1. DonationQuirine
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    10 Feb '05 06:06
    In a game white plays 1.e4
    On the 5th move a knight takes a rook checkmate.
    Who gets mated is open.
    How did the game go?
  2. Standard memberBigDogg
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    11 Feb '05 03:31
    Originally posted by Quirine
    In a game white plays 1.e4
    On the 5th move a knight takes a rook checkmate.
    Who gets mated is open.
    How did the game go?
    How about 1.e4 Nf6 2.Qe2 (2.f3) 2...Nxe4 3.f3 (3.Qe2) 3....Ng3 4.Qxe7+ Qxe7+ 5.Kf2 Nxh1#.



    Pity about the dual on move 2. Otherwise, a nice problem.
  3. DonationQuirine
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    11 Feb '05 09:02
    Originally posted by BigDoggProblem
    How about 1.e4 Nf6 2.Qe2 (2.f3) 2...Nxe4 3.f3 (3.Qe2) 3....Ng3 4.Qxe7+ Qxe7+ 5.Kf2 Nxh1#.

    [fen]rnb1kb1r/ppppqppp/8/8/8/5P2/PPPP1KPP/RNB2BNn w kq - 0 6[/fen]

    Pity about the dual on move 2. Otherwise, a nice problem.
    Correct! How long did it take you to solve this?
    If you're interested here's a link about this problem. It states that even Kasparov and several other very strong GM's couldn't solve this.
    http://www.chessbase.com/puzzle/puzz05d.htm

    Quirine
  4. Standard memberBigDogg
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    11 Feb '05 17:06
    Originally posted by Quirine
    Correct! How long did it take you to solve this?
    If you're interested here's a link about this problem. It states that even Kasparov and several other very strong GM's couldn't solve this.
    http://www.chessbase.com/puzzle/puzz05d.htm

    Quirine
    It took me awhile to solve this. I solved in my head during slow points at work, but finally got it at home (using a board sure helps!). I'd say an hour or two of total thought.

    I'm not surprised that it stumped GM's. They are not trained to think this way at all. For them, White and Black will always be opponents. In this type of problem, White and Black must cooperate to achieve the goal.

    A similar example is http://www.chessatwork.com/board/showthread.php?threadid=19565.

  5. Joined
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    11 Feb '05 21:36
    Well a good solver of heterodox problems can be more succesful than a GM.Helpmate problems are perhaps the most challengig, even more so than the selfmate ones. But like in all problemsolving it helps to picture the possible final positions and then to discard the impossible and to distinguish the exact order of moves (so no violation of the rules occurs), and then you've got the solution. But the hardest part is to imagine the final position...
  6. Standard memberBigDogg
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    12 Feb '05 01:41
    Originally posted by ilywrin
    But the hardest part is to imagine the final position...
    In a Proof Game, the final position is given to you. A good example is:



    Position after Black's 4th move. How did the game go?
    a) diagram
    b) remove Black's Queen
  7. Joined
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    12 Feb '05 10:31
    Originally posted by BigDoggProblem
    In a Proof Game, the final position is given to you. A good example is:

    [fen]rnbqkbnr/pppp2pp/4pp2/8/8/3P4/PPP1PPPP/RNB1KBNR w KQkq - 0 5[/fen]

    Position after Black's 4th move. How did the game go?
    a) diagram
    b) remove Black's Queen
    a) 1.d3 e6 2.Qd2 Ba3 3.Qb4 f6 4.Qf8+ Bxf8

    b) 1.d3 e6 2.Bh6 Qg5 3.Qc1 Qxc1 4.Bxc1 f6
  8. Joined
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    12 Feb '05 13:393 edits
    Just laid my eyes on a beautiful problem and not too complicated at that: Mate in 1 🙄

    EDITED: FEN problems 🙁
  9. Standard memberXanthosNZ
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    12 Feb '05 15:29
    Originally posted by ilywrin
    Just laid my eyes on a beautiful problem and not too complicated at that: Mate in 1 🙄
    [fen]r1bknbrN/pppppppp/8/8/8/8/PPPPPPPP/nRBNKB1R [/fen]
    EDITED: FEN problems 🙁
    Depending on who it is to move it's Nxf7# or Nxc2#.
    Doesn't seem that beautiful to me. And nothing compared to a Babson Task.
  10. Standard memberBigDogg
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    12 Feb '05 15:591 edit
    Originally posted by XanthosNZ
    Depending on who it is to move it's Nxf7# or Nxc2#.
    Doesn't seem that beautiful to me. And nothing compared to a Babson Task.
    You're missing the point. You have to deduce whose move it is.

    This is known as a parity problem. The structure is such that neither side can lose tempo, so I can say that White has made an odd number of moves and black has made an even number of moves. Parity is different, which only happens after white has moved. Therefore, it is black's turn, and the answer is 1...Nxc2#.
  11. Joined
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    13 Feb '05 08:41
    Right. Heh...it was a problem by J.L. Tulco.
    I took it from here:
    http://www.janko.at/Retros/
    A real nice database of retrograde problems but most of them were too difficult for me 🙂
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