4481 - Ring, Ulrich
1. a4 Nc6 2. Ra3 Nd4 3. Rd3 Nxe2 4. Nc3 Nxg1 5. Qf3 Nf6 6. Qc6 dxc6 7. Nd1 Nd7 8. c3 Nb8 9. Rd7 Nh3 10. Ba6 Ng5 11. Ke2 Ne4 12. Kd3 Nf6 13. Kc4 Ng8
The two bNs swapped place.
5503 - Wolfgang Dittmann
feenschach 91, May 1989
1st Prize feenschach 1989
The first difficulty we meet is to tell what has been captured in d6? The missing wh. Bishop could not have just simply visited d6 without mandatorily capturing on c7 or e7. Perhaps it captured something on d6? But there is only one missing bl. unit: bPh7. Assuming e.g. Bc1-a3xd6 would imply a promotion of the bPh7 after hxg. Unfortunately a bPhxwPg capture cannot be arranged because Wh. would capture 1st (or immediately after in case of an en-passant h4xg3 ep.) Wh. cannot be distracted from capturing the bP by his Bxd6 capture because Bl. has to follow Bxd6 by e7xd6.
It turns out that only a wh. Knight (or a wh. King) can get captured on d6 without requiring a bl. promotion. But then the wPg2 had to promote on h8 after gxh to replace the missing unit. Such a promotion is only possible if Bl. can be distracted from capturing the wPg, e.g. by letting him capture the wh. QB just between White's g5-g6 and g6xh7. Then a promoted wh. Knight can leave the h8 corner without being forced to capture. (A promoted King could not.)
The difficulty now is to see how we can give Bl. the opportunity to capture wh. QB without forcing Bl. or Wh. to capture more units. Only possibility is bNxBb2. Can we arrange that? The Bishop can go to b2 without capturing f6 or g7 if we shield the a1-h8 diagonal. But after bNxBb2, the bN has to get out of b2 without allowing any capture. Nb2-a4 requires wPb2 already on b4 so that Nb2-a4-c5 is not possible. Nb2-a4-b6 cannot be followed by Nb6-c4 (because of wPd2) or by Nb6-d5 (because of wPb4). And a8, c8 & d7 have never been free so that the b2-a4-b6 route deadlocks.
The only possible way out for the bNb2 is via the d1-f2-h3-g5 route. This requires that f2-f3 has been played before f3-f4. So that the stipulation is answered by f2-f3.
Here is a proof game showing how all these maneuvers can be arranged:
1. Na3 Nf6 2. Nc4 Nd5 3. Nd6 exd6 4. Nh3 Nb6 5. g4 Ke7 6. b4 Ke6 7. f3 Qe8 8. Kf2 Be7 9. Ke3 Bd8 10. Kd4 Qe7 11. Nf2 Re8 12. Nd3 Qf8 13. Bb2 Re7 14. Qb1 Qe8 15. Nc1 Qf8 16. g5 Na4 17. g6 Nxb2 18. gxh7 Nd1 19. Kd3 f6 20. Kd4 Qf7 21. h8=N Qg8 22. Ng6 Re8 23. Nh4 Be7 24. Ng2 Nf2 25. Rg1 Nh3 26. Rh1 Ng5 27. f4 Nf7 28. Ne3 Nh6 29. Nd1 Qf7 30. Nc3 Rh8 31. Nd3 Bf8 32. Ne1 Qe8 33. Nf3 Qd8 34. Qd1 Ke7 35. Ke3 Ke8 36. Kf2 Rg8 37. Ke1 Rh8 38. Nb1 Ng8 39. Ng1
6490 - Betzen, Wolfgang
Suppose black can still castle. The white rook had to enter the northeast corner via the h-line, so black's last move wasn't h7xg6. It can't have been with the queen either, so black's last move was with either the rook or the king, in contradiction with the assumption he can still castle. So black can't castle.
1. Rbh1! (2. Rf8 Kf8 3. Rh8) Ra6 (OOO?) 2. Na6 Kd8/Qg8/Qh8 3. Rf8/Rg8/Rh8#
6491 - Christiaans, Frank
Suppose white can castle. Then Rd7 must be a promoted rook. If black can still castle, then his last move must've been b7xc6, but then the promoted rook must've gotten off the 8th rank via d8 or f8, thus breaking black's castling rights. So black may not castle if white can castle.
1. OOO! (Ra7?)
6492 - Marin, Emmanuel
(a) Mate in 4 with 1. Qd7 Kd7 2. Rb7 Bc7 3. Rc7 Kd8 4. Nb7#
(b) Last move must've been f7-f5, so mate in 1 with 1. gf6#.
6493 - Wasilenko, Anatolij G.
(a): 1. Kf7 Bg6 2. Kg8 f7#
(b): Last moves were f7-f5 e5xf6 ep. h#2 before these two moves: 1. Kg5 Nf6 2. Bh4 Bc1#
6494 - Wong, Peter
6495 - Kornilow, Andreij N.
The coloured position is:
The last 5 single moves were: -1. g5xf6 ep f7-f5 -2. g4-g5 Kf6-e6 -3. Qg3-h4
R1 - Günter Büsing, Markus Ott, Thomas Maeder
& Hans-Peter Reich
1. Nf3 a5 2. Nd4 a4 3. Nb3 ab3 4. e4 bc2 5. Qh5 cb1=B 6. Qh7 Be4 7. Qh6 Bh7 8. Qa6 g6 9. d4 Bg7 10. d5 Bd4 11. Bh6 Ba7 12. Bf8 b6 13. Qc8
1. d4 a5 2. Nd2 a4 3. Nb3 ab3 4. Bh6 bc2 5. Nf3 c1=B 6. e3 Be3 7. d5 Ba7 8. Qa4 b6 9. Ng5 Ba6 10. Nh7 Bd3 11. Qa6 Bh7 12. Qc8 g6 13. Bf8
So the answer is: b3, c2, b1, h7, e4, c8 or b3, c2, e3, h7, h7, f8. The first proofgame is unique, the second proofgame has duals however.
R2 - Caillaud, Michel & Wassong, Pascal
1. c4 g6 2. c5 Bg7 3. c6 Kf8 4. cd7 and now:
6557 - Plaksin, Nikita Michailowitsch
Retract Ka3-a4, and =1 with 1. Ba2.
6558 - Caillaud, Michel
1. d4 a5 2. Qd3 Na6 3. Qg6 hxg6[=wP] 4. Nf3 Rh4 5. Bh6 gxh6[=wP] 6. h7 Bh6 7. Ne5 Bc1 8. Nc6 Rf4 9. h4 bxc6[=wP] 10. Rh3 Bb7 11. Rb3 Qb8 12. Rb6 cxb6 [=wP] 13. h8=B Qd6 14. Be5 0-0-0 15. c7 Be4 16. g7 Kb7 17. c8=N Nc7 18. Na7 Ka6 19. b7 Bh7 20. b8=R f5 21. Rb3 Nf6 22. g8=Q Na8 23. Qg3 Rg8 24. Nb5 Rg4
6559 - Wilts, Gerd & Geissler, Norbert
1. d4 h6 2. Bh6 Nc6 3. Bg7 Nd4 4. Bf8 Ne2 5. Qd7 Kf8 6. Qc7 Ng1 7. Qc8 Rc8 8. Rg1 Rc2 9. Bc4 Rb2 10. Bf7 Rb1 11. Rb1 Rh2 12. Rb7 Rg2 13. Ra7 Rf2 14. Rg8 Kf7 15. Re7 Ke7 16. Rd8 Kd8 17. Kf2
6560 - Ruth, Emiliano F.
-1. .... h7xbBg8=R (bR)
-2. Ne6-c5+ c5xb6 e.p. (bP)
-3. b7-b5 b4xbQa5 (bP)
and forward 1. b5#
It is not legal to retract -3 ... b4xP/N/Ba5, as there is none of these missing on the board (white can not have made a capture with one of his knights, as all captures by white have been made by pawns). If we retract -3 ... b4xRa5, white cannot mate forward.
6561 - Ruth, Emiliano F.
The white d-pawn promoted on c8, after three captures. Black promoted twice, after captures axPb, and hxPg.
(a): Because b7-b6 was played before the promotion, and hence the bishop on a7 was there before the promotion, the white knight on c8 could never move to there, unless it is the promoted piece. So white removed one of his knights at the beginning of the game.
(b): The play was: promote the a- and h-pawn on b1 and g1; d7-d6; Bc8->h7/g8; Bb1->h7/g8; g7-g6; Bf8->a7/b8; Bg1->a7/b8; b7-b6; c6xb7; b7xc8. So black only has four retromoves, white must uncapture a piece on c8 after that. The only way that's possible is if the last moves were: d3-d2 Na5-c7 d4-d3 Ba6-f1 d5-d4 Bc8-a6 d6-d5 b7xc8=B. So white removed Bf1 at the beginning of the game.
(c): Same reasoning as in (b) leads to the retroplay: d3-d2 Qg4-d1 d4-d3 Nd8-b7 d5-d4 Qc8-g4 d6-d5 b7xc8=Q. So white removed Qd1 at the beginning of the game.
6634 - Aleksandr A. Kisljak
1. Nf3 Nf6 2. Ne5 Ne4 3. Nd7 Nd2 4. Nf8 Bd7 5. Nd7 Nf1 6. Nb8 Qb8 7. Qd8 Kd8 8. Be3 Re8 9. Bb6 ab6 10. Nd2 Qa7 11. Nf1 Rc8 12. Ng3 Qa8 13. OOO#
6635 - Gerd Wilts & Norbert Geissler
1. e4 d5 2. ed5 Qd5 3. Bd3 Qa2 4. Bh7 Qb1 5. Ra7 Qc2 6. Rb7 Bb7 7. Qc2 Bg2 8. Qc7 Bh1 9. Qb8 Rb8 10. Bg8 Rb2 11. Bb2 Rh2 12. Bg7 Rf2 13. Bf8 Rd2 14. Be7 Ke7 15. Bf7 Kf7 16. Kd2
6636 - Sergej Tkatschenko
Black has no last move, so white moved last.
1. ...Qb2/Qa2/Qb1 2. Qc5/Qd5/Qb5#
6637 - Frank Christiaans
I: Add wNc5, bNa6, and bPa4, b7, d3, d7, e4, e6.
II: Add wNc7, bNd7, and bPa2, a3, a4, a5, a7, c6. Last move must've been b7xc6, and the black pawn structure requires 14 captures, while there are only 13 white pieces missing.
However, this is cooked: Add wNa4, bNc4, and bPa6, b2, b6, b7, c3, c5.
6638 - Stefan Höning
Add wKd3, wQd7, wRe4, wBc2, sNc5, wPd2, bKa4, bRd1, bRd8, bBb1. Last move was Nb3-c5. Removal of any white piece will either result in an illegal triple check, or an illegal check by black.
6639 - René J. Millour
The imitator started in the rectangle a3-a5-h5-h3. So it made at least one more up-movement than down-movement. The white pawns, the white bishop on h2 and the white knight on g2 give five up-movements. The white rooks never left the pawn-cage, so the missing queen was captured by the black e-pawn. The black pawns on d4 and g3, and the rook on g4 give 11 down-movements. Even if you assume that the white queen was captured on d6, you get maximal 11 up-movements by white. So the missing black pieces were all captured on the 8th rank, and one black pawn promoted and got captured on the 8th rank too. This must've been the black g-pawn capturing the white h-rook on h2. So we have 11 white up-movements (3 pawn, 5 queen, 1 rook, 1 knight, 1 bishop), and 11 black down-movements (3 e-pawn, 4 f-pawn, 4 rook, -1 g-pawn). So the imitator started on the 5th rank.
The white rook on h1 is the rook originally from a1. So white made 7 east-movements with this rook, 5 east-movements with his knight from b1, and 5 east movements with his bishop from c1. Black made one west-movement with his e-pawn, and 1 west-movement with his h-rook. Black needs to make at least 17 west-movements too. His queen was captured on d8, or further to the east. The only way he could've made 17 west-movements is if both knights were captured on a8, the promoted g-pawn was captured there too (which means it promoted to a knight), and the queen and bishop were captured on d8. This also means that the imitator started on the a-line. So it started on a5.
A proofgame leading to the diagram position is: [Ia5] 1. d3 g5 2. h3 g4 3. Rh2 g3 4. Nd2 gxh2 5. Nc4 h1=N [Ic5] 6. Bd2 Nc6 7. Bf4 f5 8. Qd2 Nb4 9. Nf3 Nd5 10. 000 Nb6 [If7] 11. Bh2 Nf6 12. Qc3 Rb8 13. Qb3 Na8 14. Qa3 Ng4 15. Nb6 Bh6 [If4] 16. Nxa8 Bg5 17. Nb6 Nf6 18. Nc4 Ng3 19. Qb4 Nd5 20. Qc3 Nb6 [Ib5] 21. Qd4 Kf7 22. Qe3 Na8 23. Ng1 Rg8 24. Nb6 Nh1 25. Nxa8 Kg7 [Id6] 26. Nb6 Ng3 27. Nc4 Nh5 28. g3 Nf4 29. Nd2 Nd5 30. Bg2 Bh4 [Ie5] 31. Qf4 Kf6 32. Qf3 Rg7 33. Qh5 f4 34. Qxh7 Ke6 35. Qh6 Qe8 [Ig5] 36. Qh5 Nb6 37. Qf3 Na8 38. Bh1 Rg6 39. Nc4 Rg4 40. Nb6 Qf7 [Ib4] 41. Nxa8 Kf6 42. Nb6 Qg8 43. Nc4 Kf7 44. Qd5 Kf6 45. Qd6+ exd6 [Ib5] 46. Bg2 Qh8 47. Ne3 Qg7 48. Nf3 Qh6 49. Rh1 Ra8 50. Bf1 Qh8 [Ie6] 51. Ng2 Ke6 52. Kd2 Kd5 53. Rg1 Kc5 54. Rh1 Kb5 55. Kc3 Ka4 [Ib5] 56. Kc4 d5 57. Kd4! Qg8 58. Ke5 Qh8+ 59. Ng1 Qg8 60. Kf6! d4 [Ie4]+ 61. Ke7 Qh8 62. Kf7 Kb4 63. Ke7! Qd8+ 64. Kxd8 Rb8 65. Ke8 f3 [Ic5]! 66. Kf8 Bg5 67. Kg8 Rh4 68. Kg7 Kc3! 69. Kf8+ Kd2 70. Ke8 Be7 [Ic6] 71. Kf7 Bd8 72. Kg7 Ra8 73. Kf7 Kd1 74. Ke8 Ke1! 75. Kxd8 Rb8 [Ib6]! 76. Ke7+ Kd1 77. Kf6 Kd2 78. Ke5 Rh5 79. Kd5 Ke3! 80. Kc4 Ke4 [Ia6]! 81.Kc3+! Ke5 82.Kd2 Ke6 83.Ke1 Ke7 84.Kd1 Rh4 85.Ke1 Ra8 86.Kd1 Ke8 87.Ke1 Rg4 [Ia6]
6743 - Gerald Ettl
The only mate in 1 is with 1. Rd3#. So black moved last. Last moves were: -1. .. Bh7-g8 -2. f4xNe5 Nd3-e5 -3. Ng2-e1 Ne1-d3 -4. Nh4-g2 Bg8-h7 -5. Nf5-h4 Bh7-g8 -6. Nd4-f5 Bg8-h7 -7. Nc6-d4 Bh7-g8 -8. Nb8-c6 Bg8-h7 -9. b7-b8=N Bh7-g8 -10. b6-b7 Bg8-h7 -11. b5-b6 Bh7-g8 -12. a4xNb5 Nd4-b5 -13. a3-a4 Ne2-d4 -14. Qg2-g1 Ng1-e2 -15. Qh3-g2 Bg8-h7 -16. Qh8-h3 Bh7-g8 -17. h7-h8=Q a7-a6 -18. h6-h7 and the whole position unlocks after retracting this pawn to h3, and uncapturing h4xNg3.
6744 - Andrej Kornilow
The position, after colouring and orientation, is:
The last moves were: -1. OOO c6-c5 -2. Qf6-h4 h4-h3 -3. Rh3-h2 Rh2-h1
6745 - Günter Gla?#376;
Retract -1. .. Nd5xQb6 -2. Bd1xPb3 and forward selfmate in 4 with 1. Qc5 Ndb4 2. Qc1 b2 3. Ra2 Na2 4. Qc3 Nc3#.
6746 - Gerald Ettl
Retract -1. Kg1-h1 d3xRe2 -2. Re1-e2 c4xd3 -3. Rf1-e1 c5-c4 -4. OO c6-c5 -5. Bf1-d3 c7-c6 -6. e2xBf3 Bb7-f3 -7. a3-a4 Bc8-b7 -8. a2-a3 b7-b6 -9. b6xBa7 Bb8-a7 -10. b5-b6 Ba7-b8 -11. b4-b5 Bc5-a7 -12. b3-b4 Bf8-c5 -13. b2-b3 e7-e6 -14. e6xNf7 and the position unlocks.
6747 - Gerald Ettl
The white a-pawn captured twice, so both the black g- and f-pawn promoted. The position unlocks when black uncaptures the blacksquared bishop, and white retracts it to c1, so d2-d3 can be taken back. The last moves must've been -1. Bc6-h1 h7-h6 -2. Bb5-c6 Nb8-a6 -3. Ba6-b5. Now the two white knights and the black knight are free to move around. There are a few possibilities now:
6748 - Gerald Ettl
Black's a-pawn captured twice, and white captured twice with his d- and h-pawn. Due to the question, we should try to unlock the position without taking back g6-g5. And this is possible: Last moves were -1. .. Qd8-a8 -2. Rc8-c7 Qb6-d8 -3. Rc7-c8 Qg1-b6 -4. f5-f6 Qh1-g1 -5. f4-f5 h2-h1=Q -6. f3-f4 h3-h2 -7. f2-f3 h4-h3 -8. g3-g4 h5-h4 -9. g6-g7 h6-h5 -10. h5xQg6 Qd3-g6 -11. h4-h5 Qa6-d3 -12. h3-h4 Qa8-a6 -13. h2-h3 Qd8-a8 -14. Rc8-c7 h7-h6 -15. Ra8-c8, so it is possible that black played g7-g5.
6749 - Johan de Boer
(1): White still has all 16 pieces, so he didn't promote.
(2): Because of (1), the black pawns are all on their original file.
(3): So the white a-pawn captured axbxa, and the white b-pawn captured bxa
(4): Further white captures are exd, fxe, gxf (twice), and hxg. This accounts for all missing black pieces.
(5): None of these captures were on the h-line.
(6): Because of (2), the black d-pawn was captured with exd.
Now start counting the black moves. The black queen needed at least 1 move, the black king 4, and the black bishop from f8 needed 2. No white capture was on the 8th rank because of (3) and (4), so Ra8, Nb8, Bc8, Ng8 and Rh8 all moved at least once. Ng8 moved twice (wasn't captured on f6 because of (4), and not on h6 because of (5)), and Rh8 moved twice too (because of (5)).
The black a-pawn made at least 4 moves, the b-pawn at least 3, the c-pawn at least 1. The d-pawn wasn't captured on d7 (otherwise 32. d6 wasn't possible), so made at least one move too. The f-pawn made 1 move, the g-pawn made at least 2, due to (4), and the h-pawn made 5 for promotion, and then at least two more to get to a square where it could get captured. In total 33 black moves.
The white moves: The white king made at least 1 move, the white queen at least one. If Rb1 came from a1, then the rook on h4 had to move three times to make place for the black h-pawn. So the rooks moved at least three times. The white bishops moved 6 times, and the knights 6 times too. The white pawns moved at least 16 times, so white made at least 33 moves too.
The capture squares for white: Ra8 was captured after doing one move, and Nb8 too. The only possible capture squares for these two pieces are a6 (for the knight), and a5 (for the rook). Bc8 was captured on the c8-h3 line, but not on d7 (6) or h3 (5). g4 isn't possible too, because the black g-pawn was captured there, so it was captured on e6 or f5. The black knight from g8 wasn't captured on d7 or g4 either. d5 (black d-pawn) or e4 (no capture there) weren't possible too, so the knight was captured on f5.
Because white played 32. d5-d6, e2-e4 was played after the bishop went to b8. The promotion on h1 occurred after e2-e4, because the rook couldn't have gone to b1 until white played Ke1-e2. Because black couldn't have played b7-b5 until white captured b5xa6 (and before a4xb5), the promoted piece wasn't captured on b5, so the rook from h8 was captured on b5.
Because white had to play Rh1-b1 before the promotion, b2-b4 must've been played before this promotion too, to let Bc1 out, and Ra4-h4 was played before b2-b4, so before the promotion too. Because e4 or d5 were occupied too, the only possibility is that black promoted to a bishop, and this bishop got captured on f5, after it moved Bh1-e4-f5. So the bishop from c8 was captured on e6
Conclusion: There were two captured on f5: the h-pawn was captured on f5, after it promoted to a bishop, and the knight from g8 was captured on f5.
A proofgame to show all this is indeed possible: 1. a4 h5 2. h3 h4 3. Nc3 Rh5 4. Nf3 Rb5 5. ab5 g5 6. Ra4 g4 7. Ng5 Nh6 8. hg4 Nf5 9. gf5 h3 10. Rh4 a5 11. g4 a4 12. b4 Ra5 13. Bg2 Na6 14. ba6 b5 15. Ba8 d5 16. e4 Be6 17. Ke2 Kd7 18. Bb2 Kd6 19. Qa1 a3 20. Rb1 h2 21. Nd1 h1=B 22. ed5 Be4 23. fe6 Bf5 24. Bd4 Bh6 25. Nb2 f6 26. Bb6 c5 27. c4 Ke5 28. Nf7 Kd4 29. gf5 Bf4 30. ba5 Qb8 31. Bd8 b4 32. d6 a2 33. d7 b3.
6750 - Markus Ott and Dirk Borst
1. b4 a5 2. Bb2 a4 3. Qc1 Ra5 4. Kd1 Rh5 5. Be5 Nf6 6. Qa3 Nd5 7. Kc1 Nf4 8. Kb2 d5 9. Kc3 Qd6 10. Kd4 Qh6 11. Ke3 g6 12. Kf3 Bg4 13. Kg4 Kd7 14. Kf3 Ke6 15. Ke3 Kf5 16. Kd4 e6 17. Kc3 Bb4[=wB] 18. Kb2 Rd8 19. Kc1 Rd6 20. Kd1 Rb6 21. Qc1 c6 22. Ba3 Rb3 23. Ke1 b5 24. Qd1 b4 25. Bc1 a3 26. Beb2 ab2[=wB]
6751 - Markus Ott and Dirk Borst
1. e4 d5 2. e5 d4 3. e6 Be6[=wB] 4. Bd5 Qd5[=wQ] 5. Qc6 Nc6[=wN] 6. Nb4 000 7. Nd5 Rd5[=wR] 8. Re5 d3 9. Re2 de2[=wB]
6845 - Gerald Ettl
Last moves were: -1. Bg2-f3 Kh4-h5 -2. Ng3-f1 Kh5-h4 -3. Nh1-g3 Kh4-h5 -4. Nf2-h1. Now the black king makes waiting moves on h4 and h5, while white retracts c6-c8=B-g2; black uncaptures a rook with c7xRb6; this rook screens on g3, so the white knight can go to h1 and the bishop to c1. The rook now goes back to b1, and white uncaptures b2xPa3. The knight on h1 now goes to h5, and the white queen opens the cage by retracting Qh6-g7.
6846 - Gerald Ettl
Retract -1. .. d7-d6 -2. Kd6-e5 e5-e4 -3. Rf5-f8 e6-e5 -4. Rd5-f5 f5-f4 -5. Rd3-d5 f6-f5 -6. Rc3-d3 f7-f6 -7. Nf6-g4 g4-g3 -8. Rc1-c3 g5-g4 -9. Bg4-h5 g6-g5 -10. Rh5-h8 g7-g6 -11. Ng6-h4 h4-h3 -12. Rd5-h5 h5-h4 -13. Rd3-d5 h6-h5 -14. Rc3-d3 h7-h6 -15. Rc2-c3 Kb4-a5 -16. c3-c4 and the position unlocks.
6847 - Frank Christiaans
1. Nf3 d5 2. Rg1 d4 3. Nd4 Nd7 4. Nf5 Nb6 5. Ng3 Bd7 6. Nh1 Qb8 7. g3 Kd8 8. Bh3 Be8 9. Bc8 Nc8 10. Kf1
6848 - A.A. Kislyak
1. a4 Nf6 2. a5 Rg8 3. a6 ba6 4. Ra6 Bb7 5. Rf6 a5 6. Rf7 Na6 7. Rf8 Kf8 8. e4 Qe8 9. Qh5 Qh5 10. f3 Qf3 11. Nh3 Qa3 12. Bd3 Re8 13. OO
However, this is cooked with for instance: 1. Nh3 b5 2. a4 ba4 3. Ra4 Bb7 4. Rf4 a5 5. Rf7 Na6 6. Rf8 Kf8 7. e4 Qe8 8. Qe2 Qf7 9. f3 Qf3 10. Qc4 Qa3 11. Qg8 Rg8 12. Bd3 Re8 13. OO
6849 - Andrej Frolkin
The last moves were: -1. Re7-e8 and now:
(a): -1. .. b2xNa1=N -2. Bh2-g1 a3xNb2 -3. Bg3-h2 a4-a3 -4. Bh4-g3 a5-a4 -5. g3xNf4
. (b): -1. .. Bf8xNg7 -2. Ne8-g7 Bg7-f8 -3. Rf8-f7
In both cases, a white knight visited (and left) e8.
6850 - Gerald Ettl, Hans Gruber and Frank Müller
I: Add wKf2, wPe2, g2, g3, bKh2, bRh3, bBf1.
II: Add wKf1, wPf2, f4, h3, bKh2, bRg2, bBg3.
6851 - Gerald Ettl
Not 1. Be6[+bRa8]#?, but 1. .. Nh4#.
The last moves were: -1. b2-b3 Bd8xPb6[+wPb2] -2. b5-b6 Bc7-d8 -3. b4-b5 Bb8-c7 -4. a3xNb4 Nc2-b4 -5. a2-a3 b6xPa5[+wPa2] -6. a4-a5 Na3-c2 -7. a5-a6 Nb1-a3 -8. a2-a4 Na3xNb1 -9. a4-a5 Kc3xPd4[+wPd2] -10. Nd2-b1. So black indeed has the move.
6852 - Gerald Ettl
The intention was: -1. Kb8-a8 a7-a6 -2. Ka8xBb8[+bBf8] Kf8xRe8 -3. Rd8xQe8[+bQd8] Kg8-f8 -4. h4-h5 Qf8-e8 -5. Re8-d8.
But these moves aren't unique, since white can wait with Ka8-b8-a8-.. and black can wait with Nf1-h2-f1-...
6853 - Gerald Ettl
Retract: -1. Bf8-h6 -2. Ba3-f8 -3. Bc1-a3 -4. Ba3xBc1 -5. Bf8-a3 -6. Bh6-f8 -7. Bf8xBh6 -8. Bb4xBf8 -9. Ba3xBb4 -10. Bc1xBa3[+wBc1] -11. c2-c1=B -12. b3xPc2. Now black's last move must've been f7-f5, so mate in 1 with g5xf6 ep[+bPf7]
6854 - Rudolf Queck
I: Retract Ke4xQd5, and h=1 with 1. Kd4 Qc4=
II: Retract Ke4xRd5, and h=1 with 1. Na5 Ra5=
III: Retract g6xNf5, and h=1 with 1. Nd6 Nd6=
6855 - Sergej Tkatschenko
1. Nf3 Nc6 2. Ne5 Nd4 3. Nd7 Nf3 4. gf3 Nf6 5. Bh3 Nd7 6. Bd7 Kd7 7. OO c5 8. Nc3 Qa5 9. Nd5 Qd2 10. Nf4 Qd6 11. Qd6 ed6 12. Be3 a5 13. Rae1 Ra6
However, this is cooked: 1. Nc3 Nf6 2. Nd5 Nd5 3. Nh3 Nc6 4. Nf4 Nd4 5. Nd5 Nf3 6. gf3 c5 7. Bh5 Qa5 8. Bd7 Kd7 9. OO Qd2 10. Nf4 Qd6 11. Qd6 ed6 12. Be3 a5 13. Rae1 Ra6
6856 - Andrej Frolkin and Aleksandr Switschenko
1. d4 e5 2. d5 Qg5 3. Qd4 Kd8 4. Qe5[=bQ] Qe8 5. d6 Be7 6. de7[=bP] d6 7. g3 Bh3 8. Bh3[=bB] Bc8 9. Bg5[=bB]
6962 - Andrei Kornilov
White has four mates (Qe4, Qf3, Ne6 and g3), black has no mate.
All captures were done by the pawns. The black f-pawn captured the white e-pawn, and the a-pawn captured the missing rook on b5. The position unlocks when white can retract Qd4-d5, and a retro-screen on e4 is necessary for that. The only piece that can provide this screen is the white knight on g7. Its route was g7-e8-c7-a6-b8-d7-f8-h7-g5-e4. Black needs a tempo to let the knight pass h7. Black's only possible last move could've been a6xRb5. But since the white knight needs to go via a6, black needs to retract a7-a6 too, thus killing his last tempo. So white moved last in the diagram position, and there is no legal mate in 1.
Retro play is: -1. Nh6-g8 Bg8-h7 -2. Ne8-g7 Bh7-g8 -3. Nc7-e8 Bg8-h7 -4. Na6-c7 Bh7-g8 -5. Nb8-a6 Bg8-h7 -6. Nd7-b8 Bh7-g8 -7. Nf8-d7 Bg8-h7 -8. Nh7-f8 a6xRb5 -9. Ng5-h7 ~ -10. Ne4-g5 ~ -11. Qd4-d5 etc.
6963 - Andrei Kornilov
White can mate in 1 with Rh3, black can mate in 1 with Ne8. Which one is legal?
All captures were done by pawns. Pa2 promoted on a8, Pc2 on d8, Ph2 on h8. The only pieces that can move are the black pawn on b5, the white knight on h7, and the black bishop on g8 (after the knight moved). The position unlocks by unpromoting Nh7 on a8, and retracting axb. So black's last move couldn't have been b6-b5 or a6xb5. So white moved last. Mate in 1 with 1. .. Ne8#
Retro play is: -1. Nf8-h7 Bh7-g8 -2. Nd7-f8 Bg8-h7 -3. Nb6-d7 Bh7-g8 -4. Na8-b6 Bg8-h7 -5. a7-a8=N Bh7-g8 -6. a6-a7 Bg8-h7 -7. a5-a6 a6xQb5! -8. Qe8-b5 Bh7-g8 -9. Qh8-e8 Bg8-h7 -10. h7-h8=Q a7-a6 -11. h6-h7 Bh7-g8 -12. h5-h6 etc.
However, there is an annoying dual in the retro-play: -5. .. b6-b5 -6. a6-a7 a7xBb6 ... -9. Bc1-d2 ... -10. d2xQe3 Q~-e3 -11. e3xRf4 Rc4-f4 etc.
6964 - Peter Wong
Retract -1. .. Ra8-b8 -2. Ka1-b2 Nb8-a6 -3. Ka2xNa1 [+bNb8] a7-a5 -4. Kb3xPa2 [+bPa7] b7-b5 -5. Kc2xPb3 [+bPb7] c7-c5 -6. Kc1xPc2 [+bPc7] Nb8-d7 -7. Kd1xNc1 [+bNb8] Bc8-e6 -8. Kd2xBd1 [+bBc8] Bf8-d6 -9. Kd3xBd2 [+bBf8] d7-d5 -10. Ke2xPd3 [+bPd7] e7-e5 -11. Kf1xPe2 [+bPe7] Ke8-d8 -12. Kg1xKf1 [+bKe8] Qd8-h8 -13. Kh1xQg1 [+bQd8] Rh8-h7 -14. Kh2xRh1 [+bRa8] h7-h5 -15. Kg2xPh2 [+bPh7] g7-g5 -16. Kf2xPg2 [+bPg7] f7-f5 -17. Ke1xPf2 [+bPf7] Be1xKd2 [+wKe1] -18. Kc3xRd2 [+bRh8], and now stalemate with 1. Kc3-b2.
7088 - Thomas Volet
The whole position unlocks when a retro-screen on e8 can be built, so white can retract Rf8-f7. The retroplay is: White queen goes back to a8, black bishop goes to b8 to provide a screen, and the queen unpromotes. On its way to a8/b8, the white queen first has to form a screen for the black bishop. After unpromoting, black can retract a7xBb6, the bishop goes back to c1, white uncaptures d2xNc3, and this knight forms the retro-screen on e8.
A cyclic retro-screen (White queen screens the black bishop, and then the black bishop screens the white queen)
However, this is cooked: the black king can uncapture a white knight on d8, and the rook from c7 can escape, and either the knight or the rook can unpromote.
7089 - Thomas Volet
The black bishop first screens the black king, so the white bishop can escape. Then the white bishop screens the white king, so the black bishop can go to c8 (but first the rook from d8 has to be brought back to b8!)
A cyclic retro-screen, this time with a white and a black bishop.
7090 - Thomas Volet
White captures are a6xb7, dxexf8=B, and g2xh3. Black capture is e7xf6. The position unlocks when black can put a piece on g1, providing a screen to let out the black king. Note that white may not retract c2-c3 too soon or the king can't escape the first two ranks!
The retroplay is: The white queen forms a retroscreen on the a3-e6 diagonal to let the black bishop out. This bishop goes to g3, to form a retroscreen to let the white rook on g6 out. This rook forms a retroscreen on the b7-g2 diagonal to allow the white queen to pass b7, so it can unpromote and capture a6xNb7. This knight can be used on g1 to free the black king.
Cyclic retroscreen: wQ screens for the bB, bB screens for the wR, wR screens for the wQ.
7091 - Andrei Kornilov
The black queenside pawns captured 3 times, which is all missing white pieces. So the pawn on h3 moved there without captures, implying the white h-pawn captured twice. Together with the captures from the d- and e-pawn these account all missing black pieces. So the last move was Rf5-f6#
The white a-pawn promoted, without capture. e2xf3 can't be retracted until the white bishop is on f1, a6xb5 can't be retracted until white unpromotes, b7xc6 can't be retracted until the black bishop is on c8. The only way to free the position is to bring the white king to g6, so the white queen can escape. Unfortunately the white king needs a few extra moves due to the black pieces on a8 and a7.
Retroplay is: -1. Rf5-f6 Bb8-a7 -2. Ka6-b7 Ba7-b8 -3. Ka5-a6 Rb8-a8 -4. Kb4-a5 Ra8-b8 -5. K~ Bb8-a7 -6. Kb4-~ Ra1-a8 -7. Ka5-b4 R~-a1 and now Ka5->b7; Nc8-a7; Kb7->g6; Qg5->a8; Na7-c8; a5-a8=Q; a6xQb5; b6-b4; Kg6-g5; R->f6; Kf7-e6 etc, etc.
So in total (g5->b7 = 16 moves, e1>g5 = 4 moves) 20 moves by the white king.
7092 - Nikita Plaksin
Black may not take back g3xNh2 [-bPh2; +bPh7], because in Anti-circe, the pawns on g6 and g3 could never get past each other!
Last 10 moves were: -1. Ne8-c7 h4-h3 -2. Bd1-h5 h5-h4 -3. Qc1-h6 h6-h5 -4. Bg1-a7 Kf8-g8 -5. Ra1-a8++ Kg8-f8
7093 - A. A. Kislyak
A possible proofgame is:
1. b4 c5 2. Bb2 c4 3. Bg7 c3 4. e3 e5 5. Bc4 Qh4 6. Nf3 Ne7 7. OO cd2 8. c3 b6 9. Qa4 Bb7 10. Qa7 h6 11. Be6 de6 12. Na3 Kd7 13. Nc2 Kc6 14. Qb8 Ra3 15. Kh1 Kb5 16. Rg1 Ka4 17. Bf8 Qf2 18. h4 Rg8 19. Ncd4 Rg3 20. b5 Rh3 21. gh3 Rb3 22. Rg6 fg6 23. ab3#
There are 12 groups of pawns:
7094 - Klaus Wenda
Try: -1. Na7xBb5 h4-h3 -2. Kb7-c8 and forward 1. Bb6 Ba6 [+wRh1], but 2. Ka8!
So in the pre-plan, the white rook from a6 should become promoted, so it'll get reborn on a8:
Retract -1. Nb3-c1 h4-h3 -2. Na1-b3 g5xRh4 [-] -3. Na7xBb5 [f/h]6xRg5! -4. Kb7-c8 and now forward 1. Bb6 Ba6[+wRa8]#.
7095 - Stefan Klebes
1. a4 c5 2. Ra3 c4 3. Rh3 c3 4. Rh7 cb2 5. Rh8 bc1=N 6. Rg8 Ne2 7. Ne2 d5 8. Rg7 d4 9. Rf7 Qd5 10. Rf8 Kd7 11. Rc8 d3 12. Rb8 de2 13. d4 ed1=N 14. Bc4 Nc3 15. Nc3 Rb8 [+wRa1] 16. OOO Rd8 17. Bd5 Kc8
7096 - Stefan Klebes
1. h4 d6 2. h5 Bh3 3. h6 Bg2 [-Pf2, -Bf1, -Ng1, -Rh1] 4. fg7 [-Pf7, -Bf8, -Ng8, -Rh8, -Ph7] Bd5 5. g8=R Bg8 [+wRh1] 6. OO Ba2 [-Pb2, -Ra1, -Nb1] 7. Ba3 Bc4 8. Bd6 [-Pc7, -Pe7, +bPd6] Be2 [-Pd2, -Qd1, -Rf1, +wRh1] 9. Bb8 [-Pa7, -Pb7, -Ra8] Qb8 [+wBc1] 10. Bh6 Qg3#
7160 - Sergei Tkachenko
Last moves were -1. OO b2-b1=B -2. d2-d3 c3x[N/B]b2
7161 - Valeri Liskovets
(a): If white may castle kingside: 1. Kf4 OO 2. Kg3 Rf3#. If white may castle queenside: 1. Kd4 OOO 2. Kc3 Rd3#.
(b): 1. Ke3 Rh2 2. e4 Rh3# / 1. Kf3 Ra3 2. Kg2 Rg1#.
7162 - Valeri Liskovets
1. Rg2=wR Rb6=wR!! 2. Re8=bR! OO#! (2. .. Rf1#??, 1. .. Ra(c)8=wR??)
One of the white rooks is volage, and one of the black rooks is volage. With the castling on move 2, white proves a posteriori that Rh1 is volage, so, by the post factum rule, Rb1 becomes non-volage, so it won't change colour. And by changing colour of the rook on the first move, black proves that that rook was volage, and Rh8 was normal.
However, there is a cook: 1. Ra2=wR Rh7 2. Kf8 Ra8#.
7245 - Markus Ott
1. f3 h5 2. Kf2 Rh6 3. Ke3 Rc6 4. Kd3 Rc3 5. dc3 Nc6 6. Be3 Rb8 7. Ba7 b6 8. Kd2 Ba6 9. Ke1 Bd3 10. ed3 Rc8 11. d4 Ra8 12. Bd3 Nb8 13. Bh7 g6
7246 - Gianni Donati
1. Nc3 c5 2. Ne4 Qb6 3. Ng5 Qg6 4. Nh7 Nc6 5. Nf6 ef6 6. e4 Bd6 7. e5 Bb8 8. e6 d6 9. e7 Bf5 10. c3 Kd7 11. e8=N b5 12. Nc7 b4 13. Nb5 a5 14. Na3 a4 15. Nb1 Bb1
7247 - Sergei Tkachenko
1. Nf3 Nc6 2. Nd4 Nd4 3. Na3 Nc2 4. Nc2 Nf6 5. Nd4 Ne4 6. Qa4 Nd2 7. Qd7 Kd7 8. Ne6 Ne4 9. Bh6 Ng5 10. OOO Ke8 11. Rd8
7250 - Gerald Ettl
The position after colouring is:
Mate in 1 with 1. .. Bd2#
7252 - Mario Velucchi
(a) 1. Qf3 OOO 2. Kc3 Rd2 3. Rb4 Rc2#
(b) White may castle only if his last move was c2-c4. So the solution is: 1. dc3 e3 2. Qb5 OOO! (Rd1?) 3. Kc4 Rd4#
7253 - Theodor Steudel
Retract -1. Kb4xQa5, and forward: 1. Kc4 Qb4 2. Kd3 Qc4 3. Ke3 Qd3 4. Kf2 Qe3 5. Kf1 Qf2 6. Kf2=
7367 - Tomislav Petrovic
Setplay: 1. ... OOO 2. Bg3 Rd4#
If white may castle, then his last move was g2-g4. So h#2 with 1. hg3 OOO! (Rd1?) 2. a1=N Rd4
The Valladao theme.
7368 - Gerald Ettl
Last moves were: -1. Qh2xQh3 Qg4-h3 -2. Qg1xRh2 Rh3-h2 -3. Qf1xBg1 Bh2-g1 -4. Qg1xNf1 Nd2-f1 -5. Qf1-g1 and the whole position unlocks.
7369 - Gerald Ettl
The last moves were: -1. . Qg6-h7 -2. Kg7-h8 Qh5-g6 -3. Kh8-g7 Bh6-g5 -4. Kg7-h8 Bf4-h6 -5. Kh8-g7 Qg5-h5 -6. Kg7-h8 Ng6-h4 -7. Kh8-g7 Ne5-g6 -8. Kg7-h8 Qh4-g5 -9. Kh8-g7 Bh6-f4 -10. Kg7-h8 Bg5-h6 -11. Kh8-g7 Nh5-g3 -12. Kg7-h8 Nf4-h5 -13. Kh8-g7 Qe1-h4 -14. Kg7-h8 etc.
7511 - Theodor Steudel
Add the white king on c6. White moved last. =1 with 0. .. Kc8/Ka8/Ka7 1. Bc7/Kb6/Bb7=
7512 - Andrei Frolkin
Try: Retract -1. .. e7xPf6? -2. f5-f6 g4-g3 -3. f4-f5 f5xBg4 -4. Bf3-g4 f6-f5 -5. Bd5-f3 f7-f6 -6. Ba2-d5 h4-h3 -7. Qa1-b1 h5-h4 -8. Qd4-a1 h6-h5 -9. Bd5-a2 Ka2-a3 -10. Bg2-d5 Ka3-a2 -11. Qg1-d4 Ka2-a3 -12. ~ Ka3-a2 -13. Kb1-c1 h7-h6 -14. Ka1-b1?? -15. Qb1-g1, but this fails because black has no retro-move after retracting Ka1-b1.
Instead, the following retro-play is necessary: -1. .. e7xNf6! -2. Ne4-f6 g4-g3 -3. Nc3-e4 h4-h3 -4. Na2-c3 h5-h4 -5. Qa1-b1 h6-h5 -6. Qd4-a1 h7-h6 -7. Nc3-a2 Ka2-a3 -8. N~ and now either wN-a3; Na4-c3-d5; Ra5-a4 or wNf8; f7-f8=N;...;f4-f5 and black has 3 tempos for wKc1-a1; wQb1.
So the white pawn from f2 promoted on f8 to knight, and then was captured on f6.
7513 - Andrei Frolkin
The retroplay is: -1. Qa1-b2 e7xRd6 -2. Kc1-b1 Rb2-a2 -3. Qb1-a1 Rb3-b2 -4. f6-f7 f7xBe6 -5. Rd4-d6 Rb2-b3 -6. Ba2-e6 Rb3-b2 -7. Qa1-b1 Rb2-b3 -8. Rh4-d4 Rb3-b2 -9. Rh1-h4 Rb2-b3 -10. Kb1-c1 Rb3xNb2! -11. Rc1-h1 Rd1-d2 and everything unlocks.
Try: -4. .. Rb2-b3 -5. Rd4-d6 ... -7. .. Ra2-b2 -8. Qa1-b1 f7xe6 -9. Kb1-c1?? -10. Rc1-h1, black has no retromove on move -9!.
7514 - Andrei Frolkin
Retro play is: -1. Na4-b2 g6xBh5 -2. Bg4-h5 f7xRg6 and now:
(I): -3. Bh3-g4 e4-e3 -4. Bf1-h3 e5-e4 -5. Rg1-g6 e6-e5 -6. g2xBf3 Bd5-f3 -7. Rh1-g1 Ba2-d5 -8. Qa1-b1 Bb3-a2 -9. Qc3-a1 Ba2-b3 -10. Qc6-c3, and then Kc1-a1; bBb1; Rd2-e1; wQd1; bB out; wK out
(II): -3. Be6-g4 e4-e3 -4. Ba2-e6 e5-e4 -5. Rg1-g6 e6-e5 -6. c3xPd4 d5-d4 -7. Qa1-b1 d6-d5 -8. Kb1-c1 d7-d6 -9. Rc1-g1 Rd1-d2 -9. d2-d3 etc.
7515 - Andrei Frolkin & Iwan Soroka
1. Nc3 h5 2. Nd5 Rh6 3. Ne7 Rc6 4. Nc8 Bc5 5. Nd6 cd6 6. h3 Qa5 7. Rh2 Qa2 8. d3 a5 9. Kd2 a4 10. Kc3 Bf2 11. Kb4 Kf8 12. Kb5 Rc5 13. Kb6 Rc2 14. Kb7 Bc5 15. Kc8 Ne7
7516 - Andrei Frolkin
1. h4 Nc6 2. h5 Ne5 3. h6 Ng4 4. hg7 h5 5. c4 h4 6. c5 h3 7. c6 h2 8. cb7 Rh3 9. b8=R N8h6 10. g8=R Bb7 11. Rg6 Bf3 12. ef3 Bg7 13. Bd3 Kf8 14. Qe2 Kg8 15. Qf1 Kh7 16. Ne2 Qh8 17. Rg8 Rb8 18. Rg1 Rb4 19. Rb6 Re4 20. Rbb8 h1=B
The promoted rooks switch place
7517 - Andrei Frolkin
The retroplay is: -1. .. d5-d4 -2. f2-f3 e6xBd5 -3. Ba2-d5 d7-d6 -4. Qa1-b1 f5-f4 -5. Rb1-b2 f6-f5 -6. Qe5-a1 f7-f6 -7. Bd5-a2 Ka2-a3 -8. Bf3-d5 Ka3-a2 -9. Qh5-e5 Ka2-a3 -10. Rb3-b1 Ka1-a2 -11. Rc3-b3 Ka2-a1 -12. Rc4-c3 and now unpromote both Rc4 and Qh5 on h8, retract the pawns to h2 and h4, and further retraction requires: -25. g3xPh4 Ka2-a1 -26. Be4-f3 Ka1-a2 -27. Bd5-e4 h5-h4 -28. Ba2-d5 h6-h5 -29. Nb2-a4 h7-h6 -30. Ra3-a5 Qa5-b6.
A try is to uncapture a knight on the 2nd black move, but this will result in black retrostalemate, because the knight requires 2 moves to reach a2, instead of 1 by a bishop.
7665 - Mario Velucchi
1. d4 c5 2. Bh6 cd4 3. c4 dc3 4. Bg7 cb2 5. Bf8 ba1=B 6. Bh6 Bg7 7. Bc1 Bf8
7666 - Michel Caillaud & Dirk Borst
1. c4 d5 2. c5 d4 3. c6 d3 4. ed3(=b) Qd4 5. Qg4 Qc3 6. dc3(=b) d2 7. Ke2 d1=Q 8. Ke3 Qd8 9. bc3(=b) c2 10. Bb2 c1=N 11. Bd4 Nb3 12. ab3(=b) b2 13. Nc3 b1=Q 14. Bb5 Qc1 15. Ke4 Qe3 16. fe3(=b) e2 17. Rf1 e1=N 18. Rf4 Nf3 19. gf3(=b) f2 20. Nf3 f1=N 21. Ne1 Ng3 22. hg3(=b) g2 23. Rh6 g1=Q 24. Qh3 Qg4 25. Rd6 Qgd7 26. cd7(=b)
However, this is cooked, as found by Peter van den Heuvel: 1. c4 d5 2. Qb3 d4 3. Qh3 d3 4. c5 Qd4 5. ed3(=b) Qc3 6. dc3(=b) d2 7. Ke2 d1=B 8. Ke3 Bf3 9. bc3(=b) c2 10. Bb2 c1=N 11. Bd4 Nb3 12. ab3(=b) b2 13. Nc3 b1=Q 14. gf3(=b) Qg6 15. Ra6 Qg3 16. fg3(=b) f2 17. hg3(=b) g2 18. Nf3 g1=Q 19. Rd6 Qg4 20. Bb5 Qd7 21. Ne1 f1=B 22. c6 Bc4 23. Rf1 Be6 24. Rf4 Qd8 25. Ke4 Bed7 26. cd7(=b)
7667 - Paul Raican
1. Bb2 Rb8 2. BPg7 RNb7 3. BPh8(=Q) RNb1[bPb7] 4. Qa1 RNd1 5. Qd1[wRa1]
7668 - Paul Raican
1. Rb1 Bg7 2. RNb2 BPh8 3. RNBc4 QP-a1(=QQ) 4. Qc2 QRb1[bBa1] 5. QPb2 Qe4[bRb1] 6. RBc2[wNc4] Qd7 7. Rc1[wPc2] QPb5 8. Qb5[wPb2]
7669 - Manfred Rittirsch
1. d4 e6 2. Bg5 Bc5 3. Be7 Bd4 4. Bd6[+wPe7] cd6 5. ed8=B d5 6. Bg5 f6 7. Bd2[+wPg5] fg5 8. Bc1[+wPd2]
However, this is cooked: 1. d4 c6 2. d5 e6 3. Bg5 cd5 4. Bd8[+wPg5] f6 5. Ba5 Bc5 6. Bc3 Bd4 7. Bd2 fg5 8. Bc1(+wPd2)
7670 - Paul Raican
1. e3 a6 2. Ba6 Ra7 3. Bb7 Ra2[+bPa7] 4. Ba8[+wPb7] Rb2[+bPa2] 5. Ra2 Rc2[+bPb2] 6. Rb2[+wPa2] Rd2[+bPc2] 7. Rc2[+wPb2] Rf2[+bPd2] 8. Rd2[+wPc2] Rg2[+bPf2] 9. Rf2[+wPd2] Rg1[+bPg2] 10. Rf1[+wPf2] gh1=B 11. bc8=N Bb7 12. Na7 Rh1 13. Rh1 Bc8[+bPb7]
7671 - Udo Marks
1. b3 ab3 2. ab3 cb3 3. cb3 Qe2 4. Qe4 Qd1 5. Qd4 Qe1 6. Qd5 Qd2 7. Qc5 Rc5 8. Bd2 Rc4 9. Bb4 Kb4 10. bc4 Kc4
7672 - Paul Raican
1. Nf3 Nc6 2. Nd4 Nd4[+wNc6] 3. Nd8[+bQc6] Qc2[+wPc6] 4. f4 Qc1[+wBc2] 5. Bh7[+bPc2] cd1=N[+wQc2] 6. Bf5 Nc3 7. Kf2 Nf5[+wBd4] 8. Kf3 Qf1[+wBc1] 9. Bf2 Qf1[+wBf1] 10. Kg4 Qxg2[+wPf2] 11. Kf5[+wNg4] Qc6[+wPg2] 12. Nc6[+bQd8]
However, this is cooked: 1. c3 h6 2. Qa4 h5 3. Qb3 h4 4. Kd1 h3 5. Nh3[+bPg1=N] Nh3[+wNg1] 6. Nf3 Nf4 7. Ne5 Ne2[+wPf4] 8. Kc2 Nc6 9. Kd3 Ne5[+wNc6] 10. Ke4 Ng4 11. Kf5 Nc3[+wPe2] 12. Qc2
7673 - Udo Marks
1. Bb4[D] Be2[A] 2. Ba5[D] Be1[B] 3. Bb4[D] Bd1[A] 4. Bc5[D] Bc2[A] 5. Bd4[D] Ba2[A] 6. Be4[E] Ba2[A] 7. Bd5[E] Bb2[B] 8. Bc4[E] Bc1[B] 9. Bb5[E] Kb5[E] 10. Bd4[E] Bd2[B] 11. Kd2[B] Bb2[A] 12. Be4[D] Ba1[A] 13. Be5[E] Ba2[B] 14. Bd5[D] Bb1[B] 15. Bc4[D] Bc2[B] 16. Bb4[E] Bd2[A] 17. Ba5[E] Be1[A] 18. Ba4[D] Be2[B] 19. Ke2[B] Ka4[D] 20. Kd3[C] Ka3[C] 21. Kc4[D] Kb2[B] 22. Kc5[E] Kc1[A]
7674 - Udo Marks
1. Eb4[E] a3[D] 2. Ea3[D] b4[C] 3. Eb4[C] Kb4[D] 4. Ec5[D] Ka5[C] 5. Ed4[E] Ka5[B] 6. Ee5[D] Nd4[C] 7. Ed4[C] c3[D] 8. Ec3[D] Ka5[A] 9. Ee4[E] Nb4[C] 10. Ed5[D] e4[C] 11. Ee4[C] d3[D] 12. Ed3[D] Rab5[E] 13. Ec4[E] Rb1[E] 14. Eb5[D] Ree1[E] 15. Ea4[E] Nc2[C] 16. Ec2[C] Rb1[B] 17. Eb1[B] Eb4[E] 18. Eb4[E] Re1[B] 19. Ee1[B]
7675 - Paul Raican
A possible proofgame is: 1. Nf3 Nf6 2. Ne5 Ne4 3. Nc6 Nc3 4. Nb8 Na2 5. Ra2 h5 6. Ra6 e5 7. Re6 Qe7 8. Nc6 g5 9. Nd4 Bh6 10. Nf3 O-O 11. Rd6 c5 12. Rg6 Bg7 13. Rh6 f5 14. Rh8 Kf7 15. Rh6 a5 16. Rf6 Ke8 17. Rc6 Rh8 18. Rc7 d5 19. Rd7 b5 20. Rd8 Qd8 21. Ng1
So the first move by the black king was castling
7676 - John Beasley
Black has played 2+4 moves, 4 with his knights, and at least one with his bishop. White has played 1+3+5=9 moves, maximal 7 by knights. The black rook was captured on a8, b8, or c8. Suppose the rook was captured on a white square. This must've happened during white's second turn, because the knight couldn't have captured, and gotten back to g1 in the third white turn. The rook couldn't have gotten to c8 during black's first turn, so it was captured on a8. To reach a8, a white knight needs 4 moves, which is all the moves in the first two white turns. But black couldn't have captured the knight on a8, and bring his pieces to their current position in his second turn. So the rook was captured on b8, and the knight capturing this rook must've been the g1 knight. So the proofgame is:
1. Nf3 Nc6,Rb8 2. Ne5,Nd7,Nb8 Nf6,Nd7,Ndb8,Bd7 3. d3,Nd2,Nf3,Ng1,f3.
7677 - Unto Heinonen
1. g4 a5 2. g5 a4 3. g6 a3 4. gh7 g5 5. hg8=Va Pah6 6. h4 Pab6 7. h5 Pab1 8. h6 b5 9. h7 Maa6 10. h8=Ma Vab7 11. Mag6 fg6 12. Vad5 e6 13. f4 Ke7 14. f5 Kd6 15. f6 Kc5 16. f7 Vad6 17. f8=Pa ed5 18. Paf4 Lef8 19. Pab4 Pab4 20. e4 Pad4 21. e5 b4 22. e6 b3 23. e7 Kb4 24. e8=Le Mac5 25. LeB8 ka4 26. Lea7 Pad8 27. Leb8 Vab8
7678 - Manfred Rittirsch
1. d4 a5 2. Bh6 gh6 3. h4 Bg7 4. Rh3 Bd4 5. Re3 Nf6 6. Re6 de6 7. e3 Nbd7 8. Ba6 ba6 9. c4 Bb7 10. c5 Bg2 11. c6 Qc8 12. cd7 Bc6 13. dc8=B e5 14. Bh3 Ng8 15. Bf1 OOO
Bc6 moves the king out of check, because white can't capture, since promotion to any piece would result in a Madrasi-paralyzation, illegal in Isardam.
7679 - Manfred Rittirsch
1. d4 e5 2. de5[+bPe7] Nf6 3. ef6 Nc6[+bNb8] 4. fe7 Ne7 5. Nd2 Ng8[+bPe7]
The missing white pawn is 'hidden' under Pe2.
However, this is cooked: 1. d4 e5 2. de5[+bPe7] Nc6 3. Nd2 Ne5 4. Nb1 Nc6 5. Nd2 Nb8.
7783 - Peter Rösler
1. f3 f5 2. Kf2 Kf7 3. Kg3 Qe8 4. Kh4 g5 5. Kh5 Kf6#
7784 - Manfred Rittirsch
1. Nh3 g5 2. Nf4 gf4 3. h4 h5 4. Rh3 Rh6 5. Rb3 Rg6 6. c3 Rg3 7. a4 Rh3 8. Raa3 Rh1 9. c4 a5 10. Rf3 Ra6 11. Rae3 Rb6 12. d3 Rb3 13. Qd2 Ra3 14. Qb4 Ra1 15. d4 ab4 16. Rb3
7785 - A. A. Kislyak
1. h4 a5 2. 2. Rh3 a4 3. Rc3 a3 4. Rc7 ab2 5. Rc8 Qc8 6. a4 Kd8 7. a5 Kc7 8. a6 Kd6 9. a7 Ke5 10. f4 Kf4 11. g3 Kg3 12. ab8=A Kh2 13. Kf2 h5 14. Ke3 Ra3 15. Kd4 Rg3 16. Ra4 Nh6 17. Na3 b1=C 18. Ad6 Cc4 19. Af8 Cf5 20. Ah6 Cc6 21. Kc5 Cd3 22. Kb6 Cg4 23. Ka7 Ch1 24. Rg4 Rg1 25. Af4#
7786 - Gunter Jordan
1. h4 e6 2. Rh3 Ke7 3. Rf3 Kd6 4. Rf4 Qh4[+wPd8=R] 5. Rc8[+bBg4] Qd8 6. Rb8 Ke7 7. Rg4[+bBc8] Ke8 8. Rh4 Ne7 9. Rh1
7787 - Temur Tschchetiani
1. e3 d5 2. Qe2 Kd7 3. Qb5 Kg4 4. Ke2 Qd6 5. Kd3 Qg6 6. Ka3 e5 7. Kf8 Qd6 8. Kc8 Nd7 9. Ka8 Nf6 10. Kh8 Nf8 11. Qd7 Kd1 12. Nc3 Kf2 13. Ne4 Kh1 14. Ng3 Kf2 15. c3 Ke1 16. Nf3 Kc2 17. Nd4 Ka1 18. Nb3 Kc2 19. Na1 Ke1 20. Nh1
7854 - Gerald Ettl
Retro-play is: -1. d5-d6 d6xNc5 -2. Nd3-c5 h2-h1=B -3. Nc1-d3 h3-h2 -4. Nd2-b1 h4-h3 -5. Ne4-d2 (-5. Qb1-a1 h5-h4 -6. Kd4-c3 Na1-c2 -7. c2-c4?? fails due to the black rook on b4!) h5-h4 -6. Ng3-e4 h6-h5 -7. Nf5-g3 h7-h6 -8. Nd4xPf5 f6-f5 -9. Qb1-a1 Na1-c2 -10. Nc2-d4 f7-f6 -11. Kd4-c3 Kd2-d1 etc.
855 - Andrei Frolkin & Andrei Kornilov
Suppose black played last. Retro-play is: -1. .. c7xNd6 -2. Ne8-d6 b3-b2 -3. Ng7-e8 b4-b3 -4. Ne8xRg7 Rh7-g7 -5. Nd6-e8 Rg7-h7 -6. Nc4-d6 Rh7-g7 -7. Nb6-c4 Rg7-h7 -8. Na8-b6 Rh7-g7 -9. a7-a8=N Rg7-h7 -10. a6-a7 Rh7-g7 -11. a5-a6 Rg7-h7 -12. a4-a5 a5xRb4 -13. Rb8-b4 Rh7-g7 -14. Qh8-g8 a6-a5 -15. Kg8-f8 a7-a6 -16. Rf8-b8 Re8-e7 -17. c2-c3! (a3-a4?) e7-e6/Ra8-e8.
Why can't white retract a3-a4? The black and white b-pawn promoted both. The white pawns on the kingside captured 5 times together, so the white b-pawn promoted without a capture. So the black b-pawn had to capture a piece before promoting. The only missing white piece is the black-squared bishop, so it got captured on a3. So a3-a4 was played before black promoted.
This proves also that white needs the c2-c3 tempo, so black indeed played last. So white has the move.
7978 - Mario Velucchi
1. g4 2. g5 3. g6 4. gh7 5. hg8=N 6. Ne7 7. Nc8 8. Na7 9. Nc6 10. Nb8 11. Nd7 12. Nf8 13. Ng6 14. Nh8 15. Nf7 16. Nd8 17. Nb7 18. Nd8 Kd8
7979 - Nikita Plaksin
1. a4 b5 2. Ra3 Bb7 3. Rh3 Bg2[->c8] 4. b3 Bb7 5. ab5[->b2] Bh1[->c8] 6. Rh7[->h1] Rh2[->h8] 7. Rh8[->a1] a5 8. Ra5[->a1] Ra1[->h8] 9. Na3 Rh1 10. Nh3 Rf1[->a8] 11. Nf4 Ra3[->h8] 12. Nh5 Rh7
7980 - Manfred Rittirsch
1. d4 e5 2. de5 Nf6 3. ef6 h5 4. b4 Qf6[+wPb2] 5. b5 Qc6 6. bc6 Rh6 7. g4 Rc6[+wPg2] 8. gh5 d6 9. h6 gh6[+wPd2]
7981 - Arno Tüngler
1. Rd2[+wBd1] Nd7 2. Rd7 Bb7 3. Rc7 Qa8 4. Rb7 OOO[Ra8,Qd8] 5. Rb1 Kb7 6. Rc1[+wNb1] Ka6 7. Rd1[+wBc1]
7982 - René J. Millour
1. h4(B) a6(B) 2. h5(A) a5(A) 3. h6(B) h5(B) 4. h7(A) Na6(B) 5. hg8=Q(B) Rb8(B) 6. Qb8(A) h4(A) 7. Qc8(B) h3(B) 8. Qg8(A) h2(A) 9. Qf8(B) hg1=B(B) 10. Qd6(A) OO(B) 11. b3(B) Qh8(B) 12. Bb2(B) Qb2(A) 13. Na3(B) Qa2(B) 14. Qb1(B) Qh2(A) 15. Rh2(B) Bh2(A) 16. OOO(B) Bd6(B) 17. Rh1(A) Rf4(A) 18. Kd1(A) Bf8(A) 19. Ke1(B) Re4(B)
This is cooked: 1. h4(B) a6(B) 2. b3(B) Ra4(B) 3. Ba3(B) Rh4(A) 4. Rh4(B) Nh6(B) 5. Rh6(A) a5(A) 6. Rg6(B) Na6(B) 7. Rg8(A) h5(B) 8. Rf8(B) h4(A) 9. Rb8(A) h3(B) 10. Rc8(B) h2(A) 11. Rb8(A) hg1=B(B) 12. Bd6(A) Bh2(A) 13. Na3(B) Bd6(B) 14. Qb1(B) 00(B) 15. 000(B) Qb8(B) 16. Rh1(A) Qb3(A) 17. ab3(B) Rf4(A) 18. Kd1(A) Bf8(A) 19. Ke1(B) Re4(B)
7983 - Bernard Rothmann
1. Nf3 h5 2. Ne5 Nh6 3. Nd7 e5 4. Nb8 Bd7 5. Nd7 Rb8 6. Nf8 Qd7 7. Ng6 Rf8 8. Nh8 g6 9. Nf7 Rh8 10. Ng5 Nf7 11. Nf3 g5 12. Ng1 Ra8
7984 - Bernard Rothmann
1. Nc3 Nc6 2. Nd5 Nd4 3. Ne7 d5 4. Nc8 Ne7 5. Nd6 Kd7 6. Ne8 Ne2 7. d4 Nc1 8. Ne2 Nd3 9. Kd2 Ne1 10. Nc7 Qe8 11. Ne8 Ng2 12. Qe1 Rd8 13. Rd1 Kc8 14. Kc1
7985 - Bernard Rothmann
1. e4 Nf6 2. ef6 e5 3. Nc3 ec3 4. fg7 cb2 5. gh8=B ba1=B 6. Be5 Bd4 7. Bb7 Bg2 8. Ba8 Ba6 9. Bag2 Bf1 10. Bf1 d6 11. Ke2 Kd7 12. Kd3 Kc8 13. Be2 d5#
7986 - Anatoli Wasilenko & Andrei Frolkin
1. Nf3 Nc6 2. Nd4 Ne5 3. Nb3 Nf3 4. Kd3 Rb8 5. Qe1 Ne1 6. Kc5 e6 7. Ka7 Qh4 8. Kb8 Ke7 9. Kc8 Kf6 10. Kd8 Ke5! (Kf5? 11. Kh4 Ng2 illegal selfcheck!) 11. Kh4 Ng2 12. Kg2 Kf4
7987 - Anatoli Wasilenko & Andrei Frolkin
1. d4 e5 2. de5 d5 3. e6 d4 4. e7 d3 5. ef8=B de2 6. Be7 ef1=B 7. Bh4 Qg5 8. Ne2 Nf6 9. Qd8 Ka4 10. b4 Re8 11. b5 Re3 12. b6 Rb3 13. Nd4 Nh5 14. Nc6 Qd2 15. Ke8 Qd1 16. Na5 Kb4 17. Ba3 Ke1 18. Bf8
7988 - Thierry le Gleuher
1. c4 a5 2. Qb3(=R) a4 3. Re3(=B) Ra5(=B) 4. Bb6(=N) Bc3(=N) 5. Na8(=P) b6 6. b3 Ba6(=N) 7. Ba3(=N) Nb4(=P) 8. Nb5(=P) Na6(=P) 9. Na3(=P) Qb8(=R) 10. Rb1(=B) Rb7(=B) 11. Bf5(=N) Nb1(=P) 12. Kd1 Be4(=N) 13. Kc2 Ng3(=P) 14. e4 g6 15. Be2(=N) Bh6(=N) 16. Ned4(=P) Ng4(=P) 17. Ne2(=P) Nh6(=P) 18. Rc1(=B) Rf8(=B) 19. Bb2(=N) Bg7(=N) 20. Nd1(=P) Ne6(=P) 21. Ng7(=P)
7989 - Andrei Frolkin & Anatoli Wasilenko
1. f4 h5 2. f5 h4 3. f6 h3 4. fe7 f5 5. e4 Kf7 6. e8=G f4 7. Gc6 dc6 8. e5 Bf5 9. e6 Kf6 10. e7 Be4 11. e8=G Qe7 12. Gb5 cb5 13. a4 Nc6 14. a5 Rd8 15. Ra4 ba4 16. Bb5 Rd3 17. c4 Rb3 18. d4 hg2 19. Bd2 Rhh3 20. Bb4 Rhc3 21. h4 g5 22. h5 g4 23. Rh4 g3 24. Nh3 g1=G 25. h6 Gg4 26. h7 g2 27. h8=G g1=G 28. Ge5 Bg7 29. a6 Bh8 30. Ga5 Gc5 31. d5 Ke5 32. d6 Nf6 33. d7 Gc8 34. d8=G Nd7 35. Gb6 cb6 36. Na3 ba5 37. Nc3 Ga3 38. ba3
Unfortunately, this is cooked: 1. a4 h5 2. a5 h4 3. a6 h3 4. Ra5 hg2 5. f4 Rh3 6. f5 Rb3 7. f6 g5 8. fe7 g4 9. h4 g3 10. h5 f5 11. Rh4 Kf7 12. Rha4 Bg7 13. Nh3 g1=B 14. e8=G Bc5 15. Gc6 dc6 16. e4 g2 17. e5 f4 18. e6 Kf6 19. e7 Bf5 20. e8=G Be4 21. Gb5 cb5 22. d4 ba4 23. Bb5 g1=G 24. d5 Nc6 25. d6 Ke5 26. d7 Qe7 27. d8=G Nf6 28. Gb6 Rd8 29. h6 Gg8 30. h7 Gc8 31. h8=R Rdd3 32. Rh4 Bh8 33. c4 Nd7 34. Na3 cb6 35. Nc2 Ba3 36. Bd2 ba5 37. Bb4 Rdc3 38. ba3
7989v - Andrei Frolkin & Anatoli Wasilenko
1. f4 h5 2. f5 h4 3. f6 h3 4. fe7 f5 5. e4 Kf7 6. e8=G f4 7. Gc6 dc6 8. e5 Bf5 9. e6 Kf6 10. e7 Be4 11. e8=G Qe7 12. Gb5 cb5 13. a4 ba4 14. Bb5 Nc6 15. c4 Rd8 16. Qb3 Rd3 17. Qa2 Rb3 18. d4 hg2 19. Bd2 Rhh3 20. Bb4 Rhc3 21. h4 g5 22. h5 g4 23. Rh4 g3 24. Nh3 g1=G 25. h6 Gg4 26. h7 g2 27. h8=G g1=G 28. Ge5 Bg7 29. Ga5 Bh8 30. d5 Ke5 31. d6 Nf6 32. d7 Gc8 33. d8=G Nd5 34. Gb6 cb6 35. Ke2 ba5
8068 - Alexander Kukush
Black tries to castle, thus proving it's possible that he didn't move last. But: 1. .. fe6 2. g6 e5 3. Kg5 e6 4. g4!! and now is 4. .. Be7 mate, and other bishop moves result in stalemate. So black can't prove he didn't move last. So mate in 1 with 1. Qc8#.
8069 - Andrei Kornilov & Andrei Frolkin
Retroplay is: -1. g6-g7 e3-e2 -2. h2-h4 h3xNg2 -3. Nf4-g2 h4-h3 -4. Nh5-f4 e4-e3 -5. Ng7-h5 e5-e4 -6. Nh5xRg7 Rg8-g7 -7. Nf4-g5 Rg7-g8 -8. Ne2-f4 Rg8-g7 -9. Qh8-h7 h5-h4 -10. Kh7-h6 h6-h5 -11. Nc1-e2 Rh5-g5 and the position unlocks.
8069 - Andrei Kornilov & Andrei Frolkin
Retroplay is: -1. .. Rc2-c1# -2. Kb1xNa1 Nb3-a1 -3. Ka1-b1 Na5-b3 -4. Kb1-a1 Nc4-a5 -5. Ka1-b1 Na3-c4 -6. h3-h4 Nb1-a3 -7. h2-h3 Rc1-c2 -8. c2-c3 c3xB/Nd2 and further retroplay involves bRc1 -> a8; a7xNb6; wKa1 -> f8; Re8-e7 and the whole thing opens up.
8071 - Thomas Volet
The black bishop on a4 is the key figure: First it provides a retroscreen for the white bishop on h7, which goes back to e8. Then it uncaptures a pawn on c6, and goes to e4, so it provides a retroscreen for the white rook on f4. This rook provides a retroscreen for the black bishop on h4, so it can go back to g1. The pawn uncaptured on c6 provides the necessary tempo for this. Finally, the bishop unpromotes on h1, and retracts to h6, allowing h5xg6, opening the whole position.
8072 - Thomas Volet
The white king must pass three lines, controlled by black rooks. The black bisghop currently on a7 provides the necessary retroscreens. The white king then provides a retroscreen for the white bishop on a6, so that the bishop can leave its prison, and the position unlocks after that.
8073 - Satoshi Hashimoto
1. c4 d5 2. c5 Qd6 3. c6 Nd7 4. cb7 Qb6 5. b8=N Ba6 6. Nc6 Rd8 7. Nb8 Nb8 8. f4 Rd6 9. f5 Rg6 10. f6 e6 11. fg7 Nf6 12. g8=N Bh6 13. Ne7 Rf8 14. Ng8 Ng8
8074 - Andrei Frolkin & Mikhail Kozulya
1. g4 a5 2. g5 Ra7 3. g6 hg6 4. f4 Rh6 5. f5 gf5 6. Nh3 Rha6 7. Rg1 b6 8. Rg6 fg6 9. d3 Kf7 10. Bf4 Ke6 11. Bd6 cd6 12. e3 Qc7 13. Qh5 gh5 14. Nd2 g6 15. OOO
8075 - Michel Caillaud
1. a4 c5 2. a5 Qc7 3. a6 Qh2 4. ab7 Na6 and now:
8076 - Gianni Donati
1. c3 d6 2. Qa4 Bd7 3. b3 Nf6 4. Ba3 Ne4 5. Bc5 Nd2 6. e4 Nb1 7. Bb5 Na3 8. OOO Nc2 9. Rd5 Na1 10. Re5 d5 11. Kb2 d4 12. Ka3 d3 13. Qd4 d2 14. Kb4 d1=B 15. a3 Bg4 16. Nf3 Bc6 17. Rd1 Bc8 18. Rd3 B6d7 19. Bd7
8077 - Nikita Plaksin
8078 - Bogdan Shesherun
Try: -1: Q~-h1(=R) would leave black retropat.
Solution: -1. Ng1-e2(=P) Ke2-e1 -2. Bf2-g1(=N) Ke3-e2 -3. Rd2-f2(=B), so the position is indeed legal.
8079 - Bogdan Shesherun
(a): The last moves were: -1. Bg7-h6(=N) (-1. Bf8-h6(=N)? Kg8(x/-)f7 -2. N??-h7(=P)) Kg8(x/-)f7 -2. Nf8-h7(=P), so the position is indeed legal.
(b): The last moves were: -1. Bf8-h6(=N) (-1. Bg7-h6(=N)? retropat) Kg7-f7 -2. Rf6-f8(=B), so the position is indeed legal.
8080 - Nikita Plaksin
Last moves were: -1. Rd7-d5 Qh3xNg3(-> d8) !! -2. Nh1-g3 Qh5-h3. Seemingly, the queen can come from any square, but the capture is uniquely defined in order to explain the check by Re8.