Theorem 5.13: For vT‘, l’(v)= min{l(v), l(vk)+w(vk, v)}

Proof: Let S'=S {∪ vk}.

Suppose that l‘(v) is the length of a shortest path from v1 to v containing only vertices in S’.

(1)There are some paths from v1 to v, but these paths don’t contain the vertex vk and other vertices of T'. Then l(v) is the length of the shortest path of these paths, i.e. l'(v)=l(v) 。

(2)There are some paths from v1 to v, these paths from v1 to vk don’t contain other vertices of T', and the vertex vk adjacent edge {vk,v}. Then l(vk)+w(vk,v) is the length of

the shortest path of these paths, viz l'(v)= l(vk)+w(vk,v).

5.5 Trees5.5.1 Trees and their properties Definition 22: A tree is a connected

undirected graph with no simple circuit, and is denoted by T. A vertex of T is a leaf if only if it has degree one. Vertices are called internal vertices if the degrees of the vertices are more than 1. A graph is called a forest if the graph is not connected and each of the graph’s connected components is a tree.

Theorem 5.14: Let T be a graph with n vertices. The following assertions are equivalent.

(1)T is a connected graph with no simple circuit. (2)T is a graph with no simple circuit and e=n-1 where e is

number of edges of T. (3)T is a connected graph with e=n-1 where e is number of

edges of T. (4)T is a graph with no simple circuit, and if x and y are

nonadjacent vertices of T then T+{{x,y}} contains exactly a simple circuit. T+{{x,y}} is a new graph which is obtained from T by joining x to y.

(5)T is connected and if {x,y}E(T) then T-{x,y} is disconnected. Where T-{x,y} is a new graph which is obtained from T by removing edge {x,y}.

(6)There is a unique simple path between any two of vertices of T.

Proof ： (1)→(2): If T is a connected graph with no simple circuit, then T is a graph with no simple circuit and e=n-1. i.e prove e=n-1

Let us apply induction on the number of vertices of T.

When n=2, T is a connected graph with no simple circuit,

the result holds

Suppose that result holds for n=k n=k+1? (Theorem 5.4:Let (G)≥2, then there is a simple

circuit in the graph G.) By the theorem 5.4, and T is connected and no

simple circuit. There is a vertex that has degree one. Let the vertex

be u, and suppose that u is incident with edge {u,v}. We remove the vertex u and edge {u,v} from T, and

get a connected graph T’ with no simple circuit, and T’ has k vertices.

By the inductive hypothesis, T’ has k-1 edges. e(T)=e(T’)+1=k

(2)→(3): T is an acyclic graph with e=n-1.Now we prove T is connected and e=n-1. i.e. prove T is connected

Suppose T is disconnected. Then T have (>1) connected components T1,T2,…,T. The number of vertices of Ti is ni for i=1,2,…, and n1+n2+…+n=n.

(3)→(4): T is a connected graph with e=n-1, we prove “T is a graph with no simple circuit, and if x and y are nonadjacent vertices of T then T+{x,y} contains exactly a simple circuit”.

1)We first prove that T doesn't contain simple circuit. Let us apply induction on the number of vertices of T. T is connected with n=2 and e=1,

Thus T doesn't contain any simple circuit. The result holds when n=2 and e=1

Suppose that result holds for n=k-1 For n=k, (T)1 There is a vertex that has degree one. The

vertex is denoted by u. i.e d(u)=1. Why? 2e2k, i.e. ek( e=n-1=k-1), contradiction We has a new graph T’ which is obtained

from T by removing the vertex u and incident with edge {u,v}

By the inductive hypothesis, T' doesn't contain any simple circuit. Thus T doesn't contain any simple circuit

2) If x and y are nonadjacent vertices of T, then T+{x,y} contains a simple circuit

There is a simple path from x to y in T. (x=vi,vi1,…, vis,vj=y)。 (x=vi,vi1,…, vis,vj=y,vi=x)。 3)Next, we prove T+{x,y} contains exactly a

simple circuit. Suppose that there are two (or more than)

simple circuit in T+{x,y}.

(4)→(5): T is a graph with no simple circuit, and if x and y are nonadjacent vertices of T, then T+{x,y} contains exactly a simple circuit. We prove “T is connected and if {x,y}E(T) then T-{x,y} is disconnected”.

Suppose T is disconnected. There are vertices vi and vj such that there is not any simple path between vi and vj.

Add an edge {vi,vj} T The new graph has also no simple circuit.

Contradiction

(5)→(6): T is connected and if {x,y}E(T) then T-{x,y} is disconnected. We prove “There is a unique simple path between any two of vertices of T”.

(6)→(1): There is a unique simple path between any two of vertices of T. We prove “T is a connected graph with no simple circuit”

T is a connected graph . If T contains a simple circuit, then … contradiction

Corollary 5.1: If G is a forest with n vertices and connected components, then there are n- edges in G.

Theorem 5.15: The tree T with |V(T)|>1 contains at least 2 leaves.

Proof: e=n-1 Suppose T that contains at most 1 leaf 2e2(n-1)+1=2n-1>2(n-1) Contradiction

Definition 23: T is called spanning tree of graph G, if the spanning subgraph T of G is a tree.

Theorem 5.16: G is connected if and only if G contains a spanning tree.

Proof:1) G contains a spanning tree, we prove G is connected.

2) G is connected, we prove G contains a spanning tree

If G has not any circuit, then G is a spanning tree Suppose that C is a simple circuit of G. We remove a edge from a simple circuit, the new

graph is also connected.

NEXT :Minimum spanning trees Rooted tree and binary tree Exercise 1.Let G be a simple graph with n vertices. Show

that ifδ(G) >[n/2]-1, then G is a tree or contains three spanning trees at least