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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 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,…,en of 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-1 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: G1,G2,…,Gω

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)

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},

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

(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 uV,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 V∪ 2, V1∩V2= Let vV1∩V2, there is an odd simple circuit in G such that

these edges of the simple circuit p1 p∪ 2

each edge joins a vertex of V1 to a vertex of V2

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 y2ui(1i2n) There is uj so 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

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.

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.

(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 em e=m+1 ， (G)≥2.By the theorem 5.4, there is a simple circuit C in the graph G

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 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. Hi

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)=5The 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

5.3.2 Hamilton paths

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.

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|=2The graph G has not any Hamilton circuit, if there is a nonempty purely subgraph S of V(G) so that (G-S)>|S|.

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

any Hamilton circuit

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

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|

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 pathHamilton 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

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