Author: Robert Hyatt
Date: 19:38:41 06/16/00
Go up one level in this thread
On June 16, 2000 at 20:39:07, Ricardo Gibert wrote: >On June 16, 2000 at 14:32:35, Robert Hyatt wrote: > >>On June 16, 2000 at 09:43:56, leonid wrote: >> >>>Hi! >>> >>>Is the number between minimax nodes/second, devided on number of nodes/second >>>for alpha-beta, equal to around five? >>> >>>If, for instance, one position in minimax give 1000000 NPS, the same position >>>but solved by alpha-beta logic will indicate around 200000 NPS speed? >>> >>>Leonid. >> >> >>Speed is not an issue. NPS between alpha/beta and minimax is meaningless. >>Minimax searches way more nodes than alpha/beta, but it also searches faster >>because no moves are tossed out by backward pruning. >> >>The formula you are looking for is this: >> >>for minimax, the total nodes (N) is >> >>N = W^D (W=branching factor, roughly 35-38 for chess, D=depth of search). >> >>For alpha/beta, the total nodes (N) is >> >>N= W^floor(D/2) + W^ceil(D/2) >> >>or when D is even >> >>N=2*W^(D/2) >> >>alpha/beta searches roughly the SQRT(minimax) nodes. Which is significant. >>In effect, alpha/beta will reach a depth approximately 2x deeper than minimax >>alone will reach. >> >>But NPS really isn't part of the issue, only total nodes searched. > >I would just like to add that the above formula assumes perfect move ordering >for alpha/beta. In other words, the formula above for alph/beta gives the number >of nodes for the minimal search tree. In practice, it should do less well. > >Just out of curiousity, what would alpha/betas formula look like with random >move ordering? Difficult to say and it would have to have too many assumptions. IE do you assume that for any fail-high node, there is only one move that will fail high? Or could there be several different moves that would be good enough to cause the fail-high? The formula is derivable, but probably isn't worth the effort.
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