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Quotient graph

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## PowerPoint Slideshow about ' Quotient graph' - charles-carney

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- Quotient graph
- Definition 13: Suppose G(V,E) is a graph and R is a equivalence relation on the set V. We construct the quotient graph GR in the follow way. The vertices of GR are the equivalence classes of V produced by R. If [v] and [w] are the equivalence classes of vertices v and w of G, then there is an edge in GR between [v] and [w] if some vertex in [v] is connected to some vertex in [w] in the graph G.

5.2 Paths and Circuits

- 5.2.1 Paths and Circuits
- Definition 14: Let n be a nonnegative integer and G be an undirected graph. A path of length n from u to v in G is a sequence of edges e1,e2,…,en of G such that e1={v0=u,v1}, e2={v1,v2},…,en={vn-1,vn=v}, and no edge occurs more than once in the edge sequence. When G is a simple graph, we denote this path by its vertex sequence u=v0,v1,…,vn=v. A path is called simple if no vertex appear more than once. A circuit is a path that begins and ends with the same vertex. A circuit is simple if the vertices v1,v2,…,vn-1 are all distinct

(e6,e7,e8,e4,e7) is not a circuit;

(e1,e6,e7,e8,e4,e5) is a circuit

(e1,e8,e4,e5) is a simple circuit

(e6,e7) is a simple circuit

- (e6,e7,e8,e4,e7,e1) is not a path;
- (e6,e7,e1) is a path of fromv2to v1
- (e8,e4,e5) is a simple path of from v2to v1

- Theorem 5.4:Let (G)≥2, then there is a simple circuit in the graph G.
- Proof: If graph G contains loops or multiple edges, then there is a simple circuit. (a,a) or (e,e').
- Let G be a simple graph. For any vertex v0 of G,
- d(v0)≥2, next vertex, adjacent, Pigeonhole principle

- 5.2.2 Connectivity
- Definition 15: A graph is called connectivity if there is a path between every pair of distinct vertices of the graph. Otherwise , the graph is disconnected.

G1,G2,…,Gω

- A graph that is not connected is the union of two or more connected subgraphs, each pair of which has no vertex in common. These disjoint connected subgraphs are called the connected components of the graph

- Example: Let G be a simple graph. If G has n vertices, e edges, and ω connected components , then

Proof: e≥n-ω

Let us apply induction on the number of edges of G.

e=0, isolated vertex，has n components ，n=ω,

0=e≥n-ω=0，the result holds

Suppose that result holds for e=e0-1

e=e0, Omitting any edge ,G',

(1)G' has n vertices, ω components and e0-1 edges.

(2)G' has n vertices, ω+1 components and e0-1 edges

2. edges, and ω

- Let G1,G2,…,Gωbe ω components of G. Gi has ni vertices for i=1,2,…, ω, and n1+n2+…+nω=n，and

If G is connected, then the number of edges of G has at least n-1 edges.

Tree.

Euler paths and circuits, P296 8.2

Hamiltonian paths and circuits, P304 8.3

- Exercise P128 11; least n-1 edges.
- P295 11,17,19,22,23,28
- 1.Prove that the complement of a disconnected graph is connected.
- 2.Let G be a simple graph with n vertices. Show that ifδ(G) >[n/2]-1, then G is connected.
- 3.Show that a simple graph G with n vertices are connected if it has more than (n-1)(n-2)/2 edges.

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