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Subject: Re: Is the NPS for minimax devided by NPS in alpha-beta = 5 ?

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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