Author: Peter W. Gillgasch
Date: 06:18:55 01/26/00
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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 ? 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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