53

Here is an example of a 12-move game after which White (to move) is stalemated. All 32 units (pieces and pawns) are still on the board. The original version of this concept game was created by Charles Henry Wheeler, and published in Sunny South in 1887, according to Edward Winter's C.N. 3679. Samuel Loyd is often, and wrongly, given credit. [Title ""] [...


50

Initial Analysis White is clearly in a dire situation, since Black is threatening mate in one with Qh7# or Qh8# if White doesn't do anything. But the White king can't move move nor can White get rid of any of the pawns that surround the king. White's rook, the only White piece, is too far away to do anything. The move Ra8? to try and stop Qh8# fails to Qh7#....


48

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 ...


46

If you promote to a queen with 1. b8Q, black has: 1...Re2+ 2. Kd1 Rb2 attacking the queen and hinting at mate with Rb1++. If white takes the rook 3. Qxb2 it is stalemate. Because of the mate threat white does not have any other good square for the queen either (no good check and no square that would defend b1). If you promote to a rook, black does ...


45

As a starter here is a solution in 7 moves: Can we do better ? [FEN ""] 1.g3 e5 2.Bh3 Ke7 3.Bxd7 Kxd7 4.Kf1 Ke7 5. Kg2 Ke6 6.Kh3 Kf6+ 7.Kh4 Kg6# If this is best, interestingly, the play is almost forced. The only variations are 4...Kd6 or 4...Ke7 in the line above, or 4...Ke6 and 5...Kf5, or 3.Kf1 Kd6/Ke8 4.Bxd7 Kxd7 5.Kg2 Ke6 etc. Only five possible ...


41

It's quite a fun problem to think about, before getting to the calculation of long variations, try to first spot the key idea needed to crack the problem. Here are the first observations that come to my mind which eventually led me to spot the solution, let's break them down step by step: a) With our bishop eyeing g7 and our doubled pawn formation on g6-g7,...


39

It's impossible to checkmate faster than 7 turns (handicap on black) or 8 turns (handicap on white). Proof by elimination I'm going to argue from the perspective that white is helping and black is handicapped, since otherwise it would take 1 more move. Only Queen, Bishops and Rooks can be used by the black player to threaten the white king, as only they ...


39

White is one tempo short of catching the pawn - if White could make two moves immediately it would be a draw as white would just take the black pawn. But they can't, so white has to find a threat which black has to respond to which gains them that move. The only threat they can make is to queen their pawn, and apparently black can stop that with their bishop ...


34

Use the fact that the rook is pinned, and that the king has few squares left; I'm thinking that 1.Rd8 Kd3 (only possible move) 2.Nc5# should be the solution. [White "NN"] [Black "NN"] [FEN "7R/1B1N4/8/3r4/1K2k3/8/5Q2/8 w - - 0 1"] 1.Rd8 Kd3 2.Nc5#


29

White to play checkmates in 2 with 1 d4+! exd3 e.p. 2 Qbf4#. Black to play cannot checkmate in 2 but should win after Bxf7. 7r/1p3Q1p/2q5/3bk3/1Q2p3/2P5/r2P2PP/2KR4 w - - 0 1 1. d4+! exd3 2. Qbf4


27

Here is an example: [fen "8/4P3/8/4p1p1/2p3Pp/p4p1K/k1p2P1P/8 w - - 0 1"] 1. e8 leads to stalemate next move, while all legal promotions lose to 1... c1=Q followed by 2... Qf1#.


25

It is a nice little puzzle: 3k4/3P4/3Q4/8/8/8/8/4K3 w - - 3 13 1. Qd5 Kc7/Ke7 2. d8=Q# 1-0


25

Depending on whether occupied squares need to be covered as well, the number is: [Title " 12 knights, Without Covering Occupied Squares"] [FEN "8/5N2/1NN1NN2/2N5/5N2/2NN1NN1/2N5/8 w - - 0 1"] [Title " 14 knights, With Covering Occupied Squares"] [FEN "8/2N1NN2/2N1N3/2N3N1/2N1N3/1NN1NNN1/8/8 w - - 0 1"] Problems like this are called domination problems and ...


24

Assuming you allow promoted material (since you didn't say anything :-), this (on Page 13 of the PDF) is the (unfortunately, extremely unknown) finite record since ages. It shows the record for the longest sequence of only 1 legal move for each side, with use of promoted material. [Title "Karl Scherer, Feenshach 1980, Page 13"] [FEN "BQ4R1/2Q5/3Q4/4Q1pp/...


24

On the first and third pages of the post that seemed to make the world go crazy on the subject, there are a few given historical problems that feature the "legal" triple check along with historical details. It even made a recent appearance in Episode 20 of "The Chess Pit" at the 12:13 minute mark. The dubious wording of the law in ...


23

It's probably a trick problem with a promotion to a black knight. Such promotions to the wrong colour are not allowed, and never were. In the official rules it is now specifically pointed out that the new piece has to have the same colour as the promoted pawn. FIDE's laws of chess, Article 3.7 e: When a pawn reaches the rank furthest from its starting ...


23

As @BrianTowers showed, it can be done. Here’s a shorter proof game for the fun of it. [FEN ""] 1. h4 g5 2. hxg5 f5 3. g4 Nf6 4. gxf6 d6 5. gxf5 Be6 6. f4 Bb3 7. axb3 c5 8. e4 c4 9. bxc4 d5 10. cxd5 e6 11. dxe6 Na6 12. b4 Nc5 13. bxc5 Qd5 14. c4 Bd6 15. cxd6 Rf8 16. cxd5 Rh8 17. d4


22

4 moves, as far as I can tell. [FEN ""] 1. e3 e5 2. Ke2 Qh4 3. Kf3 d6 4. Qe2 e4# Another one: [FEN ""] 1. d3 d5 2. Kd2 e5 3. Kc3 Be6 4. Qd2 d4# And one more: [FEN ""] 1. e3 e5 2. Ke2 d5 3. Kd3 Qf6 4. Qe2 e4# Same theme for all of them, really.


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 ...


22

[FEN "1B1Q1Q2/2R5/pQ4QN/RB2k3/1Q5Q/N4Q2/K2Q4/6Q1 w - - 0 1 "] 105 mates — Nenad Petrovic, Sahovski Vjesnika 1947 (Chess Problem Database) In this position any check is mate. There are 3 knight mates (c4, g4, f7), 23 discovered mates (14 moves for the rook on c7, 9 for the bishop on b5), and 79 queen mates: 1 on a1, 2 on b2, 3 on c3, 4 on c5, 6 on d4, 3 on ...


22

[fen ""] 1. a4 b5 2. b4 bxa4 3. c4 c5 4. d4 cxb4 5. e4 d5 6. f4 dxc4 7. g4 e5 8. h4 exd4 9. Ba3 g5 10. Nc3 gxh4 11. Qb3 f5 12. Kf2 h3 13. Kg3 h2 14. Bd3 Qa5 15. Kh4 dxc3 16. Kg5 h5 17. Kg6 cxb3 18. g5 Rh7 19. Ra2 Rc7 20. Bc4 Rxc4 21. Kh7 Bd6 22. g6 fxe4 23. Kg7 hxg1=Q 24. Kh7 h4 25. Kg7 h3 26. Kh7 h2 27. Kg7 bxa3 28. Kh7 bxa2 29. Kg7 Qgb6 30. Rg1 hxg1=Q 31. ...


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 knights. ...


20

It should be easily possible to get 18 queens. If white captures four enemy pieces, that's enough to get doubled pawns on four files (a, c, e and g, for instance). And black captures four times to get his pawns on the b, d, f and h files. Then they can all advance and promote, and it should be easy to avoid mate by storing them all in some corner. Here, ...


20

The Matt Bengtson problem Prof. Elkies mentions is: [Title "Matt Bengtson, Chess Braintwisters (Burt Hochberg), no. 103. White to move & draw."] [FEN "4kn2/3p1pPp/4pPpK/6P1/8/2p5/1b6/8 w - - 0 1"] However, the problem is actually cooked with a win for Black starting with “3... f6!”, and there is no stalemate for White. The solution and the cooking line ...


20

Here is one simple solution: [fen ""] 1.c4 d5 2.cxd5 e6 3.dxe6 f6 4.g4 g6 5.g5 Ne7 6.gxf6 Nd5 7.e4 Nc3 8.bxc3 Qd4 9.cxd4 b5 10.d5 b4 11.a3 c6 12.axb4 c5 13.bxc5 Bd6 14.cxd6 g5 15.h3 g4 16.hxg4 Rg8 17.f3 Rg5 18.f4 Rf5 19.gxf5 Na6 20.d4 More moves can be added to reach the exact position with no other pieces other than kings.


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

1...R8xe3 wins a piece (the knight on e3), because either recapture leads to worse things for White, since the d2 bishop is pinned to the white queen, and the f-pawn recapture opens the possibility of mate on g2: [fen "4rbk1/pp3pp1/1nq4p/8/1PP5/P3N1P1/Q2BrP1P/3R1RK1 b - - 0 1"] 1...R8xe3 2.Bxe3 (2.fxe3 Qg2#) Rxa2


17

Anthony Stewart Mackay Dickens found another solution, also with 105 mating moves, but with only 17 units in the diagram (16 white and the black king): [title "Anthony Stewart Mackay Dickens, The Problemist, Jan 1970. 105 mates"] [fen "2Q1Q3/2Q4Q/Q4Q2/3k4/Q5Q1/1R6/B1NBQ3/K2R1N2 w - - 0 1"] This may be found here on PDB. Black's last move must have been ......


17

Sir Jeremy Morse, in Chess Problems: Tasks and Records, Introduction, cites this construction by E. Luukonen: [Title "E. Luukonen, Uusi Suomi, 1936"] [fen "kr6/pnPPPPPPPP/P2N4/R7/3BBN2/1Q6/3K3P/6R1 w - - 0 1"] [StartFlipped "0"] All of White's 117 moves lead to mate in 2. White has no mate in 1, but if Black's rook were to move, or if (after White captures ...


17

An unsorted list of arguments for the vitality of chess problems (probably my answer is a little bit biased because I am an enthusiastic chess composer ...) Computers: A lot of new ideas have been developed during the last decades. One reason for it is the rise of computer programs which help to check the correctness of a chess problem. This simplified the ...


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