Author: Edward Seid
Date: 22:10:12 09/30/03
Go up one level in this thread
On September 30, 2003 at 18:05:50, Gian-Carlo Pascutto wrote: >If you only have 3 possible scores, I don't think the issue of having to >research actually plays. You center around 0. If you get a score back, it's 0. >If you fail high, the score must necessarily be +9999. If you fail low, it's >necessarily -9999. No need to resolve anything. > >The original poster was correct there will be more cutoffs (I think), and if >the goal is to solve the game, this may be a good optimization. > >But for a beginner, things like this should be avoided since it gets complicated >fast. After doing some pencil-and-paper calculations using a NegaMax with alpha-beta algorithm, I've come to the conclusion that setting alpha = -9999 and beta = +9999 is a good optimization IN MY PARTICULAR APPLICATION. The only valid scores in my application are +9999, zero, or -9999. Therefore, it isn't possible for a returned score to be out of the range [-9999, +9999]. In a typical fail-low or fail-high, there's a possibility that the score will fall outside of the [alpha, beta] range, but that doesn't exist in my application. Consider a 1-ply search with alpha = -9999 and beta = +9999: 1- If the first candidate move is a win for the root node, +9999 is returned. Since this score >= beta, then the rest of the candidates aren't considered. 2- If the first candidate move is a draw for the root node, 0 is returned. Since this score > alpha, then alpha = score (0), and search continues with the next candidate. 3- If the first candidate move is a loss for the root node, -9999 is returned. This has no effect on alpha or beta, and the search continues with the next candidate. I emphasize that this is good for my application, but may not be good for a real chess program that requires a heuristic evaluation at a non-terminal node.
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