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Regarding finding a sequence of twists that requires a maximal number of

repetitions to restore the original state, it is indeed true that 1260 is the

maximum, but the sequence suggested by ED (on July 18) is not it. (He

thought its period was 216; ALAN thought it was 1260. It's actually 315.)

This assumes that we don't consider the orientation of the center-face cubies

nor the orientation of the cube as a whole.

First off, note that the transformation given by ED was ill-specified, since he

didn't say which way to rotate the cube relative to the rotations of the faces.

In Singmaster's notation the two operations would be, say, LBRF and LFRB.

It turns out these are each period-315, though it's not immediately obvious

that the two should be at all similar. Having done (LBRF)^1, you'll find

there are 7 edge cubies that have cycled through each other's positions, and

the other 5 edge cubies have done likewise, 5 corner cubies have rotated 120

degrees in place, and the other 3 corner cubies have cycled through each

other's positions and one of them has rotated. Hence, if we do (LBRF) a

multiple of 9 times, the corner cubies will be back where they started, and if

we do it a multiple of 35 times, the edge cubies will be restored, so if we do

(LBRF)^315, all of the cubies will be back to the solved state.

Perhaps the reason ALAN thought this was period 1260 was because it takes

1260 twists to restore the original cube. But since the transformation that we

are repeating is a sequence of four twists, it actually has period 315. If you

count total twists, you can obviously get "identity" sequences with lengths

much greater than 1260. Or perhaps ALAN was considering the transformation

to be LJ (twist left face, rotate cube about vertical axis). This indeed has

period 1260, but it "cheats" in that we're not supposed to consider reorienting

the entire cube as a legitimate operation. If we do, then it is again possible

to get periods greater than 1260.

To get a period-1260 transformation, we need to observe whence the period

arises. If we do a sequence of twists, and then look at the cycles of the form

"face X moved to where face Y used to be, face Y moved to where face Z was,

. . . , face <mumble> moved to where face X was", we take the least common

multiple of the lengths of those cycles. If we repeat the sequence of twists

that many times, the faces will moved around those cycles an integral number

of times and end up where they started; if we stop short of that many

repetitions, at least one of the cycles will be left stuck in the middle.

(Apologies to any readers who are familiar with group theory and are bored

by this verbosity.)

To get a period of 1260, we need to have a cycle of 7 edge cubies, another

cycle of 2 edge cubies where one of them is also rotated (so that the four

faces on those cubies form a single period-4 cycle), a cycle of 5 corner cubies,

and a cycle of 3 corner cubies where again one of them is rotated (forming

a face-cycle of length 9). If, instead of the 2 edge cubies, we could get 4

edge cubies cycling with one of them rotated, that would be a period-8 cycle

and the overall transformation would have period 2520, but due to the fact

that it's impossible to swap exactly two cubies it turns out you can't get that

combination of subcycles.

Based on this analysis, it's not hard to construct a period-1260 sequence,

although this one is almost certainly not the shortest such (it's 24 twists):

F U F^2 u r f b l r f D^2 F b R U (R^2 F^2)^3 u r B