Here is a Sudoku-like puzzle, only with chess pieces instead of numbers.
Starting with this position, fill the board with two complete chess sets of white and black pieces (2 kings, 2 queens, 4 bishops, 4 knights, 4 rooks, and 16 pawns of each color, 64 pieces total) in such a way that:
No king is in check. (Intervening pieces block horizontal, vertical, and diagonal checks, of course.)
There are no pawns on the first or eighth ranks.
There are exactly two white pawns and exactly two black pawns on each file, with each of the two white pawns occupying a lower-numbered rank than each of the two black pawns.
There are at most five white pawns on any rank, and at most five black pawns on any rank.
All pieces on the first and second ranks are white, and all pieces on the seventh and eighth ranks are black.
Of the four white bishops, two are on light squares and two are on dark squares, and similarly for the four black bishops.
No white rook is on the same rank or file as any black rook or black queen, and no black rook is on the same rank or file as any white rook or white queen.
No white bishop is on the same diagonal as any black bishop or black queen, and no black bishop is on the same diagonal as any white bishop or white queen.
No white queen is on the same rank, file, or diagonal as any black queen.
No white knight is a knight’s move away from any black knight.
Unless I’ve messed up, this puzzle has one, and only one, solution.
Taking into account the no checking gives
The five pawns (of one color) in a rank rule means that a black pawn cannot be on h5 and gives us
There are still four white pawns to place, with only one more allowed on each of the 2nd through 4th ranks. That means that at least one will have to go on the 5th rank (none can go on the 6th or 7th rank because then it would be behind a black pawn on the same file). G5 is out because that would be checking a black king and h5 is out because there is already a black pawn on h4. That leaves a5 and requires thus requires that each of the 2nd through 4th ranks have five white pawns to fill all 16 in.
That then forces the two black a-pawns, the five-pawn rule (and the existing fifth rank pawns) forces the two black g-pawns and the white g-pawn has only one spot.
Now the remaining 20 pieces can be placed.
I, most likely, made the task more difficult for myself. I figured out where the pieces went first. I didn’t get around to the pawns until later. (My Sudoku skills are rusty. I haven’t solved a Sudoku puzzle in years.)
Later this summer I will attempt to create a similar type of puzzle. Alas, work and school keep me pretty busy right now.
b2 and d2 must be white (5) cannot be Q or B (check) or N (last rule).
b7 and d7 must be black (5), cannot be Q or B (check) or R (7).
that uses up all four black knights, and rule 7 means no black Q or R can be on the a through d files, so they are all bishops.
Thus h1 cannot be a B or Q (rule 6 or 8 ) and the white rooks are all in play, so it must be a knight.
Since rooks are the only pieces black has left, they have to finish filling the eighth rank (5).
That leaves one black rook, four white bishops, two white queens and one white knight to still be placed.
Rules 7 and 9 mean that the two white queens cannot be on a3 or the efgh files. That leaves b1 and d1.
rules 1, 6 and 9 mean that the only two squares for the dark-squared white bishops are e1 and g1.
The only two squares left for the light-squared white bishops are f1 and h5.
Bill Brock already pointed out that the remaining black rook has to be on g5, since the remaining white pieces on that square would have violated rules 1 or 10.
The only piece left to put on a3 is the last white knight.
How about asking beginners to find the mate in one for black using the final position of this puzzle? Maybe this would generate questions about how the position was placed. Congratulations for creating such a challenging puzzle!
