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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，

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. 5.2.3 Connectivity in directed graphs

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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，

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  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 e1,e2,…,enof G such that e1=(v0=u,v1), e2=(v1,v2), …, en=(vn-1,vn=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 v0,v1,…,vn-1are all distinct.

  3. (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) • (e1,e2,e7,e1,e2,e9)is not a path • (e1,e2,e7,e6,e9)is a path from ato e • (e1,e2,e9)is a path from ato e, is a simple path. • (a,b,c,e)

  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: G1,G2,…,Gω

  6. V ={v1,v2,v3,v4,v5,v6,v7, v8} • V1={v1,v7,v8}, V2={v2,v3,v5,v6}, V3={v4}, • strongly connected components : • G(V1),G(V2),G(V3)

  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 V1 and V2 such that every edge in the graph connects a vertex in V1 and a vertex in V2. (so that no edge in G connects either two vertices in V1 or two vertices in V2).The symbol Km,n denotes a complete bipartite graph: V1 has m vertices and contains all edges joining vertices in V2, and V2 has n vertices and contains all edges joining vertices in V1. • K3,3, K2,3。 V1={x1,x2,x3,x4}, V2={y1,y2,y3,y4,y5}, or V'1={x1,x2,x3,y4,y5}, V'2={y1,y2,y3,x4},

  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=(v0,v1,…,vm,v0) 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 V1={x|l(u,x) is even simple path} ,and V2={y|l(u,y) is odd simple path} • 1)We prove V(G)=V1∪V2, V1∩V2= • Let vV1∩V2, • there is an odd simple circuit in G such that these edges of the simple circuit p1∪p2 • each edge joins a vertex of V1 to a vertex of V2

  10. 2) we prove that each edge of G joins a vertex of V1 and a vertex V2 • If it has a edge joins two vertices y1 and y2 of V2 • odd simple path • (u,u1,u2,,u2n,y1,y2),even path • y2ui(1i2n) • There is ujso that y2=uj. The path (u,u1,u2,,uj-1, y2,uj+1,,u2n,y1,y2) from u to y2, • Simple path (u,u1,u2,,uj-1,y2),simple circuit (y2,uj+1,,u2n,y1,y2) • 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. • (v0,v1,…,vi, …,vk),v0=vk • First note that an Euler circuit begins with a vertex v0 and continues with an edge incident to v0, say {v0,v1}. The edge {v0,v1} contributes one to d(v0). • 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 • ①If H is connected, by the inductive hypothesis, H has an Euler circuit C1， • C=(v0, v1,…,vk-1, v0) • ②When H is not connected, H has lcomponents, 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. Hi • 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. 5.3.2Hamilton 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 :v1,v2,v3,…vn,v1 Hamilton path:v1,v2,v3,…vn, 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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