Chapter 10.5 Euler and Hamilton Paths Slides by Gene Boggess
Computer Science Department Mississippi State University Based on
Discrete Mathematics and Its Applications, 7 th ed., by Kenneth H.
Rosen, published by McGraw Hill, Boston, MA, 2011. Modified and
extended by Longin Jan Latecki [email protected] 1
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Euler Paths and Circuits The Seven bridges of Knigsberg,
Prussia (now called Kaliningrad and part of the Russian republic) a
b c d A B C D The townspeople wondered whether it was possible to
start at some location in the town, travel across all the bridges
once without crossing any bridge twice, and return to the starting
point. The Swiss mathematician Leonhard Euler solved this problem.
His solution, published in 1736, may be the first use of graph
theory. 2
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Euler Paths and Circuits An Euler path is a path using every
edge of the graph G exactly once. An Euler circuit is an Euler path
that returns to its start. A B C D Does this graph have an Euler
circuit? No. 3
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Necessary and Sufficient Conditions How about multigraphs? A
connected multigraph has an Euler circuit iff each of its vertices
has an even degree. A connected multigraph has an Euler path but
not an Euler circuit iff it has exactly two vertices of odd degree.
We will first see some examples before we discuss the algorithm.
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Example Which of the following graphs has an Euler circuit? e d
a c b e d a c b e c a d b yes no no (a, e, c, d, e, b, a) 5
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Example Which of the following graphs has an Euler path? e d a
c b e d a c b e c a d b yes noyes (a, e, c, d, e, b, a ) (a, c, d,
e, b, d, a, b) 6
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Euler Circuit in Directed Graphs NO(a, g, c, b, g, e, d, f,
a)NO 7
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Euler Path in Directed Graphs NO (a, g, c, b, g, e, d, f, a)
(c, a, b, c, d, b) 8
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THEOREM 1. A connected multigraph with at least two vertices
has an Euler circuit if and only if each of its vertices has even
degree. Proof: We first show that if a connected graph has an Euler
circuit, then every vertex must have even degree. To do this, first
note that an Euler circuit begins with a vertex a and continues
with an edge incident with a, say {a, b}. The edge {a, b}
contributes one to deg(a). Each time the circuit passes through a
vertex it contributes two to the vertexs degree, because the
circuit enters via an edge incident with this vertex and leaves via
another such edge. Finally, the circuit terminates where it
started, contributing one to deg(a). Therefore, deg(a) must be
even, because the circuit contributes one when it begins, one when
it ends, and two every time it passes through a (if it ever does).
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THEOREM 1. A connected multigraph with at least two vertices
has an Euler circuit if and only if each of its vertices has even
degree. The above proof is constructive and leads to the following:
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APPLICATIONS OF EULER PATHS AND CIRCUITS Euler paths and
circuits can be used to solve many practical problems. For example,
many applications ask for a path or circuit that traverses each
street in a neighborhood, each road in a transportation network,
each connection in a utility grid, or each link in a communications
network exactly once. Finding an Euler path or circuit in the
appropriate graph model can solve such problems. For example, if a
postman can find an Euler path in the graph that represents the
streets the postman needs to cover, this path produces a route that
traverses each street of the route exactly once. If no Euler path
exists, some streets will have to be traversed more than once. The
problem of finding a circuit in a graph with the fewest edges that
traverses every edge at least once is known as the Chinese postman
problem in honor of Guan Meigu, who posed it in 1962. 14
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Hamilton Paths and Circuits A Hamilton path in a graph G is a
path which visits every vertex in G exactly once. A Hamilton
circuit is a Hamilton path that returns to its start. 15
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Hamilton Circuits Is there a circuit in this graph that passes
through each vertex exactly once? Dodecahedron puzzle and it
equivalent graph 16
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Dodecahedron is a polyhedron with twelve flat faces 17
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Hamilton Circuits Yes; this is a circuit that passes through
each vertex exactly once. 18
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Finding Hamilton Circuits Which of these three figures has a
Hamilton circuit? Or, if no Hamilton circuit, a Hamilton path?
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Finding Hamilton Circuits G 1 has a Hamilton circuit: a, b, c,
d, e, a G 2 does not have a Hamilton circuit, but does have a
Hamilton path: a, b, c, d G 3 has neither. 20
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Finding Hamilton Circuits Unlike the Euler circuit problem,
finding Hamilton circuits is hard. There is no simple set of
necessary and sufficient conditions, and no simple algorithm.
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Properties to look for... No vertex of degree 1 If a node has
degree 2, then both edges incident to it must be in any Hamilton
circuit. No smaller circuits contained in any Hamilton circuit (the
start/endpoint of any smaller circuit would have to be visited
twice). 22
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Show that neither graph displayed below has a Hamilton circuit.
There is no Hamilton circuit in G because G has a vertex of degree
one: e. Now consider H. Because the degrees of the vertices a, b,
d, and e are all two, every edge incident with these vertices must
be part of any Hamilton circuit. No Hamilton circuit can exist in
H, for any Hamilton circuit would have to contain four edges
incident with c, which is impossible. 23
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A Sufficient Condition Let G be a connected simple graph with n
vertices with n 3. If the degree of each vertex is n/2, then G has
a Hamilton circuit. 24
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Travelling Salesman Problem A Hamilton circuit or path may be
used to solve practical problems that require visiting vertices,
such as: road intersections pipeline crossings communication
network nodes A classic example is the Travelling Salesman Problem
finding a Hamilton circuit in a complete graph such that the total
weight of its edges is minimal. 25
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The best algorithms known for finding a Hamilton circuit in a
graph or determining that no such circuit exists have exponential
worst-case time complexity (in the number of vertices of the
graph). Finding an algorithm that solves this problem with
polynomial worst-case time complexity would be a major
accomplishment because it has been shown that this problem is
NP-complete. Consequently, the existence of such an algorithm would
imply that many other seemingly intractable problems could be
solved using algorithms with polynomial worst-case time complexity.
Time Complexity 26
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Summary PropertyEulerHamilton Repeated visits to a given node
allowed? YesNo Repeated traversals of a given edge allowed? No
Omitted nodes allowed?No Omitted edges allowed?NoYes 27