ALL DE FR ES IT

Logo

StrateGems

Solutions

R0001 - Thomas Volet
The last moves were: -1. Ra3-a8 e5xRd4 -2. Ra6xPa3 a4-a3 -3. Rh6-a6 e6-e5 -4. Rh8-h6 a5-a4 -5. h7-h8=R a6-a5 -6. g6xRh7. Further retractions are: h3xBg4-g6; Ra8-h7; Bc8-g4; c6xPb7-b8=B-g1.


R0002v - Henrik Juel

The parity of the position establishes that Black moved last, yet black has no possible last move. Removing any unit legalises the position.


R0003 - Gianni Donati

Last moves are: -1. ... c4xb3 ep -2. b2-b4 Rb4-a4 -3. Kd4-c5 Nd6-b5 -4. Kc5-d4.

Black's pawns have made 12 captures. White has four units left. Therefore, black's last move was not a capture, and must have been a non-capturing discovered check. Five discovered checks appear to be possible. Which occurred? White's immediately preceding last move could not have been by the pawn at a6, because black has fifteen pieces and white's pawns at g4 and g5 account for the only capture made by white. Nor could it have been by the white king since b1 and b2 are triple-attacked and the white king would have been in retro-check on those squares. Therefore, white's last move was by the g4 or g5 pawn. The capture made by the g4 or g5 pawn was not of a black pawn, since black's h-file has no capture available for it to reach the g-file. Therefore, one of white's pawns captured a black piece. That capture cannot have been recent, for after the capture (which must have been from the h-file as otherwise black's a-file pawn would not be accounted for), black's h-file pawn promoted to replace whatever piece was captured. Therefore, white's last move was a move, not a capture, by the g4 or g5 pawn. Only the pawn at g4 could have moved last, and to make the move available, g3 could not have been occupied. Therefore, white played g3-g4, and black played Nf1-g3.


R0004 - Paul Raican
Add wNd2 and bKc4. Last move after removing Rb4, Bd5 or Nd2 were Nf1-d2, Rb1-b4 or Bh1-d5.


R0005 - Alessandro Cuppini
Retract Kc5-b6. White's pawn structure requires 11 captures. So black's last move can't have been c4xBb3, because this would require an extra capture on b3. So black's last move was b7-b5. Mate in 1 with ab6ep.


R0007 - Stanilav Vokal
(a)i Black's Rh8 was captured in the northeast prison. So on b3 and g3 white captured two bishops. On c6 black captured the missing white rook. a2xb3 was played after black captured on c6, and this capture is needed to free the a1-rook. So on c6, black captured the white h1-rook, and the rook currently on that square is the rook originally from a1. So white may not castle.
(b) Now the bishop from c8 could get captured on b3 before the black d-pawn captured on c5. So white may still castle.


R0008 - Henrik Juel
Add pawns on a3,b2,c3,d2,e2,f3,g2,h2. The white rook can't have escaped to h3. Removing any pawn will result in a legal position. Note that a3 may not be on a2, since removing the rook will still result in an illegal position, since the king can't have passed the pawn barrier then!


R0009 - Michel Caillaud
The white pawns captured g6xf7, f5xg6 and bxa. White promoted twice, on a8 and h8. The last moves were: -1. .. Qa5-c3 -2. a7-a8=N Rb8-b7 -3. a6-a7. The black rook goes back to e4 via g8, g7, h7, h4, e4. Then the pawn on g2 and the bishop on h1 can be freed. The bishop goes to b5, and the black queen can be freed. The blacksquared white bishop now goes to b4, and the black rook goes to a4 via b8, b5, a5, and a4. The white bishop now moves away from b4, and the black rook goes to c4. The white king now can escape via a4.
So the black rook necessarily visited a4, a5, b5, b7, b8, c4, e4, g7, g8, h4 and h7.


P0001 - Mark Kirtley
1. b3 g6 2. Bb2 Bg7 3. Ba3 Na1 4. Bc1 Bc3 5. dc3 b6 6. Kd2 Ba6 7. Qe1 Bb7 8. Kd3 Bc8 9. Kc4


P0002 - David P. Moulton

1. h4 a5 2. h5 a4 3. Rh4 Ra5 4. Rb4 Rh5 5. d4 d5 6. Qd3 Be6 7. Qe4 de4 8. Bg5 Ba2 9. Bf6 Bd5 10. Ra3 Rh1 11. Rf3 h5 12. Rf5 h4 13. f4 h3 14. Kf2 h2 15. Kg3 Rh3 16. Kg4 Ra3 17. e3 Ra1 18. Be2 a3 19. Bd1 a2

However, this is cooked: 1. h4 a5 2. h5 Ra6 3. h6 Rh6 4. Rh5 d5 5. Rf5 Rh1 6. d4 h5 7. Qd3 h4 8. Qe4 de4 9. Bg5 Be6 10. f4 h3 11. Kf2 Ba2 12. Kg3 Bd5 13. Kg4 h2 14. Ra4 Rh3 15. Rb4 Ra3 16. e3 Ra1 17. Be2 a4 18. Bd1 a3 19. Bf6 a2


P0004 - Olli Heimo
1. a4 b5 2. Ra3 b4 3. Re3 b3 4. a5 bc2 5. b4 h5 6. Ba3 c1=Q 7. b5 Qc6 8. b6 Qg3 9. hg3 c5 10. Bh3 c4 11. Kf1 c3 12. Kg2 c2 13. Kg3 c1=Q 14. Kf4 Qc7 15. Ke4 Qg3 16. Be6 d6 17. Bc4 Be6 18. hg3 Nd7 19. Rh4 Rc8 20. Rg4 h4 21. b7 h3 22. b8=Q h2 23. Qb2 h1=Q 24. Qf6 Qh6 25. Nh3 Qf4 26. gf4 ef6 27. Ng5 Bf5 28. Kd5 Qe7 29. Re7


P0006 - Mario Velucchi
1. Nc3 d5 2. Nd5 g5 3. Nf4 g4 4. Nfh3 g3 5. f4 gh2 6. Kf2 hg1=R 7. Qe1 Rf1 8. Rf1 e6 9. Kg1 e5 10. Kh1 ef4 11. Ng1


P0007 - Michel Caillaud
1. h4 g5 2. hg5 h5 3. g6 Rh6 4. g7 Rf6 5. Rh3 Nh6 6. g8=N Rf4 7. Ne7 f6 8. Nc6 dc6 9. Rd3 Bh3 10. g4 Qd5 11. Bg2 Qf3 12. Rd7 Be7 13. d4 Bd8 14. Bd2
1. h4 e6 2. h5 Qf6 3. h6 Qf3 4. hg7 h5 5. Rh4 Nh6 6. g8=Q f6 7. Qe6 Be7 8. Qc6 dc6 9. Rd4 Bh3 10. Rd7 Rg8 11. d4 Rg4 12. Bd2 Rf4 13. g4 Bd8 14. Bg2


P0008v - Thierry le Gleuher
\ 1. a4 f5 2. a5 Nf6 3. a6 ba6 4. Ra3 Bb7 5. Rg3 Bf3 6. d3 Nc6 7. Bf4 Rb8 8. Bd6 ed6 9. gf3 Qe7 10. Bh3 Qf7 11. Bg4 Be7 12. Bh5 Bd8 13. Bg6 h5 14. Bh7 Qa2 15. Bg8 Rb3 16. Bc4 d5 17. Qd2 Kf7 18. Qb4 Re8 19. Qb8 Re3 20. Bb5 Ne5 21. Bc6 Ne4 22. Bb7


P0009 - Paul Raican
1. d4 e6 2. d5 ed5(=N) 3. h4 Ne3(=P) 4. Rh3(=B) ef2(=N) 5. Bd7(=R) Nd1(=B) 6. Rd1(=Q) Ne7(=P) 7. Nd2(=P)


P0010 - C.C. Frankiss
1. Nf3 h5 2. Ne5 Rh6 3. Nd7 Rc6 4. Nb8 Be6 5. Nc6 Ba2 6. Ne7 Bb1 7. Ng8 Be7 8. Ne7 Qe7


P0011v - Mark Kirtley
1. a4 h5 2. a5 Rh6 3. a6 Ra6 4. Nf3 Ra1 5. Ne5 Rb1 6. Nd7 Nd7 7. g3 Nb6 8. Bg2 Qd5 9. Bh3 Qh1 10. Bf1 Qc6 11. Bg2 Kd7 12. Bh3 Kd6 13. Bf1 Qd7


P0013 - Michel Caillaud
1. Nc3 Nc6 2. Nd5 Nd4 3. Ne7 Ke7 4. Nh3 Kf6 5. Nf4 Kg5 6. Nd5 Qf6 7. Nc3 Qf3 8. ef3 Bd6 9. Bb5 Bg3 10. Bc6 bc6 11. Ne2 Ba6 12. Ng1 Re8 13. Qe2 Re3 14. d3 Ne7 15. Bd2 Ra8 16. OOO


Michel Caillaud
1. g3 f5 2. Bh3 f4 3. Bf5 f3 4. Nh3 fe2 5. Nf4 ed1=N 6. Nd5 Nc3 7. dc3 h6 8. Bg5 hg5 9. c4 Rh4 10. c5 Rf4 11. h4 d6 12. h5 Be6 13. h6 Kd7 14. h7 Kc8 15. h8=Q Nd7 16. Qh5 Nh6 17. Qd1 Bg8 18. f3
1. g3 f5 2. Bh3 f4 3. Bf5 f3 4. Nh3 fe2 5. Rf1 ef1=B 6. f3 Bc4 7. d3 h6 8. Bg5 hg5 9. Nf4 Rh4 10. Nd5 Rf4 11. h4 d6 12. h5 Be6 13. h6 Kd7 14. h7 Kc8 15. h8=R Nd7 16. Rh1 Nh6 17. dc4 Bg8 18. c5


Michel Caillaud
1. e4 d5 2. e5 Bh3 3. Qg4 d4 4. Qc8 d3 5. g4 dc2 6. d4 cb1=B 7. Bg5 Bf5 8. Bf6 gf6 9. ef6 e5 10. OOO e4 11. Kb1 e3 12. Ka1 e2 13. d5 ef1=B 14. Rd4 B1d3 15. Ne2 Bb1 16. Rc1 Bf1 17. Rc7 Nc6 18. Qb8 Bc8

Michel Caillaud
1. d3 a5 2. Be3 a4 3. Bb6 a3 4. e3 ab2 5. a4 and now:
(a) Ra5 6. Ra3 Rf5 7. Rc3 Rf3 8. Na3 b1=R 9. Ke2 Rb5 10. Kf3 Ra5 11. Kg3 Ra8
(b) d6 6. Ra3 Bg4 7. Rc3 Be2 8. Na3 b1=B 9. Ke2 Ba2 10. Kf3 Be6 11. Kg3 Bc8
(c) d5 7. Ra3 Qd6 8. Rc3 Qa3 9. Na3 b1=Q 10. Ke2 Qb5 11. Kf3 Qd7 12. Kg3 Qd8

Michel Caillaud & Thierry le Gleuher
1. e4 f5 2. e5 Nf6 3. ef6 Kf7 4. fg7 Kg6 5. gh8=B Bg7 6. c4 Qg8 7. c5 Qc4 8. Ne2 Qc1 9. Nec3 Qb1 10. Be2 Qa2 11. OO Qf7 12. Ra5 d5 13. Qa4 Be6 14. Ra1 Nd7 15. Nb1 Bc3 16. Bf6 Rh8 17. Bg5 Nf6 18. Kh1 Ng8

Michel Caillaud
1. e4 Nc6 2. Qg4 Na5 3. Qd7 Kd7 4. d4 Kc6 5. Bg5 Kb6 6. Be7 Ne7 7. e5 Nec6 8. e6 Bd6 9. e7 Bg3 10. e8=Q Qh4 11. Qe2 Re8 12. d5 Re6 13. d6 Rh6 14. d7 g6 15. d8=B Be6 16. Bg5 Rd8 17. Bc1 Rd2 18. Qd1 Re2

Michel Caillaud
1. Nc3 e5 2. Ne4 Qh4 3. Ng3 Ke7 4. f3 Kf6 5. f4 Be7 6. f5 Kg5 7. f6 Bd8 8. fg7 Ne7 9. g8=N d6 10. Nf6 Rg8 11. Nd5 Rg6 12. Nc3 Rf6 13. Nb1 Rf2 14. Nh5 Bf5 15. Nf6 Bg6 16. Ng8

Goran Forslund
1. d3 Nf6 2. Bh6 gh6 3. g4 Bg7 4. g5 Ng4 5. g6 Bc3 6. bc3 OO 7. g7 h5 8. gf8=B Kh8 9. Bh8 Qg8 10. Be3 Qg5 11. Ba7 Qc5 12. Bb6 Ra6 13. Ba5 Rg6 14. Bb4 Rg8 15. Ba3 Rf8 16. Bc1 Kg8

Thierry le Gleuher
1. Nc3 Nf6 2. Nd5 Ne4 3. Nf6 gf6 4. h4 Bh6 5. Rh3 Kf8 6. Rb3 Kg8 7. Rb6 Qf8 8. b4 Qg7 9. Bb2 Qg3 10. Be5 Qb3 11. ab3 a6 12. Ra5 Ra7 13. Rab5 a5 14. e3 a4 15. Be2 a3 16. Bh5 a2 17. g4 a1=Q 18. Bg3 Qe5 19. Ke2 Qg5 20. Kf3 Qg7 21. g5 Qf8 22. Kf4 Qd8 23. Qg4 Kf8 24. Qe6 Ke8 25. Kf5 Bf8

Cook: 6....Kg7 7...Qg8 8...Kf8 9....Qg3 ...

Corrected version:

1.Nc3 Nf6 2.Ne4 Nd5 3.Nf6+ gxf6 4.h4 Bh6 5.Rh3 Kf8 6.Rb3 Kg7 7.Rb6 Qg8 8.b4 Kf8 9.Bb2 Qg3 10.Be5 Qb3 11.axb3 a6 12.Ra5 Ra7 13.Rab5 a5 14.e3 a4 15.Be2 a3 16.Bh5 a2 17.Qg4 a1=Q+ 18.Ke2 Qd4 19.Qe6 Qg4+ 20.Kd3 Qg8 21.g4 Kg7 22.g5 Qd8 23.g6 Kf8 24.g7+ Ke8 25.g8=N Bf8 26.Nh6

Michel Caillaud
1. a4 c6 2. Ra3 Qc7 3. Rb3 Qh2 4. Rb6 ab6 5. Nc3 Ra5 6. Ne4 Rf5 7. a5 Qd6 8. a6 Qb4 9. a7 d6 10. a8=R Be6 11. Ra1 Ba2 12. c4 Rf3 13. Nf6
1. c4 c6 2. Qb3 Qc7 3. Qb6 ab6 4. Nc3 Ra3 5. Nd5 Rf3 6. a4 Qh2 7. a5 Qd6 8. a6 Qb4 9. a7 d6 10. a8=Q Bf5 11. Qa4 Bb1 12. Qd1 Ba2 13. Nf6

R0011 - Henrik Juel
wKe8, wPa6, wPb5, wPd5, wPe6, bNc7
wKc8, wPa4, wPc4, wPd5, wPd7, bNb6
wKd6, wPa6, wPb7, wPc6, wPc7, bNa7
wKc6, wPa7, wPb7, wPc5, wPc7, bNa6

R0013 - Andrei Frolkin
The last moves were: -1. .. d5-d4 -2. g3-g4 e6xBd5 -3. Ba2-d5 e7-e6 -4. Qa1-b1 h6-h5 -5. Qg7-a1 h7-h6 -6. Bf7-a2 Ka2-a3 -7. Bh5-f7 and the whole position unlocks. White unpromotes his queen on g8, and uncaptures the missing g-pawn on g4, and then brings his king to a1 and his bishop to b1 to let out the black rook.

P0014 - Joost de Heer
Intention was: 1. g4 d5 2. Bg2 Bg4 3. Bd5 Be2 4. Bb7 Bd1 5. Ne2 Bc2 6. OO Bb1 7. Rd1 Ba2 8. Kf1 Bc4 9. Ra7 Ra7 10. Ke1 Rb7 11. Nc3 Rb2
Unfortunately, this is cooked, as Michel Caillaud pointed out: 1. c4 d5 2. Qb3 Bg4 3. Qb6 Be2 4. Qa7 Bf1 5. Qb7 Bg2 6. Qd5 Bd5 7. Ne2 Ra2 8. Nec3 Ra1 9. Ke2 Rb1 10. Rd1 Rb2 11. Ke1 Bc4

P0015 - C. C. Frankiss
1. h4 d6 2. h5 Qd7 3. h6 Kd8 4. hg7 h5 5. f4 Nh6 6. g8=N h4 7. Nf6 h3 8. Nh5 hg2 9. Rh4 Qh3 10. f5 Qh2 11. Nh3 g1=N 12. Bg2 Nf3 13. Kf1 Ng5 14. Bh1 Nh7

P0016 - John Meyers
1. e3 h6 2. Ba6 ba6 3. a4 Bb7 4. Ra3 Be4 5. Rb3 Nc6 6. Rb8 Bh7 7. Rc8 Rb8 8. a5 Rb6 9. ab6 g6 10. b7 Bg7 11. b8=B Bc3 12. bc3 Nf6 13. Ba3 Rg8 14. Bc5 Rf8 15. Bb6 ab6 16. Ba7

P0017 - Andrei Frolkin
1. Nf3 a5 2. Nd4 a4 3. Nb3 ab3 4. a4 Ra5 5. Ra2 ba2 6. Rg1 ab1=N 7. Rh1 Nc3 8. Rg1 Nd1 9. Rh1 Ne3 10. Rg1 Nf1 11. Rh1 Ng3 12. hg3 Rg5 13. Kf1 e5 14. Kg1 Bc5 15. Kh2 Ne7 16. Rf1 OO 17. Kg1

P0018 - Mikhail Kozulya
1. b4 f5 2. b5 Kf7 3. b6 ab6 4. e4 Ra3 5. e5 Rg3 6. hg3 Ke6 7. Rh6 Kd5 8. Rg6 hg6 9. Qg4 Rh4 10. Qh3 Ra4 11. d4 Ra8 12. a4 Ke4 13. a5 d4 14. a6 Be6 15. a7 Nd7 16. Ra6 Rc8 17. a8=R ba6

P0019 - Michel Caillaud & Noam Livnat
1. c4 b5 2. Qa4 b4 3. Qb5 g6 4. a4 Bg7 5. Ra3 Kf8 6. Rd3 d6 7. Rd6 a6 8. d4 Ra7 9. Bf4 Rb7 10. Be5 Rb6 11. f4 Rc6 12. Kf2 Rc5 13. Kg3 Rd5 14. Kh4 Rd4 15. g3 Re4 16. Bh3 Re3 17. Bg4 Rf3 18. Nh3 Rf2 19. Rf1 Rg2 20. Rf3 Rg1 21. Re3 Rh1 22. Re4

P0020 - Unto Heinonen
1. Nc3 Nc6 2. Nd5 Nd4 3. Ne7 Ke7 4. h4 Kf6 5. Rh3 Bd6 6. Rb3 Bf4 7. Rb6 ab6 8. Nf3 Ra3 9. Nh2 Rg3 10. h5 Kg5 11. a4 Nf6 12. Ra3 Re8 13. Rf3 Re3 14. Ng4 Ra3 15. e3 Qg8 16. Qe2 Nf8 17. Qa6 ba6 18. Nf6 Bb7 19. Nd5 f6 20. Nc3 Be4 21. Ne2 Qd5 22. Ng1

H0085 - Dr. Tomislav Petrovich
Suppose white can castle. The two pawns on h5 and h6 need 5 captures to get there, so white's last move wasn't a capture. Only b2-b4 is possible then. So if white may castle, black may capture en passant.
1. ab3 OOO! 2. a1=B cb3#

0023 - Noam Livnat
(a): Black pawns captured 10 times, white pawns on the board captured 5 times. The missing a- and h-pawn were captured by the black pawns, this requires two more captures. The white a-pawn was captured on b3. Where was the white b-pawn captured? d2xc3 was played after the black pawn arrived on b2, so the white bishop from c1 was constantly trapped there if the b-pawn was captured on b2. But the only captures on black squares were done by the black pawn currently on b2. So the white b-pawn had to make one capture too. But this results in 8 captures by white, and there are only 7 missing pieces. So the position is illegal.
(b): The white b-pawn was now captured on b3, and the white a-pawn captured a6xb7, and promoted on b8. This requires only 7 captures in total, so the position is legal.

R0024 - Alessandro Cuppini
1. e4! (1. h7? OOO!) Bc1 2. h7 Bh6 3. Kd6 Bc1 4. hg8=Q#.
This seems to be cooked though: 3. h8=Q/R Bc1 4. Q/Rg8#.

R0025 - Stanislav Vokal
(a): 1. Rf5! (2. Rf8#) OOO 2. Qb7#
(b): 1. OO! ~ (OOO?) 2. Rf8#. Why may black not castle? White may castle, so Rd5 is promoted. To get a pawn to a promotion square where it wouldn't break black's castling rights, white needs 8 captures (exdxcxb, fxexdxc, and gxfxe). Black misses 8 pieces, but Bf8 was captured on its homesquare. So white didn't promote on b8, so either the pawn had to pass d7 or f7, or the rook had to pass f8, and in both cases, black lost the right to castle.

R0026 - Valery Liskovets
Bf3 and Bb8 are promoted. Suppose black may castle. Then the bishop on f3 was created on g8 with h7xg8=B. This implies 5 captures by white pawns: axbxc, dxe, fxg, and h7xg8. Black misses 7 pieces, and the bishop on f8 was captured on its homesquare. One of the a-, d-, or h-pawns promoted to bishop, and one other was captured by the white pawns. The black h- and d- pawn couldn't have reached g1 without disturbing the white king. The black a-pawn could've reached c1 with two captures. But the captured pawn needs to make a capture too to go to a field where it could've been captured. This requires too many captures. So white may not castle. So castlings are mutually exclusive.
Is the white castling legal? Yes: Bf3 was created on e8 or c8. This requires 5 captures, axbxc, d7xc(e)8=B, fxe and hxg. In this case, the black bishop could've been created on g1 with only one capture. Together with the capture, needed to get one of the other pawns to a square where it could get captured, and Bf1, these are all 3 missing pieces. So white may castle
Solution: 1. OOO? Ne2! 2. Be2/Kb2 bc6/Nd4!. 1. Ng3! OO 2. Ne7 Kh8 3. Ra4! (OOO? illegal) ~ 4. Rh4#. 1. .. bc6 2. OOO! (2. Ra8? OO, 2. Bc6? Kf8 3. OOO/Ra8 Rh1) Rh1/~ (OO? illegal) 3. Rh1/Bc6 (4. Rh8#/Rd8#) 1. .. Rh4! 2. Ra8! (2. OOO Rd4) Rh1 3. Nh1. 1. .. Rh1 2. Nh1 Ba7/~ 3. Ra7/Ra8 (4. Ra8#/Rb8#).

R0027 - Leonid Borodatov
Suppose white may castle. Then the extra rook must've been created by a promotion on c1. This requires 6 captures, together with the missing Bf1 all missing white pieces. So the f-pawn and the h-pawn were captured by a black pawn too. To preserve as many castling rights for black as possible, this occurred by promoting the h-pawn on h8, and fxe. So white's and black's kingside castling are mutually exclusive. So what was white's last move? It could only have been with a7 (d4xc5 isn't possible). This pawn captured the missing black bishop, so it must've played b6xBa7 last. The only retromove allowing that is b7-b5. So if white may castle, black played b7-b5 as last move. h#2.5 with 1. .. cb6 2. Ba6 OO! 3. OOO a8=Q

R0028 - Andrei Kornilov & Andrei Frolkin
Retroplay is: -1. Ba5-c3 d5-d4 -2. Bd8-a5 d6-d5 -3. c7xNd8=B Nc6xQd8! -4. Qe8-d8 Nb4-c6 -5. c6-c7 c7xNd6 -6. Qd8-e8 N~ -7. Ne8-d6 N~ -8. Ng7-e8 N~ -9. Ne8xRg7 Rh7-g7 -10. Ng7-e8 N~ -11. Qh8-g8 N~ -12. Kg8-f8 N~ -13. Qd8-f8 Re8-e7 -14. c5-c6 Rd8-e8 -15. Ne8-g7, and further retracting involves Qf8->a8; a2xb->b6xRa7-a8-Q; h7->h3xNg2-g1=R.

P0029 - Pascal Wassong
1. g4 Nf6 2. g5 Ne4 3. g6 Nd2 4. gh7 Nb3 5. Bh6 g5 6. e3 g4 7. Be2 g3 8. Bh5 g2 9. Ne2 g1=N 10. ab3 Nh3 11. Ra4 Ng5 12. Rh4 Ne6 13. b4 Ng7 14. hg7

P0030 - Thierry le Gleuher
1. d4 g5 2. Bf4 g4 3. Kd2 g3 4. hg3 Nc6 5. Rh6 Ne5 6. Rb6 Nf3 7. ef3 ab6 8. Bc4 Ra3 9. Qf1 Rc3 10. Na3 h5 11. Re1 h4 12. Re6 h3 13. Rc6 dc6 14. Ke2 Bf5 15. Be6 Qa8 16. Bc8 e6 17. d5 Ne7 18. d6 Ng6 19. d7 Ke7 20. d8=R Rh4 21. Rd1 Nh8 22. Ra1 Kd8 23. Nb1 Ba3 24. Bc1 Rb4 25. Bd7 Be4 26. Qd1

P0031 - Thierry le Gleuher
1. a4 h5 2. Ra3 h4 3. Rg3 hg3 4. h3 Rh4 5. Rh2 gh2 6. Nc3 h1=N 7. Nd5 Ng3 8. Ne7 Nf1 9. Nd5 Nh2 10. Nc7 Qc7 11. Nf3 Nf3 12. Kf1 Qh2 13. Qe1 Bd6 14. Qd1 Bc7 15. Qe1 d6 16. Qd1 Bd7 17. Qe1 Bb5 18. Qd1 Nd7 19. Qe1 Rc8 20. Qd1 Bb8 21. Qe1 Rc5 22. Qd1 Rg5 23. Qe1 f5 24. Qd1 Kf7 25. Qe1 Kg6 26. Qd1 Kh5 27. Qe1 g6 28. Qd1 Ne1

P0032 - Michel Caillaud
1. h4 a5 2. h5 a4 3. h6 a3 4. Rh5 ab2 5. a4 b5 6. a5 b4 7. a6 b3 8. Rha5 f5 9. hg7 h5 10. a7 h4 11. ab8=R h3 12. Rb7 Rh4 13. Rba7 Ra4 14. c4 Nh6 15. Nc3 b1=R 16. g8=R Rb2 17. Rg6 Rba2 18. Rga6 b2 19. Qb3 b1=R 20. Qb8 Rb3 21. Bb2 Rba3

P0033 - Michel Caillaud
1. c4 Nf6 2. c5 Nd5 3. c6 Nb4 4. cb7 c5 5. d4 c4 6. d5 c3 7. d6 c2 8. de7 d5 9. ed8=N Bd6 10. bc8=N Be5 11. Nb6 Ke7 12. Nb7 Rc8 13. Nf3 Rc3 14. Nd4 Rh3 15. Nb5 d4 16. Be3 d3 17. Bd4 c1=N 18. e3 Nb3 19. Qc1 d2 20. Ke2 d1=N 21. Rg1 Nb2

P0034 - Michel Caillaud
1. d4 c5 2. d5 c4 3. d6 c3 4. de7 cb2 5. c4 d5 6. c5 d4 7. c6 d3 8. c7 de2 9. cb8=B ed1=B 10. Bc4 Bc2 11. Ke2 ba1=B 12. Kf3 Bc3 13. Kg3 Qd3 14. Kh4 Kd7 15. e8=B Kd8 16. Bc6 Bc5 17. Bc7

P0035 - Michel Caillaud
1. d4 a5 2. d5 a4 3. d6 a3 4. dc7 ab2 5. a4 h5 6. a5 h4 7. a6 h3 8. a7 hg2 9. ab8=Q gh1=Q 10. Bh3 Rh4 11. Be6 de6 12. Ra7 Bd7 13. c8=Q Bc6 14. Qbd6 Bg2 15. Qc3 Bf1 16. Nh3 Qd5 17. f3 Re4 18. Nf4 g5 19. h4 g4 20. h5 g3 21. h6 g2 22. h7 g1=Q 23. Ng2 Qgd4 24. Be3 Bh6 25. Bg1 Bc1 26. Qcd2 Qd3 27. Nc3 b1=Q 28. h8=Q Qbb5 29. Qd4 Qbd7

P0036 - Satoshi Hashimoto
1. h4 a5 2. h5 Ra6 3. h6 gh6 4. Nc3 Bg7 5. Na4 Bc3 6. dc3 Re6 7. Be3 Re4 8. Bb6 cb6 9. e3 Qc7 10. Qf3 Kd8 11. OOO Qd6 12. Ne2 Kc7 13. Ng3 Kc6 14. Bb5 Kb5 15. Rd4 Nc6 16. Rc4 Ne5 17. Rc8 Rc4 18. Ra8 Rc8 19. Ra6 Ra8

P0037 - Mikhail Kozulya
1. a4 f5 2. Ra3 f4 3. Rc3 f3 4. Rc7 fg2 5. f4 a6 6. Nf3 g1=B 7. Bh3 Ba7 8. Rg1 Nf6 9. Rg6 Nd5 10. Rd6 Nb6 11. Kf2 g5 12. Ke3 g4 13. Kd4 Bg7 14. Kc5 Bc3 15. Nd4 Ba5 16. c3 g3 17. Qb3 Nd5 18. Qb6 Nf6 19. b4 Ng8 20. Ba3 B7xb6

R0030 - Andrei Frolkin
(a): No (b): The captures were: a4xBb5, b6xNc7; a6xQb5, e6xBd5, h4xBg3. (wQb1 was born on h8)
The last moves were: -1. ... d5-d4 -2. g4-g5 e6xBd5 -3. Ba2-d5 f5-f4 -4. Qa1-b1 f6-f5 -5. Qe5-a1 f7-f6 -6. Bd5-a2 Ka2-a3 -7. Be4-d5 etc. Later in the retroplay wB screens, either on b1 for wKa1 or on a2 for bKa1.

R0031 - A. A. Kislyak
The last moves were: -1. ... Sh7-f8# -2. a7xNb8=R Na6-b8 -3. h4-h5 Nb4-a6 -4. h3-h4 Nd3-b4 -5. h2-h3 Nb2-d3 -6. a6-a7 Nc4-b2 -7. a5-a6 Nd6-c4 -8. a4-a5 Nc4xNd6 -9. Ne8-d6 Kf8-f7 -10. a3-a4 Bf7-g8 -11. a2-a3 Kg8-f8 -12. Nd6xQe8 Qf8-e8 -13. Qe8-d8 etc.

R0032 - Thomas Volet
The last moves were: -1. Rf3-f2 Rd1-b1 -2. Bg1-h2 Rd4-d1 -3. Bf2-g1 Rb4-d4 -4. Be1-f2 Q~ -5. Bc3-e1 Q~ -6. Be5-c3 Rd4-b4 -7. R~ Rd6-d4 -8. R~ Rb6-d6 -9. Bc7-e5 Qc8-a8 -10. Be5-c7 Qd7-c8 -11. R~ Rd6-b6 -12. R~ Rd4-d6 -13. R~ Rb4-d4 -14. Bc3-e5 Qd1-d7 -15. R~ Qc1-d1 -16. Be5-c3 c2-c1=Q -17. R~ c3-c2 etc.
The white bishop needs to screen the white king so the black queen can unpromote on c1. But for this, the black rook needs to screen the black king from two sides too!

P0039 - Michel Caillaud & Mark Kirtley
1. g3 h5 2. g4 Rh6 3. gh5 Rc6 4. h6 d5 5. h7 Bg4 6. hg8=N e6 7. Nf6 Qf6
1. g4 e5 2. g5 Bg4 3. g6 e6 4. gh7 Qf6 5. hg8=N Rh3 6. Ne7 Rc3 7. Nc6 Rc6

P0040 - Michel Caillaud & Mark Kirtley
1. g3 e5 2. g4 Be7 3. g5 Bf6 4. gf6 h5 5. fg7 Ke7 6. gh8=N Ke6 7. Ng6 fg6
1. g4 h5 2. g5 Rh6 3. gh6 e5 4. hg7 Ke7 5. gf8=Q Kf6 6. Qg7 Ke6 7. Qg6 fg6

P0041 - Satoshi Hashimoto
1. e4 d5 2. Qh5 Nd7 3. Qh6 gh6 4. e5 Bg7 5. e6 Kf8 6. ed7 Qe8 7. d8=Q Bh3 8. Qd7 Rd8 9. Qg4 Rd6 10. Qd1 Bc8 11. h4 Qd8 12. Rh3 Ke8 13. Rg3 Bf8

P0042 - Satoshi Hashimoto
1. b4 d5 2. b5 Bf5 3. b6 ab6 4. Nf3 Ra3 5. Ne5 Rf3 6. Nd7 e5 7. ef3 Ne7 8. Bb5 Ng6 9. Bc6 bc6 10. Nc3 c5 11. Ne2 Nc6 12. Ng1 Qa8 13. Nf6 Kd8 14. Ne4 Kc8 15. Nc3 Kb8 16. Nb1

P0044 - Unto Heinonen
1. d4 a5 2. Bg5 a4 3. e3 a3 4. Bb5 ab2 5. a4 Ra6 6. a5 Rf6 7. a6 h5 8. a7 h4 9. ab8=B h3 10. Ra8 hg2 11. Na3 b1=B 12. Ba7 Ba2 13. Bb6 Be6 14. Ba5 b6 15. h4 Bb7 16. h5 Bf3 17. h6 Bh5 18. h7 Nh6 19. Rh4 Rg8 20. Nh3 g1=B 21. hg8=B Bh2 22. Bh7 Be5 23. Be4 Bf5 24. Bd5 e6 25. Re4 Bc5

R0033 - Pal Benko
The mate is legal, and the last six moves were: -1. c5xd6 ep d7-d5 -2. Kf3xPg3 f4xg3 ep -3. g2-g4 Bg6-h5.

R0034 - Pal Benko
Retract Kd2xNd1, and h#1 with 1. Kc3 Qb4#
Retract d3xRe2, and h#1 with 1. Kd1 Qf1#
Retract d3xBe2, and h#1 with 1. d2 Qf1#

R0036 - Klaus Wenda
Retract -1. Na7xPb5 -2. Nc8xPa7 -3. Nd6-c8 -4. Gd5xPd7. Now black's only last move could've been b7-b5. So s#2 with 1. ab6 ep Ka5 2. Ga6 ab6#.

P0045 - Satoshi Hashimoto
1. d4 g6 2. d5 Bh6 3. Qd4 Kg8 4. Qf6 ef6 5. d6 Qe7 6. de7 Kg7 7. e8=Q Be3 8. Qd7 Bc5 9. Qd1 Bf8

P0046 - Satoshi Hashimoto
1. Nf3 d5 2. Ne5 Bg4 3. Nd7 Bf3 4. ef3 e5 5. Bb5 Qf6 6. Bc6 bc6 7. Nb6 c5 8. Na4 Qb6 9. Nac3 Nc6 10. Ne2 OOO 11. Ng1

P0047 - Satoshi Hashimoto
1. f3 Nc6 2. Kf2 Ne5 3. Ke3 Ng4 4. Kd3 e5 5. Nh3 Qh4 6. Nf2 Qf2 7. h4 Qf1 8. h5 Qf2 9. Qg1 Qb6 10. Qc5 Qb3 11. cb3 a5 12. Kc2 a4 13. Kd1 a3 14. Qc2 Bc5 15. Ke1 Bb6 16. Qd1

P0048 - Michel Caillaud
1. e3 b6 2. Qf3 Ba6 3. Qb7 Nc6 4. Bb5 Nb4 5. Bc6 dc6 6. Nf3 Qd7 7. Ne5 Rd8 8. f3 Qc8 9. Kf2 Rd3 10. Kg3 Rb3 11. ab3 Bd3 12. Ra4 Na6 13. Rh4 Bf5 14. d4 Qd8 15. Nd2 Bc8 16. Nd7 Nb8

P0049 - Noam Livnat
1. h4 a5 2. Rh3 Ra6 3. Ra3 Rh6 4. Ra5 Rh4 5. a4 h5 6. Ra3 Rh6 7. Re3 Rd6 8. Ree5 Rdd4 9. e4 d5 10. Qe2 Qd7 11. Qa6 Qh3 12. Qa8 Qh1 13. Ba6 Bh3 14. d3 e6 15. Bf4 Bc5 16. Kd2 Ke7 17. Ne2 Nd7 18. Nec3 Ndf6 19. Nb5 Ng4 20. Na7 Nh2 21. g4 b5 22. Bg3 Bb6 23. f4 c5
However, Israel Tzur gave the following cook: 1. h4 a5 2. Rh3 Ra6 3. Ra3 Rh6 4. Ra5 Rh4 5. a4 h5 6. Ra3 Rh6 7. Re3 Rd6 8. Ree5 Rdd4 9. e4 d5 10. Qe2 Bh3 11. Qa6 e6 12. Qa8 Qf6 13. Ba6 Qf3 14. Ne2 Bc5 15. Nec3 Ke7 16. Nb5 Nd7 17. Na7 Ndf6 18. d3 Ng4 19. Bf4 Nh2 20. g4 Qh1+ 21. Ke2 b5 22. Bg3 Bb6 23. f4 c5

R0038 - Noam Livnat
Consider (a) the blocked structure on the 7th and 8th ranks and (b) the last 6 moves:

(a) The block:

(b) The last 6 moves:

So the bRa8 was the last piece to visit b8 before the bK.

R0039 - Paul Raican

Try Solution
[8/1p5P/b2p1PP1/P2RP3/k2P4/2P4N/1Pp5/1N3K2] [8/7P/5PP1/PN2Pp2/k2PN3/1RPpK3/1Pp5/2b5]

In the try: Last moves were: -wP: d3xPc2, -Nh3: d3xNc2, -Nb1: d3xBc2 Bb1-c2, -bPd6: Kb5-a4 Rd8-d5. But the position is legal! Last moves were Kb5xNa4(+wNb1) Nc5-a4.

P0050 - Michel Caillaud
1. h4 a5 2. h5 a4 3. h6 a3 4. hg7 ab2 5. a4 h5 6. Ra3 Rh6 7. Rf3 Rc6 8. Na3 Nh6 9. g8=R b1=R 10. Rg4 Rb5 11. Re4 Rd5 12. g4 b5 13. g5 b4 14. g6 b3 15. g7 b2 16. g8=R b1=R 17. Rg2 Rb7

P0051 - Mark Kirtley
1. e3 d5 2. Be2 Bh3 3. Bh5 Qd7 4. Qg4 Qa4 5. Nf3 Nc6 6. OO Rd8 7. Kh1 Rd6 8. Rg1 Rf6 9. Qe6 b6 10. g4 Bf1 11. Rg3 Ba6 12. Rh3 Bc8

P0052 - Mark Kirtley
1. h4 Nf6 2. h5 Ne4 3. Rh4 Nd2 4. Rg4 Ne4 5. Qd6 Nc6 6. Qh2 d6 7. Bf4 d5 8. e3 d4 9. Bb5 d3 10. Nf3 d2 11. Ke2 d1=R 12. Nd4 Rd3 13. Nd2 Rc3 14. Rh1 Rc5 15. Qg1 Rd5 16. Qd1 Rd7 17. Bd6 Nb8 18. Bd7
However, this move order isn't unique: 1. h4 Nf6 2. h5 Ne4 3. Rh4 Nd2 4. Nf3 Ne4 5. Qd6 Nc6 6. Qh2 d6 7. Bf4 d5 8. e3 d4 9. Bb5 d3 10. Rg4 and the rest goes as in the intention.

P0053 - Satoshi Hashimoto
1. Nc3 e5 2. Nd5 e4 3. Nf6 Qf6 4. Nh3 Qf3 5. Nf4 f5 6. Nd5 Kf7 7. Nc3 Kg6 8. Nb1 Kh5

R0040 - Tivadar Kardos & ?#8364;rpàd Molnàr
Black has no legal last move, so he has the move. Mate in 1 with 0. .. Bf4/Be7/Nf6/Qf8/Qg7/Qh5/Qg6/Qh7/ab6/ba6/bc6/cb6/Nb6 1. Nf6/Re7/Nf6/ef8=[QR]/Ng7/Ng7/Bg6/gh7/Ra8/Qb8/Qc6/Nd6/Nc7#

R0041 - A.A. Kislyak
The last moves were: -1. .. b6xNa5 -2. Nc4-a5 e6-e5 -3. Ne5-c4 f7xBe6 -4. Ra5-a6 Kc8-b8 -5. a6xRb7, and the position unlocks. White captured e2xd3xc4xb5xPa6. The captures on d3, c4, and b5 aren't uniquely determined, but since the black a-pawn was captured on his own file, the capture on a6 must be of this pawn. This explains too why white must uncapture a rook on b7: If white uncaptures it on d3, c4 or b5 it can't get back to a8 or h8.
So the last four captures were: b6xNa5, f7xBe6, a6xRb7 and b5xPa6.

R0042 - ?#8364;rpàd Molnàr
The last moves were: -1. .. h4xNg3 -2. Rf6-g6 g6xNh5 -3. Rf8-f6 f7xNg6 -4. Bf6-g5 g5xNh4 -5. Nh8-g6 g6-g5 -6. Nf3-h4 Kh4-g4 -7. Ng5-f3. The four promoted knights all originated on d8.

R0043 - Nikita Plaksin
The intention was: -1. .. e2xQd1=B(+bBc8) -2. Qh1xRg2(+wQd1). But this is cooked as Michel Caillaud showed: -1. .. e2xQd1(+bBc8) -2. Qh6xBf8(+wQd1), or -1. .. Qa1-a4 -2. Kb4-a5 c4xb3(+bPb7!)

P0054 - Andrei Frolkin & Valeri Gorbunov
1. a4 Nc6 2. a5 Ne5 3. a6 Ng4 4. ab7 Nh2 5. b8=N Nf1 6. Nc6 dc6 7. g3 Bh3 8. Rh3
1. a4 Nc6 2. a5 Na5 3. g3 Nc4 4. Bh3 Ne3 5. Bd7 Nf1 6. Bc6 bc6 7. h3 Bh3 8. Rh3

P0055 - Andrei Frolkin & Mikhail Kozulya
1. e3 h5 2. Qg4 hg4 3. Ne2 Rh3 4. Rg1 Rf3 5. gf3 e5 6. Rg2 e4 7. fe4 d5 8. ed5 Nc6 9. dc6 Bb4 10. cb7 Bc3 11. bc3 Qd4 12. cd4 Bf5 13. Nc3 Kd7 14. Rb1 Re8 15. Rb2 Re5 16. de5 Nf6 17. ef6 Bh7 18. fg7

P0056 - Michel Caillaud
1. a4 e5 2. a5 e4 3. Ra4 e3 4. Re3 Ne7 5. b4 Nc6 6. Bb2 Nd4 7. Qc1 Ne2 8. Bf6 Ng3 9. d4 Ngf5 10. Ke2 Nh6 11. Kf3 Nhg8 12. Kg4 e2 13. Qh6 e1=N 14. Ne2 Nd3 15. Nc1 Nc5 16. Bd3 Na6 17. Rhe1 Nb8 18. R1e3 a6 19. Re7

P0057 - Mark Kirtley
1. h4 Nf6 2. Rh3 Ne4 3. Rc3 Nd2 4. Nf3 Ne4 5. Qd6 Nc6 6. Qa3 d6 7. Bf4 d5 8. e3 d4 9. Bb5 d3 10. Be5 d2 11. Ke2 d1=R 12. Nbd2 Rh1 13. Rd1 Rh3 14. Nb1 Rg3 15. Nfd2 Rg6 16. Kf3 Re6 17. Kg4 Rd6 18. Kh5 Rd7 19. g4 Nb8 20. Bd7

P0058 - Thierry le Gleuher
1. Nc3 h5 2. Ne4 h4 3. Ng3 hg3 4. Nf3 gh2 5. Rg1 hg1=B 6. c3 Bh2 7. Qa4 Bf4 8. Qa7 Be3 9. Qd4 Ra4 10. Qg4 Rc4 11. a4 Ba7 12. a5 b6 13. a6 Bb7 14. Ra5 Be4 15. Rh5 Nc6 16. Rh1 Bb8 17. a7 Bh7 18. a8=R g6 19. Ra3 Bh6 20. Rb3 Be3 21. de3 Kf8 22. Nd2 Kg7 23. Nb1

P0059 - Unto Heinonen
The intention was: 1. a4 c5 2. a5 c4 3. a6 c3 4. ab7 a5 5. g4 Na6 6. b8=Q cb2 7. Qe5 d6 8. c4 Be6 9. c5 Bb3 10. c6 e6 11. c7 Ne7 12. c8=Q Rg8 13. Qcc3 Nc6 14. g5 Qf6 15. g6 OOO 16. gh7 g5 17. e3 g4 18. Bb5 g3 19. Ne2 g2 20. Rf1 g1=Q 21. Ra4 Qgg7 22. Rg4 a4 23. d4 a3 24. d5 a2 25. Qdd4 a1=Q 26. Nd2 b1=Q 27. h8=Q Qbb2
This problem is cooked: 1. g4 c5 2. g5 c4 3. g6 c3 4. a4 cb2 5. gh7 g5 6. a5 g4 7. a6 g3 8. ab7 a5 9. c4 Na6 10. b8=Q g2 11. Qe5 d6 12. c5 Be6 13. c6 Bb3 14. c7 e6 15. c8=Q Ne7 16. Qcc3 Nc6 17. e3 Qf6 18. Bb5 OOO 19. Ne2 Rg8 20. Rf1 g1=Q 21. Ra4 Qgg7 22. Rg4 a4 23. d4 a3 24. d5 a2 25. Qdd4 a1=Q 26. Nd2 b1=Q 27. h8=Q Qbb2, with many possibilities to interchange two moves (for instance white's fourth and fifth)

H0313 - Dr. Tomislav Petrovic
If black has the move in the diagram position, what was white's last move? Because white has 8 pawns, Bh6 is the original bishop from c1, so the last move wasn't d2-d4. It can't have been c3xd4 either, because this would imply 4 white captures, one too many. So white's last move was f2-f4.
If white has the move in the diagram position, what was black's last move? His pawn structure requires 4 captures, and none of them could've been on the last move. So the only possible last move is b7-b5.
h#1 with 1. gf3 Qf3 and duplex with 1. ab6 ab6#.

R0044 - ?#8364;rpàd Molnàr
The last moves were: -1. .. f4xRg3 -2. Rh3-g3 g5xRf4 -3. Rf3-f4 g6-g5 -4. Bb8-c7 Bc7-d8. (White captures: a6xb7=B; bxc->c8R; White promotions: b8B, c8R, c8R; White capture: hxg->g5xh6->h7; Black captures g6->g5xf4->f4xg3; c7xd6)

R0045 - Joshua Green
If the position is as it "appears", then the bBd8 must be promoted. Black has only made one capture, so the only place where the bB could have been created is on g1. On g1, however, the bB could not have escaped. Therefore, the position is illegal unless Black is really playing up the board and the bK is on h1!
Now, assume that the board is set up correctly. If it is White's turn, then Black must have just moved. Neither the bR nor the bB could have just moved, so Black must have just moved a pawn. Any pawn move must have been a capture. The only possibilities are axb2, fxe5, and fxg2. The answer can be found by examining the pawn structure around the bK. With White playing up the board, this is an ordinary structure. But with Black playing up the board (and having only made one capture), this structure is difficult to arrange. In fact, the only way to arrange this structure is for Black to have captured White's h-pawn and then to have run his own pawn to h2 so that White could capture gxh3 and Black could move his g-pawn to g2. Could the White h-pawn have captured off of the h-file and promoted, to be captured elsewhere? White has made seven captures and at least one is required to play gxh3. If the White h-pawn captured off the file, then that leaves White with five captures and Black with one capture to arrange the pawn structure to the left of the f-file. If Black has just captured, then this is impossible. Similarly, if the White h-pawn was captured on the h-file, then the position is legal, but Black cannot have just captured. It is clear that Black cannot have just played a capture and, since he cannot have played any other moves, Black cannot have just moved. Therefore, it is Black's turn, and all of Black's moves allow immediate mate:
1. ... Rf1/Bf2/Bg3 2. Rf1/Nf2/Ng3#

R0047 - George P. Sphicas
Tries in 11: 1. Nc2 2. Nb4 3. Ka5 4. Ba7 5. Kb6 6. a5 7. a4 8. Ka5 9. Qe6 10. Qa6 11. Bb6 ab4#, or 1. Nc2 2. Nb4 3. Bc6 4. Kb5 5. Kc5 6. Bb5 7. Qe6 8. Qc6 9. Rd5 10. Rf6 11. Rfd6 ab4#
In the diagram position, b2xa3 is illegal: it would require 7 captures, and there are only 6 missing black pieces. The four legal last moves in the diagram position are a2-a3, b2-b4, Kb2-c3 and g4xh5. Three of those are eliminated by the first two moves:
1. Qg4 2. Ra2. Now the last move could've been Kc2-c3 or Kd2-c3, before that was b2xa1=N. 3. Rdd2. But now only b2-b4 could've been the last move. 4. cb3 ep The last move was b2xa3. 5. Ka5 5. b2? would leave white without a last move. 6. b2 7. b1=B 8. Bc2 9. Ba4 10. Qb4 ab4#.

P0060 - Satoshi Hashimoto
1. b4 g5 2. Bb2 Bh6 3. Bf6 ef6 4. h4 f5 5. Rh3 Nf6 6. Rc3 OO 7. Rc6 dc6 8. e3 Qd6 9. Bb5 cb5 10. c3 c5 11. Nd5 Nc6 12. Nc7 Rb8 13. Na6 ba6

P0061 - Mark Kirtley
1. 1. e4 a5 2. Bc4 a4 3. Bb3 ab3 4. f4 ba2 5. f5 and now:
I: 5. .. ab1N 6. Ra6 Nd2 7. Rg6 e6 8. f6 Be7 9. fe7 Nf6 10. h3 Rf8 11. ef8B Nb3 12. Be7 Nc1 13. Bd8 Kd8 14. Qc1
II: 5. .. ab1=Q 6. Ra6 Qc1 7. Rg6 e6 8. f6 Be7 9. fe7 Nf6 10. h3 Rf8 11. ef8=R Ke7 12. Rd8 Qd2+ 13. Qd2 Kd8 14. Qc1

P0062 - Thierry Le Gleuher
1. Nc3 h5 2. Ne4 h4 3. Ng3 hg3 4. Nf3 gh2 5. Rg1 hg1B 6. d3 Bh2 7. Qd2 Bd6 8. Qf4 Bc5 9. Nd2 d6 10. Qf7+ Kd7 11. Qf8 Kc6 12. Qf6 Be6 13. Qc3 Bb3 14. ab3 Nd7 15. Ra4 Nf6 16. Rh4 Qd7 17. Rh1 Rh4 18. Nb1 Nh6 19. Bf4 Rh8 20. Kd2 Nfg8

P0063 - Michel Caillaud
1. f4 Nf6 2. f5 Nd5 3. f6 Nc6 4. fe7 f5 5. ed8=N Bd6 6. a4 Bg3 7. hg3 f4 8. Rh5 f3 9. Rg5 h5 10. a5 h4 11. a6 h3 12. ab7 h2 13. ba8=N h1=N 14. Nb7 Kf7 15. c4 Re8 16. Nc3 Re3 17. Ne4 Rc3 18. e3 f2 19. Ke2 Rc2 20. Qe1 fe1=N 21. Nf3 Ng2

P0064 - Michel Caillaud
1. g4 c5 2. g5 c4 3. g6 c3 4. gh7 g5 5. h4 g4 6. h5 g3 7. Rh4 g2 8. Nh3 g1=B 9. f4 Bd4 10. e3 cb2 11. Ba6 ba6 12. d3 Bb7 13. Nd2 Be4 14. Nf1 Bg6 15. f5 ba1=B 16. fg6 f5 17. g7 f4 18. gh8=B f3 19. Be5 f2 20. Ke2 Kf7 21. Qe1 fe1=B 22. h8=B Bec3 23. Bhf6 Bg7 24. h6 Qf8 25. h7 Nh6 26. h8=B Kg8 27. Bb2

R0048 - Joost de Heer
Add white king on a1, and pawns on a2, b2, c2, d2, e2, f2, g2 and h2. The white king couldn't have gotten past the queen unless one of the pawns is removed.

R0049 - Andrei Frolkin & Andrei Kornilov
Last moves were: -1. Qg8-g7 Ba7-b8 -2. c6-c7 Bb8-a7 -3. c5-c6 Be5-b8 -4. c4-c5 Ba1-e5 -5. c3-c4 a2-a1=B -6. c2-c3 a3-a2 -7. b4-b5 a4-a3 -8. b3-b4 a5-a4 -9. b2-b3 a7-a5 -10. a6xNb7! (-10. a6xQb7? Qd5-b7 -11. a5-a6 Qd1-d5 -12. a4-a5 d2-d1=Q -13. a3-a4 d3-d2 -14. a2-a3, and white needs this tempo, as shall be obvious from the solution) Nd6-b7 -11. a5-a6 Ne8-d6 -12. a4-a5 Ng7-e8 -13. a3-a4 Ne8xBg7 -14. Bh8-g7 Nd6-e8 -15. h7-h8=B Ne4-d6 -16. Qh8-g8 Nc3-e4 -17. Kg8-f8 Nd1-c3 -18. Kf8-g8 d2-d1=N -19. Kg8-f8 d3-d2 -20. Kf8-g8 c4xBd3 -21. Bf1-d3 c5-c4 -22. e2xNf3 Ng1-f3 -23. Rd3-g3 Nf3-g1 -24. Kg8-f8 Ng1-f3 -25. Rd6-d3 Nf3-g1 -26. Rc6-d6 Ng1-f3 -27. Rc8-c6 Nf3-g1 -28. Rf8-c8 Re8-e7 -29. a2-a3 and the position unlocks. So at least 56 single moves have passed since the first move of the white pawn!

R0050 - ?#8364;rpàd Molnàr
Last moves were: -1. .. d4xQc3 -2. Qb6-c5 c5xQd4 -3. Qb8-b6 b6xQc5 -4. Qf3-c3 b7-b6 -5. Re3-b3 Kb3-a4 -6. Nd3-c1.

R0051 - Gianni Donati
Intention: White is in the middle of castling. Since this move would result in mate, black can't resign under the U.S. rules.
But this is cooked: Black can have the move too, and the last moves were -1. Nf1-g3 h7-h6 -2. g3xQh4.

P0065 - Mark Kirtley
This problem is cooked: 1. a4 d6 2. Ra3 Bd7 3. Rh3 Ba4 4. b3 Bb3 5. e3 Bc4 6. Qf3 f6 7. Ne2 Be2 8. Qf6 Bf1 9. Qd6 Nd7 10. Qc7 Nc5 11. Qc5 Rc8 12. Qc8 Qc8 13. Kf1 a5

P0066 - Jorge Lois, Julio Pancaldo and Jorge Kapros
1. a4 Nc6 2. a5 Rb8 3. a6 ba6 4. c3 Bb7 5. Qb3 Qc8 6. Qf7 Kd8 7. Qa2 Ke8 8. Kd1 Qd8 9. Kc2 Bc8 10. Kd3 Rb2 11. Ke3 Rb8 12. Ba3 Ra8 13. Bb4 Nb8

P0067 - Joost de Heer
1. a4 h5 2. a5 h4 3. a6 h3 4. ab7 hg2 5. h4 Rh6 6. h5 Ra6 7. h6 d6 8. h7 Nd7 and now:

P0068 - Mario Parinello
1. g4 c5 2. g5 Qb6 3. g6 Qb3 4. ab3 Nh6 5. Ra6 Nf5 6. Rf6 gf6 7. g7 a5 8. g8=R a4 9. Rg4 a3 10. Ra4 a2 11. c4 a1=Q 12. Qc2 Qa3 13. Qc3 Qb4 14. Ra1 Qb6 15. Qa5 Qd8

P0069 - Theodore Hwa
1. e3 Nc6 2. Bd3 Nd4 3. Bg6 hg6 4. h4 Rh5 5. Rh3 Ra5 6. Rf3 e5 7. Rf6 Ba3 8. Rd6 cd6 9. ba3 Qb6 10. Bb2 Qb3 11. Bc3 b5 12. Bb4 Bb7 13. Bc5 OOO 14. Bb6 Kb8 15. Bc7 Ka8 16. Bb8 Rb8

P0070 - Paul Raican
1. a4 h5 2. a5 h4 3. a6 h3 4. ab7 a5 5. d4 a4 6. Bg5 a3 7. e3 a2 8. Qf3 ab1=B 9. Qf6 ef6 10. Be2 Bd6 11. Bg4 Bf4 12. Nf3 d6 13. O-O hg2 14. h4 Bf5 15. h5 Kd7 16. h6 Kc6 17. h7 Nd7 18. hg8=R Rh3 19. Rh8 Rg3 20. Rh1 Qh8 21. b8=R Qh5 22. Rh8 Rg8 23. Ra8 Ba2 24. Ra1

P0071 - Unto Heinonen
1. c4 Nc6 2. c5 Nd4 3. c6 Nf6 4. cb7 c5 5. b8=N c4 6. Nc6 c3 7. Ne7 c2 8. Nf5 Bd6 9. h4 Bb8 10. h5 d6 11. h6 Be6 12. hg7 Ba2 13. gh8=N Be6 14. Ra6 Kf8 15. Rc6 a5 16. Nf3 a4 17. Ne5 a3 18. f3 a2 19. Kf2 Ra3 20. Kg1 Rf3 21. b3 Qa5 22. Ba3 c1=N 23. Ng7 Nd3 24. Qc1 Nb2 25. d3 Qe1 26. Nc3 a1=N

P0072 - Unto Heinonen
1. c4 a5 2. c5 a4 3. c6 a3 4. cb7 ab2 5. a4 d5 6. a5 d4 7. a6 Qd5 8. a7 Nd7 9. b8=Q g5 10. Qb7 Rb8 11. a8=Q g4 12. Ra7 g3 13. Na3 gh2 14. g4 Bg7 15. g5 Be5 16. g6 Ngf6 17. g7 Rf8 18. g8=Q b1=Q 19. Qg2 Qbe4 20. d3 c5 21. Bg5 c4 22. e3 c3 23. Qdf3 c2 24. Be2 c1=Q+ 25. Bd1 Qcc6 26. Ne2 Nc5 27. Rf1 h1=Q

P0073 - Noam Livnat
1. Nc3 Nf6 2. Nd5 Ne4 3. Ne7 Rg8 4. Ng8 Nc3 5. Rb1 Nb1 6. Ne7 Nd2 7. Nd5 Ne4 8. Nc3 Nf6 9. f3 d5 10. Nb1 Bh3 11. Kf2 Ng8 12. Kg3 Qc8

P0074 - ?#8240;tienne Dupuis
1. a4 d6 2. Ra3 Ba2 3. Ra3! e6 4. h3 Be7 5. Nh2 Bf6 6. Nb3 Bc3 7. dc3 h5 8. Kd2 h4 9. Kd3 Rh5 10. Bh6 Rh5! 11. Nc1

P0075 - ?#8240;tienne Dupuis
1. Nh3 c5 2. Nb4 cb4 3. Na3 ba3 4. ha3 Qa5 5. Rh4 b6 6. f4 Bf3 7. f5 h5 8. Rh4! Nh7 9. Rh4! OOO 10. Rh4! Kb7 11. Rh4! Rhc8 12. Rh4! Rc3 13. dc3 a6 14. Qd3 Na7 15. Bd2 Ng6 16. OOO