Author: Robert Hyatt
Date: 12:23:28 05/31/00
Go up one level in this thread
On May 31, 2000 at 13:22:34, blass uri wrote: >On May 30, 2000 at 18:11:51, Robert Hyatt wrote: > >>On May 30, 2000 at 15:24:36, Ed Schröder wrote: >> >>>On May 30, 2000 at 00:28:47, Robert Hyatt wrote: >>> >>>>On May 28, 2000 at 16:37:32, Gian-Carlo Pascutto wrote: >>>> >>>>>On May 28, 2000 at 10:02:05, Georg v. Zimmermann wrote: >>>>> >>>>>>From my tests it shows that it sticks with the hash-move about 50% of the time. >>>>>>Should this number be higher ? >>>>> >>>>>Hmm...if this number is also effectively your 'move ordering percentage', >>>>>which I assume it is, it is quite low. I'd expect it to be at least about 75%. >>>>> >>>>>> >>>> >>>> >>>> >>>>The classic definition of a "strongly-ordered tree" is this: If, for every >>>>node where you fail high, you fail high on the first move at least 90% of the >>>>time, then your move ordering is good." If you are much below 90% and already >>>>have a serious problem that is not hard to fix. The traditional ordering ideas >>>>holds Crafty at 92% and better for most of the game. >>> >>>I can't understand the 92%. A perfect mini-max search requires many many >>>nodes an alpha-beta cutoff will not work and you are forced to search all >>>the nodes of the ply in question. And this number is certainly much higher >>>than 8%. >> >>You have to re-read the definition again, _very carefully_ to avoid the semantic >>trap you just fell into. >> >>For every position where you fail high, if you fail high on the first move you >>try, you increment a counter "right++". You always increment a counter "fh++". >>When you finish the search, you compute percent=right/fh. That number needs to >>be over 90% to consider your tree strongly ordered. Notice that this 92% number >>(in crafty) simply says this: >> >> "if we look at _all_ the positions in the tree where the search fails high, >> then 92% of those fail highs happen on the first move searched in that >> position, which is known as 'optimal move ordering'. > > >I do not agree that failing high on the first move is optimal move ordering. > >Here is an example: That particular idea isn't open to debate. Alpha/beta is all about minimizing the number of nodes searched. It is easy to prove mathematically that if I get the best move first every time, and you don't, I am going to search fewer total nodes than you are to get the exact same score. > >[D]8/6k1/rp3ppp/8/N7/8/4RPPP/6K1 w - - 0 1 > >My understanding of optimal move ordering is that after the moves Nxb6 or Nc5 >the first move to search will be Ra1+(at least in cases that you are going to >search more than few plies after these moves because Ra1+ Re1 Rxe1# is the >faster way to prove that Nxb6 or Nc5 is wrong) > >If you start with taking the knight than your first move may fail high but you >waste more time to prove that Nxb6 or Nc5 are wrong. > >Uri No... that is the wrong way to think about alpha/beta. In any given position, the 'best' move is the one which produces the best score _in that position_. It doesn't matter a dime what has happened in similar positions, or at shallower search depths. Ie it doesn't matter if a move looks "best" to a human, in the context of alpha/beta, or anything else. It is all about the move that produces a move that causes a cutoff. It is _not_ necessary that alpha search the "best" move first, ever. It is only necessary that alpha/beta searches a move good enough to cause a cutoff... My original statement is still on target: If, at every move where you get a cutoff, you get it on the _first_ move, you are searching the "minimal tree" which is the goal of alpha/beta.
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