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Using only the one colour set, is it possible to set up a chess board where every square on the board is only controlled once? Occupying a square does not mean controlling it.

If it is possible, with how few pieces?

Piece Cost:

  • Pawn: 1

  • Bishop: 3

  • Knight: 3

  • Rook: 5

  • Queen: 9

You are not allowed to use the king.

All pawns must face the same direction.

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  • $\begingroup$ Is this really not close enough to your other question? Just adding "pawns facing the sanme way" doesn't add that much to the question, does it? What's next, same thing but limited to one full set of pieces? $\endgroup$ Commented Oct 18, 2014 at 13:42
  • $\begingroup$ @Joe This question actually limits the possible answers a lot. Using only one colour, and pawns in only one direction, makes it quite a bit more challenging. Look at the answer in my previous question, it would not work here. $\endgroup$ Commented Oct 18, 2014 at 13:45
  • $\begingroup$ I can manage 34 points with 2 colors and one full set of pieces. I haven't figured out anything using a single color full set of pieces. $\endgroup$ Commented Oct 18, 2014 at 19:33

2 Answers 2

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I can manage 34. I haven't gotten better than that. 4 rooks, 14 pawns.

34.

34 Points

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    $\begingroup$ You are a genius $\endgroup$ Commented Oct 18, 2014 at 16:58
  • $\begingroup$ Aren't you only allowed one color set? How would this work with two rooks and eight pawns? I like this answer it just got me thinking... $\endgroup$ Commented Oct 18, 2014 at 21:16
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    $\begingroup$ Also I think you could stagger the rooks and remove the first 2 pawns from the front of the board. $\endgroup$ Commented Oct 18, 2014 at 21:17
  • $\begingroup$ Not just one color set. I haven't come close with that. $\endgroup$ Commented Oct 18, 2014 at 21:18
  • $\begingroup$ Can't stagger the rooks because then they both control the same square. $\endgroup$ Commented Oct 18, 2014 at 21:19
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If you ignore the number of pieces, except that kings are not allowed, integer linear programming yields a minimum cost of

34, as in the other answer,

and a maximum cost of

54, with 23 pawns, 2 bishops, and 5 rooks: \begin{matrix}&P&.&.&R&P&P&P&B\\&.&.&.&.&.&.&P&P\\&P&P&.&.&.&.&P&.\\&P&P&.&.&.&P&B&.\\&P&P&.&.&.&P&P&.\\&.&.&.&P&.&.&.&P\\&.&.&.&P&.&.&P&P\\&R&R&.&P&P&.&R&R\\\end{matrix}

If you restrict to one set of pieces (without a king),

the problem is infeasible.

If you restrict to one set of pieces (without a king) and relax to controlling each square at least once (instead of exactly once), the minimum cost is

33: \begin{matrix}&R&.&.&.&.&.&.&.\\&.&.&.&.&.&.&.&.\\&.&P&.&.&.&P&.&.\\&.&.&P&B&.&Q&.&.\\&.&.&.&P&.&.&.&.\\&.&.&P&B&.&.&.&.\\&.&P&.&.&.&P&.&.\\&.&.&P&.&.&.&.&R\\\end{matrix}

and the maximum cost is

39 (use all 15 pieces): \begin{matrix}&.&.&N&.&.&.&.&.\\&.&.&.&P&P&.&.&.\\&.&.&P&.&P&.&.&.\\&R&.&.&.&B&P&.&.\\&.&.&.&.&.&.&.&.\\&.&.&B&.&N&.&.&.\\&P&.&P&P&.&.&Q&.\\&.&.&.&.&.&.&.&R\\\end{matrix}

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  • $\begingroup$ In you maximum cost position, 4 pawns are on the last rank. I don't know if we care of the illegality of that placement, but they obviously control no square, and only the Pe8 is necessary for the solution. $\endgroup$ Commented Sep 2, 2025 at 9:48
  • $\begingroup$ @Evargalo I assumed that a pawn can be placed in any square. Note that the accepted answer has pawns in the first rank, which is also not attainable in a real game. $\endgroup$ Commented Sep 2, 2025 at 16:15

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