Another Problem the Computer Couldn't Solve
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Both bxc5 and dxc5 could have been black's last move, besides c7-c5. You can fix it with white pawns on b6 and d6, but then you're giving it away already.
White is missing seven members of his army. Black pawns have captured seven times. Given White's remaining forces, Black must have played c5. White would have one less piece, had Black captured onto c5!
benws wrote: Easy! 1. dxc6 e.p. and 2. exf7#. or if 1...fxe6 2. Bxe6#. but only if the last move was c5. If Black's last move was bxc5 or dxc5, there is no mate. so don't brag that the computer didn't get the problem, there is no solution!
Refer to my post directly above yours. Black's kingside pawns have already captured all seven of White's missing pieces. There was nothing left for Black's pawn on c5 to have captured.
Jan 19, 2008
0
#25
silentfilmstar13 wrote:
White is missing seven members of his army. Black pawns have captured seven times. Given White's remaining forces, Black must have played c5. White would have one less piece, had Black captured onto c5!
You are absolutely right! :) Nice observation :) Therefore from all of the info that we came up with we can use the en passant rule to capture to make this a mate in 2. We worked well as a team. :)
silentfilmstar13 wrote:
White is missing seven members of his army. Black pawns have captured seven times. Given White's remaining forces, Black must have played c5. White would have one less piece, had Black captured onto c5!
wow, this thing is solid. only possible move black could have played was c5.
Good work, team!
Jan 19, 2008
0
#28
I would think this is one of the most trickiest puzzles out of all puzzles since it took a lot of brainstorming to figure out what blacks last move was. :)
wow, I really enjoyed thinking about this puzzle and reading all the comments, this is really amazing!
i see it
Solved very easy
silentfilmstar13 wrote:
White is missing seven members of his army. Black pawns have captured seven times. Given White's remaining forces, Black must have played c5. White would have one less piece, had Black captured onto c5!
Ah yes, you're right :)
Finding the solution is easy. Proving the solution works is the hard part.
You can also say "The puzzle says White to move and mate in two. So there must be a mate in two. If white can't take en passant, there's no mate in two. Therefore dxc6 must be a legal move and that's the solution." The reasoning may be a little flawed, but it does work when you know there is a solution.
Cool puzzle but i must say i hate the EP puzzles now. They`ve gotten old at this stage.
I see it and you have to figure out black's last move in order to solve.
I solved
Solved. There are many mate in 3 variations, as already discussed, but only 1 (that I can see) mate in 2 variation.
I get it now! Ha ha!
Since pawns can never move backwards then the pawns on f7 and g7 never moved through out the whole game. How does black have a pawn on h6? It must have been the g7 starting pawn and at one point it captured diagonally to move to h6. Could this have been the previous move? No because white is already occupying the g7 square. What about the black pawn on g5? This has to be the e7 starting pawn because we have already determined the f7, g7, and h7 starting pawns and it is not possible for a black pawn on the d-file to reach g5. And since the white bishop occupies f6 then this pawn did not move on the previous turn. There is a black pawn on h3. This pawn is the d7 starting pawn since the c7 starting pawn would not be able to reach that square. This pawn did not move on the previous turn because the white rook occupies g4. What is left? The black pawn on c5.
If I'm not mistaken this doesn't exclude that black moved bxc5 the previous move.
Yep :) I missed the possibilities of bxc5 or dxc5. I think we need more info about the pawn on c5.