A cube side 3 is divided into 27 unit cubes. The unit cubes are arbitrarily labeled 1 to 27 (each cube is given a different number). A move consists of swapping the cube labeled 27 with one of its neighbours. Is it possible to find a finite sequence of moves at the end of which cube 27 is in its original position, but cube n has moved to the position originally occupied by 27-n (for n = 1, 2, ... , 26)?
Solution
Answer: no.
We label the available positions of the cubes as black or white in a checkerboard fashion. Thus we might label the hidden central position as black. Then the middle of each visible edge is black, and the eight corners and the centers of each visible face are all white. Now a move necessarily changes the color of the location of cube 27. It ends up in its original position, so the total number of moves must be even.
We can also regard each move as a transposition of the numbers 1, 2, ... , 27. The required final position is a permutation which can be derived from the original position by 13 transpositions, namely (1, 26), (2, 25), (3, 24), ... , (13, 14). But the parity of the number of transpositions in a permutation is independent of the transpositions, so the final position cannot be derived from the initial position by an even number of transpositions, and in particular by an even number of moves.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002