Applied Combinatorics, 4th Ed.
Alan Tucker
Section 2.2
Hamilton Circuits
Prepared by:
Nathan Rounds and
David Miller
4/13/2015
Tucker, Sec. 2.2
1
Definitions
• Hamilton Path – A path that visits each
vertex in a graph exactly once.
F
Possible Hamilton Path:
A
D
B
A-F-E-D-B-C
C
E
4/13/2015
Tucker, Sec. 2.2
2
Definitions
• Hamilton Circuit – A circuit that visits each
vertex in a graph exactly once.
F
A
D
Possible Hamilton Circuit:
B
A-F-E-D-C-B-A
C
E
4/13/2015
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
D
B
Edges FE and ED must be
included in a Hamilton Circuit if
one exists.
C
E
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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
D
B
Edges FE, FD, and DE
cannot all be used in a
Hamilton Circuit.
C
E
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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
D
B
If edges FA and FE are
required in a Hamilton
Circuit, then edge FD can be
deleted in the circuit building
process.
C
E
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Tucker, Sec. 2.2
6
Example
• Using rules to determine if either a
Hamilton Path or a Hamilton Circuit exists.
A
B
D
C
E
G
F
H
I
J
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K
Tucker, Sec. 2.2
7
Using Rules
• Rule 1 tells us that the red edges must be
used in any Hamilton Circuit.
A
B
D
Vertices A and G are the only
vertices of degree 2.
C
E
G
F
I
H
K
J
4/13/2015
Tucker, Sec. 2.2
8
Using Rules
• Rules 3 and 1 advance the building of our
Hamilton Circuit.
A
B
D
C
•If we choose IJ, Rule 3 lets us eliminate
IK making K a vertex of degree 2.
E
G
F
•Since the graph is symmetrical, it doesn’t
matter whether we use edge IJ or edge IK.
H
•By Rule 1 we must use HK and JK.
I
J
4/13/2015
K
Tucker, Sec. 2.2
9
Using Rules
• All the rules advance the building of our
Hamilton Circuit.
A
B
D
C
E
G
F
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.
H
I
J
4/13/2015
K
Tucker, Sec. 2.2
10
Using Rules
• Rule 2 tells us that no Hamilton Circuit
exists.
A
B
D
C
E
G
F
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
I
J
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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
n=6
n/2 = 3
Possible Hamilton Circuit: A-
B
B-E-D-C-F-A
E
D
4/13/2015
F
Tucker, Sec. 2.2
12
However, not “if and only if”
F
A
D
B
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.
C
E
4/13/2015
Tucker, Sec. 2.2
13
Theorem 2
X2
• Let G be a connected graph X4
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
X3
deg(xk) > k or deg(xn-k)  n-k,
then G has a Hamilton
Circuit.
4/13/2015
Tucker, Sec. 2.2
X5
X6
X1
n/2 = 3
k = 3,2,or 1
Possible Hamilton Circuit:
X1-X5-X3-X4-X2-X6-X1
14
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 ri be the number of regions outside the circuit
bounded by i edges. Then numbers ri and ri ' satisfy the
following equation.
'
(
i

2)(
r

r

i
i)0
i
4/13/2015
Tucker, Sec. 2.2
15
Use of Theorem 3
Planar Graph G
'
(
i

2)(
r

r

i
i)0
4
6
i
6
2(r4  r4' )  4(r6  r6' )  0
6
No matter where a Hamilton
Circuit is drawn (if it exists), we
know that r4  r4'  3 and
6
4
6
r6  r6'  6.
4
Therefore, r and r
must have the same parity and
6
4/13/2015
Tucker, Sec. 2.2
| r4  r4' | 3 .
16
'
Use of Theorem 3 Cont’d
2(r4  r4' )  4(r6  r6' )  0
Eq. (*)
•Consider the case r6  r6'  0.
•This is impossible since then the equation would require that
is impossible since r4  r4'  3.
•We now know that | r6
r4  r4'  0 which
 r6' | 2, and therefore | 4(r6  r6' ) | 8 .
•Now we cannot satisfy Eq. (*) because regardless of what possible value is taken
on by 2(r4  r4' ) , it cannot compensate for the other term to make the
equation equal zero.
•Therefore, no Hamilton Circuit can exist.
4/13/2015
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
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.
C
4/13/2015
(K4, 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.
Advantages for using Grey Code:
F=011
I=001
G=111
H=101
-Very useful when plotting positions in space.
-Helps navigate the Hamilton Circuit code.
Example of an Hamilton Circuit:
D=010
C=110
A=000
4/13/2015
000-100-110-010-011-111-101-001-000
B=100
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.
4/13/2015
A
B
D
F
Tucker, Sec. 2.2
C
E
G
H
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
F
4/13/2015
E
G
Tucker, Sec. 2.2
H
21