Longest possible game of Horde Chess

Question

The longest possible normal game of chess has been calculated, but I want to know what the longest possible game of Horde Chess is. What makes this an interesting case is the crazy amount of pawn moves and captures there are that allow the game to be extended for a very long time.

For the sake of the question, as to avoid answers of "infinity," the 75-move will be in effect. Captures and pawn moves will be labeled as "marks" for simplicity.

As for calculating the longest possible game, White needs to promote a pawn to gain waiting moves. This takes 4 moves at the least.

[FEN "rnbqkbnr/pppppppp/8/1PP2PP1/PPPPPPPP/PPPPPPPP/PPPPPPPP/PPPPPPPP w kq - 0 1"]

1. b6 a6 2. b5 Ra7 3. bxa7 Nc6 4. a8=Q

The promotion is our first mark. After this, White has 184 marks in pawn moves. All but one piece can be captured for 31 more marks. (184+31+1)*75 gives us 16200 moves for the White piece alone.

Each Black pawn should be able to get at least two moves given that White captures properly, so that's 16 marks from all Black pawns. Additionally, each Black piece (minus a rook from the opening) is a mark, minus the king. (14+16)*75=2250. After all of the marks are done, the Black king and the last White piece can duke it out for another 75 moves.

16200+2250+75=18525 moves, give or take a ply or two for parity switches, as a lower bound. I'm sure that there are ways to give Black more pawn moves.

Can anyone refine my starter pack calculations and find the longest possible Horde Chess Game?

Answer 1

EDIT: Final calculation matches actual proof game exactly.

That’s a nice new chess-math question, which deserves wider circulation.

Begin by promoting a White pawn as quickly as possible, because until that happens, White is just haemorrhaging pawn moves. Now can promote a second pawn more slowly. This promoted piece is captured by Black pawn to open a file, and the first file's promotions can happen.

Scheme: axb to let all wPs from the a-file promote, then bxa+bxa to let all remaining wPs from the b-file promote, then cxb to let all wPs from the c-file promote, and so on.

One player is the active one, moving pawns. This player can also capture spare officers with officers. But pawns can never be captured before they promote (hugely expensive) and there are only 2 captures by wPs and 9 captures by bPs.

There are in principle 188+36+48+15=287 marks (captures & pawn moves). However, we lose 9 for Black pawn captures, 2 for White pawn captures, and 5 for pawn moves leading to the first White promotion. So there are really just 271 marks. So if the count begins after 4. a8=?, we have a total length of 20328.5 moves minus 0.5 for each parity switch.

So how many parity switches must there be? White is the first active player, then need two switches to promote the pawns from any file. So that's 16 switches, and White has promoted all his h-file pawns. A 17th switch is required for Black to capture these.

What impact does dead position have? If it applies literally, then Black cannot win if he has lost all his units, because he can't capture the last White unit. So let's suppose that Black winning "counts" as checkmate for DP purposes. That gives a total of 20320.0 moves.

The final proposed total is hence: 20320.0 moves.

An actual animated proof game has been constructed which achieves this total: https://pdb.dieschwalbe.de/P1370701