Unformatted text preview: 1 ∪ V 2 , it is still connected through v 2 . That means v 1 is not an articulation point of V 1 ∪ V 2 , which holds for v 2 and other vertices in V i ∩ V j in general. Therefore, V i ∪ V j does not contain any articulation points, which means that V i ∪ V j is actually a biconnected component. A contradiction! Theorem 3. If R = { ( v,u )  there exist paths from v to w and w to v } then R is an equivalence relation. Proof. 1. Reﬂexivity: Since in a graph, there exists a path for a vertex itself, ( a,a ) ∈ R . 2. Symmetry: If ( a,b ) ∈ R , there exist paths from b to a and a to b . Therefore, ( b,a ) ∈ R . 3. Transitivity: If ( a,b ) ∈ R and ( b,c ) ∈ R , then there exist paths from a to b and b to c , there exist paths from a to c . Also since there exist paths from c to b and b to a , there exist paths from c to a . Therefore, ( a,c ) ∈ R . 1...
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This note was uploaded on 02/04/2010 for the course COMPUTER S cs300 taught by Professor Unkown during the Spring '08 term at Korea Advanced Institute of Science and Technology.
 Spring '08
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