16
$\begingroup$

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?

Source

Improve this question
$\endgroup$

2 Answers 2

Reset to default
22
$\begingroup$

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}

Improve this answer
$\endgroup$
2
  • 7
    $\begingroup$ Could you tell us how you came to the solution. Is it unique? $\endgroup$
    – Bernardo Recamán Santos
    Commented Feb 8, 2021 at 19:10
  • 1
    $\begingroup$ Very cool! I didn't know it was possible. $\endgroup$
    – Dmitry Kamenetsky
    Commented Feb 9, 2021 at 6:48
7
$\begingroup$

I can manage to get

exactly one unattacked square:
enter image description here

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

Methodology:

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.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ 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
    Commented Feb 8, 2021 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.