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by jakeinspace
1742 days ago
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Let's assume the 50-move rule is automatically applied. We can have 49 moves per pawn move, followed by 49 moves per piece captured. A maximum of 8 pawns (4 white and 4 black, for example) can possibly be advanced over the course of the game to the opposite side (6 squares ahead from their home square). Therefore, it seems to me that the maximum number of pawn moves could involve one player advancing their pawns 4 spaces, followed by the opposing player capturing all those pawns (using pieces rather than pawns, as both moving a pawn and capturing a pawn would be a waste), and then advancing their pawns to the back rank. This would be 8x4 + 86 = 80 pawn moves, plus 8 pawn captures. That's 88 allowances for 49-move knight sequences, plus the initial 49-move knight sequence made from move one. So, with no pawns remaining, and all of black's pawns converted to pieces on the back rank, we have 8949=4361 moves. Now, there are 22 pieces and 2 kings remaining on the board. That puts an upper limit of 22 captures, leading to an additional 2249 allowable moves. All together, that is 11149=5439 total moves. I'm probably off by a few. |
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Not so! Pawns move diagonally via capturing, which theoretically could allow white to promote all 8 pawns just in the d column, and black to promote all 8 in the e column (at the expense of various other pieces).
edit: of course, you'd run out of pieces first, but you can accomplish something similar by using multiple columns.
Sample game start that accomplishes a subset of this (2 pawns promoted from each side, 6 pawns remaining from each side):
1. Nf3 Nc6 2. Ng5 Nb4 3. Ne6 Nd3+ 4. exd3 dxe6 5. Ke2 Qd7 6. Kf3 Qc6+ 7. Kg3 e5 8. d4 e4 9. d5 e3 10. d6 e6 11. d7+ Ke7 12. d4 Kf6 13. d8=R e2 14. Rd5 e1=R 15. Ra5 Re5 16. d5 Rh5 17. d6 e5 18. d7 e4 19. d8=R e3 20. Rdd5 e2 21. Rdb5 e1=R
https://lichess.org/wtJoIdT3#42