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Subject: Re: Would this work?

Author: Ralf Elvsén

Date: 01:37:58 09/13/99

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On September 12, 1999 at 20:46:26, Robert Hyatt wrote:

>On September 12, 1999 at 19:51:14, Ralf Elvsén wrote:
>
>>I have read about some experiments where you let a
>>program play against itself, but one version is searching
>>one or more ply deeper than the other. From this
>>you can estimate how much stronger a program gets
>>given more time to search.
>>
>>Is it possible to perform an experiment like this on
>>one computer according to the following?
>>
>>You take the principal variation given by the
>>computer which looks like
>>
>>move1  move2  move3 ...
>>
>>i.e. the computer plays move1 and assumes the reply will be move2.
>>
>>Then you consistently make move2 (just like when you're cheating
>>and look at the info from the computer)
>>in reply to move1.  Move2 is based on a search one ply
>>more shallow than move1, so I figure this would work OK
>>to simulate the abovementioned experiments.
>>
>>On the other hand, both moves "aim" for the same position
>>(the one after the last move in the line).
>>
>>I haven't really been able to decide whether this would work or not.
>>Presently I take the position that this would indeed give
>>the same results as an ordinary experiment, but I sway back
>>and forth, so if you think otherwise,
>>you will not have a hard time to convince me :)
>>
>>Ralf
>
>
>It won't quite work.  Because (at least for me) I treat things at the root
>differently than at ply > 1.  For example, if I am in check at ply=1 I don't
>extend because that would extend _every_ move and it makes no sense.  But at
>ply=2 it makes perfect sense to extend.  This means that taking the first move
>off the PV and assuming that the rest would be seen by a 1-ply shallower search
>is not a good assumption...

Ok, it won't work if you want to do a high quality experiment.
But I guess that in general move2 is one ply "weaker" than move1?
How much is a ply, a factor 3-4? That is about 100 rating points.
The effect should be quite easy to see.

Ralf




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