TITLE: Why is it impossible to move from the corner to the center of a $3 \times 3 \times 3$ cube under these conditions?
QUESTION [5 upvotes]: There is a $3 \times 3 \times 3$ cube block, starting at the corner, you are allowed to take 1 block each time, and the next one must share a face with the last one. Can it be finished in the center?
This is a question about graph theory (I think), and obviously it is impossible to finished in the center. I start to consider about the Hamiltonian path and the degree of each vertex , but it is different because you have to start and end with a specific vertex. Can anyone tell me the reason why it is impossible?

REPLY [7 votes]: Set a coordinates system, where every block corresponds to a triplet $(a,b,c)$, for example, the starting cornere will be $(1,1,1)$ and the center ($2,2,2)$. Every time you move from a block to another who shares a face with the previous you add or subtract $1$ from the previous triplet. You start from $(1,1,1)$ and you want to finish in $(2,2,2)$ with $26$ steps. Call $a$ the number of $+1$ and $b$ the number of $-1$, you must have
$$a+b=26$$
and
$$a-b=2+2+2-1-1-1=3$$
and this is not possible.