I've never seen a decisive game where the lichess computer analysis determines no mistake or blunder by either side. Is it possible to have such a game where the worst moves are inaccuracies?
I doubt it. Here is why. The computer analysis starts from the end. It sees the game ended in mate. Hence, as it backs up it will find a move that it calculates caused the mate (because other moves give better points) and then that move will be marked as a mistake.
@brochess have you checked games from engine tournaments?
I have watched such a game on lichess recently.
only inaccuracies- no mistakes or blunders - on both sides.
Ah, it might be argued that the game was not decisive. The one party resigned, but it was not that obvious to me that the game was lost. But then, they are in a different league...
only inaccuracies- no mistakes or blunders - on both sides.
Ah, it might be argued that the game was not decisive. The one party resigned, but it was not that obvious to me that the game was lost. But then, they are in a different league...
@xK4LIBUR Did it end in mate? That's what I thought OP meant by "decisive". Could you post a link?
I found one possible hole in my argument. Suppose White gets mated. So at the mate the eval is some really large (in absolute value) negative number; say -10,000. Now if *every* move, as the computer walks back up the game, also gives a -10,000, then White made no "mistake"; there was no better move. So on reaching the first move every move loses. I don't know what the program does in that case. Funny would be, "1.e4 ?! there is nothing better." :-)
See www.chesshistory.com/winter/extra/breyer.html
See www.chesshistory.com/winter/extra/breyer.html
@anton_webern is there a link to those games on lichess?
@xK4LIBUR yeah by decisive I meant ending in checkmate.
@xK4LIBUR yeah by decisive I meant ending in checkmate.
mmm, on some levels 2 pawns up might be considered decisive, but I do get your point. :)
I would still have blundered that position, even with 2 pawns up.
I would still have blundered that position, even with 2 pawns up.
So now I see another hole in my argument. If every move as the engine backs up is only slightly less bad than -10,000 (in my example) then it might accumulate down to close to 0 by move 1. But that would require either a different number than my -10,000 which is closer to 0, or a *lot* of moves in the game.
@xK4LIBUR Thanks for the post.
@xK4LIBUR Thanks for the post.
This topic has been archived and can no longer be replied to.