I was thinking about a situation on a chessboard. You can call it a riddle, but I am not sure whether it is solvable or not.

Let's say that there are more than 5 pieces for both Black and White, including the kings, on a chessboard and it is White's turn.

The Riddle: Is it possible to arrange all the pieces on the chessboard, in such a manner, so that, for every possible move by White, the Black king gets checkmated? Can you construct such a position?

What I've thought so far:

Since King can't give a checkmate (apart from exceptional cases of a two queens/two rook mate), the White king should be immovable. Otherwise, the criteria of the riddle can't be fulfilled. But thinking about it this way, I can't set up any position that meets the criteria of my riddle.=

  • 5
    No reason I can see that the White King has to be immobile if when it moves a discovered check gives mate. Tim Krabbe's site might be a good place to search for this.
    – Ian Bush
    Commented Apr 12, 2021 at 22:02

2 Answers 2

[FEN "K3k2n/7n/rr1R3N/7n/8/8/4BQ2/4R3 w - - 0 1"]

1. Bxa6#

Here is a much more simple situation where White only has only one legal move, which is a discovered checkmate.


This is a very easy concept to construct, and it has already been done on Tim Krabbe's site as noted by Ian Bush in a comment. Here are the current records, i.e. the highest that has been constructed by chess problemists so far.

I found two cases of 55 forced mates with promoted pieces in Tim Krabbe’s Diary Entry #383.

[Title "Ludwig Zagler, Feenschach 1972, 55 Forced Mates"]
[FEN "1rB1Q1Q1/1PRQ3Q/2b2k2/2pQ1N1Q/4KRPp/5N2/1R6/B3nn1b w - - 0 1"]

[Title "Sampsa Lahtonen, Tim Krabbe's Website Diary Entry #383 2008, 55 Forced Mates"]
[FEN "B3n1bb/1R6/4pN2/4KRPp/2pQ1N1Q/2b2k2/1PRQ3Q/1rB1Q1Q1 w - - 0 1"]

Without promoted pieces, the record is 29, as Krabbe reported in his Diary Entry #334.

[Title "Harold Holgate Cross, The Problemist FCS 4/1936, 2251"]
[fen "8/1b6/p1N5/P1r5/P3KPr1/QBk1NRP1/P2R1P2/4B3 w KQkq - 0 1"]
  • 2
    Just for fun, I mention this (surely minimal) position with only 5 men and two forced mates: Ka1 Qb1 Rb2 - Ka3 Qa2. Commented Apr 13, 2021 at 7:37

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