Running through all permutations of a Rubik's cube According to Wikipedia a $3 \times3\times3$ Rubik's cube has $43252003274489856000$ permutations. 
I never tried solving one myself (too tedious), however I wondered, if one could miraculously "solve" the cube in color blind mode (all faces of the cube have the same color to him or the person is just blind folded). (Yes, it is tedious, but I guess sooner or later he will just hit the right permutation and hopefully someone tells him, that it's good and he may stop.)
I wondered, if there were some kind of "best" approach to run all through those  permutations without too many repetitions, for example the algorithm makes sure one gets any permutation at most $n$ times. 
I don't know the answer yet and probably won't have it anytime soon. However I would be quite interested in seeing this algorithm and thus I hope the question has been asked before and been solved. Any constructive comment/answer is appreciated. 
 A: In this link, it is claimed (and I believe them!) that one can achieve every single permutation of the cube without ever revisiting a single vertex twice. The details are, not surprisingly, pretty devilish. 
The basic idea is to imagine creating a graph with one vertex for each permutation of the cube. We'll call the solved state's vertex $1$, and give a name to each vertex depending on the moves that need to be taken to get there from $1$. If you perform the move $URU^3$ starting from $1$, you'll call that vertex $URU^3$ (Note that we have to have some convention about names; they're not unique). It's not especially important for this question, but the graph is called the Cayley graph of the Rubik group -- this is what let me search for an answer, by knowing the magic password.
Your question could now be phrased as, "Are there any circuits (walks that return to where they started, also called cycles) of this Rubik graph that visit each vertex at least once, without too many repeats?" The best kind of circuits are called Hamiltonian circuits, and they visit each vertex exactly before returning to where they started.
What the webpage shows, in full technical jargon, is that the Cayely graph of the Rubik's cube group admits a Hamiltonian circuit. 

It's my understanding that it would be completely ridiculously unmanageable to use the circuit they found. It was constructed by "opening up" more and more states of the cube in a very controlled manner. 
For example, if you can only use the move $UR$ (not $U$ and $R$ separately, yet), then by repeating the move $UR$ $105$ times, you'll wind up where you started. If you can use any combination of $U$'s and $R$'s, there are a lot more possibilities -- $73{,}483{,}200$, to be precise! But you can string together a bunch of walks that look like $UR$ repeated $105$ times, with some appropriate shifting between walks, to get all $73{,}483{,}200$ moves accessible with only $U$ and $R$.
Apparently the author of the page played that game with increasingly less limited movesets, slowly adding the moves $D,\ L$, and $F$ to their repertoire. So I think the cycle they obtained looks a lot like a bunch of those little $UR$-repeated-$105$-times walks, but with various move combinations thrown in between the walks. 
It is definitely not something that could be committed to memory! But probably something a very, very dumb computer could implement to solve the Rubik's cube with absolutely no strategy besides executing an unfathomably large sequence of the basic quarter-turn moves.
