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Die Schwalbe

Solutions

Die Schwalbe Heft 55, Feb. 1979

S370 - Wolfgang Dittmann
-1. Qb1-e1! (Threat: -2. Qb8-b1 and forward Kb7/Kc7#) and now:
-1. .. Bb3-e6 -2. Qg1-b1 (Threat: -3. Qb6-g1 ~ -4. Qb8-b6 and forward Kb7/Kc7#) Rd4-d6 -3. Qh1-g1 (Threat: -4. Qa8-h1 and forward Kb7/Kc7#) Bd5-b3/Rd5-d4 -4. Qh3-h1/Qh5-h1 and forward Qd7/Qf7#.
-1. .. Rb6-d6 -2. Qb5-b1 Rb7-b6 -3. Qb6-b5 B~ and forward Qd7/Qf7#, or -2. .. Rc6-b6 -3. Qb8-b5 and forward Kb7/Kc7#, or -2. .. Rc6xb6 -3. Qa6-b5 ~ -4. Kb8-c8 and forward Qc8#.

2695 - Frank Christiaans
Tries:
Last move was h3-h2. The sequence of captures on the g- and h-line was: g7-g4; h2-h4xg5; h7-h3; g2-g3xh4-h6. But this means that the white rook could never have gotten to h5 after the white pawn went to h6, since all the pawns on h6, g5, g4 and h3 were already there.
Last move was anything, and white played a5xb6 before that. This implies that the white captures were h4xg5, gxh and a5xb6. One of the missing pieces is the black bishop on white squares. So the undetermined capture is g2xBh3. But this bishop could never have gotten to h3 after the black pawn went to h2, since g7-g4 and h4xg5 were played before.
So the only move that'll give white a retromove is f7-f5. So mate in 3 with 1. ef6 ep [2. Qb8 Kd7 3. Qe8#] Bd6 2. Qd6 [3. Re8#] Nc7 3. Qc7#

2696 - Alexander Kislyak
White has captured on his last move (or black has no retromoves). So the white captures are d2xe3, g6xf7xg8=R and the last capture. Only capture that'll result in a position in which white can mate in 1 is Qg6xf7 (and #1 with 1. Qg3#). All 3 missing black pawns promoted, and the promotions happened after d2xe3. Only queens could've gotten from d1 to f7 (twice) and g8. So retract Qg6xQf7.
Ceriani-Frolkin theme with 3 black queens.

2697 - Wolfgang Dittmann
Try: -1. Rg5-b5 Kb5-a6 -2. Bf5xRh7 Rh8xXh7 -3. d2-d3 and forward 1. Bd3 Kc6!
Try: -1. Bg8xRh7 Rh8xBh7! and now is retracting f5xe6ep illegal (the white pawns need all 9 missing black pieces for captures)
Solution: -1. f5xe6 ep! e7-e5 -2. Bg8xRh7 Rh8xXh7 (X can be anything except a bishop) -3. Re5-b5 Kb5-(x)a6 -4. Bd5-g8 and forward 1. Bb7#

2698 - Klaus Wenda
(a): Tries: Retract d3-d4, and 1. Ke3 Rd3#, but 1. Bg7#. Retract Nf6/g5xQe4 and forward 1. Rd1 Qd3#, but Nf7#. Solution: Retract Ke3xNd2, and r#1 with 1. Bf3 Nc4#
(b): Retract Ke1-e2, and r#1 with 1. OOO a1=Q#

Die Schwalbe Heft 56, April 1979

2755 - Werner Keym
Last moves were: Bg8xQh7 Q~-h7 h7xRg8 (because of this uncapture, the last move can't have been uncapturing a rook).

2759 - Wolfgang Dittmann
OO is illegal (last move was with either rook or king). Solution: -1. Kf2-e1 (Threat: Ke1xNf2 and forward OO) Bd2-e3 -2. Rc1xNh1 (Ke1xNf2? Be3xPd2, or Ra1xNh1? Bc1-d2) Ng3-h1/h2-h1=N/g2xh1=N -3. Ke1-f2 Be3-d2 -4. Kd2-e1! (Threat: -5. Ke1xPd2 ~ -6. Ra1-c1 and OOO) Bf2-e3 -5. Ke1xNd2! B~-(x)f2 -6. Ra1-c1 and forward OOO.

2760 - Günter Lauinger
Retract: -1. OO b7-b6/c7-c6 -2. Rh6-d6 c7-c6/b7-b6 -3. f4xPg5, and now the last black move must've been g7-g5.

2761 - Wolfgang Dittmann
(a): Retract -1. h7xBg8=N[B-] a2xBb1=N[B-] and now h-ep in 9 with 1. Bh7[+wPh2] h3 2. Bg4 hg4[+bBc8] 3. Bf5 gf5[+bBc8] 4. Be6 fe6[+bBc8] 5. Be6[+wPe2] Ba2[+bPa7] 6. Bd3 ed3[+bBc8] 7. Bc4 dc4 8. Rb5 cb5[+bRa8] 9. a5 ba6ep[+bPa7]
(b): Retract -1. g7-g8=N a2xBb1=N[B-] and now h-ep in 9 with 1. Kf7 Ba2[+bPa7] 2. Kg7[+wPg2] g4 3. Bf5 gf5[+bBc8] 4. Be6 fe6[+bBc8] 5. Bd7 ed7[+bBc8] 6. Bd7[+wPd2] d3 7. Rc4 dc4[+bRa8] 8. Bb5 cb5[+bBc8] 9. a5 ba6ep[+bPa7]

Die Schwalbe Heft 57, June 1979

2816 - Nikita Plaksin
(a): Last moves were: -1. .. Bh2xNg1 -2. Kg3-f2 Bg1xNh2[+wNg1] -3. Kh4-g3 f2xNg1=B[N-] -4. Kg5-h4 g3xNf2[N-] -5. Nd1-f2 h4xQg3 -6. Kf6-g5 g5xNh4[+wNg1] -7. Rf2-f1 g6-g5 -8. Ke7-f6 g7-g6 -9. d2xPg3[P-].
(b): The position is illegal, since no black pawn can be uncaptured with the white king on h7.

S.441 - André Hazebrouck
The black pawns have made 4 captures. The promoted black bishop is the f-pawn which promoted on f1 without captures. So the white captures are: fxe, gxf, and hxgxf. So black can't take back c4xb3 too soon, the white c-pawn needs to get back to c2 first.
The last moves were: -1. c7-c8=B Ke7-d7 -2. d7-d8=B. The position unlocks when black takes back e7xd6. Because this implies that the bishop on f8 should return (and the rook on g8/h8), white needs to unpromote the four kingside pawns. The only free pieces for these unpromotions are the four bishops on a1, g1, f7 and g8. The bishop on a8 is clearly promoted too.
So all seven bishops in the diagram are promotees!

2872 - Wolfgang Dittmann
(a): Retract -1. Ne3-d1 Ke8-d8 -2. Nf5-e3 Kd8-e8 -3. Nd6xNf5! N~-f5 -4. Ne8-d6 N~ -5. Nd6xNe8 and forward 1. Nf7#. Not: -1. Nc3-d1 Ke8-d8 -2. Nb5-c3 Kd8-e8 -3. Nd6xNb5? Nc3-b5 -4. Ne8-d6 NxPc3! and now is Nd6xNe8 illegal, since the white pawn on c3 captured this knight.
(b): Retract: -1. Kc3-d4 Ke8-d8 -2. Kb2-c3 Kd8-e8 -3. Kc1-b2 Ke8-d8 -4. OOO! Kd8xNe8 -5. Nd6xNe8 and forward 1. Nf7#.

2874 - Nikita Plaksin & Alexander Klibanski
If the white king never moved, then black's last move must've been c7-c5 (and white played d3xPc4[+bPc7] before that). So white may capture en-passant, but only if he validates it by castling later on.
h#4.5 with 1. .. bc6ep[+bPc7] 2. Rh8 cd7 3. Rc8 dc8=R[+bRa8] 4. Rc8[+wRh1] OO! 5. Ka5 Ra1#

2875 - Werner Frangen
1. g3 b6 2. h4 a5 3. Nf3 Nc6 4. Bg2 Bb7 5. Kf1 Ra6 6. Kg1 Qa8 7. Kh2 Kd8 8. Kh3 Kc8 9. Qg1 Kb8 10. Nc3 Nf6 11. Nd1 Ne8 12. Ne3 Bc8 13. Kg4 Qa7 14. Kg5 Nd8 15. h5 a4 16. Nh2 Ne6+ 17. b4 Nxg5 18. Ba3 Ne6 19. Nf3 Kb7 20. Rh3 Kc6 21. Bh1 Bb7 22. Qg2 Ba8 23. Rg1 Qb8 24. Qf1 Rg8 25. Rg2 Rh8 26. Ng1 Rg8 27. Nd1 Rh8 28. Bc1 Kd5 29. b5 g5 30. bxa6 Nd6 31. Qe1 Bh6 32. Qf1 c6 33. Qe1 Rc8 34. Qf1 Qc7 35. Qe1 Qd8 36. Qf1 Nf8 37. Bb2 Ke6 38. Bd4 b5 39. Ba7 Bg7 40. Ne3 Bb2 41. Nd1 Ba3 42. Qe1 Ke5 43. Qf1 Ne8 44. Qe1 Bd6 45. Qf1 b4 46. Nb2 a3 47. c4 axb2 48. Qd1 Bc7 49. Qe1 Kd6 50. Nf3 g4 51. a4 gxh3 52. Rg1 b3 53. g4 Ke6 54. Be3 Bd6 55. Rf1 Ba3 56. Rg1 Kf6 57. Bf4 Ne6 58. Bh6 Rb8 59. a7 Rb6 60. Rf1 N6c7 61. Rg1 Na6 62. Bf4 Ke6 63. g5 Ng7 64. h6 f5 65. hxg7 Nb4 66. Rf1 Qc8 67. c5 h5 68. cxb6 Kd5 69. g6 c5 70. a5 c4 71. Bh6 Bc6 72. Be3 Ba4 73. Rg1 d6 74. Rf1 Ke6 75. Rg1 Kd7 76. Ng5 h2 77. Rg3 Kd8 78. Nh7 Qd7 79. a8=B Na6 80. Bag2 Qc8 81. f3 Ke8 82. Bh6 Nc5 83. Qf1 Nb7 84. a6 Nc5 85. a7 Na6 86. Ng5 f4 87. d4 fxg3 88. Nh7 Qd8 89. b7 Qa5 90. Bh3 h4 91. Bg4 Bb5 92. Bh5 Bb4 93. a8=B Qa2 94. Bg2 Ba4 95. Bh1 Ba5 96. Bg2 Bc7 97. Bh1 Nb8 98. Bg2 c3 99. e4 c2 100. Bd2 Bd8 101. Qe2 b1=B 102. Qe3 b2 103. Qh6 Qa1 104. Be3 Ba5 105. f4 Bb4 106. f5 Ba3 107. Bd2 Bb5 108. Bh1 Bf1 109. B1f3 Nc6 110. d5 e5 111. dxc6 Bb5 112. c7 Ba4 113. c8=B Bc5 114. Be6 Bg1 115. Bc4 Kd8 116. Ba6 Ke8 117. Bfe2 Kd8 118. f6 Kd7 119. f7 Ke6 120. f8=B Kf5 121. exf5 d5 122. Ba5 e4 123. Ng5 e3 124. Qh8 g2 125. Nf7 Bf2 126. Nd6 g1=B 127. b8=B d4 128. Bd5 h3 129. Be6 h1=B 130. Bac7 Bac6 131. Bac8 Ba8 132. Bb5 Bhg2 133. Bbd7 d3 134. Be7 h2 135. Qd8 h1=B 136. g8=B c1=B 137. g7 Bc2 138. f6 d2 139. Bhe8 d1=B 140. f7 e2 141. f8=B e1=B 142. Bgf7 b1=B 143. g8=B Bf1

Die Schwalbe Heft 59, Oct. 1979

4v - Andrei Frolkin
Last moves were: -1. .. c4xNb3 -2. Bc6-b5 b5xNc4 -3. N~-c4 a6xNb5 -4. c4-c5 Kc5-b4 -5. N~-b3.
Ceriani-Frolkin theme with three white knights.

9 - Andrei Frolkin
Black captures are: b3xa2xb1, Ba5xb6 and h7xRg6. White captures are: e6xd7, f6xe7, and gxh. Promotions on e8 and twice on h8. Only a rook and a queen can leave e8/h8, and only a queen can get to a2, b1 and b6.
Ceriani-Frolkin with 3 white queens.

12 - Andrei Frolkin
Last moves were: -1. .. Qb6xBd8 -2. Bc7-d8 Nd8-e6. White captures: exd and gxh. Black captures: d3xc2, b3xa2xb1, Qb6xBd8 and h7xg6. Only queens could've gotten from h8 to a2/b1.
Ceriani-Frolkin with 2 white queens and one white bishop.

2930 - Boris Ostruh
Retract OO!. Black only misses the h-pawn, so it promoted. Because white castled, the black h-pawn must've captured to get to the g-line. So black captures are a5xb4, gxf and hxg. White capture is c6xb7. White misses his h-pawn too. None of the captures of black was on the h-line, and black misses only one piece, so the white h-pawn promoted on h8, without capturing. So black may not castle. The only move that'll give white a retromove is c7-c5 (and white played c6xb7 before that), because white can't retract g2-g3 (Bf1 was one of the captured pieces, on g2 or f5).
Mate in 2 with 1. bc6 ep! (Threat: b8=Q/R) Kd8 (OO? illegal) 2. b8=Q#

2933 - Nikita Plaksin
The position in the right corner unlocks after retracting Bf1-g2 g2xf3; Rf6-f2; f2-f1=B. Then the black king can go out via a3 and a4, so white played b2-b3. So the white bishops were captured on their homesquares. All other missing white pieces were captured by the pawns now on g5 and c2. So the last moves could only have been -1. .. e7-e6 -2. Kc5-d5 d7-d6. So the black bishops were captured on their homesquare too!

Die Schwalbe Heft 60, Dec. 1979

2986 - Luigi Ceriani, Luis A. Garaza & Hans-Heinrich Schmitz
1. .. Be1 2. Qf2 Ng1 3. Qe2 Bf2 4. Qe1 Ne2 5. Bg1 Rh2 6. Bh3 Rg2 7. Bh2 Rg1 8. Rg2 Qg3 9. Bg4 Qh3 10. Rg3 Bg2 11. Qf1 Be1 12. Qf2 Rf1 13. Bg1 Qh2 14. Rh3 Qg3 15. Rh2 Bh3 16. Rg2 Qh2 17. Rg3 Qg2 18. Bh2 Qg1 19. Bg2 Qh1 20. Qg1 Rf2 21. Bf1 Bg2 22. Rh3 Ng3 23. Be2 Bf1 24. Qg2 Qg1 25. Qh1 Qg2 26. Bg1 Qh2 27. Qg2 Nh1 28. Rg3 Qh3 29. Qh2 Rg2 30. Bf2 Rg1 31. Rg2 Qg3 32. Bh3 Qg4 33. Rg3 Rg2 34. Bg1 Rf2 35. Bg2 Qh3 36. Rg4 Ng3 37. Bh1 Qg2 38. Qh3 Qh2 39. Bg2 Qh1 40. Bh2 Qg1 41. Bh1 Bg2 42. Bf1 Ne2 43. Qg3 Bh3 44. Bfg2 Rf1 45. Qf2 Ng3 46. Qe2 Rf2 47. Qf1 Ne2 48. Rg3 Bg4 49. Bh3 Rg2 50. Qf2 Qf1 51. Bg1 Rh2 52. B1g2 Rh1 53. Bh2 Qg1 54. Bf1 Qg2 55. Bg1 Qh2 56. Rg2 Qg3 57. Rh2 Qg2 58. Qg3 Qf2 59. Bhg2 Bh3 60. Qg4 Qg3 61. Bf2 Rg1 62. Bh1 Bg2 63. Qh3 Qg4 64. Qg3 Qh3 65. Qg4 Ng3 66. Be2 Bf1 67. Bg2 Nh1 68. Qg3 Qg4 69. Bh3 Rg2 70. Bg1 Rf2 71. Qg2 Qg3 72. Bg4 Qh3 73. Qg3 Qg2 74. Rh3 Qh2 75. Qg2 Ng3 76. Qh1 Qg2 77. Bh2 Qg1 78. Qg2 Qh1 79. Qg1 Bg2 80. Bf1 Ne2 81. Rg3 Bh3 82. Bg2 Rf1 83. Qf2 Qg1 84. Bh1 Qg2 85. Bg1 Qh2 86. Rg2 Qg3 87. Rh2 Bg2 88. Rh3 Qh2 89. Rg3 Qh3 90. Bh2 Rg1 91. Qf1 Bf2 92. Qe1 Bf1 93. Rg2 Qg3 94. Bh3 Qg4 95. Rg3 Rg2 96. Bg1 Rh2 97. B3g2 Rh3 98. Bh2 Ng1 99. Qe2 Be1 100. Qf2 Ne2 101. Qg1 Bf2

4832v - Leonid Borodatow
Retract -1. Qh3-f5+ Kg6xRg5+ -2. Rf5-g5+ Kg5-g6 -3. Rf3-f5+ Kg6xRg5+ -4. Rf5-g5+ Kg5-g6 -5. Rf4-f5+ Kg6xBg5+ -6. Rf5-f4 Kh5-g6 -7. Ng4-f6+ Bg6-h7 -8. Nh6-g4+ and the position unlocks.

7643v - Leonid Borodatow
Retract -1. 0-O-O# Ke4-d4 -2. e5xf6ep f7-f5 -3. Rg6-b6+ Kf5-e4 -4. c7-c8=B+ and the position unlocks.

8493v - Andrej Kornilow
Kings in check can only be explained by -1. d5xe6ep+ e7-e5 -2. Ne5-c6+. Thus bKd6, wKe3, wQd4, wBf4, wPe6, wNc6, further entailing wRd3, wRg3, bRd7, wNc2.

Further colorings are obtained by minimizing captures and promotions, e.g. a bPc7?? would require a promoted wBd8.

We get bPa7, bNb2, bPb7, bBd8, bPf7, bPg5, bRg6, bNg8, bPh5, bQh6, wPa4, wPc7, wPd2, wPe4, wPf6, wPg2, wPg7, wBh7, where required captures are wPbxPc, wPcxPdxPe6ep, wPh5xBg6.

All captures are accounted for. Last moves are -1. d5xe6ep+ e7-e5 -2. Ne6-c6+ Ke6-d6 -3. c4xPd5+.

Die Schwalbe Heft 148, Aug. 1994

8595 - Kornilow, Andrej
This is a picture problem. Bl. has no last move, so that mate in 1 with 1. Rg3#?? is not legal.

Bl. has the move: the shortest mate is 1 ... h1=Q+ 2. Kxh1 a1=Q/R+ 3. Rg1#

8596 - Plaksin, Nikita and Zolotarew, Alexandr
Authors intention was:
-1 ... Ra4xNa5+
-2. b6xRc7 Rc6-c7
-3. h5-h6 Rc7-c6
-4. Qe4-b7 Rb7-c7+
-5. Qe8-e4 c6-c5
-6. e7-e8=Q c7-c6
-7. f6xNe7 Nc6-e7
-8. h4-h5 Qb2-b3
-9. Nb3-a5+ Na5-c6+
-10. f5-f6 Qc1-b2
-11. f4-f5 Qh1-c1
-12. f3-f4 h2-h1=Q
-13. f2-f3 g3xNh2
-14. Nf3-h2 g4-g3
-15. Ng1-f3 h5xBg4
-16. Ne2-g1 h6-h5
-17. Nc1-e2 Qb2-a2
-18. Na2-c1+ Qc1-b2
-19. Bf3-g4 Qe1-c1
-20. Bg4-f3 e2-e1=Q
-21. Bf3-g4 e3-e2
-22. Be2-f3 e4-e3
-23. Bf1-e2 e5-e4
-24. e2xQd3
with three Q promotions and a complex pattern of pins and captures.

Unfortunately, it is also possible to unlock with
-7. f6xRe7 Re1-e7
-8. f5-f6 Rf1-e1
-9. f4-f5 f2-f1=R
-10. f3-f4 g3xNf2
-11. Ne4-f2 g4-g3
-12. h4-h5 h5xBg4
-13. Ng3-e4 h6-h5
-14. Ne2-g3 h7-h6
-15. Nc1-e2 Qb2-a2
-16. Na2-c1+ Qc1-b2
-17. f2-f3 Qe1-c1
-18. Bh5-g4 e2-e1=Q
-19. Bf3-h5 e3-e2
-20. Be2-f3 e4-e3
-21. Bf1-e2 e5-e4
-22. e2xNd3 etc.

8597 - Plaksin, Nikita and Zolotarew, Alexandr
-1 ... Ne6-g5
-2. Ng5-f7+ Bf7-g6+
-3. g6-g7 Ng7-e6+
-4. h5xNg6 e2-e3
-5. e6-e7 Ne7-g6+
-6. d5xBe6 e4-e3
-7. d4-d5 Bc4-e6
-8. b4-b5 Ba6-c4
-9. d3-d4 Bc8-a6
-10. d2-d3 b7xBc6
-11. Bb5-c6 e5-e4
-12. Bf1-b5 e6-e5
-13. e2xRf3 etc.

8599v - Andrej Frolkin
Author's intention: 1. a4 Nf6 2. Ra3 Ne4 3. Rg3 Nxd2 4. Kxd2 h5 5. Kc3 Rh6 6. Kb4 Rf6 7. Nc3 Rf3 8. exf3 h4 9. Bc4 h3 10. Nge2 hxg2 11. Rf1 g1=Q 12. Rg2 Qh1 13. Ng3 Qg1 14. Nce2 Qh1 15. c3 Qg1 16. Qb3 Qh1 17. Qa2 Qg1 18. Bb3 Qh1

Unfortunately, this corrected version is cooked too: 1. a4 Nf6 2. Ra3 Ne4 3. Rg3 Nxd2 4. Nc3 Nf3!! 5. exf3 h5 6. Kd2 h4 7. Bc4 h3 8. Nge2 hxg2 9. Rf1 g1=Q 10. Rg2 Rh7 11. Ng3 Rh6 12. Nce2 Rb6 13. Kc3 Rb4 14. Kxb4 Qh1 15. c3 Qg1 16. Qb3 Qh1 17. Qa2 Qg1 18. Bb3 Qh1

6453v - Borodatow, Leonid N.
The last moves were: -1. .. Qc2xPb3 -2. b2-b3 Qe4xNc2 -3. c3-c4 Qf4xPe4 -4. e3-e4 Qg4xPf4 -5. f3-f4 Q?xPg4 -6. f2-f3! (for this tempo, black needed to uncapture Qf4xPe4!)

Die Schwalbe Heft 149, Oct. 1994

R1 - Myllyniemi, Matti Arvo
Last moves were: -1. ... O-O-O -2. Ke5-d6 e4xd3 ep -3. d2-d4 b2-b1=B

8643 - Borodatow, Leonid N.
Suppose black can still castle. The white pawns on the queenside captured 8 times, and the pawns on the kingside captured 2 times. The white a-pawn didn't capture, because this would require too many captures. So the black pawn on a2 made 2 captures. The only other capture by black was g3xh2. So to give black a retro-move, white must've played g2-g4 last (and black played g3xh2 before that).
1. hg3 a7 2. OOO Be6#

8644 - Frolkin, Andreij
Last moves were: -1. .. g4xf3 ep -2. f2-f4 g5-g4 -3. Na4-c5 c4xd3 ep -4. d2-d4 c5-c4 -5. Kd3-e3 b4xc3 ep -6. c2-c4 b5-b4

8645 - Zolotarew, A. & Plaksin, Nikita Michailowitsch
Last moves were: -1. .. Qg3xNg2 -2. Ne1-g2 Bg2-f1 -3. Nc2-e1 f5xRe4 -4. Rc4-e4 g6xRf5 -5. Rc8-c4 h7xRg6 -6. c7-c8=R Ke8-d7 -7. b6xRc7 and further retracting involves: bringing the black king to g8 and the black rook to f8, unpromoting Rf5 and Rg6 on c8, uncapturing c7xBd6, unpromoting Nc2 on d8, bringing back the d-pawn to d2, and uncapturing d3xNe2, and uncastling by black. So the minimal number of king moves by black is 4 (Ke8-g8-f8-e8-d7).

8648 - Keym, Werner
Last move was g7xRh8=N

Die Schwalbe Heft 150, Dec. 1994

R1 - Keym, Werner
If white did his last move with his king, there is no solution. If he moved the a1-rook, the solution is 1. Kc2 Ra2 2. Kc1 OO#, if he moved the h1-rook, the solution is 1. Rf2 OOO 2. Ke2 Rhe1#.

R2 - Gruber, Hans
(a) Last moves were -1. ... Kh6xPg6 -2. h5xg6 ep g7-g5 -3. Ba3-c1. So #1 with 1. Rh6#.
(b) If black just moved, the last moves must've been like in (a), but then the white king couldn't have passed the pawn barrier. So white moved last. So #1 with 1. Ra8#.
(c) Same reasoning as in (b): Black has the move. #1 with 1. ... Kf6 2. Rh6#.

8696 - Wilts, Gerd
(a) 1. h4 Nf6 2. Rh3 Nh5 3. Re3 f6 4. Re6 dxe6 5. e3 Kd7 6. Ne2 Kc6 7. Nec3 Bd7 8. Ke2 Be8 9. Qe1 Bf7 10. Nd1 Kb5
(b) 1. h4 Nf6 2. Rh3 Nh5 3. Re3 f6 4. Re6 dxe6 5. e3 Kd7 6. Ne2 Kc6 7. Nec3 Bd7 8. Ke2 Qc8 9. Kd3 Be8 10. Qe2 Bf7 11. Nd1 Kb5 12. Qe1 Qd8 13. Ke2

8699v - Andrej Kornilow
Only caps are dxe & hxg. This implies wPa4, b5, c4, f5 & bPa5, b6, c7, f7. There has been no promotions so Bishops on d8 and h8 imply wPe7 (and then wPe5, bPd7, e5) and wPg7 (and then wPg5, bPg6, h2). Precisely, the captures were h6xNg7 and d5xBe6, entailing wBc6.

We cannot have a bKf6 because he would be in check by both wPg5 and one of the 2 wN's. Hence wKf6 and bKa7, entailing bQa8, bRa6, b7, wQh6, wRg3, h3. Further, only a bNb8 will let the NW cage unlock, so that wNd5, h5.

Now, Bl. is retropat (whatever the uncolored Bd8 & h8). Last moves were necessarily
-1. Rh4-h3 h3-h2
-2. Rg4-h4 h4-h3
-3. Nf4-h5 h5-h4
-4. Qh7-h6 h6-h5
-5. Qg8-h7 h7-h6
-6. h6xNg7 Ne6-g7
and Bl. retropat is avoided.

This closing sequence requires wBh8 & bBd8.

8700 - Lubkin, Juri
Retract -1. Qb6xBe6. #1 with 1. Qa7#, and h#1 with 1. Qa5 Bc8#.

8701 - Borodatow, Leonid N.
Setplay: Black's last move was with Pg4. The only last move it could've made was f5xg4. It had to capture a rook there. To get this rook out of the a1-corner, and to let the king behind the pawns, white had to lose his castle rights. So setplay is 1. Kf2!#.
If white moved last, there is no problem to get the king out of the southwest corner, while keeping the castle rights. So h#1 with 1. g3 OO!#.

8702 - Keym, Werner
All the captures were done by the pawns, so the white pieces can't uncapture anything. The only move that'll give black a retromove was g4xPh5.

8703 - Schmitz, Hans Heinrich
1. h4 Nf6 2. h5 Nxh5 3. f3 Nf6 4. Rxh7 Ng8 5. Rh1 f6

Die Schwalbe Heft 151, Feb. 1995

R3 - Dunsany, Edward John Moreton Drax Plunk
The board is upside down! #4 with either Nc6 or Nd7.

8756 - Keym, Werner
Add a white knight on e4. Black has no last move, so he has the move. 0. .. Kg4 1. Ng5#

8757v - Zolotarew, A.
Last 26 moves were: -1. .. g6xRf5 -2. Rg5-f5 Rh3-g3 -3. Rg3-g5 Rh5-h3 -4. c2-c3 Ra5xPh5 -5. b5-b6 Ra1-a5 -6. b4-b5 a2-a1=R -7. b3-b4 a3-a2 -8. a2xNb3 Nd4-b3 -9. h4-h5 Nf5-d4 -10. e6-e7 Ne7-f5 -11. d5xBe6 Bc8-e6 -12. d4-d5 d7xBc6 -13. Bb5-c6 a4-a3 -14. Bf1-b5 and the position unlocks easily.

8758 - Haas, Josef
White's captures were cxb, dxc, exd, hxg, and the f-pawn on its own file. So white's last move can't have been a capturing move by a piece. Furthermore, the black king could never have passed the 3rd row. So the only place on the board, on which the black king could legally be added, and on which it's mated, is f5.

8760 - Wong, Peter
1. a3! Nf6 2. a4 Nd5 3. a5 Nb6 4. ab6 a5 5. bc7 Ra6 6. cb8=R Rh6 7. Rc8 Rh3 8. gh3 h6! 9. Bg2 h5 10. Bf3 h4 11. Bh5 b6! 12. Nf3 b5 13. Rg1 b4 14. Rg6 b3 15. Rf6 ef6 16. Kf1 Bc5 17. Kg2 Be3 18. de3 OO 19. Nbd2 Qc7! 20. Nf1 Qb6 21. Bd2
Unfortunately, this is cooked: 1. Nf3 a5 2. a4 Ra6 3. Rg1 Rh6 4. Ne5 Rh3 5. gh3 b5 6. Rg6 Nf6 7. Rf6 ef6 8. Bg2 Bc5 9. Kf1 Be3 10. de3 b4 11. Bf3 Ba6 12. Kg2 Bb5 13. ab5 h5 14. b6 00 15. bc7 Nc6 16. c8=R h4 17. Bh5 Qb6 18. Nd2 b3 19. Nf1 Nd4 20. Bd2 Nf3 21. Nf3

8761 - Bachmann, Karlheinz
1. Na3 Na6 2. Nc4 Nc5 3. Ne5 Ne4 4. c4 c5 5. Qa4 Qa5 6. b3 b6 7. Bb2 Bb7 8. Rc1 Rc8 9. Rc3 Rc6 10. Rh3 Rh6 11. e3 e6 12. Be2 Be7 13. Bh5 Bh4 14. Ngf3 Ngf6 15. 00 00 16. Rc1 Rc8 17. Rc3 Rc6 18. Rd3 Rd6 19. Rd4 Rd5 20. d3 d6 21. Qd7 Qd2 22. Qe7 Qe2 23. Nd7 Nd2 24. Rg4 Rg5 25. Be5 Be4 26. Bf4 Bf5 27. Nfe5 Nfe4 28. f3 f6 29. Be8 Be1 30. Bg3 Bg6 31. Bf2 Bf7 32. Rhg3 Rhg6 33. h3 h6 34. Kh2 Kh7

8762 - Haas, Josef
1. Ng2 Rh1#
1. Rg2 Qh6#
1. Bg2 Rh7#
Not a h#1 because white has no last move.

Die Schwalbe Heft 152, Apr. 1995

8817v - Wong, Peter
1. h3 g5 2. Rh2 g4 3. hg4 a6 4. Rh6 a5 5. Ra6 e6 6. Ra7 Ba3 7. ba3 b6 8. Bb2 Ba6 9. Bg7 Bd3 10. ed3 Nf6 11. Be2 Rf8 12. Bf3 Rg8 13. Bc6 Nc6 14. Qf3 Ne5 15. Qe2 c6

8818 - Haas, Josef
White captures are c2xb3xa4, d2xc3xb4xa5, e2xd3, and h2xg3xf4xe5xd6xc7. So white's last move wasn't a piece capturing a black piece. The black king couldn't have gotten past the third rank either. The only square on which the black king can legally be added, and on which it is mate, is g6.

8819 - Haas, Josef
White's pawns captured 6 times. One of the missing black pieces is the f-pawn, but none of the captures took place on the f-line. So the f-pawn promoted. So white's last move couldn't have been castling. axb happened before the black pawn arrived on a2, and cxb happened before the black pawn arrived on c3, unless the capture was c2xb3. So white's last move couldn't have been [ac]6xb7 either. So the only square where the king can be added legally, and on which he is mated, is c2.

820 - Klebes, Stefan
1. h4 d6 2. h5 Bh3 3. h6 Bg2[-wPf2, -wPh2, -wBf1, -wNg1, -wRh1] 4. hg7[-sBf8, -sNg8, -sRh8, -sPf7, -sPh7] Bd5 5. g8=R Bg8[+wRh1] 6. 00 Ba2[-wRa1, -wNb1, -wPb2] 7. c4 e5 8. c5 e4 9. Qa4 b5 10. cb6ep[-sPb5, +sPb7, -sPa7, -sPc7] Nc6 11. Qa8[-sPb7] e3 12. b7 ed2[-wBc1, -wPe2] 13. b8=R d1=R 14. Qc6[-sPd6, +sPd7, +sNg8] Rf1[+wRh1] 15. Kf1[+sRa8] Bf7 16. Rb6 Qb6[-wQc6, +wRa1, +wQd1] 17. Ke1 000

8821 - Tkatschenko, Sergej N.
Qh6/Qa3/Kc4/e8=N? Black has no last move. So #1 with:
1. ... Ba6 2. Qh6#
1. ... Bc6 2. Qa3#
1. ... Bc8 2. Kc4#
1. ... Ba8 2. e8=N#
After each black move, only one of the tries works.

8822 - Keym, Werner
Add a white pawn on b4, and #1 with 1. c5#
Add a black pawn on c7. Black has the move. #1 with 0. .. c5 1. bc6#

8823 - Tkatschenko, Sergej N.
Author's intention is: 1. h4 d5 2. Rh3 d4 3. Ra3 Qd5 4. Ra7 Qa2 5. Ra8 Qa7 6. Ra3 Nc6 7. Rh3 Qb8 8. Ra1 Na7 9. Rh1 with a Platzwechsel between the two white rooks.
However, this is cooked: 1. Na3 Nc6 2. Nb5 d5 3. Na7 Bd7 4. Nb5 Ra2 5. Ra2 Na7 6. Ra1 Qb8 7. Na3 Bc8 8. Nb1 d4 9. h4

8824 - Christiaans, Frank
The bishop on a3 is promoted. The only square it could've promoted on was d8. This requires 11 captures. If the black a-pawn was among those captures, black needs 6 captures (Bc1, axbxc, bxcxd and cxd, these last captures are needed since only 1 capture took place on the c-line). This isn't possible. Last move was Ka5xPa4, and before that, black played b5xa4. This requires only 5 captures (axbxa4, bxc, cxd and Bc1).

8825 - Christiaans, Frank
Last move was Rb3xPb5.

8826 - Christiaans, Frank
Last moves were: -1. Qa5xPb6 b7-b6 -2. Bd4-a7 Ka7-a8 -3. Kc5-b5.

Die Schwalbe Heft 153, June 1995

8880 - Wasilenko, Anatolij G. and Frolkin, Andrej Nikolajewitsch
Position is legal, last moves were -1. c4xd3 ep+ d2-d4 -2. c5-c4+.
(a) Mate in 2 with 1. Kxd3+ g5 2. hg6ep #.
(b) In the position after 1. Kxd2 g5, hg6ep is not legal. Rather mate in 2 with 1. Bb2 gh4 2. Bc1#.
(c) In the position before -1. c4xd3ep+ d2-d4, mate in 2 with 1. Kf4 (threat. 2. g5#) 1 ... g5+ 2. hg6#.

8881 - Christiaans, Frank
Bl. caps.: f7xg6 (locking the NE cage) and h7xg6xf5 or d7xe6xf5, accounting for all missing wh. units, except the dark-squared wQB. Wh. caps are f2xg3xh4; exfxgxh; cxd; dxe (7 caps). Wh. promoted his Pa2 & Pb2 (for balance) using bxa to pass through the bPa7 & b7. bPb7 and c7 promoted for balance. We accounted for all wh. caps and all bl. caps, save the wQB.

Black has no last move. White cannot retract Kg4-h5. Only possibility to avoid bl. retro-stalemate is by retracting -1. Pg3xPh4 Ph5-h4 -2. Pf2xQ/R/Bg3. (So that Bl. did play d7xe6xf5.)

White played last, so that Black now has the move. Next move can only be 1 ... gxh6 mate, which will be the "last" move.

8882 - Christiaans, Frank
Wh. 7 caps are fxexd; fxexd7; dxc; gxf and hxg for a promotion into Bg8. Bl. Ps b7 & c7 could not have both been part of these 7 caps. One of them has been taken on the b file (or the a or c file if Bl. played bxNa/c).

Black has no last move. White cannot retract anything unlocking Black, except captures. -1. BxPa6?? b7xNa6 fails. -1 NxPc5?? also fails.

-1. Kb5xPa5!! b6xNa5 etc. is the only possibility. A bB is the bPg7 promoted on g1. Other Bl. promotion is the bPh7 on h1.

White played last, so that Black now has the move. Next move can only be 1 ... Kxc6 mate, which will be the "last" move.

8883v - Ettl, Gerald
All 3 wh. caps are required for promotions: hxg6, gxf and cxb/d8. 4th promotion is wPf2 on f8. Thus we must unlock the position without uncapturing bl. units. Observe that retracting bPfxe is not allowed, so that avoiding Bl. retropat is the main difficulty.

Solution: retract
-1. Bc5-d6+ Kc6xNc7
-2. Be6-f7,g8 e6-e5
-3. Ne8-c7 Kc7-c6+
-4. Nf6-e8+ Kc6xQc7
-5. Qf4-c7+ Kc7-c6+
-6. Re5-d5+ Kc6xNc7
-7. Nd5-c7+ Kc7-c6+
-8. Nb6-d5+ Kc6-c7
and everything unlocks (bPc7 promoted on d1).

8884 - Kornilow, Andrej Nikolajewitsch
Pawn structure requires exf and hxg. The other h pawn promoted straight away to h1 or h8. We already deduce wPa2, b3, c2, d2; bPa7, b7, c7, d7.

Assuming bPg7?? gives promoted wBh8 as the h-promotion. However, the Kh3 is under check by one of the 2 Qs, requiring a last move among bPh2-h1=Q, bPh2xg1=N, wPh4xg5, wPh5xg6, all of them illegal. So that wPg7!!, entailing bNg6, wPg5 and wh. did play h6xg7 earlier.

Assuming bKh3?? requires bQh1 and bQh6 so that the wQ was taken by exf. Then the white-squared h6xg7 capture had to concern the unaccounted for bB, clearly an impossibility. So that wKh3.

This entails bQh1, wQh6, wRg2, wRf3, bKe7, bRd7, bRe6, wNf2, wNg1, bNa1, bNe8, bBd8, wBh8, bPf6. Now only bPe5 is possible, entailing wPf4, wPf7, wBh7.

Black has not many last moves. Only possibility is -1. h2-h1:Q+ Kg3-h3 -2. h3-h2 Kh2-g3 -3. h4-h3 g4-g5 -4. h5-h4 Qg5-h6 -5. h6-h5 Bg8-h7 -6. h7-h6 and then -6 ... h6xQg7 is possible, now or "next" move.

8885 - Frolkin, Andrej Nikolajewitsch
1. h4 Nh6 2. h5 Nf5 3. h6 Nh4 4. hxg7 h6 5. g8=N Bg7 6. Nf6 exf6 7. a4 O-O 8. a5 Bh8 9. a6 Kg7 10. axb7 a5 11. Nc3 Na6 12. b8=N a4 13. Nc6 dxc6 14. Nd5 Bf5 15. Ne7 Bh7 16. Ng8 Qe7 17. Nf3 Rad8 18. Ne5 Rd4 19. Nd7 Kg6 20. Nb8

Two men (here 2 wNs) on promotion squares really are the original men.

8886 - Frolkin, Andrej Nikolajewitsch
1. f4 Nf6 2. f5 Ne4 3. f6 Ng5 4. fxe7 f5 5. Nh3 Kf7 6. e8=R Bd6 7. Re6 Be5 8. Rb6 axb6 9. Nf4 Ra3 10. Nd5 Rg3 11. hxg3 f4 12. Rh6 f3 13. Re6 h5 14. a4 h4 15. a5 h3 16. a6 h2 17. a7 h1=R 18. a8=Q Rg1 19. Qa4 Rhh1 20. Qc6 bxc6 21. Re8 Ba6 22. Ne7 Bd3 23. cxd3 Qc8 24. Qa4 Qb7 25. Qa8

Two men (here wQ + wR) on promotion squares really are the original men.

8887 - Heinonen, Unto
Author's intention is 1. h4 a5 2. h5 a4 3. h6 Ra5 4. hxg7 h5 5. g4 h4 6. g5 h3 7. g6 Rg5 8. e4 Rg2 9. Qg4 h2 10. Ne2 Rh3 11. Rg1 h1=N 12. f4 Nf2 13. f5 Nd1 14. f6 Nc3 15. fxe7 f5 16. a3 Nf6 17. g8=B Bg7 18. Ba2 Bh8 19. g7 Nb5 20. g8=B Na7 21. Bgb3 d5 22. e5 Kd7 23. e8=N Kc6 24. Nd6 Kb6 25. Nc4 dxc4 26. e6 cxb3 27. e7 bxa2 28. e8=B Rb3 29. Bb5 c6 30. d4 cxb5 31. d5 Ka6 32. d6 Nd5 33. d7 Qf6 34. d8=B Bd7 35. Ba5 b6 36. Rh1 bxa5 displaying multiple Frolkin Theme. 5 promoted men (4 Bs and 1 N) are captured.

Unfortunately, the problem is cooked, e.g. with 1. h4 f5 2. h5 Nf6 3. h6 a5 4. hxg7 h5 5. g8=Q Ra6 6. a3 Rb6 7. Qa2 h4 8. Rh3 e5 9. Rb3 h3 10. g4 h2 11. g5 h1=N 12. g6 Ng3 13. g7 Nge4 14. g8=Q Nc3 15. Qc4 d5 16. f4 dxc4 17. fxe5 cxb3 18. e6 bxa2 19. e4 Rb3 20. e7 Kd7 21. Qg4 Kc6 22. e8=Q Kb6 23. Qa4 Nb5 24. e5 Na7 25. e6 Rh2 26. e7 Rg2 27. e8=B Bg7 28. Beb5 c6 29. Ne2 cxb5 30. d4 bxa4 31. d5 Ka6 32. d6 Nd5 33. d7 Qf6 34. d8=R Bd7 35. Rh8 b5 36. Rh1 Bh8.

8888 - Lubkin, Juri
Sol1: Add wKg4 & 1. Bd4#
Sol2: Add wKg5 (threat. Bd4#) and now Bl. has the move: 0 ... b6/c5/e3 1. Nc6/Bb8/Rxe3#

8889 - Samilo, Wladimir
(a) Bl. has the move: 0 ... Kxa5 1. Nc4+ Ka6 2. Bb5#
(b) Bl. has the move: 0 ... Kxa5 1. Rb5+ Ka6 2. Bc8#

8890 - Keym, Werner
-1. b7xRa8=B+ (not -1. a7-a8=B+?? or -1. b7xQ/B/Na8=B+??)

8891 - Borodatow, Leonid N.
-1. Rc6xPb6 and Black can follow with -1 ... Pb7-b6.

Die Schwalbe Heft 154, Aug. 1995

8945 - Gerald Ettl
Bl. captures are c7xd6, f7xe6xd5. Bh2 is wPh2 promoted on f8 or h8 after 2 captures h6xg7xf/h8=B. Other Wh. promotion is a2-..-a6xb7-b8.

The position is released by retracting -1. Qh7-h8+ Rc7-b7 -2. e6xd5 b7-b8=Q -3. f7xe6 a6xRb7. Then everything unlocks if we extract the wK before retracting c7xBd6.

8946 - Gunter and Raúl Jordan
1. a4 e5 2. Ra3 e4 3. Rf3 exf3 4. Nh3 fxg2 5. Nf4 g1=R 6. Ng6 Rg4 7. Nxh8 Rxa4 8. e4 Na6 9. Bxa6 Rxa6 10. Ng6 Re6 11. Ne7 Rxe7 12. O-O

8947 - Andrej and Alex Frolkin
1. h4 h5 2. Rh3 Rh6 3. Rb3 Rd6 4. c3 Rd3 5. exd3 Nc6 6. Qf3 Ne5 7. Qf6 gxf6 8. Ne2 Bh6 9. Ng3 Be3 10. Nh1 Bc5 11. g3 Ba3 12. Bg2 Bxb2 13. Bc6 Bxa1 14. Ba3 Bb2 15. Bc5 Ba3 16. Be3 Bc5 17. Bh6 Be3 18. Bg7 Bh6 19. Bh8 Bf8 20. Ba4 Nh6.

Very nice trip of the Bf8.

8948 - Josef Haas
Wh. caps are c2xb3, d2xc3, e2xd3, hxg. So that Bl. a & f Pawns have been taken on their home file while bPh7 promoted to h1 (and we can't retract wPh4xg5 right now).

Only places where a bK would be checkmated and not under an impossible double (triple/quadrule) check are:

Solution: +bKe6! (with -1. Bg8xPf7 mate).

8949 - Andrej Kornilow
Last move must be -1. c2xd1=R+ entailing wKc1, bKd3, bQc3, bRd1 and then wPb2, bPe2, bBe4, bNe1, bRf3.

Because of the wPb2, we have bBa1 & bPa2 allowing a promotion a2-a1=B. This already requires two promotions and there remains 14 Ps (8 wh + 6 bl), thus we have original Bs, entailing wBh1 & bPg2.

Last move is -1. c2xNd1=R# because xQ or xR implies an illegal check, and xB implies one more promotion. Thus the wPs captured (at most) bR, bN, bKB; while the bPs captured (at most) 2wR, wQ, wN, wQB (and wNd1 as last move).

Only coloring of remaining Ps respecting inventory is wPb3, b4, d4, e5, f4, f5, h2 and bPa2, a3, e6, g2, h3. This accounts for all captures.

After -1. c2xNd1=R#, we can't retract g3xf4??, or g4xf5??, or e3xd4?? so that -1 ... Ne3-d1.

8950 - René J. Millour
The bK could only reach e1 when Wh. removed some Wh. unit occupying e1 (thus giving check through the wK). Now Bl. cannot retract Ke1.

The bK is in check (from Rb8). Last moves are -1. b7-b8=R a5-a4 -2. Ka7-a6! a6-a5 -3. b6-b7 b2(b7)xBa6 and Wh. may retract B..(f1)xN/Q/Ba6.

The bPb2 may not retract anything as long as wPb6 is behind it. Indeed Wh. must drive Pb6 back to b3 so that a/cxb3 can be undone.

8951 - Werney Keym
Answer is "castling". You can't have both sides castling while giving check.

Die Schwalbe Heft 155, Oct. 1995

9003 - Josef Haas
There are 31 squares where it is tempting to put the bK ... But not all of these tries are legal.

Wh. caps are c2xb3, d2xc3xb4, e2xd3 & hxg. bPa and bPf have been captured on their file, not by wPs.

+ bKa(c-h)7/8 would require -1. Rxb7/8 one capture too many. + bKd6/f6 leaves no last move. + bKh6 requires -1. f4xg5, conflicting balance. + bKe5/f4 requires -1. Bh2xg3, conflicting balance. Finally, it is not possible to add the bK on the 1st or 2nd rank because there is no way he could have passed the barrier of wPs.

Solution: add bKf5 explained with -1. Nd4#.

9004 - Thomas Brand
1. Nf3 a5 2. Nd4 a4 3. Nb3 axb3 4. a3 Ra4 5. Ra2 bxa2 6. e3 axb1=Q 7. Bd3 Qxc1 8. O-O Qxd1 9. Re1 Qh5 10. Kf1 Rg4 11. Ke2 d6 12. Kd1 Rg3 13. Kc1 Bg4 14. Rd1

White castled. King-side !!

9005v - Andrej Frolkin
1. h4 Nf6 2. h5 Nd5 3. h6 Nf4 4. hxg7 h5 5. g8=N h4 6. Nxe7 h3 7. Ng8 Rxg8 8. b4 Rg3 9. b5 Qg5 10. b6 Be7 11. bxc7 Kf8 12. cxb8=N Kg7 13. Nxd7 Kh6 14. Nb8 Rxb8

The two promoted Knights are captured on promotion square (after they move).

9006 - Andrej Kornilow
Here is the colored position:

[BBNrrkRK/pPP1b1pp/1ppPppQN/2P1pPRP/6P1/888]

Last moves were -1. f7xNg8=R# Rd7-d8.

9007 - Werner Keym
(a) -1. Kd3xPc3+ d4xc3 ep+ -2. c2-c4 Rb5-/xb7+
(b) -1. e5xf6 ep+ f7-f5 -2. e4-e5+ Rd4-/xd7+

9008 - Juri Lubkin
Wh. has 20 mates in one, but Black has the move: 1 ... h6#

Die Schwalbe Heft 156, Dec. 1995

9063 - Juri Arefjew
Bl. caps are e7xd6 and f/hxg. Wh. caps are bxa and e2xPf3. Balance requires bPb7 promoted to b1, and wPh2 to h8.

Retract -1. Rb1-b8 a5-a6 -2. b2-b1=R a4-a5 -3. b3-b2 a3-a4 -4. b4-b3 b2xRa3 -5. Re3-a3 Nf5-e7 -6. Re7-e3+ Ne3-f5 -7. b5-b4 Nf1-e3 -8. Nf5-g3 Ng3-f1+ -9. Nh6-f5 Bc4-f7 -10. Nf7-h6+ Bf1-c4 -11. b6-b5 e2xPf3 -13. Ke4-f4 etc. Last 23 single moves precisely determined.

Forward mate is 1. Ne2# and not 1 ... Qxh8#.

9064 - Josef Haas
Two tries are -1. Kh7xQg7? Qf6xNg7 & forward 1. Qh6#, or -1. Kf8xQg7? Qf6xBg7 & forward 1. Qd8#. But none of these is legal. Indeed, the bRh1 is a promotee requiring two bl. caps. Assuming bNg1 is also a promotee would require 1 capture too many.

Thus bNg1 is an original N so that we must extract it through f3, after we retracted f2-f3, after we reconducted the wK home. But the wK must stay on d1 when we retract bNf3-g1 so that the wQ has been captured at home earlier. Then any later wQ is wPa2 promoted on d8, requiring 3 wh. caps which is impossible because, again, it would involve another bl. promotion for balance.

Solution: -1. Kh7xRg7! Rg6xNg7 & forward 1. Rh6#

9065 - Alexandr Zolotarew
It is possible to assume wQRa1 never moved and wKRh1 only played once with Rh1-e1.

Retract: -1. e2xBf1=N+ Kc1-d2 -2. f7xQe6 d2-d3 -3. d3xBe2 Qe5(d6)-e6 -4. d4-d3 Qb8-e5 -5. d5-d4 b7-b8=Q -6. d6-d5 a6xNb7 and now we must unpromote wBf1 & wBe2 on c8, reconduct the corresponding wPs so that bPc7xQ/Bd6 can be retracted. Then this Q/Bd6 is unpromoted on h8, allowing the retraction of bPh7xNg6 and everything unlocks.

This displays a four-fold Ceriani theme with wB, wB, wQ, wQ or B.

9066 - Alexandr Zolotarew
(The problem was printed with a typo and had wBg1 instead of wN.)

9067 - Pascal Wassong
The black queen was captured on h6 by a white knight.

She was needed on h7 to break the mutual paralysis between wRh6 and bRh8, and after white played Rh6-g6 she could not get out of the NE corner. Black pawns captured a4xNb3xRa2-a1=N and d7xNe6; the white e-pawn captured exNf-f5xNg6xBf7xNe8=R.

Because of the Madrasi condition the retroplay cannot go like -1.a3 Rg8 -2.g4??, but must be something like
-1.a3 Rg8
-2.f7xNe8=R Nd6
-3.Ke8 Nc4
-4.Qd4 Na5
-5.Qg4 Nb3
-6.Qd1 Kb6 -7.Bd7 Na1
-8.Bb5 a2-a1=N
-9.Bf1 b3xRa2
-10.Kd8 a4xNb3
-11.Nd4 Kc6
-12.Nf5 Kb6
-13.Nh6 Rhh8
-14.Nf5xQh6 Qh7
-15.Rh6
and now wPf7 and wK may be released via g6 and f5.

Early in the game the white played his g-pawn to g5, and then black played h7-h5. Later, wK reached f7 and white played f5xNg6, and then black played d7xNe6. [from HJ]

9068 - Rolf Uppström
White last move must have been with his K or R, so that 1. 0-0# is forbidden and 1. Rf1# is the only solution. (Indeed, Bl. cannot retract -1. f7xe6? because of his Bh7.)

Unfortunately, this problem is completely anticipated by N. A. Bakke in Hamar Arbeiderblad 1970.

9069 - W. Samilo
(a) Bl. has the move. 0 ... Kxa1 1. Rb1+ Ka2 2. Rc2#
(b) Bl. has the move. 0 ... Kxa1 1. Nc2+ Ka2 2. Rb2#

9070 - Wassili Mercowzi and Oleg Kovbessa
Bl. caps are fxg, c7xd6xe5 and axb for promotion of bPa7 in Bb1 (now on e8). So that retracting -1. Nd8xf7? would be one capture too many. Really, last move is -1. g7-g5! and mate in 3 is with 1. fg6 ep!! (threat. 2. g7 3. g8=N#) hg6+ 2. Kxg4 g5 3. Nh5#.

9071 - Juri Lubkin
(a) add wh. Kc4 & 1. Bxd7 #
(b) add wh. Kg4 & 1. Bf7 #
(c) add wh. Kg5 but now 1. Bf7?? is illegal. Black has the move and mate in 1 is with 0 ... Qg7/f6+ 1. Bxg7/Qxf6#

Die Schwalbe Heft 157, Feb. 1996

9129 - Gerald Ettl
No capture at all so that last move is d7-d6+. Position unlocks by reconducting both wRs behind the wPs so that c3-c4 can be retracted.

-1. d7-d6+ Kd6-e5 -2. e5-e4 Rf5-f8 -3. e6-e5 Rd5-f5 -4. f5-f4 Rd3-d5 -5. f6-f5 Rc3-d3 -6. f7-f6 Nf6-g4 -7. g4-g3 Rc1-c3 -8. g5-g4 Bg4-h5 -9. g6-g5 Rh5-h8 -10. g7-g6 Ng6-h4 -11. h4-h3 Rd5-h5 -12. h5-h4 Rd3-d5 -13. h6-h5 Rc3-d3 -14. h7-h6 Rc2-c3 -15. Kb5-a6 c3-c4+ and everything unlocks.

9130 - Alexandr Zolotarew
Bl. caps are g7xf6, cxdxe & dxexfxg. Wh. caps are axb for promotion on b8, bxc for promotion on c8 & hxg. Further wh. promotions required for balance are wPc2 to c8 and 1 or 2 promotions on g8.

Position unlocks by retracting -1. Bh3-g4+ d6xNe5 -2. Nc6/d7-e5+ f4xNg3 -3. Nb8-c6/d7 e5xNf4 -4. b7-b8=N b6-b5 -5. a6xNb7 and bl. retropat is avoided. Further retro-play consists in:
(a) retract bPb7-b6,
(b) unpromote wNf4 & wNg3 on c8.
(c) retract wPbxbKB (e.g. on c7),
(d) reconduct the bKB on its f8 home square,
(e) retract bPc7xNd6,
(f) unpromote this Nd6 on g8,
(g) retract wPd5-d8=R,
(h) retract bPd6xNe5,
(i) unpromote this Ne5 on g8,
(k) retract bPg7xNf6.

Then the wK is free to move to g5 so that the whole position can be unlocked and all Rooks can be driven to their home squares through a succession of 5 consecutive rooks moves. The nature of all uncaptured promotees is precisely determined (e.g. a wQ from e5 cannot reach g8) so that we have a quintuple Ceriani task with 5 wNs !

In the (b) twin, you have to start with -1. Bh3-g4+ d5xQe4 -2. Qe5-e4+ f4xNg3 -3. Qb8-e5 e5xNf4 -4. b7-b8=Q d7-d6 and go on as previously seen. Here we have a quintuple Cerianitask with 4 wNs + 1 wQ.

In the (c) twin, there is no way for the wNs to reach g8 (they had to visit e3) but wQs can now go there (visiting h4). So that we must now uncapture 2 wQs in steps (e) & (h) instead of wNs. Here we have a quintuple Ceriani task with 3 wNs + 2 wQs.

The (d) twin combine twins (b) & (c) so that we have a quintuple Ceriani task with 2 wNs + 3 wQs.

9131 - Alexandr Zolotarew
Wh. caps are b2xc3, exdxcxb, f3xe4 and last move -1. Be6xd5. Bl. caps are hxg for promotion on g1, gxf for promotion on f1 and axb. Further bl. promotions are bPf7 on f1, and 1 or 2 promotions on b1.

Position unlocks by retracting -1. Be6xNd5+ Nf4-e5 -2. Nd5-b4+ Ne2-f4 -3. b6-b7 Ng1-e2 -4. c5xQb6 g2-g1=N -5. h2-h3 h3xQ/Ng2 and wh. retropat is avoided. Further retro-play consists in:
(a) unpromote Qb6 on f1 (via c8),
(b) retract bPgxwQB (e.g. on f4),
(c) retract wPd4xQc5,
(d) unpromote this Qc5 on f1,
(e) retract wPf3xQe4, then bPe4-e3 and then wPe3xQd4,
(f) unpromote these 2 wQs on b1,
(g) reconduct the wQB on c1,
(h) retract wPb2xQc3,
(i) retract wPf2-f3 and wPe2-e3.

Then the bK is free to reach e4 so that the whole position can be unlocked and all Rooks can be driven to their home squares through a succession of 9 consecutive rooks moves. The nature of all uncaptured promotees is precisely determined (e.g. a wN from c5 could not reach f1 without the retraction of g2-g3, which would forbid further retraction of e2-e3 in step (i)) so that we have a quintuple Ceriani task with 1 bN and 4 bQs !

In the (b) twin, you have to start -1. Be6xQd5+ Qb5-d5+ -2. Nd5-f4+ Qb1-b5 -3. b6-b7 Qg1-b1 -4. c5xQb6 g2-g1=Q and go on as previously seen. Here we have a quintuple Ceriani task with 5 bQs.

9132 - Juri Arefjew
Bl. caps are axb & hxg, allowing promotions of wPa2,h2 on a8,h8. Wh. caps are d2xe3 & e2xf3. Position unlocks by unpromoting the wRd5. There are two possible promotion squares.

9133 - Andrej Frolkin
Author's intention: Wh. may castle if the game was 1. h4 Na6 2. h5 Nc5 3. h6 a6 4. hxg7 h5 5. f4 h4 6. f5 h3 7. f6 h2 8. fxe7 Nf6 9. g8=B Bh6 10. Bxf7 Kxf7 11. e8=R Kg6 12. Re6 Qg8 13. Rc6 Qb3 14. axb3 dxc6 15. Ra4 Bf5 16. Re4 Rag8 17. Re8 hxg1=Q 18. e4 Qd4 19. Bc4 Rg7 20. Bg8 Qd8

Two anti-phoenix wh. promotions combined with one bl. phoenix.

Unfortunately, this intention is cooked e.g. by 1. h4 Nf6 2. h5 Na6 3. h6 Nc5 4. hxg7 h5 5. g8=N h4 6. Nh6 h3 7. Nxf7 h2 8. Ne5 hxg1=Q 9. Nc6 Bh6 10. Nxe7 Kf7 11. Nc6 dxc6 12. f4 Qe3 13. f5 Qb3 14. axb3 Bxf5 15. Ra4 Qd3 16. Re4 Rag8 17. Re8 Rg7 18. e4 Qd8 19. Bc4+ Kg6 20. Bg8 a6

9134 - Rolf Uppström
Bl. used all 14 caps. for his Pawn structure. Therefore his last move is not a capture, so that Wh. last move must have been made by King or Rook.

1. O-O#?? illegal
1. Rf1# !!

9135 - Juri Lubkin
+wKh6 & 1. Be5 #
+wKh4 & 1. Bd3#
+wKh5 & Bl. has the move: 0 ... f3 1. Rxf3#

9136 - Juri Lubkin
Retract -1. Qb5xNd7. Now
(a) 1. Qxb7#
(b) 1 ... Ra7#
(c) 1. Qa5 Nb8#

Die Schwalbe Heft 158, Apr. 1996

9189 - Werner Keym
Bl. only cap. is bxc. Wh. caps are dxe, g2xh3, hxgxh & axb (for a promotion on b8). Last moves must be -1. d7-d5 Rc6-e6+ etc.

Mate in 2 with 1. cd6 ep+ c6 2. Bxc6#.

Any Bl. unit can be added without disrupting the balance of captures. (But then the bPf7 had to promote on f1). It is possible to do so without breaking the analysis concluding to the legality of the en-passant key:
- add a bP (e.g. on g3) & 1. cd6 ep+,
- add a bB (e.g. on a1 or h8) & 1. cd6 ep+,
- add a bR (e.g. on h8) & 1. cd6 ep+,
- add a bQ (e.g. on h8) & 1. cd6 ep+,
- add a bN (e.g. on d1) & 1. cd6 ep+.

9190 - Andrej Frolkin
Wh. caps are a/c6xb7 & d2xe3. Bl. made his 7 captures with Pawns: hxg, gxh, a/c6xb5, exd and there are several possibilities like cxd, fxg, gxh and one bl. promotion on g1, or fxexd, gxh and one bl. promotion on a1 or c1. In both cases, Wh. had to promote on a8 or c8.

The try -1. Ke6-f5?? fails because Bl. retracts -1 ... Be4-d5+ and plays forward 1 ... Rg6# before Wh.

Solution: retract -1. Kf6-f5 !! forcing -1 ... Bc5-d4+ (with no forward mate by Bl.) -2. c7-c8=N and play forward 1. c7-c8=Q#.

In the (b) twin, it is not possible to unpromote the Nc8. But now -1. Kg6-f5 !! forcing -1 ... Bg2-d5+ (because -1 ... g2-g1=R+ is not possible in this twin) continues with -2. c6xPb7 and forward 1. g8=Q#.

9191 - Andrej Frolkin
(a) 1. h4 h5 2. Rh3 Rh6 3. Re3 Rc6 4. Re6 dxe6 5. a4 Qd5 6. a5 Kd7 7. a6 Kd6 8. axb7 a5 9. Nh3 Na6 10. b8=R a4 11. Rb7 Rb8 12. Ra7 Bb7

(b) 1. a4 h5 2. a5 Rh6 3. a6 Rc6 4. axb7 a5 5. h4 Na6 6. b8=Q a4 7. Qb3 Rb8 8. Qe6 dxe6 9. Rh3 Qd5 10. Rb3 Kd7 11. Rb7 Kd6 12. Ra7 Bb7 13. Nh3

9192 - Gerald Ettl
Bl. caps are e7xf6 and hxgxf. Wh. caps include a2xb3 & b2xc3. bBc1 is the original bKB so that we must reconduct it on f8 before unlocking with e7xf6. Wh. is under a retropat threat.

Retract: -1. d7-d6 a2xRb3 -2. Rb8-b3 Re1xBe2 -3. Bb5-e2 Re2-e1+ -4. b3-b2 Re1xQe2 -5. Qc4-e2 Re2-e1+ -6. Ba3-c1 Re1xNe2 -7. Nc1-e2 re2-e1+ -8. Bf8-a3 Kg5-f5 -9. e7xf6+ etc.

Here Q/R/B/N are uncaptured on determined squares and three of them are only used to get retro-tempi.

9193 - Thomas Volet
Bl. cap. is c7xNb6. Wh. caps are e2xd3 & f2xPe3xd4xc5. Bl. promoted his bPf7 on f1. Position unlocks with the retraction of e2xd3. This requires reconducting the wKB (now on a6) to f1 & the wKR (now on b8) to h1, and then unpromoting the bPf7 on f1. The wKR will use the b2 cross-point to get to h1. But this trip will need reconducting the bPa3 to a6 so that the bQR (now on g8) must be reconducted to a8, again through b2. The bQh8 is the promoted bl. unit.

Here is the unlocking maneuver:
(1) bQh8 to c3,
(2) wN gets away, freeing b2,
(3) bRg8 to a4,
(4) wPb5 to b4,
(5) wBa6 to e2,
(6) wRb8 to b5,
(7) bRa4 to a8,
(8) bPa3 to a6,
(9) wRb5 to h1,
(10) wN back to b2,
(11) bQc3 to f1,
(12) retract f2-f1=Q, Bf1-e2, e2xQ/Nd3
and eveything unlocks peacefully.

9195 - Dr. Niels Hansen
Bl. only cap is axb. His last move must have been -1. f7-f5 so that 1. g5xf6 ep !! is legal. The threat 2. Qxe6# leads to 1 ... Qe2/Qxf1 2. Bxe2/Qxf1#.

9196 - Stanislaw Pestunow
The position is legal. Last moves were -1. d5xe6 ep c7-c5 -2. d4-d5+ Kg8-g7 -3. g7xh8=R+ Kf8-g8 -4. f6xg7+ etc.

Too bad the captures are not determined.

9197 - Juri Lubkin
Three solutions:
(1) Add wPd7 & forward 1. d8=Q/R #,
(2) Add wPe7 & forward 1. e8=N #,
(3) Add wPe6. Now Bl. has the move. Forward play is 0 ... d4/dxe4/cb6/fe6 1. Rxd4/Nxe4/Rd7/Re6#.

Die Schwalbe Heft 159, June 1996

9256 - Andrej Kornilow
Both black's a- and h-pawn promoted, so white can't retract a3xb4 or h4xg5. First one of these pawns has to get unpromoted. Black has only 2 retromoves, so there's some time pressure. Last moves were -1. Ng2-e3 c6-c5 -2. Nh4-g2 c7-c6 -3. Ng6-h4 Nc6-e5 -4. Ne5-g6 and further Nc6-a7-b5-a3-b1-d2-f1-e3-d1-f2-h1-h2(=P)-h3-h4-h5, and white can uncapture h4xg5. So black has the move. Mate in 1 with 1. .. Nh5#. (1. Be5?)

9257 - Josef Haas
Tries:
- Retract b3xa4, forward 1. Qa4 and 2. Qd7#. But this results in an illegal pawn structure (7 captures needed, only 6 available)
- Retract e4-e5, forward 1. Kg8 OOO 2. Rf8#. But this results in an illegal pawn structure (e5 comes from h2, f3 from e2).
- Rc2-b2 gives black a flight to b8 after 1. Kg8 OOO 2. Rf8.
- Nd3/e4-c5 gives black a flight to d7 after 1. Kg8 OOO 2. Rf8.
- c2-c4, and black has Qa2 as defence after 1. Kg8
- d3xc4 results in an illegal pawn structure.
- Rf7xPf5, and no mate after 1. Kg8 OOO!
Solution: Rf7xBf5! The only way the check can be explained is by g6-g5, and that implies the king went to h6 via f7/f8, so the black king moved, so the black castling isn't allowed. Mate in 2 with 1. Kg7! ~ (OOO?) 2. Rf8#

9258 - Andrej Frolkin, Jewgeni Reitsen & Alexander Schwichenko
White captures are axb, bxc, and the capture on g8 on his last move. Black captures are bxa, cxb, dxcxb, fxg, hxg and e3xd2.
(a): Last move was h7xNg8=B
(b): Last move was Bh7xBg8
(c): h7xBg8=B
(d): Bh7xNg8

9259 - Nikita Plaksin and Alexander Zolotarew
Bl. last move is -1. e7-e6+ (not -1. d7xe6??). Bl. caps are g7xf6, dxexfxg, cxd & a2xb1=B. Thus wh. promoted wPc2 on c8, wPg2 on g8, wPh2 on g8 after hxg, wPa2 on (a/c)8 after axbx(a/c), accounting for all wh. caps.

It is possible to assume that there have only been one Q move: bQd8-a8. Then no Q-promotion ever took place. Retract -1. .. e7-e6 -2. e3-e4 f4xRg3 -3. Rg8-g3+ e5xNf4 -4. Rc8-g8 d6xRe5 -5. c7-c8=R Qd8-a8 -6. b6xBc7 and bl. retropat is avoided. Further unlocking is by (1) unpromoting wRe5 on g8, wNf4 on c8, (2) reconducting wPc8 to c3, (3) retracting bPc4xRd3, (4) unpromoting wRd3 on g8, (5) retracting wPd2-d3, (6) reconducting bBc7 on home square f8, (7) retracting bPg7xBf6 and the position unlocks. We can now see that a wN has been captured on b1, and bNs have been captured by axb and hxg.

Quadruple Ceriani task with rich determination of promotion and captures squares.

9260 - Chris Patzke
(a)1. d3 a5 2. Bg5 a4 3. Be7 Ke7 4. Kd2 Kd6 5. Kc3 Ne7 6. Kb4 Nec6 7. Ka3 Be7 8. Nd2 Re8 9. Ndf3 Bf8 10. Qd2 Re2 11. Rd1 Re5 12. Be2 Ra5 13. Ne5 Qf6 14. Bg4 Qf3 15. Be7 Kd5 16. b4 ab3
(b)1. b3 a5 2. Ba3 a4 3. Be7 ab3 4. Bb4 Qf6 5. Ba5 Ke7 6. e3 Kd6 7. Bb5 Ne7 8. Bd7 Nec6 9. d3 Be7 10. Nd2 Re8 11. Nf3 Bf8 12. Qd2 Re5 13. OOO Rc5 14. Ne5 Kd5 15. Kb2 Qd6 16. Ka3 Ra5

9261 - Andrej Frolkin
1. h4 a5 2. Rh3 a4 3. Rb3 axb3 4. a4 Ra5 5. Ra2 bxa2 6. Nf3 axb1=B 7. Ne5 Ba2 8. Nxd7 Re5 9. Nf6 exf6 10. a5 Bc5 11. a6 Ne7 12. a7 O-O 13. axb8=Q Ba7 14. Qa8 c5 15. Qb8 Qc7 16. Qa8 Rd8 17. Qb8 Rd3 18. Qa8 Bd5 19. Qb8 c4 20. Qa8 Qb8

The promoted wh. Queen loses 6 tempi on a8 & b8.

Die Schwalbe Heft 160, Aug. 1996

9313 - Nikita Plaksin
(a): 1. Ne3? (2. Qd1#). But given is that there is at least one promoted piece on the board. Black claims that this is Rc5, and proves at least 49.5 moves have passed since the last capture, pawn move, or castling. The critical position is:

[8/pp2pp1p/1nkp4/4b3/2PqQB1b/1Pp1N1PP/B1PPPP2/R3KRrn]

The rook on g1 is a promoted rook. Play now goes: (Moves not mentioned are waiting moves) 0. OOO 1. Kb1 2. Ka1 3. Rb1 4. Rb2 5. Bb1 6. Ka2 7. Ka3 8. Ba2 9. Ra1 10. Bb1 11. Kb4 12. Ra6 13. Ba2 Ra1 14. Rb1 15. Rg1 16. Bb1 Ra5 17. Rg2 Rc5 18. Ra1 19. Ba2 20. Rag1 Rb5 21. Ka3 22. Bb1 23. Ka2 Qb4 24. Ka1 Qa4 25. Ba2 Kc5 26. Kb1 Kb4 27. Kc1 Ka3 28. Kd1 Kb2 29. Ke1 Qa3 30. Kf1 Na4 31. Rh2 32. Kg2 33. Kf3 34. Kg4 35. Qg2 36. Qf1 37. Rgg2 38. Qg1 39. Nf1 Kc1 40. Kh5 Kd1 41. Kh6 Ke1 42. Kg7 43. Kf8 Bg7 44. Ke8 45. Kd8 46. Kc8 Rc5 47. Kb8 48. Ka8 49. Bh6, and black can make his tempos in such a way that either white or black has the move in the diagram position. If black has the move, 49.0 moves have passed since the last capture, pawn move or castling. So not 1. .. Nb6 (2. .. Rc8#) because white can claim a draw after 2. Kb8!, but 1. .. Rc8 2. Ka7 (a capture, so no draw claim possible) Nb6#.
(b): Now there is no problem, since the black rook doesn't have to escape from the southeast corner. Mate in 2 with 1. Ne3

9314 - Wernew Keym
White pawns on c5 and f5 captured 4 times. Since there are 8 white pawns, the promotion piece must be black.
If the black knight on a1 is promoted, black could've moved last, namely a2-a1=N or b2xQa1=N, but not e7-e5. White may castle, since the white rook could've gotten out of the southwest corner via pawn captures on a3 and b3. So the solution is 1. OO Kg3 2. Be5#.
If the black bishop is promoted, black's only last move could've been e7-e5. Because the black bishop was captured on f8, the white rook couldn't have gotten out without breaking white's castling, so mate in 2 with 1. fe6 d6 2. Bd6#
If the black queen is promoted, or if both the black queen and the knight on a1 are promoted, black has no last move. So then mate in 2 with 1. .. Bg3 2. Kd1/Kf1 Qh1#

9315 - Frank Christiaans
(a): If a promotion piece is present, then black's last move was g7-g5 (and he may still castle). Mate in 3 with 1. hg6! (1. Nc4? OOO) Rd8 2. f7 Ke7 3. Nf5#
(b): If no promotion pieces are present, then black's last move must've been with Ra8 or Ke8. Mate in 3 with 1. Nc4 (2. f7 Kd8/Ke7 3. f8=Q/Bf6#)

9316 - Alexander Zolotarew
(a): Last moves were: -1. Rf8-c8 e6-e5 -2. Rf4-f8 e7-e6 -3. Rb4xPf4 Nd3-b2 -4. Rb2-b4 Nb4-d3 -5. Bc4-a6 Na6-b4 -6. Bg8-c4 f5-f4 -7. g7-g8=B f6-f5 -8. g6-g7 f7-f6 -9. h5xBg6 Bf5-g6 -10. h4-h5 Bc8-f5 -11. Kb5-a5 d7x[QRN]c7 and the position is unlocked. So black has the move.
(b): Last moves were: :-1. Nb6-c8 e6-e5 -2. Na4-b6 Nd3-b2 -3. Nb2-a4 Nb4-d3 -4. Be2-a6 Na6-b4 -5. Bf3-e2 e7-e6 -6. Bd5xPf3 f4-f3 -7. Bg8-d5 f5-f4 -8. g7-g8=B f6-f5 -9. g6-g7 f7-f6 -10. h5xBg6 Bf5-g6 -11. h4-h5 Bc8-f5 -12. Kb5-a5 d7x[QRN]c6 and the position is unlocked. So black has the move.

9317 - Josef Haas
Possible last moves: Kb7/c6(x/-)c7 (12), Qa2..a7(x/-)a8 (28, because QxNa8 isn't possible), Rg2/f3(x/-)f2 (12), Re4/f3(x/-)f4 (12), Nd5/g2(x/-)e3 (12), Ne4/f3(x/-)g5 (12), a7-a8=Q (1), b7xa8=Q (3), b7xc8=B (3, because b7xQc8 isn't possible), d5-d6 (1), d5xe6 (3, because d5xBe6 (both the white and black a-pawn need to promote without capture, or the black a-pawn needs to promote on b1), d5xNe6 (illegal check) and d5xPe6(ep) (Which implies Bf8 was captured on f8, resulting in an illegal white pawn structure)), e5xPd6(ep). In total 12+28+12+12+12+12+1+3+3+1+3+1=100 last moves. Only retracting Rf3xBf4! and playing forward Rf3xg3# fulfills the stipulation.

9318 - Josef Haas
Possible last moves [Illegal moves are easy to filter out]: K(x/-)a2 (14), Q(x/-)c8 (40), R(-/x)d7 (30), B(-/x)a8 (11), B(-/x)b2(36), N(x/-)d8 (20) Ne7(x/-)g8 (5), P(x/-)d6 (11), f2-f3 (1), P(-/x)f6 (11), a7-a8=B, b7xa8=B (2), c7xd8=N (4), e7xd8=N (4), f7xg8=N (4), c7-c8=Q (1), b7xc8=Q (4), a total of 199 possible last moves. Only retracting f7xNg8=N, and playing forward f7-f8=N fulfills the stipulation.

9319 - Olli Heimo
1. h4 d5 2. h5 d4 3. h6 d3 4. hg7 de2 5. gh8=R ef1=N 6. Ne2 f6 7. Nec3 Kf7 8. Ke2 Kg7 9. Qf1 Kh8 10. Nd1 Bg7 11. c3 Qf8 12. Kd3 a5 13. Kc2 a4 14. Qc4 a3 15. Re1 ab2 16. Re7 ba1=R 17. Rc7 Nd7 18. Kb2 Rb8 19. Ka1

9320 - Andrej Frolkin
1. h4 a5 2. h5 Ra6 3. h6 Rf6 4. hg7 h5 5. Nc3 Nh6 6. g8=Q h4 7. Qh3 h3 8. Qd6 cd6 9. g4 Qb6 10. g5 Qb3 11. ab3 Kd8 12. Ra4 Kc7 13. Rh4 a4 14. g6 a3 15. g7 a2 16. g8=Q a1=Q 17. Qg2 Qa5 18. Qc6 bc6 19. Ne4 Ba6 20. Ng3 Bd3 21. ed3 Kb7 22. Qg4 Qd8 23. Qg8 Bg7 24. Be2

9321 - Andrej Kornilow
The coloured position is:

[7n/5pp1/6kr/4nqr1/8/1PPPPPpK/Pppppp1P/Bb4bR]

The last 5 moves were: -1. .. h4xg3 ep -2. g2-g4 Ng4-e5 -3. Kg3-h3 h5-h4

Die Schwalbe Heft 161, Oct. 1996

9371 - Franck Christiaans
(a): Mate in 2 with 1. Qf3 (2. Qd1#) Kd3 2. OOO#. Because white can still castle, white must've played f2xe3, to let the king pass d3 and the bishop pass e3. So there was no en-passant possibility for black.
(b): White may not castle, since castling requires the pawn on e3 to come from f2 (see (a)). Mate in 2 with 1. Qg4 (2. Qd1/Qc4) Kb3/Kd3 2. Qd1/Qc4#

9372 - Andrej Frolkin
Intention was: Try: -1. .. Re8xNd8 -2. Nf7-d8 Rd8-e8 -3. Ne5-f7 Re8-d8 -4. Nf3-e5 Rd8-e8 -5. Ne5xPf3 Re8-d8 -6. Rd8-c8 f4-f3 -7. Kc8-c7 f5-f4/g5xf4 -8. c7xNb8=R and black can't retract g6-g5/g6xf5 because the white king would be trapped on the 7th rank. So last moves were -1. .. Re8-d8 -2. Rd8xNc8 Nd6-c8 -3. Kc8-c7 N~-d6 and further c7xNb8=R; c7xNd8=R; K-c7; R-b8; N-c8; king out. The black h-pawn promoted to a knight on d1.
However, this is cooked: -1. .. Re8xNd8 (!) -2. Nf7-d8 Rd8-e8 -3. Ng5-f7 Re8-d8 -4. Nh3-g5 Rd8-e8 -5. Nf4xPh3 Re8-d8 -6. Rd8-c8 h4-h3 -7. Kc8-c7 h5-h4 -8. c7-Nd8=R Nf7xNd8 -9. c6-c7 Ng5-f7 -10. c7xNb8=R N~ -11. d5xQc6 Qa4-c6 -12. c6-c7 ~ -13. Kc7-c8 ~ -14. Nf7-d8 ~ -15. c5-c6 Nc6-b8 -16. ~ Rb8xQe8 -17. Qc8-e8 and further wQ/wK out; bR->h8; bK->e8; c7xRb6; f7xRe6.

9373 - Gerald Ettl
White capture is axQb. White's promotions are b8=B and c8=R. Black captures are bxc, c3xd2, and h5xg4.
Last moves were: -1. Rf2-g2 b5xNc4 -2. Ne5-c4 a5-a4 -3. Nf3-c5 Nf5-h4 -4. Nh4-f3 Nd4-f5 -5. Rg2-f2 Nf3-d4 -6. Ba7-g1 Ng1-f3 -7. Bb8-a7 a6-a5 -8. b7-b8=B a7-a6 -9. a6xQb7 Qd5-b7 -10. Rf2-g2 Qf3-d5 -11. a5-a6 Qg3-f3 -12. Rf3-f2 b6-b5 -13. Re3-f3 b7-b6 -14. Re5-e3 Qe3-g3 and the whole position unlocks after unpromoting Re5 on c8, retracting the pawn to c2, and uncapturing c3xQd2.

9374 - Andrej Frolkin, Anatoli Wasilenko & Sergej Komarow
1. e4 g5 2. Qf3 g4 3. Ke2 gf3 4. Ke3 fg2 5. Nf3 g1=N 6. Nh4 Nf3 7. Bc4 Nd2 8. Rd1 Nb3 9. Rd7 h5 10. Rc7 Rh6 11. Re7 Ke7 12. ab3 Bd7 13. Ra6 Qa5 14. Kd4 b6 15. Bf4 Bg7

9375 - Thomas Volet
The position unlocks when the black rook is brought back to h8, and the white rook is brought back to h1. White then can uncapture h2xBg3, and the whole position unlocks. The optimal play, starting from the following position, is:

[NB5r/1Kpp1Ppp/1p2p3/2p1k3/N7/2P3b1/1PP1PPPP/5B1R]

1. hg3 Rg8 2. Rh4 Rh8 3. Rb4 Rg8 4. Rb5 Rh8 5. Ra5 Rg8 6. Ra7 Rh8 7. Ka6 Rg8 8. Rb7 Rh8 9. Ba7 Rg8 10. Rb8 Rh8 11. Rf8 Rg8 12. Kb7 Rh8 13. Kc8 Rg8 14. Kd8 Rh8 15. Ke7 Rg8 16. Rb8 Rd8 17. Rb7 Rc8 18. Bb8 Rd8 19. Ra7 Rc8 20. Ra5 Rd8 21. Ba7 Rb8 22. Rb5 Rb7 23. Bb8 Ra7 24. Rb4 Ra5 25. Rc4 Rb5 26. Rd4 Rb3 27. Rb4 Kf5 28. Rb5 Ke5 29. Ra5 Rb5 30. Ra7 Ra5 31. Rb7 Kf5 32. Ba7 Ke5 33. Rb8 Kf5 34. Kf8 Ke5 35. Kg8 Kf5 36. Re8 Ke5 37. Bb8 Ra7 38. Kh8 Rb7 39. Ba7 Rb8 40. Rf8 Re8 41. Kg8 Re7 42. Rb8 Kf5 43. Rb7 Ke5 44. Bb8 Kd5 45. Ra7 Ke5 46. Ra5 Kf5 47. Rb5 Ke5 48. Rb4 Kf5 49. Rh4 Ke5 50. Rh1 Ke4 51. Kf8
So at least 50 full-moves have been played after the last capture, so the game is a draw.
However, this is cooked. By starting from the position [N5K1/B1pp1Ppp/1p2p3/2p5/N3k3/1rP3b1/1PP1PPPP/5B1R], the diagram position can be reached in 26.5 moves.

9376 - Gianni Donati
Intention was: The P is black, the S is white. Last move must've been b2xa1=R. The white pawns captured 5 times, the black pawns 6 times. Together with the bishop from f1 these are all captures. But the white a-pawn couldn't have left its column, and no capture was on the a-line, apart from the one on a1. Removing a black piece gives white the chance to bring the a-pawn to the b-line, removing a white piece removes the necessity for the white pawn to get captured by the black pawns.
Unfortunately there are cooks like: black: Ka6, Ra1, Ra2, Pa7, b7, c5, g6, g3, with the same reasoning as the author solution.

9377 - Werner Keym
The last moves were -1. c6xQb7xBc8 h7xXg6 -2. c5-c6 Q~-b7. The last black move was h7xg6. Black has 8 pawns, so on c8 and b7, original pieces had to get captured. The bishop is the only piece that won't give check on c8, and the queen had to get uncaptured to give black a retromove after c5-c6.

9378 - Klaus Wenda
Try: Retract -1. Qa7xa6 B~-a6 -2. Kb7-c8, and forward 1. Bb6 Ba6#. But black defends with 1. .. B~xPa6!.
(a): Retract -1. Rg1-c1 e7xNd6 -2. Rb1-g1 c4xNb3 -3. Qa7xBa6! Bb5-a6 -4. Kb7-c8 and forward 1. Bb6 Ba6#. Black can't uncapture a pawn, because all pawns are (in original or promoted form) on the board.
(b): Retract -1. Ra1-c1 e7xRd6 -2. Rh1-a1 c4xRb3 -3. Qa7xBa6! Bb5-a6 -4. Kb7-c8 and forward 1. Bb6 Ba6#.

9379 - Klaus Funk
(a): The black pawns captured 7 times. Together with the bishop on f1, these are all captures black did. So the g-pawn got captured on its own row, and the missing bishop was captured on e4. So f4xe5, c7-c8=R, and Rc7-c8 weren't white's last move, so b2-b4 was.
1. cb3 Ka7 2. Qb6 Kb6#
(b): Now white's last move could've been Kc7-b8, so the ep-capture isn't legal.
1. Qc3 Ka7 2. Qb3 Kb6#

9380 - Josef Haas
Retract -1. .. Kc6xPd6 -2. c5xd6ep, and play forward 1. Bd5#. Retracting -1. .. Kc6xRd6 -2. Rd5xBd6, and playing forward Rd5-b5# isn't possible. Bd6 is a promoted bishop too. g2-g3 was played after this bishop got out, and f2xe3 before. So Bf1 was captured on f1 (which results in too little captures available for both promotions), because Nf1 had to be on that square before g2-g3.

9381 - Stanislav Vokàl
White retracts d2-d4. Now what was black's last move? The only piece that could've played is the g-pawn or the e-pawn. But both the e-pawn and the g-pawn couldn't have made a capture, since that would result in too many black captures. So black's last move must've been g7-g5. White mates in 1 with 1. hg6#.
But of course retracting h4-h5, and playing forward h4xg5# is a terrible cook ...

9382 - Pavel Kameník
1. Nf3 e5 2. Ne5 Nc6 3. Nd7 Rb8 4. Nb8 Qd4 5. Nc6 Bd7 6. Nd4 Kd8 7. Nf3 Kc8 8. Ng1 Kb8

Die Schwalbe Heft 162, Dec. 1996

9437 - Andrej Kornilow
Last moves were: -1. h7xNg8=N Nh6-g8 -2. a3-a4 Qg8xNf8 -3. Ng6-f8 Qf8-g8 -4. a2-a3 Ng8-h6 -5. Rh6-h5 a7-a6 -6. Rh5-g5 Rg5-f5 -7. Kg3-h4 f5-f4 -8. Rh4-h5 Rh5-g5
However, there is an alternative way to untangle the position: -1. Nh6-g8 Qg8-f8 -2. Qh7-h8 Qf8-g8 -3. g6xNf7 Nh8-f7 -4. Nf7-h6 Qg8-f8 -5. Rh6-h5 h5xBg4 -6. Be2-g4 Bg4-h3 etc.

9438 - Alexander Zolotarew
The position can be released by moving the white queen only once: Uncapture a black knight on f5, bring it to b7 to free the white knight, unpromote the white knight on h8, and bring the pawn back to h3, uncapture a rook on g3, and bring it to b3, so white can retract Ba2-b1, freeing the black rook. This rook goes to b3, so white can play Bb1-a2 once again to bring his rook to a1. Black has h5-h4 as tempo. Then bring the black rook to e7, so white can retract e5xBd6-d7. Bring this bishop to f8, uncapture the white bishop on f6, bring it back to c1, after bringing back the white queen to d1, and finally the whole position unlocks by uncapturing a knight on c3.

9439 - Alexander Zolotarew
Retroplay is: uncapture a knight on e5, unpromote this knight on b8, and uncapture a6xQb7. Now uncapture a queen on g3 with the black pawn, unpromote this queen on c8, and retract the pawn to c2. Uncapture a queen on f4, unpromote this queen on c8 too, and retract the pawn to c3. Uncapture a black bishop on c3. Uncapture a queen on d6, unpromote this queen on g8, and retract the pawn to g6. Uncapture a queen on e5, and bring it to g8. Retract d7-d6, retract the black bishop to f8, unpromote the queen on g8 and bring the pawns to g5 and g6. Uncapture a queen on f6. Now move everything out of the way, and the rook moves can be retracted. These rook moves are: For black: Ra8-d8, Rh8-h6-d6-d4. For white: Rh1-h3, Ra1-d1-d2.

9440 - Peter Wong
1. Nf3 Nc6 2. Ne5 Nd4 3. Nd7 Nf3 4, ef3 Nf6 5. Bb5 Nd7 6. d3 f6 7. Bh6 gh6 8. Nc3 Bg7 9. Ne2 OO 10. Ng1 Nb8

9441 - Unto Heinonen
1. h4 a5 2. h5 a4 3. h6 Ra5 4. hxg7 h5 5. Rh3 h4 6. Rc3 h3 7. g4 h2 8. g5 h1=N 9. g6 Rg5 10. f4 Rg2 11. f5 Ng3 12. f6 Rh1 13. fxe7 f5 14. a3 Nf6 15. g8=B Bg7 16. Ba2 Bh8 17. g7 Na6 18. g8=B Nc5 19. Bgb3 d5 20. e4 Kd7 21. e8=N Kc6 22. Nd6 Kb6 23. Nc4+ dxc4 24. e5 cxb3 25. e6 bxa2 26. e7 Be6 27. e8=B Bb3 28. Beb5 c6 29. d4 cxb5 30. d5 Ka6 31. d6 Nd5 32. d7 Qh4 33. d8=B Qh2 34. Ba5 b6 35. Nh3 bxa5

Five Frolkin promotions !

Unfortunately, K. Bachmann cooked this problem: 1. h4 f5 2. h5 Nf6 3. h6 e5 4. hg7 h5 5. g8=Q h4 6. a3 a5 7. Qa2 Na6 8. Rh3 Nc5 9. Rb3 h3 10. g4 h2 11. g5 h1=N 12. g6 Ng3 13. g7 Rh1 14. g8=B Bg7 15. Bc4 d5 16. f4 Be6 17. d4 dc4 18. d5 cb3 19. d6 ba2 20. fe5 Bb3 21. e4 Qd7 22. e6 O-O-O 23. e7 Rdh8 24. e8=B R8h2 25. e5 Bh8 26. e6 Qh7 27. Ba4 b5 28. d7 Kb7 29. d8=R Rg2 30. Rd3 Nd5 31. e7 Qh2 32. e8=B ba4 33. Bb5 c6 34. Rc3 cb5 35. Nh3 Ka6

9442 - Werner Keym
Last move was c6xNb7xBa8 (and black captured a piece on b7 before that).

9443 - Andrej Frolkin
White captured cxb, the h-pawn on the h-line, and the bishop on f8. Black captured dxe. The white d-pawn promoted on d8. The last repetition-free moves were: -1. .. Rd8-d3 -2. g4-g5 Rf8-d8 -3. Rd8xBf8(=b) Ne4-d2 -4. Rd2-d8 Nf6-e4 -5. g3-g4 Nh5-f6 -6. Nf4xPh5(=b) a7-a6 -7. Nd3-f4 Nd1-b2 -8. Nb2-d3 Rc1-b1 -9. Ka2-a3 Kb4-c3 -10. Nc4-b2. The pinned black pieces on b2 and d2 are switched, both in place and in colour!

9444 - Leonid Borodatow
Black captures were c7xd6 and f7xe6xd5. What was black's last move? The only possible last moves were g7-g6 or b7-b6. In the first case is Ba7 promoted on a1, so no queenside castling allowed, in the second case is Bf5 promoted on h1, so no kingside castling allowed. So only 1 castling is allowed, but we can't tell which one.
1. OO/OOO (but not both!) Re4 2. c8=Q Re1 3. Qb7#.

9445 - Andreas Witt
Black has all his pawns, so Ra5 is original. This implies that the black a- and b-pawn cross-captured. So the bishop on a7 is promoted. It must've promoted on b8, and this requires 6 pawn-captures. So white's last move wasn't a capture. The only move that'll give black a retro-move is Bf1-b5!, so retract this, and #1 with 1. Nf2#

9446 - Pavel Kamenik
1. d4 h5 2. d5 h4 3. d6 h3 4. de7 Rh4 5. Bh6 gh6 6. ed8=B h5 7. Bg5 Bh6 8. Bc1 Bd2

Die Schwalbe Heft 163, Feb. 1997

9501 - Andrej Frolkin
(a): Black could have moved last: Retro-play goes -1. .. b6-b5 -2. b2xPa3 a4-a3 -3. Ba6-c4 a5-a4 -4. Bc8-a6 a6-a5 -5. c7-c8=B a7-a6 -6. c6-c7 c7xNd6. The whole cluster of pieces in the northeast corner now opens up. The white bishop originally from f1 was captured on g6 by the black h-pawn.
Mate in 1 with 1. Rh3#
(b): White must have played last. Retro-play is: -1. b2xPa3 a4-a3 -2. Bb3-c4 a5-a4 -3. Bd1-b3 a6-a5 -4. Be2-d1 a7-a6 -5. Bf1-e2 b6-b5 -6. e2xRf3. The black f-pawn now uncaptures the promoted pawn from the c-line.
Mate in 1 with 1. .. Nf5#!

9502 - Juri Arefjew
Black captures are axb, e6xd5, and f6xe5. White captures are c6xb7 and h5xg6. Intention: Retroplay is -1. Nb3-a5 Rc7-c6 -2. Nd4-b3 Rc6-c7 -3. Nf5-d4 Rf7-g7 -4. Ng7-f5 Rc7-c6 -5. Nc6-e7 Re7-f7 -6. Nb8-c6 Rc6-c7 -7. Na6-b8 Rc7-c6 -8. g2-g3 Rc6-c7 -9. Nc7-a6 h5-h4 -10. Na8-c7 Rc7-c6 -11. a7-a8=N etc. So black has the move, and mate in 1 with 0. .. Be7/~ 1. de7/Qg8#
But this is cooked, white can have the move: -1. .. Rc7-c6 -2. Nf5-e7 Ne7-g8 -3. Bg8-h7 Rc6-c7 -4. Qh7-h8 Rc7-c6 -5. Bf7-g8 Rg8-g7 -6. Ng7-f5 Rh8-g8 -7. Nb3-a5 Rg8-h8 -8. Rc6-c5 Rh8-g8 -9. Nc5-b3 Rg8-h8 -10. Na6-c5 Rh8-g8 -11. Rc5-c6 Rc6-c7 -12. Nc7-a6 Rg8-h8 -13. Na8-c7 Rh8-g8 -14. a7-a8=N. Further retracting requires unpromoting a white piece on f8, for instance the bishop on h6. So white may mate in 1 with 1. Qg8#.

9503 - Chris Patzke
(a): Black's last move was g2-g1=N. There is no problem in resolving the position, while keeping white's castling rights: The black a-pawn captured a white officer, and promoted on b1. The white a-pawn then promoted to replace that captured officer on a8
1. OOO! Kc5/Kc3 2. Na4/Qe1#
(b): Suppose white can castle. Then the black knight went to g1 via h3, so g2xh3 was played after this knight-move. But then the rook from h1 was always trapped in the g1/h1 corner, so the black a-pawn didn't have any capture-possibility. But this pawn was captured by one of the white pawns. It can't have promoted on a1 either, since the white rook was always there (because of the assumption). Contradiction, so white can't castle. Now what was black's last move? The only possibility is -1. .. e7-e5 -2. Rf6-b6, so the enpassant capture is legal.
1. fe6 Kc5/Be5 2. Na4/Qe5#.

504 - Leonid Borodatow
(a): White's pawns captured all missing black pieces, including the black h-pawn. Since the last move by black was f5xBe4 or d5xBe4, this pawn must've promoted without a capture, so the white castling is illegal. Mate in 3 with 1. Qa3! Qb2 2. Ke2 Kc2/Qc1 3. Nb4/Rc1# 1. .. Qa2 2. Ke2 Kc2 3. Rc1# 1. .. Kc2 2.Na1 Kd3/Kb1 3. Qa6/Ke2#
(b): Black has the move, and white has to prove it by castling in the solution. 0. .. Kb2 1. Bc1 (Na1? Bf7 2. Qc2 Ka3 3. Nc1#, but no castling) Kb1 2. OO Qc3/Qa2 3. Nc3/B[b2/a3]. 0. .. Kc2 1. Na1 Kb2/Kd3 2. Qb3/OO Ka1/Bf7 3. OO Nc1# 0. .. Qa2 1. OO Kc2 2. Rc1 Kb2/Kd3 3. Be1/Qb5 0. .. Qb2 1. OO Qc1 2. Rc1 Kb2 Be1# 0. .. Bf7 1. OO Kc2 2. Bc1 Kb1/Kd3 3. Nd2/Rd1#

9505 - Josef Haas
Retract -1. .. Ka7xRb7 -2. Rb5xb7 and mate in 1 with 1. Ra5#. The black g-pawn captured the two knights on its way to g1, and a knight captured the queen on d1, to give the king some room, so the knight on g1 can enter the southeast cage via f3.

9506 - Josef Haas
Retract -1. .. Kb5xQc6 -2. Qe4xQc6, and play forward 1. Qb4#.

9507 - Alexander Zolotarew
Retroplay is -1. .. f7xQe6 -2. Qd6-e6 f4xQg3 -3. Qb8-d6 e5xQf4 -4. b7-b8=Q d6xQe5 -5. a6xQb7. Further retracting involves c3-c8=Q-e5; b3xNc4-c8=Q-f4; c7xQd6; h4xBg5-g8=Q-d6; g6-g8=Q-g3; Bf8-g5; g7xQf6; and now move everything out of the way for the 6 rook moves: 3 white ones (Ra1-d1, Rh1-h2-g2) and 3 black ones (Rh8-h6-e6-e4).
The restriction on the rook moves forces black to uncapture 5 promoted queens!

9508 - Per Olin
1. h4 Na6 2. h5 Nc5 3. h6 Na4 4. hg7 Nh6 5. g8=B Bg7 6. Bh7 Bd4 7. Be4 Bb6 8. Bc6 dc6 9. e3 Bh3 10. e4 f5 11. e5 Kd7 12. Nc3 Ke8 13. Nb1 Kd7 14. Nc3 Rc8 15. Nb1 and now black can claim a draw after Ra8, because this position, with the same possibilities, occurred after the 11th and 13th move too. Errors possible: Switch f5 and Kd7: then white has lost the ep-right, so no 3-fold repetition. Rc8 instead of Kd7: Then black has still kingside castling rights, so no repetition there either.

9509 - Olli Heimo
1. h4 d5 2. h5 d4 3. h6 d3 4. hg7 de2 5. gh8=R ef1=N 6. Ne2 f6 7. Nec3 Kf7 8. Ke2 Kg7 9. Qf1 Kh8 10. Nd1 Bg7 11. c3 Qf8 12. Kd3 a5 13. Kc2 a4 14. Qc4 a3 15. Re1 ab2 16. Re7 ba1=R 17. Rc7 Nd7 18. Kb2 Rb8 19. Ka1

9510 - Leonid Borodatow
Intention was that either promotions occurred on a1 and b8, or on b1 and a8, so the castlings are mutually exclusive. But since black's last move must've been with his king or rook, 1. Rd1 works in all cases.

9511 - Leonid Borodatow
(a): Last moves couldn't have been b7-b5, or f3xg2, so black may not castle. 1. Qb3 ~ (OOO?) 2. Qg8#
(b): Last moves could've been -1. f2-f4 f3xg2, so black may both castle and capture en passant. 0. .. ba4 1. Qc4 ~ 2. Qg8# 0. .. gf3 1. Qf3 Bc8/OOO 2. Qf8/Qb7#

9512 - Manfred Seidel
1. Nc3 Nf6 2. Ne4 Nd5 3. Nf6 ef6 4. h3 Ke7 5. h4 Kd6 6. h5 Kc5 7. h6 Bd6 8. hg7 Re8 9. g8=N Re4 10. Ne7 Qe8 11. Nc6 Qe5 12. Na7 Be7 13. Nb5 Ra3 14. Nc3 Na6 15. Nb1 Nc3 16. dc3

Die Schwalbe Heft 164, Apr. 1997

9570 - Andrej Kornilow
Only the knights are available for the retro play. The position unlocks when a knight goes to e4, and the only knight that can go to that square is the black knight. But it has to pass the white knights. This can only be achieved with the following retroplay: -1. Ng7-e8 Nf8-h7 -2. Nb8-d7 Nd7-f8 -3. Na6-b8 Nb8-d7 -4. Nc7-a6 Na6-b8 -5. Na8-c7 Nc7-a6 -6. Ne8-g7 a6xRb5! -7. Ng7-e8 Ne8-c7 -8. Nc7-a8 a7-a6 -9. Na6-c7 Nc7-e8 -10. Nb8-a6 Na6-c7 -11. Nd7-b8 Nb8-a6 -12. Ne8-g7 Na6-b8 -13. Nc7-e8 Nb8-a6 -14. Na8-c7 Na6-b8 -15. Nf8-d7 Nc7-a6 -16. Nd7-f8 Ne8-c7 -17. Nf8-d7 Ng7-e8 -18. Nd7-f8 Nh5-g7 -19. Nf8-d7 Ng3-h5 -20. Nd7-f8 Ne4-g3 -21. Qd4-d5 Ng3-e4 -22. Qg1-d4 and the position unlocks. So black has the move. Mate in 1 with 1. .. Rh5# and not with 1. g3#?.

9571 - Juri Arefjew
(a): The bishops on h7 and g1 are white, the bishops on b6 and g8 are black. Retroplay is -1. .. Ba5-b6 -2. c5-c6 Bb4-a5 -3. c4-c5 Bf8-b4 -4. c3-c4 e7-e6 -5. c2-c3 Be6-g8. Mate in 1 with 1. Rh4#.
(b): The bishops on h7 and b6 are white, the bishops on g1 and g8 are black. Retroplay is -1. Ba5-b6 d5-d4 -2. Bd2-a5 d6-d5 -3. Bc1-d2 d7-d6 -4. d2xNe3. Mate in 1 with 1. .. Qh7#.

9572 - Chris Patzke
(a): A promotion piece is present. It could only be the promoted e- or f-pawn, so black may not castle anymore. So mate in 1 with 1. Rf8#, since the king or rook could've moved last.
(b): Suppose black may castle. Last moves could only have been -1. a5-a6 Na6-c7 -2. Kc7-c6 Nb8-a6. Releasing the position while keeping black's castling rights is fairly easy, so black can mate in 1 with 1. .. OOO#

9573 - Josef Haas
The only possibility to retract a legal move, so a position remains in which white can mate in 1 is retracting g5xPf6 (ep), and mate in 1 with 1. Rg7#.

9574 - Waleri Liskowets
(a) 1. Bb3 2. Be6 3. Ke3 4. Kd3 5. Bc4 OOO# (not 4. Kf3 5. Bg4 OO#, since one of white's castlings after 4. Kf3 is forbidden, and there's no solution if OO is forbidden)
(b) 1. Bh3 2. Be6 3. Ke3. Now only one of the castlings is legal. Dependant on which one is, black continues with 4. Kd3 5. Bc4 OOO# or 4. Kf3 5. Bg4 OO#.

9575 - Andrej Kornilow
Intention: Last moves were -1. .. Nh6-g8 -2. a5-a6 Ra8-c8 -3. a4-a5 Qb8-d8 -4. a3-a4 Nc8-e7 -5. Re7-f7 Bf7-e8 -6. a2-a3 Bg8-f7 -7. Qf7-g7 Rg7-h7 -8. Qe8-f7.
But there are two flaws:
1) The position unlocks too with -1. .. Nh6-g8 2. Qg8-g7 Rg7-h7 and the queen has waiting moves.
2) The position is illegal! a7xb6, b6xc5, g5xf4 and f6xg5 all occurred on black squares, the h-pawn was captured on its own row, and the d-pawn was captured by the black e-pawn. So no piece could've captured the white bishop from f1!

9576 - Alexander Zolotarew
Retroplay is: Uncapture a rook on e4, and unpromote it on b8. Uncapture a bishop on b6 with the pawn. The pawn on h2 uncaptures a knight on h2 (which came from f1), a knight on g3, a knight on f4. Unpromote the knight on f4 and g3 both on c8, and bring the knight on f1 to d8. Retract Qc8-d7. Unpromote the knight, uncapture a rook on d5, and bring it to g8, after retracting the white pawn to d2. Umpromote the rook and bring the pawn to g6. Uncapture a rook on e5, and bring it to g8. Bring the black bishop to f8, unpromote the rook, and bring the pawns back to g5 and g6. Uncapture a knight on f6. Now the whole position can be freed. The bishop on h3 is the promoted b-pawn. The remaining queen moves are Qh1-h4, and Qd8-c8.
However, it's possible to release the position easier, with only 3 queen moves too: Last moves were -1. .. g7xNf6 -2. Ke5-f5 Ra6-e6. Now bring the white king back to e1, the knight on f6 to a3, retract e5-e4, bring the black king to e8, Bh5 to b1, retract b5-b1=B, bring Qh4 to b1 with 2 moves, and retract c3xPb2-b1=Q, d6xPe5, d4xNe5.

9577 - Andrej Frolkin
1. e3 Nh6 2. Ba6 ba6 3. h4 Bb7 4. Rh3 Bd5 5. Rf3 Ba2 6. Rf7 Bb1 7. Rf8 Kf8 8. f4 Qe8 9. f5 Qh5 10. f6 g6 11. f7 Kg7 12. f8=R Nf7 13. Rb8 Rab8 14. b4 Rb6 15. b5 Rd6 16. b6 h6 17. b7 Rh7 18. b8=R

9578 - Andrej Frolkin & Alexander Schvichenko
The intended solution is:

[5bqR/4ppqB/2ppr1pR/6Np/ppk1Pn2/K1P2b1N/PP1P1PPP/6n1]

But C. Patzke found the following cook:

[3RrbqN/ppK1ppkp/2pprR1P/4P1p1/6N1/1P1P1P2/P1P1B1P1/2n1nb2]

Last moves were -1. h5-h6 g6-g5 and Re6 is imprisoned.

9579 - Stefan Klebes
1. d4 a5 2. Bh6 Ra6 3. Bg7 [-Ph7, -Pf7, -Bf8, -Ng8, -Rh8] Rh6 4. Be5 Rh2 [-Ng1, -Rh1, -Pg2] 5. Bh2 [+bRh8] OO 6. Bc7 [-Pb7, -Pd7, -Nb8, -Bc8, -Qd8] Rf2 [-Bf1, -Pe2] 7. Bg3 Rc2 [-Nb1, -Qd1, -Pb2] 8. Bf2 Rb2! 9. OOO Rf2
But this is cooked: 1. d4 a5 2. Bh6 Ra6 3. Bg7 [-Ph7, -Pf7, -Bf8, -Ng8, -Rh8] Rc6 4. Be5 Rc2 [-Nb1, -Qd1, -Pb2] 5. Bc7 [-Pb7, -Pd7, -Nb8, -Bc8, -Qd8] Rc3 6. Bb6 Rh3 7. OOO Rh2 [-Ng1, -Rh1, -Pg2] 8. Bh2 [+bRh8] OO 9. Bg1 Rf2 [-Bf1, -Bg1, -Pe2]

9580 - Otto Kerekes
If promoted pieces are allowed, then black's last move was f2xg1=N. If no promoted pieces are allowed, black has no last move.
(a): 1. Bd3 Nh3/Nf3/c1/fg3 2. Be2/Rf3/Rc1/Ng3#
(b): 1. .. e1=Q 2. Kd3 Qe2#

9581 - Andreas Witt
If promoted pieces are allowed, both castlings are allowed, if promoted pieces aren't allowed, Rh6 and Rg4 are original rooks, so both white and black may no longer castle.
(a): 1. OOO (2. g8=R/g8=Q#) OOO 2. Qc7#
(b): 1. Qg4 (2. g8=R/g8=Q#)

Die Schwalbe Heft 165, June 1997

9645 - Friedrich Ziak
If white may still castle, then his last move was with the pawn on e4. The only missing black pieces are the two bishops, and the one on the white squares was captured on c8. So white's last move was e2-e4.
1. fe3 f8=N 2. Qf5 Nd7 3. Re4 OOO!#

9646 - Josef Haas
Retract -1. .. Kb7xNb8 -2. Nc6xNb8 and mate with 1. Na5#.

9647 - Josef Haas
Retract -1. .. Ka6xBb7 -2. Bc6xBb7 and mate with 1. Bb5#.

9648 - Alexander Zolotarew
The white pawns captured 3 times, so the pawn that needs to be added is white, and it needs to be the a-pawn. First try is a7 to give white as many retro-tempos as possible. And the following retro-play shows that white needs all those tempos: -1. .. Bh5xNg4 -2. a6-a7 d6xNe5 -3. Nd3-e5 c7-c6 -4. Ne1-d3 Bd5-h2 -5. Ng4-h2 Rg4-g2 -6. Ng2-e1 Bc3-e5 -7. a5-a6 Bd2-c3 -8. a4-a5 Bc1-d2 -9. a3-a4 c2-c1=B -10. a2-a3 b3xRc2 -11. Rc1-c2 b4-b3 -12. Ra1-c1 a5xBb4 -13. Bd2-b4 a6-a5 -14. Bc1-d2 a7-a6 -15. d2x[N/B]e3.

9649 - Tom Volet
Last moves were: -1. Qd5-h5 g3-g2 -2. Qa8-d5 g4-g3 -3. a7-a8=Q g5-g4 -4. a6-a7 g6-g5 -5. a5-a6 a6xBb5 -6. Bc4-b5 a7-a6 -7. Be6-c4 b4-b3 -8. h2-h3 b5-b4 -9. Rh3-h6 b6-b5 -10. Rg3xPh3 h4-h3 -11. Rg1-g3 h5-h4 -12. Bh3-e6 h6-h5 -13. Bf1-h3 h7-h6 -14. g2xBf3.
However, this is cooked: The retroplay could be -1. h2-h3 b6-b5 -2. Qd5-h5 g3-g2 -3. Rh3-h7 g4-g3 -4. Rh5xPh3 g5-g4 -5. Qa8-d5 g6-g5 -6. a7-a8=Q b4-b3 -7. a6-a7 b5-b4 -8. a5-a6 a6xBb5 -9. Bc4-b5 a7-a6 -10. Be6-c4 h4-h3 -11. Rg5-h5 h5-h4 -12. Bh3-e6 h6-h5 -13. Bf1-h3 h7-h6 -14. Rg1-g5 Kg4-f4 -15. g2xBf3.

9650 - Andrej Frolkin
(a) 1. d4 e5 2. Bf4 Qg5 3. e3 f6 4. Ba6 bxa6 5. h4 Bb7 6. h5 Bxg2 7. h6 Bh3 8. hxg7 h5 9. Ne2 Nh6 10. g8=B Bg7 11. Bd5 Rf8 12. Bb7 Nc6 13. Rf1 Rd8 14. Bc8 Kf7

(b) 1. h4 e5 2. h5 Qg5 3. h6 f6 4. hxg7 h5 5. d4 Nh6 6. g8=B Bg7 7. Bc4 Rf8 8. Ba6 bxa6 9. Bf4 Bb7 10. e3 Bxg2 11. Ne2 Bh3 12. Bg2 Kf7 13. Bb7 Nc6 14. Rf1 Rad8 15. Bc8

9651 - Unto Heinonen
1. a4 d5 2. Ra3 Qd6 3. Rh3 Qa3 4. b4 Nf6 5. Bb2 Ne4 6. Bxg7 Rg8 7. Ba1 Bg7 8. c4 Bb2 9. d4 Nd2 10. e4 Rg3 11. Qf3 Bg4 12. Qf6 Rc3 13. f4 Bf3 14. g4 Nd7 15. Rg3 O-O-O 16. h4 Rh8

9652c - Karlheinz Bachmann
TlG cooks this corrected version with e.g. 1. g4 Na6 2. Bg2 Rb8 3. Bc6 bxc6 4. Nc3 Rb3 5. Nd5 Rc3 6. dxc3 Nb8 7. Qd3 Ba6 8. Be3 Bc4 9. Kd2 Bb3 10. Qb5 a6 11. Ba7 Bc4 12. Re1 Bd3 13. exd3 Nh6 14. Re6 Ng8 15. Rg6 hxg6 16. Ke3 Rh3+ 17. Kd4 Rg3 18. fxg3 Nh6 19. Ne2 Ng8 20. Rf1 Nh6 21. Kc5 Ng8 22. Rf4 Nh6 23. Rb4 Ng8 24. Rb3 Nh6 25. d4 Ng8 26. Ndf4 Nh6

9653 - Andrej Frolkin
1. h4 a5 2. h5 a4 3. h6 Ra5 4. hxg7
(a) 4... h5 5. Rh3 h4 6. Rc3 h3 7. g4 h2 8. g5 Rh3 9. g6 Nh6 10. g8=N Rf5 11. g7 e5 12. Ne7 Rd3 13. Nc6 bxc6 14. g8=Q Bb7 15. Qg2 Be7 16. Qd5 cxd5 17. Nh3 Nc6 18. Nf4 Qb8
(b) Rf5 5. Rh6 e5 6. Rc6 h5 7. Nh3 h4 8. Nf4 h3 9. g4 h2 10. g5 Rh3 11. g6 Nh6 12. g8=N bxc6 13. Ne7 Bb7 14. Ned5 cxd5 15. g7 Nc6 16. g8=R Qb8 17. Rg3 Be7 18. Rc3 Rd3
19. Bh3 Bd8 20. Kf1 Ne7 21. Kg2 Bc6 22. Qh1 Qb3 23. axb3 a3 24. Ra2 d4+

9654 - Valery Liskovets
Only one of the white rooks is volage, and only one black rook is volage. There are four cases:
(1): Rd3 and Rc2 are volage: 1. Rf5? Rf2!; 1. Rb5 (2. Rb8#) Rc6 2. Nc6 Kc8 3. Rb8#; 1. .. Re6 2. Ke6 Kc7 3. Rd7#
(2): Rd5 and Rc2 are volage: 1. Rf3? Rf2!; 1. Rd3 (2. Rb8#) Rc6 2. Nc6 Kc8 3. Rb8#; 1. .. Re6 2. Ke6 Kc7 3. Rd7#
(3): Rd3 and Re2 are volage: 1. Rb5? Rb2!; 1. Rf5 (2. Rf8#) Rc6 2. Kc6 Ke7 3. Rd7; 1. .. Re6 2. Ne6 Ke8 3. Rf8#
(3): Rd5 and Re2 are volage: 1. Rb3? Rb2!; 1. Rf3 (2. Rf8#) Rc6 2. Kc6 Ke7 3. Rd7; 1. .. Re6 2. Ne6 Ke8 3. Rf8#

9655 - Leonid Borodatow
Suppose black has both castling rights. What was black's last move then? It was either b7-b6 or g7-g6. The white pawns in the center made 5 captures. The missing white pawns were captured by the black e-pawns (or those pawns captured white pieces, and the missing white pawns replaced those pieces). If black may still castle both queenside and kingside, those pawns had to make a capture each. But the black a- and h-pawn had to get promoted too, in order to get captured. If black played b7-b6 as last move, the white and black a-pawn couldn't get past each other. The same story goes for the h-pawns if the last move was g7-g6. So assuming black may castle both king- and queenside results in a contradiction.
If black can't castle kingside anymore: #2 with Nc6!. If black can't castle queenside anymore: #2 with Ng6!

9656 - Mario Velucchi
In (a) and (b) does white have no last move. So only in (c) is a legal h=5 possible:
1. c3 h5 2. c2 h6 3. c1=B h7 4. Bg5 h8=Q 5. Bh4 Qh5=

Die Schwalbe Heft 166, Aug. 1997

9720 - Alexander Zolotarew
Retract
-1. Bf1xNg2+ d7xRe6
-2. Re4-e6 a5-a4
-3. Rg4-e4 Nf1-g2
-4. Rg2-g4+ Nd3-e1
-5. c5-c6 Ne5-d3
-6. c4-c5 Ng4-e5
-7. Bf4-h6 Nh6-g4+
-8. Bb8-f4 a6-a5
-9. b7-b8=B a7-a6
-10. a6xQb7 Qd5-b7
-11. c3-c4 Qd1-d5
-12. c2-c3 d2-d1=Q
-13. a5-a6 c3xBd2
-14. Bc1-d2 c4-c3
-15. d2xBe3 Bc5-e3
-16. a4-a5 Bf8-c5
-17. Kg5-h5 e7xQ/Nf6+.

Olli Heimo sent me this cook: -1.Bf1xNg2 c7xBb6 -2.Ba5-b6 d7xRe6 -3.Bd2-a5 a5-a4 -4.Bc1-d2 a6-a5 -5.d2xBe3 Bc5-e3 -6.Re3-e6 Bf8-c5 -7.Kg5-h5 e7xQf6 -8.Ra3-e3 Ng3-h1 etc.

9721c - Unto Heinonen
1. h4 d5 2. h5 d4 3. h6 d3 4. hxg7 dxc2 5. d4 Bf5 6. d5 Kd7 7. d6 Kc6 8. dxe7 Kc5 9. exd8=Q Ne7 10. g8=Q Nbc6 11. Qg4 Rg8 12. Bg5 Rg6 13. e3 Bg7 14. Qg8 Rf8 15. Qd8 Nd4 16. Ba6 Rxa6 17. Qd1 c6 with a beautiful rotation between the 3 wh. Queens.

9722 - Andrej Frolkin
1. h4 a5 2. h5 a4 3. h6 a3 4. hxg7 h5 5. Rh3 h4 6. Rb3 h3 7. g4 h2 8. g5 Rh3 9. g6 Nh6 10. g8=N Rd3 11. Nf6+ exf6 12. g7 Bd6 13. g8=N Kf8 14. Ne7 Qe8 15. Nc6 dxc6 16. Nc3 Be6 17. Nd5 Nd7 18. Ne7 Rd8 19. Ng8

9723 - Andrej Frolkin and Oksana Kartsewa
1. Nf3 b5 2. Nd4 b4 3. Nb5 b3 4. axb3 c5 5. Rxa7 Qc7 6. N1a3 Qg3 7. hxg3 c4 8. Rh6 c3 9. dxc3 Nc6 10. Qd6 Nd8 11. Qb8 d5 12. Rc7 Bh3 13. gxh3 d4 14. Bg2 d3 15. cxd3 gxh6 16. Nc2 Ra3 17. bxa3 Bg7 18. Bb2 Bd4 19. Ba1 Be3 20. fxe3 Nf6 21. Kf2 Ne4+ 22. Kf3 Nd2+ 23. Kg4 f5+ 24. Kh5 f4 25. Be4 f3 26. exf3 O-O

Fantastic 8 cross-captures by the White Pawns.

Unfortunately cooked: 1. e3 b5 2. Ke2 b4 3. Nc3 b3 4. ab3 c5 5. Ra7 Qc7 6. Nb5 Qg3 7. hg3 c4 8. Rh6 c3 9. dc3 Nc6 10. Qd6 f5 11. Qb8 f4 12. Rc7 f3 13. Kf3 Nf6 14. Bd3 Ne4 15. Kg4 Nd2 16. Kh5 Ra3 17. ba3 gh6 18. Bb2 d5 19. Ba1 d4 20. Be4 d3 21. cd3 Bg7 22. Ne2 Bd4 23. Ned4 Nd8 24. Nc2 Bd7 25. f3 Bh3 26. gh3 OO

9724 - Jon Bang
Last move was -1. Bf5-e4++ (and then the three Rooks on c6, h6 & g5 have same color) or -1. Rf5-g5++ (and then the two Bishops on g8 & e4 have same color). Thus we have one promotion to account for. Also, because of the Kg6, we cannot assume both wPh5 and bPh7.

Position is 1B1Qq1B1/1pp3Pp/PPr3kr/3K2rp/2pRbpP1/3PpP2/2P1P1n1/4n3 (13+15) and, with the following captures: wPhxbBg3/5/7, bPdxc, bPgxh, bPaxb and promotion on b1, the position unlocks easily.

Sol: The White King is mate.

9725 - Klaus Wenda
Retract -1. Ba6xBf1 [+bBc8] Rg3xBb3. This makes Be4 and Ba6 promoted pieces. h#1 with 1. Ba6 [+wBa8] Be4 [+bBe1]#.

9726 - Juri Arefjew
The white pawns on the d- and e-line cross-captured, as did the pawns on the g- and h-line. The black pawns require all 7 captures. So black's last move wasn't a capture. This means that white's last move was g2xh3. So black's captures were: axbxa2, bxa2, gxh, fxe, dxe, and cxd. This also implies that white first captured on the d-line, and then on the e-line, and first on the g-line, and then on the h-line. So the position is indeed legal.

9727 - Juri Arefjew
Intention: Unpromoting of a queen, bishop, rook on a8 or h8 don't work, and unpromoting a knight on h8 doesn't work either. Only unpromoting a knight on a8 works.
But: Last move could've been h3xNg2, and any piece can unpromote on a8 (except a bishop). White's pawns in the diagram all captured towards the left, and the white a-pawn was captured by one of the black pawns currently on the b-line.

9728 - Stanislav Vokál
(a) White captures are dxc and hxg, black captures are gxh and b7xc6. The only piece that white could've captured on the c-line was the black rook originally from h8, since before that capture, Ra8 and Bc8 are still imprisoned. After that capture, the black bishop gets captured on g4, and the white rook from h1 gets captured on h4. So white may castle, black may not
(a) Same analysis as the previous case shows that the rook originally from h8 was captured on the c-line. But for that rook to escape, g7xh6 must've occurred. The only piece which could've gotten captured there is the rook originally from a1, since the white h-pawn is still on h2 or h3. So the move sequence was: d2-d4; Ra1 ->h6; g7xRh6; Rh8 ->c; dxRc; b7xPc6; Bc8 ->g4; h3xBg4; Rh1 ->a1, Ra8 ->h8. So neither side can castle
(c) Now the black bishop from f8 could've gotten captured on the c-line. But for this bishop to escape, g7xh6 must've occurred first. Again, the only piece which could've gotten captured there was the rook originally on a1. So the move sequence was: d2-d4; Ra1 ->h6; g7xRh6, Bg8 ->c5; d4xc5-c6; b7xPc6; Ra8 ->g-line; hxRg; Rh1 ->a1. So black may castle, white may not
(d) Now black could've played g7-g6 to let out the bishop. Move sequence was: g7-g6; Bf8 ->c; dxBc-c6; b7xPc6; Ra8 -> g; hxRg; Rh1-h5; g6xRh5. So both players may castle

Die Schwalbe Heft 167, Oct. 1997

9792 - Alexander Zolotarew
Retro-play is: -1. Qd6-g3 c4xBd3 -2. Qd8-d6 c5-c4 -3. d7-d8=Q c6-c5 e6xQd7 and further retractions are: Bd3->a8; Ba7->b8; a6->a8=B; a7xQb6; b6->b8=B; b7xQc6; Qc6->g8; Qb6->g8; h4xBg5->g8; Bg5->f8; g6->g8=Q; g7xQf6.
Mate in 1 with 1. .. Rg3#.

9793 - Alessandro Cuppini
The white pawns on the board captured 15 times, so the missing pawn has to get added on the h-line.
Add Ph7 and Kh6, and #1 with 1. h8=N#.

9794 - Andrej Frolkin & S. N. Tkatschenko
1. a4 h5 2. a5 h4 3. a6 h3 4. ab7 hg2 5. ba8=Q gh1=Q 6. Bh3 Qh2 7. Qh1 d5 8. Bc8 Qe5 9. Qh8 Qb2 10. Nh3 Qa1 11. Bb2 Qa3 12. Bf6 Qc1 13. Bh4 g5 14. Qa1 Qd1

9795 - Unto Heinonen
1. g3 Nf6 2. Bg2 Rg8 3. Bc6 bc6 4. f3 Ba6 5. Kf2 Bc4 6. Kf1 Bg3 7. ab3 Ne4 8. Ra5 f6 9. Rh5 Kf7 10. Rh6 gh6 11. Na3 Rg5 12. Nc4 Rc5 13. Na5 Rc3 14. bc3 Bg7 15. Ba3 Kf8 16. Bc5 Ke8 17. Bb6 c5 18. Ke1 Nc6

9796 - Andrej Frolkin & S. I. Tkatschenko
1. f3 Nc6 2. Kf2 Nd4 3. Kg3 Ne2 4. Kg4 Ng3 5. hg3 c6 6. Rh5 Qa5 7. Nh3 Qa2 8. Na3 Qb3 9. cb3 Kd8 10. Nc2 Kc7 11. Ra5 Kd6 12. Na1 Ke6 13. Qe1 Kf6 14. Rf5 Kg6 15. Bb5 Rb8 16. d3 Ra8 17. Bf4 Rb8 18. Bb8 cb5 19. Qe5 b4 20. Rhg5 Kh6 21. Rh5 Kg6 22. Rhg5 Kh6 23. Rh5 Kg6
Draw by 3-fold repetition.

9797 - Gunter Jordan & Manfred Seidel
1. g4 g6! 2. g5 h5 3. gh6 e5 4. h7 Qg5 5. hg8=N Qg1 6. Nh6 Qg5 7. Ng4 Rh2 8. Bh3 Qd2 9. OO g5

Die Schwalbe Heft 168, Dec. 1997

9861 - Werner Keym
If a white queen or rook is added, then black must've done a capture on a black square, so black has no last move. If a bishop is added, black could've played a4-a3 as last move.
(a): 0. .. Rh2 1. Qg7 Kh1 2. Rh2 Kh2 3. Kf2 Kh3 4. Qg3#
(b): 0. .. Rh2 1. OOO#
(c): 1. Bd4#
A stronger piece on a1 lengthens the mate!

9862 - Gianni Donati
Suppose both white and black may still castle. The rook on b2 came from the f-pawn, which promoted on b1. This requires 4 captures, together with hxg and d7xc6 all black captures. The white g- and h-pawn cross-captured to let the black g- and h-pawn pass. To give black a retro-move, white must've played f2-f4 as last move (and black played his knight to g3). So black may capture en-passant, if white and black both castle in the solution.
1. gf3 2. Qg4 3. OOO 4. Re8 5. Kd8 6. Qh5 OOO!#

9863 - Gerald Ettl
The position unlocks after retracting: -1. .. Kb2-c1 -2. Na3-b1 h6-h5 -3. Rc1-a1 Ka1-b2 -4. Nb1-a3 Nb2-d1 -5. Qd1-e1 h7-h6 -6. Ke1-f1 Qg2-h1.
-5. Rd1-d2 h7-h6 -6. d2-d3 Re3-c3 -7. c3-c4 Be6-b3 seems to be a cook though.

9864 - Johannes Quack
1. ab7 (2. Nd4 Ra1#) Ra1 2. Na1 Ra1#; 1. .. Nf3 2. gf3 Nf3#; 1. .. Bf1 2. OOO#. But if castling isn't possible, then Bf1 isn't a defense against 2. Nd4. So we'll have to prove white can no longer castle.
Suppose white can still castle. Then the black d- and e-pawn were captured on their own file. The a-pawn was captured on its own file too. Together with the 5 captures, needed for white's current pawn structure, these are all captures. The captured piece on the g-line is the missing pawn. The black c- and h-pawn promoted. Last black move can't have been Kc4-d3 (c2xb3??), or Kd4-d3 (N~-c2 is illegal, because Rd8# was possible, and white had to play that due to the reflex-condition). So black's last move was Bf3-e2 or Bd1-e2. White couldn't have given this check with Re2-e8 (because Rd2# was possible), so he had to deliver this check with Ne2-g3 (no capture possible). To give black a retromove before this, Bf3-e2 must've been black's last move. But white could castle, and Ne2-g3 was illegal, because OOO# was possible.
So white may not castle.

9865 - Cuppini
Author solution is: Both castlings are legal. 1. Bd6 Rb7 2. OOO Nd6# and 1. OOO Ra2 2. Nb2 Ra1#.
But there are lots of cooks for the second solution, for instance 1. OOO Ba3 2. N~ Rb3#, or 1. Rd1 Ng3 2. c3 Re2#.

9866 - Christian Poisson
(a): 1. g3 c5 2. Bg2 Qb6 3. Bb7 Qc6 [+bPc7] 4. Bc8 Qd6 [+bBd8] 5. Bb7 Qa6 6. Bg2 Qc8 7. Bf1
(b): 1. g3 d5 2. Bh3 Qd6 3. Bc8 Qe6 [+bBd8] 4. Bb7 Qg6 [+bPd7] 5. Bd5 Qa6 6. Bg2 Qc8 7. Bf1

9867 - Jacques Rotenberg & Wassong Pascal
1. e3 h5 2. Qh5 d5 [+bPh3] 3. Qd5 Rh4 [+bPd1=R] 4. Ke2 Rc1 5. Nh3 [+wBd3] Rf1 6. Qa5 [+wBc1] Rh1 7. Kf3 Rf1 8. Bf1
The white bishops switched places!

9868 - Leonid Borodatow
Black has no last move. #2 with:
1. .. g6 2. de8=Q Kc7 3. Qf4
1. .. gh6 2. Qd5 (Bg7/Kc7)/(Re7/Be7) 3. de8=N/Rc8#
1. .. Kc7 2. Qb7 Kd8/Kb7/Kd6 3. Nc6/Rc2/de8=N#
1. .. Be7 2. de8=Q Kc7 3. Qe7#

9869 - Mario Velucchi
Since there are 8 pawns, the rook on d8 is original. Nh8 was on that square before h7xg6 was played, so the rook had to escape the northeast corner via e8, so black may not castle. Solution: 1. f6 gf6 2. Rd7 f7 3. Kd8 f8=Q#, and not 1. Rb8 Kc6 2. Rb6 ab6 3.OOO b7#.

Die Schwalbe Heft 169, Feb. 1998

9919 - Nikita Plaksin & Alexander Zolotarew
Last moves were -1. Nc6-d4 d5xRe4 -2. Nb8-c6 c6xRd5 -3. b7-b8=N e7-e5 -4. a6xRb7 and further: unpromote Re4 on g8, and bring the pawn back to g3, uncapture b7xRc6, unpromote Rd5 on g8, unpromote Rc6 on g8, bring the pawns back to g5 and g4, and uncapture h3xRg4 and f4xBg5.
However, the position can be unlocked in a different way too: -1. Nc6-d4 d7-d6 -2. Nb8-c6 d5x[RN]e4 -3. b7-b8=N c6x[QRN]e5 -4. a6xRb7 and further: unpromote R/Ne4, Q/R/Nd5 and Nc4 on g8, uncapture a rook on g4 and a bishop on g5, and the rest is easy.

9920 - Andrej Kornilow
Last move was g2-g4. Author's intention is: 1. hg3 Bb1 2. Kf5 fe4 3. Kg6 e5 4. f5 ef6#. But there are several cooks: 1. hg4 Bb3 2. Kf5 Bf6 3. Kg6 Bf5 4. Kh5 fg4# or 1. hg3 fe4 2. Nf6 e5 3. Kf5 ef6 4. Kg6 Bb1#.

9921 - Tom Volet
White captures are c2xd3, hxgxfxe-e6xf7. Black captures are cxbxa, and d7xe6. The position unlocks after retracting -1. .. b5xPa4 -2. Rf1-d1 c6xNb5 -3. Na3-b5 c7-c6 -4. Nc2-a3 Kd1-e2 -5. Ne1-c2 Kc2-d1 -6. Nf3-c1. Further retracting involves getting the king to the third rank via a2, uncapturing c2xBd3, bringing Rg5 and Bd3 back to a8 and c8, uncapturing d7xQe6, and uncapturing e6xRf7.
Last 5 captures are precisely determined.

9922 - Alexander Zolotarew
Last moves were: -1. Ba7-b6 b6xNc5 -2. Nb7-c5 c4xBb3 -3. Nd8-b7 c5-c4 -4. d7-d8=N c6-c5 -5. e6xRd7. Further retraction requires: Bb8-a7; a6->a8=B->d3; a7xNb6; b8=B->b5; b7xNc6; g5->g8=N>b6; g6>g8=N>c6; h4xBg5; Bf8>g5; g7xNf6 and the whole position unlocks.
So the game could have started with 1. e2-e4.

9923 - Peter Wong
1. e4 a5 2. Ke2 a4 3. Kf3 a3 4. Kg4 ab2 5. a4 Ra6 6. a5 Ra7 7. a6 Ra8 8. a7 Nc6 9. Ra6 Nb8 10. Rh6 g6 11. ab8=N Bg7 12. Na6 Bc3 13. dc3 Kf8 14. Qd4 Qe8 15. Qa4 Kg7 16. Bc4 Qf8 17. Ne2 Qd8 18. Rd1 Kf8 19. Rd6 Ke8 20. Rf6 e6 21. Nd2 Ne7 22. Nf3 Nc6 23. Nfg1 Nb8

9924 - Wladimir Chutornoj
Retract Kc3-d3, and mate in 1 with Kc3-c2.

9925 - Christian Patzke
Retract Nd5-e3, and mate in 1 with Rg3-a3.

9926 - Wladimir Chutornoj
Retract Kc3-d3, and mate in 1 with Kc3-c2.

9927 - Christian Patzke
Retract Nd7-c5, and mate in 1 with Rd5-a5.

9928 - Wladimir Chutornoj
Retract b2-b3, and mate in 1 with Qd5-a2. Last moves were Nb3-c1 Ka2-a1.

9929 - Andrej Kornilow
The white pawns currently on the board captured 6 times. Together with the bishop from c8, these are all captures by white. The black pawns captured 7 times, all on the black squares. White misses 8 pieces. The missing bishop on the white squares wasn't captured by the pawns, so the a-pawn was. It couldn't have captured itself, so it promoted on a8, and then got captured. So black's last move wasn't a7xb6. The only other legal last move could've been g7-g5, so white may capture en-passant.
1. hg6 hg6 2. Kf4 b5 3. cb5 b6 4. Ke3 Kg5 5. Kf3 Kh6 6. Kf4 Kh7 7. g5 etc.

1 - A. Kornilov
1. h4 a5 2. h5 a4 3. h6 a3 4. Rh5 Ra4 5. Ra5 Rh4 6. b4 g5 7. Bb2 Bg7 8. Bd4 Be5 9. Nc3 Nc6 10. Qb1 OO 11. Qb3 Kh8 12. OOO Rg8 13. Kb1 Rg6 14. Ka1 Qg8 15. Rb1 Ne8 16. Nd1 Rd6 17. Qd3 Qg6 18. Rb3 Qe6 19. Qg6 Bf6 20. Re3 Qb3 21. Bc3 Rdd4 22. Ree5 Nd6 23. Ne3

2 - A. Kornilov
1. Nf3 Nc6 2. Ne5 Nd4 3. Nd7 Ne2 4. Nb6 Ng3 5. Bc4 Bf5 6. Qe2 Qd7 7. Kd1 Rd8 8. Re1 Qc8 9. Qf1 Rd3 10. Re6 Ra3 11. Rh6 gh6 12. ba3 h5 13. a4 Bh6 14. Ba3 Be3 15. Bd6 Nh6 16. Na3 Rg8 17. Rb1 Rg6 18. Rb3 Re6 19. Rd3 f6 20. c3 Kf7 21. Kc2 Kg6 22. Kb3 Kg5 23. Kb4 Bg6 24. Bb3 Nhf5 25. Nac4 h6 26. a3

3 - A. Frolkin & A. Kornilov
1. a4 h5 2. a5 h4 3. a6 h3 4. ab7 hg2 5. h4 a5 6. h5 a4 7. h6 a3 8. h7 a2 9. Rh6 Ra3 10. Nh3 Na6 11. b8=Q g1=Q 12. Qb4 Qg5 13. Qh4 Qa5 14. b4 g5 15. b5 g4 16. b6 g3 17. b7 g2 18. b8=Q g1=Q 19. Qbb4 Qgg5 20. Qbg4 Qgb5 21. c4 f5 22. Rf6 Rc3 23. dc3 ef6 24. Be3 Bd6 25. Nd2 Ne7 26. Rb1 Rg8 27. h8=N a1=N 28. Ng6 Nb3 29. Ngf4 Nbc5 30. Ng2 Nb7 31. f4 c5 32. Bg1 Bb8 33. e3 d6 34. Be2 Bd7 35. Nf1 Nc8 36. Qd2 Qe7 37. Kd1
However, there is a dual, as pointed out by Olli Heimo: 21. c4 Rc3 22. dc3 f5 23. Rf6 ef6, and the rest of the solution is the same.

Die Schwalbe Heft 170, Apr. 1998

9985 - Leonid Borodatow
Suppose black can still castle. Then his last move must've been b7-b5. So #4 with 1. cb6 c6 2. Bc6 Kf8 3. Qf4 Kg8 4. Qf7#. In the other case, 1. OOO! leads to a forced mate in 4.

9986 - Alexander Hölle
Suppose black may castle. The rook originally from h8 was captured on g3, so the pawns on the g- and h-file must have cross-captured. This means that the pawns made 4 captures. So either the a- or the e-pawn must have promoted on their own file, because they can't capture to go to the b- or d-file. But Ra8 and Ke8 never moved, according to the assumption. This is a contradiction, so black may no longer castle.

9987 - Nikita Plaksin & Alexander Zolotarew
Retroplay is: -1. Ne2-c1 f4xNg3 -2. Ne4-g3 Qd6-c5 -3. Nc5-e4 Qd8-d5 -4. Nd7-c5 e5xNf4 -5. Nb8-d7 d6xRe5 -6. b7-b8=N h4-h3 -7. a6xNb7 and further: Unpromote Nf4 on b8, and retract that pawn to b3. Uncapture b4xRc3. Unpromote Re5 and Rc3 on g8, bring Bh2 to c1, retract the pawns to g4 and g3, and uncapture h2xBg3. Bring the bishop back to f8, and uncapture g7xRf6.
So only 2 queen moves are needed.

9988 - Alexander Zolotarew
(a): The last moves were: -1. .. Rh8xNh7 -2. b5xBc6 Be4-c6 -3. Ng5-h7 Bh7-e4 -4. b4-b5 Qf5-h3 -5. Nh3-f5 Qb1-f5 -6. a3-a4 b2-b1=Q -7. c2-c3 b3-b2 -8. a2-a3 a4xBb3 -9. Bc4-b3 a5-a4 -10. Bf1-c4 Kg4-h4 -11. e2x[QR]f3 and everything unlocks.
(b): The last moves were: -1. .. Rh8xBh7 -2. b4xNc5 Ne4-c5 -3. a3-a4 Ng5-e4 -4. Bd3-h7 Nh7-g5 -5. Bf1-d3 b7-b6 -6. e2xBf3 Bh5-f3 -7. a2-a3 Kg4-h4 -8. c2-c3 Qa3-h3 -9. Rh3-h2 and everything unlocks.

9989 - Gianni Donati
1. Nc3 b5 2. Nd5 Ba6 3. Nxe7 Nxe7 4. e4 Ng6 5. e5 Bd6 6. e6 Kf8 7. e7+ Kg8 8. e8=N Bg3 9. Nd6 Qh4 10. Nf5 Qa4 11. Nd4 b4 12. Nde2 Bxe2

9990 - Alexander Hölle
White captured b6xa7 and e2xf3. The bishop on a2, or the bishop just captured on f1, was promoted, and the pawn made no captures towards promotion. So the g-pawn promoted on g8. This means that the black f- and g-pawn cross-captured. Because no black pawn was captured, and white still can castle, the d-pawn captured once on the c-file and promoted on c1. This, and the capture on b6, accounts for all captures of black. Because the d-pawn captured to promote on c1, the white c-pawn had left c2 already. So the only last move white has was f2xe3. So the answer is The white pawn on e3 moved last

Die Schwalbe Heft 171, June 1998

10046 - Werner Keym
Add a white bishop on d5, and #1 with 1. Re7#
Add a black bishop on g8. Last move must've been g7-g5. Mate in 1 with 1. fg6#
Add a white bishop on g6. Black has the move. Mate in 1 with 0. .. Kg8 1. Re8#
Add a black bishop on f3. Mate in 1 with 1. Kg5 Rg8#.

10047 - Alexander Zolotarev
Last moves were: -1. .. f4xQg3 -2. d2-d3 d5xRe4 -3. Rc4-e4 a7-a6 -4. Rc8-c4 e5xNf4 -5. c7-c8=R c6xNe5 -6. b6xNc7. The further retroplay looks like (black moves between brackets): b4-b6; Nc8-e5; a3xBb4; c4-c8=N; Nc8-f4; c6-c8=N; b5xBc6; (c7xRd6); Re4-d6; Qh8-g3; Bg3-h4; (Rh4-g4); Rg4-e4; (Rg7-h4); h4-h8=Q; (Rh8-g7); (h7xRg6); Rh6-g6; (g6-g5); (Bg5-b4); Bg3-f4; Rh5-h6; (Bh6-g5); Bg5-f4; (Bf8-h6); (g7-g6); g6xQf7; and now the 6 king moves in a row: (Ke6-f5) Kf1-f2 (Kd7-f6) Kf2-f1 (Ke8-d7) Ke1-f2.
But there is a whole different way to release the position: -1. .. f4xNg3 -2. Nf1-g3 e5xQf4 -3. Qc1-f4 d6xNe5 -4. Bg3-h4 Rh4-g4 -5. Ng4-e5 and both sides can move freely. Further retroplay is: Qg7-c1; (Rh7-h4) Qh8-g7 (Rg7-h7) h7-h8=Q (d5xRe4) h6-h7 (Rh7-g7) h5-h6 (h6xQg5) Qc1-g5 and further Nd7-f1; Qc8-c1; c5-c8=Q; Rd4-e4; b4xBc5; Nc8-d7; (Rh8-h7); (h7-h6); Bf8-c5); (g7-g6); g6xQf7 and now the same king maneuver.

10048 - Unto Heinonen
1. Nc3 e5 2. Nd5 Bd6 3. Ne7 e4 4. Ng8 Bh2 5. Nh6 gh6 6. e3 00 7. Qa4 Kh8 8. Qa6 Rg8 9. a4 Rg2 10. Ra3 Bg3 11. Rh5 ba6 12. Rb5 d5 13. Rb8 d4 14. Rb6 Rb8 15. Rd6 Rb2 16. Rb3 Rd2 17. Kd2 Qg5 18. f4 ef3 19. Kd3 Bf5 20. e4 de3 21. Kd4 Bg6 22. Bc4 c5#
But this is cooked: 1. d4 Nh6 2. Bh6 d5 3. Qd3 Na6 4. Qa6 ba6 5. Nd2 Rb8 6. OOO Rb2 7. Kb2 e5 8. Ne4 de4 9. Bc1 ed4 10. h4 Bd6 11. h5 OO 12. h6 gh6 13. Rh3 Kh8 14. Rb3 Rg8 15. e3 de3 16. Kc3 Rg2 17. Bc4 Bg3 18. Rd6 Qg5 19. Kd4 Bf5 20. f3 ef3 21. c3 Bg6 22. a4 c5#

Die Schwalbe Heft 172, Aug. 1998

10097 - Alexander Zolotarew
The intended way to release the position, using only one queen move is: -1. Bh6xRg5 Rb5-g5 -2. Rg5-g6 Rb1-b5 -3. g6xNh7 Nf8-h7 -4. f5xNg6 b2-b1=R -5. e4xNf5 b3-b2 -6. a2-a3 a4xBb3 and further: (black moves between brackets) (Nc1-f5); (c7-c1=N); (Nc1-g6); (c6-c1=N); c2xRd3; (Rg3-d3); (Nd1-f8); (d4-d1=N); d3xRe4; (Bg6-g2); (g3-g1=R); (Re4-g1); Bh3-b3; (g4-g3); Bf1-h3; (g2-g1=R); d2-d3; (g3-g2); g2xNf3; (Ke4-f4); Ra1-g5; Bb8-h4; b3-b8=B; (d5-d4); (Kd7-e4); (Ng5-f3); Ke1-f6; (Nd1-g5); (Nc3xQd1); (Nb8-c3); Bc1-h6; (Bc8-g6); (Ra8-g7); (Bg7-h8); Qh8-h5; (b7xRc6); Rh1-c6; h2-h8=Q; Nb1-g8; (h5xNg4) etc.
However, there's another possible way to release the position with only one queen move: -3. g6xRh7! b2-b1=R -4. f5xPg6! b3-b2 -5. e4xRf5! Rb5-f5 -6. d2-d3! a4xBb3 and further: B~-b3; d3xR/Se4; (R/Sc1-e4); (c7-c1=R/S); (Bc8-g2); (Rg1-b5); (g3-g1=R); Bf1-~; c2xPd3; g2xNf3; (Ne5-f3); (Ka6-f4); Ke1-f6; Bc1-h6; (Nc6-e7); (Ra1-e8); b2-b8=B; (b3xRa2-a1=R); etc.

10098 - Alexander Kisljak
The intention was: Retroplay is: -1. Bc8xNb7 Nd8-b7 -2. Ka6-a5 Nb7-d8 -3. Bb8-a7 Na5-b7 -4. Ka7-a6 Nb3-a5 -5. Ka8-a7 Nc1-b3 -6. Ba7-b8 Ne2-c1 -7. Ba6-c8 Ng1-e2 -8. Kb8-a8 Ne2-g1 -9. Kc8-b8 Ng1-e2 -10. Kd8-c8 Ne2-g1 -11. Bc8-a6 Ng1-e2 -12. Bb8-a7 Ne2-g1 -13. Nh6-g8 Ng1-e2 -14. Nf5-h6 Ne2-g1 -15. Nh4-f5 Ng1-e2 -16. Ng6-h4 Ne2-g1 -17. Nh8-g6 Ng1-e2 -18. h7-h8=N Ne2-g1 -19. h6-h7 Ng1-e2 -20. h5-h6 Ne2-g1 -21. h4-h5 Ng1-e2 -22. h3-h4 g2-g1=N -23. h2-h3 h3xRg2 -24. Ra2-g2 h4-h3 -25. Ra8-a2 h5-h4 -26. a7-a8=R h6-h5 -27. a6-a7 h7-h6 -28. a5-a6 a6x[QRN]b5 and the position unlocks.
But unlocking goes much easier with -1. Bc8-b7 a6xPb5 -2. a4xXb5. Unpromote X on h1. Black captures are e7xf6, fxe, bxc and the h-pawn, white captures are dxc and exd.

10099 - Alexander Zolotarew
The intention was: Retroplay is: -1. .. f7xNg6 -2. Nh4-g6 Ne1-g2 -3. Ng2-h4 Nd3-e1 -4. b4-b5 Ne5-d3 -5. b3-b4 Ng4-e5 -6. Qf4-h6 Nh6-g4 -7. Qb8-f5 c6-c5 -8. b7-b8=Q c7-c6 -9. a6xQb7 Qd5-b7 -10. a5-a6 Qd1-d5 -11. a4-a5 d2-d1=Q -12. a3-a4 d3-d2 -13. d2xBe3 Bc5-e3 -14. a2-a3 Bf8-c5 -15. Kg5-h5 e7xBf6. Further retractions involve Ba3-f6; Kb4-g5; (Kh4-h3); Ne3-g2; (Ke8-h4); (Ra8-h2); (Nb8-g1); Ra1-h1; Ke1-b4; (Ng4-h6); (Rg6-h7); Rh1-h8; h7xQg8=B; h2-h7; (Qd8-g8); (g6xQf5); Qd1-f5; (h7xBg6); Bf1-g6; g2xBh2 etc.
But also possible is: -1. .. f7x[RN]g6; -2. b4-b5 b7-b6 -3. a3xRb4. The rook now unpromotes on d1 while the piece on g6 makes waiting moves. After retracting d3-d2, white uncaptures d2xBe3, this bishop goes to f8, and black uncaptures e7xRf6. This rook replaces bNg2 when Ne1-g2 is retracted. The whole position now unlocks easily.

10100 - Werner Keym
If there are promoted pieces on the board, they must be black. There are three cases:
0 promoted pieces on the board: Then black has no last move (e7-e5 and a2-a1=N both result in a promoted piece on the board). There is no mate in 2 with black to move.
1 promoted piece on the board: If the black rook is a promotion piece, then black has no last move, so no mate in 2 for white in all cases.
2 promoted pieces on the board: If the black rook and black knight are promotion pieces, or if the black bishop and black knight are promotion pieces, then mate in 2 with 1. OO Kg3 2. Rf3# (Retroplay: a2xXb3; Ra1-g6; a7-a1=B; b7xBa6-a2; b2xXa3; Ba1-h4; a2-a1=N, or a2xXb3; Ra1-g6; a7-a1=R; Ra1-g4; b7xBa6-a2; b2xXa3; a2-a1=N, in both cases black's last move was a2-a1=N). If the rook and bishop are promotion pieces, then black's last move must've been e7-e5. White doesn't have enough captures to crosscapture on the a- and b-line, so Ke1 had to move to let the rook pass, so no castling possible. Mate in 2 with 1. de6 d6 2. Bd6#.
3 promoted pieces on the board: The retroplay in this case looks like: a2xXb3; Ra1-g6; a7-a1=B; b7xBa6-a2; b2xXa3; c7-c3xQb2-b1=R; Rb1-g4 (over e1, so white may not castle in the diagram position); last move was a2-a1=N. So white may not castle, and white may not capture en passant either, so no mate in 2.

10101 - Olli Heimo
1. h4 h5 2. Rh3 Rh6 3. Ra3 Re6 4. Ra5 Re2 5. Ke2 e5 6. a4 Qf6 7. Ra3 Ba3 8. Kd3 Be7 9. Qf3 c5 10. Qc6 bc6 11. g4 d5 12. g5 Be6 13. g6 Kd7 14. gf7 g6 15. Kc3 e4 16. d4 ed3 17. Kb3 d4 18. c4 dc3 19. Ka3 c4 20. b4 cb3
But this is cooked: 1. h4 h5 2. Rh3 Rh6 3. Ra3 Rc6 4. g4 Rc2 5. e4 g6 6. e5 d5 7. Qf3 d4 8. Kd1 Bg4 9. e6 c5 10. ef7 Kd7 11. Kc2 e5 12. Qc6 bc6 13. Ra5 e4 14. a4 Qf6 15. Ra3 Be7 16. Rc3 dc3 17. d3 ed3 18. Kb3 Be6 19. Ka3 c4 20. b4 cb3

10102 - Peter van den Heuvel
1. b3 Nh6 2. Bb2 Rg8 3. Bf6 gf6 4. c4 Rg3 5. hg3 Ng4 6. Rh5 Bh6 7. Ra5 Be3 8. Ra3 a5 9. de3 a4 10. Kd2 Ra5 11. Kc3 Rd5 12. Kb4 c5 13. Kb5 Qc7 14. Nd2 Qe5 15. Kb6 Qb2 16. Kc7 Kf8 17. Kd8 Kg7 18. Ke8 Kg6 19. Kf8 Kf5 20. Kg7 Ne5 21. Kh6 Nc6 22. Kh5 Nd8 23. Kh4 h5 24. Kh3 h4 25. Kh2 h3 26. Kh1 h2 27. Ndf3 hg1=B 28. Qd2 Bh2 29. Ng1 Ke4 30. f3

10103 - Peter Schäfer
(a) Suppose white may castle. Then the last moves were -1. ... Ka6xPb6 -2. a5xb6ep b7-b5 -3. e2xf3. This means that the white rook on h3 is a promoted one! But there are 7 pawns on the board, and Ba8 is clearly a promoted one, so this isn't possible. So white may not castle.
(b) If white may castle, the last moves are the same as in (a). The white pawn structure and the promoted Ba8 require 8 captures, and the two black bishops are captured on their homesquares. These account for all missing black pieces. In particular, the black h-pawn has to get captured. It can't have promoted, since it can't pass f2, and other routes to promotion require too many captures. So it was captured either on the f- or the e-line. In the first case, the f-pawn was captured on the e-line. Both options require 3 captures, but since Rh1 was captured on h1 or g1, this is one capture too many. So white may not castle.
(c) Now there are no problems in getting 3 white pieces captured, so the f- and h-pawn can get captured on the e- and f-line, so white may castle.

10104 - Alexander Hölle
The rook on c2 is clearly promoted, so the c-pawn made 2 captures to promoted. Neither of those two captures was the rook from h8, or the bishop from c8, because b6 was played after the promoted white rook left the a8-corner. So the pawn on f7 got captured by the pawn on his route to a7. If white may still castle, this pawn needs 3 captures to promote on g1. But Ra1 was captured in his own corner, and Bf1 was captured on his homesquare, so black can only capture 2 pieces before promoting. Contradiction. So white may not castle.

10106 - Frank Genenncher
The stipulation implies that white shuffles only around with his bishop, since moving his king would give black the opportunity to capture the b-pawn, thus winning.
Let B(n) be the number of series of n moves, with the bishop ending on e1 or h2, and C(n) be the number of series of n moves, with the bishop ending on e2 or g1. Then A(n) = B(n) + C(n).
It can be seen easily that B(n+1) = C(n), C(n+1) = B(n) + C(n), B(0) = C(1) = 1, C(0) = B(1) = 0, or A(0) = 1, A(1) = 1, A(n+1) = A(n) + A(n-1). This is the Fibonacci series.

Die Schwalbe Heft 173, Oct. 1998

10156 - Peter Schäfer
Retract d5-d6. White captures were cxdxe, a4xb5, and bxc. Black captures were the rook in the southeast corner, c6xBb5 and gxh. The white b-pawn promoted on c8. So white may still castle. Mate in 5 with:
1. OOO! Kg2 2. d6 Bc6 3. Bc6 Nd5 4. Bd5 Kh3 5. Rh1#.

10157 - Günther Weeth & Werner Keym
In the first case, the white pawn on a6 uncaptures the whitesquared black bishop on a6. This bishop is retroscreen on d1, so black can retract Kb1-c1; Rc1-c2. White then retracts c2-c3. Black captured h7xBg6, white promoted on h8, and this promotion piece was captured by the black g-pawn on its way to b2. So the castling was indeed legal.
In the first case, the bishop has to be uncaptured on a4. This has to be done before the black rook enters the cage. The only way to release the position is to use a white bishop on d1 as screen for the white king. The black g-pawn captures 5 pieces to b2, then white plays c2-c3, black plays his rook to c2, the white bishop then sacrifices itself on g6, and the white h-pawn promotes. So the promotion piece is still on the board. A knight can't escape h8 (f7 and g6 are occupied) so Rh1 is a promoted piece. So white's castling was illegal. So white has to retract his castling, and play another king-move. The only other possible king-move is Ke1-e2, and this is mate too! So black is lost after all....

10158 - Alexander Zolotarew
White has waiting moves with his rook, black has waiting moves with his queen. Retroplay is: Bb1-a2; uncapture a knight on a3, a bishop on b4; bring the bishop to f8 and the knight to g8, while black makes waiting moves with his queen. Unpromote the bishop, and uncapture a rook on f7. Unpromote the knight, and uncapture a bishop on e5. Bring this bishop back to f8, and uncapture two knights on c5 and d6. Now release the queen via a3 and d6. Bring the white rook to a2, and uncapture a knight on c5 and a rook on b4 with the pawn on c6. Bring the uncaptured rook to a8, uncapture a7xQb6 and the whole position is freed.

10159 - Nikita Plaksin & Alexander Zolotarew
Intention: Retroplay is: -1. Rg2-h2 b4xRc3 -2. Ra3-c3 b5-b4 -3. Ra8-a3 b6-b5 -4. Rd8-a8 c4xBd3 -5. d7-d8=R b5xQc4 -6. e6xQd7 and further Qg8-c4; g5-g8=Q; e4-e6; Ba8-d3; a4-a8=B; (a7xQb6); h4xBg5; (Bf8-g5); Qg8-b6; g6-g8; (g7xNf6) etc.
But there is a different way to release the position: -1. Rg2-h2 b4xQc3 -2. Qd2-c3 c4xPd3! -3. Qh6-d2 b5xQc4 -4. Qc3-c4 b6-b5 -5. Qd2-c3 b5-b4 -6. Qg5-d6 d7-d6 -7. e3xQd4. Both queens unpromote on g8, and one pawn uncaptures a bishop with h4xBg5. One piece unpromotes on a8, after which the position completely unlocks.

10160 - Tom Volet
The last moves were: -1. Bc5xPa3 a4-a3 -2. Bd4-c5 a5-a4 -3. Be5-d4 a6-a5 -4. Bf4-e5 a7-a6 -5. Bh6-f4 Qh7-g8 -6. Bg5-h6 Qf5-h7 -7. Bh4-g5 Qh3-f5 and the black queen can go to e2, opening the position. The white bishop shields the king against checks by the black queen from three sides!

10161 - Andrej Frolkin & Anatoli Wasilenko
Retroplay was: -1. Qb1-a1 g3-g2 -2. Bd4-c5 g4-g3 -3. Bh8-d4 g5-g4 -4. h7-h8=B g6-g5 -5. h6-h7 h7xBg6 -6. Bf7-g6 e5-e4 -7. Ba2-f7 e6-e5 -8. Qa1-b1 e7-e6 -9. Rb1-c1 h3-h2 -10. Rb2-b1 h4-h3 -11. Qc1-a1 h5-h4 -12. Bg8-a2 d4-d3 -13. g7-g8=B d5-d4 -14. g6-g7 d6-d5 -15. g5-g6 g6xNh5 -16. Nf4-h5 d7-d6 -17. Nd3-f4 Ka2-a3 -18. Rb1-b2 Ka3-a2 -19. Nb2-d3 Ka2-a3 -20. Nc4-b2 f4-f3 -21. Na3-c4 f5-f4 -22. Nb2-a4 f6-f5 -23. Ra4-a5 f7-f6 -24. a5xNb6 and everything unlocks, just in time!

10162 - Mikhail Kozulya
1. f3 g5 2. f4 Bg7 3. Nf3 Bc3 4. dc3 a5 5. Be3 Ra6 6. Ba7 Rb6 7. Nfd2 Rb4 8. cb4 b6 9. Nc3 Ba6 10. Rc1 Bd3 11. ed3 Na6 12. Be2 Qa8 13. OO Qf3 14. Kh1 d5 15. Rg1 Qf1 16. Bh5 Kd7 17. Qg4 Kc6 18. Ne2 Kb5 19. c3 Ka4 20. b3

10163 - Andrej Frolkin & Oksana Kartseva
1. Nf3 b5 2. Nd4 b4 3. Nb5 b3 4. ab3 c5 5. Ra7 Qc7 6. N1a3 Qg3 7. hg3 c4 8. Rh6 c3 9. dc3 gh6 10. Qd6 Bg7 11. Qb8 Bd4 12. Rc7 Be3 13. fe3 d5 14. Kf2 d4 15. Kf3 d3 16. cd3 Kf8 17. Nc2 Ra3 18. ba3 Kg7 19. Bb2 Bh3 20. gh3 f5 21. Bg2 f4 22. Kg4 f3 23. ef3 Nf6
Unfortunately, this correction of 9723 is cooked: 1. e3 d5 2. Ke2 Qd6 3. Kf3 Qg3 4. hg3 d4 5. Rh6 Bh3 6. gh3 f5 7. Bg2 f4 8. Kg4 f3 9. Nf3 d3 10. cd3 b5 11. Nd4 b4 12. f3 b3 13. Qb3 a5 14. Qb8 Kf7 15. b3 a4 16. Bb2 a3 17. Na3 Ra5 18. Rc1 Rb5 19. Nab5 gh6 20. Rc7 Bg7 21. Nc2 Bc3 22. dc3 Kg7 23. a3 Nf6

10164 - Franz Aistleitner
Retract -1. Kg5-g6 f2-f1=Q -2. Kf6-g5 h2-h1=N -3. Kg7-g6 e2-e1=B -4. Kg8-g7 a2-a1=R -5. b2-b4, and forward b7-b8=Q#.

Die Schwalbe Heft 174, Dec. 1998

10218 - Alexander Zolotarew
Last moves were: -1. Ne3-g4 Rg4-g2 -2. Ng2-e3 Ra4xPg4 -3. a6-a7 Ra1-a4 -4. a5-a6 a2-a1=R -5. a4-a5 b3xNa2 -6. Nc3-a2 b4-b3 -7. Ne4-c3 b5-b4 -8. Nf6-e4 b6-b5 -9. Ng8-f6 g6-g5 -10. Kh8-h7 h7-h6 -11. g3-g4 Qh6-f8 -12. Kg7-h8 Qf4-g6 etc. So mate in 1 with 1. .. Qg8# and not with 1. Rh2?

10219 - I. Avroutine
(a) Retract c2-c4, and #1 with 1.d5#.
(b) Retract b4xPc5, and #1 with 1.dc6ep#.
(c) Retract b5xc6ep, and #1 with 1. d7#.
(d) Retract c6-c7, and #1 with 1. d8=N#

10220 - Andrej Frolkin
(a): Last moves were: -1. .. Bb1xNa2 -2. Nb4-a2 Ba2-b1 -3. N~ Bb1xRa2 -4. N~ c2xBb1=B -5. N~ c3-c2 -6. Bc2-b1 c4-c3 -7. Kb1-a1 c5-c4 -8. Ra1-a2 c6-c5 -9. a2xPb3 b4-b3 -10. Bb3-c2 Nc2xNe1 -11. B~ Nd4-c2 -12. Nc2xQe1 Qf1-e1 -13. Re1-d1 and everything unlocks.
(b): Now the retro-play becomes -1. .. e7xRd6 -2. R~ Bb1xBa2 -3. R~ c2xNb1 -4. R~ c3-c2 -5. R~ c4-c3 -6. Nc3-b1 c5-c4 -7. Kb1-a1 Nc2xNe1 etc.

10221 - Andrej Frolkin
Retroplay is: -1. .. Qe1-f1 -2. d3xNe4 Rg7-h7 -3. Bh7-g8 Nd6-e4 -4. Bg8-h7 Nf5-d6 -5. Bh7-g8 Nh4-f5 -6. Bg8-h7 Ng2-h4 -7. Bh7-g8 Qf1-e1 -8. Bg8-h7 Rh7-g7 -9. Re1xNe2 Nc1-e2 -10. Re2-e1 Ba1-b8 -11. Re1xBe2 Rh8-h7 -12. Bh7-g8. Black's promotions are: queen on b1, bishops on a1 and e1. White's promotion is on a8 or c8.

10222 - Gianni Donati
Author's intention was: 1. Na3 b5 2. Nc4 b4 3. a3 b3 4. Ra2 ba2 5. h4 a1=N 6. Rh3 Nb3 7. Re3 Nc1 8. Nh3 Nd3 9. ed3 f6 10. Ke2 Kf7 11. Kf3 Kg6 12. Re1 Kh5 13. Qe2 Kh4 14. Ra1 Kh5 15. Qd1 Kg6 16. Ke2 Kf7 17. Ke1 Ke8
But there are several cooks, for instance 9. .. Nh6 10. Qf3 Nf5 11. Kd1 Nh4 12. Re1 Nf5 13. Ke2 Nh6 14. Ra1 Ng8 15. Ke1 f6 16. Qd2 Kf7 17. Qd1 Ke8.

10223 - Unto Heinonen
1. h4 Nf6 2. h5 Rg8 3. h6 gh6 4. Nf3 Rg2 5. Rg1 Rh2 6. Rg8 Rh1 7. Rh8 Ng8 8. b3 Bg7 9. Ba3 Bb2 10. Nd4 e5 11. Bf8 c5 12. f4 Qa5 13. Kf2 Qa2 14. Bh3 Qb1 15. Qg1 Qd1 16. Ra7 Bc1 17. Rb7 Ra1 18. Ra7 Bb7 19. Qg5 Bg2 20. Ra8 d5 21. Bc8 Bf1 22. Qd8

10224 - Gianni Donati
The author's intention was: 1. d4 c6 2. d5 Qb6 3. d6 Qc7 4. dc7 d5 5. c4 Be6 6. c8=N Nd7 7. Nb6 Ndf6 8. Nd7 Nh5 9. Nf6 ef6 10. c5 Bd6 11. cd6 d4 12. d7 Ke7 13. d8=N Bd5 14. Ne6 d3 15. Ng5 fg5 16. b3 Kf6 17. b4 Ne7 18. b5 Rhc8 19. b6 Rc7 20. bc7 b5 21. c8=N b4 22. Nd6 b3 23. Nf5 b2 24. Nh6 gh6
But this is unfortunately cooked: 1. d4 Nf6 2. Bg5 Nh5 3. Bf6 ef6 4. d5 Be7 5. d6 OO 6. de7 d5 7. e8=Q d4 8. Qe3 Nc6 9. b4 Ne7 10. b5 c6 11. b6 Qc7 12. bc7 Be6 13. c8=Q Bd5 14. Qf5 Rfb8 15. Qh6 gh6 16. Qg5 fg5 17. c4 b5 18. c5 Rb6 19. cb6 d3 20. b7 b4 21. b8=B b3 22. Bf4 b2 23. Bd2 Kg7 24. Bc1 Kf6

Die Schwalbe Heft 175, Feb. 1999

10283 - Alexander Kisljak
Mate in 1 with 1. Qe3#. But the real question is of course: Is the position legal? And yes, it is: -1. .. a2-a1=N -2. Rb5-b3 a3-a2 -3. Rh5-b5 a4-a3 -4. Rh8-h5 a5-a4 -5. Rb8-h8 a6-a5 -6. b7-b8=R a7-a6 -7. a6xNb7 Nd6-b7 -8. a5-a6 Nf5-d6 -9. a4-a5 Ng3-f5 -10. a3-a4 Nh1-g3 -11. a2-a3 Ng3xRh1 -12. Rg1-h1 Nf5-g3 -13. Kh1-h2 g3-g2 -14. Kh2-h1 h4xNg3 and the whole position unlocks.

10284 - Werner Keym & Günther Weeth
Add the white king on d3, and retract -1. Kc3xPd3! c4xd3 -2. d2-d4 (Be1 is also promoted) f6-f5 (Bf8(x/-)g7 3.?? and forward Qf8#) 3. h2-h4! (Be1 entered the SE cage after f2xe3, but before g2-g3. The bishop on f1 couldn't leave f1 until after g2-g3, but g2-g3 was played after the knight on f1 got there. so Ba4 is promoted). The promotions occurred on a8 and b8, so the black rook moved. So black may not castle. Forward mate in 2 with 1. Qg7 (2. Qg8/Qf8#) e5/Bd5 (OOO?) 2. Qf8#

10285 - Alexander Zolotarew
Try: -1. Ra1xBa2? g4xNf3 -2. Nh4-f3 e7-e6 -3. Nf5-h4 g5-g4 -4. Ne3xPf5 f6-f5 -5. Nd5-e3 f7-f6 -6. Nb4-d5 Be6-a2 -7. Na2-b4 Bc8-e6 -8. f3-f4 d7xNc6 -9. Nb4-c6 g6-g5 -10. Kb5-a5 and retropat. (black may not retract b6xc5, because Bb8 would be trapped).
So the retroplay is: -1. Ra1xNe2 g4xBf3 -2. Bd5-f3 Nb4-a2 -3. Ba2-d5 Nd5-b4 -4. f2-f4! Nb4-d5 -5. Bc4-a6 Na6-b4 -6. Bd5-c4 e7-e6 -7. Bf3-d5 g5-g4 -8. Bd5xPf3 f4-f3 -9. Bg8-d5 f5-f4 -10. g7-g8=B f6-f5 -11. g6-g7 f7-f6 -12. h5xBg6 Bf5-g6 -13. h4-h5 Bc8-f5 -14. h3-h4 d7xNc6 -15. Kb5-a5 Nb4-a6 -16. Kc4-b5 etc. So black had the possibility of an en-passant capture!

10286 - Andrej Frolkin
Retroplay is: -1. .. g4xBh3 -2. h5-h6 g5-g4 -3. Be6-h3 g6-g5 -4. Ba2-e6 d5-d4 -5. Qa1-b1 d6-d5 -6. Qf6-a1 d7-d6 -7. Be6-a2 Ka2-a3 -8. Bh3-e6 Ka3-a2 -9. Qg5-f6 Ka2-a3 -10. Qg1-g5 Ka3-a2 -11. Qe1-g1 Ka2-a3 -12. Bg2-h3 Ka3-a2 -13. Bf1-g2 Ka2-a3 -14. g2xPf3 Ka3-a2 -15. Kb1-c1 f4-f3 -16. Ka1-b1 f5-f4 -17. Qb1-e1 Rd1-d2 -18. h4-h5 Re1-d1 -19. Qd1-b1 f6-f5 -20. Kb1-a1 f7-f6 -21. Kc1-b1 Ka2-a3 -22. Kd2-c1 and everything unlocks.

10287 - Mikhail Kozulya
1. a4 d5 2. a5 Qd7 3. a6 Qh3 4. axb7 Bf5 5. Ra6 Nd7 6. Rf6 exf6 7. g4 Bd6 8. Bg2 Bg3 9. fxg3 a5 10. Kf2 a4 11. Ke3 a3 12. Kd4 a2 13. Kc3 a1=R 14. Kb3 R1a4 15. Na3 Ne7 16. Ka2 Nc6 17. Kb1 Ke7 18. Nc4 Rag8 19. Ne5 Nf8 20. Ng6 hxg6 21. b8=N Rh4 22. gxh4 Qb3 23. Nf3 Bxc2

10288 - Unto Heinonen
1. g4 b5 2. g5 Bb7 3. g6 Be4 4. gh7 g5 5. hg8=B Rh6 6. h4 Ra6 7. h5 Ra2 8. h6 a5 9. h7 a4 10. h8=N a3 11. Ng6 fg6 12. Bd5 e6 13. f4 Bc5 14. f5 Ke7 15. f6 Kd6 16. f7 Qf6 17. f8=R Qf3 18. Rf4 Bf5 19. Rc4 ed5 20. e4 dc4 21. e5 Kd5 22. e6 Be3 23. e7 c5 24. e8=Q Nc6 25. Qh8 Rg8 26. Qc3 Rg7 27. Qb4 cb4

Die Schwalbe Heft 176, Apr. 1999

10342 - Valeri Liskovets
Suppose white may castle. Ba2 is promoted. If it was promoted on c8 or e8, black may not castle. If it promoted on g8 via h7, the white pawns captured 6 pieces (cxdxe, d2xe3xf4xg5, h7xg8). Among those captures is the black a-pawn. But that pawn could never have promoted (since white may castle, Ra1 was always there, and the black pawn could never pass the white a-pawn, since that pawn hasn't made a capture) So white can't have promoted on g8, so black may not castle.
1. OOO Rh1 (OO? illegal) 2. Rh1 and 3. Rh8#; 1. .. bc6 2. Ra5 and Ra8#
But black can try to prove he has the move by castling. White can't castle then. 0. .. OO 1. Ne7 Kh8/Kh7 2. Kf2! (OOO?) and 3. Rh1#

10343 - Andrei Kornilov & Andrei Frolkin
Last moves without repetition were: -1. .. Re7-e8 -2. Kf8xRg8! Rg7-g8 -3. Kg8-f8 Rh7xQg7! -4. Qf8xPg7! Re8-e7 -5. a4-a5 Be7-d8 -6. a3-a4 Rd8-e8 -7. Qe8-f8 Bf8-e7 -8. Qe7-e8 Re8-d8 -9. Qd8-e7. The extra rook must be unpromoted on a1, and the pawn then uncaptures on a2 and on the b-line. Surprising that only by throwing even more pieces into the cage, the position can be unlocked!

10344 - Michel Caillaud
The retroplay starts with -1. .. Qa2xNb3 -2. g4-g5 Now there are some tries:
- -2. .. f6-f5 -3. g3-g4 Nf5-h4 -4. g2-g3 Nd6-f5?? but the white king is in check.
- -2. .. Ng2-h4 -3. g3-g4 Nf4-g2 -4. g2-g3 Ne2-f4 -5. g5-g6 Nc1-e2 -6. Nc5-b3 Nb3-c1 -7. Ne6-c5 f6-f5 -8. Nd8-e6 and black can't retract any move.
Solution: -2. .. f6-f5 -3. g3-g4 Nf5-h4 -4. g2-g3 Ng3-f5 -5. g5-g6 Nh1-g3 -6. g4-g3 h2-h1=N -7. g3-g4 h3-h2 -8. h2xNg3! Ne2-g3 -9. f2-f3 Nc1-e2 -10. Nc5-b3 Nb3-c1 -11. Ne6-c5 h4-h3 -12. Nd8-e6 h5-h4 -13. d7-d8=N h6-h5 -14. d6-d7 h7-h6 -15. e5xNd6 Nc8-d6 -16. e4-e5 Na7-c8 -17. Nd6-b5 Nb5-a7 and the whole position unlocks.

10345 - Werner Keym
(a) Last move was Ke4-d5. #1 with 1. c4#
(b) Last move was Ke4-d5. #2 with 1. e4 fe4 2. Ne3#
(c) Last moves were Kc6-d5 d7-d8=S. #3 with 1. Bb1!
(d) Last moves was Ke4-d5. #4 with 1. bc7 Bc2 2.Kc3 Bd3/Ba4 3. c8=S (4. Se7#)
(e) Black has no last move. #5 with 0. .. Bc2 1. Kc3 Ba4 2. ba7 c5 3. a8=N cb4 4. Kb4 (5. Nb6/Nc7#) 1. .. ab6 2. a7 c5 3. a8=N cb4 4. Kb4 (5. Nb6/Nc7#)

10346 - Andrei Kornilov & Andrei Frolkin
Retroplay is: Nh5-f4; (c7xNd6); (Ng7-b1); Nd6-g7; Ne8xRg7; (Rh7-g7); Qh8-g8; Kg8-f8; Na8-e8; (b2-b1=N); a7-a8=N; (b5-b2); a4-a7; (a6xRb5); Rb8-b5; (a7-a6); Rf8-b8; (Re8-e7); a3-a4; (Ra8-e8); Rb8-f8 and the whole position unlocks. Other promotions were b7-b8=R and b3xa2-a1=R.

10347 - Radovan Tomasevic
Intentions:
(a): 1. Nc3 d5 2. Nd5 Nc6 3. Ne7 Qd3 4. Ng8 Be7 5. Ne7 Qc2 6. Nc8 Qb1 7. Na7 Ke7 8. Nb5 Ra3 9. Na3 Ra8 10. Nb1 Nb8
(b): 1. c4 b5 2. cb5 a6 3. ba6 Bb7 4. ab7 Qc8 5. bc8=N d5 6. Ne7 d4 7. Ng8 Be7 8. Ne7 Rg8 9. Ng8 d3 10. Ne7 Ke7
But in b) there is a dual at the second move: a5 instead of a6, and a change at the end too: 7. .. d3 8. Nh6 Rg8 9. Rg8 Be7 10. Ne7 Ke7. And in a) there are other possibilities: 1. c4 d5 2. cd5 e5 3. ed6 Qd7 4. ed7 Ke7 5. dc8=Q Kf7 6. Qf8 Ne7 7. Qh8 Nec6 8. Qb8 Ke6 9. Qa7 Ke7 10. Qb8 Nb8.

10348 - Thierry le Gleuher
1. h4 g5 2. hg5 Nf6 3. gf6 d6 4. fe7 Be6 5. ef8=B Ba2 6. Bd6 Bb1 7. Bc7 Bc2 8. Bb8 Bd1 9. Ba7 Ra7 10. Ra7 Be2 11. Rb7 Bf1 12. Rf7 Bg2 13. Rfh7 Bh1 14. Rh1

10349 - Paul Raican
Retract -1. Nc4xNd2 [+wNd3] Be7xNc5 [+bBa2] and forward 1. Bf6 [+bBd3]# or 1. Nc4 [+bBb6] Nd7 [+wNd6]#.

10351 - Vladimir Chutornoi
The castlings result in 16 different combinations. There are 12 different possibilities for an ep-capture (5 for white, 5 for black, an no ep possibility at all, both for white and for black to move). So in total there are 12*16 = 192 different positions.

But this has been published by O. Riihimaa in British Chess Magazine, 1954 too.

Die Schwalbe Heft 177, June 1999

10408 - Alexander Kislyak
Last moves were: -1. Be2xPh5 h6-h5 -2. Bf1-e2 h7-h6 -3. e2xNd3 Nc5-d3 -4. c3-c4 Na6-c5 -5. c2-c3 Nc5xRa6 -6. Ra5-a6 Ne4-c5 -7. Ra6-a5 Ng3-e4 -8. Ra5-a6 Nh1-g3 -9. Ra6-a5 Ng3xRh1 -10. Ra5-a6 Ne4-g3 -11. Ra6-a5 Nc5-e4 -12. Ra5-a6 Na6-c5 -13. Ka8-a7 Nb8-a6 -14. Ra7-a5 Ra6-b6 -15. Rg1-h1 Kb6-c7 -16. Nc5-d7 and everything unlocks.

10409 - Stanislav Vokal
Suppose white can castle. Then Bh6 promoted on g1, after capturing 3 pieces. Together with the other 2 captures on the g-line, these are all captures. The a- and h-pawn were among those captures too. The a-pawn promoted, after 1 capture, on b8. Together with Bf8 and the 2 captures on the e-line, these are all captures white did, so the h-pawn must've promoted without capture, so black's kingside castling is illegal, if white may castle.
1. OOO? ab4!; 1. Rd1? OO!; 1. Na6! (2. Nc7#) Ra6 2. OOO (3. Rd8#) Ra8 3. Rfd2 ~ (OO?) 4. Rd8 Rd8 5. Rd8#.

10410 - Alexander M. Yarosh
Retract Nc8xBa7. How did the rook go to b7? If it came via g8 and b8, black can't castle kingside anymore. But could it have come from the f-pawn, making 4 captures? Yes, but at the moment white plays b7-b8=R, black has no retromove, apart from his king or rook on h8. So black may not castle anymore. Mate in 3 with 1. Qc7 ~ (OO?) 2. Qd7 Kf8 3. Qd8#

10411 - Andrei Frolkin
Add a black bishop on a8. The position unlocks by the following maneuvers: Qb1-b2; (g2-g1=B); Qg8-d7; (d7xNc6); Nc5-c6; (Bb3-a4); Na4-c5; Bh8-h2; h4-h8=B; (h5xBg4); Ba2-g4; Qa1-b1; Kb1-c1; g7-g8=Q; (h2-h1=B); g3-g7; (g4xNh3); Nc1-h3; (Rd1-d2); h3-h4; (Rh1-d1); Nb3-c1; (Rh2-h1); Kc1-b1; and everything unlocks. White captures are c2xb3 and a5xb6 (both on black squares, that's why a knight on a8 won't work), black capture is a5xBb4.

10412 - Tom Volet
Last moves were: -1. .. Bh2-g1 -2. Kc2-c1 Bg3-h2 -3. Kc1-c2 Bh4-g3 -4. Kc2-c1 Bg5xPh4 -5. h3-h4 Bc1-g5 -6. h2-h3 Ba3-c1 -7. Kc1-c2 Bb4-a3 -8. Kc2-c1 Ba5-b4 -9. Kc1-c2 Bb4aPa5 -10. a4-a5 Kb3-c4 -11. Ra5-b5 Ka2-b3 -12. Kc2-c1 and further Bf8-b4; Bf1-a6; Rb5-c5; c5-c3; Rh8-b5; g7xQf6; Ke1-c2; Rh1-b6; g2xf3; c4xd5; b6xc7; etc.

10413 - Michel Caillaud
1. b3 a5 2. Bb2 a4 3. Bg7 a3 4. Bd4 Bg7 5. Ba7 b6 6. Nc3 Bb7 7. Rb1 Bg2 8. Rb2 ab2 9. Bg2 b1=B 10. Ba8 Ba2 11. Nb1 Ba1

10414 - Mark Kirtley
The intention was: 1. e3 a5 2. Qh5 Ra6 3. Qa5 Rh6 4. Qa8 Nc6 5. Qc8 Nd4 6. Qa8 Nf5 7. Qa6 Qa8 8. Qg6 Kd8 9. Qh7 Kc8 10. Qg8 Kb8 11. Qf8 Ka7 12. Qa8 Kb6 13. Qa5 Kc6 14. Ba6 Kd6 15. d3 Ke6 16. Bd2 Kf6 17. Bc3 Kg6 18. Nd2 Kh7 19. OOO Kg8 20. Kb1 Kf8 21. Ka1 Ke8
But this has been cooked by Satoshi Hashimoto: 1. Nc3 h5 2. Nb5 Rh6 3. Na7 Rg6 4. Nc8 Raa6 5. e3 Rae6 6. Nb6 Na6 7. Qh5 Qa8 8. Qh8 Nf6 9. Ba6 Kd8 10. Qf8 Ne8 11. Qh8 Qa7 12. Nc4 Qb6 13. Qh5 Qa5 14. Qa5 Nd6 15. d3 Ke8 16. Bd2 Rh6 17. Bc3 Rh8 18. OOO Reh6 19. Kb1 Nf5 20. Ka1 Rh5 21. Nd2 R5h6

10415 - Andrei Frolkin
1. Nf3 a5 2. Nd4 a4 3. Nb3 ab3 4. a4 Ra5 5. Ra2 ba2 6. h4 ab1=B 7. Rh3 Ba2 8. Rg3 Be6 9. Rg7 Bh3 10. Rg6 Rf5 11. Rf6 ef6 12. a5 Bc5 13. a6 Ne7 14. a7 Rg8 15. ab8=Q Ba7 16. Qa8 c5 17. Qb8 Qc7 18. Qa8 Qh2 19. Qb8 d6 20. Qa8 Kd7 21. Qb8 Kc6 22. Qa8 Bd7 23. Qb8 Nc8
But this is cooked: 1. h4 a5 2. Rh3 a4 3. Rb3 ab3 4. a4 Ra5 5. Ra2 ba2 6. Nf3 ab1=B 7. Ng5 Ba2 8. Ne4 Be6 9. Ng3 Bh3 10. Ne4 Rf5 11. Nf6 ef6 12. a5 Bc5 13. a6 Ne7 14. a7 Rg8 15. a8=Q Ba7 16. Qb8 c5 17. Qg3 Qc7 18. Qg7 Qh2 19. Qg6 d6 20. Qg7 Kd7 21. Qf8 Kc6 22. Qd8 Bd7 23. Qb8 Nc8

10416 - Unto Heinonen
1. Nf3 e5 2. Nd4 ed4 3. Nc3 dc3 4. dc3 Nf6 5. Bh6 gh6 6. a4 h5 7. a5 Bh6 8. a6 Bc1 9. ab7 a5 10. g3 a4 11. g4 Ra5 12. g5 Rf5 13. g6 d5 14. g7 Kd7 15. g8=N Ne8 16. Nf6 Kc6 17. Ne4 Kb5 18. Nd2 c5 19. Nb1 Nc6 20. b8=N Ng7 21. Nd7 Re8 22. Ne5 Bd7 23. Nf3 Re3 24. Ng1 Be8 25. f3 Qh4#

9438v - Alexander Zolotarev
Retroplay is: -1. .. g4-g3 -2. Ng3-h1 Ra1-a2 -3. Ne2-g3 Ra2-a1 -4. Nf4-e2 Ra1-a2 -5. Ng6-f4 Ra2-a1 -6. Nh8-g6 Ra1-a2 -7. h7-h8=N Ra2-a1 -8. h6-h7 Ra1-a2 -9. h5-h6 Ra2-a1 -10. h4-h5 h5xRg4 -11. Rg3-g4 Ra1-a2 -12. Rf3-g3 Ra2-a1 -13. Rf5-f3 Ra1-a2 -14. Rb5-f5 Ra2-a1 -15. Rb3-b5 Ra1-a2 -16. Ba2-b1 Rg1-a1 -17. Bb1-a2 Rg3-g1 -18. Rb5-b3 Rf3-g3 -19. Re5-b5 Rf5-f3 -20. Re1-e5 Rb5-f5 -21. Rc1-e1 Rb3-b5 -22. Ba2-b1 d4-d3 -23. Ra1-c1 Rb5-b3 -24. Bb1-a2 Re5-b5 -25. g4-g5 Re7-e5 -26. d6-d7 Rd7-e7 -27. e5xBd6 Bf8-d6 -28. h2-h4 e7xBf6 -29. Bg5-f6 d5-d4 -30. Bc1-g5 Kd4-c4 -31. d2xNc3 and everything unlocks.

10098v - Alexander Kislyak
Try: -1. Bc8-b7 a6xPb5 -2. b4-b5 b5xXc4 -3. Bb8-a7 a7-a6 -4. Ka6-a5 and black has no further retractions. White can't uncapture anything with his king either, since the white pawns captured all missing black pieces.
Retroplay is: -1. Bc8xNb7 Nd8-b7 -2. Ka6-a5 Nb7-d8 -3. Bb8-a7 Na5-b7 -4. Ka7-a6 Nb3-a5 -5. Ka8-a7 Nc1-b3 -6. Ba7-b8 Ne2-c1 -7. Ba6-c8 Ng1-e2 -8. Kb8-a8 Ne2-g1 -9. Kc8-b8 Ng1-e2 -10. Kd8-c8 Ne2-g1 -11. Bc8-a6 Ng1-e2 -12. Bb8-a7 Ne2-g1 -13. Nc3-a4 Ng1-e2 -14. Ne2-c3 Ne2-g1 -15. Nf4-e2 Ng1-e2 -16. Ng6-f4 Ne2-g1 -17. Nh8-g6 Ng1-e2 -18. h7-h8=N Ne2-g1 -19. h6-h7 Ng1-e2 -20. h5-h6 Ne2-g1 -21. h4-h5 Ng1-e2 -22. h3-h4 g2-g1=N -23. h2-h3 h3xRg2 -24. Ra2-g2 h4-h3 -25. Ra8-a2 h5-h4 -26. a7-a8=R h6-h5 -27. a6-a7 h7-h6 -28. a5-a6 a6x[QRN]b5 and the position unlocks.

10099v - Alexander Zolotarev
Last moves were: -1. .. Rh7xQh6 -2. Nh4-g6 Ne1-g2 -3. Ng2-h4 Nd3-e1 -4. c3-c4 Ne5-d3 -5. c2-c3 Ng4-e5 -6. Qg5-h6 Nh6-g4 -7. Qb5-g5 f5-f4 -8. Qb8-b5 b5-b4 -9. b7-b8=Q b6-b5 -10. a6xQb7 Qd5-b7 -11. a5-a6 Qd1-d5 -12. a4-a5 d2-d1=Q -13. a3-a4 d3-d2 -14. d2xBe3 Bc5-e3 -15. a2-a3 Bf8-c5 -16. Kg5-h5 e7xBf6 and everything unlocks.
Unfortunately there's a dual: -3. ... Nc2-e1 -4. c3-c4 Nd4-c2 -5. c2-c3 Nf5-d4 -6. Qg5-h6 Nh6-f5 and further as in the intention.

Die Schwalbe, Heft 178, Aug. 1999

10483 - Gerald Ettl
The position is released when white can retract his e-pawn, resulting from unpromoting the second queen, to e2. This requires the following retro-play: -1. .. d7xNe6 -2. Nf4-e6 Ne3-g2 -3. Ng2-f4 Nc2-e3 -4. b5-b6 Na1-c2 -5. b4-b5 a2-a1=N -6. b3-b4 a3-a2 -7. a2xNb3 Nc1-b3 -8. d4-d5 c2-c1=N -9. d3-d4 c3-c2 -10. c2xNd3 Nf4-e3 -11. Qg6-h5 Nh5-f4 -12. Qe6-g6 a4-a3 -13. Qe8-e6 a5-a4 -14. e7-e8=Q c4-c3 -15. e6-e7 c5-c4 -16. e5-e6 c6-c5 -17. e4-e5 c7-c6 -18. e2-e4 e3xNf2 and everything unlocks. This position, and the moves beginning from Nh5-f4, are unique.

10484 - Alexander Zolotarev
Last moves were: -1. .. Rh8xQh7 -2. b5-b6 Nd5-b4 -3. b4-b5 Nf6-d5 -4. Qd3-h7 Nh7-f6 -5. Qd8xPd3 d4-d3 -6. Qa8-d8 d5-d4 -7. a7-a8=Q d6-d5 -8. a6-a7 c5-c4 -9. a5-a6 c6-c5 -10. a4-a5 c7-c6 -11. a3-a4 a4xNb3 -12. Nd4-b3 a5-a4 -13. Ne6-d4 a6-a5 -14. Nf4-e6 Ng1-h3 -15. Nh3-f4 Kh1-h2 -16. Nb3-a1 Kh2-h1 -17. Nd4-b3 Kh1-h2 -18. Nf3-d4 a7-a6 -19. Nh2-f3 Nf3-g1 -20. Ng1-h3 N~ -21. Rh3-h4 N~ -22. Rh4-h5 N~ -23. Kh5-h6 N~ -24. Bh6-g7 Bg7-f8 -25. a2-a3 Nf8-h7 -26. Bh7-g8 and the position unlocks.

10485 - Andrei Frolkin
Last moves were: -1. Nf2-d1 d6-d5 -2. Be8-f7 c7xRd6 -3. Qf7-g7 and while the bishop waits on h6 and g7, unpromotes white the knight from f2 and the bishop from e1 on b8, after bringing the uncaptured rook to g3. When the pawns are both brought back to b3 and b4, and the black bishop is on h6, retract further -23. Bg7-f8 a4-a3 -24. Qf8-f7 a5-a4 -25. Bf7-e8 a6-a5 -26. Qa8-f8 a7-a6 -27. Bf8-g7 (-27. a3xPb4 b5-b4 -28. Be8-f7 b6-b5 -29. Nf8-h7?? -30. Kf7-g6) Bg7-h6 -28. a3xPb4 Bh6-g7 -29. Be8-f7 Bg7-h6 -30. a2-a3 Bh6-g7 -31. Bg7-f8 b5-b4 -32. Nf8-h7 b6-b5 -33. Kf7-g6 Bh7-f8 -34. Rg6-g5 and the whole position unlocks.

10486 - Mark Kirtley & Michel Caillaud
I: 1. d3 e6 2. d4 Ke7 3. d5 Kd6 4. de6 Kc5 5. ed7 Qe8 6. d8=N Be7 7. Nf7 Bg5 8. Nh8 Nf6 9. Ng6 Qg6
II: 1. d4 e5 2. de5 Be7 3. Qd7 Kd7 4. e6 Kc6 5. ef7 Bg5 6. f8=Q Nf6 7. Qh8 Qd3 8. Qd8 Qg6 9. Qd1 Kc5

10487 - Thierry le Gleuher
1. Nc3 d5 2. Nd5 a5 3. c3 a4 4. Qa4 b5 5. Kd1 h5 6. Kc2 h4 7. Qh4 c5 8. e4 c4 9. Bc4 f5 10. d3 f4 11. Bf4 e5 12. Be5 b4 13. Bg7 b3 14. Bb3

10488 - Satoshi Hashimoto
1. d4 Na6 2. d5 Rb8 3. d6 cd6 4. c4 Nc7 5. c5 Na8 6. c6 dc6 7. a4 Kd7 8. a5 Kc7 9. a6 ba6 10. g4 Kb7 11. g5 Qc7 12. g6 hg6 13. b4 Rh4 14. b5 Rb4 15. b6 ab6 16. h4 Ka7 17. h5 Bb7 18. h6 gh6 19. f4 Bg7 20. f5 Ba1 21. f6 ef6 22. e4 Ne7 23. e5 Nc8 24. e6 fe6

Die Schwalbe Heft 179, Oct. 1999

10538 - Günther Weeth & Werner Keym
The only possibility for a mate in 1 is to retract Rh1(x/-)h4, and forward OO#. If white doesn't uncapture anything, or he uncaptures a knight, black has no last move. If he uncaptures a pawn, it results in an illegal pawn structure. If he uncaptures a bishop, the c-pawn needs to many captures to reach g1, if he uncaptures a rook, the c-pawn captured twice, but the rook had to escape the promotion row via e1, so no castling allowed. Only uncapturing a queen gives black a retromove. So: Retract Rh1xQh4, and mate in 1 with 1. OO#.

10539 - Alexander Zolotarev
Intended solution is: Retroplay is: -1. .. Nc2-e1 -2. f4xNe5 Nd7-e5 -3. c6-c7 Nf6-d7 -4. Ng5-h7 Nh7-f6 -5. Ne4-g5 Na1-c2 -6. Ng5-e4 a2-a1=N -7. Ne4-g5 a3-a2 -8. Ng5-e4 a4-a3 -9. Ne4-g5 a5-a4 -10. Ng5-e4 Bf5-h3 -11. Nh3-g5 Be4-f5 -12. c5-c6 Bc6-e4 -13. c4-c5 Ba4-c6 -14. c3-c4 Bd7xPa4 -15. a3-a4 Bc8-d7 -16. a2-a3 d7xBe6 -17. Bb3-e6 g4-g3 -18. Bd1-b3 g5-g4 -19. Be2-d1 g6-g5 -20. Bf1-e2 Kg4-h4 -21. e2xQf3 and further Qf3-d1; c2-c3; d2-d1=Q; c3xRd2.
But it's also possible to release the position with the following retroplay: -1. f4xRe5 Rg5-e5 -2. Ne8-g7 Rg7-g5, then play the knight on e1 back to f6, and then -i. Ng5-h7 Nh7-f6 -(i+1). N~ Bf5-h3 -(i+2). Nh3-g5, bring the freed bishop to c8, uncapture a bishop on e6, bring this bishop to f1, and uncapture a piece on f3. Promotions occurred on d1 and a1. Even c6-c7 could be the last move.

10540 - Alexander Kislyak
Last moves were (~ means 'waiting move'): -1. .. a7-a5! -2. a6xRb7 Rb5-b7 -3. a5-a6 Rd5-b5 -4. a4-a5 Rd3-d5 -5. a3-a4 Rd5xRd3! -6. Rd4xNd3! Ne1-d3 -7. Re4-d4 ~ -8. Kf1-f2 ~ -9. Bf2-g1 ~ -10. Kg1xNf1! Nd3-e1 -11. ~ Qe1-d1 -12. ~ Kd1-d2 -13. Nb3-a1 Rd2-c2 -14. Na5-b3 Kc2-d1 -15. Nb7-a5 Kb3-c2 etc. The promotions occurred on h1 and d1.

10541 - Andrei Frolkin
1. f4 a5 2. f5 a4 3. f6 a3 4. fg7 f5 5. h4 Nf6 6. g8=B Bg7 7. Bb3 Rg8 8. h5 Bh8 9. h6 Rg7 10. hg7 h5 11. g8=B h4 12. Bgc4 d5 13. Nh3 dc4 14. Nf2 cb3 15. Rh3 ba2 16. Rb3 Qd3 17. ed3 h3 18. Qg4 h2 19. Qb4 f4 20. Be2 Bf5 21. Bf3 h1=Q 22. Ke2 Qc1 23. Bd5 Qg1 24. Nd1 Qd4 25. Bg8 Qd8 26. Nbc3 f3
But this is unfortunately cooked: 1. h4 a5 2. h5 a4 3. h6 a3 4. hg7 h5 5. Rh4 Rh6 6. Rb4 Rd6 7. Nc3 Rd3 8. ed3 f5 9. Be2 Nf6 10. g8=Q h4 11. Qb3 h3 12. Bf3 h2 13. Bd5 h1=Q 14. Bc4 d5 15. Qg4 dc4 16. Ke2 cb3 17. Nh3 ba2 18. Rb3 Qg1 19. Qb4 Qh1 20. f4 Bg7 21. Nf2 Qh4 22. Nfd1 Qg5 23. fg5 Bh8 24. g6 f4 25. g7 Bf5 26. g8=B f3

10542 - Unto Heinonen
1. c4 a5 2. c5 a4 3. c6 a3 4. cb7 ab2 5. a4 h5 6. a5 h4 7. a6 h3 8. a7 hg2 9. ab8=Q Rh3 10. Ra7 Ra3 11. h4 c5 12. h5 c4 13. h6 c3 14. h7 c2 15. Nc3 b1=Q 16. Bb2 c1=Q 17. Rh6 Qf5 18. Rc6 e6 19. e4 Bd6 20. Nge2 g1=Q 21. f3 Qgc5 22. d4 Qcg5 23. h8=Q Bf4 24. Qh5 Ne7 25. Qe5 Ng6 26. b8=Q Ba6 27. Qdb3 Bd3 28. Q8b5 Nh4 29. Rb7 R8a4 30. Qbd5 Qa5

10543 - Alexander Yarosh
Solution:

[8/1ppp2p1/p7/8/4P1P1/PP1PrbRB/K1PPkbQn/2qRBNrN]

Without a6 or c7, the last moves were -1. Ka1xBa2 Bb1-a2 -2. a2-a3 Qa3-b1, and later a3/c3xb2-b1=B. Without b7, d7, or g7, the last moves were -1. b2-b3 a7-a6 -2. Kb3-a2.

Die Schwalbe Heft 180, Dec. 1999

10598 - Theodor Steudel
(a): Retract Kg2xRh1, and ser-h#8 with 1. Kf3 2. Ke4 3. Kd3 4. Kc4 5. Kb3 6. Ka2 7. Kb1 8. Kc1 OO#
(b): Retract Kh2xBh1, and ser-h=8 with 1. Kh3 2. Kg4 3. Kf5 4. Ke5 5. Kd6 6. Kc5 7. Kc4 8. Kd3 Bd5=

10599 - Alexander Zolotarev
The intended retroplay is: Last moves are: -1. .. Ba5xNb4 -2. Na6-b4 Rb4-b6 -3. Nb8-a6 e6xNd5 -4. Nf6-d5 d2xBc1=Q -5. Ng8-f6 e3xBd2 -6. g7-g8=N h7-h6 -7. h6xNg7. Further retractions are: Bf8-d2; f3-f8=B; Bf8-c1; f5-f8=B; g4xBf5; (Ba6-f5); (f4xRe3); b5-b8=N; Rb8-e3; (Bc8-a6); b6-b8=R; (b7xPc6); f2-f3; (Qd8-e7); (e7-e6); Kd5-c5; (Rh8-d6); Ke1-d5; (Rf5-b4); Rh1-a4; (Bf8-a5); (Ra8-f5); Ng1-a2; (Qa1-a3); (a7-a1=Q); (Kb4-b3); Bc1-c3; Ra1-c2; Bf1-d1; Qd1-d3; Ke8-b4; a4xNb5 etc.
However, there is a different way to release the position with only 3 queen moves: -3. .. e6xRd5 -4. Rf5-d5 d2xBc1=Q-5. Rf8-f5 e3xNd2! -6. Nf1-d2 e4-e3 -7. Rc8-f8 Qd8-e7 -8. b7-b8=N e7-e6 -9. b6-b7 h7-h6 -10. Bh6-c1 and further: f5-f8=B-h6; g4xBf5; Rb8-c8; (Bc8-f5); b5-b6; b6-b8=R; b7xc6 etc.

10600 - Alexander Zolotarev
The position unlocks after retracting -1. Nc3-d1 d7xNe6 -2. Nd4-e6 a6-a5 -3. Nf3-d4 Ng4-h2 -4. Nh2-f3 Ne5-g4 -5. Nd5-c3 Nc4-e5 -6. Ne7-d5 Na3-c4 -7. Nc8-e7 Nb1-a3 -8. Ne7-c8 Nc3-b1 -9. Nc8-e7 Nd1-c3 -10. c7-c8=N d2-d1=N -11. c6-c7 d3-d2 -12. c5-c6 d4-d3 -13. c4-c5 c5xNd4 -14. Nc2-d4 a7-a6 -15. Ne1-c2 Nh4-g2 -16. Ng2-e1 Nf5-h4 -17. f3-f4 Nh4-f5 -18. g4-g5 Nf5-h4 -19. Bf4-h6 Nh6-f5 -20. Be5-f4 c6-c5 -21. Bc3-e5 c7-c6 -22. Bd2-c3 b4-b3 -23. Bc1-d2 b5-b4 -24. d2xBe3 Bc5-e3 -25. c3-c4 Bf8-c5 -26. c2-c3 e7xRf6 and the position is unlocked. So black has the move, and wins with 1. .. Rh6 2. gh6 Qh6#.

10601 - Andrei Frolkin
Solutions are:

[8/2BB1B1B/1BrQqNn1/1rpKbkr1/nRP5/r2prr2/1RrPbN1r/8]   [8/2pp1p1p/1pRqQnN1/1RPkBKR1/Nrb5/R2PRR2/1rRbBn1R/8]

For (a) the promotions are b2-b8=B, a7-a1=R, b7xPa6-a1=R, e2-e8=B, f7-f1=R, e7xPf6-f1=R, g2-g8=B, h7-h1=R, g7xPh6-h1=R. In (b), the white rooks promoted on a8, e8 and g8, on each square twice.

10602 - Mario Velucchi
1. e4 f5 2. ef5 g5 3. fg6 Bg7 and now:
I: 4. gh7 Bb2 5. hg8=B Bg7 6. Bgc4 d5 7. Bfd3 OO 8. Bh7 Kh7
II: 4. Bc4 Bb2 5. Bg8 Bg7 6. Bb3 d5 7. Bc4 OO 8. gh7 Kh7

10605 - Jouli Avroutine
Remove from the second, the third, and the fourth diagram the piece on d5. Add them in the first diagram in the following way: wQc4, wPe4, wKd7. Remove wNc6, bNe6 and wBe7 from the first diagram, and add them on e5 in the other diagrams, the white knight in the second diagram, the black knight in the third, and the white bishop in the fourth. We now see the diagrams represent "1999".

Now play in the first diagram 1. Be7 Ke7#, and in the others, move the piece on d6 to c5, resulting in a mate. We now see the diagrams represent "2000"!

Die Schwalbe Heft 181, Feb. 2000

10663 - Nikita Plaksin & Alexander Zolotarev
The intention was: Last moves were: -1. b4-b5 b6xNc5 -2. Ne6-c5 c4xBd3 -3. Nd8-e6 h5-h4 -4. d7-d8=N b5xQc4 -5. e6xQd7 Qd8-d7, and further: Re4-e6; Qg8-c4; g5-g8=Q; h4xBg5; Ba8=d3; a2-a8=B; Bf8-g5; a7xQb6; Qg8-b6; g5-g8=Q; g7xBf6; d7-d6 etc.
However, this is cooked: the last moves could've been: -1. Ra6-c6 b6xNc5 -2. Nb7-c5 c4xQd3 -3. Qd2-d3 c5-c4 -4. Nd8-b7 c6-c5 -5. d7-d8=N h5-h4 -6. e6xQd7. Further retractions are: Unpromote the rook now on a6 on a8, and take the pawn back to a6; Uncapture a white queen on b6 with a7xQb6; Unpromote both the uncaptured queens on g8; uncapture a bishop on g5 with h4xBg5; bring this bishop back to f8; uncapture a bishop on f6 with g7xf6.
Note that black can't take back b7-b6 or b7xc6, because the white king needs to escape its cage over e5, forcing black to take back d7-d6.

10664 - Gianni Donati
1. Na3 b5 2. Nc4 b4 3. a3 b3 4. Ra2 ba2 5. Nf3 a1=N 6. Nh4 Nb3 7. f3 Nc1 8. Kf2 Nd3 9. ed3 e6 10. Be2 Ke7 11. Re1 Kf6 12. Bf1 Kg5 13. Qe2 Kh6 14. Ra1 Bc5 15. Ke1 Bb6 16. Qd1

10665 - Satoshi Hashimoto
1. a4 e6 2. Ra3 Bc5 3. Rb3 Be3 4. Rb6 ab6 5. de3 Ra5 6. Nd2 Re5 7. a5 d5 8. a6 Bd7 9. a7 Bb5 10. a8=R Nd7 11. Ra1 Qa8 12. Nb3 Qa2 13. Bd2 Qb1 14. Ra8 Ke7 15. Rg8 c5 16. Ra8 c4 17. Ra1 Ra8 18. Bb4 Ke8 19. Bd6 Nb8

10666 - Andrei Kornilov & Alexander Kislyak
1. h4 a5 2. h5 a4 3. h6 a3 4. hg7 ab2 5. Rh6 Ra3 6. Rc6 Rf3 7. a4 h5 8. a5 h4 9. a6 h3 10. a7 h2 11. a8=Q h1=Q 12. Qa5 Qh4 13. Qe5 Qd4 14. Ra5 Rh4 15. Na3 Nh6 16. g8=Q b1=Q 17. Qgg3 Qbb6 18. Qh2 Qa7 19. g4 b5 20. g5 b4 21. g6 b3 22. g7 b2 23. g8=Q b1=Q 24. Qgg2 Qbb7 25. Bb2 Bg7 26. Qa1 Kf8 27. Kd1 Kg8 28. Kc1 Kh7 29. Kb1 Qh8 30. Ka2 Bf8 31. Bc1

10667 - Oliver Sick
1. e4 c5 2. e5 d5 3. ed6 Qb6 4. d7 Qh6 5. d8=N Be6 6. Ne6 fe6 7. a4 Nc6 8. a5 b5 9. ab6 OOO 10. b7 Rd5 11. b8=N Rh5 12. Nd7 g5 13. Ne5 Bg7 14. Ng6 hg6
Black's sixth and seventh move can be interchanged.

10668- Tomislav Petrovic
If white may castle, then the last moves must've been -1. b2-b4 Rc3-a3. So black may capture e.p., but white has to castle in the solution to validate this en-passant capture.
h#3 with 1. cb3 Rb4 2. Ba6 Ba6 3. Kf3 OO#.

10669 - Tomislav Petrovic
What was black's last move? The only capture was h6xBg5. Only possibility is d7-d5, so the ep capture is allowed.
1. ed6 gf3 2. Qe4 Ke4#; 1. .. f6 2. f4 ed6#; 1. .. fe6 2. Qe6 Ke6#

10670 - Alexander Yarosh
Add white bishops on a5 and e4. The white pawns currently on the board captured 5 times. The black g- and h-pawn didn't capture, so the white g- and h-pawn had to capture twice together to get past them. So all missing black pieces were captured by the white pawns, so white's last move wasn't a capture. No white retraction move gives black a last move.

10224c - Gianni Donati
1. d4 c6 2. d5 Qb6 3. d6 Qc7 4. dc7 d5 5. c4 Be6 6. c8=N Nd7 7. Nb6 Ndf6 8. Nd7 Nh5 9. Nf6 ef6 10. c5 Bd6 11. cd6 d4 12. d7+ Ke7 13. d8=N Bd5 14. Ne6 Rd8 15. Ng5 fg5 16. e4 Kf6 17. e5 Kg6 18. e6 Rd6 19. e7 Rf6 20. e8=N Ne7 21. Nd6 Rd8 22. Nf5 Rd7 23. Nh6 gh6

10414c - Mark Kirtley
1. e4 a5 2. Qh5 Ra6 3. Qa5 Rh6 4. Qa8 Na6 5. Qc8 Rh5 6. Qa8 h6 7. Qa6 Qa8 8. Qg6 Kd8 9. Qh7 Kc8 10. Qg8 Kb8 11. Qf8 Ka7 12. Qa8 Kb6 13. Qa5 Kc6 14. Ba6 Kd6 15. d3 Ke6 16. Bd2 Kf6 17. Bb4 Kg6 18. Kd2 Kh7 19. Kc3 Kg8 20. Nd2 Kf8 21. Re1 Ke8

Die Schwalbe Heft 182, Apr. 2000

10728 - Alexander Zolotarev
The author's intention was: The last moves were: -1. .. Bh3xRg4 -2. Re4-g4 Rg4-g3 -3. Re6-e4 c4xNd3 -4. Rd6-e6 c5-c4 -5. Rd8-d6 c6-c5 -6. d7-d8=R c4-c3 -7. d6-d7 b5xQc4 -8. e5xBd6 B~-d6. Further retroplay is: Qc1-c4; Ne3-d3; Ra1-d2; Na8-e3; a5-a8=N; a6xRb5; Rb8-b5 b5-b8=R; b7xQc6; e4-e5; Qg8-c1; g6-g8=Q; Qg8-c6; Bh6-~; g5-g6; Bf8-h6; g7-g8=Q; a7-a6; g6-g7; g7xRf6 etc. So the geometrically shortest route is 3.
However, Karl-Dieter Schulz found the following cook: -1. Bh3xRg4 -2. Rg8-g4! Rg4-g3 -3. g7-g8=R c4xNd3 -4. Nf4-d3 c5-c4 -5. Ne6-f4 c6-c5 -6. Nd8-e6 c4-c3 -7. d7-d8=N b5xQc4 -8. d6-d7 b6-b5 -9. e5xBd6 Ba3-d6 -10. e4-e5 Bd6-a3 -11. Qa2-c4 Bf4-d6 -12. Qa1-a2 Be3-f4 -13. Rd3-d2 d6-d5 -14. Ra3-d3 Bh6-e3 -15. Ra8-a3. And now the queen on a1 goes to g8 via c1 and h6, and the black bishop goes to h6. Then retract -(i). g6-g7 Bg7-h6 -(i+1). g5-g6 Bf8-g7 -(i+2). g7-g8=Q d7-d6 -(i+3). g6-g7 g7xRf6. The rook on a8 unpromotes, black uncaptures a7xRb6, white retracts Rb8-b6, and unpromotes this rook.

10729 - Andrei Frolkin
White's last move was queenside castling. Black captured e7xf6 and hxg, so the bishop on a2 is an original one. The position unlocks when black can retract e7xf6. This requires 4 white retro-moves. The pawn on b7 is the only piece that can make those retro-moves. It can't have come from b3, because the bishop on a2 would've been incarcerated, so it had to come from f3. This pawn captured 5 times, and the black a-pawn was captured on its own file, to let the white a-pawn promote.
Last moves were: -1. OOO Nf1-g3 -2. c6xPb7 Bg3-h4 -3. d5xPc6 Bd6-g3 -4. e4xPd5 Bf8-e6 -5. f3xRe4 e7xf6 and the whole position unlocks.
So the route of the pawn on b7 was: g2xRf3xRe4xPd5xPc6xPb7.
Unfortunately, this is cooked: Last moves could've been: -1. OOO Nf1-g3 -2. b6-b7 Bg3-h4 -3. b5-b6 Bd6-g3 -4. b4-b5 Be6-a2 -5. b3-b4 Bf8-d6 -6. a2xPb3 e7xNf6 etc.

10730 - Alexander Kislyak
The last moves were: -1. c5xb6 ep b7-b5 -2. c4-c5 Kb6-a6 -3. d5xc6 ep c7-c5 -4. d4-d5.
Rd8 is clearly a promoted rook. How did the white knight get to g8? First black must've played h7-h6, then the white queen went to g8 via h7, then the black rook went to h7, after that the queen went to h8, and only then the knight went to g8 over f6. In order to avoid an illegal check, the black queen must've been captured on d8. The white promotion rook arose after e6xNf7xNe8, since the missing black rook couldn't have gotten to f7 or e8. The black queen is the promoted a-pawn. Since the white a-pawn didn't capture, the black one must've made one capture, and the only possibility is a3xRb2.
So the places of the captures were: d8 (black queen), g3 (black rook), b2 (white rook), f7 (black knight), e8 (black knight), b5 (black pawn, was captured en-passant), and c5 (black pawn, was captured en-passant).

10731 - Thomas Volet
The position unlocks when white can retract g2xNh3. This knight can then go to d5, and the kings can escape their mutual pinning. So a rook has to be brought back to g1 or h1 (the bishop can go without difficulty to f1).
First, the white queen goes to f7 via a1 and a2. Then the bishop on d8 goes to e7, allowing the white queen to go to c6 via e8 and c8. The black rook now can leave the northwest corner via a7, c7, c8, e8, and it goes to e6. The white rook on g8 can leave its corner now too. It goes via g7, e7 to e8, the bishop plays Bd8-e7, and the rook can escape. This maneuver requires one tempo, and black has as tempo c4-c3. The rook now can go to the southeast corner, and white can finally uncapture g2xh3.

10732 - Thierry le Gleuher
1. Na3 Nc6 2. Nc4 Nd4 3. Na5 Ne2 4. Ne2 e5 5. Ng3 Ba3 6. Bb5 Bb2 7. Bc6 dc6 8. Bb2 Qd2 9. Kd2 Bf5 10. Kc1 Bc2 11. Qc2 OOO 12. Qg6 Rd2 13. Bd4 Ra2 14. Ra2 hg6 15. Rb2 Rh2 16. Rh2
But Unto Heinonen gave the following cook: 1. e4 e5 2. Bb5 Ba3 3. Na3 Nc6 4. Bc6 dc6 5. Nc4 Bf5 6. Na5 Be4 7. Ne2 Bc2 8. Qc2 Qd5 9. Qg6 Qa2 10. Ng3 Qb2 11. Bb2 OOO 12. OOO Rd2 13. Rd2 hg6 14. Bd4 Rh7 15. Rb2 Rh2 16. Rh2.

10733 - Peter Wong
1. Nc3 f5 2. Ne4 f4 3. Ng3 fg3 4. Nh3 gh2 5. Rg1 h1=Q 6. g3 Qf3 7. Rg2 Qa3 8. ba3 h6 9. Bb2 h5 10. Bd4 Nh6 11. Bb6 ab6 12. c3 Ra5 13. Qa4 Rb5 14. Qa6 ba6 15. Kd1 Bb7 16. Kc2 Qc8 17. Rd1 Be4 18. Kc1 Bh7 19. Ng1 Nf5
But this is cooked too: 1. Nc3 f5 2. Ne4 f4 3. Ng3 fg3 4. Nh3 gh2 5. Rg1 h1=Q 6. g3 Qf3 7. Rg2 Qa3 8. ba3 Nh6 9. Bb2 Nf5 10. Bd4 h5 11. Bb6 ab6 12. c3 Ra5 13. Qa4 Rb5 14. Qa6 ba6 15. Kd1 Bb7 16. Kc2 Bd5 17. Rd1 Bg8 18. Kc1 Qc8 19. Ng1 Bh7

10734 - Unto Heinonen
1. h4 g5 2. hg5 Nc6 3. Rh6 Rb8 4. Rf6 ef6 5. a4 Bd6 6. Ra3 Nge7 7. Rb3 OO 8. Rb7 Kh8 9. Rb3 Rb3 10. a5 Rd3 11. c3 Ba6 12. Qa4 Qa8 13. Qh4 Nb4 14. f3 Qf3 15. Kd1 Qf1 16. Qe1 Qf5 17. Qh4 Qa5 18. Ke1 Nbd5 19. Qa4 Rb8 20. Qd1 Nc8
But this has been cooked too: 1. f4 Nc6 2. f5 Rb8 3. f6 ef6 4. a4 Bd6 5. Ra3 Nge7 6. Rb3 OO 7. Rb7 g5 8. Rb3 Rb3 9. h4 Ba6 10. Rh3 Qa8 11. Rd3 Rd3 12. c3 Rb8 13. Qb3 Nc8 14. Kd1 N6e7 15. Kc2 Qf3 16. a5 Qf1 17. hg5 Qf5 18. Kd1 Qa5 19. Ke1 Nd5 20. Qd1 Kh8

10735 - Tomislav Petrovic
Suppose white may castle. What was his last move, before black's last one? No black retromove will free a piece, except c7-c5 (and white played c6xb7 before that). So if white can castle, then he may capture en passant.
h#1.5 with 1. .. bc6 2. gf1=B OOO# (2. .. Rd1?)
But black's last move could've been d6xNc5! too. Other black captures were: e7xPd6, c6xBb7 (after black retracted his pawn from c3 to c7), a5xPb4, and g2xPh3.

10736 - Stefano Spinelli
The intention was: Retract -1. Kd8xNe7 Ka6xNb7, play forward Ba5, allowing a s#1 with 1. Nc5 bc5#
But also possible is: Retract -1. Bh5xQe8 Qg6xQe8, play forward Bg7, allowing a s#1 with 1. Qc8 Kc8#, or: retract -1. Kd8xNe7 Bf6xQc3, play forward Kb8, allowing a s#1 with 1. Qc8 Nc8.

10737 - Vladimir Chutornoi
The last moves could've been: by the king (11, not Kc3xQd3?), queen (101), Rd5 (12), Re6(12), Rf5(12), Na4 (11, not Nc3-a4?), Nb5 (11, not Nc3-b5?), Nc4 (12), Nc6 (12), Nd1 (9, not Nc3-d1?), Ne2 (9, not Nc3-e2?, Nc3xBe2?, Nd4xBe2?), Ne4 (9, not Nc3-e4?, Nc3xBe4?, Nf6xBe4?), Nf3 (12), a3 (5). There were 238 possible last moves. Only retracting Kc3-d3 results in a forward #1 with 1. Kc2#.

Die Schwalbe Heft 183, June 2000

10794 - Alexander Kislyak
The bishop on h7 is a promoted one. White's captures are axb-b6xRa7, d2xe3 and h2xg3. The black h-pawn promoted. The black king was on g4 when [or went to there the move after] h2xg3 was played. This means that the h-pawn promoted on g1.
The last moves were: -1. .. Nd4xPc6 -2. c5-c6 Nf3-d4 -3. c4-c5 Ng1-f3 -4. c3-c4 h2xBg1=N -5. Bf2-g1 Bb8-a7 -6. Be1-f2 Ba7-b8 -7. Bd2-e1 Bb8-a7 -8. Bc1-d2 Ba7-b8 -9. d2xNe3 Nf5-e3 -10. Rf7-g7 Ng7-f5. Now bring back the rook to f1, retract f2-f4, and the queen on h6 can escape. The white queen has to go back to d1 over c2, that's why white couldn't retract c2-c4 earlier.
So the first move of the white c-pawn is c2-c3.

10795 - Alexander Zolotarev
Retroplay is: -1. .. Be8xRd7 -2. a3xNb4 Nd3-b4 -3. g4-g5 Nf2-d3 -4. g3-g4 Nh1-f2 -5. g2-g3 h2-h1=N -6. f5-f6 h3-h2 -7. f4-f5 h4-h3 -8. f3-f4 h5-h4 -9. f2-f3 h7-h5 -10. h6xQg7. Now further: (Qg7->c1); Bf8->g5; (g6->g7); (Nh8->c5); retract Rd5-d7 (Nd7-c5); Rd5->b3; Bb1->a2; (Qc1->a1); Ba2->b1; Rb3->d1; (Qa1->b3); (d5xQe4); Qe1-e4; Ba2-b1; R->a1; Bb1-a2; B->c1; d2xBc3, and the whole position unlocks.
However, it's also possible to uncapture a queen on d7, and a rook on e4.

10796 - Mikhail Kozulya
Intentions were:

 

[3B1r2/1n3pp1/2b1Pq2/Nnn3N1/1kB4Q/NNRn2P1/3NPKn1/2b1nn1r] [3N1b2/1r3pp1/2n1Pk2/Rrr3R1/1qN4K/RRBr2P1/3RPQr1/2n1rr1b]
-1. .. Nf4-d3 -1. f5xe6 ep e7-e5 -2. Ne5-c4 Nd4-c6

But there are other possibilities, for instance

[4Q2r/1b1N1Pn1/rbp3Kn/2P3Pn/B4nN1/2qnBR1q/1n1nkN2/3N1N2]

And the black queen, rook and bishop can be interchanged too!

10797 - Thierry le Gleuher
1. Nc3 d6 2. Nd5 Bf5 3. c3 Bc2 4. d3 Kd7 5. Bg5 Kc6 6. e3 Kc5 7. Qh5 c6 8. f3 Qb6 9. Kf2 Qb3 10. ab3 Nh6 11. Ra6 Nf5 12. Rb6 Nd4 13. Kg3 ab6 14. Kh4 Ra1 15. g3 Re1 16. Bh3 Re2 17. Bd7 e6 18. Be7 f6 19. Qf7 g6 20. Nh3 Bh6 21. Ra1 Bf4 22. Ng5 Na6 23. h3 Nc7 24. Ra6 Ncb5 25. Nc7 Na3 26. ba3 ba6
But this is cooked: 7. .. Nc6 8. f3 Nd4 9. Kf2 c6 10. Kg3 Qb6 11. Kh4 Qb3 12. ab3 Nf6 13. Ra6 Ne4 14. Rb6 ab6 15. g3 Ra1 16. Bh3 Re1 17. Bd7 e6 18. Be7 f6 19. Qf7 g6 20. Nh3 Re2 21. Ra1 Bh6 22. Ra6 Bf4 23. Ng5 Nd2 24. h3 Nc4 25. Nc7 Na3 26. ba3 ba6

10798 - Stefano Spinelli
The author's intention was: Retract -1. Qa6xPg6 Qg3xPa3, play forward 1. .. e6, and now s#1 with 1. Rd5 ed5#.
But this is cooked: Retract -1. Qa6xNg6 Qh3xPa3, play forward 1. .. e6, and now s#1 with 1. Rd5 ed5#. Or retract -1. Ke5xBe4 Qf3xNa3, play forward 1. .. Bd5, and now s#1 with 1. Qd6 ed6#

10799 - Oliver Sick
1. e4 h5 2. e5 d5 3. ed6 Bf5 4. d7 Qc8 5. d8=R Bh7 6. Rd6 Qh3 7. Rf6 gf6 8. d4 Bh6 9. d5 Bd2 10. d6 Nh6 11. d7 Rg8 12. d8=R Rg3 13. Rd4 Bg8 14. Rg4 hg4 15. f4 gf3 16. Bd3 f2 17. Ne2 Rf3 18. OO Be1 19. Bd2

10801 - Anonymous
Black has no last move (-1. .. Ka7-a8 -2. b7-b8=B??), so he has the move. Mate in 1 with 1. .. Qg6#

10802 - Tomislav Petrovic
What was black's last move? Can't have been a pawn capture, so it must've been f7-f5.
1. gf6 ep Kg4/Nf3/Ne2/hg2/Rg4/Rh5 2. Qg6/Qf3/Ne2/Rg2/Bf2/Nh5#

10803 - Günther Weeth
Retract Kb2xBa1. This bishop is promoted. The black e-pawn needs 4 captures to get to a2. So all white pieces had to promote. If black may castle, this requires 5 captures. Together with Bc8, these are all captures. So white either has to unpromote a piece, or retract b2-b3, to give black more retromoves. However, black may not rectract h7-h6 (the white h-pawn needs to promote on h8) or f7-f6 (the white rook on f7 couldn't have gotten there without checking then). The piece that can reach the promotion line the fastest is the knight on b5, but that knight needs 3 moves, 1 move too many. So black may not castle.
Mate in 3 with 1. Ka1! ~ 2. Rh5 ~ 3. Rh8#

10804 - Andrei Kornilov
Last moves must've been -1. .. Rd8xNg8 -2. Nh6-g8 OOO. The white king could only escape the first rank via a2. Before a2xb3, the white knight was already on a1, and the rook currently on b1 was trapped on the first rank too. The only way the white king could get past the rook is by castling too. So there's twice a OOO.

10805 - Wilfried Neef
1. Na3 e6 2. Nc4 Ba3 3. ba3 d6 4. Bb2 Bd7 5. Qc1 Ba4 6. Qb1 Bb3 7. ab3 Nc6 8. Qa2 Qb8 9. OOO Nd8 10. Qa1 c6

10806 - Mario Velucchi
Qg7, Qd6, Qf5, Qe4, Nc1.
This problem was posed (and solved) before in the book "Schach und Mathematik" ("Chess and mathematics") by J. Gik.

10099vc - Alexander Zolotarev
The last moves were: -1. .. Rh7xQh6 -2. Ne5-g6 a6-a5 -3. Nd3-e5 a7-a6 -4. Ne1-d3 Nh4-g2 -5. Ng2-e1 Nf5-h4 -6. Qg5-h6 Nh6-f5 -7. Qb5-g5 f5-f4 -8. Qb8-b5 b5-b4 -9. b7-b8=Q b6-b5 -10. a6xQb7 Qd5-b7 -11. a5-a6 Qd1-d5 -12. a4-a5 d2-d1=Q -13. a3-a4 d3-d2 -14. d2xBe3 Bc5-e3 -15. a2-a3 Bf8-c5 -16. Kg5-h5 e7xBf6 and everything unlocks.

10415c - Andrei Frolkin
1. Nf3 a5 2. Nd4 a4 3. Nb3 ab3 4. a4 Ra5 5. Ra2 ba2 6. h4 ab1=B 7. Rh3 Ba2 8. Rg3 Be6 9. Rg7 Bg4 10. Rg6 Rf5 11. Rf6 ef6 12. a5 Bc5 13. a6 Ne7 14. a7 Rf8 15. ab8=Q Ba7 16. Qa8 c5 17. Qb8 Qc7 18. Qa8 Qh2 19. Qb8 d6 20. Qa8 Kd7 21. Qb8 Kc6 22. Qa8 Bd7 23. Qb8 Nc8

10541c - Andrei Frolkin
1. f4 a5 2. f5 a4 3. f6 a3 4. fg7 f5 5. h4 Nf6 6. g8=B Bg7 7. Bb3 Rg8 8. h5 Bh8 9. h6 Rg7 10. hg7 h5 11. g8=B h4 12. Bgc4 d5 13. Nh3 dc4 14. Nf2 cb3 15. Rh3 ba2 16. Rb3 Qd3 17. ed3 h3 18. Qg4 h2 19. Qb4 f4 20. Be2 Bf5 21. Bf3 h1=Q 22. Ke2 Qc1 23. Bd5 Qg1 24. Nd1 Qd4 25. Bg8 Qd8

Die Schwalbe Heft 184, Aug. 2000

10859 - Andrei Kornilov
White captures are cxdxe and dxe, black captures are bxc, exd, fxe, gxf and hxg. The white b- and h-pawn promoted without capture, so black can't take back bxc before white unpromoted a piece on b8, nor can he take back hxg before white unpromoted a piece on h8.
The last moves were: -1. Rg6-h6 a5-a4 -2. Bd4-b2 c3-c2 -3. Ba7-d4 c4-c3 -4. Bb8-a7 c5-c4 -5. b7-b8=B c6-c5 -6. b6-b7 b7xNc6 -7. Nd4-c6 a6-a5 -8. Nf5-d4 a7-a6 -9. Nh6-f5 Bh7-g8 -10. Ng8-h6 f5-f4 -11. Rh6-g6 g6-g5.

10860 - Alexander Zolotarev
Last moves were: -1. .. c7xRd6 -2. Rg6-d6 h6-h5 -3. Rg8-g6 h7-h6 -4. g7-g8=R e6-e5 -5. g6-g7 d5xNc4 -6. h5xBg6 and further: (Bf1-g6); (Bg2xQf1); Qf8-f1; f2-f8=Q; Nf8-c4; f3-f8=N; (f7xRe6); Rb8-e6; e2-e7; e6xRd5; b6-b8=R; Rb8-d5; (Ba6-g2); b5-b6; (Bc8-a6); (Qd8-h4); b7-b8=R; (e7-e6); b6-b7; (b7xPc6) etc. The other queen move was Qd1-e1. So the longest queen move was Qf1-f8.

10861 - Andrei Kornilov & Andrei Frolkin
Last moves were: -1. Nh7-f8 Nf2-h1 -2. c5-c6 Nd3-f2 -3. c4-c5 Nb4-d3 -4. c3-c4 Nc6-b4 -5. c2-c3 Nd8-c6 -6. b4-b5 Nf7-d8 -7. b3-b4 Nd8xQf7 -8. Qg8-f7 Na6-b8 -9. Qf8-g8 Qg8-g7 -10. Bg7-h8 Qh8-g8 -11. Qg8-f8 Nb4xPa6 -12. Kf8xBe8 Nc6-d8 -13. a5-a6 Rd8-d7 -14. a4-a5 Rb8-d8 -15. a3-a4 Nd8-c6 -16. a2-a3 Bc6-e8 -17. Ke8-f8 Be4-c6 etc.

10862 - Stefano Spinelli
The intention was: Retract -1. Kc3xBd3 Ra8xQf8, play forward Nf3 and s#1 with 1. Qa3 Ra3#.
But this is cooked: Retract -1. Kc2xQd3 Qd4xQd3, play forward Nc4 and s#1 with 1. Qa3 Na3#.

10863 - Anatoli Vasilenko & Andrei Frolkin
1. h4 f5 2. h5 f4 3. h6 f3 4. Rh5 fg2 5. f4 a5 6. Kf2 a4 7. Ke3 a3 8. Kd3 ab2 and now:
(a) 9. Na3 b1=Q 10. Bb2 Qa2 11. Bf6 Qc2 12. Nc2
(b) 9. Bb2 Ra2 10. Bf6 Rc2 11. Na3 Ra2 12. Nc2 Ra8

10864 - Satoshi Hashimoto
1. a4 d6 2. Ra3 Be6 3. Rc3 Bb3 4. Rc7 d5 5. Rc3 Qb6 6. Na3 Qe3 7. de3 Nc6 8. Qd4 Rd8 9. Rd3 Rd6 10. Rd1 Rg6 11. Bd2 e6 12. Ra1 Bc5 13. Bc1 Nce7 14. Qd1 Bd4 15. Nb1

Die Schwalbe Heft 185, Oct. 2000

10914 - Andrei Frolkin & Andrei Kornilov
Last moves were: -1. Sh5-g3 Qh7-h8 -2. Sf4xSg6 Se5-g6 -3. g6-g7 Sc6-e5 -4. Se2-f4 Sb8-c6 -5. Sf4xSh5 Sg3-h5 -6. h5xBg6 Bh7-g6 -7. Sg6-f4 Se4-g3 -8. g7-g8=Q Bg8-h7 -9. Sg1-e2 Kh7-h6.

10915 - Alexander Zolotarev
Play forward from the following position:

[Qs3bb1/rSppppp1/qpk5/s6p/1KP1P3/rRpR4/SP1P1PPP/2B5]

1. Rd6 ed6 2. h3 Be7 3. h4 Bg5 4. hg5 f6 5. g6 Bf7 6. gf7 h4 7. f8=B h3 8. Be7 h2 9. Bf8 h1=N 10. Be7 Ng3 11. Bf8 Nh5 12. Be7 Nf4 13. Bf8 Ne6 14. Be7 Nd8 15. Nc5 Ndb7 16. Ne6 f5 17. Ng5 f4 18. Nf7 and the diagram position (with the added bishop) is reached. So the bishop has to be added on e7.

10916 - Alexander Yarosh
All pieces are white, except Bd1 and Kh8. Last move was d2-d1=B. Mate in 6 with 1. Nb3! and white mates ultimately on the 6th move. Unfortunately none of the variants is dualfree. Reversing colours doesn't work, since white would have no last move, and after Bc2/Be2 there is no mate in 6.

10917 - Roberto Osorio
1. d4 b5 2. d5 b4 3. Qd4 b3 4. Kd2 bc2 5. b4 cb1=R 6. Ba3 Rf1 7. b5 Rg1 8. b6 Rc1 9. b7 Rc6 10. Qb6 ab6 11. dc6

10918 - Andrei Frolkin & Mikhail Kozulya
1. e3 b5 2. Be2 b4 3. Bg4 b3 4. ab3 h5 5. Ra5 h4 6. Rh5 a5 7. Nf3 a4 8. OO a3 9. ba3 e6 10. Bb2 Bd6 11. Bd4 Bg3 12. Nc3 d6 13. Qa1 Bd7 14. Rb1 Ba4 15. ba4 Nd7 16. Rb5 Rb8 17. Rbg5 Rb4 18. ab4
But this is cooked: 1. e3 d6 2. Be2 Bg4 3. Bg4 a5 4. Nf3 a4 5. OO a3 6. ba3 e6 7. Bb2 Be7 8. Bd4 Bh4 9. Nc3 Bg3 10. Rb1 b5 11. Rb5 h5 12. Qa1 h4 13. Rh5 Nd7 14. Rb1 Rb8 15. Rbb5 Ke7 16. Rbg5 Rb4 17. ab4 Ke8 18. a4

10919 - Unto Heinonen
1. d3 Nh6 2. Bh6 e6 3. e3 Ba3 4. Qg4 O-O 5. f3 g6 6. Bf8 f6 7. Bc5 d6 8. Kf2 Bd7 9. Kg3 Ba4 10. Kh4 c6 11. g3 Qa5 12. Bh3 Qe1 13. c3 Bd1 14. b3 Bc1 15. Na3 Na6 16. Nc2 Nb4 17. a3 Na2 18. Qb4 a6 19. Bg4 Ra7 20. Ba7 h6 21. Bd4 b6 22. h3

10920 - Gerald Irsigler
1. Qa6 Bb6 2. Qb6 Bb6 3. Rb6 Nc6 4. Rc6 Qd6 5. Nbd5 Rd5 6. Ncd5 Kg5 7. Re5 Qe5 8. Re5 Be5 9. Qe5 N6f5 10. Qf5 Nf5 11. Nf5 Rd2 12. Bcd2 Rd2 13. Bd2 Qhe3 14. Nge3 Qhg4 15. Rg4 Qg4 16. Ng4

10921 - Werner Keym
Adding only a white queen (and mate with 1. Qe6) isn't possible: either the white bishop on f1, or the black piece which captured this bishop, should be added. If one does the latter: This black piece is a promoted rook or queen, and captured b7-b5xPa4xPb4 (en passant)xQa2xRb1. So this promotion piece captured the white queen, so this isn't possible. But adding only the white bishop on f1 isn't enough either! The only two pieces which could capture are the white queen and the black king, both on white squares. So an extra, third piece should be added, on the black squares.
The most economical way however is adding a black queen on a3 and a white bishop on f1.
[threat: 1. .. Qa1/Qc1#] 1. f4 Qg3#

10922 - Arnold Beine
(a): X=15, Y=black pawn. A possible position is:

[B5Q1/2S1S3/1S1p1S2/R5R1/1SPKPS2/2S1S3/8/7k]

and after d6-d5, 15 white pieces can capture this pawn.
(b): X=9, Y=black pawn. A possible position is:

[6Q1/1B1p4/8/R5R1/1SPKPS2/8/8/7k]

and after d7-d5, 9 white pieces can capture this pawn. Chnaging the two rooks to pawns on c5 and e5 (which can capture en passant) works too. And a black knight on c7 instead of the black pawn works too.

10732c - Thierry le Gleuher
1. Na3 Nc6 2. Nc4 Nd4 3. Na5 Ne2 4. Ne2 e5 5. Ng3 Ba3 6. Bb5 Bb2 7. Bc6 dc6 8. Bb2 Qd2 9. Kd2 Bf5 10. Ke3 Bc2 11. Qc2 OOO 12. Qg6 Rd2 13. Bd4 Ra2 14. Ra2 hg6 15. Rb2 Rh2 16. Rh2

10733c - Peter Wong
1. Nc3 f5 2. Ne4 f4 3. Ng3 fg3 4. Nh3 gh2 5. Rg1 h1=Q 6. g3 Qf3 7. Rg2 Qa3 8. ba3 h6 9. Bb2 h5 10. Bd4 h4 11. Bb6 ab6 12. c3 Ra5 13. Qa4 Rb5 14. Qa6 ba6 15. Kd1 Bb7 16. Kc2 Qc8 17. Rd1 Be4 18. Kc1 Db7

Die Schwalbe Heft 186, Dec. 2000

10978 - Andrei Kornilow
Last moves were: -1. Sc7-a8 Sb8-a6 -2. Se8-c7 Sa6-b8 -3. Sg7-e8 Sc7-a6 -4. Se8-g7 Sa8-c7 -5. Sc7-e8 c4-c3! -6. Sa6-c7 Sc7-a8 -7. Sb8-a6 Se8-c7 -8. Sa6-b8 Sg7-e8 -9. Sb8-a6 Se8xRg7! -10. Rh7-g7 Qh8-g8 -11. S~ Kg8xBf8! -12. Re8-e7 and the position unlocks. So black has the move.
Mate in 1 with 1. .. Qg6# and not with the illegal 1. Re8?

10979 - Alexander Kislyak
Last moves were: -1. .. Ra6xQb6 -2. e4xSf5 Sd6-f5 -3. e3-e4 Sc3-d5 -4. Qd4-b6 Sb6-c4 -5. Qg7-d4 h5-h4 -6. Qg8-g7 h6-h5 -7. g7-g8=Q h7-h6 -8. h6xBg7 Bc3-g7 -9. h5-h6 Bd2-c3 -10. h4-h5 Bc1-d2 -11. h3-h4 c2-c1=B -12. h2-h3 c3-c2 -13. c2xSb3! (c2xPb3?) and the position unlocks.
So the mate is indeed legal.

10980 - Alexander Zolotarev
Last moves were: -1. .. f7xBe6 -2. Bc8-e6 e2xBf1=Q -3. c7-c8=B d3xBe2 -4. b7xRc7. Further retroplay is: Bc8-e2; c4-c8=B; Bc8-f1; c5-c8=B; a5xBb6; b4xSc5; e2-e3; (e3xSf2); (e4-e3); (d5xRe4); (c6xRd5); Rg8-e4; g5-g8=R; (Rh8-c7); Rg8-d5; (Bf8-b6); g7-g8=R; h6xSg7; etc.
The two queen moves were Qd8-d4 and Qd1-h1.

10981 - Alexander Zolotarev
(a): Add a white rook on g3. Last moves were: -1. .. Qh8xSg7 -2. d3-d4 Se6-g5 -3. Sf5-g7 Sg7-e6 -4. Se3-f5 Bc8-g4 -5. Sg4-e3 d7-d6 -6. d2-d3 Ba3-f8 Re8-g8 etc.
(b): Add a black knight on g3. Last moves were: -1. .. Qh8xSg7 -2. d2-d3 Se6-g5 -3. Sf5-g7 Sg7-e6 -4. Se3-f5 e5-e4 -5. Sf1-e3 Sf5-g3 -6. Sg3-f1 Se7-f5 -7. Bd3-g6 Sg6-e7 -8. Bf1-d3 e6-e5 -9. e2xRf3 Rb3-f3 -10. Sf3-g1 etc.

10982 - Radovan Tomasevic
(a): 1. Nf3 e5 2. Ne5 Qh4 3. Nd7 Qh2 4. Nb8 Bd7 5. Nd7 Rd8 6. Nf8 Rd5 7. Nh7 Rg5 8. Ng5 Rh3 9. Nh3 Qg1 10. Ng1
(b): The intention was: 1. h4 g5 2. hg5 h5/h6 3. gh6 Bg7 4. hg7 d5 5. gh8=Q d4 6. Qd4 Kf8 7. Qd8 Kg7 8. Qc8 Nd7 9. Qa8 Nf8 10. Qf8 Kf8. But this is cooked: 1. h4 g5 2. hg5 h5 3. gh6 Bg7 4. hg7 d5 5. gh8=Q d4 6. Qd4 Nd7 7. Qd7 Kf8 8. Qc8 Kg7 9. Qa8 Qf8 10. Qf8 Kf8

10983 - Jevgeni Reitsen & Andrei Frolkin
1. d4 h6 2. Bh6 a5 3. Bg7 Ra6 4. Bf8 Rah6 5. Be7 Nc6 6. Bd8 Nge7 7. d5 OO 8. Be7 Nd4 9. Bf8 Ne2 10. Bh6 Nc1 11. Ba6 ba6 12. Bc1 Bb7 13. Nd2 Ba8. But this is cooked too: 1. d4 e5 2. de5 Nf6 3. ef6 a5 4. fg7 Qg5 5. gh8=Q Qd5 6. Qh7 Bh6 7. e3 Kf8 8. Qh6 Kg8 9. e4 Na6 10. Ba6 Ra6 11. Qa6 ba6 12. Nd2 Bb7 13. ed5 Ba8

10984 - Roberto Osorio
1. a4 h5 2. a5 h4 3. a6 Na6 4. Ra5 Nb8 5. Rb5 a5 6. b3 a4 7. Ba3 Ra5 8. Bc5 a3 9. Qc1 a2 10. Qb2 a1=B 11. Qf6 Be5 12. Qa6 Bh2 13. Qa8 Be5 14. Rh3 Rh5 15. Rg3 h3 16. Rg5 h2 17. g3 h1=B 18. Bh3 Bd5 19. Bf5

10985 - Yves Cheylan
[r7/1R4PK/7P/8/8/6B1/P5p1/Rb4k1]

White's mate, since both Rab1 and Rbb1 are illegal under the exclusive condition. But black's last move was illegal too, since promotion to queen would've been mate too. Removal of any unit results in a legal position.

10986 - Werner Keym
(a): The position is legal. The missing black pieces from the white squares were captured by the queen, and the missing black pieces from the black squares were captured by the rook, except the knight from b8 and the queen. They were captured by the white king.
(b): The position is illegal. History would be: (g7xBf6); d2-d4xQe5xf6xBg7xRh8=QxSb8; (h7-h5xg4xh3(ep)xBg2xRh1=QxSb1); Qd1xb1; KxQe6. But the promoted white queen couldn't get captured.
(c): The position is legal. History: d2-d4xQc5xb6(ep)xa7xSb8=Q; QxBf8; (Rh8xQf8); (Rf8-f2); Ke1xRf2, and mirrored the same for black.
(d): The position is legal. History: d2-d4; e2-e4; (d7-d5); (e7-e5); wBB out, wK out, bBB out, bK out, wQxSg8; (bQxSg1); d4xe5xRd6; (c7xd6); (d5xe4xRd3); c2xd3; b7-b5xQa4xb3(ep); a2xb3; f2-f4xQg5xf6(ep); (g7xf6); wBxRa8; wBxSb8; (bBxRa1); (bBxSb1); g2xBf3; h2xBg3; (a7xBb6); (h7xBg6).

10987 - Mario Velucchi
Place the pieces as follows: Sa2, Ba3, Ka5, Rb1, Qc6, Rd4, Be2, Se3, Pe5.

10097v - Alexander Kislyak
The last moves were: -1. Ba6xRb5# Rg5-b5 -2. Rb5-b6 Rg1-g5 -3. b6xRa7 g2-g1=R -4. c5xNb6 g3-g2 -5. d4xNc5 h4xBg3, and further: (Nf1-c5); (f7-f1=N); (Nf1-b6); (f6-f1=N); f2xRe3; (Rb1-e3); e3xRd4; (b3-b1=R); (Rb1-d4); (Bf8-b2); Bc1-~; (b4-b3); Qd1-h5, (b2-b1=R); e2-e3; (b3-b2); b2xNc3. The extra white bishop is the g-pawn which promoted on g8, and the other white queen move is Qa8-a5, and this extra queen is the promoted a-pawn.

10163vx - Andrei Frolkin & Oksana Karseva
1. Nf3 b5 2. Nd4 b4 3. Nb5 b3 4. ab3 c5 5. Ra7 Qc7 6. N1a3 Qg3 7. hg3 c4 8. Rh6 c3 9. dc3 Nc6 10. Qd6 Nd8 11. Qb8 d5 12. Rb6 Bh3 13. gh3 e6 14. Bg2 Bc5 15. Bh1 Be3 16. fe3 d4 17. Kf2 d3 18. cd3 Nh6 19. Nc2 OO 20. Re7 Ra3 21. ba3 f5 22. Bb2 f4 23. Ba1 f3 24. ef3

10539v - Alexander Zolotarev
The retroplay is: -1. .. Nc6-d8 -2. f4xNe5 Nd7-e5 -3. d5-d6 Nf6-d7 -4. Ng5-h7 Nh7-f6 -5. Ne4-g5 Nb4-c6 -6. Ng5-e4 Nc2-b4 -7. Ne4-g5 Na1-c2 -8. Ng5-e4 a2-a1=N -9. Ne4-g5 a3-a2 -10. Ng5-e4 a4-a3 -11. Ne4-g5 a5-a4 -12. Ng5-e4 Bf5-h3 -13. Nh3-g5 Bc2-f5 -14. d4-d5 Ba4-c2 -15. c2-c3 Bd7xPa4 -16. a3-a4 Bc8-d7 -17. a2-a3 d7xBe6 -18. Bc4-e6 c7/c6-c5 -19. Bf1-c4 Kg4-h4 -20. e2x{Q/R/N}f3 and the whole position is unlocked.

10663v - Nikita Plaksin & Alexander Zolotarev
Last moves were: -1. Ne2-c1 b6xNc5 -2. Ne6-c5 c4xBd3 -3. Nd8-e6 h5-h4 -4. d7-d8=N b5xQc4 -5. e6xQd7 Qd8-d7, and further: Re4-e6; Qg8-c4; g5-g8=Q; h4xBg5; Ba8=d3; a2-a8=B; Bf8-g5; a7xQb6; Qg8-b6; g5-g8=Q; g7xBf6; d7-d6 etc.

Die Schwalbe Heft 188, Apr. 2001

10729c - Andrei Frolkin
The last moves were: -1. OOO R~-g3 -2. c6xPb7 Bg3-h4 -3. d5xPc6 Bd6-g3 -4. e4xPd5 Bf8-d6 -5. f3xSe4 e7xSf6. So the route of the pawn on b7 is g2xSf3xSe4xPd5xPc6xPb7.

10795c - Alexander Zolotarev
Last moves were: -1. .. Be8xRd7 -2. a3xSb4 Sd3-b4 -3. g4-g5 Sf2-d3 -4. g3-g4 Sh1-f2 -5. e2-e3 h2-h1=S -6. f5-f6 h3-h2 -7. f4-f5 h4-h3 -8. f3-f4 h5-h4 -9. f2-f3 h7-h5 -10. h6xQg7 and now: (Qd1-g7); Bg5-f8; (g7-g6); (Sb3-h8); -n. Ba2-b1 Qa1-d1 -(n+1). Bb1-a2 Sc5-b3 -(n+2). Rd1-d7 Sd7-c5; (Qb3-a1); -m. Ba2-b1 d6xQe5 -(m+1). Ra1-d1 Qb4-b3 -(m+2). Bb1-a2 etc.

10797c - Thierry le Gleuher
1. Nc3 d6 2. Nd5 Bf5 3. c3 Bc2 4. d3 Kd7 5. Bg5 Kc6 6. e3 Kc5 7. Qh5 c6 8. f3 Qb6 9. Kf2 Qb3 10. ab3 Nh6 11. Ra6 Nf5 12. Rb6 Nd4 13. Kg3 ab6 14. Kh4 Ra1 15. g3 Rd1 16. Bh3 Rd2 17. Bd7 e6 18. Be7 f6 19. Qf7 g6 20. Nh3 Bh6 21. Ra1 Bf4 22. Ng5 Na6 23. h3 Nc7 24. Ra6 Ncb5 25. Nc7 Na3 26. ba3 ba6