IBM Ponder This #298 · February 2023

In the eight queens problem, eight queens must be placed on a chessboard in such a way that no pair of queens is currently threatening one another. One possible solution to the eight queens problem is this:

In any solution to the eight queens problem, all the empty squares on the chessboard are threatened by at least two queens. We refer to a square that is threatened by exactly two queens as safe. These squares are marked here in green:

For the given configuration of queens, there is only one way to put four kings on the safe squares such that no two kings are currently threatening one another:

With apologies to standard chess notation, we can describe the locations of the queens and kings on this board using the following two lists of coordinates:
[(0, 2), (1, 5), (2, 3), (3, 1), (4, 7), (5, 4), (6, 6), (7, 0)]
[(1, 7), (3, 0), (7, 1), (7, 3)]

For $n=14$, an $n\times n$ board and the following queen configuration:
[(0, 1), (1, 9), (2, 7), (3, 5), (4, 3), (5, 12), (6, 10), (7, 13), (8, 11), (9, 6), (10, 4), (11, 2), (12, 0), (13, 8)]

Is it possible to find arrangement of $n$ kings. One such arrangement is
[(0, 0), (1, 10), (2, 0), (4, 0), (4, 2), (9, 2), (9, 9), (11, 0), (11, 7), (11, 11), (13, 5), (13, 9), (13, 11), (13, 13)]

There are 41 possible arrangements of $n$ kings in total.

Your goal: Find a placement of $n=20$ queens on an $n\times n$ board such that no pair of queens threatens each other, and the number of different ways to place $n$ kings on the safe squares without any pair of kings currently threatening one another is 48.

Supply your answer in a list of the queen positions, as in the above example.

A bonus "*" will be given for finding a placement of $n$ queens on an $n\times n$ board such that no pair of queens is currently threatening one another, and there is no way to place $n$ kings on the safe squares without any pair of kings threatening each other, for $n\ge 26$.

Solution

Best opened after a real attempt

To be added.