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Consider a 3 by 3 by 3 cube — in essence, a Rubix cube — like the one pictured in the following image:

3 by 3 by 3 cube

Start from the cube in the middle, enclosed on all sides. Moving to only cubes that are directly adjacent, is it possible to visit all small cubes exactly once?

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  • $\begingroup$ Just to clarify, by center small cube you mean the inner cube or one of the center cubes on a face? $\endgroup$ – gabbo1092 Nov 29 '18 at 18:45
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    $\begingroup$ It is the inner one, the one whose faces are all adjacent to other small cubes. $\endgroup$ – nafhgood Nov 29 '18 at 18:46
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No

The reason is to do with parity:

Color the little cubes in an alternating pattern, black and white. Suppose the corners are white. Then there are 14 white and 13 black cubes. You always go from a white to a black and from a black to a white. But the center cube is black. So if you start from black, you'll have covered either one more black than white or the same number of each at every step. You'll never get to one more white than black.

Here's an interesting video: 3x3x3 snake cube puzzle. Note how this cube has green corners and when it is stretched out it has green on either end. This is inevitable for the same reason.

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    $\begingroup$ Was just about to post this; well done. $\endgroup$ – a guy Nov 29 '18 at 18:53
  • $\begingroup$ I knew I'd have to be quick! It's a very old trick! $\endgroup$ – Dr Xorile Nov 29 '18 at 19:01
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I think the answer is

No

because,

the cubes can be represented as nodes on a graph, connected by edges to adjacent cubes. Each of the corner cubes is adjacent to 3 others, and for a graph to be traversible there must be 0 or 2 nodes with odd degree (that is, an odd number of edges to other nodes).

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    $\begingroup$ This does not apply. You're not trying to visit every edge. It's a hamiltonian path not an euler path. $\endgroup$ – Dr Xorile Nov 29 '18 at 18:50
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    $\begingroup$ Oh. Guess it's been too long since I've dealt with graph theory. $\endgroup$ – Emnepho Nov 29 '18 at 18:52

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