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lidden,
Unfortunately it won't - at least, not as I understand it. In the problem posed by neversaint, all paths between two nodes is desired. It is never said that cycle avoidance is required but I have assumed it for my purposes. In other words, each node visited along the path may only be visited once. Any algorithm designed to find the shortest or cheapest path or paths will not produce the correct result - even if every edge has the same cost. Assume the two nodes are directly connected. Imagine taking a Hamiltonian path around the graph with the start and end points being the two nodes in question - this can certainly be proven not to be among the shortest/cheapest paths.

Cheers - L~R


In reply to Re^2: Short Circuiting DFS Graph Traversal by Limbic~Region
in thread Short Circuiting DFS Graph Traversal by Limbic~Region

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