# A 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 each of the $54$ little squares exactly once, and
• such that the path does not go through any corner of these little squares?
• By "not go through any corner" do you mean non-diagonal? – JonTheMon Nov 20 '15 at 17:04
• Reminds me of the "Attach 3 houses to 3 wells without crossing pipes" puzzle – APrough Nov 20 '15 at 17:08

An observation is..

that you can traverse a cube's face from one corner and end up diagonally opposite or in same column (or row).

Using this observation, one of many possible solutions is (left image is front of cube, right image is inside view):

• very nice, what did you use to make these drawings? – Jonah Nov 20 '15 at 19:26
• @Jonah: I used Microsoft Powerpoint. It has a draw button for cube and path (in red), but for the 3x3 grid I drew individual lines. – beginner 101 Nov 22 '15 at 0:47

The minimum degree of the graph involved is 4: each of the 56 squares has exactly four neighbors. Thus, the graph is 4-connected.

It is also planar: we can obviously draw it on a sphere (imagine a "deformed" Rubik's cube.)

• Another argument that it's planar is that it can clearly be drawn on a sphere, which is equivalent to being planar. – xnor Nov 21 '15 at 2:30
• That's a very elegant argument, @xnor. Thank you! – Lynn Nov 21 '15 at 22:45

Yes:

As can be seen:

(Similar path on opposite three faces)

There are actually sticker mods that does exactly what you describe, called Maze Cubes, and they are pretty hard to solve. There are a lot of different lay-outs for Maze Cubes. Here is a potential path - the original one from 1982 (available for sale at OliverStickers.com):

Yes, you can

On the base, draw an S shaped path going from near-left to far-right
On the right and front faces, do an S shape so you end up on the front face, top left.
On the top, draw an S shape, going from front-left to back-right
On the left and back, do an S shape, so you end up on the left face, bottom-near.

Yes, it is possible. Each and every square in the Rubik's cube has four non-diagonal neighbors, including those neighbors on adjacent faces. There are no squares with an odd number of neighbors, so the parity checks out, and a closed path can be drawn.

I have just sketched out a couple of these paths. If requested, I can attach.

• The argument you are thinking only works for Eulerian cycles, while this problem is asking for a Hamiltonian cycle. The former uses every edge once, the latter must use every vertex once. – Mike Earnest Nov 20 '15 at 17:59
• @Sleafar: The edge squares of a 3x3 square have three neighbors each, not an even number. – ard Nov 20 '15 at 18:01
• @ard Right, my fault. But can you please provide an example? – Sleafar Nov 20 '15 at 18:04