22 Nov 08
In the thread Thread 104039: "2 Queens", in the Forum "Only Chess", AttilaTheHorn wrote: "Theoretically, it's possible for each side to have 9 queens."
Is it? Or not?
Problem: Show a game, from move 1 and onward, where every pawn is promoted, so the result will be 18 queens on the board.
22 Nov 08
1 edit
Originally posted by FabianFnasAnd if I did read the thread to its end, I would discover that one solution is already there...
In the thread Thread 104039: "2 Queens", in the Forum "Only Chess", AttilaTheHorn wrote: "Theoretically, it's possible for each side to have 9 queens."
Is it? Or not?
Problem: Show a game, from move 1 and onward, where every pawn is promoted, so the result will be 18 queens on the board.
I've sent a message to the moderator to delete the thread.
22 Nov 08
This is not possible in 40 moves (which would be the minimum required by length of time it takes a pawn to promote alone)
Here's my first attempt. It's 60 moves, which means a whole 20 moves of "waste" It's the shortest posted on this forum so far though, and gives people something to beat 🙂
22 Nov 08
Originally posted by doodinthemoodHmm this one is cool, but its not 18 queens. So I guess we have 2 problems. Least possible moves for 16 queen promotions and the other having 18 queens on the board 😉
This is not possible in 40 moves (which would be the minimum required by length of time it takes a pawn to promote alone)
Here's my first attempt. It's 60 moves, which means a whole 20 moves of "waste" It's the shortest posted on this forum so far though, and gives people something to beat 🙂
[pgn]1.a4 b5 2.a5 b4 3.a6 Bb7 4.axb7 b3 5.Ra2 bxa2 6.b ...[text shortened]... 54.Qa2 d1=Q 55.Q8a3 Kg4 56.Qd8a5 f4 57.Qac3 f3 58.Qhf7 f2 59.Qfa7 Qfe2 60.Qa7a5 f1=Q [/pgn]
22 Nov 08
1 edit
Originally posted by doodinthemoodI think we can prove that the lower bound must be higher than 40.
This is not possible in 40 moves (which would be the minimum required by length of time it takes a pawn to promote alone)
Each pawn needs to make 5 moves to reach the other side of the board, so that is 16x5 half-moves. Also, the a-file pawns will bump into each other, so at l ...[text shortened]... (I think - I am a bit shaky on this one).
Now I just need to find _which_ 45 moves will do it!
Nice problem!
23 Nov 08
Originally posted by crazyblueThose surviving Knights cost me a move each, unfortunately. They're in the way of the promoting pawns and there's no way to take them with the promoting pawns.
very nice. i think will be difficult to top that. maybe a task for those proof game experts? (which i am not)
also funny that a knight survives on each side 😉
I'm now working on a scheme where the promoting pawns have no obstacles on the back ranks.
24 Nov 08
Originally posted by SwissGambitHere's a much better scheme with even some bonus captures available on the back ranks to save even more time:
Those surviving Knights cost me a move each, unfortunately. They're in the way of the promoting pawns and there's no way to take them with the promoting pawns.
I'm now working on a scheme where the promoting pawns have no obstacles on the back ranks.
49.0 moves now.
24 Nov 08
Originally posted by SwissGambitSee the thread in the first post for an update. This task has been done in 3 different ways in 48.0 moves.
Here's a much better scheme with even some bonus captures available on the back ranks to save even more time:
[pgn]
1. f4 h5 2. f5 h4 3. g4 h3 4. g5 a5 5. c4 a4 6. b4 a3 7. b5 d5 8. c5 d4
9. Bb2 e5 10. Bg2 e4 11. Nc3 dxc3 12. Nf3 exf3 13. e4 axb2 14. a4 hxg2 15. h4
Rh6 16. gxh6 Ra6 17. bxa6 Bd6 18. cxd6 Be6 19. fxe6 f5 20. e7 Kf7 21. e5 c5
22. d4 ...[text shortened]... e7 b2 46. d8=Q bxa1=Q 47. Qdd5 Qfa6 48. Qeh5 f1=Q 49. e8=Q c1=Q+
[/pgn]
49.0 moves now.