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A standard Rubik's cube is initially unscrambled (say per picture below, green facing observer)

Rubik's cube

The same manipulation is repeated:

  1. rotation of front face by 1/4 turn (say clockwise)
  2. rotation of the whole cube 1/4 turn around vertical axis (say anticlockwise seen from top)

(with the proposed orientations, the center of the rotated face at 1 will cyclically be green, orange, blue, red; in standard notation these moves cycle between F L B R; the center of the upper face always remain white).

After how many manipulations will the cube be first unscrambled again? (note: by symmetry, this is independent of the direction of the rotations, as long as they remain the same across manipulations).

Is there a simple argument to tell which face is facing the observer at that point (equivalently, to determine the answer modulo 4)?

Note: I'm interested in the reasoning to solve that kind of problems, rather than in the answer for that particular manipulation.

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    $\begingroup$ I did 420 manipulations on a real cube of mine amd found that it resulted in a twist of three corners at the bottom. The answer to the original problem must therefore be 3 × 420 = 1260. $\endgroup$ – Parcly Taxel May 13 '17 at 15:18
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    $\begingroup$ In fact, 1260 is the largest possible order for any move sequence on the cube. $\endgroup$ – Parcly Taxel May 13 '17 at 15:19
  • $\begingroup$ @Parcly Taxel: indeed the maximum order of any transformation is 1260. But this only tells us that the order of the transformation resulting from the 4 moves F L B R is at most 1260 (and a divisor of that); when the desired number of moves is 4 times that order. $\endgroup$ – fgrieu May 15 '17 at 8:34
  • $\begingroup$ Wrong. The manipulations I did above already prove that the order is 1260 - 420 rounds twist three corners, so 1260 rounds and only 1260 rounds restore the solved state. $\endgroup$ – Parcly Taxel May 15 '17 at 8:35
  • $\begingroup$ @Parcly Taxel: I fully agree that the 420 manipulations in your first comment is a valid proof that 1260 manipulations/moves is the right answer, and the first posted. And that the observation in your second comment is valid. What I pointed is that the maximum order 1260 in this second comment is a number of sequences of manipulations, not of individual manipulations. Here, the order is 315 sequences of 4 manipulations, giving 1260 manipulations/move. $\endgroup$ – fgrieu May 15 '17 at 12:09
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Answer for part $2$ with suggestion for general idea:

The face facing the observer at the end must be green, so the solution is

$$0\mod 4\,.$$

Consider the piece starting at the bottom right of the green face. At the end of each manipulation, it remains at the bottom right of the face facing the observer, so in order to be in the correct place at the end, the face facing the observer must be green.

Moreover after $4$ manipulations the piece has rotated, so for the piece to be in the correct orientation, we must have performed some multiple of $12$ manipulations - that is the solution is

$$0 \mod 12\,.$$

In general, look at when specific pieces or collections of pieces (say corners) return to their original position. You may also cluster manipulations. Knowing the solution is $0 \mod 12$, my next step would be to observe what happens to another piece after $12$ manipulations.

The top right piece (of the green face) for example is moved to the bottom right of the green face, left alone for to manipulations then moved back up, but rotated, hence after $12$ manipulations it is left unchanged.

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(Answering my own question, after a day of thought)
As pointed in Robert Chamberlain's answer, the green/red/yellow piece is first back to its position w.r.t. the center pieces (albeit with a different orientation) after a sequence of 4 manipulations. It follows that the desired number of manipulations is a multiple of 4, after which the face in front of the observer is the same as initially.

From then on, we can consider only iterations of the transformation comprising the 4 rotations F L B R (hereafter a sequence); our desired result is 4 times the order of that sequence, thus no more than 4×1260 = 5040.

We can find experimentally after how many sequences each piece (or group of pieces) is first back to its original position, including orientation, w.r.t. the center pieces; then compute the Least Common Multiple of these for all pieces, obtaining the period in sequences; and finally multiply by 4 to get the number of manipulations. We get, with only 9 sequences and observations of the cube after each:

  • 3 sequences for 5 corners (the 4 bottom ones, and green/red/white)
  • 5 sequences for 5 sides (the 4 bottom ones, and green/red)
  • 7 sequences for the 7 other sides
  • 9 sequences for all 8 corners

We thus get that 5×7×9 = 315 sequences = 1260 manipulations are necessary to get back to the original position. That confirms the value given in Parcly Taxel's comment.

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