4
$\begingroup$

This question: How can 3 queens control the white squares? got me thinking...

What is the fewest number of queens needed to attack every white square?
Rules:

  • Only queens allowed
  • Every white square is attacked — not just occupied
  • No queen attacks another queen
$\endgroup$
6
  • $\begingroup$ What do you mean by "attacked — not just occupied"? Every white square must be attacked, regardless of whether it's occupied or not? Or white squares can't be occupied? $\endgroup$ Feb 8, 2021 at 19:39
  • $\begingroup$ That was to clarify vs the linked question. In that question, it was enough for a queen to be on a white square (occupy). This question requires attacking all white squares. $\endgroup$ Feb 8, 2021 at 19:41
  • $\begingroup$ Correct me if I'm wrong (I probably am), but rule #2 and #3 makes it seem like queens can only be on black squares. All white squares must be attacked (not just occupied), meaning if a queen was on a white square, another queen would have to attack it, which breaks rule #3. If a queen is on a black square and we only care about attacking white squares, we only need to care about it's diagonal movement if it's attacking another queen, which would yield a possible solution like the naive symmetrical approach like this $\endgroup$ Feb 8, 2021 at 19:47
  • $\begingroup$ Your interpretation of rules #2 and #3 are correct. It seemed like explicitly stating that made the starting point too easy. $\endgroup$ Feb 8, 2021 at 19:48
  • $\begingroup$ @LukasRotter I was thinking that too ... but then I think there's no difference between queens and rooks, for the purposes of this problem? $\endgroup$ Feb 8, 2021 at 19:52

1 Answer 1

4
$\begingroup$

The minimum is

4 queens: \begin{matrix}&.&.&.&.&.&.&.&.\\&.&.&.&.&Q&.&.&.\\&.&.&.&.&.&.&.&.\\&Q&.&.&.&.&.&.&.\\&.&.&.&.&.&.&.&.\\&.&.&.&.&.&.&Q&.\\&.&.&.&.&.&.&.&.\\&.&.&Q&.&.&.&.&.\\\end{matrix}

$\endgroup$
3
  • $\begingroup$ Damn it, was just about to post my answer xD +1 $\endgroup$ Feb 8, 2021 at 20:06
  • $\begingroup$ The reason I actually made the question is that this is the only solution I could find. $\endgroup$ Feb 8, 2021 at 21:10
  • $\begingroup$ The solution is unique up to symmetry. $\endgroup$
    – RobPratt
    Feb 8, 2021 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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