alt Hacker NewsSorry I was in the car and read the headline and tried to see how I would do it as an exercise. I actually think I undercounted because for the king and rooks you have to store whether they have been moved yet, and for the pawns whether they just jumped two spaces (so you know if en passant is a valid move).
So I wasn't saying the article is wrong, just engaging in a bit of intellectual exercise.
My count was: there are 16 pawns, each could be in one of 64 places or not on the board, so 6+1 bits per pawn for that, and each could be promoted to a knight a bishop a rook or a queen, so 4+1 bits per pawn for that, and one of them might have just moved two spaces, so 3+1 bits for that. And the other 16 pieces don't change their nature so we only need to know where they are, so 6+1 bits for each of those, plus 3 bits for each of the rooks and the king to tell if they have already been moved. That's.. 40 bytes actually. Im really curious how they got it in 26.
Update: I see! They used illegal positions to store all the special statuses. Very clever!
You have a good starting point, but using 6+1 bits is a bad way to encode 65 possibilities. If you use base 65, you'll see that 65^32 possibilities only require one more bit to store than 64^32.
And four promotion possibilities wouldn't be 4 bits, it would be 2. But even better is 5^16 squeezing into 38 bits.
Combining those cuts your strategy down to 192.7+37.2+4+6 bits which is 30 bytes.
The main savings the article has is being smarter about how to store promotions. And then it uses some tricks by swapping pieces to encode extra bits.
(But it turns out that storing which tiles are occupied, then listing each piece in order, works better than encoding locations.)