Graph Theory

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Chapter 10

• Graph Theory

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• 10.2 Eulerian Cycle and the property of graph
  theory
• 10.3 The important property of graph theory and
  its representation
• 10.4 Shortest path

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10.2 Eulerian Cycle and the property of graph
theory
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• Example 1 What is Königsberg Bridge problem?
• Ans.
• Two islands A and B are reduced to nodes.
• On the side of the river C and D are also reduced
  to nodes, it can be modeled as a graph and it is
  shown in Fig. 10.2.1.
• G(V,E), V denotes Node Set, and E denotes Edge
  Set. VA,B,C,D and E(A,C),(A,C),(A,D),(A,D),(A
  ,B),(B,C),(B,D).

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Fig.10.2.1 Königsberg Bridge
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Edge(A,C) appears two times, it is
called Multiple Edges. The degree of node B is 3,
it can be denoted by d(B)3 which means that
there are 3 edges incident to the node B.
Fig.10.2.2 The model of Fig. 10.2.1
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• Example 2 From Fig. 10.2.2, Can we find an
  Eulerian cycle from node A?
• Ans.
• Node C will consume two adjacency edges when we
  cross. Since d(C) is odd, there will remain at
  last an adjacency edge. The discussion on node C
  is also suitable for node A, since d(A) is also
  odd. We are unable to find an Eulerian cycle in
  Fig.10.2.2.

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• Example 2.1 Let G(V,E) be a loop-free connected
  undirected graph, and let (a, b) be an edge of G.
  Prove that (a, b) is a part of a cycle if and
  only if remove the vertices a and b does not
  disconnect G.
• Ans.
• (?) If (a, b) is a part of a cycle and let
  a-b-v1-v2-vi-a be a cycle. Suppose remove edge
  (a, b) from G. The path b-v1-v2--vi-a connects
  (a, b), thus G still connect.
• (?)It is still a connected image when
  removing the edge (a,b) from G, thus there exists
  a path a-v1-v2--vi-b. If we replace edge (a, b),
  we can obtain image G with a cycle
  a-v1-v2--vi-b-a, thus (a, b) is a part of a
  cycle.

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• Example 3 Let G(V,E) be a connected graph.
  Prove that G has an Eulerian cycle if and only if
  all degree of nodes in G are even?
• Ans.
• (?) If G has an Eulerian cycle, it means
  that for all nodes in the cycle, the number of
  in-edges are equal to out-edges. Thus, for all
  nodes in G, the degree of them are all even.
• (?) Assume that we select two different node
  x, y?V from G, and it forms a longest Eulerian
  Chain C1 from x to y. At this moment, we must
  select an edge (y, z) not be a part of C1 from G.
  Then, add an edge (y, z) to C1, we have a longer
  Eulerian chain . In fact, node z is node x .
  Assume that C1 has included all the edges in G,
  we get the proof.

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If there are still edges in G which extend from
node u?C1 but are not a part of C1, let G be
discarded the edge set of C1 in G. Since G is a
connected graph and every degree of node is even,
the degree of nodes in G are still even. Assume
we can find a cycle in G, we merge this cycle
into C1, we have Eulerian cycle larger than C1,
this is a contradiction.
Discard multiple edges (A,D) and (A,C) in
figure 10.2.2, we can also find an Eulerian
chain.
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10.3 The important property of graph theory and
its representation
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Example 1 Given a simple planar graph , are
there upper bounds to limit the degree of nodes?
Ans. A face at least includes three edges. Let
represents the number of faces that are
formed by i edges, then the total number of faces
F in G will satisfy Consider an edge shared by
two faces, we obtain We have (10.3.1) We
prove it by contradictionthere is a degree of
node less than 6. Assume every node in V
satisfy 6, we obtain (10.3.2)
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represents the degree of nodes greater than
6. From (10.2.1) we induce that
(10.3.3) We have (10.3.4) M
erge (10.3.1) and (10.3.4) , we have (10.3.1)
multiplied by 2 and then merge (10.3.4) again, we
obtain (10.3.5) From Euler formula we
know (10.3.5) is wrong, since .
Thus we prove it is impossible that every
degree of nodes is greater than 6 in simple
planar graph.
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• Example 2 Prove that there are at most
  edges in simple planar graph G.
• Ans.
• From (10.3.1) we have
• (10.3.6)
• Substitute Euler formula
  into the left pattern of (10.3.6), we have
• Thus, we prove that there are at most
  edges.

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Example 3 Is a simple planar graph? Ans.
In , there are 10 edges, there are at most
edges in simple planar graph.
Thus, is not a simple planar graph.
Fig. 10.3.2
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Example 3.1 For every tree , if
, T has at least two pendant vertices .
i.e. vertices of degree 1. Ans. (1) If T
doesnt have pendant vertices and T is connected
graph, then , , and
E V ? 1, we have Induce that ,
it is a contradiction. (2) If T has one pendant
vertices, then Induce that , it is a
contradiction. Combine the discussion (1) and
(2), thus T has at least two pendant vertices.
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Example 3.2 Suppose G is an arbitrary digraph
with n vertices. What is the largest possible
number of distinct subgraphs with k vertices that
G may have?(Isomorphic subgraphs are distinct.
Choose G to maximize this number.) Ans. When G
is a complete graph which the number of nodes is
n, it will have the largest possible number of
distinct subgraphs. When the number of nodes is
k, there are at most edges, and these
edges have two choices choose or not choose,
it will compose all the digraph, there are
different digraph. There are
choices and we need to consider about which the
number of nodes is k, thus the answer
.
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Example 4 What is the representation of
adjacency list? Ans.
Fig. 10.3.3 A little example
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• Example 5 What is the representation of
  adjacency matrix?
• Ans. The representation of adjacency matrix
• From above, the representation of adjacency
  matrix in figure 10.3.3 is

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Example 6 What the problem is Warshall algorithm
used to solve? Ans. If , but
node i can reach node j through node k , we say
that node i can reach node j and write down
. As shown in Figure 10.3.5.
Fig. 10.3.5
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represents that node I can
reach node j through m edges. Warshall algorithm
is the check of reachability between any two
nodes in graph G, that is to compute (10.3.7
) Here, . We solve the transitive
closure problem in graph G.
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Let Then we obtain
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• Example7 Before introduce Warshalls algorithm,
  can you use brute force method to obtain ?
• Ans.
• AM
• BA
• for i2 to (V-1)
• begin
• AA?M
• BB?A
• end
• It will take by brute force method.

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• Example8 What is Warshall method?
• Ans.
• Warshalls method is easy, the following is the
  three for loops included in Warshalls method
• BM
• for k1 to V
• begin
• for i1 to V
• begin
• for j1 to V
• Bi, jBi, j?(Bi, k?Bk, j)
• end
• end
• The above procedure only need time

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10.4 Shortest path
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• Example1 How many types does the shortest path
  have ?
• Ans.
• Let the source node in is S, and the target node
  is T. The four combinations of the shortest path
  are
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• Fig. 10.4.1 four combinations of
  the shortest path
• The symbol I denotes the set of inner nodes.

S T
1 S0 T0
2 I T0
3 S0 I
4 I I
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• Fig. 10.4.2 an example of a direct graph
• The first type of the shortest path is the
  problem of walking from node 1 to node 6. Let
  I4, then the second type of the shortest path is
  the problem of walking from node 4 to node 6.

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• Example2 How is Floyd-Warshall algorithm used to
  solve the fourth type of the all pairs shortest
  path problem? And what is its time complexity?
• Ans.
• The fourth type of shortest path problem
  Floyd-Warshall algorithm is used to solve all
  pairs shortest path problem.

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• Let adjacency matrix is W, Floyd-Warshall
  algorithm can bedesigned as follow
• BW
• for k1 to V
• begin
• for i1 to V
• begin
• for j1 to V
• Wi, jmin(Wi, j,Wi, kWk, j)
• end
• end
• Its time complexity is .

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• Example3 Can Floyd-Warshall algorithm solve the
  all pairs shortest path problem in figure 10.4.2?
• Ans.
• From figure 10.4.2, we can get the source
  adjacency matrix is
• Here, represents that node 1
  and node 4 not connect each other. After running
  k1, adjacency matrix not change, that is

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• After running k2, we obtain

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• After running k3, will change into

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• Here, pay attention!
• . When we complete k4, 5 and 6, finally we
  obtain
• Thus, the solve of the all pairs shortest path in
  figure 10.4.2 hides in . For example
  tells us, the shortest path between node 2 and
  node 6 is 2.

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• Example4 How Dijkstra solve the all pairs
  shortest path problems?
• Ans.
• Given fig. 10.4.3
• Fig. 10.4.3 An example

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• Select node A and place it into Labeled Node
  Set (LNS), and the other four nodes B?C?D and E
  are placed into Unlabeled Node Set (UNS).
• Find the edge node set which connects LNS
  and UNS and denote it as (LNS, UNS). In the edge
  node set (LNS,UNS), we find the smallest weighted
  edge (A,C) and know its weighted edge value is 2.
  Fig. 10.4.4 is the present condition. Direct at
  node C in the figure, we record A,2.
• Fig. 10.4.4 The condition of LNS after include
  node C
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• Check the edges joined in LNS and UNS, then
  we discover that (A,B), (A,D), (C,B) and (C,D) is
  four edges to join LNS and UNS. We can calculate
  from figure 10.4.4
• (10.4.1)
• Similar, can calculate
• Choose node B! Then the record on node B of
  C,3. Figure 10.4.5 is a sketch map after adding
  node B.
•  

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• Fig. 10.4.5 Adding B to LNS
•  

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• And then, we choose node B. Fig. 10.4.6 is a
  sketch map of LNS adding node D. Finally, we
  choose edge (B,E ) and node E, then pursued the
  final result of the Fig. 10.4.7.
• Fig. 10.4.6 Adding node D into LNS

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• Fig. 10.4.7 Final result
• By Fig. 10.4.7, we can get the shortest path and
  the length of shortest path from node A to all
  other nodes.

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• Example5 What is the time complexity of Dijkstra
  algorithm in Example4?
• Ans.
• As to any node in LNS, at least (V-k) edges
  can link to any node in UNS. If consider all
  extreme point in LNS, then there are at most
  k(V-k) edges link between LNS and UNS. Time
  complexity is
• (10.4.2)