Beyond the 8-Queens Puzzle: The 12-Piece Peace Treaty (4R, 4B, 4N) on 8x8!

[Nazdik1975]

Bravo!

The two positions you discovered are indeed rare gems in this fascinating challenge. With the third configuration I am sharing below, I strongly doubt any other valid configuration exists for this framework.

My campaign is drawing to a close: unveiling a uniquely harmonious placement of chess's finest pieces on the board—where complete peace reigns and zero mutual threats exist.

Here is the third distinct configuration (2 Queens, 2 Rooks, 4 Bishops, 4 Knights):

Position 3: The a7-Queen / d3-Rook Setup
* Queens: a7, c1
* Rooks: b5, d3
* Bishops: g2, h2, g4, h4
* Knights: e8, f8, g8, h8

FEN: 4NNNN/Q7/8/1R6/6BB/3R4/6BB/2Q5 w - - 0 1

Code: Select all

.  .  .  .  N  N  N  N
Q  .  .  .  .  .  .  .
.  .  .  .  .  .  .  .
.  R  .  .  .  .  .  .
.  .  .  .  .  .  B  B
.  .  .  R  .  .  .  .
.  .  .  .  .  .  B  B
.  .  Q  .  .  .  .  .

Notice how the Queen on a7 operates completely conflict-free: its downward diagonal (b6 -> c5 -> d4 -> e3 -> f2 -> g1) glides cleanly through the empty gaps without disturbing a single bishop, knight, or rook.

Thank you very much for sharing this journey and elevating the puzzle!

[Ajedrecista]

Hello Chris:

Reverting to a simpler puzzle, how many ways can you arrange eight rooks on a chess board so that there are never two rooks on the same rank or file, I figured the answer is (8!)^2. Logic being that the first rook can go on any of 8x8 squares, and this takes away 16 squares for the next one, leaving 7x7 squares and so on, 6x6 down to 1x1 for the 8th rook. eg 8! squared.
Gemini however thinks the answer is 8!
Am I being dumb, or is Gemini gaslighting me?

I agree with 8! (or n! in a squared chessboard with n > 1). I was going to give an explanation which happened to be the one of Wikipedia article (link), which is taken from a book of 1917 (problem 295 of the book and its solution):

295.—THE EIGHT ROOKS.—solution

Obviously there must be a rook in every row and every column. Starting with the top row, it is clear that we may put our first rook on any one of eight different squares. Wherever it is placed, we have the option of seven squares for the second rook in the second row. Then we have six squares from which to select the third row, five in the fourth, and so on. Therefore the number of our different ways must be 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320 (that is 8!), which is the correct answer.

[...]

Regards from Spain.

Ajedrecista.