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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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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.

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(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)

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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.

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

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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 )

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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 },

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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

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(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

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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

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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.

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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.

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(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

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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

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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.

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

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Hamilton paths

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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.

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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|.

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

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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

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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|

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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

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

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