How do you use a Bidirectional BFS to find the shortest path?

Question

How do you use a Bidirectional BFS to find the shortest path? Let\'s say there is a 6x6 grid. The start point is in (0,5) and the end point is in (4,1). What is the shortest

Answer 1

How does Bi-directional BFS work?

Simultaneously run two BFS's from both source and target vertices, terminating once a vertex common to both runs is discovered. This vertex will be halfway between the source and the target.

Why is it better than BFS?

Bi-directional BFS will yield much better results than simple BFS in most cases. Assume the distance between source and target is k, and the branching factor is B (every vertex has on average B edges).

• BFS will traverse 1 + B + B^2 + ... + B^k vertices.
• Bi-directional BFS will traverse 2 + 2B^2 + ... + 2B^(k/2) vertices.

For large B and k, the second is obviously much faster the the first.

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In your case:

For simplicity I am going to assume that there are no obstacles in the matrix. Here is what happens:

iteration 0 (init):
front1 = { (0,5) }
front2 = { (4,1) }

iteration 1: 
front1 = { (0,4), (1,5) }
front2 = { (4,0), (4,2), (3,1), (5,1) }

iteration 2:
front1 = { (0,3), (1,4), (2,5) }
front2 = { (3,0), (5,0), (4,3), (5,2), (3,2), (2,1) }

iteration 3:
front1 = { (0,2), (1,3), (2,4), (3,5) }
front2 = { (2,0), (4,4), (3,3), (5,3), (2,2), (1,1), }

iteration 4:
front1 = { (0,1), (1,2), .... }
front2 = { (1,2) , .... }

Now, we have discovered that the fronts intersect at (1,2), together with the paths taken to get there from the source and target vertices:

path1: (0,5) -> (0,4) -> (0,3) -> (0,2) -> (1,2)
path2: (4,1) -> (3,1) -> (2,1) -> (1,1) -> (1,2)

We now just need to reverse path 2 and append it to path 1 (removing one of the common intersecting vertices of course), to give us our complete path:

(0,5) -> (0,4) -> (0,3) -> (0,2) -> (1,2) -> (1,1) -> (2,1) -> (3,1) -> (4,1)