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3N2R1/pp3p2/3K2p1/8/2pk4/3pp2p/3P1pB1/1Q4BR w - - 0 1

As stated in the title, white must mate black in 1 move. But first you must solve the mystery behind this puzzle.

This puzzle was created by Hieronymus Fischer (1843-1927).

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    $\begingroup$ Why does the Rook at h1 has a dot on it? Unless it is relevant for the solution (looks like not given the nice answers and already one accepted) the it probably should not be there. My bet would be that "dotted rooks" can fly; so R*h4++ $\endgroup$
    – user10179
    Apr 17, 2016 at 22:23
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    $\begingroup$ @RolazaroAzeveires A dot in the lower right square is commonly used to indicate whose move it is. $\endgroup$
    – f''
    Apr 17, 2016 at 22:46
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    $\begingroup$ @AntonyD'Andrea No, black can place a pawn between the queen and the king: 1. Qb2 c3. $\endgroup$
    – Sleafar
    Apr 18, 2016 at 16:29
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    $\begingroup$ @Kevin The board is numbered. But even if it was missing or you suspected it was lying, you could look at some specific square to know which side started at the bottom. For example, the bottom right square, or the queen's starting square, indicates the side. $\endgroup$
    – Reti43
    Apr 19, 2016 at 12:28
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    $\begingroup$ @Kevin White pawns are moving upwards, black pawns downwards. $\endgroup$
    – Sleafar
    Apr 19, 2016 at 19:56

4 Answers 4

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The mystery

There are 9 pawns on the board (black supremacy propaganda?). If we remove any of them to make it legal, we can mate in 1.

The mate

Remove a7, 1. Qb6#
Remove b7, 1. Nc6#
Remove c4, 1. Qb4#
Remove d3, 1. Qe4#
Remove e3, 1. Bxf2#
Remove f7, 1. Ne6#
Remove f2, 1. Bxe3#
Remove g6, 1. Rg4#
Remove h3, 1. Rh4#

Additional note

We note the double pawns occur on the f file. A potential question could be whether we could remove any of the pawns and still have the board position reachable legally. The answer is that no pawn removal matters. For example, if we removed the a7 pawn, the pawn that is currently at f2 originally started at a7 and with sequential captures traversed a7, b6, c5, etc to reach f2. We can similarly show that for any other pawn from any file, the pawn at f2 could have started from that file and reached f2 legally.

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  • $\begingroup$ I don't see how you get to "fix" the problem this way. BaSzAt gives a perfectly good answer--this position isn't illegal in tandem chess. $\endgroup$
    – Loren Pechtel
    Apr 18, 2016 at 3:33
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    $\begingroup$ @LorenPechtel I like BaSzAt's answer, too, but it seems the answer I went for is the intended solution to the puzzle, i.e., removing any of the 9 pawns results to a unique mate in 1. We should also note that while this position in bughouse is legal, it makes the implicit assumption that your ally is in a position to provide you with a piece you can checkmate with. And if clocks are involved, you also have the time luxury to wait on your turn until that happens. While these might be reasonable assumptions, they are assumptions nevertheless, which might taint the beauty of the solution. $\endgroup$
    – Reti43
    Apr 18, 2016 at 3:45
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    $\begingroup$ Having said that, chess puzzle composers don't have to abide by any rules. You can have impossible compositions (board crammed with pieces), illegal solutions (promoting to an opponent's piece), solutions that require assumptions (en passant/castling are available) or cooked up solutions (helpmates). In most cases the point of such puzzles is homing in on such absurdities because there are no other viable solutions. $\endgroup$
    – Reti43
    Apr 18, 2016 at 4:23
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    $\begingroup$ The promoting to an opponent's piece puzzle was legit--at the time the rules didn't specify that you could only promote to your own piece, nobody had considered that someone might want to promote to an enemy piece. $\endgroup$
    – Loren Pechtel
    Apr 18, 2016 at 4:55
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    $\begingroup$ I looked at the OP and noticed the extra pawn. But I don't understand what gives you the right to "remove any of them". It simply looks to me like the board is set up in error and there is no solution. Or perhaps it is mate in 0 because black has cheated and has an invalid setup. $\endgroup$
    – Paul
    Apr 18, 2016 at 19:13
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A possible explanation to the mystery:

This is tandem chess.

In which case:

You might have a queen or a rook in your deck. Drop it on d5... quickly!

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    $\begingroup$ A knight on b5 or bishop on c5/e5 would also do the trick. So as long as your ally can force a minor/major piece capture before getting mated, you can play for the win. Cool thinking by the way. $\endgroup$
    – Reti43
    Apr 17, 2016 at 19:18
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    $\begingroup$ A pawn at c3 also mates, I believe, which I think covers all the pieces. $\endgroup$
    – Tim C
    Apr 18, 2016 at 6:04
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Alternate solution that does not notice or take care of the excess of black pawns

The mystery:

The board is oriented opposite the usual direction: white started at the top and white pawns advance toward the bottom of the board.

The mate:

Queen to a1 or b2. Black can't interpose the pawn because his pawns advance up the board.

Note:

We have to assume that the black pawn at d3 got there by capturing something after the white pawn at d2 obtained that position. No explanation for the pure number of pawns that have not already queened.

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    $\begingroup$ The orientation of the board is well-defined because of the included labeling. White always starts in the 1st and 2nd row, black in the 7th and 8th row. $\endgroup$
    – Sleafar
    Apr 17, 2016 at 19:59
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    $\begingroup$ The rules of how things are "always" done are often ignored in puzzling.SE. $\endgroup$
    – Peregrine Rook
    Apr 18, 2016 at 4:22
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    $\begingroup$ @Sleafar Yet a well defined rule of the amount and type of pieces on the board has been broken. Black always has no more than 8 pawns yet here we are with Black having 9. $\endgroup$
    – Captain Man
    Apr 18, 2016 at 21:19
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Another alternate solution: The pawns are Berolina pawns (which move diagonally and capture straight ahead; thus, the pawns on d2 and d3 are threatening each other); and Black gave odds by replacing his queen with another pawn on d8.

This allows Qb2 (or Qa1) mate.

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