Question

Problem: Given a tree T = (V,E) . Add minimum number of edges so that even if one edge is removed from newly created graph still there is path from any vertex to any other vertex.

I believe the problem can be reduced to increasing the size of min-cut of graph to 2 from current min-cut of 1. But what will be an efficient algorithm to do so.

Answer 1

Here is algorithm, solving this problem for any undirected graph. It may be applied to a tree after some simplifications (step 1 is not needed).

1. Find all bridges in the graph with DFS or with Bridge-finding algorithm by Robert Tarjan.
2. Create a graph (in fact, it is a tree), where each bridgeless subgraph is substituted with a single vertex.
3. Collapse every chain in the tree into a single edge.
4. Find a path between two leaves (of length at least 3, when possible).
5. Pick any two vertices in subgraphs of the original graph, corresponding to both ends of this path, and connect them.
6. Collapse this path into a single vertex.
7. While there is more than one vertex in the tree, repeat from step 3.

Step 4 guarantees that we’ll not get a new unneeded leaf after step 6, which is key to optimality.