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Subject: Re: DB NPS (anyone know the position used)?

Author: Ed Schröder

Date: 23:08:14 01/26/00

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On January 26, 2000 at 18:31:21, leonid wrote:

>On January 26, 2000 at 13:24:21, Ed Schröder wrote:
>
>>>Posted by Peter W. Gillgasch on January 26, 2000 at 09:18:55:
>>>
>>>In Reply to: Re: DB NPS (anyone know the position used)? posted by Ed
>>>Schröder on January 26, 2000 at 03:07:42:
>>>
>>>On January 26, 2000 at 03:07:42, Ed Schröder wrote:
>>>
>>>>On January 25, 2000 at 23:57:33, Ernst A. Heinz wrote:
>>>>
>>>>>> In a one by one setting it does not matter at all.
>>>>>
>>>>>Still not convinced: a quiescence node that produces a direct
>>>>>"stand pat" cutoff obviously generates less work than one
>>>>>which fails to do so -- even in hardware!  *** QED ***
>>>>>
>>>>>Or am I missing something?
>>>>>
>>>>>=Ernst=
>>>>
>>>>Something else... I always wondered about this free 4-ply evaluation. I
>>>>can understand that evaluation for the current position done in hardware
>>>>is possible in a few cycles. I can't understand this also to be true for
>>>>4 plies as it should involve: search, hash table, q-search etc. In other
>>>>words a complete chess program.
>>>
>>>Well of course they have a complete chess program for interior nodes
>>>in hardware as you know. The idea why I think that the position does
>>>probably not matter too much is because something like 0.07 percent
>>>of the nodes they do are calculated on the SP and the remaining
>>>99.93 percent of the nodes are done on the hardware where the transition
>>>from father to sibling and back has a fixed cost regardless of move
>>>ordering. I am not saying that the size of the tree is not influenced
>>>by the position, I am also not saying that the time it takes to complete
>>>a 4 ply search on the chips does not depend on the position.
>>>
>>>You have experience with one by one move generators since your ARM
>>>program did that. What is your gut feeling, assuming that all moves
>>>spend the same time in MakeMove/UnmakeMove (hypothetical) and all
>>>your move  generators need the same time to produce the next move
>>>(only a little hypothetical) and you have no instruction count
>>>differences between the usual case versus the "get out of check" case,
>>>would you see any major NPS differences between different positions ?
>>
>>I think you mixed me up with somebody else. I always do and have done
>>a full move generation and then sort the move list first based on a fast
>>static evaluation. I have tried the one by one approach but it was not
>>superior.
>
>I am curious to see if I understood correctly what you have said. You find for
>each ply all the moves and aline them in the way that the most promissing goes
>first. What I would like to know is if all the moves that you generate for the
>ply are legal.

Yes.

Ed

>In my logic I use almost everywhere legal moves and I find all the moves for the
>each ply before using them. Found that very few do the same. Now I see that
>probably you do the same. This is why I make this question.

>Leonid.
>
>
>
>>
>>I suspect the reason is Rebel's expensive evaluation function. If you have
>>a fast eval NPS will drop considerable doing a full move generation plus a
>>quick-sort. Having a slow eval like Rebel you hardly see the NPS drop and
>>you can afford such time consuming things.
>>
>>Ed
>>
>>>For me it is pretty much constant, ups and downs by maybe 1/6 which
>>>I attribute to the varying execution times of MakeMove/UnmakeMove and
>>>the differences between "in check" and "not in check" nodes.
>>>
>>>-- Peter



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