I would like to determine all the biconnected components of a graph $G$. Let $G'$ is a subgraph of $G$ which does not contain the edges belonging to the articulation points of $G$. Are the connected components of $G'$ the biconnected components of $G$?
Could I determine the biconnected components with leaving edges belonging to the articulation points?
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graph-theory
algorithms
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0I think, they constitute a subset of all possible biconnected graphs. There might be a $G'$, with an articulation point, which is biconnected. – 2017-02-13
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0So, is it a right way to get the biconnected components? – 2017-02-13
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0I assume that when you want to get a subgraph of $G$, you delete a vertex and the connected edges. Then, your method does not always give a biconnected subgraph. Consider two $C_4$ graphs that are attached to each other, at one vertex. Call it $v$. $v$ is an articulation point. This is the only articulation point that the graph has. Now , you can see the graph $C_4-v$, which is line graph, is not biconnected. – 2017-02-13
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0You are right, but try to perform a slight modification. Call the edges of $C_{4}-v$ $w_{1}$, $w_{2}$, $w_{3}$. If we consider the subgraph of $G$ which contains all the edges connected to $w_{1}-w_{3}$, do we get a biconnected graph? Could it be generalized? – 2017-02-14