Can you place 3 queens on a 6x6 chess board such that they can attack every square?
Good luck!
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asked Sep 15, 2019 at 4:34
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$\begingroup$ Related: puzzling.stackexchange.com/questions/22/… $\endgroup$– sunfishhoSep 15, 2019 at 5:10
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1 Answer
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5
Here's the solution:
It's interesting to note that the domination problem has very few solutions for 3 queens on a 6x6 chessboard as compared to other possibilities. (Neither of those links contains the answer! Just background info.)
Perhaps the key realisation is that
some squares which you'd intuitively expect to be attacked diagonally can be attacked orthogonally. With one queen at $a1$ covering the central white diagonal, it'd seem natural to use two more to cover the black diagonals to either side, including those difficult squares $b3/c2$ and $e6/f5$. But putting the other two queens on the next diagonals leaves not enough space to reach out to the edges ... and we realise there's a nice configuration to cover $b3/c2$ and $e6/f5$ orthogonally.
Before that, starting with
a queen at the corner is a counterintuitive move in which I was inspired by the 5 queens on an 8x8 chessboard problem.
answered Sep 15, 2019 at 7:52
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$\begingroup$ Nice! I was trying to prove that you can't :) $\endgroup$ Sep 15, 2019 at 7:55
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1$\begingroup$ @ArnaudMortier It turns out "proving that you can't" is surprisingly difficult for domination problems. For the 8x8 chessboard, there is no known 'nice' proof that it can't be done with 4 queens; that was proved by brute-forcing every case. $\endgroup$ Sep 15, 2019 at 7:59
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$\begingroup$ In this case it seemed that counting the appropriate kind of squares would do the trick (e.g. side squares that are of the opposite colour to the queen's square, etc). $\endgroup$ Sep 15, 2019 at 8:29
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$\begingroup$ Great solution. I wasn't able to find any other solutions. Can we prove that this is the only one? $\endgroup$ Sep 15, 2019 at 9:29
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$\begingroup$ @Dmitry According to the MathWorld link, there is exactly one arrangement in which all the queens are 'protected' by each other. Which isn't this one, so it's not unique. $\endgroup$ Sep 15, 2019 at 9:32