4
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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
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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? $\endgroup$ – Rand al'Thor Feb 8 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. $\endgroup$ – Joel Rondeau Feb 8 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 $\endgroup$ – Lukas Rotter Feb 8 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. $\endgroup$ – Joel Rondeau Feb 8 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? $\endgroup$ – Rand al'Thor Feb 8 at 19:52
4
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The minimum is

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

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

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