Traveling-Salesman Problem

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# Traveling-Salesman Problem - PowerPoint PPT Presentation

Traveling-Salesman Problem. Ch. 6. Hamilton Circuits. Euler circuit/path => Visit each edge once and only once Hamilton circuit => Visit each vertex once and only once (except at the end, where it returns to the starting vertex) Hamilton path => Visit each vertex once and only once

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### Traveling-Salesman Problem

Ch. 6

Hamilton Circuits
• Euler circuit/path => Visit each edge once and only once
• Hamilton circuit => Visit each vertex once and only once (except at the end, where it returns to the starting vertex)
• Hamilton path => Visit each vertex once and only once
• Difference: Edge (Euler)  Vertex (Hamilton)

A

B

E

D

C

Examples of Hamilton circuits
• Has many Hamilton circuits:
• A, B, C, D, E, A
• A, D, C, E, B, A
• Has many Hamilton paths:
• A, B, C, D, E
• A, D, C, E, B
• Has no Euler circuit, no Euler path => 4 vertices of odd degree

Graph 1

Hamilton circuits can be shortened into a Hamilton path

by removal of the last edge

Examples of Hamilton circuits

A

B

• Has no Hamilton circuits:
• What ever the starting point, we are going to have to pass through vertex E more than once to close the circuit.
• Has many Hamilton paths:
• A, B, E, C, D
• C, D, E, A, B
• Has Euler circuit => each vertex has even degree

E

D

C

Graph 2

Examples of Hamilton circuits

F

A

B

• Has many Hamilton circuits:
• A, F, B, E, C, G, D, A
• A, F, B, C, G, D, E, A
• Has many Hamilton paths:
• A, F, B, E, C, G, D
• A, F, B, C, G, D, E
• Has Euler circuit => Every vertex has even degree

E

D

C

G

Graph 3

Examples of Hamilton circuits

G

F

Has no Hamilton circuits:

Has no Hamilton paths:

Has no Euler circuit

Has no Euler path => more than 2 vertices of odd degree

A

B

E

D

C

I

H

Graph 4

A

B

D

C

Complete graph
• A graph with N vertices in which every pair of vertices is joined by exactly one edge is called the complete graph.
• Total no. of edges = N(N-1)/2

In K4, each vertex

has degree 3 and

the number of

edges = 4 (3)/2 = 6

The six Hamilton circuits of K4

A

B

D

C

Rows => 6 Hamilton circuits

Cols=> same Hamilton circuit with different reference points

Graph

Reference point A

Reference point B

Reference point C

Reference point D

Complete graph
• The number of Hamilton circuits in a complete graph can be computed by using factorials.
• N! (factorial of N) = 1x 2x3x4x … x(N-1)x N
• The complete graph with N vertices has

(N-1)! Hamilton circuits.

• Example: The complete graph with 5 vertices has 4! = 1x2x3x4 = 24 Hamilton circuits
Factorial

Which of the following is true?

n! = n! x (n-1)!

n! = n! + (n-1)!

n! = n x (n-1)!

n! = n + (n-1)!

No. of edges

No of edges in K10 is

• 10
• 10!
• 90
• 45
Complete graph

In a complete graph with 14 vertices (A through N), the total number of Hamilton circuits (including mirror-image circuits) that start at vertex A is

• 14!
• (14x13)/2
• 15!
• 13!