Chapter 12

The pairs that are not adjacent are p and s, and r and t.

Vertex q

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.

Graph E, 34

Graph D, 18

higher

124,750 introductions

Triangle: (a, b, e or a, d, e)
Quadrilateral: (b, c, d, e or a, b, e, d)
Pentagon: (a, b, c, d, e)

(B, D, C, E, F)

286

Graph C₁ has 6 edges, but Graph C₂ has 8.
Graph C₁ has 4 vertices and Graph C₂ has 5.
Graph C₁ has no vertex of degree 4, but Graph C₂ has one vertex of degree 4.
They do not have the same cycles. For example, Graph C₂ has a pentagon cycle, but Graph C₁ does not.

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.

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.

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.

The degrees are 0. There are no adjacent vertices in Graph N since all vertices are adjacent in Graph M.

b Only V → C → B → G → C is a walk.

walk, path, and trail

walk and a trail, but not a path

none of these

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.

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

Yes. The chromatic number is 4 or less.

3

3 or 4

Graphs 1 and 3

Answers may vary. Here is an example of a 3-coloring:

The chromatic number is 3. Here is a 3-coloring of the graph:

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}.

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

The multigraph of the neighborhood is shown:

Yes, the graph is connected. It has only one component.

There are four corner vertices of degree 2 and the remaining twelve vertices are all degree 4.

Yes

Yes, an Eulerian graph has an Euler circuit; so, it is possible.

Graph X does not have an Euler circuit because it is disconnected.

a → b → g → d → f → c → g → e → a

Yes.

No. It doesn’t cover every edge.

No. It is not a walk, because the vertices b and e are not adjacent.

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.

m

qn, qt; qn bridge

rs, st, tq, qn

nm, np

om, mp, pn, nm

s → q → r → s → t → q → n → o → m → p → n → m

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

both

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

$5!=5\cdot <!--\; \cdot \; -->4\cdot <!--\; \cdot \; -->3\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->1=120$

120

26

No

No

Yes

C → A → B → F → D → E

p → m → o → n → q → t → s → r

none

none

Hamilton path

Euler trail

neither

brute force algorithm

1. V → W → X → Y → Z → V
2. V → W → X → Z → Y → V
3. V → W → Y → X → Z → V
4. V → W → Y → Z → X → V
5. V → W → Z → X → Y → V
6. V → W → Z → Y → X → V
7. V → X → W → Y → Z → V
8. V → X → W → Z → Y → V
9. V → X → Y → W → Z → V
10. V → X → Y → Z → W → V (reverse of 6)
11. V → X → Z → W → Y → V
12. V → X → Z → Y → W → V (reverse of 4)
13. V → Y → X → W → Z → V
14. V → Y → X → Z → W → V (reverse of 5)
15. V → Y → W → X → Z → V
16. V → Y → W → Z → X → V (reverse of 11)
17. V → Y → Z → X → W → V (reverse of 2)
18. V → Y → Z → W → X → V (reverse of 8)
19. V → Z → X → Y → W → V (reverse of 3)
20. V → Z → X → W → Y → V (reverse of 15)
21. V → Z → Y → X → W → V (reverse of 1)
22. V → Z → Y → W → X → V (reverse of 7)
23. V → Z → W → X → Y → V (reverse of 13)
24. V → Z → W → Y → X → V (reverse of 9)

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.

D → B → E → A → C → F → D; 550 min or 9 hrs 10 min.

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.

star topology

tree topology

none

Vertices k, l, and m are in one component, and vertices g, h, i, and j are in the other.

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.

a quadrilateral

a True

b False

a True

a True

Answers may vary. Here are three possible spanning trees of Graph J:

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

There are two possible minimum spanning trees, each with a total weight of 174:

a True

b False

a True

b False

b False

b False

a True

b False

a True

b False

a True

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.

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.

always true

never true

sometimes true

never true

always true

sometimes true

sometimes true

never true

sometimes true

always true

always true

sometimes true

always true

sometimes true

always true

always true

sometimes true

always true

sometimes true

always true

always true

has no repeated egdes

is closed

has no repeated vertices

has no repeated edges

$n$

at least

never true

sometimes true

never true

sometimes true

always true

always true

sometimes true

never true

always true

sometimes true

edge

Fleury’s

circuit, trail

components

local bridge

bridge

vertex

complete

is

is

is not

is

is

is not

is not

is not

different from

the same as

different from

different from

b False

b False

b False

a True

a True

b False

b False

b False

a True

b False

b False

a True

a True

b False

b False

a True

a True

a True

a True

b False

b False

a True

a True

a True