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Subject: Re: may I ask you, Mr. R. Hyatt?

Author: Robert Hyatt

Date: 14:57:12 12/03/02

Go up one level in this thread


On December 03, 2002 at 12:54:00, Ingo Lindam wrote:

>On December 03, 2002 at 11:29:56, Robert Hyatt wrote:
>
>>Within the framework of alpha/beta there is no way to do this.
>>
>>Why?
>>
>>Because alpha/beta doesn't search the entire tree.  It searches the first branch
>>at the
>>root, for the score, then it uses that score to prove that the other root moves
>>are worse,
>>without proving exactly how much worse they are.
>>
>>And therein lies the problem.  To get the second best move, you don't just prove
>>that it
>>is worse than the best, you have to search it completely to see exactly how much
>>worse
>>it is, and that drives up the time exponentially.
>
>
>Sorry, but to me is not clear...
>
>WHAT drives up
>WHAT TIME
>exponentially
>in comparisson to the
>TIME it takes to
>OBTAIN WHAT?
>


You are going to have to search a tree that you would normally dismiss
instantly.  And
the size of that tree is exponential with respect to search depth...

In the normal case, finding the second-best move will take almost exactly as
much time
as finding the best move did, a 2x time increase.  To find the best N moves,
will require
N times as long.  But, in cases where a move or two is really better than the
rest, it is
possible that finding the second move will take much more than 1x as long as the
first,
and I have seen cases where it can take 10x or 20x in rare cases.  The cost of
this can be
seen clearly by using Crafty's "annotate" command with N-best mode set to
something like
5, and then comparing how long it takes to annotate a game that way vs
annotating with just
the best move given at each ply...




>or in different words:
>
>It takes time t to obtain               ............(single best variation?)
>then it takes aprox. time t^n to obtain ............(n best variations?)
>where n is                              ............(n(best variations)?)
>assume t = const. (?)
>
>or exponentially with number of plys...?

Yes...  I wasn't very clear.  Since we are comparing two exponential values, the
ratio is somewhat linear although even that isn't true in "interesting"
positions...


>
>(don't need an exact formula, just want to understand what you mean / think of)
>
>Internette Gruesse,
>Ingo



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