Chat with us, powered by LiveChat Does each of these graphs have an Euler circuit? If so, find it Does - Writeedu

Does each of these graphs have an Euler circuit? If so, find it Does

11.  Does each of these graphs have an Euler circuit? If so, find it.

12.  Does each of these graphs have an Euler circuit? If so, find it. 

13.  Eulerize this graph using as few edge duplications as possible. Then, find an Euler circuit. 

17.  Does each of these graphs have at least one Hamiltonian circuit? If so, find one. 

19.  A company needs to deliver product to each of their 5 stores around the Dallas, TX area. Driving distances between the stores are shown below. Find a route for the driver to follow, returning to the distribution center in Fort Worth: 

a. Using Nearest Neighbor starting in Fort Worth 

b. Using Repeated Nearest Neighbor 

c. Using Sorted Edges 

21. When installing fiber optics, some companies will install a sonet ring; a full loop of cable connecting multiple locations. This is used so that if any part of the cable is damaged it does not interrupt service, since there is a second connection to the hub. A company has 5 buildings. Costs (in thousands of dollars) to lay cables between pairs of buildings are shown below. Find the circuit that will minimize cost: 

a. Using Nearest Neighbor starting at building A 

b. Using Repeated Nearest Neighbor 

c. Using Sorted Edges 

23.  Find a minimum cost spanning tree for the graph you created in problem #3 

Graph Theory 117

© David Lippman Creative Commons BY-SA

Graph Theory and Network Flows In the modern world, planning efficient routes is essential for business and industry, with

applications as varied as product distribution, laying new fiber optic lines for broadband

internet, and suggesting new friends within social network websites like Facebook.

This field of mathematics started nearly 300 years ago as a look into a mathematical puzzle

(we’ll look at it in a bit). The field has exploded in importance in the last century, both

because of the growing complexity of business in a global economy and because of the

computational power that computers have provided us.

Graphs

Drawing Graphs

Example 1

Here is a portion of a housing development from Missoula, Montana1. As part of her job, the

development’s lawn inspector has to walk down every street in the development making sure

homeowners’ landscaping conforms to the community requirements.

1 Sam Beebe. http://www.flickr.com/photos/sbeebe/2850476641/, CC-BY

118

Naturally, she wants to minimize the amount of walking she has to do. Is it possible for her

to walk down every street in this development without having to do any backtracking?

While you might be able to answer that question just by looking at the picture for a while, it

would be ideal to be able to answer the question for any picture regardless of its complexity.

To do that, we first need to simplify the picture into a form that is easier to work with. We

can do that by drawing a simple line for each street. Where streets intersect, we will place a

dot.

This type of simplified picture is called a graph.

Graphs, Vertices, and Edges

A graph consists of a set of dots, called vertices, and a set of edges connecting pairs of

vertices.

While we drew our original graph to correspond with the picture we had, there is nothing

particularly important about the layout when we analyze a graph. Both of the graphs below

are equivalent to the one drawn above since they show the same edge connections between

the same vertices as the original graph.

You probably already noticed that we are using the term graph differently than you may have

used the term in the past to describe the graph of a mathematical function.

A

B

C

D

E

F

G

H

I

J

K L

M

N

O P Q

R

A

B

C

D

E F

G

H

I

J

K

L

M

N O

P Q

R

A B C

D E F

G

H

I J

K

L

M

N

O

P

Q

R

Graph Theory 119

Example 2

Back in the 18th century in the Prussian city of

Königsberg, a river ran through the city and seven

bridges crossed the forks of the river. The river

and the bridges are highlighted in the picture to

the right2.

As a weekend amusement, townsfolk would see if

they could find a route that would take them

across every bridge once and return them to

where they started.

Leonard Euler (pronounced OY-lur), one of the

most prolific mathematicians ever, looked at this problem in 1735, laying the foundation for

graph theory as a field in mathematics. To analyze this problem, Euler introduced edges

representing the bridges:

Since the size of each land mass it is not relevant to the question of bridge crossings, each

can be shrunk down to a vertex representing the location:

Notice that in this graph there are two edges connecting the north bank and island,

corresponding to the two bridges in the original drawing. Depending upon the interpretation

of edges and vertices appropriate to a scenario, it is entirely possible and reasonable to have

more than one edge connecting two vertices.

While we haven’t answered the actual question yet of whether or not there is a route which

crosses every bridge once and returns to the starting location, the graph provides the

foundation for exploring this question.

2 Bogdan Giuşcă. http://en.wikipedia.org/wiki/File:Konigsberg_bridges.png

North Bank

Island

South Bank

East Bank

EB

NB

SB

I

120

Definitions

While we loosely defined some terminology earlier, we now will try to be more specific.

Vertex

A vertex is a dot in the graph that could represent an intersection of streets, a land

mass, or a general location, like “work” or “school”. Vertices are often connected by

edges. Note that vertices only occur when a dot is explicitly placed, not whenever two

edges cross. Imagine a freeway overpass – the freeway and side street cross, but it is

not possible to change from the side street to the freeway at that point, so there is no

intersection and no vertex would be placed.

Edges

Edges connect pairs of vertices. An edge can represent a physical connection between

locations, like a street, or simply that a route connecting the two locations exists, like

an airline flight.

Loop

A loop is a special type of edge that connects a vertex to itself. Loops are not used

much in street network graphs.

Degree of a vertex

The degree of a vertex is the number of edges meeting at that vertex. It is possible for

a vertex to have a degree of zero or larger.

Path

A path is a sequence of vertices using the edges. Usually we are interested in a path

between two vertices. For example, a path from vertex A to vertex M is shown below.

It is one of many possible paths in this graph.

Degree 0 Degree 1 Degree2 Degree 3 Degree 4

A B C D

E F G H

J K L M

Graph Theory 121

Circuit

A circuit is a path that begins and ends at the same vertex. A circuit starting and

ending at vertex A is shown below.

Connected

A graph is connected if there is a path from any vertex to any other vertex. Every

graph drawn so far has been connected. The graph below is disconnected; there is no

way to get from the vertices on the left to the vertices on the right.

Weights

Depending upon the problem being solved, sometimes weights are assigned to the

edges. The weights could represent the distance between two locations, the travel time,

or the travel cost. It is important to note that the distance between vertices in a graph

does not necessarily correspond to the weight of an edge.

Try it Now 1

1. The graph below shows 5 cities. The weights on the edges represent the airfare for a

one-way flight between the cities.

a. How many vertices and edges does the graph have?

b. Is the graph connected?

c. What is the degree of the vertex representing LA?

d. If you fly from Seattle to Dallas to Atlanta, is that a path or a circuit?

e. If you fly from LA to Chicago to Dallas to LA, is that a path or a circuit?

Seattle

LA

Chicago

Dallas

Atlanta

$165

$150

$120

$85

$75

$100

$70

$145

$140 $170

A B C D

E F G H

J K L M

122

Shortest Path When you visit a website like Google Maps or use your Smartphone to ask for directions

from home to your Aunt’s house in Pasadena, you are usually looking for a shortest path

between the two locations. These computer applications use representations of the street

maps as graphs, with estimated driving times as edge weights.

While often it is possible to find a shortest path on a small graph by guess-and-check, our

goal in this chapter is to develop methods to solve complex problems in a systematic way by

following algorithms. An algorithm is a step-by-step procedure for solving a problem.

Dijkstra’s (pronounced dike-stra) algorithm will find the shortest path between two vertices.

Dijkstra’s Algorithm

1. Mark the ending vertex with a distance of zero. Designate this vertex as current.

2. Find all vertices leading to the current vertex. Calculate their distances to the end.

Since we already know the distance the current vertex is from the end, this will just

require adding the most recent edge. Don’t record this distance if it is longer than a

previously recorded distance.

3. Mark the current vertex as visited. We will never look at this vertex again.

4. Mark the vertex with the smallest distance as current, and repeat from step 2.

Example 3

Suppose you need to travel from Tacoma, WA (vertex T) to Yakima, WA (vertex Y).

Looking at a map, it looks like driving through Auburn (A) then Mount Rainier (MR) might

be shortest, but it’s not totally clear since that road is probably slower than taking the major

highway through North Bend (NB). A graph with travel times in minutes is shown below.

An alternate route through Eatonville (E) and Packwood (P) is also shown.

Step 1: Mark the ending vertex

with a distance of zero. The

distances will be recorded in

[brackets] after the vertex name.

96 79

27

76 96

57

20

36 104

T

A

NB

MR

E P

Y

96 79

27

76 96

57

20

36 104

T

A

NB

MR

E P

Y [0]

Graph Theory 123

Step 2: For each vertex leading to

Y, we calculate the distance to the

end. For example, NB is a

distance of 104 from the end, and

MR is 96 from the end.

Remember that distances in this

case refer to the travel time in

minutes.

Step 3 & 4: We mark Y as visited, and mark the vertex with the smallest recorded distance

as current. At this point, P will be designated current. Back to step 2.

Step 2 (#2): For each vertex leading to P (and not leading to a visited vertex) we find the

distance from the end. Since E is 96 minutes from P, and we’ve already calculated P is 76

minutes from Y, we can compute

that E is 96+76 = 172 minutes

from Y.

If we make the same computation

for MR, we’d calculate 76+27 =

103. Since this is larger than the

previously recorded distance from

Y to MR, we will not replace it.

Step 3 & 4 (#2): We mark P as visited, and designate the vertex with the smallest recorded

distance as current: MR. Back to step 2.

Step 2 (#3): For each vertex

leading to MR (and not leading to

a visited vertex) we find the

distance to the end. The only

vertex to be considered is A, since

we’ve already visited Y and P.

Adding MR’s distance 96 to the

length from A to MR gives the

distance 96+79 = 175 minutes

from A to Y.

Step 3 & 4 (#3): We mark MR as visited, and designate the vertex with smallest recorded

distance as current: NB. Back to step 2.

96

79

27

76 96

57

20

36 104

T

A

NB [104]

MR [96]

E P [76]

Y [0]

96

79

27

76 96

57

20

36 104

T

A

NB [104]

MR [96]

E [172] P [76]

Y [0]

96

79

27

76 96

57

20

36 104

T

A [175]

NB [104]

MR [96]

E [172] P [76]

Y [0]

124

Step 2 (#4): For each vertex

leading to NB, we find the

distance to the end. We know the

shortest distance from NB to Y is

104 and the distance from A to

NB is 36, so the distance from A

to Y through NB is 104+36 = 140.

Since this distance is shorter than

the previously calculated distance

from Y to A through MR, we

replace it.

Step 3 & 4 (#4): We mark NB as visited, and designate A as current, since it now has the

shortest distance.

Step 2 (#5): T is the only

non-visited vertex leading to

A, so we calculate the

distance from T to Y through

A: 20+140 = 160 minutes.

Step 3 & 4 (#5): We mark A as visited, and designate E as current.

Step 2 (#6): The only non-visited vertex leading to E is T. Calculating the distance from T

to Y through E, we compute 172+57 = 229 minutes. Since this is longer than the existing

marked time, we do not replace it.

Step 3 (#6): We mark E as visited. Since all vertices have been visited, we are done.

From this, we know that the shortest path from Tacoma to Yakima will take 160 minutes.

Tracking which sequence of edges yielded 160 minutes, we see the shortest path is T-A-NB-

Y.

Dijkstra’s algorithm is an optimal algorithm, meaning that it always produces the actual

shortest path, not just a path that is pretty short, provided one exists. This algorithm is also

efficient, meaning that it can be implemented in a reasonable amount of time. Dijkstra’s

algorithm takes around V2 calculations, where V is the number of vertices in a graph3. A

graph with 100 vertices would take around 10,000 calculations. While that would be a lot to

do by hand, it is not a lot for computer to handle. It is because of this efficiency that your

car’s GPS unit can compute driving directions in only a few seconds.

3 It can be made to run faster through various optimizations to the implementation.

96

79

27

76 96

57

20

36 104

T

A [140]

NB [104]

MR [96]

E [172] P [76]

Y [0]

96

79

27

76 96

57

20

36 104

T [160]

A [140]

NB [104]

MR [96]

E [172] P [76]

Y [0]

Graph Theory 125

In contrast, an inefficient algorithm might try to list all possible paths then compute the

length of each path. Trying to list all possible paths could easily take 1025 calculations to

compute the shortest path with only 25 vertices; that’s a 1 with 25 zeros after it! To put that

in perspective, the fastest computer in the world would still spend over 1000 years analyzing

all those paths.

Example 4

A shipping company needs to route a package from Washington, D.C. to San Diego, CA. To

minimize costs, the package will first be sent to their processing center in Baltimore, MD

then sent as part of mass shipments between their various processing centers, ending up in

their processing center in Bakersfield, CA. From there it will be delivered in a small truck to

San Diego.

The travel times, in hours, between their processing centers are shown in the table below.

Three hours has been added to each travel time for processing. Find the shortest path from

Baltimore to Bakersfield.

While we could draw a graph, we can also work directly from the table.

Step 1: The ending vertex, Bakersfield, is marked as current.

Step 2: All cities connected to Bakersfield, in this case Denver and Dallas, have their

distances calculated; we’ll mark those distances in the column headers.

Step 3 & 4: Mark Bakersfield as visited. Here, we are doing it by shading the corresponding

row and column of the table. We mark Denver as current, shown in bold, since it is the

vertex with the shortest distance.

Baltimore Denver Dallas Chicago Atlanta Bakersfield

Baltimore * 15 14

Denver * 18 24 19

Dallas * 18 15 25

Chicago 15 18 18 * 14

Atlanta 14 24 15 14 *

Bakersfield 19 25 *

Baltimore

Denver

[19]

Dallas

[25]

Chicago Atlanta Bakersfield

[0]

Baltimore * 15 14

Denver * 18 24 19

Dallas * 18 15 25

Chicago 15 18 18 * 14

Atlanta 14 24 15 14 *

Bakersfield 19 25 *

126

Step 2 (#2): For cities connected to Denver, calculate distance to the end. For example,

Chicago is 18 hours from Denver, and Denver is 19 hours from the end, the distance for

Chicago to the end is 18+19 = 37 (Chicago to Denver to Bakersfield). Atlanta is 24 hours

from Denver, so the distance to the end is 24+19 = 43 (Atlanta to Denver to Bakersfield).

Step 3 & 4 (#2): We mark Denver as visited and mark Dallas as current.

Step 2 (#3): For cities connected to Dallas, calculate the distance to the end. For Chicago,

the distance from Chicago to Dallas is 18 and from Dallas to the end is 25, so the distance

from Chicago to the end through Dallas would be 18+25 = 43. Since this is longer than the

currently marked distance for Chicago, we do not replace it. For Atlanta, we calculate 15+25

= 40. Since this is shorter than the currently marked distance for Atlanta, we replace the

existing distance.

Step 3 & 4 (#3): We mark Dallas as visited, and mark Chicago as current.

Step 2 (#4): Baltimore and Atlanta are the only non-visited cities connected to Chicago. For

Baltimore, we calculate 15+37 = 52 and mark that distance. For Atlanta, we calculate 14+37

= 51. Since this is longer than the existing distance of 40 for Atlanta, we do not replace that

distance.

Baltimore

Denver

[19]

Dallas

[25]

Chicago

[37]

Atlanta

[43]

Bakersfield

[0]

Baltimore * 15 14

Denver * 18 24 19

Dallas * 18 15 25

Chicago 15 18 18 * 14

Atlanta 14 24 15 14 *

Bakersfield 19 25 *

Baltimore

Denver

[19]

Dallas

[25]

Chicago

[37]

Atlanta

[40]

Bakersfield

[0]

Baltimore * 15 14

Denver * 18 24 19

Dallas * 18 15 25

Chicago 15 18 18 * 14

Atlanta 14 24 15 14 *

Bakersfield 19 25 *

Graph Theory 127

Step 3 & 4 (#4): Mark Chicago as visited and Atlanta as current.

Step 2 (#5): The distance from Atlanta to Baltimore is 14. Adding that to the distance

already calculated for Atlanta gives a total distance of 14+40 = 54 hours from Baltimore to

Bakersfield through Atlanta. Since this is larger than the currently calculated distance, we do

not replace the distance for Baltimore.

Step 3 & 4 (#5): We mark Atlanta as visited. All cities have been visited and we are done.

The shortest route from Baltimore to Bakersfield will take 52 hours, and will route through

Chicago and Denver.

Try it Now 2

Find the shortest path between vertices A and G in the graph below.

Euler Circuits and the Chinese Postman Problem In the first section, we created a graph of the Königsberg bridges and asked whether it was

possible to walk across every bridge once. Because Euler first studied this question, these

types of paths are named after him.

Euler Path

An Euler path is a path that uses every edge in a graph with no repeats. Being a path,

it does not have to return to the starting vertex.

Baltimore

[52]

Denver

[19]

Dallas

[25]

Chicago

[37]

Atlanta

[40]

Bakersfield

[0]

Baltimore * 15 14

Denver * 18 24 19

Dallas * 18 15 25

Chicago 15 18 18 * 14

Atlanta 14 24 15 14 *

Bakersfield 19 25 *

7

4 4

5

2

1 2

6

6

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