chap. 11 graph theory and applications 1. directed graph 2
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Chap. 11 Graph Theory and Applications
1
Directed Graph
2
(Undirected) Graph
3
Vertex and Edge Sets
4
Walk
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Closed (Open) Walk
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Trail, Path, Circuit, and Cycle
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Comparison of Walk, Trail, Path, Circuit, and Cycle
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Theorem 11.1
Observation:
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Theorem 11.1
1. It suffices to show from a to b, the shortest trail is the shortest path.2. Let be the shortest trail
from a to b.3.
4. 10
Connected Graph
connected graph disconnected graph11
Multigraph
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Subgraph
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Spanning Subgraph
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Induced Subgraph
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Which of the following is an induced subgraph of G? O
Induced Subgraph
O X
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Components of a Graph
1 2
connected sugraph
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G-v
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G-e
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Complete Graph
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Complement of a Graph
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Isomorphic Graphs
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Isomorphic Graphs
Which of the following function define a graph isomorphism for the graphs shown below?
OX
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Isomorphic Graphs
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Isomorphic Graphs
Are the following two graphs isomorphic?
In (a), a and d each adjacent to two other vertices.In (b), u, x, and z each adjacent to two other vertices.
X
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Vertex Degree
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Theorem 11.2
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Corollary 11.1
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a
b
c
d
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Euler Circuit and Euler Trail
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Theorem 11.3
(⇒) 1.2.3.4.5.6.
7.32
Theorem 11.3
8.
9.
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Theorem 11.3
(⇐) 1.2.3.
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Theorem 11.3
4.
5.6.
7.8.
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Theorem 11.3
9.10.11.12.
13.
14.36
Corollary 11.2
(⇐) 1.
2.3.4.(⇒) The proof of only if part is similar to that of
Theorem 11.3 and omitted.
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Incoming and Outgoing Degrees
2
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Theorem 11.4
The proof is similar to that of Theorem 11.3 and omitted.
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Planar Graph
Which of the following is a planar graph?
O O40
Euler’s Theorem
v =e =r =v – e + r = 2
783
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Euler’s Theorem• Proof. 1. Use induction on v (number of vertices).• 2. Basis (v = 1):
– G is a “bouquet” of loops, each a closed curve in the embedding.
– If e = 0, then r = 1, and the formula holds.– Each added loop passes through a region and cuts
it into 2 regions. This augments the edge count and the region count each by 1. Thus the formula holds when v = 1 for any number of edges.
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Euler’s Theorem• 3. Induction step (v>1):
– There exists an edge e that is not a loop
because G is connected.– Obtain a graph G’ with v’ vertices, e’ edges, and r’
regions by contracting e.– Clearly, v’=v–1, e’=e–1, and r’=r.– v’– e’+ r’ = 2. – Therefore, v-e+r=2.
e
(induction hypothesis)
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Corollary 11.3
1. It suffices to consider connected graphs; otherwise, we could add edges.
2. If v 3, every region contains at least three edges (L(Ri) 3r).
3. 2e=L(Ri), implying 2e3r.
4. By Euler’s Theorem, v–e+r=2, implying e≤ 3v– 6.
If also G is triangle-free, then e ≤ 2v–4.
(L(Ri) 4r)
(2e4r)
(e≤ 2v–4)
If G is a simple planar graph with at least three vertices, then e≤3v–6. (A simple graph is not a multigraph and does not contain any loop.)
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Bipartite Graph
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Nonplanarity of K5 and K3,3
K5 (e = 10, n = 5) K3,3 (e = 9, n = 6)
• These graphs have too many edges to be planar. – For K5, we have e = 10>9 = 3n-6.
– Since K3,3 is triangle-free, we have e = 9>8 = 2n-4.
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Subdivision of a Graph
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Subdivision of a Graph
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Hamilton Cycle
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Hamilton Cycle
Does the following graph contain a hamiltion cycle? X
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Theorem 11.8
1.2.
3.4.
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Theorem 11.8
5.6.
7.8.9.10.11.12.
13.54
Theorem 11.814.15.
16.
17.18.
19.
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Theorem 11.8
17.18.
19.
20.
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Theorem 11.9
1.2.
3.
4.5.
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Theorem 11.9
6.
7.
8.9.
10.
11.58
Proper Coloring and Chromatic Number
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Counting Proper Colors
1.
2.
3.4.
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Theorem 11.10
1.2.3.4.5.6.
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Example 11.36
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Example 11.37
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