GeekDad Puzzle of the Week Solution – Eight Queens

Geek Culture

Here is last week’s puzzle:

max_nora_chessGrowing up playing chess with my Dad, it has been my honor and privilege to teach Max and Nora to play chess. Yes, I can still beat them both… for now at least.

When showing him how each piece moves, I challenged Max to a puzzle that I remember from both my Dad teaching me, as well as a high-school level computer science book.

The puzzle is simple, and is the core for this week’s GeekDad Puzzle of the Week:

Given that you have eight queens pieces, each of which can traverse the whole chessboard in the vertical, horizontal, and both diagonal directions, is it possible to place all eight on a standard 8×8 chessboard such that no queen can immediately capture another? If so, is there just one way, or are there multiple ways? And if there are indeed multiple ways, how many are truly distinct – and not simple rotations, reflections, or inversions of another solution?

For example, on a 5×5 chessboard, here are two solutions:

board_5x5_01 board_5x5_02
On each board, any given queen is in its own row and column, and no two queens share the same of the nine diagonals in either direction.

While the two solutions above appear different, further examination will lead you to the second being a translation or interpretation of the first. Specifically, the second board is simply the first rotated through 180° and then reflected across the upper-left to lower-right diagonal. These two would not be considered as “distinct” solutions for the purposes of this puzzle, as one can be flipped, rotated, or inverted to become the other.

Writing code to solve will produce a total of 92 different looking solutions, 12 of which are unique and cannot be made out of rotations, flips, or reflections of another. While the puzzle could be solved with a chessboard and 8 pieces, finding each of the possible iterations would take on the order of an hour or so.

The twelve distinct solutions appear as follows:

Congratulations to John Schram for entering a solution and being on the winning end of the random drawing of all reasonable solutions. A $50 Gift Certificate from the fine folks at ThinkGeek will soon be on its way to him!

Thanks to everyone that submitted a solution, and happy puzzling!

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