Theorem 5.13: For v?T

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• Theorem 5.13 For v?T, l(v) minl(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 dont 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 dont 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).

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5.5 Trees

• 5.5.1 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 graphs connected
  components is a tree.

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• 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 en-1
  where e is number of edges of T.
• (3)T is a connected graph with en-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
  Tx,y contains exactly a simple circuit.
  Tx,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.

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• Proof(1)?(2) If T is a connected graph with no
  simple circuit, then T is a graph with no simple
  circuit and en-1. i.e prove en-1
• Let us apply induction on the number of vertices
  of T.
• When n2, T is a connected graph with no simple
  circuit,

the result holds
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• Suppose that result holds for nk
• nk1?
• (Theorem 5.4Let ? (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)1k

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• (2)?(3) T is an acyclic graph with en-1.Now we
  prove T is connected and en-1. i.e. prove T is
  connected
• Suppose T is disconnected. Then T have ?(?gt1)
  connected components T1,T2,,T?. The number of
  vertices of Ti is ni for i1,2,?, and
  n1n2n?n.

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• (3)?(4) T is a connected graph with en-1, we
  prove T is a graph with no simple circuit, and
  if x and y are nonadjacent vertices of T then
  Tx,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 n2 and e1,

Thus T doesn't contain any simple circuit. The
result holds when n2 and e1
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• Suppose that result holds for nk-1
• For nk, ?(T)?1
• There is a vertex that has degree one. The vertex
  is denoted by u. i.e d(u)1.
• Why?
• 2e?2k, i.e. e?k( en-1k-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

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• 2) If x and y are nonadjacent vertices of T, then
  Tx,y contains a simple circuit
• There is a simple path from x to y in T.
• (xvi,vi1,, vis,vjy)?
• (xvi,vi1,, vis,vjy,vix)?
• 3)Next, we prove Tx,y contains exactly a
  simple circuit.
• Suppose that there are two (or more than) simple
  circuit in Tx,y.

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• (4)?(5) T is a graph with no simple circuit, and
  if x and y are nonadjacent vertices of T, then
  Tx,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

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• (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

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• 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)gt1 contains
  at least 2 leaves.
• Proof en-1
• Suppose T that contains at most 1 leaf
• 2e?2(n-1)12n-1gt2(n-1)
• Contradiction

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• 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.
• Proof1) 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.

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• NEXT Minimum spanning trees
• Rooted tree and binary tree
• Exercise
• 1.Let G be a simple graph with n vertices. Show
  that ifd(G) gtn/2-1, then G is a tree or
  contains three spanning trees at least