 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

 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 },

 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

 (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

 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 0,u 1,u 2, ,u 2n,y 1,y 2 ),even path  y 2  u i (0  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

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.

 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.

 (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

 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  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

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.

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

 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 1,2,3,5,6  P306 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, P  Shortest-path problem