In the last post, I examined the number
of times a set of moves would loop back
on itself, on a 2x2x2 Rubik's Cube.What
this means, is that if I take a solved Rubik's
Cube, and I do a particular set of moves on
it, then that set of moves' Cube-Length was
defined as the number of times you would
have to do that set of moves until the Rubik's
Cube goes back to its original form. So I took
a number of random set of moves, and I tried it
out on a 2x2x2 Cube, and I got the results.
In this post, I'm going to use a particular set of
moves, which you need in order to actually
solve the Cube.
moves, which you need in order to actually
solve the Cube.
The results are those done on a 2x2x2 cube, not
a 3x3x3 Cube.
Algorithm Cube-Length
U, R, U', R', U', F', U, F (The Second-Layer algorithm.) 4 (Times to loop back)
F, R, U, R', U', F' (The 'Cross' algorithm.) 6
U, R, U', L', U, R', U', L (The corner-position algorithm.) 3
R', D', R, D 6
As clearly seen here, the length of the algorithm has
nothing to do with its Cube-Length. It's only the level of
complexity, or simplicity that matters. Here, the latter
makes the Cube-Length go up.
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