A 8-by-8 chessboard has two mirrors added on its left and right margin. The mirrors reflect the queen moves so that a queen threatens additional squares on the board. A queen threatens

  • all squares in the same row or column
  • all squares in the same diagonals
  • all squares in the same diagonals reflected at one of the mirrors.

For example a queen in e2 threatens the e-column, the 2-row, the diagonal d1-to-h5, the reflection h5-to-e8 of this diagonal on the right mirror, the diagonal f1-to-a6 and the reflection a6-to-c8 of this diagonal on the left mirror.

What is the largest number of queens that can be placed on the 8-by-8 chessboard with mirrors so that no queen threatens another queen?

  • 1
    $\begingroup$ If you place a mirror on the margin of the board, the reflected diagonals for a queen on e2 would be h6-e8 and a7-b8. See my comment to @BaSzAt 's answer. Which interpretation is correct? $\endgroup$ – Sleafar Apr 17 '16 at 15:02

If you look at the possible solutions of the classical problem, you can see none of them works here.

(Solution 1: | mirror: g-queen hits e-queen, _ mirror: g-queen hits b-queen)

(Solution 2: | mirror: b hits c, _ mirror: b hits h)

(Solution 3: | mirror: b hits e, _ mirror: b hits h)

(Solution 4: | mirror: f hits e, _ mirror: a hits f)

(Solution 5: | mirror: b hits f, _ mirror: a hits g)

(Solution 6: | mirror: b hits d, _ mirror: a hits h)

(Solution 7: | mirror: b hits c, _ mirror: c hits h)

(Solution 8: | mirror: c hits d, _ mirror: a hits f)

(Solution 9: | mirror: b hits f, _ mirror: a hits h)

(Solution 10: | mirror: b hits e, _ mirror: e hits h)

(Solution 11: | mirror: c hits g, _ mirror: b hits h)

(Solution 12: | mirror: b hits d, _ mirror: b hits e)

For 7 queens, I think the following configuration works:

enter image description here

EDIT: If the mirrors are, as interpreted by Sleafar, placed along the outer edges of the marginal squares, a 7-queen solution could be achieved by removing the g5 queen from Solution 2 and rotating the board by 90°. I think the 8-queen solutions still fail in this case.

  • 1
    $\begingroup$ Doesn't the mirror of a8 threaten b6? $\endgroup$ – IanF1 Apr 17 '16 at 14:37
  • $\begingroup$ Doesn't b6 threaten e7 with the _ mirror? Wait... that mirror doesn't exist. $\endgroup$ – Kevin Brown Apr 17 '16 at 14:39
  • $\begingroup$ @IanF1 I don't think so. The example provided by Haobin implies that the diagonals and their reflections are always the same color. $\endgroup$ – BaSzAt Apr 17 '16 at 14:56
  • $\begingroup$ @Sleafar If the reflection of d1h5 is h5e8, this means the mirrors are actually 1/2 square inner than you think. $\endgroup$ – BaSzAt Apr 17 '16 at 15:02
  • $\begingroup$ @BaSzAt You are right in both cases. $\endgroup$ – Sleafar Apr 17 '16 at 15:48

For even $n$ placing $n$ independent queens on an $n$ times $n$ board is impossible given the mirrors.

Proof: Note first there are exactly $2n-1$ positive (or negative) diagonals. Next note that every square not on either of the main diagonal are adjacent to at exactly two positive and two negative diagonals. This implies that if we don't place in these diagonals we can have at most $\frac{2n-1}{2}$ independent queens, thus contradicting any possibility of $n$ independent queens.

Next note that we can place at most 2 queens along these main diagonals. In fact, if only one is placed along these diagonals, say the main, negative diagonal (without loss of generality), we again arrive at the upper bound of $\frac{2n-1}{2}$ by considering the positive diagonals and the queens that occupy them.

Next, if there are exactly two queens placed in the main diagonals, then these queens occupy exactly 3 positive and 3 negative diagonals. Thus considering the total number of either positive or negative diagonals yields a total of $2+\frac{2n-4}{2}=n$ queens. However, since this bound assumes all diagonals are filled, including the extreme diagonals which have only corner squares, then we know 2 of the 4 corners must be filled. However, these 2 queens aren't independent.



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