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?

  • 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 – DrCapablasker Aug 9 at 10:37
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    "There obviously exists a mating sequence." Except for stalemates. – TheSimpliFire Aug 9 at 12:41
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    @TheSimpliFire The solution to that is not to stalemate. – Studoku Aug 10 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. – Andreas Blass Aug 11 at 3:54

The longest K + Q + Q vs K mate whilst maintaining optimality is a mate in 4.

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    Wouldn't 1... Kd4 give one more movement to the black? – fedorqui Aug 10 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 at 8:07
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    Good one, thanks @justhalf – fedorqui Aug 10 at 8:19

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