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Using only the one colour set; Is it possible to setup 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 little 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? $\endgroup$
    – Joe
    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. $\endgroup$
    – warspyking
    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. $\endgroup$ 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 $\endgroup$
    – skv
    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... $\endgroup$
    – yurisich
    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. $\endgroup$
    – yurisich
    Oct 18, 2014 at 21:17
  • $\begingroup$ Not just one color set. I haven't come close with that. $\endgroup$ Oct 18, 2014 at 21:18
  • $\begingroup$ Can't stagger the rooks because then they both control the same square. $\endgroup$ Oct 18, 2014 at 21:19
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Does this count?

Only one knight because the Knight's tour.

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    $\begingroup$ Here, warspyking is asking for a solution where every square is being attacked. A knight's tour doesn't qualify, since at any given moment only eight (or less) squares are being attacked by the knight. $\endgroup$
    – user46002
    Feb 12, 2019 at 20:24

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