1. 期中测验时间和地点： 11 月 4 日, 上午 9:40—11 ： 40 地点： 教室 2. 答疑时间和地点： 1)11 月 1 日 ( 周五 )13:00—15:00 软件楼 319 2)11 月 2 日和 3 日， 14:00—17:00 软件楼 3 楼 机房讨论室

 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.  A graph that is not connected is the union of two or more connected subgraphs, each pair of which has no vertex in common. These disjoint connected subgraphs are called the connected components of the graph

 Example: Let G be a simple graph. If G has n vertices, e edges, and ω connected components, then 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=e 0 -1 e=e 0, Omitting any edge, G', (1)G' has n vertices, ω components and e 0 -1 edges. (2)G' has n vertices, ω+1 components and e 0 -1 edges

2.  Let G 1,G 2,…,G ω be ω components of G. G i has n i vertices for i=1,2,…, ω, and n 1 +n 2 +…+n ω =n ， and

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

 Connectivity in directed graphs  Definition 16: Let n be a nonnegative integer and G be a directed graph. A path of length n from u to v in G is a sequence of edges e 1,e 2,…,e n of G such that e 1 =(v 0 =u,v 1 ), e 2 =(v 1,v 2 ), …, e n =(v n-1,v n =v), and no edge occurs more than once in the edge sequence. 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 v 0,v 1,…,v n are all distinct.

 (e1,e2,e7,e1,e2,e9)is not a path  (e1,e2,e7,e6,e9)is a path from a to e  (e1,e2,e9)is a path from a to e, is a simple path.  (a,b,c,e) (e1,e2,e7,e1,e2,e7)is not a circuit (e1,e2,e7,e6,e12) is a circuit (e1,e2,e7) is a simple circuit. (a,b,c,a)

 Definition 17: A directed graph is strongly connected if there is a path from a to b and from b to a whenever a and b are vertices in the graph. A directed graph is connected directed graph if there is a path from a to b or b to a whenever a and b are vertices in the graph. A directed graph is weakly connected if there is a path between every pair vertices in the underlying undirected graph.

 (a)strongly connected  (b)connected directed  (c)weakly connected  strongly connected components: G 1,G 2,…,G ω

 V ={v 1,v 2,v 3,v 4,v 5,v 6,v 7, v 8 }  V 1 ={v 1,v 7,v 8 }, V 2 ={v 2,v 3,v 5,v 6 }, V 3 ={v 4 },  strongly connected components :  G(V 1 ),G(V 2 ),G(V 3 )

 Bipartite graph  Definition18: A simple graph is called bipartite if its vertex set V can be partioned into two disjoint sets V 1 and V 2 such that every edge in the graph connects a vertex in V 1 and a vertex in V 2. (so that no edge in G connects either two vertices in V 1 or two vertices in V 2 ).The symbol K m,n denotes a complete bipartite graph: V 1 has m vertices and contains all edges joining vertices in V 2, and V 2 has n vertices and contains all edges joining vertices in V 1.  K 3,3, K 2,3 。 V 1 ={x 1,x 2,x 3, x 4 }, V 2 ={y 1, y 2, y 3, y 4, y 5 }, or V' 1 ={x 1,x 2,x 3, y 4, y 5 }, V' 2 ={y 1, y 2, y 3, x 4 },

 The graph is not bipartite  Theorem 5.5:A graph is bipartite iff it does not contain any odd simple circuit.  Proof:(1)Let G be bipartite, we prove it does not contain any odd simple circuit.  Let C=(v 0,v 1,…,v m,v 0 ) be an simple circuit of G

 (2)G does not contain any odd simple circuit, we prove G is bipartite  Since a graph is bipartite iff each component of it is, we may assume that G is connected.  Pick a vertex u  V,and put V 1 ={x|l(u,x) is even simple path},and V 2 ={y|l(u,y) is odd simple path}  1)We prove V(G)=V 1 ∪ V 2, V 1 ∩V 2 =   Let v  V 1 ∩V 2,  there is an odd simple circuit in G such that these edges of the simple circuit  p 1 ∪ p 2  each edge joins a vertex of V 1 to a vertex of V 2

 2) we prove that each edge of G joins a vertex of V 1 and a vertex V 2  If it has a edge joins two vertices y 1 and y 2 of V 2  odd simple path  (u=u 0,u 1,u 2, ,u 2n,y 1,y 2 ),even path  y 2  u i (0  i  2n)  There is u j so that y 2 =u j. The path (u,u 1,u 2, ,u j-1, y 2,u j+1, ,u 2n,y 1,y 2 ) from u to y 2,  Simple path (u,u 1,u 2, ,u j-1,y 2 ),simple circuit (y 2,u j+1, ,u 2n,y 1,y 2 )  j is odd number  j is even number

5.3Euler and Hamilton paths  Euler paths  Definition 19: A path in a graph G is called an Euler path if it includes every edge exactly once. An Euler circuit is an Euler path that is a circuit  Theorem 5.6: A connected multigraph has an Euler circuit if and only if each of its vertices has even degree.

 Proof:(1)Let connected multigraph G have an Euler circuit, then each of its vertices has even degree.  (v 0,v 1,…,v i, …,v k ),v 0 =v k  First note that an Euler circuit begins with a vertex v 0 and continues with an edge incident to v 0, say {v 0,v 1 }. The edge {v 0,v 1 } contributes one to d(v 0 ).  Thus each of G’s vertices has even degree.

 (2)Suppose that G is a connected multigraph and the degree of every vertex of G is even.  Let us apply induction on the number of edges of G  1)e=1,loop The graph is an Euler circuit. The result holds 2) Suppose that result holds for e  m e=m+1 ，  (G)≥2. By the theorem 5.4, there is a simple circuit C in the graph G

 If E(G)=E(C), the result holds  If E(G)-E(C) , Let H=G-C, The degree of every vertex of H is even and e(H)  m  ① If H is connected, by the inductive hypothesis, H has an Euler circuit C 1 ，  C=(v 0, v 1,…,v k-1, v 0 )  ② When H is not connected, H has l components, The degree of every vertex of components is even and the number of edges less than m. By the inductive hypothesis,each of components has an Euler circuit. H i  G is connected

the puzzle of the seven bridge in the Königsberg d(A)=3. The graph is no Euler circuit. Theorem 5.7: A connected multigraph has an Euler path but not an circuit if and only if it has exactly two vertices of odd degree. d(A)=d(D)=d(C)=3, d(D)=5 The graph is no Euler path.

 d(A)=d(B)=d(E)=4, d(C)=d(D)=3,  Euler path:C,B,A,C,E,A,D,B,E,D

 Exercise P140 (Sixth) OR 11;  P317 (Sixth) OR P302 (Fifth) 1,2,3,5,6  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.  Next: Hamiltonian paths and circuits, P318 (Sixth) OR P304(Fifth) 8.3,  Shortest-path problem,  1. 期中测验时间和地点：  11 月 4 日, 上午 9:40—11 ： 40  地点： 教室  2. 答疑时间和地点：  1)11 月 1 日 ( 周五 )13:00—15:00 软件楼 319  2)11 月 2 日和 3 日， 14:00—17:00 软件楼 3 楼机房讨论室