Shortest two disjoint paths; two sources and two destinations

Question

We\'re given an unweighted undirected graph G = (V, E) where |V| <= 40,000 and |E| <= 10⁶. We\'re al

Answer 1

This may be reduced to the shortest edge-disjoint paths problem:

1. (Optionally) Collapse all chains in the graph into single edges. This produces a smaller weighted graph (if there are any chains in the original graph).
2. Transform undirected graph into digraph by substituting each edge by a pair of directed edges.
3. Split each node into the pair of nodes: one with only incoming edges of the original node, other with only its outgoing edges. Connect each pair of nodes with a single directed edge. (For example, node c in the diagram below should be split into c1 and c2; now every path containing node c in the original graph should pass through the edge c1 -> c2 in the transformed graph; here x and y represent all nodes in the graph except node c).

Now if a = b or a' = b', you get exactly the same problem as in your previous question (which is Minimum-cost flow problem and may be solved by assigning flow capacity for each edge equal to 1, then searching for a minimum-cost flow between a and b with flow=2). If a != b, you just create a common source node and connect both a and b to it. If a' != b', do the same with a common destination node.

But if a != b and a' != b', minimum-cost flow problem is not applicable. Instead this problem may be solved as Multi-commodity flow problem.

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My previous (incorrect) solution was to connect both pairs of (a, b) and (a', b') to common source/destination nodes, then to find a minimum-cost flow. Following graph is a counter-example for this approach:

Answer 2

How about this? Do BFS (breadth first search) traversal from a1 -> a2 and remove the path and compute BFS b1 -> b2. Now reset the graph and do same with b1->b2 first and remove path and then a1->a2. Whatever sum is minumum is the answer.