# Computer Chess Club Archives

## Messages

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

Author: Ricardo Gibert

Date: 17:39:07 06/16/00

Go up one level in this thread

```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?

```