No. 6, Apr-Jun 1999
R0023 - Noam Livnat
StrateGems 6, Apr-Jun 1999
a) Black pawns captured 10 times, white pawns on the board captured 5 times. The missing a- and h-pawn were captured by the black pawns, this requires two more captures. The white a-pawn was captured on b3. Where was the white b-pawn captured? d2xc3 was played after the black pawn arrived on b2, so the white bishop from c1 was constantly trapped there if the b-pawn was captured on b2. But the only captures on black squares were done by the black pawn currently on b2. So the white b-pawn had to make one capture too. But this results in 8 captures by white, and there are only 7 missing pieces. So the position is illegal.
b) The white b-pawn was now captured on b3, and the white a-pawn captured a6xb7, and promoted on b8. This requires only 7 captures in total, so the position is legal.
R0024 - Alessandro Cuppini
1. e4! (1. h7? OOO!) Bc1 2. h7 Bh6 3. Kd6 Bc1 4. hg8=Q#.
This seems to be cooked though: 3. h8=Q/R Bc1 4. Q/Rg8#.
R0025 - Stanislav Vokal
a) 1. Rf5! (2. Rf8#) OOO 2. Qb7#
b) 1. OO! ~ (OOO?) 2. Rf8#. Why may black not castle? White may castle, so Rd5 is promoted. To get a pawn to a promotion square where it wouldn't break black's castling rights, white needs 8 captures (exdxcxb, fxexdxc, and gxfxe). Black misses 8 pieces, but Bf8 was captured on its homesquare. So white didn't promote on b8, so either the pawn had to pass d7 or f7, or the rook had to pass f8, and in both cases, black lost the right to castle.
R0026 - Valery Liskovets
Dedicated to Anatoly Kuznetsov
Bf3 and Bb8 are promoted. Suppose black may castle. Then the bishop on f3 was created on g8 with h7xg8=B. This implies 5 captures by white pawns: axbxc, dxe, fxg, and h7xg8. Black misses 7 pieces, and the bishop on f8 was captured on its homesquare. One of the a-, d-, or h-pawns promoted to bishop, and one other was captured by the white pawns. The black h- and d- pawn couldn't have reached g1 without disturbing the white king. The black a-pawn could've reached c1 with two captures. But the captured pawn needs to make a capture too to go to a field where it could've been captured. This requires too many captures. So white may not castle. So castlings are mutually exclusive.
Is the white castling legal? Yes: Bf3 was created on e8 or c8. This requires 5 captures, axbxc, d7xc(e)8=B, fxe and hxg. In this case, the black bishop could've been created on g1 with only one capture. Together with the capture, needed to get one of the other pawns to a square where it could get captured, and Bf1, these are all 3 missing pieces. So white may castle
Solution: 1. OOO? Ne2! 2. Be2/Kb2 bc6/Nd4!. 1. Ng3! OO 2. Ne7 Kh8 3. Ra4! (OOO? illegal) ~ 4. Rh4#. 1. .. bc6 2. OOO! (2. Ra8? OO, 2. Bc6? Kf8 3. OOO/Ra8 Rh1) Rh1/~ (OO? illegal) 3. Rh1/Bc6 (4. Rh8#/Rd8#) 1. .. Rh4! 2. Ra8! (2. OOO Rd4) Rh1 3. Nh1. 1. .. Rh1 2. Nh1 Ba7/~ 3. Ra7/Ra8 (4. Ra8#/Rb8#).
R0027 - Leonid Borodatov
Suppose white may castle. Then the extra rook must've been created by a promotion on c1. This requires 6 captures, together with the missing Bf1 all missing white pieces. So the f-pawn and the h-pawn were captured by a black pawn too. To preserve as many castling rights for black as possible, this occurred by promoting the h-pawn on h8, and fxe. So white's and black's kingside castling are mutually exclusive. So what was white's last move? It could only have been with a7 (d4xc5 isn't possible). This pawn captured the missing black bishop, so it must've played b6xBa7 last. The only retromove allowing that is b7-b5. So if white may castle, black played b7-b5 as last move. h#2.5 with 1. .. cb6 2. Ba6 OO! 3. OOO a8=Q
R0028 - Andrei Kornilov & Andrei Frolkin
Retroplay is: -1. Ba5-c3 d5-d4 -2. Bd8-a5 d6-d5 -3. c7xNd8=B Nc6xQd8! -4. Qe8-d8 Nb4-c6 -5. c6-c7 c7xNd6 -6. Qd8-e8 N~ -7. Ne8-d6 N~ -8. Ng7-e8 N~ -9. Ne8xRg7 Rh7-g7 -10. Ng7-e8 N~ -11. Qh8-g8 N~ -12. Kg8-f8 N~ -13. Qd8-f8 Re8-e7 -14. c5-c6 Rd8-e8 -15. Ne8-g7, and further retracting involves Qf8->a8; a2xb->b6xRa7-a8-Q; h7->h3xNg2-g1=R.
P0028 - Radovan Tomasevic
P0029 - Pascal Wassong
1. g4 Nf6 2. g5 Ne4 3. g6 Nd2 4. gh7 Nb3 5. Bh6 g5 6. e3 g4 7. Be2 g3 8. Bh5 g2 9. Ne2 g1=N 10. ab3 Nh3 11. Ra4 Ng5 12. Rh4 Ne6 13. b4 Ng7 14. hg7
P0030 - Thierry le Gleuher
Dedicated to Sam Loyd
1. d4 g5 2. Bf4 g4 3. Kd2 g3 4. hg3 Nc6 5. Rh6 Ne5 6. Rb6 Nf3 7. ef3 ab6 8. Bc4 Ra3 9. Qf1 Rc3 10. Na3 h5 11. Re1 h4 12. Re6 h3 13. Rc6 dc6 14. Ke2 Bf5 15. Be6 Qa8 16. Bc8 e6 17. d5 Ne7 18. d6 Ng6 19. d7 Ke7 20. d8=R Rh4 21. Rd1 Nh8 22. Ra1 Kd8 23. Nb1 Ba3 24. Bc1 Rb4 25. Bd7 Be4 26. Qd1
P0031 - Thierry le Gleuher
1. a4 h5 2. Ra3 h4 3. Rg3 hg3 4. h3 Rh4 5. Rh2 gh2 6. Nc3 h1=N 7. Nd5 Ng3 8. Ne7 Nf1 9. Nd5 Nh2 10. Nc7 Qc7 11. Nf3 Nf3 12. Kf1 Qh2 13. Qe1 Bd6 14. Qd1 Bc7 15. Qe1 d6 16. Qd1 Bd7 17. Qe1 Bb5 18. Qd1 Nd7 19. Qe1 Rc8 20. Qd1 Bb8 21. Qe1 Rc5 22. Qd1 Rg5 23. Qe1 f5 24. Qd1 Kf7 25. Qe1 Kg6 26. Qd1 Kh5 27. Qe1 g6 28. Qd1 Ne1
P0032 - Michel Caillaud
1. h4 a5 2. h5 a4 3. h6 a3 4. Rh5 ab2 5. a4 b5 6. a5 b4 7. a6 b3 8. Rha5 f5 9. hg7 h5 10. a7 h4 11. ab8=R h3 12. Rb7 Rh4 13. Rba7 Ra4 14. c4 Nh6 15. Nc3 b1=R 16. g8=R Rb2 17. Rg6 Rba2 18. Rga6 b2 19. Qb3 b1=R 20. Qb8 Rb3 21. Bb2 Rba3
P0033 - Michel Caillaud
1. c4 Nf6 2. c5 Nd5 3. c6 Nb4 4. cb7 c5 5. d4 c4 6. d5 c3 7. d6 c2 8. de7 d5 9. ed8=N Bd6 10. bc8=N Be5 11. Nb6 Ke7 12. Nb7 Rc8 13. Nf3 Rc3 14. Nd4 Rh3 15. Nb5 d4 16. Be3 d3 17. Bd4 c1=N 18. e3 Nb3 19. Qc1 d2 20. Ke2 d1=N 21. Rg1 Nb2
P0034 - Michel Caillaud
1. d4 c5 2. d5 c4 3. d6 c3 4. de7 cb2 5. c4 d5 6. c5 d4 7. c6 d3 8. c7 de2 9. cb8=B ed1=B 10. Bc4 Bc2 11. Ke2 ba1=B 12. Kf3 Bc3 13. Kg3 Qd3 14. Kh4 Kd7 15. e8=B Kd8 16. Bc6 Bc5 17. Bc7
P0035 - Michel Caillaud
1. d4 a5 2. d5 a4 3. d6 a3 4. dc7 ab2 5. a4 h5 6. a5 h4 7. a6 h3 8. a7 hg2 9. ab8=Q gh1=Q 10. Bh3 Rh4 11. Be6 de6 12. Ra7 Bd7 13. c8=Q Bc6 14. Qbd6 Bg2 15. Qc3 Bf1 16. Nh3 Qd5 17. f3 Re4 18. Nf4 g5 19. h4 g4 20. h5 g3 21. h6 g2 22. h7 g1=Q 23. Ng2 Qgd4 24. Be3 Bh6 25. Bg1 Bc1 26. Qcd2 Qd3 27. Nc3 b1=Q 28. h8=Q Qbb5 29. Qd4 Qbd7
P0036 - Satoshi Hashimoto
1. h4 a5 2. h5 Ra6 3. h6 gh6 4. Nc3 Bg7 5. Na4 Bc3 6. dc3 Re6 7. Be3 Re4 8. Bb6 cb6 9. e3 Qc7 10. Qf3 Kd8 11. OOO Qd6 12. Ne2 Kc7 13. Ng3 Kc6 14. Bb5 Kb5 15. Rd4 Nc6 16. Rc4 Ne5 17. Rc8 Rc4 18. Ra8 Rc8 19. Ra6 Ra8
P0037 - Mikhail Kozulya
1. a4 f5 2. Ra3 f4 3. Rc3 f3 4. Rc7 fg2 5. f4 a6 6. Nf3 g1=B 7. Bh3 Ba7 8. Rg1 Nf6 9. Rg6 Nd5 10. Rd6 Nb6 11. Kf2 g5 12. Ke3 g4 13. Kd4 Bg7 14. Kc5 Bc3 15. Nd4 Ba5 16. c3 g3 17. Qb3 Nd5 18. Qb6 Nf6 19. b4 Ng8 20. Ba3 B7xb6