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I understand the basics of minimax and alpha-beta pruning. In all the literature, they talk about the time complexity for the best case is O(b^(d/2)) where b = branching factor and d = depth of the tree, and the base case is when all the preferred nodes are expanded first.

In my example of the "best case", I have a binary tree of 4 levels, so out of the 16 terminal nodes, I need to expand at most 7 nodes. How does this relate to O(b^(d/2))?

I don't understand how they come to O(b^(d/2)).

Please, can someone explain it to me? Thanks a lot!

1 Answer

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In the worst case, where there is no node to be pruned, the full tree will be examined (or the complete tree up to the cutoff at a depth d). Alpha-beta pruning saves, but how much? In the best-case scenario, each node will examine 2b-1 grandchildren to decide on its value. But in the worst-case scenario, the node will examine b^2 grandchildren. Logically this means that the overall algorithm examined O( b^(d/2) ) nodes, the same as a worst-case algorithm whose cutoff is half of d.

For more information regarding the Best-Case Analysis of Alpha-Beta Pruning, refer the following link: http://www.cs.utsa.edu/~bylander/cs5233/a-b-analysis.pdf

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