PR5 - 1 ∪ V 2 it is still connected through v 2 That...

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Theorem 1. Each subgraph consisting of the edges in an equivalence class and the incident vertices forms a biconnected component. Proof. Assume that the subgraph does not form a biconnected component. Then by definition, the subgraph contains at least one articulation point. By deleting one of these articulation points, the subgraph is split into two or more parts. Take two of them ( S 1 and S 2 ), and we know that there is no cycle containing both edges from S 1 and S 2 , which is a contradiction to our assumption that the subgraph consists of the edges in an equivalence class. Therefore, it holds. Theorem 2. For 1 i k , let G i = ( V i , E i ) be the biconnected components of a connected undirected graph G = ( V, E ) . Then, for all i 6 = j , V i V j contains at most one vertex. Proof. Assume that there are more than one vertices in V i V j and let us call two of them v 1 and v 2 . Since V i is biconnected, it does not contain any articulation points and the situation is same for V j . Then, if we remove v 1 from V 1
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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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