7.2 Eulerian Paths
Definition 1. An Eulerian path is one that uses every edge in a graph exactly once. An
Eulerian circuit is an Eulerian path that is a cycle, i.e. it ends at the same vertex that it
starts at.
Not every graph has an Eulerian path or circuit. There is a simple criterion for
determining if a graph has an Eulerian path or circuit and in this section we discuss an
algorithm to find a one if there is one.
Example 1. A company has offices in the
six cities that are the vertices in the graph at
the right. These are connected by a
computer network that has direct data links
between some of the offices. These are
represented by the edges in the graph below.
Grand
Rapids
Chicago
East
Lansing
Battle
Creek
Detroit
Every five minutes the server in the home
office of Detroit does a test of the network
by sending out a test message that will go
through every direct link. The message can
Indianapolis
specify the sequence of links in order that
the message will travel. Is there a way it can traverse each link exactly once and start and
end in Detroit? If so, what is a the sequence in edges in this route?
This is an example of a situation in which one wants to find an Eulerian circuit in a graph.
First, let's discuss the criterion for determining if a graph has an Eulerian circuit or path.
B
Consider a path that doesn't use any edge more
than once. It might not use some edges, but it
should not use an edge more than once. At the
right is a typical path of this sort in some graph.
Let's count the number of times, d, the path
touches each vertex. We get the table of values
at the right.
A
C
H
F
G
D
Note that for each vertex except the start and finish the path touches
the vertex an even number of times. This is because every time the
graph arrives at the vertex it also leaves the vertex, which produces
and even count. At the starting and finishing vertex it touches the
vertex an odd number of times. This is because every time the graph
arrives at the vertex it also leaves the vertex except for the start or
finishing edge which has no companion edge. This produces an odd
count.
7.2 - 1
E
Vertex
A
B
C
D
E
F
G
H
d
1
1
6
2
2
4
4
2
If the path were a cycle then it would touch every vertex an even number of times.
There is some terminology related to this.
Definition 2. The degree of a vertex in a graph is the number of edges touching it.
The argument above proves the following Theorem.
Theorem 1. If a graph has an Eulerian circuit then every vertex has even degree. If a
graph has an Eulerian path that is not a circuit, then every vertex except two have even
degree. The Eulerian path must start and end at the vertices with odd degree.
If one looks at the graph in Example 1, then the vertices have
the following degrees. So there is no Eulerian circuit. If there
is an Eulerian path it would start and end in Detroit and Battle
Creek.
The following converse of Theorem says there is an Eulerian
path or circuit if the condition in Theorem 1 is satisfied.
Vertex
Chicago
Grand Rapids
Battle Creek
Indianapolis
East Lansing
Detroit
Degree
2
4
5
4
4
3
Definition 3. A graph is connected if there is a path between every pair of vertices.
Theorem 2. Suppose a graph is connected. If every vertex has even degree there is an
Eulerian circuit. If every vertex except two has even degree there is an Eulerian path.
Proof. First consider the case where every vertex has even degree.
We pick any vertex, call it v, and start out from that vertex traversing edges making sure
that we don't traverse any edge more than once. Continue on until we get stuck, i.e. reach
a vertex where there are no more unused edges. This creates a certain path p.
Claim. We must be stuck at the vertex v that you started at. The reason for this is that
the path p uses up an even number of edges at every vertex except at the start and finish if
they are different. So if the start and finish were different then an odd number of edges
would be used up at the finish. Since every vertex has even degree we would not be
stuck at the finish. So we must be stuck at the vertex v that we started at. In other words
the path p is a circuit.
If this path uses every edge we are done. If not we extend this initial path as follows.
First note that the initial circuit touches every vertex an even number of times. This
leaves an even number of unusded edges at every vertex. Pick an unused edge that
touches some vertex w on the original circuit and start out from that vertex traversing
unused edges making sure that we don't traverse any edge more than once. Continue on
until we get stuck again, i.e. reach a vertex where there are no more unused edges. This
7.2 - 2
creates another path q. By the same arugment q must be a circuit, i.e. it must return to the
vertex w that we started from.
Now splice the second circuit q into the original circuit p by starting out at the starting
vertex v of the original path p and going until we reach w. Then traverse the second
circuit q until we come back to w. Then continue on the original circuit p until we come
back to v. This creates a larger circuit that uses all the edges of both the original cirucit p
and the second circuit q.
We can continue in this fashion until we have used up all the edges and have created an
Eulerian circuit.
Now consider the case where every vertex except two have even degree. The proof of
this case case is very similar to the first case. This time we start out at one of the vertices
v with odd degree and traverse edges until we get stuck. An argument similar to the
above shows we must be stuck at the other vertex of odd degree. If this doesn't use up all
the edges, then the unused edges much touch every vertex an even number of times. At
this point the argument is the same as the first case. //
Grand
Rapids
Example 2. Consider the graph in Example 1.
Let's find an Eulerian path the starts in
Detroit and ends in Battle Creek.
If we start out in Detroit and traverse edges
until we get stuck, one possibility is
(1)
Battle
Creek
Chicago
Detroit
D – BC – EL – GR – BC – C – I – BC
where D = Detroit, BC = Battle Creek, etc.
This is pictured at the right.
Indianapolis
The unused edges are pictured at the right
below the initial path. Note that the unused
edges touch every vertex an even number of
times. Now we can start out at East Lansing
and traverse unused edges giving the circuit
(2)
East
Lansing
Grand
Rapids
Chicago
EL – C - GR – I – D - EL
We splice the circuit (2) into the path (1) by
starting at Detrot, traversing the path (1) until
we get to East Lansing, then following the
circit (2) until we get back to Lansing and
then following the remainder of the circuit (1).
East
Lansing
Battle
Creek
Indianapolis
This gives the following Eulerian path for the original graph in Example 1.
7.2 - 3
Detroit
D – BC – EL – C - GR – I – D - EL – GR – BC – C – I – BC
Grand
Rapids
East
Lansing
It is pictured at the right.
Battle
Creek
Chicago
Detroit
Directed Graphs. In the above we have
been assuming the graphs are undirected.
The same problem of finding an Eulerian path
or circuit is of interest in a directed graph
Indianapolis
where in the path or circuit we must traverse
each edge in its given direction. Pretty much the same discussion goes through.
However, now it is necessary to distinguish an edge going into a vertex from an edge
going leaving a vertex. This leads to the following modified definition of the degree of a
vertex.
Definition 3. Consider a vertex v in a directed graph.
In(v) = the incoming degree of v = the number of edges coming into v
Out(v) = the outgoing degree of v = the number of edges going out of v
With this definition we have the following analogue of Theorems 1 and 2.
Theorem 3. Suppose a directed graph is connected. Then there is an Eulerian circuit if
and only if In(v) = Out(v) for every vertex v. There is an Eulerian path if and only if the
following is true.
i.
Out(s) = In(s) + 1 for some vertex s.
ii.
In(f) = Out(f) + 1 for some vertex f.
i.
Out(v) = In(v) for every vertex v other than s and f.
The proof is similar to the proofs of Theorems 1 and 2.
Example 3 (an Analogue to Digital Converter). One has a dial whose rim turns and
whose center is stationary. The center
0
1
is conducting and the rim is divided up
into eight segments, some of which are
1
0
conducting and some are not. Thus
there are eight different positions for
1
0
1
1
the dial.
0
0
+
There is a wire leading from a power
supply to the center and there are wires
- power supply
7.2 - 4
1
leading from three positions adjcent to three of the positions on the rim on the upper right.
The center touches the rim and the wires on the upper right touch whatever three
segments in the rim happen to be next to the wires. So one of three wires carries a high
voltage (which represents a 1 bit) if it happens to be touching a conducting segment in the
rim. Otherwise the wire is grounded (and represents a 0 bit). Thus for each dial position
the three wires represent a three bit sequence.
We would like to adjust the conducting segments on the rim so each dial position has a
different corresponding three bit seqence in the wires.
In the configuration above this is not true, since the three bit sequences corresponding to
the eight positions on the dial are
101, 010, 001, 100, 110, 011, 101, 010
Here we start with dial in the current position and rotate the dial counter-clockwise. We
read the three bits on the dial counter clockwise so we are reading what is displayed on
the right.
Note that 101 and 010 appear twice while 000 and 111 don't appear, so this assignment of
conducting segments on the rim is not what we want.
There is a solution to this problem involving Eulerian circuits. We make the following
directed graph. There are four vertices corresponding to the four two bit sequences 00,
01, 10 and 11. There are eight edges labeled with the eight three bit sequences
000, 001, 010, 011, 100, 101, 110, 111
01
011
001
The edge xyz goes from the vertex xy to the vertex yz.
The graph looks like this.
000 00
101
010
11
111
A path corresponds to a way of labeling the segments
100
110
on the rim. We start out with the three bits of the first
10
edge and then add on the left most bit of each
successive edge. An Eulerian circuit corresponds to the desired way of labeling the
segments on the rim.
This graph meets the condition for an
Eulerian circuit since each vertex has
incoming and outgoing degree 2. It is
not hard to construct an Eulerian circuit.
One possibility is
1
0
1
0
1
0
0
000, 001, 010, 101, 011, 111, 110, 100
1
+
- power supply
where we have used the edge labels.
The corresponding dial is at the right.
7.2 - 5
0
0
0
Hamiltonean Paths and Circuits.
Definition 4. A Hamiltonean path is one that visits every vertex in a graph exactly once.
A Hamiltonean circuit is a Hamiltonean path that is a cycle, i.e. it ends at the same vertex
that it starts at.
Not every graph has a Hamiltonean path or circuit. For
example, the graph at the right doesn't have a Hamiltonean
path since it would have to include each of the four edges in
order that the path visit each of the vertices A, B, C and D.
However, this would mean that it would visit E more than
once.
Unfortunately there is no simple criterion for determining if
a graph has an Hamiltonean path or circuit.
7.2 - 6
B
A
E
D
C