th

### Alan Tucker

Section 2.2

Hamilton Circuits Prepared by: Nathan Rounds and David Miller 1 4/30/2020 Tucker, Sec. 2.2

### Definitions

Hamilton Path

– A path that visits each vertex in a graph exactly once.

A Possible Hamilton Path: A-F-E-D-B-C 4/30/2020 Tucker, Sec. 2.2

2

### Definitions

Hamilton Circuit

– A circuit that visits each vertex in a graph exactly once.

F A B Possible Hamilton Circuit: A-F-E-D-C-B-A E 4/30/2020 D C Tucker, Sec. 2.2

3

### Rule 1

• If a vertex

x

has degree 2, both of the edges incident to

x

must be part of any Hamilton Circuit.

F A B Edges FE and ED must be included in a Hamilton Circuit if one exists.

D E 4/30/2020 C Tucker, Sec. 2.2

4

### Rule 2

• No proper subcircuit, that is, a circuit not containing all vertices, can be formed when building a Hamilton Circuit.

F A B Edges FE , FD , and DE cannot all be used in a Hamilton Circuit.

D E 4/30/2020 C Tucker, Sec. 2.2

5

### Rule 3

• Once the Hamilton Circuit is required to use two edges at a vertex

x

, all other (unused) edges incident at

x

can be deleted.

F A B If edges FA and FE are required in a Hamilton Circuit, then edge FD can be deleted in the circuit building process.

D E 4/30/2020 C Tucker, Sec. 2.2

6

### Example

• Using rules to determine if either a Hamilton Path or a Hamilton Circuit exists.

B D A C 4/30/2020 F J I E G Tucker, Sec. 2.2

H K 7

### Using Rules

• Rule 1 tells us that the red edges must be used in any Hamilton Circuit.

B D A C Vertices A and G are the only vertices of degree 2.

E G H F I J 4/30/2020 K Tucker, Sec. 2.2

8

### Using Rules

• Rules 3 and 1 advance the building of our Hamilton Circuit.

F B D A E G C H •Since the graph is symmetrical, it doesn’t matter whether we use edge IJ or edge IK . •If we choose IJ , Rule 3 lets us eliminate IK making K a vertex of degree 2.

•By Rule 1 we must use HK and JK .

I J 4/30/2020 K Tucker, Sec. 2.2

9

### Using Rules

• All the rules advance the building of our Hamilton Circuit.

B D A C E Rule 2 allows us to eliminate edge EH and Rule 3 allows us to eliminate FJ . Now, according to Rule 1, we must use edges BF , FE , and CH .

G F H I J 4/30/2020 K Tucker, Sec. 2.2

10

### Using Rules

• Rule 2 tells us that no Hamilton Circuit exists.

B D A F E G C H Since the circuit A-C-H-K-J-I-G-E-F-B-A that we were forced to form does not include every vertex (missing D ), it is a subcircuit. This violates Rule 2.

I J 4/30/2020 K Tucker, Sec. 2.2

11

### Theorem 1

• A graph with

n

vertices,

n

> 2, has a Hamilton circuit if the degree of each vertex is at least

n

/2.

A C B

n

= 6

n

/2 = 3 Possible Hamilton Circuit: A B-E-D-C-F-A E F D 4/30/2020 Tucker, Sec. 2.2

12

E 4/30/2020

### However, not “if and only if”

A Theorem 1 does not necessarily have to be true in order for a Hamilton Circuit to exist. Here, each vertex is of degree 2 which is less than

n

/2 and yet a Hamilton Circuit still exists.

Tucker, Sec. 2.2

13

### Theorem 2

X 2 • Let

G

be a connected graph X 4 with

n

vertices, and let the vertices be indexed

x 1

,

x 2

,…,

x n ,

so that deg(

x i

 deg(

x i+

1 ).

• If for each

k

deg(

x k

) > k then

G

n

/2, either or deg(

x n

-

k

has a Hamilton 

n

-

k

, X 3 Circuit.

X 6 X 5

n

/2 = 3

k

= 3,2,or 1 Possible Hamilton Circuit: X 1 -X 5 -X 3 -X 4 -X 2 -X 6 -X 1 X 1 4/30/2020 Tucker, Sec. 2.2

14

### Theorem 3

• Suppose a planar graph

G

, has a Hamilton Circuit

H

.

• Let

G

be drawn with any planar depiction.

• Let

r i

denote the number of regions inside the Hamilton Circuit bounded by

i

edges in this depiction. bounded by

i

edges. Then numbers

r i

following equation.

r i

' and satisfy the 

i

(

i

 2)(

r i

r i

' )  0 4/30/2020 Tucker, Sec. 2.2

15

Planar Graph

G

4 6 4 6

### Use of Theorem 3

i

(

i

 2)(

r i

r i

' )  0 6 2(

r

4 

r

4 ' )  4(

r

6 

r

6 ' )  0 6 4 6 6 No matter where a Hamilton Circuit is drawn (if it exists), we

r

4 

r

4  3

r

6

r

6 6

r

| must have the same parity and

r

4

r

4 | 3 .

r

' 4/30/2020 Tucker, Sec. 2.2

16

### Use of Theorem 3 Cont’d

2(

r

4 

r

4 ' )  4(

r

6 

r

6 ' )  0 Eq. (*)

r

6 6 0

r

4 

r

4  3 |

r

6 

r

6 |  2 | 4(

r

6 

r

6 '

r

4   4 0 •Now we cannot satisfy Eq. (*) because regardless of what possible value is taken 2(

r

4 

r

4 equation equal zero.

' ) •Therefore, no Hamilton Circuit can exist.

4/30/2020 Tucker, Sec. 2.2

17

### Theorem 4

• • Every tournament has a directed Hamilton Path.

Tournament

– A directed graph obtained from a (undirected) complete graph, by giving a direction to each edge.

A 4/30/2020 C B The tournaments (Hamilton Paths) in this graph are: A-D-B-C, B-C-A-D, C-A-D-B, D-B-C-A, and D-C-A-B .

(K 4 , with arrows) D Tucker, Sec. 2.2

18

### Definition

Grey Code

uses binary sequences that are almost the same, differing in just one position for consecutive numbers.

F=011 I=001 H=101 D=010 G=111 C=110 Advantages for using Grey Code: -Very useful when plotting positions in space.

-Helps navigate the Hamilton Circuit code.

Example of an Hamilton Circuit: 000-100-110-010-011-111-101-001-000 B=100 A=000 4/30/2020 Tucker, Sec. 2.2

19

### Class Exercise

• Find a Hamilton Circuit, or prove that one doesn’t exist.

Rule’s: •If a vertex

x

has degree 2, both of the edges incident to

x

must be part of any Hamilton Circuit.

•No proper subcircuit, that is, a circuit not containing all vertices, can be formed when building a Hamilton Circuit.

•Once the Hamilton Circuit is required to use two edges at a vertex

x

, all other (unused) edges incident at

x

can be deleted.

A F D B G E C H 4/30/2020 Tucker, Sec. 2.2

20

### Solution

• By Rule One, the red edges must be used • Since the red edges form subcircuits, Rule Two tells us that no Hamilton Circuits can exist.

A B C D E H F G Tucker, Sec. 2.2

4/30/2020 21