# Fewest Chess Pieces

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):

• So only pieces from one side (black or white) and only those present at the start of the game (i.e. no promoted pieces)? – Jens Nov 12 '19 at 21:39
• We can do it with 6 pieces, but we have to use the two queens. – Alain Remillard Nov 12 '19 at 21:47
• Only pieces at the start of the game from one side. And yes, as mentioned, occupied squares count as covered. – ThatOneNerdyBoy Nov 12 '19 at 21:59
• @AlainRemillard we can do it with 5 pieces, if all of them are queens (a2, c4, d5, e6, g8 for example) – trolley813 Nov 22 '19 at 10:44
• @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). – trolley813 Nov 22 '19 at 10:46

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:

Here's the board

• I'll accept your answer as less pieces are required! Nice work! – ThatOneNerdyBoy Nov 21 '19 at 22:43

This one only requires

8 pieces

Here's the board

• I am assuming your answer accounts for the fact that g3 is covered by the pawn... – ThatOneNerdyBoy Nov 12 '19 at 22:56
• Not sure I understand your question? g3 is indeed covered by the pawn. – Jens Nov 12 '19 at 22:59
• 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.... – ThatOneNerdyBoy Nov 12 '19 at 23:02
• OK, I did a little rearrangement. – Jens Nov 12 '19 at 23:10
• 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! – ThatOneNerdyBoy Nov 12 '19 at 23:16

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.

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.

• Good answer. But you're right, eliminating at least one piece is possible. – ThatOneNerdyBoy Nov 12 '19 at 22:25
• 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. – Danagon Nov 15 '19 at 15:10
• @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 :)) – Danagon Nov 15 '19 at 15:11
• @Danagon Actually, this one above has nine. So indeed, the answer I accepted having 8 is the least (that I know of). – ThatOneNerdyBoy Nov 15 '19 at 21:49