44

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 ...


36

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 ...


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 ...


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

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/5B1P/...


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

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 ...


16

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 ...


15

Just to prove that this is possible, here's an opening bid of 9 moves. It's ugly because black makes several time wasting moves, but white needs to get his king to somewhere where it can be mated and also remove black's pawn (I don't see how to get a mating position with 7 black pawns on their original squares), so it takes time. [FEN ""] 1.e4 e5 2.Ke2 Ke7 ...


14

Inspired by Ed Dean's answer, here is another "infinite loop": [FEN "8/6p1/1p3pPk/1P3Pp1/1Pp3p1/KpP3P1/1P6/8 - - - 0 0 "]


12

I think I found a double-self-smothered-stalemate that could occur via sequence of legal moves from the normal starting position. [FEN "5bnk/4ppbp/4P2p/5P1P/p1p5/P2p4/PBPP4/KNB5 w - - 0 1"] 1. c3 f6 Each player has seven pawns and two dark-square bishops, so one pawn from each player must have been promoted. Once White's b pawn has captured to a3 and ...


12

Very nice puzzle! Looks like the solution is as follows: [FEN "8/8/6p1/5pP1/5P1K/5PpP/4p2p/2Q2Bkr w - - 0 1"] 1. Qe1 g2 (1... exf1=Q 2. Kxg3 Qxe1#) (1... exf1=R 2. Qxg3#) (1... exf1=N 2. Qf2+ Kxf2 (2... gxf2)) (1... exf1=B 2. Kxg3) 2. Bxe2# 1-0 Interestingly, in the different lines, following results are achieved: White gives a checkmate Black gives a ...


12

After 1.Ra2, black's rook and bishop are both stuck; 1...Bb5, defending the rook on a6 while attacking the rook on a2 is no longer effective since 2.cxb5 defends the rook on a2. After this, black's play is completely shut down due to the pin on the bishop, while white is able to make improvements in his position. In particular, white can get a very strong ...


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 ...


11

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.


11

You can find this chess problem [FEN "8/8/8/8/4k3/8/8/2BQKB2 w - - 0 1"] on page 582 of Pal Benko's book "My Life, Games and Compositions" 2003 ISBN:1-890085-08-1 which also states it was first published in "Chess life" in 1968, so ChessBase is correct.


11

I noticed a simplification of @RosieF's solution: Label the squares with two-number labels as follows: 4,2 0,3 1,4 2,0 3,1 1,0 2,1 3,2 4,3 0,4 3,3 4,4 0,0 1,1 2,2 0,1 1,2 2,3 3,4 4,0 2,4 3,0 4,1 0,2 1,3 Then two knights may occupy two squares x,y and X,Y iff those squares' first numbers x and X are different and their second numbers y and Y are ...


10

What would make you play Nc7? It is the only move with which White can give check mate in two moves. This results in checkmate in 2 if, as the computer does, Rxe5 is played, but why wouldn't the computer just play Nxg5, winning the queen and preventing checkmate? On 1. ...Nxg5 White plays 2.Rd5 which check mates. Or d1 and promote to a queen? On ...


10

I don't know how famous it is, but here is a study shown to be incorrect by computers: [FEN "8/5n2/5N2/3K4/8/3p4/Rn6/1k6 w - - 0 1"] 1. Ra3 d2 2. Rd3 Nxd3 3. Ne4 d1=N! 4. Ke6 Nd8+ (4... Nh6!) 5. Kd7 Nb7 6. Kc6 Na5+ 7. Kb5 Nb3 8. Nc3+ Nxc3+ 9. Kc4 Kc2 1/2-1/2 White to play and draw (Jan van Reek, 1987). The intended solution is given in the diagram. It ...


10

If you're looking for the puzzle with the longest forced sequence, it would be this position, which contains a win for white in 545 moves. [FEN "QN4n1/6r1/3k4/8/b2K4/8/8/8 b - - 1 1"] It was found by computers from the generation of 7-man endgame tablebases, and there are several similar positions with the same distance to mate. If that seems a little ...


10

Oliver Sick listed many good examples. Here are some of my thoughts: Andrew Buchanan's "Dead Reckoning" (a consequence of Article 5.2b of the Laws of Chess) Helpselfmates Parry series mates Goals other than (stale)mate, e.g. check, capture, or White to move a unit to a specified square. Further developments in genres that are more than 20 years old, e.g.: ...


10

This is a proposed solution, I wonder if we can find more patterns like this with more pieces on the board


9

CHESS PROBLEMS MADE EASY - How to solve, how to compose How to compose chess problems and why (You have to register for free to read it) Chess Wizardry: The New ABC of Chess Problems Compose Like Mozart What are chess problems Additional links Chess Problems The Chess Portal: Chess theory/Chess Problems (that list has many recourses on Chess composition)


9

Here's a simpler mutual smothered stalemate (9 men on a side, no promoted pieces): [Title "Mutual smothered stalemate"] [fen "5brk/4p1pb/4P1p1/6P1/1p6/1P1p4/BP1P4/KRB5 w - - 0 1"]


9

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, ...


9

In this famous problem by H. Hultberg (1944), the white king castles to prevent Black from castling: [FEN "r3k2/1p1p2p1/2pP3p/8/8/5R2/PPPP4/4K2R - - - 0 0 "] White to mate in two moves. According to chess problem conventions, castling is presumed to be legal unless it's provably illegal. In this position, you can prove that at least one player has lost ...


9

This is my answer :) Spoiler alert: Answer is in a comment in the last line of the code block using System; namespace Toroidal5Knights { class Program { static int solution = 0; static bool[,] board = new bool[5, 5]; static int[] dx = { -1, 1, 2, 2, 1, -1, -2, -2 }; static int[] dy = { -2, -2, -1, 1, 2, 2, 1, -1 }; ...


8

You mean like this? [FEN "K1k5/P1Pp4/p2P4/Pp6/P1p5/2P5/8/8 - - - 0 0 "] 1. axb5 axb5 a6 b4 cxb4 c3 b5 c2 b6 c1=Q b7# 1-0 White mates in 6. I guess that's 9 consecutive forced moves. It would be eleven except for black's choice of promotion piece on his fifth move. I don't know if it's a record, and I don't know who composed this classic chess problem. ...


8

Reduce number of units from Noam's record of 18 to a new record of 15. (8+7 units, no promoted units necessarily on the board, pawn captures all achievable): [Title "Minimal mutual smothered stalemate"] [fen "5brk/4p1pb/4P1p1/6P1/6PB/6PK/6PP/8 w - - 0 1"] Is this the best possible? It's interesting that this solution is asymmetric. If you wrap both kings ...


8

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


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