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What is the least amount of chess pieces you can use to cover the board with their possible moves and what is their placement? You can only use the pieces that are found on the chessboard (from the one side) at the start of the game (for example, you only have one queen, two knights, two rooks, etc.). Similar to this, but there is a piece requirement. Here is an example of the "coverage" of a knight and a queen (however, the pieces themselves would be covered as well):

enter image description here

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    $\begingroup$ So only pieces from one side (black or white) and only those present at the start of the game (i.e. no promoted pieces)? $\endgroup$
    – Jens
    Commented Nov 12, 2019 at 21:39
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    $\begingroup$ We can do it with 6 pieces, but we have to use the two queens. $\endgroup$
    – Alain Remillard
    Commented Nov 12, 2019 at 21:47
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    $\begingroup$ Only pieces at the start of the game from one side. And yes, as mentioned, occupied squares count as covered. $\endgroup$
    – ThatOneNerdyBoy
    Commented Nov 12, 2019 at 21:59
  • $\begingroup$ @AlainRemillard we can do it with 5 pieces, if all of them are queens (a2, c4, d5, e6, g8 for example) $\endgroup$
    – trolley813
    Commented Nov 22, 2019 at 10:44
  • $\begingroup$ @ThatOneNerdyBoy Replaced "at any given time" with "at the start of the game" (since the pawns can promote, so at some times you can see several queens of the same color, for example). $\endgroup$
    – trolley813
    Commented Nov 22, 2019 at 10:46

3 Answers 3

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The lone pawn in the previously accepted answer bugged me, so I tried replacing it with a knight and moving some pieces around. I now have:

A 7 piece answer

Here's the board

Seven Pieces

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    $\begingroup$ I'll accept your answer as less pieces are required! Nice work! $\endgroup$
    – ThatOneNerdyBoy
    Commented Nov 21, 2019 at 22:43
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This one only requires

8 pieces

Here's the board

enter image description here

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    $\begingroup$ I am assuming your answer accounts for the fact that g3 is covered by the pawn... $\endgroup$
    – ThatOneNerdyBoy
    Commented Nov 12, 2019 at 22:56
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    $\begingroup$ Not sure I understand your question? g3 is indeed covered by the pawn. $\endgroup$
    – Jens
    Commented Nov 12, 2019 at 22:59
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    $\begingroup$ Technically that would be an invalid move as a piece doesn't occupy that space and therefore the pawn cannot move there. This can be overcome by placing your remaining knight somewhere.... $\endgroup$
    – ThatOneNerdyBoy
    Commented Nov 12, 2019 at 23:02
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    $\begingroup$ OK, I did a little rearrangement. $\endgroup$
    – Jens
    Commented Nov 12, 2019 at 23:10
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    $\begingroup$ I believe that this would be the best solution, but I welcome someone to prove that less is possible, or that another solution with the same number of pieces exists! $\endgroup$
    – ThatOneNerdyBoy
    Commented Nov 12, 2019 at 23:16
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Here is an upper bound. Since is has been proven that 8 pieces, with opposite colored bishops since the question asks for starting pieces only, that only 63 squares can be covered. Addding a single pawn works since occupied sqaures count as covered.

enter image description here

I say upper bound because all of the sqaures occupied by the pieces are also attacked, which is not needed under the confines of this question. Perhaps it could be down with one or two less pieces.

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  • $\begingroup$ Good answer. But you're right, eliminating at least one piece is possible. $\endgroup$
    – ThatOneNerdyBoy
    Commented Nov 12, 2019 at 22:25
  • $\begingroup$ You can also use two bishops on the same color (if that helps) because the question asks for the starting pieces, but not their starting positions or colors. I know we refer in chess to a light-squared bishop, but it is still just a bishop. $\endgroup$
    – Danagon
    Commented Nov 15, 2019 at 15:10
  • $\begingroup$ @ThatOneNerdyBoy you say at least 1 less piece than 8 is possible, but you have accepted an answer with 8 pieces. Is 7 possible? (because I will keep looking :)) $\endgroup$
    – Danagon
    Commented Nov 15, 2019 at 15:11
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    $\begingroup$ @Danagon Actually, this one above has nine. So indeed, the answer I accepted having 8 is the least (that I know of). $\endgroup$
    – ThatOneNerdyBoy
    Commented Nov 15, 2019 at 21:49

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