 Graph. Undirected Graph Directed Graph Simple Graph.

Presentation on theme: "Graph. Undirected Graph Directed Graph Simple Graph."— Presentation transcript:

Graph

Undirected Graph

Directed Graph

Simple Graph

Walk, Trail, Path

Maximal Path, Maximum Path

Connected Graph, Component

Cut-Edge, Cut-Vertex

Degree

Forest, Tree

Lemma 2.1.3 1.There exists a maximal nontrivial path. 2.The endpoints of a maximal nontrivial path are leaves. 3.Let v be a leaf of a tree G, and let G’=G- v. (1) G’ has n-1 vertices. (2) G’ is connected. (3) G’ is acyclic.

Lemma 2.1.4 1.(A=>B) Use induction on n. There exists a leaf v of G. Let G’=G-v. 2.G is connected and has no cycles => G’ is connected and has no cycles => G’ is connected and has n-1 edges => G is connected and has n edges.

Lemma 2.1.4 3. (B=>C) Use induction on n. There exists a leaf v of G. (since n vertices has 2n-2 degree). Let G’=G-v. … 4. (C=>D) Use induction on n. There exists a leaf v having a neighbor with degree greater than 1. Let G’=G-v. … 5. (D=>A) Use induction on n. There exists a leaf v having a neighbor with degree greater than 1. Let G’=G-v. …

Lemma 2.1.4 6. (B=>C) Deleting edges from cycles of G one by one until the resulting graph G’ is acyclic. 7. No edge of a cycle is a cut-edge => G’ is connected => G’ has n-1 edges => G=G’ => G is acyclic. 8. (C=>A) Let G_1, G_2, …, G_k be the components of G. \sum_{i} n(G_i)=n. 9. e(G_i)=n(G_i)-1 => \sum_{i} e(G_i)=n-k => k=1.

Lemma 2.1.4 10. (A=>D) G is connected => each pair of vertices is connected by a path. 11. Let P, Q be the shortest (total length) pair of distinct paths with the same endpoints => P and Q are disjoint => P \union Q is a cycle. 12. (D=>A) Each pair of vertices is connected by a path => G is connected. 13. If G has a cycle, then G has two u,v-paths for u,v \in V(C).

Proposition 2.1.6 1.Every edge of T is a cut-edge. Let U and U’ be two components of T-e. 2.T’ is connected (since e \notin T’) => T’ has an edge e’ with endpoints in U and U’ => T-e+e’ is connected and has n(G)-1 edges => T-e+e’ is a spanning tree of G.

Proposition 2.1.7 1.T’+e contains a unique cycle C. 2.T is acyclic => There is an edge e’ in E(C)- E(T). 3.Deleting e’ breaks cycle C => T’+e-e’ is connected and has n(G)-1 edges => T’+e-e’ is a spanning tree of G.

Proposition 2.1.8 1.Use induction on k. Let v be a leaf of T, and let u be its neighbor. Let T’=T-v. 2.\delta (G) \ge k => G contains T’ as a subgraph. 3.Let x be the vertex in this copy of T’ that corresponds to u. 4.T’ has only k-1 vertices other than u and \delta (G) \ge k => x has a neighbor y in G that is not in the copy of T’. 5.Adding the edge xy expands this copy of T’ into a copy of T in G, with y playing the role of v.

Example 2.1.10

Theorem 2.1.11

Definition 2.1.12

Theorem 2.1.13

Wiener Index

Theorem 2.1.14

Lemma 2.1.15

Corollary 2.1.16

Homework 2.1.19, 2.1.29, 2.1.39, 2.1.49, 2.1.65 –Due 10/2, 2006 The first paper presentation –10/5, 2006 ~ 11/9, 2006 The first paper report –Due 11/9, 2006