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Given an undirected connected graph, give an efficient algorithm for generating a path that traverses each edge exactly twice
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Answer #1

The algorithm is given as:

Perform a DFS of the graph G starting at an arbitrary vertex. The path required by the problem can be obtained from the order in which DFS explores the edges in the graph. When exploring an edge (u, v) that goes to an unvisited node the edge (u, v) is included for the first time in the path.

Now, run the DFS algorithm once more in the backward direction from the last vertex visited in the last DFS run. In this case, the edge (u, v) is included for the 2nd time in the path, this time in the opposite direction (from v to u). When DFS explores an edge (u, v) that goes to a visited node we add (u, v)(v, u) to the path. In this way each edge is added to the path exactly twice.

The runtime of this algorithm is O(V + E) where V and E are the number of vertices and edges, respectively.

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