According to most of the reading I have done, a bidirectional search algorithm is said to terminate when the “forward” and “backward” frontiers first intersect. However, in Section 3.4.6 of Artificial Intelligence: A Modern Approach, Russel and Norvig state:
Bidirectional search is implemented by replacing the goal test with a check to see
whether the frontiers of the two searches intersect; if they do, a solution has been found.
It is important to realize that the first solution found may not be optimal, even if the
two searches are both breadth-first; some additional search is required to make sure there
isn’t a shortcut across the gap.
I have considered this statement for quite some time, but am unable to find an example of this behavior. Can anyone provide an example graph where the first intersection between the forward and backward frontiers of a bidirectional BFS or A* search is not the shortest path?
Edit: Clearly BFS will not find the shortest path in a weighted graph. It sounds like this excerpt is referring to bidirectional BFS on a undirected graph. Alternatively, I would be interested in seeing a counterexample using bidirectional A* on a weighted graph.
I don’t know if this is what Russel and Norvig had in mind, but if the graph is weighted (i.e. some edges are longer than others), the shortest path might have more steps than another which would be found sooner in a BFS. Even if the number of steps is the same, the best might not be found first; consider a graph with six nodes:
(A->B) = (A->C) = 0
(B->D) = 2
(C->E) = 1
(D->F) = (E->F) = 0
After one step forward from A and one step backward from F, the forward frontier is {B,C} and the backward frontier is {D,E}. The searcher now expands B and hey! intersection! (A->B->D->F) = 2. But it should still search a little farther to discover that (A->C->E->F) is better.