Come to my arms, my beamish boy!
Is it possible to arrange a Rubik’s cube such that no two adjacent facelets have the same color? I’m having a lot of trouble getting it to happen. If it’s impossible, I’m sure there’s a clever proof of that fact.
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Does “adjacent” include tiles that meet at the edges of faces, or is it just “all faces must have no two identically-coloured tiles that are adjacent”?
Tiles can’t meet at the edges of faces; there’s no piece on the cube that’s built like that. Unless I’m confused as to what you mean.
I mean, two tiles of the same color can’t meet that way.
You know, 10 seconds of careful thought before posting my inane comment would have been a really good idea. I haven’t actually played with a real Cube in years, and was just visualizing it as an abstract cube with coloured faces.
That’s what’s kind of fun about thinking about these kinds of problems on the cube. Obviously it’s possible to arrange 54 tiles of 6 colors across the surface of an arbitrary cube without any same-color adjacencies; that’s the 4-color problem. On the cube you’re limited by which pieces exist, and where they can move in relation to one another.
Hrm…one would think it would be a simple matter of repeating the same pattern with different color layout on each side.
like
1 | 2 | 3
———
4 | 5 | 6
———
2 | 3 | 1
now making that happen on the cube itself is up to you. i know how to do it the “wrong” way.