2
$\begingroup$

Consider an odd chessboard with an odd number of squares. A king is placed on each square of the board, then the kings are picked up and placed on each square of the board again. Can this be done in a way that every king is in a square next to its original position?

Note: we do not follow the rules of chess in this question, it is practical

$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

No, because every pawn will change the color of the square it stays on, and there is different number of light and dark squares on the odd-sized board.

Edit: Adding a new point into the question is a bad practice, see https://meta.stackexchange.com/questions/43478/exit-strategies-for-chameleon-questions

$\endgroup$
2
  • 3
    $\begingroup$ If the board is even, it is possible to split it into pairs of adjacent squares and just swap the pawns on them, so the answer is positive in this case. $\endgroup$
    – Wolfram
    Feb 1, 2017 at 20:01
  • $\begingroup$ Prove what?​ $\endgroup$
    – Wolfram
    Feb 1, 2017 at 22:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .