# Computer Chess Club Archives

## Messages

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

Author: Ralf Elvsén

Date: 01:09:59 06/17/00

Go up one level in this thread

```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) .