we have an Euler circuit, we must add edges connecting the odd valenced vertices so that at the
end all vertices have even valence. The most economic way to do this is to join the 6 odd valenced
vertices in pairs. We need three new edges for this. The correct answer is
(d)
.
✷
Question 9.
Label the consequtive edges of an Euler circuit in the following graph:
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View Full DocumentAnswer 9.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
✷
Question 10.
True or False. Give a reason or a counterexample.
(1)
There exists a connected planar graph with 6 vertices, 12 edges, and 6 faces.
(2)
A planar graph consisting of 3 connected pieces must have Euler characteristic 3.
(3)
If a connected planar graph has 8 edges and 3 vertices, then it splits the plane into 7 regions.
(a)
(1)
(2)
(3)
T
T
T
(b)
(1)
(2)
(3)
T
F
T
(c)
(1)
(2)
(3)
F
T
F
(d)
(1)
(2)
(3)
F
F
F
(e)
(1)
(2)
(3)
F
F
T
(f)
(1)
(2)
(3)
T
T
F
Answer 10.
(1)
The Euler characteristic of a connected planar graph is equal to two. The
Euler characteristic of a graph with 6 vertices, 12 edges, and 6 faces is 0. So there can not be such
a connected planar graph. Thus
(1)
is
False
.
(2)
By the Euler characteristic theorem, a planar graph with 3 corrected pieces is 3 + 1 = 4. So
(2)
is also
False
.
(3)
If a connected planar graph has 8 edges, 3 vertices, and
F
faces, then by the Euler characteristic
theorem we have 3

8 +
F
= 2 or
F
= 7. So
(3)
must be
True
.
The correct answer is
(e)
.
✷
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 Spring '08
 schneps
 Math, Graph Theory, Correct Answer, Planar graph, g2, Euler characteristic

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