4 ms·
If you're going to go that far, you might as well use 64 bits to say whether each square contains a piece. There are a maximum of 32 pieces, and for each (poten
by JoshuaDavid 9y ago
If you're going to go that far, you might as well use 64 bits to say whether each square contains a piece. There are a maximum of 32 pieces, and for each (potentially) occupied square there are 12 possibilities for what might be there (not counting "nothing" as a possibility in a square since that's already covered by the 64 bits indicating whether each square is occupied), which can be represented in 4 bits per square. So that's 64 bits to say which squares have pieces, plus 32 * 4 = 128 bits to say what is in each occupied square, for a total of 192 bits to store the chessboard.
You can take it even further, too. For the 32 possibly occupied spaces, you can encode what is in them as a 32 digit base 12 integer, then convert to binary, which is guaranteed to fit in 115 bits. So you can get away with 115 + 64 = 179 bits.