Algorithm to find lowest common ancestor in directed acyclic graph?

Question

Imagine a directed acyclic graph as follows, where:

• \"A\" is the root (there is always exactly one root)
• each node knows its parent(s)
• the no

Answer 1

If the graph has cycles then 'ancestor' is loosely defined. Perhaps you mean the ancestor on the tree output of a DFS or BFS? Or perhaps by 'ancestor' you mean the node in the digraph that minimizes the number of hops from E and B?

If you're not worried about complexity, then you could compute an A* (or Dijkstra's shortest path) from every node to both E and B. For the nodes that can reach both E and B, you can find the node that minimizes PathLengthToE + PathLengthToB.

EDIT: Now that you've clarified a few things, I think I understand what you're looking for.

If you can only go "up" the tree, then I suggest you perform a BFS from E and also a BFS from B. Every node in your graph will have two variables associated with it: hops from B and hops from E. Let both B and E have copies of the list of graph nodes. B's list is sorted by hops from B while E's list is sorted by hops from E.

For each element in B's list, attempt to find it in E's list. Place matches in a third list, sorted by hops from B + hops from E. After you've exhausted B's list, your third sorted list should contain the LCA at its head. This allows for one solution, multiple solutions(arbitrarily chosen among by their BFS ordering for B), or no solution.