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An $n \times n \times n$ cube is divided into $n^3$ unit cubes, of which $n^2$ are black and the remaining ones are white. A move consists of selecting two parallel $1 \times 1 \times n$ columns (i.e., blocks of $n$ unit cubes aligned in the same direction) and swapping their positions without rotating them. Is it always possible, by a sequence of such moves, to arrange the unit cubes so that all black unit cubes lie on a single face of the big cube?

(I do not know the source of the problem.)

The statement is true for $n=2$, but checking bigger values of $n$, is hard. I tries multiple colorings to find something invariant, which would help me find a counterexample but I failed. I am not even sure whether the answer is "no".

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    $\begingroup$ When you say "two parallel $1×1×n$ columns (i.e., blocks of n unit cubes aligned in the same direction)" you presumably could have said "two $1×1×n$ columns or two $1×n×1$ columns or two $n×1×1$ columns" $\endgroup$ Commented 14 hours ago

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It seems to me that it is not always possible, even when $n=2$. Every possible state is some rotation/reflection of one of the six states below. The states in the top row cannot reach the states in the bottom row.

  W---W      W---B      W---B
 /|  /|     /|  /|     /|  /|
W---W |    W---B |    B---W |
| B-|-B    | B-|-W    | B-|-W
|/  |/     |/  |/     |/  |/
B---B      B---W      W---B  


  W---W      W---B      W---W
 /|  /|     /|  /|     /|  /|
W---B |    W---B |    B---W |
| W-|-B    | W-|-W    | B-|-B
|/  |/     |/  |/     |/  |/
B---B      B---B      W---B
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