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 without any
edges crossing, as shown in Figure 5.
We can
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 minimum total weight that
visits every vertex of a weighted g
that K3,3 Is Nonplanar. P1: 1 CH10-7T
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May 13, 2011 16:18 720 10 / Graphs
Next, note that there is no way to place
the final vertex v6 without forcing a
crossing. For if v6 is in R1, then the
edge between v6 and v3 cannot b
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 then construct the
circuit using multiple layers. (See 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 Regions. COROLLARY 1
If G is a connected planar simple gra
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 graph? Once a starting
point is chosen, there are (n 1)!
d
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 produce a solution at most k
times a best solution. If suc
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 attend two
meetings at the same time if the
committees are
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) Boston and Los Angeles b)
New York and San Francisco c) Da
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 Michigan is
one region. Consider the vertices
representi
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 find the length of a
shortest path between a and z in
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 discovered for the
traveling salesperson problem, many
oth
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 now proceed
by induction. The relationship r1 = e1
v1
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 edges.
However, the inequality e 3v 6 is
not satisfied
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 between two vertices if the
variables they represent must b
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) Kazimierz
Kuratowski, the son of a famous
Warsaw lawyer,
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 of vertices, and
the number of edges of a planar graph.
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(H ).
38. What is k(G) if G is a bipartite
graph and k i
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 of
reasoning turned out to be the basis of
the success
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 colors, assign red to a, blue to b,
and green to c. Then, d
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 cities on these
roads; part (b) shows the tolls. Trenton
30
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 drawn in the plane without
edges crossing. In particula
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 empty
sequence of elementary subdivisions.
To obtain G2 f
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 cfw_a, f , and
deleting cfw_i, j and adding cfw_g, i and
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
deleting h, j , and k and all edges
incident with these ve
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 graph is the same
as asking for the minimum number of
c
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 divides the plane into
regions, say r of them. The degree
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 that correspond to the maps
shown in Figure 1. The problem
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 graphs. a) K5
b) K6 c) K7 d) K3,4 e) K4,4 f ) K5,5
28. Fin
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 show that the chromatic
number of a graph is k. First,