Graph. Undirected Graph Directed Graph Simple Graph

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 correspond

s to u.4. T’ has only k-1 vertices other than u and \delta (G) \g

e 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.

Distance, Diameter, Eccentricity, and Radius

Example 2.1.10

Example 2.1.10

Theorem 2.1.11

Definition 2.1.12

Theorem 2.1.13

Theorem 2.1.13

Wiener Index

Theorem 2.1.14

Theorem 2.1.14

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