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.
Originally posted by FabianFnas 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.
And if I did read the thread to its end, I would discover that one solution is already there...
I've sent a message to the moderator to delete the thread.
Ive recorded it once. On move 99 queen #18 comes on the board. It would be interesting to see what is the lowest possible moves. Or is there a solution for that too already?
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 🙂
Originally posted by doodinthemood 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 🙂
Hmm 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 😉
Originally posted by doodinthemood This is not possible in 40 moves (which would be the minimum required by length of time it takes a pawn to promote alone)
I think we can prove that the lower bound must be higher than 40.
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!
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 😉
Originally posted by crazyblue 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 😉
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.
Originally posted by SwissGambit 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.
Here's a much better scheme with even some bonus captures available on the back ranks to save even more time:
22 Nov '08 08:05
YouTube
Eighteen queens
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