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On the board, there are 2 queens (of the same color) placed on random squares. The opposing side has only a king, placed on a random square (and not in check). There obviously exists a mating sequence. What is the biggest possible 'n' such that there is a forced Mate in 'n'? In other words, given the above situation, what is the upper bound on the minimum number of moves required to give a checkmate?

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  • I don't have an answer to your question but I have a comment that may be relevant to your research. This link has mate in 2 puzzles with two queens vs a lone king: apronus.com/chess/puzzles/mate-in-2/?KQQvK Aug 9, 2020 at 10:37
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    "There obviously exists a mating sequence." Except for stalemates. Aug 9, 2020 at 12:41
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    @TheSimpliFire The solution to that is not to stalemate. Aug 10, 2020 at 1:58
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    @You'rebadandshouldfeelbad I think the stalemate comment was intended to point out the possibility that the "random" starting position in the question might be stalemate. Aug 11, 2020 at 3:54

1 Answer 1

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The longest KQQ-K mate whilst maintaining optimality is a mate in 4.

[FEN "8/8/8/3k4/8/5K2/8/4Q2Q w - - 0 1"]
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    Wouldn't 1... Kd4 give one more movement to the black?
    – fedorqui
    Aug 10, 2020 at 7:35
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    @fedorqui, in that case, 2. Qc7 (Kd5 3. Qe4#) (Kd3 3. Qe3#) mate in 3.
    – justhalf
    Aug 10, 2020 at 8:07
  • Yes, this seems right. I played a lot in an analysis board and couldn't find a position which requires mate in 5. Nov 12, 2021 at 16:36

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