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An entry in the 19th fortnightly challenge...

9x9 grid

Above is a 9x9 "chess" board with some pieces already laid out, your goal is to fill the board completely, following standard Sudoku rules (i.e. every cell should be filled, such that a specific piece of a given colour appears only once per row, column and 3x3 block). The only additional rule being that black kings may not be in check by any white pieces (in the finished grid). This "check" rule applies across region boundaries (for example, you can't put a black king on the bottom row of the top left 3x3 because of the white queen in the row below).

The nine unique pieces that you need to fit to the grid are:

9 pieces

Finally, note that I've added the tag. It's not a particularly brutal Sudoku, so you shouldn't need any solvers to help you.

Credits: component images come from lichess, though I believe they just use freely available images anyway.

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  • 8
    $\begingroup$ Your title is in iambic tetrameter! :P $\endgroup$ – Deusovi Nov 11 '16 at 2:33
  • 1
    $\begingroup$ (Also, cool idea for a puzzle!) $\endgroup$ – Deusovi Nov 11 '16 at 2:33
  • $\begingroup$ I think you could better phrase "a piece of a given colour appears only once per row, column and 3x3 block" as "a specific piece of a given colour appears only once per row, column and 3x3 block". When I first read that it read as if you have a piece of a given colour in a row you cannot have another piece of that colour in that cell. I then got very confused about you having two black pieces in several of the cells... $\endgroup$ – Chris Nov 12 '16 at 1:53
  • $\begingroup$ This puzzle was great fun - would love to see more puzzles in this format. $\endgroup$ – Jakob Pamp Bengtsson Jan 27 '17 at 8:15
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If I didn't make any mistakes, this should do:

enter image description here

Insights into the solving process:

The first thing I noticed was that no white queen and black king could be adjacent. This observation is enough to place every white queen and black king on the board.

With every black king on the board, the location for the other white pieces is considerably restricted: I started by placing a few white knights, because their checks can't be obstructed, followed by a few white rooks. At this point it can be solved almost like a normal sudoku; because I'm not very well versed on how to do this properly, I had to use the Poor Man's Sudoku Solver:

enter image description here

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Sorry but spoilering everything takes effort :/ (why did you scroll down here if you didn't want to read the answer tho?)

I will use chess coordinates (1-9, a-i), so you can follow along if you'd like.

  1. Consider the black kings on c1 and i3. They reduce the possibilities of a black king to either e2 or f2. However, e2 is guarded by the white rook. Thus, there is a black king on f2.

  2. In the bottom 3x3 square, the black kings restrict the position of the white queen to d3.

  3. We set up a Ghost. Consider the bottom left 3x3 square. The middle column cannot contain white queens, and the right column is restricted by the white queen on d3 and the black king. Thus, there is a white queen on the left column. Now consider the top left 3x3 square. By our previous deduction, and other known white queens, we deduce that there is a white queen on c7.

  4. We set another Ghost. Consider the top left 3x3 square. Notice that the black king can only reside in either a9 or a8. Now consider the left middle 3x3 square. By the previous deduction, there are no black kings in the left column. The white queens thus restrict the black king to b4.

  5. We set up a third flippin Ghost. Consider the bottom right 3x3 square. Notice that the white queen can only reside on either g1 or h1, eg. the bottom row. Now consider the bottom left row. The previous deduction, and the known white queens, prove that there is a white queen on a2.

  6. Consider the top middle square. Notice that the black king must be on e7.

  7. Now, notice that there must be a black king on d5.

  8. Are you ready for... you guessed it... ANOTHER GHOST??? First of all, say it with me: "THE WHITE QUEEN WILL NEVER BE IN THE CENTER SQUARE!!!". Good. Consider the middle right 3x3 square. Notice that the white queen it on either g5 or g4. Thus, there is no white queen on g8. By our wonderful revelation, there can't be a white queen on h8. So, there must be a white queen on i8, considering the top right 3x3 square.

  9. The same Ghost from the previous step shows that there is no white queen on g1. Consider the bottom right 3x3 square, and notice that the white queen must be on h1.

  10. Consider the top right 3x3 square. Notice that the black king is definitely on g8.

  11. Notice that there must be a black king on h6 as well.

  12. Hey, there must be a black king on a9.

WE HAVE ALL THE BLACK KINGS!!!

  1. Note that there must be a black knight on b7.

  2. The remaining white queens are on f5 and g4, for trivial reasons.

WE HAVE ALL THE WHITE QUEENS!!!

  1. Look at the middle right 3x3 square. Please notice how the white rook must be on g5.

  2. Look at the same set of squares. Oh hey, there must be a white knight on i6!

  3. There must be a black bishop on i4...

  4. There must be a white rook on h2.

  5. We set up another Ghost. Please note that in the left middle square, the rooks must be either in a6 or c6. Based on this, we can conclude that there is a white rook on f4.

  6. Look on row 7. THe white rook can only be on i7.

  7. There must be a black knight on h3.

  8. The white bishop is either on i1 or i2, thus a Ghost. We now see that there must be a white bishop on h8.

  9. White bishop on g6.

  10. Black rook on h5.

WE COMPLETED AN ENTIRE SQUARE!!!

  1. We have a Ghost of a white knight on g1 and g3. Thus, there is a white knight on h9.

  2. Note the b file. The white rook can only be on b8.

  3. White rook on d9.

  4. White bishop on d7.

  5. White bishop on e5.

  6. White bishop on a4.

  7. White knight on d4.

  8. White knight in e8.

  9. Black rook on a6.

  10. Black rook on e4.

  11. Black knight on c4.

  12. Black knight on a1.

  13. Black knight on e2.

  14. Black Knight on d6.

WE HAVE ALL THE BLACK KNIGHTS!!!

  1. White rook on a3, and c6.

WE HAVE ALL THE WHITE ROOKS!!!

  1. Black queen on a8.

  2. Black bishop on d8.

  3. Black queen on d1.

  4. White knight on c5 (by b1-b2 Ghost).

  5. Black queen on b5.

  6. White knight on b2.

  7. White bishop on f3 (row elimination).

  8. Black queen on e6 (column!).

  9. Black bishop on e3.

  10. White knight on f1.

  11. White knight on g3.

  12. Black queen on g7, f9, i2, c3.

WE HAVE ALL THE BLACK QUEENS!!!!

  1. Black bishop on b1, c9, f6.

ALL THE BLACK BISHOPS!!!

  1. White bishop on c2, b9, i1.

ALL THE WHITE BISHOPS!!!

  1. Black rooks on the remaining squares.

YAYYYYYYYYYYYYYYY

CHASSS

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  • $\begingroup$ Thank you for your analysis. Please could you explain "ghost"? $\endgroup$ – Rosie F Nov 11 '16 at 7:28
  • $\begingroup$ step 1 you say king on h3 rather than i3 (probably you are too used to a standard chess board). $\endgroup$ – Jonathan Allan Nov 11 '16 at 7:32
  • $\begingroup$ step 3 you say c6 where I think you mean c7 too. $\endgroup$ – Jonathan Allan Nov 11 '16 at 7:39
  • $\begingroup$ I realized that it wasn't an 8x8 board halfway through and I swill didn't fix all of it >.< ty ._. @RosieF Not sure what it's called formally, but I learned it as a "ghost". Basically, ghosting is using information from a number (in this case, piece) that you aren't 100% sure where it is yet. Let's say in one of the 3x3 squares, there are two squares A and B, in which you know one or the other must be a certain number. If A and B are on the same row, then all the other cells shared in that row cannot be that number. $\endgroup$ – greenturtle3141 Nov 11 '16 at 7:52
  • $\begingroup$ Spoilering the final grid would be sufficient I think. :P $\endgroup$ – Alenanno Nov 11 '16 at 12:47

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