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Bobby Fischer liked to play the following game on a standard $8\times8$ chessboard:

  • In his first step, Bobby placed a white rook and a black rook somewhere on the chessboard (on two different squares, of course).
  • In every further step, Bobby picked one of the rooks, and then moved it away from its current square to a horizontally or vertically adjacent (currently empty) square.

Is it possible that after $64\cdot63=4032$ steps, one has generated each of the $4032$ possible positions for the two rooks exactly once?

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Answer:

No, this is not possible.
Call a position even if the two rooks are on squares of the same color squares, odd if they are on squares of different colors. Every move changes the position from even to odd or from odd to even. There are a total of $31\cdot 64=1984$ even positions and $32\cdot 64=2048$ odd positions, so it is not possible to reach them all while alternating between them.

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  • $\begingroup$ Is this answer valid? I don't see any reason why this 'parity' has to change every move. It's perfectly fine for a rook to move from a white space to another white space. $\endgroup$
    – Phylogenesis
    Commented Feb 12, 2015 at 9:30
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    $\begingroup$ @Phylogenesis In the question, the rook is moved to an adjacent square. All squares adjacent to white squares are black. $\endgroup$
    – Trenin
    Commented Feb 12, 2015 at 15:22
  • $\begingroup$ Guess who missed the word 'adjacent'? $\endgroup$
    – Phylogenesis
    Commented Feb 12, 2015 at 15:27
  • $\begingroup$ Neat puzzle! What about K/Q/B/N? $\endgroup$
    – Hauke Reddmann
    Commented Apr 6, 2022 at 13:58

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