Networks - Cambridge University Press€¦ · The graphs are connected because, starting at any vertex, say A,you can always ﬁnd a path ... TI-Nspire & Casio ClassPad material prepared

P1: FXS/ABE P2: FXS0521672600Xc10.xml CUAU034-EVANS March 10, 1904 23:21

C H A P T E R

10Networks

How are graphs used to represent networks?

How do we analyse the information contained in graphs?

How do we use graphs to represent everyday situations?

10.1 Graph theory basicsThe Bridges of Konigsberg problemThe problem that began the scientific study of networks is known as the Konigsberg Bridges

problem. The problem began as follows:

The centre of the old city of Konigsberg in Germany was on an island in the middle of the

Pregel River. The island was connected to the banks of the river and to another island by five

bridges. Two other bridges connected the second island to the banks of the river, as shown

below.

A view of Konigsberg as it was in Euler’s day.

407Cambridge University Press • Uncorrected Sample Pages • 978-0-521-74049-4 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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408 Essential Standard General Mathematics

A simplified view of the situation is shown in the

diagram opposite.

The question was: Can a continuous walk be planned so that all bridges are crossed only once?

Whenever someone tried it, they either ended up missing a bridge or crossing one of the

bridges more than once, as in the two routes shown below.

Enter the mathematicianThe Konigsberg Bridges problem was a well-known problem in 18th century Europe and

attracted the attention of the Swiss mathematician Euler (pronounced ‘Oil-er’). He started

analysing the problem by drawing a simplified diagram to represent the situation, as shown

below on the right.

A

B

C

D

A

B

C

DEuler's diagram

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Chapter 10 — Networks 409

This diagram is now what we call a network or a graph.

The dots are called vertices (plural of vertex).

The lines are called edges.

A

B

C

D

Edge

Vertex

Euler found that the answer to the question depended

on the degree of each vertex. The degree of a vertex

is given by the number of edges attached to the

vertex. The number (degree) may be odd or even.

For Euler’s graph:A

B

C

D

deg(B) = 3

deg(A) = 5

1

1

2

2 3

3

45vertex A is of degree 5: we write deg(A) = 5;

vertices B, C and D are all of degree 3:

we write, for example, deg(B) = 3.

All four vertices are of an odd degree.

Euler was able to prove that a graph with all odd vertices cannot be traced or drawn without

lifting the pencil or going over the same edge more than once. The problem was solved. The

seven bridges of Konigsberg could not be crossed in a single walk without either missing a

bridge or crossing one bridge more than once.

With this analysis, a new area of mathematics was developed, which has many practical

applications in today’s world. These include: analysing friendship networks, scheduling airline

flights, designing electrical circuits, planning large-scale building projects, and many more.

This relatively new area of mathematics is now called graph theory, some aspects of which

we will explore in this chapter.

Exercise 10A

1 a With a pencil, or the tip of your finger,

see whether you can trace out a

continuous walk that crosses each

of the bridges only once.


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b An eighth bridge has been added to the

diagram as shown.

With a pencil, or the tip of your finger,

see whether you can trace out a

continuous walk that crosses each of

the bridges only once. Such a walk exists.

2 For each of the graphs shown, complete the associated statements by filling in the boxes.

A B

D C

a i The graph has vertices.

ii The graph has edges.

iii deg(A) = iv deg(C ) =v The graph has odd vertices.

A B

D C

b i The graph has vertices.


iii deg(A) = iv deg(C ) =v The graph has odd vertices.

c A B

D C

E

i The graph has vertices.


iii deg(B) = iv deg(D ) =v The graph has odd vertices.

3 The game of Sprouts is played between two people and involves drawing a network.

Rules for the game of SproutsTwo or more points are drawn on a piece of paper. These are network vertices. Players then

take turns adding edges according to the following rules:

1 Each edge must join two vertices or itself.

2 Every time a new edge is drawn, a new vertex must be added somewhere on the edge.

3 Edges cannot cross nor pass through a vertex.

4 No vertex may have a degree greater than 3.

5 The last player able to add a new edge wins.

For more information see: http://www.madras.fife.sch.uk/maths/games/sprouts.html

A sample game of Sprouts is played on the following page.


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Sample game of SproutsThe starting vertices

The starting vertices Step 1: Player 1 Step 2: Player 2

Step 3: Player 1 Step 4: Player 2 Step 5: Player 1 wins

Player 1 wins because Player 2 cannot draw in a new edge without creating a vertex of

degree greater than 3.

Will a game starting with two points always end in five steps? Can you find a way of ending the

game more quickly? Does the first player always win? What happens if you start with three,

four or more points? Explore.

10.2 Isomorphic and connected graphsIsomorphic graphsOne of the things that you will find when working with networks is that quite different-looking

graphs actually contain the same information. When this happens, we say that these graphs are

equivalent or isomorphic.

For example, the three graphs below appear to be different, but, in graphical terms they are

isomorphic.

A B

D C

Graph 1

A

BC D

Graph 2

A

B

C

D

Graph 3

This is because they contain the same information. Each graph has the same number of edges

(5) and vertices (4), corresponding vertices have the same degree and the edges join up the

vertices in the same way (A to C, A to D, B to C, B to D, and D to C ).


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However, the three graphs below, although having the same numbers of edges and vertices,

are not isomorphic. This is because corresponding vertices do not have the same degree and

the edges do not connect the same vertices.

A B

D C

Graph 1

A

BC D

Graph 2

A

B

C

D

Graph 3

Isomorphic graphsTwo graphs are said to be isomorphic (equivalent) if:

they have the same numbers of edges and vertices

corresponding vertices have the same degree and the edges connect the same vertices.

Connected graphsSo far, all the graphs we have met have been connected. That is, every vertex in the graph can

be reached from every other vertex in the graph.

For example, the three graphs shown below are all connected.

A B

D C

Graph 1

A BC D

Graph 2

A

B

C

D

Graph 3

The graphs are connected because, starting at any vertex, say A, you can always find a path

along the edges of the graph to take you to every other vertex. For example:

In Graph 1 you can get from vertex A to vertex B by travelling along edge AB. A similar

statement can be made about all the other vertices.

In Graph 2 you can get from vertex A to vertex B by travelling along edge AC to vertex C,

then edge CD to vertex D and finally edge DB to vertex B. All other vertices are accessible

from vertex A in a similar manner.

In Graph 3 you can get from vertex A to vertex B by travelling along edge AC to vertex C,

then edge CB to vertex B. All other vertices are directly accessible from vertex A.

However, the three graphs below are not connected, because there is not a path along the

edges that connects vertex A (for example) to every other vertex in the graph.

A B

D C

Graph 1

A B

D

C

Graph 2

A

B

C

D

Graph 3


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Connected graphs have application in

a range of problems such as planning

airline routes, communication systems

and computer networks, where a

missing connection can lead to an

inoperable system.

Connected graphsA graph is connected if every vertex in the graph is accessible from every other vertex in

the graph along a path formed by the edges of the graph.

Exercise 10B

1 In each of the following sets of three graphs, two of the graphs are isomorphic (equivalent).

In each case, identify the isomorphic graphs.

aA

B

C

Graph 1

C

A

B

Graph 2

C

A

B

Graph 3

b

C

A

B

Graph 1

C

A

B

Graph 2

C

A

B

Graph 3

c A

B

C

Graph 1

C

A

B

Graph 2

C

A

B

Graph 3


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d A

B

C

D

Graph 1

A

B

CD

Graph 2

A

B

C

D

Graph 3

e A

B

C

D

Graph 1

A

B

CD

Graph 2

A

B

CD

Graph 3

2 Which of the following graphs are connected?

AA

B

C

D

BA

B

CD

CA

B

C

DF

D

C

A

B

EA

B

C

D

F

C

A

B

3 Draw a connected graph with:

a three vertices and three edges b three vertices and five edges

c four vertices and six edges d five vertices and five edges.

4 Draw a graph that is not connected with:

a three vertices and two edges b four vertices and three edges

c four vertices and four edges d five vertices and three edges.

5 What is the smallest number of edges that can form a connected graph with four vertices?


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10.3 Planar graphs and Euler’s formulaPlanar graphsA planar graph can be drawn on a plane (page

surface) so that no edges intersect (cross), except

at the vertices.

AB

C

DA planar graph: no intersecting edges

Some graphs do not initially appear to be planar; for example, Graph 1 shown below left.

However, Graph 2 (drawn below right) is equivalent (isomorphic) to Graph 1. Graph 2 is

clearly planar.

A

B C

D

EF

Graph 1: non-planar graph as drawn

A

B C

D

EF

Graph 2: planar form of Graph 1

Not all graphs are planar. For example, the graph

opposite cannot be redrawn in an equivalent planar

form, no matter how hard you try.A

B

C

DE

Non-planar graph

Example 1 Redrawing a graph in planar form

Redraw the graph shown opposite in a planar form. A B

CD


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Solution (there are others)

1 Redraw the graph with edge DB removed.Note: We have removed edge DB because itintersects edge AC.

A B

CD

2 Replace edge DB as a loop that avoids

intersecting with the other three edges.

The graph is now in an equivalent

planar form: no edges intersect,

except at vertices.

A B

CD

Faces of a graphThe graph opposite can be regarded as dividing the

paper it is drawn on into two regions. In the language

of graphs, these regions are called the faces of the

graph. One face, f1, is bounded by the graph. The

other face, f2, is the region surrounding the graph.

Note that this outside face is infinite in extent.

The graph opposite divides the paper into four

regions, so we say that it has four faces: f1, f2, f3

and f4. Here f4 is an infinite face.

A

B

f2

f1

C

A

D

E

F

B

f1 f2f3

f4

C

Euler’s formulaEuler discovered that, for connected planar graphs, there is a relationship between the number

of vertices, v, the number of edges, e, and the number of faces, f. This relationship can be

expressed as

v − e + f = 2

This is known as Euler’s formula.

For the graph opposite:

v = 6, e = 8 and f = 4.

So v − e + f = 6 − 8 + 4 = 2

confirming Euler’s formula.

A

D

E

F

B

f1 f2f3

f4

C


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Euler’s formulaFor a connected planar graph,

v − e + f = 2

where v = number vertices, e = number of edges and f = number of faces.

Example 2 Verifying Euler’s formula

Consider the connected planar graph shown opposite.

a Write the number of vertices, v, the number of

edges, e, and the number of faces, f.

b Verify Euler’s formula. A

D

B

C

Solution

a

1 There are four vertices: A, B, C, D, so v = 4. Number of vertices : v = 4

2 There are six edges: AB, AC, AD, BC (× 2) and BD,

so e = 6.

Number of edges : e = 6

3 There are four faces, so f = 4.

A

D

B

C

f1

f2

f3

f4

Tip: Mark the faces in on the diagram. Do not forget theinfinite face f4 that surrounds the graph.

Number of faces : f = 4

b

1 Write down Euler’s formula.

2 Substitute the values of v, e, and f. Evaluate.

3 Write your conclusion.

Euler’s formula : v − e + f = 2

v − e + f = 4 − 6 + 4 = 2

∴ Euler’s formula is verified.

Example 3 Using Euler’s formula to determine the number of faces of a graph

A connected planar graph has four vertices and five edges. Determine the number of faces.

Solution

1 Write v and e. v = 4, e = 5

2 Write Euler’s formula. Euler's formula :

v − e + f = 2

3 Substitute the values of v and e. 4 − 5 + f = 2

4 Solve for f. −1 + f = 2

f = 3

5 Write your answer. The graph has three faces.


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Exercise 10C

1 Which of the following graphs are drawn in planar form?

AA

D

BC

BA

D

B

EC

CA

D

B

E

C

DA

DBE

C

EA

D

BE

F

C

FA

B

C

D

2 Show that each of the following graphs is planar by redrawing it in an equivalent planar

form.

aA B

CD

b

D

C

A

B

c

D

A

C

EB

d

D

A

B

C

e

D

E

FA

B

C

f

D C

A

E F

H G

B

3 For each of the following graphs:

i state the values of v, e and f ii verify Euler’s formula.

a b


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c d

4 For a planar connected graph, find:

a f given v = 4 and e = 4 b v given e = 3 and f = 2

c e given v = 3 and f = 3 d v given e = 6 and f = 4

e f given v = 4 and e = 6 f v given e = 6 and f = 4

g f given v = 6 and e = 11 h e given v = 10 and f = 11.

10.4 Traversable networksThe Konigsberg Bridges problem of planning a walk that crosses each of the seven bridges

only once is an example of what is called traversing a network.

Traversing a network means finding a route

through the network, along the edges, that uses all

the edges, but once only.

A

B

C

Start

Finish

Network 1: traversable

An example is shown opposite. Start at vertex A

and follow the arrows to vertex C. This network is

traversable.

Not all networks can be traversed.

For, example, no route can be found through the

network opposite that uses all the edges only once.

A B

CD

Network 2: non-traversable

Try it and see. This network is not traversable.

One way of finding out whether a network is traversable is to try a number of routes through

the networks and see whether they work. This is known as ‘trial and error’. However, for all

but the simplest networks, this can become very tedious after a while.

Fortunately, we do not have to use ‘trial and error’ to see whether a network can be traversed.

The solution has to do with the order of the vertices and whether they are of odd or even

degree.

Rules for determining whether a network is traversableFor a network to be traversable, it must first be connected.

A connected network is traversable if:

all vertices are of even degree; or

exactly two vertices are of odd degree and the rest are of even degree.

If a network has more than two vertices of odd degree, it is not traversable.


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Using these rules, we can see why Network 1 on the previous page is traversable: it has two

odd vertices

(A and C) and the remaining vertex (B) is even. Likewise, Network 2 is not traversable because

it has more than two odd vertices (A, B, C and D).

Example 4 Traversing a network

For each of the following networks:

Determine whether the network can be traversed, and state why.

If traversable, check by identifying a path that traverses the network.

a b c

Solution

a Traversable: all even

vertices

Start/Finish

b Traversable: two odd vertices,

the rest even

Start

Finish

c Not traversable: more

than two odd vertices

Note: In each case, more than one path is possible.

Exercise 10D


Determine whether the network can be traversed, and state why.

If traversable, check by identifying a path that traverses the network.

1 2 3


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4 5 6

7 8 9

10.5 Euler paths and circuitsPaths and circuits

A path is a sequence of edges, linking successive vertices, that connects two different

vertices in a network.

For example, in the graph opposite:

A-C-B is a path. It starts at vertex A, passes

through vertex C and ends at vertex B.

C-B-D-A is another path. It starts at vertex C,

passes through vertices B and D, and ends

at vertex A.

A B

D

C

Paths in networks can take many forms. They can

involve just one edge, or up to all the edges.

A circuit is like a path but starts and finishes at the same vertex.

For example, in the graph opposite:

A-C-B is a circuit. It starts at A, passes through

vertices C and B, and returns to vertex A.

A circuit can involve just two edges, or up to all

the edges in a network.A B

C

D


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Euler paths

An Euler path passes along every edge in a connected network, but uses each edge only

once. It may pass through any vertex more than once.

If a network is traversable, it has Euler paths.

A real-life example of an Euler path is the route used to collect the garbage in a housing estate.

Ideally, it can be planned so that the garbage truck has to travel down each street only once.

For example, the graph above is traversable. It has several Euler paths. Two of these paths,

C-B-C-A-B-D-A and A-B-C-A-D-B-C, are shown below. They connect vertices A and C.

(Follow the arrows.)

A B

C

D

Start

Finish

A B

C

D

Start

Finish

Euler circuits

An Euler circuit is an Euler path that starts and finishes at the same vertex.

A real-life example of an Euler circuit is the route followed by a country road inspector.

Ideally, a route can be planned so that the inspector can start and finish in their hometown, but

travel along each road only once.

For example, in the graph opposite,

A-B-D-C-A is an Euler circuit. It starts and

finishes at vertex A (follow the arrows).

An Euler circuit has as many start/finish

points as there are vertices in the network.

Start/FinishA

B

CD

Conditions for Euler paths and Euler circuitsTo have an Euler path or Euler circuit, a network must first be connected.

To have an Euler path but not an Euler circuit, the connected network must have

exactly two vertices of odd degree, with the remaining vertices having even degree.

An Euler path will start at one of the odd vertices and finish at the other.

To have an Euler circuit, the connected network must have all vertices of even degree.

An Euler circuit starts and finishes at the same vertex. It can be any vertex in the

network.

If there are more than two vertices of odd degree, the network is not traversable, so it is

not possible to have either an Euler path or an Euler circuit.


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Example 5 Identifying Euler circuits and paths


Determine whether the network has an Euler path only, an Euler circuit, or neither, and

state why.

If the network has an Euler path only, or an Euler circuit, show one example.

a b c

Solution

a Euler circuit: all even

vertices

b Euler path only: two odd

vertices, the rest even

c Neither: more than two

odd vertices

Start/FinishFinish

Start

Note: In each case, more than one path is possible.

Application of Euler paths and circuitsEuler paths and circuits have many practical applications. You have already met one

application of an Euler path: the Konigsberg Bridges problem.

In everyday life, Euler paths relate to situations such as delivering mail, inspecting roads,

picking up garbage and cleaning streets in a city. In all these situations there is a need to travel

along each road or street in an area, but to do so no more than once. Euler circuits apply to the

same situations, but with the added condition of wanting to return to the starting point.

Exercise 10E

1 For each of the following networks:

Determine whether the network has an Euler path only, an Euler circuit, or neither, and

state why.


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If the network has an Euler path only, or an Euler circuit, show one example.

a b c

d e f

g h i

2 A road inspector lives in Town A and is

required to do an inspection of the roads

connecting the neighbouring towns B, C, D

and E. The network of roads is shown on

the right.

Town C

Town D Town BTown E Town A

a Can the inspector set out from Town A, carry out his road inspection by travelling over

every road linking the five towns only once, and return to Town A? Why?

b If he can, show a possible route.

3 A postman has to deliver letters to

the houses located on the network of

streets shown on the right.

a Is it possible for the postman to

start and finish his deliveries at

the same point in the network

without retracing his steps at

some stage? If not, why not?

b It is possible for the postman to

start and finish his deliveries at different points in the network without retracing his steps

at some stage. Why is this so? Identify one such route.

A

H

G

B

C

F

D

E


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4 Two islands in a river are connected to

the banks of the river and to each other

by bridges as shown. There is also

another bridge connecting the two banks

of the river, as shown.

a Draw a graph to represent this

situation. Label the vertices A, B, C

and D to represent the banks of the

river and the two islands. Use the edges of the graph to represent the bridges.

b It is not possible to plan a walk that passes over each bridge only once. Why not?

c Show where another bridge can be added to make such a walk possible. Explain why it

can now be done by drawing a graph that represents the new situation.

A

B

C

D

10.6 Hamilton paths and circuitsEuler paths and circuits focus on edges.

Hamilton paths and circuits focus on vertices.

A Hamilton path passes through every vertex in a connected network once and once only. It

may or may not involve all the edges.

A Hamilton circuit is a Hamilton path that starts and finishes at the same vertex.

A Hamilton path involves all the vertices

but not necessarily all the edges.


A-B-D-C is a Hamilton path. It starts at

vertex A and ends at vertex C. (Follow

the arrows.) Note that it does not involve

all edges.

Start

B

DC Finish

A

A Hamilton circuit is a Hamilton path

that starts and finishes at the same vertex.


A-C-F-E-D-B-A is a Hamilton circuit. It starts

and finishes at vertex A. (Follow the arrows.)

Note that it does not involve all edges.

Unfortunately, unlike Euler paths and

circuits, there are no simple rules for

determining whether a network contains a

Hamilton path or circuit. It is just a matter

of ‘trial and error’.

Start/Finish

A

C

B

D

E

F


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Applications of Hamilton paths and circuitsHamilton paths and circuits have many practical applications.

In everyday life, Hamilton circuits relate to situations

like the following:

A courier leaves her depot to make a succession

of deliveries to a variety of locations before

returning to her depot. She does not like to go

past each location more than once.

A tourist plans to visit all the historic sites in a

city without visiting each site more than once.

You are planning a trip from Melbourne to visit

Shepparton, Wodonga, Bendigo, Swan Hill,

Natimuk, Warrnambool and Geelong before

returning to Melbourne.

A Hamilton path would apply to situations like the

following:

You plan a trip from Melbourne to Mildura,

with visits to Bendigo, Halls Gap, Horsham,

Stawell and Ouyen on the way.

In all these situations, there would be several suitable paths, and other factors, such as time

taken or distance travelled, would need to be taken into account in order to determine the best

route. This is an issue addressed in the next section: weighted graphs.

Exercise 10F

1 List a Hamilton path for the network shown:

a starting at A and finishing at D

b starting at F and finishing at G.

A

F

B

G

E H

D C

2 Identify a Hamilton circuit in each of the following networks (if possible), starting at A each

time.

a B C

A

FE

D

b

AC

B

D

E

c

E

D

G

C

B

F ACambridge University Press • Uncorrected Sample Pages • 978-0-521-74049-4 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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d A B C

FE

H IG

D

e

A

B

E F C

D

f C B A

ED

H G F

10.7 Weighted graphs and the shortest path problemWe plan a trip starting at Town A and

travelling through all the other towns

shown on the network below, before

returning to Town A. We do not want

to pass through any town more than

once.

Town B

Town E

Town F

Town D

Town C

Town A

There are many routes we could choose. Any Hamilton circuit is appropriate. Two possible

routes, Route 1 and Route 2, are shown below.

Town B

Town E

Town F

Town D

Route 1

Town C

Town A

Town B

Town E

Town F

Town D

Town C

Town A

Route 2

But which is the ‘best’ route?

There are several ways in which a route could be regarded as ‘best’. It could be the route

with the shortest distance, the route that takes the shortest time, the route that costs the least

amount of money, or even the most scenic route. In this instance, we will take the ‘best’ route

to mean the shortest distance route.

To see which is the shortest distance

route, we need to add the distances

between the towns to the graphs. We do

this by writing the distances (in km)

beside the relevant edges on the graph.

When we do this we say we are adding

weights to the graph, and we call the

resulting graph a weighted graph.

5

546

6 7

8

9

7

Town B

Town E

Town F

Town D

Town C

Town A


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Now we add up the distances along Route 1 and Route 2.

6

7

78

9

4 5 6

5Town B

Town E

Town F

Town D

Town C

Town A

Route 1 (A-B-C-D-E-F-A):

Length = 7 + 5 + 6 + 9 + 7 + 6

= 40 km

6

7

78

9

4 5 6

5Town B

Town E

Town F

Town D

Town C

Town A

Route 2 (A-D-C-B-E-F-A):

Length = 8 + 6 + 5 + 4 + 7 + 6

= 36 km

Thus Route 2 is the shortest path through the network that will enable you to start and finish at

Town A, visit every other town on the way and not pass through any town more than once.

Shortest pathThe shortest path that passes through each vertex once only is the shortest Hamilton circuit.

While there are systematic ways of determining the shortest path in networks, they are

complicated to apply and beyond the scope of this book. Consequently, finding shortest paths

in networks is a matter of ‘trial and error’.

Example 6 Finding the shortest path

Find the shortest path through the network

shown that:

starts and finishes at the same point,

passes through each vertex once only.

State its length. 5

2

2

33

A

C

D

EB

94

Strategy: The path we want is the shortest Hamilton circuit in the network.

Solution

1 Identify the Hamilton circuits in

the network.

The Hamilton circuits are:

A-B-C-D-E-A and A-C-D-E-B-A.

For A-B-C-D-E-A.2 Determine the length of each

circuit. Length = 3 + 9 + 2 + 5 + 2 = 21 units

For A-C-D-E-B-A :

3 Choose the shortest path and

write your answer and its length.

Length = 4 + 2 + 5 + 3 + 3 = 17 units

The shortest path is

A-C-D-E-B-A and its length

is 17 units.


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Exercise 10G

1 The following graphs represent the roads between a number of towns labelled A, B, C, D and

E. The length of each road is given in kilometres.

Identify the shortest route, starting and finishing at A, that would enable each town to be

visited once only, and determine its distance. The graphs are not drawn to scale.

a

AC D

E

B

2020

15

15

10

2535

b AB

C

D

E

2

66

4

8

7

105

2 The graph below shows a mountain bike rally course. Competitors must pass through each

of the checkpoints, A, B, C, D, E and F. The average times (in minutes) taken to ride

between the checkpoints are shown on the edges of the graph.

Competitors must start and finish at

checkpoint A. They can pass through the

other checkpoints in any order they wish.

What is the route that would have the

shortest average completion time?

A

C

D

F

E

B

10

1111

7

12

12

9

8

15

Start/Finish

3 Find the shortest path from vertex A

to vertex E in this network.A

C

DF

E

B7

83

3 5

64

4 Find the shortest Hamilton path for the

graph on the right, starting at vertex A.3

3

3

2

5

5 5

6

644

4

A

C

DG

F E

B


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10.8 Minimum spanning treesTrees

A tree is a connected graph that contains no circuits, multiple edges or loops. It may be part

of a large graph.

For example, Graphs 1 and 2 below are trees. Graph 3 is not a tree.

Graph 1: a tree Graph 2: a tree

Loop

Circuit

Graph 3: not a tree

Multipleedges

A tree with n vertices has (n − 1) edges. Graph 1 above (a tree) has eight vertices and seven

edges, but Graph 3 (not a tree) has 10 vertices and 12 edges.

Every connected graph contains one or

more trees. Two trees are shown on the

graph opposite.

Tree 1 Tree 2

Spanning trees

A spanning tree connects all the vertices in a connected graph but has no circuits, multiple

edges or loops.

An example of a spanning tree is shown

on the graph opposite. There are several

other possibilities.

Spanning tree

For a network of n vertices, a spanning tree will have n vertices and (n − 1) edges.

For example, the network above has eight vertices. A spanning tree for this network will have

eight vertices and seven edges.


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Example 7 Finding a spanning tree in a network

Find two spanning trees for the network

shown opposite.

A

C

D

EB

Solution

1 The network has five vertices and

seven edges. A spanning tree will have

five (n) vertices and four (n − 1) edges.

Required vertices = 5

Required edges = 4

2 To form a spanning tree, remove any

three edges, making sure that:� all the vertices remain connected.� there are no multiple edges or loops.

Spanning tree 1 is formed by removing

edges EB, ED and CA.

Spanning tree 2 is formed by removing

edges EA, AC and CD.Note: Several other possibilities exist.

A

C

D

E

B

Spanning tree 1

A

C

D

E

B

Spanning tree 2

Minimum spanning trees and Prim’s algorithmFor weighted graphs, it is possible to

determine the ‘length’ of each spanning

tree, by adding up the weights of the

edges in the tree.

6

4

31

1 1 4

5

78

2

5Spanning tree

For the spanning tree shown opposite:

Length = 5 + 4 + 2 + 1 + 5 + 4 + 1

= 22 units


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The minimum spanning tree is the spanning tree of minimum length (may be minimum

distance, minimum time, minimum cost, etc.). There may be more than one minimum

spanning tree in a graph.

Minimum spanning trees have applications such as planning the layout of a computer network,

or a water supply system to a new housing estate. In these situations, we might want to

minimise the amount of cable or water pipe needed to do the job. Alternatively, we might want

to minimise the time needed to complete the job, or the cost.

Prim’s algorithm is a set of rules to determine a minimum spanning tree for a graph.

Prim’s algorithm for finding a minimum spanning tree1 Choose a starting vertex (any will do). Inspect the edges starting from this vertex and

choose the one with the lowest weight. (If two edges have the same weight, the choice

can be arbitrary.) You now have two vertices and one edge.

2 Next, inspect the edges starting from the two vertices. Choose the edge with the lowest

weight, provided it does not form a cycle. (If two edges have the same weight, the choice

can be arbitrary.) You now have three vertices and two edges.

3 Repeat until all the vertices are connected, then stop.

The result is a minimum spanning tree.

Example 8 Applying Prim’s algorithm

Apply Prim’s algorithm to obtain a minimum

spanning tree for the network shown, and

calculate its length.

A

F

C

D

E

B6

66

3 5

7

8

2

2

5

Solution

1 Start at A:

AB is the lowest-weighted edge (2).

Draw it in.A

B

2

2 From A or B:

AD is the lowest-weighted edge (5).

Draw it in.A

B

2

5 D


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3 From A, B or D:

DC is the lowest-weighted edge (3).

Draw it in.

A

B

C

2

5 D

3

4 From A, B, C or D:

CE is the lowest-weighted edge (5).

Draw it in.

A

B

C

2

5

53

D

E

5 From A, B, C, D or E:

EF is the lowest-weighted edge (2).

Draw it in.

All vertices have now been joined.

The minimum spanning tree is determined. A

B

C

2

2

5

53

D

E

F

6 Find the length of the minimum spanning

tree by adding the weights of the edges.

Minimum spanning tree

Length = 2 + 5 + 3 + 5 + 2 = 17 units

Exercise 10H

1 A connected graph has eight vertices and ten edges. Its spanning tree has vertices and

edges.

2 Which of the following graphs are trees?

A B C


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D E F

3 For each of the following graphs, draw three different spanning trees.

a b c

4 For each of the following connected graphs, use Prim’s algorithm to determine the minimum

spanning tree and its length.

a A

B

C

D

E

2

6

86

4

7

5 10

b A

B

C

D

E

2

2

6

45

9

3

c

A

B

C

D

E

64

53

78

3

F

d

A

B

C D

E

15

15

1020

20

3525

e

A

B C

F

GD

E

2

6

6

74

4 55

5 3

3

3

f A

B

C

D

E

2

2

2

45

9

33

1

5 Water is to be piped from a water tank

to seven outlets on a property. The

distances (in metres) of the outlets from

the tank and from each other are shown

in the network below. Starting at the

tank, determine the minimum length of

pipe needed.

Outlet A Outlet B

Outlet C

Outlet DOutlet E

Outlet FOutlet G Tank

11

11

13

612

7

9

8

6

510

2


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6 Power is to be connected by cable from

a power station to eight substations

(A to H). The distances (in kilometres) of

the substations from the power station

and from each other are shown in the

network opposite. Determine the minimum

length of cable needed.

21

11 11

10

19

8 1531

17

12

9 1818

623

H

BC

DE

FG

A

Power station


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P1: FXS/ABE P2: FXS0521672600Xca10.xml CUAU034-EVANS March 10, 1904 23:23

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iew


Key ideas and chapter summary

Graph or network A graph or network consists

of a set of points called vertices

and a set of lines called edges.

Each edge joins two vertices.A

B

C

D

E

F

Vertices and edges In the graph above, A, B, C, D, E and F are the vertices and

the lines AB, AE, AF, BC, BF, CD, CE, and FD are the edges.

Degree of a vertex The degree of vertex A, written deg(A), is given by the

number of edges attached to the vertex.

For example, in the graph above:

deg(A) = 3 and deg(E) = 2.

Isomorphic graphs Two graphs are said to be isomorphic (equivalent) if:� they have the same number of edges and vertices� corresponding vertices have the same degree and the

edges connect the same vertices

For example, the two graphs below are isomorphic or

equivalent.

A B

CD

A

BC D

Multiple edges and loops

CB

A

D

multipleedges

loop

The graph above is said to have multiple edges, as there are

two edges joining A and D.

C has one edge, which links C to itself. This edge is called a

loop.


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Path A path can be thought of as a sequence of edges. For

example, in the graph below:

A-B-C, G-F-E-D and

B-G-E-D-C are all

examples of paths.A

BC

D

E

F

G

Connected graph A graph is connected if there is a path between each pair of

vertices.

This graph is connected.

This graph is not connected.

Circuit A circuit is a sequence of edges linking successive vertices

that starts and finishes at the same vertex.

For example, in the graph below: A-B-C-D-F-A and F-D-E-F

are both circuits

A

B C

D

E

F

Planar graph A graph that can be drawn in such a way that no two edges

intersect, except at the vertices, is called a planar graph.

This graph is planar.

This graph is not planar.


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Euler’s formula For any connected planar graph, Euler’s formula states:

v − e + f = 2

v = the number of vertices.

e = the number of edges.

f = the number of faces.

For example:

f3

v = 3, e = 4, f = 3

v – e + f = 3 – 4 + 3 = 2

f1

f2

Euler path A path that includes every edge just once (but does not start

and finish at the same vertex) is called an Euler path.

Condition for an Euler path To have an Euler path (but not an Euler circuit), a network

must be connected and have exactly two vertices of odd

degree, with the remaining vertices having even degree.

The Euler path will start at one of the odd vertices and finish

at the other.

For example, the graph opposite is

connected. It has two odd vertices and

three even vertices. It has an Euler path

that starts at one of the odd vertices and

finishes at the other odd vertex.

Start

Finish

Euler circuit An Euler path that starts and finishes at the same vertex is

called an Euler circuit.

Condition for an Euler circuit To have an Euler circuit, a network must be connected and

all vertices must be even.

For example, in the network

opposite, all vertices are even.

It has an Euler circuit.

The circuit starts and finishes at

the same vertex.

Start

Finish

Hamilton path A Hamilton path is a path through

a graph that passes through each

vertex exactly once, but does not

necessarily start and finish at the

same vertex.

Start

Finish


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Hamilton circuit A Hamilton circuit is a Hamilton

path that starts and finishes at the

same vertex.

Start

Finish

Weighted graphs A weighted graph is one where

a number is associated with each

edge. These numbers are called

weights.

A

C

D

E

2

9

3 2

3

5 4

Tree A tree is a connected graph that contains no circuits,

multiple edges or loops.

A tree with n vertices has n−1 edges.

The tree above has 8 vertices and 7 edges.

Spanning tree A spanning tree is a graph that contains all the vertices of a

connected graph, without multiple edges, circuits or loops.

Spanning tree

Minimum spanning tree A minimum spanning tree is a spanning tree for which the

sum of the weights of the edges is as small as possible.

Prim’s algorithm Prim’s algorithm is a systematic method for determining a

minimal spanning tree in a connected graph.


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Skills check

Having completed this chapter you should be able to:

recognise and determine the number of vertices, edges and faces of a graph

recognise isomorphic graphs, connected graphs, planar graphs, weighted graphs,

trees and spanning trees

use Euler’s formula for connected planar graphs

know and apply the definitions of Euler paths and circuits

know the conditions for a graph to have an Euler path or Euler circuit

locate an Euler path or circuit in a graph

know and apply the definitions of Hamilton paths and circuits

locate a Hamilton path or circuit in a graph

determine the shortest path in a weighted graph

know and apply the definition of a spanning tree

use Prim’s algorithm to determine a minimum spanning tree and its length.

Multiple-choice questions

The following graph relates to Questions 1 to 6

A

B C

E

F

D

1 The number of vertices in the graph above is:

A 3 B 5 C 6 D 7 E 9

2 The number of edges in the graph above is:

A 3 B 5 C 6 D 7 E 9

3 The number of faces in the graph above is:

A 3 B 4 C 5 D 6 E 7

4 The degree of vertex B in the graph above is:

A 1 B 2 C 3 D 4 E 5

5 The number of even vertices in the graph above is:

A 1 B 2 C 3 D 4 E 5

6 The graph above has:

A an Euler path but not an Euler circuit B several Euler paths but no Euler circuits

C an Euler circuit D several Euler circuits

E neither an Euler path nor an Euler circuit.


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Review


7 For which one of the following graphs is the sum of the degrees of the vertices equal

to 14?

A B C

D E

8 For the graph on the right, the sum of the degrees of the vertices is:

A 22 B 23 C 24

D 25 E 26

9 For the graph in Question 8:

A v = 9, e = 13, f = 5 B v = 9, e = 13, f = 6

C v = 10, e = 13, f = 5 D v = 9, e = 14, f = 6

E v = 10, e = 13, f = 5

10 Euler’s formula for a planar graph is:

A v − e = f + 2 B v − e + f = 2 C v + e + f = 2

D e − v + f = 2 E v − e = f − 2

11 A connected graph with 10 vertices divides the plane into five faces. The number of

edges connecting the vertices in this graph will be:

A 5 B 7 C 10 D 13 E 15

12 For a connected graph to have an Euler path but not an Euler circuit:

A all vertices must be odd B all vertices must be even

C there must be exactly two odd vertices and the rest even

D there must be exactly two even vertices and the rest odd

E an odd vertex must be followed by an even vertex

13 For a connected graph to have an Euler circuit:

A all vertices must be odd B all vertices must be even

C there must be exactly two odd vertices and the rest even

D there must be exactly two even vertices and the rest odd

E an odd vertex must be followed by an even vertex


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14 Which one of the following graphs has an Euler path but not an Euler circuit?

A B C

D E

15 Which one of the graphs in Question 14 has an Euler circuit?

16 Which one of the following graphs has an Euler circuit?

A B C

D E

17 For the graph shown, which additional edge

could be added to the network so that the graph

formed would contain an Euler path?C

A

B

E

DF

A AF B AD C AB

D CF E BF

18 The length of the shortest path from F to B

in the network shown is:

C

AB

E

D

F

7

73

3

8

8

2

613

A 17 B 18 C 19

D 20 E 21


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Review


19 Which one of the following paths is a

Hamilton circuit for the graph shown here?

A F-E-D-F

B F-E-D-A-B-C-E-F

C F-E-D-A-B-C-F

D F-C-A-B-D-E-F

E F-D-E-C-A-B-C-F

C

AB

E

D

F

20 Of the following graphs, which one has both Euler and Hamilton circuits?

A B C

D E

21 Which one of the following graphs is a tree?

A B C

D E

22 Which one of the following graphs is a spanning

tree for the network shown on the right?

6

1

4 5

7

2

3


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A

6

1

4 5

7

2

3B

6

1

4 5

7

32

C

6

1

4 5

7

23

D

6

1

4 5

7

2

3E

6

1

4 5

7

23

23 A park ranger wants to check all of the trails in a national park, starting at and

returning to the Park Office. She would like to do it without having to travel over the

same trail more than once. If possible, the route she should follow is:

A an Euler path B an Euler circuit C a Hamilton path

D a Hamilton circuit E a minimum spanning tree

24 A park ranger wants to check all of the camp sites in a national park, starting at and

returning to the Park Office. The campsites are all interconnected with trails. She

would like to check the campsites without having to visit each camp site more than

once. If possible, the route she should follow is:



25 The park authorities plan to pipe water to each of the camp sites from a spring

located in the park. They want to use the least amount of water pipe possible. If

possible, the water pipes should follow:



26 Each week, a garbage run starts at the Park Office and collects the rubbish left at

each of the camp sites and dumps the rubbish at a tip outside the park. The plan is to

visit each camp site only once. If possible, the garbage run should follow:




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Review


Short-answer questions

1 Draw a connected graph with:

a four vertices, four edges and two faces

b four vertices, five edges and three faces

c five vertices, eight edges and five faces.

2 Redraw each of the following graphs in a planar form.

a AB

CD

b A

B

CD

c

A

B

C

D

E

3 For the network shown, write down:

a the degree of vertex C

b the numbers of odd and even vertices

c the route followed by an Euler path

starting at vertex B.

AB

CD

4 For the network shown, write down:

a the degree of vertex C

b the numbers of odd and even vertices

c the route followed by an Euler circuit

starting at vertex A.

AB

EC

D

5 For the network shown, determine:

a the length of the shortest Hamilton path

connecting vertex D to vertex F

b the length of the minimum spanning tree.

A

B

CD

F

10

9

85

7

7

6

6

6 For the network shown, determine:

a the length of the shortest Hamilton circuit

starting and finishing at vertex A

b the length of the minimum spanning tree.

A

B

CD

F

12

8

10

7

11

6 5

5

4

3


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Extended-response question

1 The diagram opposite shows the network of

walking tracks in a small national park.

These tracks connect the camp sites to each

other and to the Park Office. The lengths of

the tracks (in metres) are also shown. Park office

140

350250

160

600

150

280

80

110

130400120

230200C1

C2

C3

C4

C5

C6C7

C8

a The network of tracks is planar. Explain what this means.

b Verify Euler’s rule for this network.

c A ranger at camp site C8 plans to visit camp sites C1, C2, C3, C4 and C5 on her

way back to the Park Office. What is the shortest distance she will have to walk?

d How many even and how many odd vertices are there in the network?

e Each day, the ranger on duty has to inspect each of the tracks to make sure that

they are all passable.

i Is it possible for her to do this starting and finishing at the Park Office? Explain

why.

ii Identify one route that she could take.

f Following a track inspection after wet weather, the Head Ranger decides that it is

necessary to put gravel on some walking tracks to make them weatherproof.

i What is the minimum length of track that will need to be gravelled to ensure

that all camp sites and the Park Office are accessible along a gravelled track?

ii Show the tracks to be gravelled on a diagram.

g A ranger wants to inspect each of the camp sites but not pass through any camp

site more than once on his inspection tour. He wants to start and finish his

inspection tour at the Park Office.

i What is the technical name for the route he wants to take?

ii With the present layout of tracks, he cannot inspect all the tracks without

passing through at least one camp site twice. Suggest where an additional track

could be added to solve this problem.

iii With this new track, write down a route he could follow.


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Networks - Cambridge University Press€¦ · The graphs are connected because, starting at any vertex, say A,you can always ﬁnd a path ... TI-Nspire & Casio ClassPad material prepared