04/21/23 Tucker, Sec. 2.2 1

Applied Combinatorics, 4th Ed.Alan Tucker

Section 2.2

Hamilton CircuitsPrepared by: Nathan Rounds and David Miller

04/21/23 Tucker, Sec. 2.2 2

Definitions

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

B

CD

E

F Possible Hamilton Path: A-F-E-D-B-CA B

CD

E

F

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Definitions

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

Possible Hamilton Circuit: A-F-E-D-C-B-A

A B

CD

E

F

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Rule 1

• If a vertex x has degree 2, both of the edges incident to x must be part of any Hamilton Circuit.

A B

CD

E

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

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Rule 2• No proper subcircuit, that is, a circuit not

containing all vertices, can be formed when building a Hamilton Circuit.

A B

CD

E

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

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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.

A B

CD

E

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

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Example

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

AD

E

GF

I

H

CB

J K

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Using Rules

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

H

AD

E

GF

I

CB

JK

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

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Using Rules

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

AD

E

GF

I

H

CB

J K

•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.

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Using Rules

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

AD

E

GF

I

CB

J

H

K

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.

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Using Rules

• Rule 2 tells us that no Hamilton Circuit exists.

AD

E

GF

I

CB

J

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.

H

K

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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

B

C

DE

F

n = 6 n/2 = 3 Possible Hamilton Circuit: A-B-E-D-C-F-A

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However, not “if and only if”

E

B

CD

FA B

CD

F 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.

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Theorem 2

• Let G be a connected graph with n vertices, and let the vertices be indexed x1,x2,…,xn, so that deg(xi) deg(xi+1).

• If for each k n/2, either deg(xk) > k or deg(xn-k) n-k, then G has a Hamilton Circuit.

n/2 = 3

k = 3,2,or 1 Possible Hamilton Circuit: X1-X5-X3-X4-X2-X6-X1

X5

X1

X6

X3

X4

X2

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Theorem 3

• Suppose a planar graph G, has a Hamilton Circuit H.

• Let G be drawn with any planar depiction.

• Let ri denote the number of regions inside the Hamilton Circuit bounded by i edges in this depiction.

• Let be the number of regions outside the circuit bounded by i edges. Then numbers ri and satisfy the following equation.

'ir

'ir

'( 2)( ) 0i ii

i r r

04/21/23 Tucker, Sec. 2.2 16

Use of Theorem 3'( 2)( ) 0i i

i

i r r 4

4 4

6 6

6

6

6

6

Planar Graph G

' '4 4 6 62( ) 4( ) 0r r r r

No matter where a Hamilton Circuit is drawn (if it exists), we know that and

. Therefore, and must have the same parity and

.

'4 4 3r r

'6 6 6r r r 'r

'4 4| | 3r r

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Use of Theorem 3 Cont’d

'4 4 0r r

Eq. (*)

•Consider the case .

•This is impossible since then the equation would require that which is impossible since .

•We now know that , and therefore .

•Now we cannot satisfy Eq. (*) because regardless of what possible value is taken on by , it cannot compensate for the other term to make the equation equal zero.

•Therefore, no Hamilton Circuit can exist.

'6 6 0r r

'4 4 3r r

'6 6| | 2r r '

6 6| 4( ) | 8r r

'4 42( )r r

' '4 4 6 62( ) 4( ) 0r r r r

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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 B

C D

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.

(K4, with arrows)

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Definition

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

A=000 B=100

C=110D=010

F=011G=111

H=101I=001

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

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Class Exercise• Find a Hamilton Circuit, or prove that one

doesn’t exist.

A B C

D E

F G H

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.

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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

F G H