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周二下午 1 : 30—4 : 15 在软件楼 4 楼密码与信息安全实验室答疑 周三下午 1 : 15 到 3 : 15 期中测验. 5.2.4 Bipartite graph - PowerPoint PPT Presentation
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周二下午 1 : 30—4 : 15在软件楼 4楼密码与信息安全实验室答疑
周三下午 1 : 15 到 3 : 15期中测验
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=u0,u1,u2,,u2n,y1,y2),even path y2ui(0i2n) 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 E(G)=E(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.
Exercise P302 1,2,3,5,6P306 3,4,5,6,18周二下午 1 : 30—4 : 15在软件楼 4楼密码与信息安全实验室答疑
周三下午 1 : 15 到 3 : 15期中测验 Next: Hamiltonian paths and circuits,
P304 8.3 Shortest-path problem