Author: Amir Ban
Date: 00:36:32 06/25/98
Go up one level in this thread
On June 24, 1998 at 21:09:40, Robert Hyatt wrote: >OK.. here's where that fails: start in a position where white has a knight >on b5, black has a rook on a8 and a king on e8. after searching some >meaningless move for black, you eventually discover at ply=2 that Nc7+ >wins at least an exchange, plus maybe wrecks his pawn structure as the knight >tries to get out. > >For the next several black moves that you try at ply=1, each falls to Nc7+. >Finally, you try (say) O-O, and after Nc7 Rc8 there is no problem for black, >so you find something better for white. That "something better" becomes your >primary killer. Back to ply=1, you try another meaningless move, and now >rather than trying the killer Nc7, you try the best move from the last search >here, and *then* discover that Nc7 is better (again). > >That's where you lose nodes. If you keep the count, you will always try >Nc7+ first, because it will have the higher count, and won't get displaced >when you occasionally stumble across a move that refutes it. Seems to me this argument is smokes and mirrors. What guarantees that Nc7+ will be the initial killer ? The order may be reversed: By your example, there is another successful (but less effective) killer in the position (say Be4), which works against O-O. It may be accepted first, and will gain popularity and count as it works against several replies. When Be4 will fail on some move, you will be forced to find Nc7+, but then the high count of Be4 will keep it out of the first place for several more iterations. Actually the chances are that it will keep it out indefinitely, since you will go back to trying Be4 first, and it may continue working for you enough to keep Nc7+ out completely. I use the same procedure Don uses: Two killers, always replace, no counters. I was surprised you guys think you have something better. Tests will decide, true, but I'm not persuaded by the verbal arguments. It seems to me that "always replace" should be equal at least. Amir
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