Quotient graph
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Quotient graph

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Quotient graph

  • 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 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

Quotient graph

(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

Quotient graph

  • 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

Quotient graph

  • 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.

Quotient graph

Quotient graph

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

Quotient graph

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

Quotient graph

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


Euler paths and circuits, P296 8.2

Hamiltonian paths and circuits, P304 8.3

Quotient graph

  • 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.