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The queen excluder is a type of barrier used in beehives which lets through small worker bees but not the larger queen bee. Extending this simple but powerful concept into the chess world, let's imagine there is a barrier between the fourth and fifth rank that the queen is not allowed to cross. Every other piece except the queen is unaffected by the excluder (marked in blue below), and all other rules stay the same.

Can the following checkmate position arise from a real game?

Checkmate position on chessboard

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2 Answers 2

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Given the queen excluder constraint, this position is

Unreachable

Reasoning:

Let's assume this position can be reached

White's original queen can't get over the excluder line, therefore white must have promoted a pawn to get a queen over the line. Counting the pieces on the board, we can see that both black and white are missing a pawn, and no other pieces. Since no move in chess can add pieces to the board, we >! can deduce there has been exactly one capture by both sides.

White's missing pawn must be the queen on f7 after promoting, so black must have taken white's original queen. We can see black has doubled d pawns, and doubled pawns in chess can only happen after a capture, therefore black's capture must have been the move e5 x Qd4.

From this we can deduce that white must have promoted his d pawn, since black hasn't taken it, and it is off the board. For white's d pawn to promote, it must have taken at least once, since the path to d8 is blocked by a black pawn. Meaning white's d pawn must have taken on the e file, and promoted on e8.

We know white has taken a black pawn as its only capture, and black's pawn that could have been captured are the e and f pawns. But we already determined black's e pawn reaches d4, therefore white has taken black's f pawn.

For this position to be reached, white's d pawn must have taken black's f pawn, but since they are more than 1 file apart, this is impossible, hence the position is unreachable.

Comment answer:

If white promotes on e8 after black takes its queen on d4, white's d pawn must end up on e4, which means capturing either on e3 or on e4. Black's only piece that can be captured is the f pawn, and it can't change file since black has no captures except on d4, white's d pawn can't reach e4.

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  • $\begingroup$ Welcome to Puzzling! Nice work, you're almost there. What if it's the white e pawn that promotes? $\endgroup$
    – Jafe
    Jun 30, 2023 at 9:40
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    $\begingroup$ You have to add something: Black's f-pawn might have promoted before been captured on e6 by wPd5... but that the wPf2 is on the way to f1, and the bP couldn't switch file for lack of White unit to be captured. $\endgroup$
    – Evargalo
    Jun 30, 2023 at 9:45
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Not that there's anything fundamentally wrong with the accepted answer, but with the several promote/capture order options and switchable identities of the pawns to rule out, the answer feels somewhat less crisp than it could be. Here's another way to solve the puzzle with (what I hope might be) somewhat more straightforward logic:

  • White has 15 pieces, and black has a doubled d-pawn, so the missing white piece was captured on the d file by black's e-pawn.
  • Therefore, black's missing f-pawn cannot have left the f file. It was captured there as a pawn, because its promotion path has been blocked the entire game.
  • Black has 15 pieces remaining, so no black pieces were captured anywhere except on the f file
  • This means white's d-pawn cannot have left the d-file. Yet it is missing; it must have been captured as a pawn, because its promotion path has been blocked the entire game.
  • White's d-pawn getting captured completes the accounting for the white pawns: the rest are still on the board
  • So the white queen at f7 must be the original one, which is forbidden by the queen excluder rule.
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