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You are given a 3 by 3 chessboard with a knight on each corner, where the knights in the top row are black and in the bottom row are white.

On each turn, you may move a knight of either color (the way it moves in chess) to an unoccupied square.

Is it possible to rearrange the knights so that they are still in the corners, but where knights in opposite corners are the same color? If so, how many turns does it take to do so?

knights

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    $\begingroup$ Nice! I assume you like chess :) $\endgroup$ Commented Jan 13, 2016 at 21:00

2 Answers 2

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Give these names to all the squares:

163
4 8
725

Each number can only be accessed by way of the numbers before and after it (where 8 wraps around to 1). That means they form a loop. Since they can never pass each other up on the loop, their relative ordering cannot change. Therefore it is impossible.

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Probably not the best explanation, but....

This is not possible. In order to change relative positioning between the pieces, you would need an additional square while making moves (a total of 9 squares) to hold a piece temporarily. (4 original + 4 to hold pieces + 1 for temporary positioning). However, despite the 3x3 grid having 9 squares, the middle square is not reachable by any knight, regardless of the number of moves it makes.

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