Re^3: [OT] Is It Possible To Serialize A Chess Board In Fewer Than 23 Bytes?

So for a pawn to be promoted, at least one other pawn has to be removed from the board

Not quite. It is that at least one other piece has to be removed from the board. A white pawn could advance past its black neighbour in the next column then take a black non-pawn piece in that column thereby side-stepping the black pawn in its original column. So a piece has to be removed, just not necessarily a pawn, and therefore that piece does not necessarily have the option of promotion. So a single pawn could be promoted without the possibility of denying promotion to any of the remaining pawns.

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Re^4: [OT] Is It Possible To Serialize A Chess Board In Fewer Than 23 Bytes? by afoken (Chancellor) on Mar 11, 2026 at 22:58 UTC
Not quite. It is that at least one other piece has to be removed from the board. Yes, bad wording. "At least" should indicate pawn or higher rank. But then, I never considered removing non-pawns. Removing a pawn removes the least number of bits from the variable length bitstream (only 2 bits). So, allowing a pawn to remove a rook, knight, or bishop removes 4 bits from the bitstream, allowing it to remove a queen removes 5 bits. That (partially) opens the way for the pawn with the opposite color from the same column to be promoted. Repeating that for a total of four pawns of each color can get us into a situation where no pawn blocks any pawn of the opposite color. Half of the columns each have two white pawns and no black pawns, and the other half each have two black pawns and no white pawns. To get there, eight pieces had to be removed from the board, each a rook, knight or bishop. I won't remove the queens, because they shorten the bitstream by one more bit, and I want it as long as possible (worst case). Each of the eight pieces frees 4 bits, so we need 32 bits less than for the initial setup. Then, we allow all 16 pawns to be promoted to queens, which need the most bits (5 bits plus color). Pawns used 2 bits plus color, so this adds 48 bits to the bitstream. In total, we are at the initial 164 bis from the fresh board + 16 bits = 180 bits = 23 bytes. Castling adds 4 bits, 184 bits total, still 23 bytes. Alexander -- Today I will gladly share my knowledge and experience, for there are no sweeter words than "I told you so". ;-)	[reply]

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Re^4: [OT] Is It Possible To Serialize A Chess Board In Fewer Than 23 Bytes?
by afoken (Chancellor) on Mar 11, 2026 at 22:58 UTC

Not quite. It is that at least one other piece has to be removed from the board.

Yes, bad wording. "At least" should indicate pawn or higher rank. But then, I never considered removing non-pawns. Removing a pawn removes the least number of bits from the variable length bitstream (only 2 bits).

So, allowing a pawn to remove a rook, knight, or bishop removes 4 bits from the bitstream, allowing it to remove a queen removes 5 bits. That (partially) opens the way for the pawn with the opposite color from the same column to be promoted.

Repeating that for a total of four pawns of each color can get us into a situation where no pawn blocks any pawn of the opposite color. Half of the columns each have two white pawns and no black pawns, and the other half each have two black pawns and no white pawns. To get there, eight pieces had to be removed from the board, each a rook, knight or bishop. I won't remove the queens, because they shorten the bitstream by one more bit, and I want it as long as possible (worst case). Each of the eight pieces frees 4 bits, so we need 32 bits less than for the initial setup. Then, we allow all 16 pawns to be promoted to queens, which need the most bits (5 bits plus color). Pawns used 2 bits plus color, so this adds 48 bits to the bitstream. In total, we are at the initial 164 bis from the fresh board + 16 bits = 180 bits = 23 bytes. Castling adds 4 bits, 184 bits total, still 23 bytes.

Alexander

--
Today I will gladly share my knowledge and experience, for there are no sweeter words than "I told you so". ;-)

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