2.1 Algorithms

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2.1 Algorithms

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Title: 2.1 Algorithms

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7.6 Shortest Path Problems

Example 1
What is the length of the shortest path between a
and z in the weighted graph shown in Figure 3?
Solution Although the shortest path is easily
found by inspection,we will develop some ideas
useful in understanding Dijkstras algorithm. We
will solve this problem by finding the length of
the shortest path from a to successive vertices
,until z is reached.

The only paths starting at a that contain no
vertex other than a (until the terminal vertex is
reached)are a,b and a,d. Since the lengths of a,b
and a,d are 4 and 2,respectively,it follows that
d is the closest vertex to a .
We can find the next closest vertex by looking at
all paths that go through only a and d (until the
terminal vertex is reached ).The shortest such
path to b is still a,b, with length 4,and the
shortest such path to e is a,d,e,with length
5.Consequently,the next closest vertex to a is b .

To find the third closest vertex to a, we need
to examine only paths that go through only
a,d,and b (until the terminal vertex is
reached).There is a path of length 7 to
c,namely,a,b,c,and a path of length 6 to z
,namely,a,d,e,z.Consequently,z is the next
closest vertex to a, and the length of the
shortest path to z is 6 .

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Algorithm 1 Dijkstras Algorithm.

Procedure Dijkstra(G weighted connected simple
graph, with all weights positive)
G has vertices a v0, v1 , ,vn z and
weights w(vi ,vj )
where w(vi ,vj )8 if vi ,vj is not an
edge in G
For I 1 to n
L(vi ) 8
L(a)0
S ø
the labels are now initialized so that the
label of a is zero and all other labels are
8,and S is the empty set

While z ? S
Begin
u a vertex not in S with L(u) minimal
SS? u
for all vertices v not in S
if L(u) w(u,v)ltL(v)L(u)w(u,v)
this adds a vertex to S with minimal label and
updates the labels of vertices not in S
End L(z)length of shortest path from a to z

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Example 2
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l1 l2 l3 l4 l5
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c 2 c
d 8 10 8 b
e 8 11 11 10 d
z 8 8 8 14 13 e

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Theorem 1

Dijkstras algorithm finds the length of a
shortest path between two vertices in a connected
simple undirected weighted graph

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Theorem 2

Dijkstras algorithm uses O(n2 )operations
(additions and comparisons )to find the length of
the shortest path between two vertices in a
connected simple undirected weighted graph.

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application

See another slide

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7.7 Planar Graphs

Definition 1. A graph is called planar if it can
be drawn in the plane without any edges crossing
(where a crossing of edges is the intersection of
the lines or arcs representing them at a point
other than their common endpoint ).Such a drawing
is called a planar representation of the graph.

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Example 1

Is K4 (shown in Figure 2 with two edges crossing
) planar?
Solution K4 is planar because it can be drawn
without crossings ,as shown in Figure 3.

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Theorem 1

Eulers formula Let G be a connected planar
simple graph with e edges and v vertices. Let r
be the number of regions in a planar
representation of G.Then re-v2.

Proof First ,we specify a planar representation
of G.We will prove the theorem by constructing a
sequence of subgraphs G1 , G2, , Ge
G,successively adding an edge at each stage. This
is done using the following inductive
definition.Arbitrarily pick one edge of G to
obtain G1.Obtain Gn from Gn-1 by arbitrarily
adding an edge that is incident with a vertex
already in Gn-1.

This construction is possible since G is
connected. G is obtained after e edges are added.
Let rn, en, and vn represent the number of
regions, edges, and vertices of the planar
representation of Gn induced by the planar
representation of G,respectively.
The proof will now proceed by induction .The
relationship r1 e1 v1 2 is true for G1,since
e11, v1 2,and r1 1.This is shown in Figure 9.

Now assume that rn en vn 2 .Let an 1,bn1
be the edge that is added to Gn to obtain Gn1.
There are two possibilities to consider. In the
first case,both an 1and bn1 and already in Gn .
These two vertices must be on the boundary of a
common region R, or else it would be impossible
to add the edge an 1,bn1 to Gn without two
edges crossing (and Gn1 is planar).The addition
of this new edge splits R into two regions.

Consequently, in this case, rn1 rn 1 , en1
en 1 ,and vn1 vn .Thus,each side of the
formula relating the number of regions,edges ,and
vertices increases by side of the formula
relation the number of regions ,edges, and
vertices increases by exactly one ,so this
formula is still true. In other words, rn1
en1 vn1 2 .This case is illustrated in
Figure 10(a)

In the second case ,one of the two vertices of
the new edge is not already in Gn. Suppose that
an1 is in Gn but that bn1 is not .Adding this
new edge does not produce any new regions,since
bn1 must be in a region that has an1 on its
boundary . Consequently, rn1 rn.Moreover, en1
en 1and vn1 vn 1.Each side of the formula
relating the number of regions ,edges, and
vertices remains the same, so the formula is
still true .In other words, rn1 en1 vn1 2
.This case is illustrated in Figure 10(b)

We have completed the induction argument .Hence
rn en vn 2 for all n. Since the original
graph is the graph Ge,obtained after e edges have
been added ,the theorem is true .

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a
n1
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n1
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n1
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Example 4

Suppose that a connected planar simple graph has
20 vertices, each of degree 3.Into how many
regions does a representation of this planar
graph split the plane ?
Solution This graph has 20 vertices ,each of
degree 3,so that v20 . Since the sum of the
degrees of the vertices, 3v 32060,is equal to
twice the number of edges, 2e, we have 2e 60,or
e 30. Consequently,from Eulers formula, the
number of regions is
r e v 2 30-202 12.

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Corollary 1

Degree of a region Suppose R is a region of a
connected planar simple graph, the number of the
edges which surround R is called the Degree of R
, which denoted by Deg(R).
The region is a closed path.

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Deg(R1)3 Deg(R1)6 Deg(R1)7
Deg(R1)12 Deg(R2)4
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Proof of Corollary 1

If G is a connected planar simple graph with e
edges and v vertices where v3, then e3v-6
Proof suppose that a connected planar simple
graph divides the plane into r regions, the
degree of each region is at least 3. (no simple
region can be surrounded by one or two different
edges)

Note that the sum of the degrees of all the
region of a connect graph G is exactly twice the
number of the edges in the graph, because each
edge occurs only on the boundary of one or two
region of G exactly twice.(one time either in two
different regions or twice in the same region).
Hence,
2e ?deg(Ri) 3r ,it imply r(2/3)e

Using Eulers formula e-v2r,we obtain
e-v2(2/3)e, this shows that
e 3v-6
The equality holds if and only every region has
exactly three edges.

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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 for this graph since e10 and
3v-69.Therefore, k5 is not planar.

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Corollary 2

If a connected planar simple graph has e edges
and v vertices with v?3 and no circuits of length
3,then e ?2v-4.
The proof of Corollary 2 is similar to that of
Corollary 1,except that in this case the fact
that there are no circuits of length 3 implies
that the degree of a region must be at least
4.The details of this proof are left for the
reader (see Exercise 13 at the end of this
section).

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Example 6

Use Corollary 2 to show that K3,3 is nonplanar.
Solution Since K3,3 has no circuits of length 3
(this is easy to see since it is
bipartite),Corollary 2 can be used . K3,3 has
six vertices and nine edges. Since e9 and
2v-48,corollary 2 shows that K3,3 is nonplanar.

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Corollary 3

Generally, if every region of a planar connected
graph has at least l edges, then
e (v-2)
Proof similarly, 2e 2e ?deg(Ri) lr , this
imply r(2/l)e, Using Eulers formula e-v2r,we
obtain
e-v2 (2/l)e, this shows that
e (v-2)

The construction of Dodecahedron(?12??)the degree
of every vertex is 3 and the degree of every
region is 5, hence 2e3v, and
e (v-2)
(v-2)(5/3)(v-2), hence
v20, e30 and r12

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KURATOWSKIS THEOREM

Elementary subdivision

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Theorem 2

A graph is nonplanar if and only if it
contains a subgraph homeomorphic to K3,3 or K5.
It is clear that a graph containing a subgraph
homeomorphic to K3,3 or K5 ,is complicated and
will not be given here.The following examples
illustrate how Kuratowskis theorem is used.
Hint homeomorphic is mainly adding or deleting a
series if vertices of degree 2

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Example 7

Determine whether the graph G shown in Figure 13
is planar.(draw in board)
Solution G has a subgraph H homeomorphic to K5.
H is obtained by deleting h,j,and k and all edges
incident with these vertices. H is homeomorphic
to K5 since it can be obtained from K5 (with
vertices a,b,c,g,and I)by a sequence of
elementary subdivisions ,adding the vertices
d,e,and f .(The reader should construct such a
sequence of elementary subdivisions.) Hence,G is
nonplanar.

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Example 8

Is the Petersen graph, shown in Figure
14(a),planar?(The Danish mathematician Julius
Petersen introduced this graph in 1891it is
often used to illustrate various theoretical
properties of graphs.)

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Solution The subgraph H of the Petersen graph
obtained by deleting b and the three edges that
have b as an endpoint ,shown in Figure 14(b),is
homeomorphic to K3,3, with vertex sets f,d,jand
e,I,h,since it can be obtained by a sequence of
elementary subdivisions,deleting d,h and adding
c,h and c,d,deleting e,f and adding a,e
and a,f,and deleting I,jand adding g,I and
g,j.Hence,the Petersen graph is not planar.

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Proof 2

In peterson graph, every simple circle has
exactly 5 edges, if it is a planar graph, then
the degree of every region is 5, hence e
(v-2),
but e15,v10, (10-2)13.333, this
leads to a contradiction.

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7.8 Graph Coloring

Definition 1
A coloring of a simple graph is the assignment of
a color to each vertex of the graph so that no
two adjacent vertices are assigned the same color

Vertex coloring and region coloring
Dual graph A region to a vertex edges
connect two vertices if the original regions are
adjacent.
G(V,E), (1)G??????Fi????vi???
(2)?Fi?Fj??,?vi?vj??
(3)?e???Fi??????,?vi?????????

v1
5
v2
1
2
v5
4
3
v4
v3
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Definition 2

The chromatic number of a graph G is the least
number of colors needed for a coloring of this
graph. Denoted by x(G)
(1) x(G1)3, (2)x(G2)2

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Theorem 1

The Four Color Theorem The chromatic number of a
planar graph is no greater than four.

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Example 1

What are the chromatic numbers of the graphs G
and H shown in Figure 3?
Solution The chromatic number of G is at least
3,since 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 can (and must )be colored red
since it is adjacent to b and c .Furthermore,e
can(and must )be colored green since it is
adjacent only to vertices colored red and blue
,and f can (and must )be colored blue since it is
adjacent only to vertices colored blue and
green.This produces a coloring of G using exactly
three colors.Figure 4 displays such a coloring.

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The graph H is made up of the graph G with an
edge connecting a and g.Any attempt to color H
using three colors must follow the same reasoning
as that used to color G ,except at the last
stage, when all vertices other than g have been
colored .Then ,since g is adjacent (in H)to
vertices colored red ,blue,and green,a fourth
color,say brown,needs to be used.Hence,H has a
chromatic number equal to 4. A coloring of H is
shown in Figure 4.

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Example 2

What is the chromatic number of Kn?
SolutionA coloring of Kn can be constructed
using n colors by assigning a different color to
each vertex.Is there a coloring using fewer
colors ?The answer is no.No two vertices can be
assigned the same color ,since every two vertices
of this graph are adjacent.Hence,the chromatic
number of Kn n.(Recall that Kn is not planar
when n?5,so this result does not contradict the
four color theorem.)A coloring of K5 using five
colors is shown in Figure 5.

X(k5)5,generally, X(Kn)n

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Example 3

What is the chromatic number of the complete
bipartite graph Km,n, where m and n are positive
integers?
SolutionThe number of colors needed may seem to
depend on m and n .However,only two colors are
needed.Color the set of m vertices with one color
and the set of n vertices with a second
color.Since edges connect only a vertex from the
set of m vertices and a vertex from the set of n
vertices,no two adjacent vertices have the same
color.A coloring of K3,4 with two colors is
displayed in Figure 6.

X(Km,n)2

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Example 4

What is the chromatic number of graph Cn ?(Recall
that Cn is the cycle with n vertices.)
SolutionWe will first consider some individual
cases.To begin, let n6.Pick a vertes and color
it red .Proceed clockwise in the planar depiction
of C6 show in Figure 7.It is necessary to assign
a second color,say blue,to the next vertex
reached. Continue in the clockwise directionthe
third vertex can be colored red ,the fourth
vertex blue, and the fifth vertex red.Finally,the
sixth vertex,which is adjacent to the first,can
be colored blue. Hence,the chromatic number of C6
is 2.Figure 7 displays the coloring constructed
here.

Next ,let n5 and consider C5. Pick a vertex and
color it red . Proceeding clockwise,it is
necessary to assign a second color ,say blue,to
the next vertex reached. Continuing in the
clockwise direction,the third vertex can be
colored red, and the fourth vertex can be colored
blue .The fifth vertex cannot be colored either
red or blue,since it is adjacent to the fourth
vertex and the first vertex.

Consequently,a third color is required for this
vertex.Note that we would have also needed three
colors if we had colored vertices in the
counterclockwise direction.Thus,the chromatic
number of C5 is 3.A coloring of C5 using three
colors is displayed in Figure 7.
In general,two colors are needed to color Cn when
n is even.To construct such a coloring ,simply
pick a vertex and color it red.Then proceed
around the graph in a clockwise direction(using a
planar representation of the graph) coloring the
second vertex blue,the third vertex red ,and so
on.The nth vertex can be colored blue,since the
two vertices adjacent to it,namely,the first and
(n-1)st.Hence,a third color must be used

X(C2n1)3, X(C2n)2

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When n is odd and ngt1,the chromatic number of Cn
is 3.To see this ,pick an initial vertex. To use
only two colors,it is necessary to alternate
color as the graph is traversed in a clockwise
direction.However,then nth vertex reached is
adjacent to two vertices of different colors
,namely,the first and (n-1)st. Hence,a third
color must be used.

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application