crossing) planar? Solution: K4 is
planar because it can be drawn
without crossings, as shown in Figure
3. EXAMPLE 2 Is Q3, shown in Figure
4, planar? Solution: Q3 is planar,
because it can be drawn wi
every vertex at least once is minimized
for a circuit that visits some vertices
more than once. [Hint: There are
examples with three vertices.] 30.
Show that the problem of finding a
circuit of minimu
that K3,3 Is Nonplanar. P1: 1 CH10-7T
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Next, note that there is no way to place
the final vertex v6 without forcing a
crossing. For i
circuit is planar. When this graph is not
planar, we must turn to more
expensive options. For example, we can
partition the vertices in the graph
representing the circuit into planar
subgraphs. We the
R R (a) (b) FIGURE 10 Adding an Edge
to Gn to Produce Gn+1. P1: 1 CH10-7T
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b d e f a g R 3 1 R2 R3 6 7 FIGURE 11
The Degrees of Reg
salesperson problem is to examine all
possible Hamilton circuits and select
one of minimum total length. How
many circuits do we have to examine to
solve the problem if there are n
vertices in the gra
complexity that works if the weighted
graph satisfies the triangle inequality
such that c = 3/2. For general weighted
graphs for every positive real number
k no algorithm is known that will
always pro
210 100 6 100 160 250 220 100
19. The mathematics department has
six committees, each meeting once a
month. How many different meeting
times must be used to ensure that no
member is scheduled to atte
Exercise 8, using the fares shown in
Figure 1. 11. Find a shortest route (in
distance) between computer centers in
each of these pairs of cities in the
communications network shown in
Figure 2. a) Bos
Which graphs have a chromatic
number of 1? 14. What is the least
number of colors needed to color a
map of the United States? Do not
consider adjacent states that meet only
at a corner. Suppose that M
distance that can be traveled to reach a
stop from another stop? c) What is the
least fare required to travel between
two stops if fares between stops are
added to give the total fare? In
Exercises 24
algorithms that solve it. However, no
algorithm with polynomial worst-case
time complexity is known for solving
this problem. Furthermore, if a
polynomial worst-case time complexity
algorithm were dis
regions, edges, and vertices of the
planar representation of Gn induced by
the planar representation of G,
respectively. a1 b1 R1 FIGURE 9 The
Basis Case of the Proof of Eulers
Formula. The proof will
e/3 v 2. This shows that e 3v 6.
This corollary can be used to
demonstrate that K5 is nonplanar.
EXAMPLE 5 Show that K5 is nonplanar
using Corollary 1. Solution: The graph
K5 has five vertices and 10
how many index registers are needed?
This problem can be addressed using a
graph coloring model. To set up the
model, let each vertex of a graph
represent a variable in the loop. There
is an edge betw
to see because it is bipartite), Corollary
3 can be used. K3,3 has six vertices and
nine edges. Because e = 9 and 2v 4 =
8, Corollary 3 shows that K3,3 is
nonplanar. KAZIMIERZ
KURATOWSKI (18961980) Ka
figure. Euler showed that all planar
representations of a graph split the
plane into the same number of regions.
He accomplished this by finding a
relationship among the number of
regions, the number
3(C5) h) 3(K4,5) P1: 1 CH10-7T
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May 13, 2011 16:18 Key Terms and
Results 735 37. Let G and H be the
graphs displayed in Figure 3. Find a)
2(G). b) 2(H ). c) 3(G). d) 3(
barrister and amateur mathematician,
Alfred Kempe. Mathematicians
accepted his proof as correct until
1890, when Percy Heawood found an
error that made Kempes argument
incomplete. However, Kempes line
graphs G and H shown in Figure 3?
Solution: The chromatic number of G is
at least three, because the vertices a, b,
and c must be assigned different
colors. To see if G can be colored with
three color
2311T MHIA017-Rosen-v5.cls May 13,
2011 16:18 10.6 Shortest-Path
Problems 717 17. The weighted graphs
in the figures here show some major
roads in New Jersey. Part (a) shows the
distances between citi
bipartite graph K3,3. The original
question can be rephrased as: Can K3,3
be drawn in the plane so that no two of
its edges cross? In this section we will
study the question of whether a graph
can be
displayed in Figure 12 are all
homeomorphic. Solution: These three
graphs are homeomorphic because all
three can be obtained from G1 by
elementary subdivisions. G1 can be
obtained from itself by an em
sets cfw_f, d, j and cfw_e, i, h, because it can
be obtained by a sequence of
elementary subdivisions, deleting cfw_d,h
and adding cfw_c, h and cfw_c, d, deleting cfw_e,
f and adding cfw_a, e and cf
Examples 8 and 9 illustrate how
Kuratowskis theorem is used.
EXAMPLE 8 Determine whether the
graph G shown in Figure 13 is planar.
Solution: G has a subgraph H
homeomorphic to K5. H is obtained by
del
least number of colors needed for a
coloring of this graph. The chromatic
number of a graph G is denoted by
(G). (Here is the Greek letter chi.)
Note that asking for the chromatic
number of a planar
deg(R). The degrees of the regions of
the graph shown in Figure 11 are
displayed in the figure. The proof of
Corollary 1 can now be given. Proof: A
connected planar simple graph drawn
in the plane div
graph is called the dual graph of the
map. By the way in which dual graphs
of maps are constructed, it is clear that
any map in the plane has a planar dual
graph. Figure 2 displays the dual
graphs tha
the plane where no three arcs
representing edges are permitted to
cross at the same point. 26. Show that
K3,3 has 1 as its crossing number.
27. Find the crossing numbers of
each of these nonplanar gra
yet been found. Note that the four color
theorem applies only to planar graphs.
Nonplanar graphs can have arbitrarily
large chromatic numbers, as will be
shown in Example 2. Two things are
required to