Author: Dave Gomboc
Date: 16:08:35 06/17/00
Go up one level in this thread
On June 17, 2000 at 04:09:59, Ralf Elvsén wrote: >On June 16, 2000 at 22:38:41, Robert Hyatt wrote: > >>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. > >Dave Gomboc has given the formula W^(2D/3) (approximately) . >I asked about the assumptions behind this but didn't get an answer. >I once made an experiment and found it to be roughly in line with >this formula. > >Ralf Hmm, maybe I missed your post, whenever that was. I don't actually ever recall seeing a derivation of the result. This estimate was discussed (in passing) in a Heuristic Search course that I took (professor: Jonathan Schaeffer). Dave
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