Author: Robert Hyatt
Date: 19:56:55 09/28/01
Go up one level in this thread
On September 28, 2001 at 12:08:37, Uri Blass wrote: >On September 28, 2001 at 10:55:46, Robert Hyatt wrote: > >>On September 28, 2001 at 00:58:15, Dann Corbit wrote: >> >>>On September 27, 2001 at 23:44:19, Robert Hyatt wrote: >>> >>>>On September 27, 2001 at 19:05:43, Dann Corbit wrote: >>>> >>>>>On September 27, 2001 at 17:48:32, Peter Fendrich wrote: >>>>>[snip] >>>>>>Yes, I buy all that. My intention was to oppose to the "it's impossible" >>>>>>statement. You are talking about some general case. There is no reason why each >>>>>>move has to be 20% because the first one is. That's why I'm talking about >>>>>>isolating cases where the other move might be better. Another question is what >>>>>>happens if the ponder move has only 10% or 5% probability. >>>>>>I have no proofs that these cases are possible to identify but I'm still open >>>>>>for it, until I know better... >>>>> >>>>>Also, it does not have to be either/or. >>>>> >>>>>We could ponder the root for 1/2 of the extrapolated opponent time slice, and at >>>>>that point, change to the pm and ponder that. >>>>> >>>>>It seems to me that there are many possibilities. >>>>> >>>>>Something that is puzzling me... >>>>>If one move is really much better than the others, then we would think that it >>>>>would fail high, re-search, and gobble most of the time anyway. If that does >>>>>not happen, then some of the alternatives must be pretty good. >>>>> >>>>>So, why does pondering root yield only a 2% gain, and pondering the pm give an >>>>>enormous one? >>>>> >>>>>It still does not make sense to me. >>>>> >>>>>I guess I'm just having a hard time understanding why it is so much better to >>>>>ponder the pm instead of the root. >>>> >>>>If by "root" you mean the position _before_ any opponent move, then the reason >>>>is obvious... you will spread your time over N moves, which means that when >>>>the opponent moves, you will have looked at the _right_ move only 1/N of the >>>>time. You still have a long time to search to meet the target time for this >>>>search. >>> >>>By the root, I mean "the root move for the opponent -- after I have made my move >>>but before the opponent returns the response. In other words, the opponent's >>>current position. >>> >>>If the search is so even that time is distributed over N moves, then the chance >>>of picking the right one is only 1/N anyway. >>> >>>If two or three moves are far better than the others, then most of the time will >>>have been spent searching them. >> >>This is not correct. We are using alpha/beta remember. The _best_ move will >>consume about 75% of the total search time. The next best move will take a >>tiny fraction of that to prove it is worse, even if it is only .01 worse. > >I believe that the truth is in the middle. >blunders are often considered for less time when moves that are worse by 0.01 >pawn considered for more time in most of the cases but it is only an average >rule and not a general rule. > >Uri The score difference is _not_ causing what you are seeing. When a move starts to take far longer to dismiss than others at ply=1, it doesn't mean it is close in score to the best move. It means something is changing in that sub-tree, and move ordering is not as efficient as it should be. Otherwise I can mathematically prove that if the score for move 1 is only .01 better than the score for move 2, that the tree will be _exactly_ the same size as if the score for move 1 is 100.00 better than move 2. The score has nothing to do with the size of the tree. That is the function of move ordering...
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