This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 682 Notes Combinatorics and Graph Theory II 1 Bipartite graphs One interesting class of graphs rather akin to trees and acyclic graphs is the bipartite graph : Definition 1. A graph G is bipartite if the vertexset of G can be partitioned into two sets A and B such that if u and v are in the same set, u and v are nonadjacent. Weve seen one good example of these already: the complete bipartite graph K a,b is a bipartite graph in which every possible edge between the two sets exists. Weve in fact also seen several other bipartite graphs. Proposition 1. Every tree (or forest) is a bipartite graph. Proof. We may use the structural theorem on trees to prove this result fairly easily by induction. Clearly when  T  = 1, T is bipartite (although one of the parts will be empty). For the inductive step, let  T  = n , and let v be a leaf; T v is a tree by the structural theorem, and by the inductive hypothesis, T v is bipartite. Thus, all vertices of T except for v may easily be classified into partitions satisfying the nonadjacency criterion. Since d T ( v ) = 1, v has one neighbor. Put v in the partition not occupied by its neighbor to satisfy the nonadjacency condition. To extend this result to forests, partition each component tree of the forest individually, and let the parts of the overall graph just be the union of the parts from each component. Its pretty easy to show that even cycles are also bipartite, but that odd cycles are not. In fact, this observiation, slightly generalized, forms the entire criterion for a graph to be bipartite. Theorem 1. A graph G is bipartite if and only if it contains no odd cycles. Proof. First, let us show that if a graph contains an odd cycle it is not bipartite. Let v 1 v 2 v 2 n 1 v 1 be the vertices of an odd cycle in G . If G were bipartite, then v 1 would be in some part; without loss of generality we may say v 1 A , so v 2 B since it is adjacent to v 1 , and v 3 A since it is adjacent to v 2 , and so forth up to showing that v 2 n 1 A . But then v 1 and v 2 n 1 would be adjacent vertices in a single part, violating the partition conditions, so such a partition cannot be valid, so G is nonbipartite. Now, to show the converse, we will, for simplicity, look at a connected graph G which lacks odd cycles. If G is not connected, we could demonstrate bipartiteness on each component, and, as in the previous proof, build an overall partition as a union of the partitions on each component. Now, we will make use of the structural theorem on connected graphs: we order the vertices of G with v 1 ,v 2 ,...,v n such that each G [ v 1 ,v 2 ,...,v k ] is connected. Then, we will step through the vertices, assigning them in order to the parts A and B , following the procedure below: Arbitrarily assign v 1 to part A ....
View
Full
Document
This note was uploaded on 01/12/2012 for the course MATH 682 taught by Professor Wildstrom during the Spring '09 term at University of Louisville.
 Spring '09
 WILDSTROM
 Combinatorics, Graph Theory, Sets

Click to edit the document details