Three queens and two rooks can be placed on a chess board so that all empty squares are under attack, as has been shown here: 3 queens and 2 rooks covering a 8x8 chess board.

What if we require that all squares are under attack, even those occupied by any of the five pieces?



I used integer linear programming to minimize the number of unattacked squares. Here is one optimal solution (unique up to symmetry), with

0 unattacked squares: \begin{matrix}&.&.&.&Q&.&.&.&.\\&.&.&.&.&.&.&.&.\\&.&.&.&R&.&.&.&.\\&.&.&.&.&.&.&.&.\\&.&.&.&.&.&.&.&R\\&.&.&.&.&.&.&.&.\\&.&Q&.&.&.&Q&.&.\\&.&.&.&.&.&.&.&.\\\end{matrix}

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    $\begingroup$ Could you tell us how you came to the solution. Is it unique? $\endgroup$ – Bernardo Recamán Santos Feb 8 at 19:10
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    $\begingroup$ Very cool! I didn't know it was possible. $\endgroup$ – Dmitry Kamenetsky Feb 9 at 6:48

I can manage to get

exactly one unattacked square:
enter image description here

(namely, the one occupied by the bottom right queen)


In hexomino's solution to the previous question, I noticed that the problem was essentially reduced to making three queens cover a $6\times6$ chessboard and then using two rooks to cover the remaining two rows and two columns. My only innovation was to shift the position of that $6\times6$ chessboard within the $8\times8$ one.

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    $\begingroup$ Welp, a few minutes after I posted this, Rob Pratt edited his answer from 5 to 0 unattacked squares. +1 to him, but I'll leave this answer in place as it has a nice non-computery method. $\endgroup$ – Rand al'Thor Feb 8 at 19:22

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