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3. Prove that every connected graph has a vertex whose removal (including all adjacent edges will not disconnect the graph by describing a DFS method that finds such a vertex.

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Answer #1

using DFS:-

We do DFS traversal of the given graph.


In DFS tree an edge (u, v) (u is parent of v in DFS tree) is vertex if there does not exit any other alternative to reach u or an ancestor of u from subtree rooted with v.


We need to define low[v] on the given graph as
low[v] = min(disc[v], disc[w])
where w is an ancestor of v and there is a back edge from
some descendant of v to w.
The value low[v] indicates earliest visited vertex reachable from subtree rooted with v.
The condition for an edge (u, v) to be a vertex is, "low[v] > disc[u]".

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