# Equal queens in every aspect

place $12$ queens on a $5$x$5$ chess board so that every queen can see the same number of empty squares.

For the purposes of this puzzle, a queen can see any square on the same rank, file, or diagonal, regardless of any pieces or empty squares in between.

For example: In this example, every queen can count $7$ empty squares which they can see.

The question is

How can you place $\mathbf{12}$ queens so that each queen sees exactly $\mathbf{6}$ empty squares?

and

Where to put $\mathbf{12}$ queens so that each queen sees exactly $\mathbf{5}$ empty squares?

and lastly,

How to place only $\mathbf{8}$ queens so that each queen sees exactly $\mathbf{10}$ empty squares?

• It may be quite clear, but I can only count 6 legal moves per queen in your example... Could you explain a bit more what is a movable square? Jul 19, 2018 at 14:22
• The queens in the middle of tops and sides; how can they move to 7 empty aquares? I only count 5, unless they can jump over existing queens. Jul 19, 2018 at 14:22
• @Untitpoi it is not legal move count, it is the count where they could originally move if all squares were empty except the queen herself.
– Oray
Jul 19, 2018 at 14:22
• @APrough please "count the empty square in all direction where queen could move if these squares were all empty."
– Oray
Jul 19, 2018 at 14:23
• In chess, moving through other pieces is expressly forbidden, and counting empty squares that can be reached while imagining some other squares being empty as well feels needlessly confusing, so maybe use another word altogether, like "A queen can see any square on the same rank, file, or diagonal, regardless of any pieces in between", and work from there?
– Bass
Jul 19, 2018 at 14:57

For 12 queens and count 5 (same as Saeïdryl's answer) For 12 queens and count 6 either of the below

For 8 queens and count 10 A brute force search confirms that these are the only solutions disregarding trivial rotations and reflections. As a bonus here is the only other solution for 12 queens and count 7 other than the one given in the question. For $12$ queens to make the count to $5$
For $8$ queens to make the count to $9$ (not in the question)