Download presentation

Presentation is loading. Please wait.

Published byAmira Biswell Modified over 2 years ago

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
5.2.3 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
5.2.4 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 5.3.1 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
5.3.2 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

Similar presentations

OK

Connectivity and Paths 報告人：林清池. Connectivity A separating set of a graph G is a set such that G-S has more than one component. The connectivity of G,

Connectivity and Paths 報告人：林清池. Connectivity A separating set of a graph G is a set such that G-S has more than one component. The connectivity of G,

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Download ppt on civil disobedience movement Ppt on area of parallelograms Ppt on power system stability using facts devices Ppt on obesity management articles Ppt on obesity management Download ppt on sectors of economy Ppt on field study 2 Ppt on water activity meter Ppt on combination of resistances organization Free ppt on personality development presentation