# Another closed path on the Rubik's cube

Is it possible to draw a closed path on the surface of a standard $3\times3\times3$ Rubik's cube

• such that the path traverses exactly one diagonal of each of the $54$ little squares, and
• such that the path does not intersect itself?
• Are two sections of the path allowed to move through the same point? Example: Imagine 4 squares in a 2*2 square. The path enters the first time through the bottom left corner, goes to the middle point and then leaves through the bottom right corner. Later the path returns through the top right corner, again goes to the middle point and then leaves through the top left corner. – The Dark Truth Nov 21 '15 at 16:52
• @TheDarkTruth: No, the path must not intersect itself. – Gamow Nov 21 '15 at 16:56
• Alright. I wasn't entirely sure wether "intersect" only meant "cross itself" or if "touch itself" was included. – The Dark Truth Nov 21 '15 at 17:00
• @TheDarkTruth, I agree the question doesn't really make sense as stated. If the path can't even "touch itself", then no, it obviously can't form a closed path because it can't end where it started. – user1717828 Nov 22 '15 at 0:01

## 1 Answer

The answer is no.

Consider the graph formed by "corners" of the squares on the Rubik's cube. These are the vertices we're moving between. Assign each vertex coordinates in $\{0 \dots 3\}^3$. We observe that diagonal moves can't change the parity of $(x+y+z \bmod 2)$: The problem asks for a Eulerian cycle on this subgraph.

Note that there are vertices of degree three: they are marked by the red spheres on the corners of the cube in the above image. This means there is no Eulerian cycle: there is an odd number of ways from/to the corners, but a Eulerian cycle needs an even number (intuitively, it needs as many ways from as it needs ways to each vertex).

• That wouldn't negate the possibility of a path which crosses each square on a diagonal and doesn't intesect itself but also traverses some non-diagonal edges. I think the possibility of such a path would depend upon the definition of "intersect", however. – supercat Nov 30 '15 at 2:56