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Subject: Re: move ordering and node count

Author: martin fierz

Date: 08:09:19 03/29/04

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On March 29, 2004 at 10:52:53, Robert Hyatt wrote:

>On March 29, 2004 at 10:18:55, martin fierz wrote:
>
>>On March 29, 2004 at 10:17:18, martin fierz wrote:
>>
>>>everybody knows that with plain alpha-beta, a fixed number of moves N per node,
>>>and perfect move ordering a search to depth D needs
>>>
>>>nodes(depth) = sqrt(N)^(D/2) nodes.
>>
>>[snip]
>>
>>sorry, typo, that should of course read either:
>>
>>nodes(depth) = sqrt(N)^D or
>>nodes(depth) = N^D/2)
>>
>>but not both... :-)
>>
>>cheers
>>  martin
>
>
>Or better:
>
>nodes = W ^ ceil(D/2) + W ^ floor(D/2) + 1
>
>or, for D even
>
>nodes = 2 * W ^ (D/2) + 1
>

ah, i didn't know about this factor 2. but of course these are just details :-)


>:)
>
>I think someone published a paper on this in the JICCA several years ago,
>Schaeffer comes to mind...  I'll try to locate it when I get to the office...
>
>It was analyzing how close to the "minimal tree" current chess programs get.  I
>think the answer was "2x".  IE we search about 2x what a perfectly ordered
>search would do...

hmm, factor 2 with roughly 90% move ordering, means a gain of roughly 10% less
nodes for every move ordering % you improve.... worth looking into then :-)

cheers
  martin



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