alt Hacker News> Since each position takes up 6 bits ($2^6 = 64$), multiplying 6 bits by 32 pieces gives us 192 bits / 24 bytes (1 byte = 8 bits).
But each position can only be used once, so you really only have 64*63*62*61*...*33 possibilities = 64!/32! = ~2^53, so you could encode this with less than 7 bytes, and then use the basic 12-byte encoding for captures, castling, en-passant, and promotions, and you are below 19 bytes in total. (Did I miscalculate this?)
Also pawns can never be on the 1st or 8th rank so you can subtract those possibilities from each of the numbers that represents a pawn.
You also don't care about the order of bishops, rooks, knights, and especially pawns - so you should be able to do even better.
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Multiplying 32 numbers each over 5 bits long obviously results in more than 160 bits. 64!/32! > 2^178.3
(Former) pawns can be on the 1st or 8th rank if they've been promoted. You can place an unpromoted pawn on the 1st or 8th rank to encode en passant, too.
Also castling can be encoded as extra squares available to the rooks only ("queen side never moved" and "king side never moved"), and that is almost free.
But without a good encoding for promotions, I doubt you can beat the encoding of the article.