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 期中测验时间:本周五上午 9 : 40  教师 TA 答疑时间 : 周三晚上 6 : 00—8 : 30  地点:软件楼 315 房间,  教师 TA :李弋老师  开卷考试.

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Presentation on theme: " 期中测验时间:本周五上午 9 : 40  教师 TA 答疑时间 : 周三晚上 6 : 00—8 : 30  地点:软件楼 315 房间,  教师 TA :李弋老师  开卷考试."— Presentation transcript:

1  期中测验时间:本周五上午 9 : 40  教师 TA 答疑时间 : 周三晚上 6 : 00—8 : 30  地点:软件楼 315 房间,  教师 TA :李弋老师  开卷考试

2  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-1 are all distinct.

3  (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)

4  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.

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

6  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 )

7  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 },

8  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

9  (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

10  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 1,u 2, ,u 2n,y 1,y 2 ),even path  y 2  u i (1  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

11 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.

12  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.

13  (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

14  If G=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

15 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.

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

17  Hamilton paths

18  Definition 20: A Hamilton paths is a path that contains each vertex exactly once. A Hamilton circuit is a circuit that contains each vertex exactly once except for the first vertex, which is also the last.

19  Theorem 5.8: Suppose G(V,E) that has a Hamilton circuit, then for each nonempty proper subset S of V(G), the result which  (G- S)≤|S| holds, where G-S is the subgraph of G by omitting all vertices of S from V(G).  (G-S)=1 , |S|=2 The graph G has not any Hamilton circuit, if there is a nonempty purely subgraph S of V(G) so that  (G- S)>|S|.

20  Omit {b,h,i} from V,   (G-S)=4>3=|S| , The graph has not any Hamilton circuit

21  If  (G-S)≤|S| for each nonempty proper subset S of V(G), then G has a Hamilton circuit or has not any Hamilton circuit.  For example: Petersen graph

22  Proof: Let C be a Hamilton circuit of G(V,E). Then  (C-S)≤|S| for each nonempty proper subset S of V  Why?  Let us apply induction on the number of elements of S.  |S|=1,  The result holds  Suppose that result holds for |S|=k.  Let |S|=k+1  Let S=S' ∪ {v} , then |S'|=k  By the inductive hypothesis,  (C-S')≤|S'|  V(C-S)=V(G-S)  Thus C-S is a spanning subgraph of G-S  Therefore  (G-S)≤  (C-S)≤|S|

23  Theorem 5.9: Let G be a simple graph with n vertices, where n>2. G has a Hamilton circuit if for any two vertices u and v of G that are not adjacent, d(u)+d(v)≥n. n=8,d(u)=d(v)=3, u and v are not adjacent, d(u)+d(v)=6<8, But there is a Hamilton circuit in the graph. Note:1)if G has a Hamilton circuit, then G has a Hamilton path Hamilton circuit :v 1,v 2,v 3,…v n,v 1 Hamilton path:v 1,v 2,v 3,…v n, 2)If G has a Hamilton path, then G has a Hamilton circuit or has not any Hamilton circuit

24  Exercise P302 1,2,3,5,6  P306 3,4,5,6,18


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