I just recently bought a Rubik’s 5x5x5 cube, and am happily working out a solution for it. I know it’s going to take a while, though I do remember the general approach from experience with the classic 3x3x3.
But while that’s going on… I was trying to reckon how many different states this 5x5x5 monster has. The only web page I’ve found on the subject is this one, but I disagree with that author’s calculations. So let me present my calculation, and maybe someone can confirm which of us is right, if either of us is.
There are 8 corner cubies which can share each other’s places, making 8! positional arrangements. Each cubie can have 3 possible orientations, although there are only 7 degrees of freedom for the set as a whole, because the last cubie’s orientation is fixed by the other 7. So in total, the corner cubies make 8! x 3[sup]7[/sup] possibilities.
There are 24 side edge cubies which can share each other’s places, and each cubie has 2 possible orientations. Again though, the orientation of any one is fixed by all the others. Thus there are 24! x 2[sup]23[/sup] possibilities for this set of cubies.
There are 12 middle edge cubies, each of which has 2 possible orientations, and the same restriction on their collective orientations. Thus there are 12! x 2[sup]11[/sup] possibilities for this set.
There are 24 inside corner cubies (face cubies nearest the corners). These have no orientation that’s interesting, or even noticeable — unless perhaps I were to draw little marks on them to tell the difference, which I’m not going to do. Also, these cubies come in quartets of the same color, and I don’t care (and couldn’t tell) if any two same-colored cubies were swapped around. So, there are 24! positional arrangements, but every 4! of those arrangements really count as one. Thus, there are 24! / 4! possibilities for this set.
A very similar computation applies to the 24 inside edge cubies (face cubies between the face centers and the middle edges). There are 24! / 4! possibilities for this set.
The 6 face center cubies have no interesting individual orientations, and are also fixed relative to each other. In fact we can put these six into some permanent orientation in space (blue on the bottom, red on the left, etc.), letting all the other cubies move around them. This saves us from having to consider orientations of the cube as a whole, which aren’t significant anyway.
So then, I make the total number of states for the 5x5x5 Rubik’s cube to be the product of the boldface numbers above, which is about 2.328 x 10[sup]73[/sup]. (Which means I can confidently scratch Exhaustive Search off my list of candidate algorithms, if I’d had any doubt. Luckily there’s a backup plan.)
Now the other guy’s number is higher than mine by about an order of magnitude. Even that would seem to be an amazing coincidence of proximity. For example, I don’t know where he’s getting his factorials raised to powers. I don’t know where he’s getting most of what he’s saying, actually. If his approach is the proper way to go about it, someone’s going to have to explain it to me.
Well, no one has to explain anything to me of course, but I sure would be grateful.