# Chapter 6 Hamilton Circuits

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7 Οκτ 2013 (πριν από 4 χρόνια και 9 μήνες)

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Chapter 6 Traveling Salesman Problem

ESSENTIAL QUESTIONS:

Section 6.1: How does Hamilton’s Circuits and Paths compare to Euler’s?

Section 6.2: What is a complete graph?

Section 6.3: What do the Trav
eling Salesman Problems (TSPs) use weighted grap
hs
?

Secti
on 6.4: What are simple strategies for solving TSPs?

Section 6.5: How does the Brute
-
Force and Nearest

Neighbor Algorithms solve TSPs?

Section 6.6: What is an approximate algorithm?

Section 6.7: How does the Repetitive Nearest

Neighbor Algorithm improv
e your ability to solve TSPs?

Section 6.8: How does the Cheapest
-
Link Algorithm differ from the Nearest
-
Neighbor Algorithm?

WORD WALL:

Brute
-
Force Algorithm

Cheapest
-

Complete Graph

Weight

Hamilton Circuit

Hamilton Path

Inefficient Algorith
m

Nearest
-
Neighbor Algorithm

Repetitive Nearest
-
Neighbor Algorithm

Traveling Salesman Problem (TSP)

Weighted Graph

Extra Review
/ Possible Homework
:

6.1

=

pp. 221
-

2 # 2, 4, 6, 7, 10, 11

6.2
= p 223 # 25

28

6.3

=
pp. 222

3 # 13

16

6.4
/
6.5
=

pp
. 223

225 # 29
-

31, 33, 35

6.7

= p. 225 #37

39, 41

6.8

=
pp. 225
-

6 #43
-
45, 47

Section 6.1: Hamilton Circuits and Hamilton Paths

EULER CIRCUITS AND PATHS

HAMILTON CIRCUITS AND PATHS

EULER = touch

each
EDGE

of a graph
exactly one time.

By Definit
ion
:

must touch every vertex

“can touch
more than once depending on degree”

HAMILTON = Touch

each
VERTEX

exactly
one time

(enter/ leave once)
.

By Definition:

Doesn’t require touching every edge

HAMILTON Circuits/Paths VERSUS EULER Circuits/Paths

For ea
ch of the following graphs, use our definitions of Hamilton and Euler to determine if circuits and
paths of each type are possible.

Graph 1

Graph 2

Graph 3

Graph 4

Graph 5

Graph 6

EULER
PATH

NO

YES

NO

NO

YES

NO

EULER
CIRCUIT

YES

NO

NO

YES

NO

N
O

HAMILTON
PATH

YES

YES

YES

YES

NO

YES

HAMILTON
CIRCUIT

YES

NO

YES

NO

NO

NO

What do you notice about the relationship between a graph having an Euler Circuit or Path and having a
Hamilton Circuit of path?

There is no relationship

IN GENERAL, THERE ARE

NO THEOREMS TO DETERMINE IF A GRAPH HAS A HAMILTON PATH
OR CIRCUIT.

#1: FOR HAMILTON CIRCUITS/ PATHS, VERTICES OF DEGREE 1 OR 2 ARE
VERY HELPFUL BECAUSE THEY REPRESENT REQUIRED EDGES TO REACH
THAT VERTEX.

#2:

EXAMPLE: Hamilton Circuit
=
A, B, C, D, A

SHIFT/SAME

Hamilton Circuits:

Follows the
SAME rotation order

for a circuit, but
with a different start.

DISTINCT

Hamilton Circuits:

A
different order of vertices

(Left to Right)
from
the same starting vertex.

(reverse order counts)

B, C,

D, A, B

C, D, A, B, C

D, A, B, C, D

A, D, C, B, A

(reverse order)

A, D, B, C, A

AND reverse = A, C, B, D, A

A, B, D, C, A
AND reverse =
A, C, D, B, A

A

B

C

D

EXAMPLE #1: List all possible Hamiltonian circuits in the following graph

1.

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

2.

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

3.

A, F, C, D, C, B, G, A

4.

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

REVER
SE EACH ONE OF THESE 4 CIRCUITS

What about the Hamiltonian circuits that start at
B
,
C, D,
etc.?

Well, since such a circuit must pass through
all

of the vertices
it doesn’t matter

.

Related to the idea of shifts of circuits.

EXAMPLE #2:

(a)

Find a Hamiltonian path that begins at A and ends at E.

(b)

Find a Hamiltonian circuit that starts at A and ends with the pair of vertices E, A.

(c)

Find a Hamiltonian path that

begins at F and ends at G.

(a)

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

(b)

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

(c)

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

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

Example 3:

The following graph has no Hamiltonian circuits or paths
--
why not?

4 bridges exist = can’t have a circuit; No path possible b
ecause
multiple vertices of degree 1

EXAMPLE #4:

(a)

Find all Hamiltonian circuits that starts at A.

(b)

Find all Hamiltonian circuits that starts at D.

(c)

Explain why your answers in a and b must have the same
number of circuits.

(a)

A, C, F, D, E, B, A

A, D, F,
E, B, C, A

A, D, E, F, C, B, A (REVERSE ORDER as well)

(b)

Re
-

(c)

Re
-
ordering concept

HOMEWORK
: pp. 221

-

22
2 # 2, 4, 6, 7, 10, 11

A

B

C

D

F

E

G

F

A

B

C

D

G

E

A

B

C

D

E

F

Section 6.2: Complete Graphs

EXAMPLE PUT 6 OR 8 DOTS ON WHITEBOARD AND START CONNECTING THEM AND ASK
STUD
ENTS TO DESCRIBE HOW I AM MAKING THE GRAPH:

“ONE DOT IS CONNECTED TO
EVERY OTHER DOT” TO START DEFINITION

COMPLETE GRAPH:

A

graph in which
every pair of distinct vertices

is joined by
exactly one edge
.

o

Notation
:

K
N

=

a complete graph of N vertices

EXAM
PLES OF COMPLETE GRAPHS

for 3, 4, and 5 vertices
:

Use the definition of a complete graph to answer the following questions:

1.

Does a complete graph have to be connected?

YES;
“every pair of vertices connected by an edge”

2.

Is a c
omplete graph allowed to have loops?

NO; “Distinct (Different) Vertices only”

3.

Is a complete graph allowed to have multiple edges?

NO; exactly

one edge

4.

Can different vertices have different degree?

NO; each vertex must be connected all remaining
ver
tices

What is the
DEGREE

of each
VERTEX

for N vertices?

K
3
: Degree = 2

K
4
: Degree = 3

K
5
: Degree = 4

1 Less than the # of vertices

Degree = N
-

1

How many
EDGES are

in K
N

(complete graph of N vertices)?

EDGES

TOTAL DEGREE

K
3
:
EDGES =

3

6 = 3*2

K
4
:
EDGES =

6

12 = 4*3

K
5
:
EDGES = 10

20 = 5*4

K
6
:
EDGES

=
15

30 = 6*5

What is the relationship between edges and degrees?

Euler’s Sum of Degree Theorem
:

(
total # of degrees
)

= 2 * (# of edges)

N(N
-
1) = 2 * (# of edges)

number of edges
in K
N

Where have you seen this formula before?

Chapter 1: the number of pairwise comparisons.

DEGREE and VE
RTICES are consecutive numbers.

SUMMARY OF
COMPLETE GRAPH
INFORMATION

Complete Graph

Number of Vertices

Degree of Each Vertex

Nu
mber of Edges

K
N

N

N

1

Connected Graph, No Loops, No Multiple Edges

K
3

K
4

K
5

K
3
= Complete Graph of 4 Vertices

K
4

= Complete Graph of 4 Vertices

1
)
How many Hamiltonian circuits does it
have?

2

1
)
How many Hamiltonian circuits does i
t
have?

6

If we were to answer the same questions for K
5

we would find the following:

1)

How many Hamiltonian circuits does it have?

(Do you notice any possible relation in how we answered this question for K
3
and K
4

that

(5

-

1) (5
-

2) ( 5
-

3) (5
-

4) = (4)(3)(2)(1) =
24

(see page 202 for exact circuits)

How many
DISTINCT HAMILTON CIRCUITS are

in K
N
?

Complete Graph

Vertices

Hamilton Circuits

K
3
:

3

2

K
4
:

4

6

K
5
:

5

24

K
6
:

6

120

(N

1)!
= (Degree) !
=

Di
sti
nct Hamilton Circuits in K
N
.

Calculator Command Reminder
:
[MATH]

[PRB]

[
4:
!]

I
N
DEPENDENT PRACTICE:
Apply the Properties
of
Complete Graphs

(1)

How many edges are there in K
20
?

20(19)/2 = 190 edges

(2)

How many distinct Hamilton Circuits exist in K
10
?
(10

1)! = 9! = 362880

(3)

How many vertices are in the K
50
graph?

N = 50

(4)

What is the degree of every vertex in K
200
?

Degree = 200
-
1 = 199

K
4

A

B

C

D

1

A, B, C, D, A

2

A, B, D, C, A

3

A, C, B, D, A

4

A, C, D,
B, A

5

A, D, B, C, A

6

A, D, C, B, A

K
3

A

B

C

1

A, B, C, A

2

A, C, B, A

(5)

How many distinct Hamilton Circuits exist in K
22
?

(22

1)! = 21! Or 5.109094217E19

(6)

How many edges are there in
K
32
?

Edges = (32)(32

1)/2 = 496 edges

(7)

What is the degree of every vertex in K
15
?

Degree = 15

1 = 14

(8)

If
K
N

has 120 distinct Hamiltonian circuits then what is
N
?

N = 6

Trial and Error:

Divide 120 by consecutive numbers starting with 2

t number used

= Number of vertices

(9)

If
K
N

has 55 edges then what is
N
?

N = 11

55 = (N)(N
-
1)/2;

HINT
:
square root trick of
DOUBLE EDGES

round up
= N and round down =
N
-
1

(10)

If the number of edges of
K
12

is x and the number of edges of
K
13

is y, what i
s
the
values of y
-

x
?

(13)(12)/2
-

(12)(11)/ 2 = (12/2) (13
-
11) = 12

(11)

For each of the following cases find the value of N.

a.

K
N
has
5040

distinct Hamilton Circuits.

5040 = 7!;
N =

7 + 1 = 8

b.

K
N

has 66 edges.

66 =
12*11/2; N = 12

c.

K
N

has 80,200 edges.

80,20
0 = 401*400/2; N = 401

(12)

If K
N

has
362,880 distinct Hamilton Circuits, then

362,880 = 6!; N = 7

a.

How many vertices are in the K
N

graph?

7 VERTICES

b.

What is the degree of each vertex are in the K
N

graph?

7
-
1 = 6

c.

How many edges are in the K
N

graph?
7*6/2 = 2
1 edges

Section 6.3: Traveling Salesman Problems

WEIGHTED GRAPH:

Any graph whose edges have
numbers associated or
assigned

to them

(NOT DRAWN TO SCALE ALWAYS)

o

Weights:

number
for

each edge

o

Examples of Weights
:

distance, time, cost…

Complete Weighted

Graph:
Complete graph that has weights associated with it

TRAVELING SALESMAN PROBLEMS:

From home, the traveling salesman must visit different destinations to sell goods and return home. This is
essentially finding a Hamilton Circuit.

GOAL:

Try to find t
he cheapest, most efficient, optimal route that is a Hamilton circuit for a given graph (find
the least total weight)

Exampl
e

The Simpsons:

Homer Simpson he has to run errands at the Kwik
-
E
-
Mart, the Retirement
Home and Moe`s. Assuming that he wants to

begin and end his day at home find the route that will allow
him to get back to napping as soon as possible.

The numbers will represent the
number of blocks between each
destination.

When we place values like this
on the edges of a graph we
refer to

them as the
weights

of
the edges.

What if Homer preferred going to see Grandpa first to get it out of the way?

A

B

C

D

A

A
-
B
-
D
-
C
-

A

12 + 20 + 17 + 8 = 57 BLOCKS

12+15+17+20 = 64 BLOCKS

What if Homer preferred going to the KWIK
-
E
-
MART las
t?

A
-
B
-
D
-
C
-

A

A

D

B

C

A

12+15+17+20 = 64 BLOCKS

8 +15 + 20 + 20 = 63 BLOCKS

Traveling Salesman Problem Steps:

(1)

Create graph models

(2)

A
pply

weights to these models

(3)

Find

optimal Hamilton Circuits.
(Not Always a practical statement)

REAL
-
LIFE EXAM
PLES:

1) Routing School Buses

2) Package Deliveries: UPS or FedEx Delivery Truck

3) Running Errands around Town

5) Grocery Shopping: Planning which aisles you need to go down

HOMEWORK
:
pp. 222

22
3 # 13

16

12

8

20

17

15

20

A

B

C

D

Section 6.4

and 6.5
:
Simple
Algorithms

for Solving T
raveling Salesman Problems

Strategy

#1:

BRUTE FORCE

(
Exhaustive Search
)

BRUTE
-
FORCE ALGORITHM

Step 1:

Make a list of
ALL

distinct Hamilton circuits of the graph
from
ONE
starting vertex

Step 2:

Calculate

its
TOTAL WEIGHT

for each circuit
.

Step 3:

Choose an
OPTIMAL

circuit. (The
re is always more than one optimal circuit because of mirror
-
images)

PRO:

Guarantee Optimal Hamilton Circuit

CON:

INEFFICIENT ALGORITHM

# steps needed to carry out algorithm grows disproportionately with the size of population (vertices)

ie TAKES A LOT

OF TIME the more vertices/ edges there are

STRATEGY

#2:
NEAREST

NEIGHBOR ALGORITHM

From the start vertex, go to the vertex that is the
shortest

to get to. From each new vertex go to the next new
city that is the
shortest

to get to. When there are
no more new vertices to go to, go to the starting vertex.

A
(12) C (9) B (16) D (15) A =
52

NEAREST
-
NEIGHBOR ALGORITHM

Start:

Start at the
designated starting vertex

(home)
. If there is no designated starting vertex, pick any vertex.

Step 1:

From the
starting vertex go to its
NEAREST NEIGHBOR

(vertex that is
the smallest weight

away)

Middle Steps:

From each vertex go to
its nearest neighbor

that
.

⡳浡汬e獴s

hee瀠摯楮朠瑨g猠畮s楬⁡汬⁴桥

Last Step:

From the last
neighbor (vertex)

RETURN

to the starting vertex

(home)
.

PRO:

EFFICIENT ALGORITHM

Amount of computational effort required to implement it grows in some reasonable proportion with the size of
populati
on (vertices)

ie
WORKS FAST/ LESS TIME CONSUMING TO COMPLETE METHOD

CON:

Not an optimal Algorithm; Answers won’t always be the OPTIMAL, but relatively close to being the

=
Circuit

Mirror Image

Weight

A, B, C, D, A

A, B, D, C, A

A, C, B, D, A

A, D, C, B, A

A, C, D, B, A

A, D, B, C, A

13 + 9 + 14 + 15 = 51

13 +
16 +
14 + 12 = 55

12 + 9 +
16 + 15 = 52

13

15

9

14

16

12

A

B

C

D

A weighted Graph may also be represented as table. Vertices alon
g the first
row and column. Weights on the interior of table.

A

B

C

D

A

AB =
12

AC =
20

8

B

BA =
12

BC =
20

BD =
15

C

CA =
20

CB =
20

CD =
17

D

DA =
8

DB =
15

DC =
17

Observations:

1)

Diagonal is blank because NO LOOPS exist.

2)

Upper Right
Triangle and Lower Left Triangle are Identical Numbers

a.

Represent Same Edges, but are in different order of starting and ending vertices

Draw the following complete wei
ghted graph described by table:

FIND THE NEAREST NEIGHBOR SOLUTION

STARTING AT D
:

D (1) E (6) A (3) B (5) C (6) D = 21

A

B

C

D

E

A

3

4

5

6

B

3

5

7

9

C

4

5

6

8

D

5

7

6

1

E

6

9

8

1

12

8

20

17

15

20

A

B

C

D

The following Complete Weighted Graph can also be represented by the table:

Does
BRUT
E FORCE

seem practical here?

NO;
(7
-
1)!= 720

POSSIBLE HAMILTON CIRCUITS

Start at Vertex A

Find the NEAREST NEIGHBOR SOLUTION:

(Use table or graph)

A

(15)

F

(13)

G

(28)

C

(30)

B

(60)

D

(20)
-

E

(35)
-

A

=
201

HOMEWORK
: pp. 22
3

225 # 29
-

31, 33, 35

A

B

C

D

E

F

G

A

75

50

28

35

15

22

B

75

30

60

80

65

50

C

50

30

40

48

35

28

D

28

60

40

20

30

29

E

35

80

48

20

40

32

F

15

65

35

30

40

13

G

22

50

28

29

32

13

A

G

F

E

D

B

C

A

B

C

G

F

E

D

75

22

50

13

40

20

40

30

65

50

80

35

29

15

48

35

60

28

30

32

28

9

10

17

11

13

15

A

B

C

D

Section 6.7: Repetitive Nearest
-
Neighbor Algorithm

The Nearest Neighbor Algorithm is an
EFFICIENT

but
NOT

always
ACCURATE

algorithm.

Is there a way we can improve it because we like its efficiency
?

Compare from different starti
ng spots

The
REPETITIVE

Nearest
-
Neighbor Algorithm:

P
erform the
Nearest
-
Neighbor Algorithm

for
every

vertex

of the
graph
as the
starting

vertex
.

Compare circuits and
CHOOSE

the circuit
with
least weight

of all
the options
.

o

If needed, Re
-
order
(shift
) that Hamilton Circuit to have the
DESI
GNATED START vertex first.

o

Re
-
ordered (shifted) circuits

have the same weight but change
its starting vertex

EX
P

#1:

Perform the nearest neighbor algorithm

for start of A
.

A
: A, B, D, C, A = 9 + 13 + 11+ 15 = 4
8

B
: B, A, D, C, B = 9 + 10 + 11 + 17 = 47

C:

C, D, A, B, C = 11 + 10 + 9 + 17 = 47

D:

D, A, B, C, D = 10 + 9 + 17 + 11 = 47

Final Answer: A,B,C,D,A or A,D,C,B,A = 47

EXAMPLE #2:

Label the graph and complete the Repetitive nearest neighbor
algo
rithm

for a start of
C
.

EXAMPLE #3
:

Perform repetitive nearest neighbor algorithm for a start of B.

Final
:
shift

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

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

HOMEWOR
K:

p. 225 #37

39, 41

A

B

C

D

E

A

\$185

\$119

\$152

\$133

B

\$185

\$121

\$150

\$200

C

\$119

\$121

\$174

\$120

D

\$152

\$150

\$174

\$199

E

\$133

\$200

\$120

\$199

A

B

C

D

E

F

G

A

75

50

28

35

15

22

B

75

30

60

80

65

50

C

50

30

40

48

35

28

D

28

60

40

20

30

29

E

35

80

48

20

40

32

F

15

65

35

30

40

13

G

22

50

28

29

32

13

: a,f,g,c,b,d,e,a

201 = 5+13+28+30+60+20+35

B
: b,c,g,f,a,d,e,b

214 =30+28+13+15+28+20+80

C
: c
,g,f,a,d,e,b,c

214 =28+13+15+28+20+80+30

D
:
d,e,a,f,g,c,b,d

221=20+35+15+13+28+50+60

E
: e,d,a,f,g,c,b,e

214= 20+28+15+13+28+30+80

: f,a,g,d,e,c,b,f

229 =15+22+29+20+48+30+65

: g,f,a,d,e,c,b,g

204=13+15
+28+20+48+30+50

A

B

C

D

E

ACEDBA = \$773

BCAEDB = \$722

CAEDBC = \$722 (Final)

DBCAED = \$722

Section 6.8: Cheapest
-

GOAL:

Piece together a Hamilton circuit by individual edges or “LINKS”
of graph
trying to choose the smallest or “cheapest”

weights first.

The Cheapest
-

Step #1:

Pick the edge with the
smallest weight first
. Mark the
edge (or otherwise note that you have chosen it).

o

Smallest weight ANYWHERE on the graph

Step #2:

Pick the next ‘cheapest’ edge

anywhere in the graph
. Mark
or note it.

Step #3, …, N
-
1:
Continue pic
king the ‘
cheapest’
available
edge

in
the graph

and as long as

(a)

it does not close a circuit and

(b)

it does not result in three edges coming out of a single vertex.

(DEGREE = 2)

(c)

Ties can be broken at random and don’t break the two rules

HINT: Draw out the
edge
s chosen helps

to double check what
is and is not available.

Step #N:
Connect the last two vertices to close the circuit

(you are forced to choose this edge so no need for “cheapest”)

REORDER CIRCUIT TO THE NECESSARY STARTING
VERTEX

(N Vertices = N edge
s chosen for circuit)

EXAMPLE #1:
Start at G

G

A

F

E

D

B

C

A

B

C

G

F

E

D

75

22

50

13

40

20

40

30

65

50

80

35

29

15

48

35

60

28

30

32

28

When table only, draw links to visualize circuit.

START IT AT G:

G

F

A

D

E

C

B

G
= 13 + 15 +
28 + 20 + 48 + 30 + 50 = 204

EXAMPLE #2:

Start at
B

HOMEWORK:

pp. 225
-

6 #43
-
45, 47

A

B

C

D

E

F

G

A

75

50

28

35

15

22

B

75

30

60

80

65

50

C

50

30

40

48

35

28

D

28

60

40

20

30

29

E

3
5

80

48

20

40

32

F

15

65

35

30

40

13

G

22

50

28

29

32

13

A

B

C

D

E

EXAMPLE #3:

Start at D

A

B

C

D

E

A

\$185

\$119

\$152

\$133

B

\$185

\$121

\$150

\$200

C

\$119

\$121

\$174

\$120

D

\$152

\$150

\$174

\$199

E

\$133

\$200

\$120

\$199

A

C = 119

E

C = 120

B

D = 150

A

D = 152

E

B = 200

741 = total cost

Start at D:

D, A, C, E, B, D or D, B, E, C, A, D
= 741

12

24

17

21

9

19

19

27

29

Start at B:

B, A, E, C, D, B

B, D, C, E, A, B