Chapter 2 Page 1 of 18
Chapter 2
Euler Circuits
Spring 2016 1/15/16
Learning Objectives: Students will demonstrate familiarity with the
terminology of graph theory and be able to model
real life problems using graphs; understand the
meaning and use of Euler circuits and paths and know
when an Euler circuit or path is appropriate to use in
solving a real life problem; apply Euler’s Theorems to
identify the presence of Euler circuits and paths in graphs;
find Euler circuits and paths in graphs and Eulerize
and semi-Eulerize graphs.
Chapter 2 Page 2 of 18
CHAPTER 2 - Routing Problems - find best route (Euler Circuit or Euler Path) through city
for garbage, mail, snowplow, police patrol, street sweeper, bus route, delivery van, etc.
Graph - think of it as a street map to be traveled by a delivery person that is required to travel each
street once.
A
B
C
Line &
vertex
graph
of above
G
D
E
F
Definitions:
B
A
F
E
C
D
AB, BC, DE, are edges (similar to streets). Edge length, shape is unimportant in this chapter.
EE is a special edge called a loop.
A, B, C, are each a vertex (street corner) - plural of vertex is “vertices”. Vertex F is disconnected.
A and B are adjacent to each other, A and E aren’t.
ABCA is a circuit (route that starts and ends at same vertex) BEDAB is another circuit.
ABC is a path (route that starts at one vertex and ends at another) CBE is another path.
Chapter 2 Page 3 of 18
degree = 4
B
Vertex B has a degree of “4” - 4 edges touch it. Vertex C has a degree of 3, E has a degree of 4, A
has a degree of 3. D has a degree of 2. F has a degree of 0.
Graph (street map)
A2
B3
C3
Note there is no vertex connecting BE
with AD. They don’t touch each other....
- one is passing over the top of the other.
E2
D4
C2
Graph (street map)
B3
Edge HA is a bridge
(if erased, graph
is disconnected)
H3
D3
G2
A3
H2
F3
G4
E3
Edge EF is a bridge
(if erased, graph is
disconnected)
Euler Circuit - a circuit is a route that travels on each edge of a graph once, and only once. This is
the most efficient route for a truck driver.
Euler’s Circuit Theorem - A graph will
have an Euler circuit if every vertex degree
is even (2,4,6,etc) degree
degree
K=2
R=2
A=2
C=2
B=4
J=2
K-A-B-C-J-B-R-K is an Euler circuit.
It starts at a vertex K, travels each edge
once and only once, and returns back
to start vertex K.
Note- Every vertex is even degree.
Chapter 2 Page 4 of 18
Euler Path - a route that starts at one vertex, which travels on each edge of entire graph, once and
only once, and ends at a vertex that is different than the start vertex. This is the most efficient route
for a truck driver if he/she wants to end someplace other than where he/she started.
Euler’s Path Theorem - A graph will have an Euler path
if every vertex except for (2) of them, is even (2,4,6,etc) degree.
And.....the (2) odd degree vertices will be the start and end points.
degree
K2
A2
C 2
J2
R3
B5
Odd degree
Fleury’s Algorithm - not covered this course
R-K-A-B-C-J-B-R-B is an Euler path
start at vertex R
travel every edge once and only once
return to some other vertex B
Note- Every vertex is even degree except
R and B
Chapter 2 Page 5 of 18
Solutions
3c.) Determine the degree of each vertex:
A(3)
B(2)
C(0)
X(4)
Y(3)
Z(1)
D(1)
20.) Draw a line-vertex model of map
North Royalton
A
B
C
South Royalton
Answer:
shape/length of edge unimportant
5 vertices
8 edges
A
NR
B
C
SR
29a.) Does graph have an Euler circuit, Euler path, neither, both?
2
4
2
4
4
4
4
4
2
Answer: graph has EULER CIRCUIT
all vertices are even degree.
No graph can have both.
4
4
2
Chapter 2 Page 6 of 18
29b.) Does graph have an Euler circuit, Euler path, neither, both?
Answer: NEITHER
Graph has no Euler Circuit because there are odd degree vertices
Graph has no Euler Path because it doesn’t have 2 odd degree vertices.
No graph can have both.
3
4
3
4
2
4
4
4
4
2
4
3
4
3
30a.) Does graph have an Euler circuit, Euler path, neither, both?
Answer: Graph has an Euler Path because there are 2 odd degree vertices.
Graph has no Euler Circuit because there are odd degree vertices.
No graph can have both.
2
4
4
2
8
3
4
3
6
Chapter 2 Page 7 of 18
35.) Simplified problem: find an Euler Circuit in this graph.
If there are all even degrees, start anywhere….travel each edge once.
I started here
must end here
1
12
4
5
2
9
10
11
3
8
6
7
…there are 100’s of correct answers
37.) Simplified problem: find an Euler Path: ...is there 2 odd degree vertices, and all the
others even? Yes. Must start at one odd degree vertex, and end at the other. Travel
each edge once.
odd
14
5
6
9
11 13
12
4
8
7
2
10
1
3
odd
…there are 100’s of correct answers
Chapter 2 Page 8 of 18
Optimal Eulerizing - modify a graph so it will have an euler circuit in it - add the MINIMUM
number of extra edges, so that each vertex is even degree.
Procedure for optimal eulerizing:
1.) count the degree of each vertex; identify all odd degree vertices
2.) change all odd degree vertices to even by adding minimum of extra edges (dashed lines).
- an extra edge must be a repeat of an existing edge. It can’t be an edge that doesn’t
already exist.
- length of line is unimportant - all lines are considered to be same length regardless how long
they appear to be drawn. Shape of line is unimportant – curved, wavy, straight are all same.
- this method is trial and error and it doesn’t tell you when you’ve reached the minimum number
of extra edges
Example:
A
C
B
D
step (1)
A3
B2
must change 2 vertices to even
C2
step (2)
A34
D3
B2
the dash tells you edge AD must be travelled twice
C2
D34
Graph will now have an Euler circuit but it will require travelling edge AD twice.
Chapter 2 Page 9 of 18
Example:
A
B
C
G
D
E
Step 1:
A3
B5
F
C2
must change 4 vertices to even
G4
D3
E4
F3
Step 2: change all odd degree vertices to even by adding minimum of extra edges (dashed lines).
A34
B56
C2
DF is impossible, doesn’t exist
G4
D34
E4
F34
Try again
A34
B56
C2
added 3 edges
G4
D34
E45
F34
Graph will now have an Euler circuit but it will require travelling 3 edges twice: AB, DE, and EF.
Chapter 2 Page 10 of 18
Another example: graph
step 1
5
must change 2 vertices to even
1
4
4
4
4
step 2
56
edge is impossible, doesn’t exist
12
4
4
4
4
4
4
Try again
56
12 added 5 edges
46
46
4
46
46
4
Chapter 2 Page 11 of 18
Try again
56
12
4
4
4
added 2 edges
46
Graph will now have an Euler circuit but it will require travelling 2 edges twice.
Another example
City Street Map: 17 Streets to sweep. Your job: Start at Garage, sweep all 17 streets once, and return
back into Garage. If you must travel some streets twice, ensure that is minimum number streets.
Street sweeper
GARAGE
B
C
D
H
E
I
J
F
W
Step 1
G2
B5
C3
must change 8 vertices to even
D3
H5
E3
I3
J4
F3
W3
Chapter 2 Page 12 of 18
Step 2:
G2
B56
C34
No: EH is an impossible edge….
D34
it doesn’t exist in original graph
H56
E34
I34
J4
F34
W34
try step 2 again
G2
B56
C34
D34
H56
E34
I34
J46
F34
Try step 2 again
added 5 edges (too many)
W34
G2
B6
C4
D4
H6
E4
I4
added only 4 edges
J4
F4
W4
There will be an Euler circuit in this graph but it requires you travel
4 streets (DE, BH, CI, FW dashed lines) twice.
Chapter 2 Page 13 of 18
Practice problems
1.) Optimally eulerize
see Notes
2.) Optimally eulerize
see Notes
43.) Optimally eulerize:
2
3 4
3 4
3 4
3 4
2
3 4
4
4
4
4
3 4
3 4
4
4
4
4
3 4
2
3 4
3 4
3 4
3 4
added 6 edges
2
Chapter 2 Page 14 of 18
Optimal Semi-Eulerizing Graphs - modify the graph by adding the MINIMUM number of extra
edges, so that each vertex, except for start and end, is even degree.
Procedure for optimal eulerizing:
1.) determine degree of each vertex, identify start and end vertices,
2.) change all odd degree vertices to even except for start and end, by adding minimum of extra edges
(dashed lines).
- length of line is unimportant - all lines are considered to be same length regardless how long
they appear to be drawn. Shape of line is unimportant – curved, wavy, straight are all same.
- an extra edge must be a repeat of an existing edge. It can’t be an edge that doesn’t
already exist.
- this method is trial and error and it doesn’t tell you when you’ve reached the minimum number
of extra edges
Example:
end here to go
to next neighborhood
Start here
A
B
C must be odd
G
D
E
must be odd
all others must be even
F
Chapter 2 Page 15 of 18
Step 1: determine degree of each vertex, identify start and end vertices
(end)
A3
B5
C 2 must be odd
G4
D3
E4
F3
(start) must be odd
Step 2: change all odd degree vertices to even except for start and end, by adding minimum of extra
edges (dashed lines).
A 34
B5
C 23
EC is impossible.
Not existing
G4
D 34
E 45
A 34
B58
F 34
Try again
C 23
added 4
edges
G4
D34
E45
A 34
B 56
F34
Try again
C 23
G4
D 34
added
3
edges
E 45
F 34
There will be an Euler path that starts at E and ends at C. 3 edges must be travelled twice
Chapter 2 Page 16 of 18
Example - Start truck at Warehouse (W), and end at a repair station (N).....
Optimal Semi-Eulerize the graph so that it will have
an euler path that starts at W and ends at N
A
H
B
G
C
W Warehouse
E
D
L
I
F
K
M
N repair station
Step 1: determine degree of each vertex, identify start and end vertices,
W1 start must be odd
A2
H3
B3
G4
C4
E5
D2
L2
I2
F3
K3
M2
N2 end must be odd
Step 2: change all odd degree vertices to even except for start and end, by adding minimum
of extra edges (dashed lines).
W1 must be odd
A2
H34
B34
G45
C4
E56
K34
I2
D2
added 5 edges each travelled twice
L2
F34
M 24
N 23 must be odd
Chapter 2 Page 17 of 18
Practice problems:
1.) Optimally semi-eulerize this graph with vertex C and E being start and end:
C
see Notes
E
2.) Optimally semi-eulerize this graph with vertex W and N being start and end:
W
see Notes
N
3.) Optimally semi-eulerize this graph with vertex X and Y being start and end:
Y
X
Ans.
12
12
Y odd
46
X
3
odd
45
56
Chapter 2 Page 18 of 18
4.) Optimally semi-eulerize this graph with vertex Q and R being start and end:
R
Q
Ans. It’s already optimally semi eulerized.
4
4
6
Q3
4
R3
4
50.) Optimally Semi-Eulerize this graph. Euler path must start at A and end at B
A
B
Ans.
4
2
2
3
odd A
4
3
2
3
4
4
4
3
4
4
4
4
4
4
4
4
2
odd
B 3
4
3
4
4
4
3
3
2
4
2
3
4
3
2
4
4
added 5 duplicates