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)
Examples of Hamilton circuits
A
B
E
D
C
Graph 1
Has many Hamilton circuits:
1) A, B, C, D, E, A
2) A, D, C, E, B, A
Has many Hamilton paths:
1) A, B, C, D, E
2) A, D, C, E, B
Has no Euler circuit, no Euler path
=> 4 vertices of odd degree
Hamilton circuits can be shortened into a Hamilton path
by removal of the last edge
Examples of Hamilton circuits
A
B
E
D
C
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.
Graph 2
Has many Hamilton paths:
• A, B, E, C, D
• C, D, E, A, B
Has Euler circuit => each vertex
has even degree
Examples of Hamilton circuits
F
A
B
E
D
C
G
Graph 3
Has many Hamilton circuits:
1) A, F, B, E, C, G, D, A
2) A, F, B, C, G, D, E, A
Has many Hamilton paths:
1) A, F, B, E, C, G, D
2) A, F, B, C, G, D, E
Has Euler circuit => Every vertex
has even degree
Examples of Hamilton circuits
G
F
A
Has no Hamilton circuits:
B
Has no Hamilton paths:
E
D
Has no Euler circuit
C
I
H
Graph 4
Has no Euler path => more than
2 vertices of odd degree
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
A
B
D
C
In K4, each vertex
has degree 3 and
the number of
edges = 4 (3)/2 = 6
The six Hamilton circuits of K4
A
B
Graph
D
Rows => 6 Hamilton circuits
Cols=> same Hamilton circuit with
different reference points
C
Reference point A Reference point B Reference point C
A,B,C,D,A
A,B,D,C,A
A,C,B,D,A
A,C,D,B,A
A,D,B,C,A
A,D,C,B,A
B,C,D,A,B
B,D,C,A,B
B,D,A,C,B
B,A,C,D,B
B,C,A,D,B
B,A,D,C,B
C,D,A,B,C
C,A,B,D,C
C,B,D,A,C
C,D,B,A, C
C,A,D,B,C
C,B,A,D,C
Reference point D
D,A,B,C,D
D,C,A,B, D
D,A,C,B,D
D,B,A,C,D
D,B,C,A,D
D,C,B,A,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!