac*and*cf*,*ad*and*af*,*ad*and*cd*,*ad*and*cf*,*af*and*cd*,*af*and*cf*, or*cd*and*cf*.
12.52

## Check Your Understanding

12.

The sum of the degrees is always double the number of edges, which means it has to be an even number, but 13 is odd.

13.

Yes. If

*n*is the number of vertices in a complete graph, the number of edges is $\frac{n(n-1)}{2}=\frac{n}{2}(n-1)$ which is half the number of vertices times one less than the number of vertices.