15
$\begingroup$

Is there a sequence of moves that can be repeated over and over again which can solve any legal position the Rubik's Cube? If so what is it, and if there's more than one, what's the shortest? If not, prove that it's impossible.

The sequence can be of any length but cannot be broken up into multiple pieces to be executed depending on Rubik's cube patterns; it must, when repeated from start to finish, end in a perfectly solved Rubik's cube.

$\endgroup$
  • 4
    $\begingroup$ Can you just do all possible (<20 move sequence> followed by <undoing previous sequence>) as a possible solution? (might not be shortest but it'll show one exists) You'll solve the cube at some point. $\endgroup$ – Sp3000 Nov 17 '14 at 0:50
  • $\begingroup$ @Sp3000 That's pretty long, the question asks for the shortest if one exists so. $\endgroup$ – warspyking Nov 17 '14 at 0:56
  • 3
    $\begingroup$ Mew, was that edit some kind of joke -_- $\endgroup$ – warspyking Nov 17 '14 at 0:56
  • 3
    $\begingroup$ @warspyking, indeed I hope you liked it :) $\endgroup$ – Kenshin Nov 17 '14 at 1:07
  • 1
    $\begingroup$ Another potentially interesting question, if anyone cared to solve it, would be the shortest primitive-recursive way of writing such a sequence, if one could define sub-sequences and then unconditionally invoke each sub-sequence in its entirety simply by including its name, so [spaces added for clarity] "1 tr 2 f 1b1 3 22r12" would define "1" as "tr", "2" as "f tr b tr", and "3" as "ftrbtr ftrbtr r tr ftrbtr". $\endgroup$ – supercat Nov 17 '14 at 18:38
17
$\begingroup$

The only way you can have a solve-all sequence is if you have a sequence of moves that goes through all 43 quintillion configurations of the Rubik's Cube. In order to do this, you need to draw a transition graph between all the states of the Rubik's Cube and find a Hamiltonian cycle through them.

This sequence of moves doesn't necessarily have to be 43 quintillion moves long - a simple sequence of 4 moves can produce a cycle of 1,260 configurations as seen in mdc32's answer, and in general a sequence of symbols in the group will produce a cycle of configurations much longer than the cycle itself. However, the sequence will still be very long, simply because 43 quintillion moves is still a lot.

Micah provided a link to a page that did construct such a Hamiltonian cycle in a comment. I haven't been able to make head or tail of its notation (or to figure out how to count the number of moves from the descriptions of the cosets), but it looks like the sequence of moves that is required is billions of moves long, which is still definitely outside of the realm of plausibility for memorization.

$\endgroup$
  • $\begingroup$ It wouldn't necessarily have to be 43 quintillion moves. Mine is only a four move sequence, technically, yet it cycles through 1260 states as you pointed out. $\endgroup$ – mdc32 Nov 17 '14 at 2:11
  • 1
    $\begingroup$ Really? That's really the ONLY way to do this? Wow... Are you sure there isn't like a 4 move sequence that can loop through them all or something? Even if it repeats a position but on a separate move? $\endgroup$ – warspyking Nov 17 '14 at 2:11
  • $\begingroup$ @mdc32 You're right. For example, in $\mathbb Z_2 \times \mathbb Z_2$, the sequence $(1, 0), (0, 1)$ does result in the whole group being traversed, so the length of the solve-all cycle doesn't necessarily have to be equal to the order of the group. And, as you pointed out, a sequence of 4 moves can generate a 1,260-state cycle. But I don't see there being a solution that's shorter than 10,000 or maybe 10,000,000 moves long. $\endgroup$ – Joe Z. Nov 17 '14 at 2:26
  • 3
    $\begingroup$ It exists. $\endgroup$ – Micah Nov 17 '14 at 4:31
  • 1
    $\begingroup$ Any sequence loops after at most 1260 repetitions. (src: Wikipedia, Rubik's Cube Group) Therefore, a lower bound on the length of warspyking's sequence is 43 quintillion divided by 1260. $\endgroup$ – Lopsy Dec 16 '14 at 23:38
5
$\begingroup$

EDIT: This only works in a certain set of configurations, as the loop only goes through 1260 states before returning to the original position. My mistake, this is incorrect, but still useful.

Great solution found here. Basically, if you rotate the right, back, left, and front faces all clockwise and in order, then it will always solve the cube - eventually. The core part of it is in one simple explanation - doing this cannot enter an inescapable loop.

If you move the front face clockwise, then counter-clockwise, nothing happens to the cube. The faces are not moved with respect to each other, and you are in the same position. In order for a loop to happen, two different layouts must lead to the same configuration with the same move. But wait - if you reverse this move, then what position does it go to? The simple answer is this will never happen. Here is a great diagram in the article itself. enter image description here

As you can see, reversing this loop would not work, as there are two possible options. This is a proof that loops will never happen.

Moving these four faces is the shortest possible configuration of moves that affects each and every cube, save for the center ones which are irrelevant anyway. This means it is the shortest possible solution that you could repeat to solve the Cube.

$\endgroup$
  • $\begingroup$ Nice! Good find! $\endgroup$ – warspyking Nov 17 '14 at 1:46
  • 4
    $\begingroup$ This doesn't work for any configuration - only starting from a solved cube and the 1260 configurations you get while performing that sequence of moves. $\endgroup$ – Joe Z. Nov 17 '14 at 1:46
  • $\begingroup$ @Joe Doesn't this mean it will eventually lead back to a solved cube, assuming the cube can be solved? You can't enter loops, so it should eventually lead back to the solved position. $\endgroup$ – mdc32 Nov 17 '14 at 1:46
  • 6
    $\begingroup$ No; as soon as you perform a move that's outside of the fixed sequence, performing that fixed sequence will never get you back to a solved cube; you'll just get another cycle of 1,260 configurations. $\endgroup$ – Joe Z. Nov 17 '14 at 1:47
  • 3
    $\begingroup$ Note, in general, that any sequence of moves, if repeated enough times, will eventually get back to the original configuration. This is a property of groups. $\endgroup$ – Joe Z. Nov 17 '14 at 2:27
-2
$\begingroup$

Is there a sequence of moves that can be repeated over and over again which can solve any position the Rubik's Cube can be scrambled in?

This actually depends on how you define "scrambling" a cube. If you mean taking the cube apart and putting it back together with the pieces in random positions, then there is no such sequence. That's simply because most of the possible configurations of a Rubik's cube are unsolvable.

I don't know about a sequence of moves that can unscramble any solvable configuration.

$\endgroup$
  • 3
    $\begingroup$ Generally, unless specified otherwise, "scrambling" refers to a random sequence of valid moves. $\endgroup$ – Joe Z. Nov 21 '14 at 1:47
-2
$\begingroup$

single move (sexy move) RUR'U' method youtube video

or 3 moves method (Peter Renzland's method)

These all very humanly intuitive that requires minimal memorization of extra algorithms, but human reasoning.

$\endgroup$
  • 2
    $\begingroup$ I think you misread the question. The question was: "Is there an algorithm you can use from any scrambled configuration and position to solve the 3x3x3 Cube?" The answer is basically Devil's Algorithm. $\endgroup$ – Kevin Cruijssen May 9 '16 at 8:52

protected by Community Mar 26 '17 at 18:08

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.