Author: GuyHaworth
Date: 01:07:23 02/10/06
Go up one level in this thread
I agree, which is why I say that Black's attempt at the swindle c8=Q?? is only _arguably_ the best defence. After c8=Q, White still needs 17 moves to capture the Rook or mate, so it's not that easy. [D] 2Q5/8/8/8/1K1r4/8/1k6/8 w With strategies SC- v SC+ (minimaxing DTC) we have: 1.Kb5' Kb3 (Rd3) 2.Qc5' Rd8' 3.Qc4+' Kb2' 4.Qe4 (Qe6) Kc3' 5.Qe5+ (Qc6+) Kb3' 6.Qe6+' Kc2' 7.Kc4' Rd2' 8.Qe1' Rd8' 9.Qe4+' Kc1' 10.Qf4+ (Qf5) Kd1' 11.Qg4+' Kc2' 12.Qf5+' Kd1' 13.Kc3' Ke2 (Ke1) 14.Qe5+ (Qe6+) Kd1' 15.Qh5+' Ke1' 16.Qh4+' K~' 17.Qxd8' ... calling for a lot of unique moves (indicated by '). The Reference Fallible Player R(c) that I've defined and written about with Rafael Andrist's help would take 17 moves as White if c = +infinity. It would take longer the less c was, and certainly - at some point - would be expected to take more than 50 moves. With c=0 for example, R(c)'s moves would be as likely to be one (winning) move as another which can be easily simulated by hand. You may feel that White's moves are harder to find than Black's: it's not clear to me. A Monte Carlo run of R(c)-R(c) games for various 'c' would show whether expected-game-length would be more or less than 17. Walter Browne's 'equivalent R(c)' was about R(19) when he played a couple of KQKR endgames against Thompson's BELLE. g
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