50

Double check is only possible by using discovered check. So either the rook check or the bishop check was discovered by moving something in between on the previous move. I don't see how that's possible with the rook check, but it is possible with the bishop -- if the board is shown with black at the bottom, contrary to what is usually done. Then there could ...


36

Even though the board is upside down, the position is still easily legal. [FEN ""] [startply "86"] [StartFlipped "1"] 1. a4 h5 2. g4 h4 3. Bg2 Rh5 4. gxh5 h3 5. h6 e5 6. Nf3 Bc5 7. Nxe5 hxg2 8. h4 Ne7 9. h5 g5 10. Rh4 Ng6 11. Rf4 g4 12. Rxf7 Qh4 13. e3 Bxe3 14. Ke2 a5 15. f4 Qf2+ 16. Kd3 Nc6 17. b4 b5 18. bxa5 Rxa5 19. Rf8+ Ke7 ...


32

Yes - the less a position looks like a real chess game, the harder it is to spot if it is illegal or not. Sometimes, retrograde analysis is needed to prove a position can be reached in a legal way. For an example, see the starting position of the Horse Concoction by Harry Goldsteen, which can be proven to be legal. [FEN "8/7P/1P5B/2B1Q1n1/3nn2P/1PRnk1nR/...


25

The task you are considered is usually called a proof game, named such because the task is to prove that the position is legal. As a genre of puzzles, there are various aesthetic constraints, most commonly that the resulting game be unique. However, this is not necessary in general, and there is even a genre of counting the number of solutions. There are ...


22

This is actually a rather typical retrograde problem, just start with the most basic observations: We see that black is missing both rooks and the f8 bishop. Given black's pawn structure it's easy to see that neither the f8 bishop nor the a8 rook could have escaped the structure, thus they must have been captured on the 8th row (by a white knight for ...


21

As @chakerian's calculations show, 40 moves is the minimum. After a bit of puzzling, I found the solution. [FEN ""] 1. a4 {First, we need to get the rooks in position. They'll be hard to swap once the board is full.} a5 2. h4 h5 3. Rh3 Rh6 4. Rg3 Rf6 5. Rg6 Rf3 6. Rh6 Rh3 7. Ra3 Ra6 8. Rc3 Rb6 9. Rcc6 Rbb3 10. Ra6 Ra3 {Now, we can start with the ...


21

Ok after playing around with a couple of lines, I finally found one line that shows it's still perfectly legal to play long castles for white at move 11, here it is: [Title "possible 11.O-O-O+"] [StartFlipped "0"] [fen "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1"] 1.e4 e5 2.Nf3 Nc6 3.c4 Rb8 4.d4 exd4 5.Nxd4 Bc5 6.Nxc6 dxc6 (6... bxc6 7.Be3 d6 ...


19

Although solving proofgames is not that difficult, the large number of moves involved may seem impressive at first. Let me share you a few tips by showing how I solve a proofgame. [Title "Étienne Dupuis, Problemesis 2001, PG in 13.5"] [FEN "rn2k1nr/ppp3pp/4p3/8/4q3/1N1bb2P/PPPP1P1P/RNBQKB1R w KQq - 0 1"] In this diagram, it is not ...


18

If we define "illegal position" as a position that cannot happen in a game using legal moves, I think that the most difficult to spot would be positions that look normal, but are impossible to achieve. [FEN "2k1rr2/pbpq2b1/1p1p1np1/nB1Pp2p/2P1Pp2/2N2NP1/PPQB1PP1/2KR3R w KQkq - 0 1"] For example this position looks legal, but indeed there is no way how this ...


18

There are dozens of problems that illustrate a potential winning moves that instead leads to a draw because of the 50 move rule. One example is the following mate in four published by Léon Loewenton in 1956 : [fen "5KBN/p2ppp1r/1p4pp/b7/RP6/1PP4P/1RpPPPkP/n1B1Q1N1 w - - 0 1"] 1. Nf3 Rg7! 2. Kxg7! (2. Qg1+?? draws) There is an apparent mate in three moves:...


17

I'm going to start from scratch even though the OP posted a partial answer in the question, so I will be covering some familiar ground. I began to break the problem down by assigning black to the upper king, then making every piece which attacked him black. Since the g4 knight attacks a king, no other piece can be giving check to either king. [title "White ...


16

White has made 6 captures. Clearly the h-pawn made at least 2, and the g-pawn 1. Meanwhile, the b-pawn made at least 1 (a7/c7), and the a-pawn at least 2 (to the b-file, then to either a7/c7). So White's c-d-e-f pawns made no captures, and thus Black's central pawn fence must have at some point had holes to allow them through. This requires 4 pawn captures, ...


16

Engines like Stockfish and Komodo are not able to work out the previous moves, because that is not what they are programmed for. However, it is vanishingly unlikely that anybody can ever program an engine that works out legal previous moves. The logic of working out whether a possible previous move is legal or not is extremely difficult. To start with, how ...


14

The solution is This works because


12

Assume the board is in it's usual position (the bottom row is the first row) then this is the solution: [FEN "1B6/8/1k6/R7/1n6/2K5/r6p/q1r1n3 - - - 0 1"] and Black's last move was b2xc1R++. I could write an explanation here about I came to this result, but it would essentially be copying @Maxwell's answer. I didn't use it, though.


12

There is only one black piece left, so the last move was with the king. Black's last move was obviously Kg1-h2. The king could not have come from h3, because there is no way the white pawn could have given check from g2 (its initial position). All other squares around h2 are occupied by white pieces. Since the black king came from g1, it was in check from ...


12

Obviously the last white move was 16. Nf3g1+. I note that black needs at least 15 moves to place the pieces as they are: 4 moves to swap queen with rook as they are (various movements possible), 2 moves for the bishop (Bf8g7, Bg7h8), 2 moves for the knight (Ng8f6, Nf6g4 or Ng8h6, Nh6g4) 1 move each for the two pawns (g7g6 and h7h5) 5 moves for the king (...


11


10

Shouldn't the minimum be 40?. I don't have any result to show yet but it seems that: Pawns (black and white): 8 moves Rooks (black and white): 10 moves Kings (black and white): 7 moves Queens (black and white): 3 moves Knights (black and white): 8 moves Bishops (black and white): 4 moves I could be wrong Something to consider is that the e pawns for ...


10

Highly non-optimal, but here's one line that reproduces the position. The basic idea is to not capture the existing pieces, but capture pawns. A single pawn capture can free 3 pawns to promote. Here's a sample game. [FEN ""] 1.h4 a6 2.h5 g5 3.hxg6 Nf6 4.g7 h5 5.g8=R h4 6.g4 h3 7.g5 h2 8.Nf3 Nd5 9. g6 Nb6 10.g7 Rh4 11.Rg1 h1=R 12.Rh8 Nd5 ...


10

The conditions you gave are incomplete. The conditions include that black (to move) can castle; it's hidden in Smullyan's original problem text. From the English book, last paragraph of the problem statement: Just then, however, news came that Black was about to castle. [...] A clearer problem statement from the Retro Corner: Schwarz (am Zuge) darf ...


9

Dag Oskar Madsen's (quite generous) hint: The last move was king takes e-eight. And here's the full solution: Pawn to e-four, pawn to e-six, bishop to b-five, king to e-seven, bishop takes d-seven, pawn to c-six, bishop to e-eight, king takes e-eight. rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1 1. e4 e6 2. Bb5 Ke7 3. Bxd7 c6 4. Be8 Kxe8


9

This puzzle was by Tibor Orban in 1976, and is a great introductory puzzle for those who have never come across a proof game before. In its miniature form, it still shows two elementary technical features: Tempo Loss and Switchback. Tempo is already discussed: there are shorter (non-unique) solutions in 3.0 & 3.5 moves. Switchback is where the Black King ...


9

This problem is by Raymond Smullyan, and is from his The Chess Mysteries of Sherlock Holmes (p.54 of the Hutchinson edition). Phonon's analysis is mostly sound, except that we don't know which black rook White's a-pawn captured. But that doesn't matter, because, before Black's h-pawn captured a White rook, neither of Black's rooks could have reached b4.


9

This was published about in the latest entry of Tim Krabbé's Chess Diary. Harry Goldsteen has put some research into this and came up with a position which is reachable only in 185 moves: [FEN ""] [StartPly "371"] 1.a4 g5 2.Ra3 g4 3.h3 g3 4.Rh2 gxh2 5.Nf3 h1=R 6.d3 d5 7.Kd2 e5 8.Nh2 a5 9.Ke3 d4+ 10.Kf3 e4+ 11.Kg3 e3 12.Qd2 exd2 13.Rb3 d1=B 14.Bf4 b5 15.Nd2 ...


9

The thoughts mentioned in user1583209's answer more or less summed up mine, but I couldn't find a way to get the king out in time without requiring extra moves. The final trick is [FEN ""] 1. Nc3 h5 2. Nd5 Rh6 3. Nxe7 Rf6 4. Nd5 Nh6 5. Nc3 Ke7 6. g4 Ke6 7. g5 Kf5 8. g6 Kg4 9. gxf7 g6 10. Nf3 Bg7 11. f8=N Bh8 12. Ne6 Qg8 13. Nf4 Rf8 14. Ng2 Kh3 15. Nb1 Ng4 ...


9

Spotting potential mates isn't too difficult. There are a couple of obvious ones involving the b and h pawns which aren't far off from promoting. An immediate b7 threatens b8=N# for example. It can only be stopped by Bc7 but then after dxc7 black has Nxf6+ and the mate is delayed beyond 3 moves. Pushing the h pawn is more promising. If it weren't for the ...


9

The reasoning is as follows: So the time series of key events:


9

The difficult part of this puzzle is the realization that the pawn on b7 can actually be black. When this is established, the rest of the puzzle can be figured out without too much difficulty. Thus, if we assume that the pawn on b7 is black, what does this entail for the other pieces? Well: The bishop on a8 must be white, since it must've gotten there by ...


8

I came up with this, which is 43 moves, but I think it could be optimized further: [fen ""] 1. h4 h5 2. Rh3 Rh6 3. Rg3 Rf6 4. Rg6 Rf3 5. Rh6 Rh3 6. Rh8 Rh1 7. a4 a5 8. Ra3 Ra6 9. Rc3 Rb6 10. Rc6 Rb3 11. Ra6 Ra3 12. Ra8 Ra1 13. e4 d5 14. Ke2 Qd7 15. Ke3 Qg4 16. Qf3 Kd7 17. Qf4 Kc6 18. Kd4 Qd1 19. Ke5 Bh3 20. g4 f5 21. Ke6 Kc5 22. Kf7 Kd4 23. Ba6 b5 24. Bc8 ...


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