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I was wondering how many queens can be put on boards without the position being a forced checkmate or a forced stalemate. The best I came up with is 53 this position.

[FEN "QQQQQRK1/QQQQQPr1/QQQQQRNk/QQQQQQpB/QQQQQQQQ/QQQQQQQQ/QQQQQQQQ/QQQQQQQQ w - - 0 1"]

Is there a way to fit 54 or more queens, given the constraints)? And if not, how would one show that more queens cannot be placed?

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  • "(forced mate) or stalemate" or "forced (mate or stalemate)"? ;)
    – Annatar
    Jun 5, 2020 at 10:27
  • 1
    Edited to make clearer
    – Lubomír Grund
    Jun 6, 2020 at 11:41

1 Answer 1

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57 Queens:

[FEN "QQQQQB1k/QQQQQ1bq/QQQQQQKB/QQQQQQQQ/QQQQQQQQ/QQQQQQQQ/QQQQQQQQ/QQQQQQQQ b - - 0 1"]

(edited FEN, White to move).
(2edited, thanks Noam D. Elkies)

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9
  • 1
    It is not clear in the OP, but I imagine the side with the queens was supposed to be on the move ?
    – Evargalo
    Jun 5, 2020 at 10:46
  • 2
    @Evargalo Then imagine White King on b3 to start. Kc2 is the only legal move.
    – Andrew Chin
    Jun 5, 2020 at 10:48
  • Sure, this solution is valid then.
    – Evargalo
    Jun 5, 2020 at 10:49
  • 2
    I think you can squeeze one more Queen out of this nice setup by changing the Ba2 to a Black Queen for a total of 57.
    – Noam D. Elkies
    Jun 6, 2020 at 17:29
  • 3
    Whoa, you're right! The question didn't say Queens of only one colour!
    – Andrew Chin
    Jun 6, 2020 at 19:14

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