@12249
"Some 0.00 positions are decisive, and some positions with high evals are draws."
++ That is what I say the whole time. The provisional, heuristic engine evaluations like +0.33 play no role, only the final objective evaluation win/draw/loss reaching the 7-men endgame table base, or a prior 3-fold repetition, or a known drawn endgame as judged that the two ICCF World Championship Finalists after days/weeks/months of consideration.
"why engines can be trusted" ++ On the contrary: engine evaluations cannot be trusted unless they reach the 7-men endgame table base or a prior 3-fold repetition.
The way to play ICCF correspondence is exactly to aim for positions engines misjudge.
@12245
"it should take a strong player less than a day to produce an example game"
++ I have played many tournaments. Once in a while the arbiter has to intervene to supply a second queen when a player wants to queen a pawn and his queen has not yet been captured. Luxury chess sets have spare queens for that.
Sometimes underpromotions to knights happen, because of the unique properties of the knight, but when that happens, at least one knight is generally captured. I never saw an arbiter intervene to supply a 3rd knight.
Underpromotions to rook or even bishop happen but only to avoid stalemate, i.e. only by the winning side. Even in the rare cases where these happen usually at least one rook/bishop is already captured. I never saw an arbiter intervene to supply a 3rd rook or bishop.
Apart from that, the vast majority of the 10^44 legal positions has multiple underpromotions from both sides.
https://github.com/tromp/ChessPositionRanking
First position: 7 white rooks, 3 black bishops
Second position: 4 white rooks, 3 black bishops
Third position: 4 white rooks, 3 black rooks
Those 3 randomly sampled positions thus can never result from optimal play by both sides: it makes no sense for both sides to avoid drawing by stalemate by underpromotion to R/B.
"you need to promote with check, for example" ++ The only example I know is the Lasker trap, but it results from non-optimal play by both sides.
1 d4 d5 2 c4 e5? 3 dxe5 d4 4 e3? Bb4+ 5 Bd2 exd3 6 Bxb4 exf2+ 7 Ke2 fxg1=N+