Rubik’s 7x7x7 Cube
I’m not about to sit down and work up a solution like we did before, but it shouldn’t be impossible to repeat the same sort of analysis. I will point out, however, that the solver in this video is making heavy use of both of our solution techniques: commutators and a tower of nested subgroups.
The nested subgroups are obvious. As the solution progresses, more and more structure becomes apparent, and is preserved as the solution continues. In particular, the solver builds up the centers of faces and then slips to the subgroup of maneuvers which leaves such “big centers” fixed in place. Near the end, almost all of the moves are twists of the outer faces, because these are assured not to affect anything but the edge and corner cubies.
The commutators take a quicker eye to spot, but they’re in there. Watch how many times he’ll do a couple twists, a short maneuver, and then undo those couple twists. Just as we used such commutators, these provide easy generalizations of basic cycles, and they form the heart of this solver’s algorithm.
Alexandre asked a question about the asymptotic growth of the “worst assembly time” for the cube. What this is really asking is for the “diameter” of the th Rubik’s group . I don’t know offhand what this would be, but here’s a way to get at a rough estimate.
First, find a similar expression for the structure of as we found before for . Then what basic twists do we have? For we had all six faces, which could be turned either way, and we let the center slices be fixed. In general we’ll have slices in each of six directions, each of which can be turned either way, for a total of generators (and their inverses). But each generator should (usually) be followed by a different one, and definitely not by its own inverse. Thus we can estimate the number of words of length as . Then the structure of gives us a total size of the group, and the diameter should be about . Notice that for this gives us , which isn’t far off from the known upper bound of quarter-turns.