After the incidents of Skyfall, 007 remembered a quote Q had said when he was accessing Silva's Omega site :

It's like solving a Rubiks cube that's fighting back.

007, intrigued by this statement, asks the following:

Starting from a completely shuffled Rubik's cube, for every 18 moves you make, the Rubik's cube makes 2 moves. The Rubik's cube is defeated if you can solve it. The Rubik's cube cannot reverse any moves you made in your last attempt, and the same holds for you. But the reverse moves can be used if the cube is at a different state than when the attempt started. Given that the cube plays optimally, can the Rubik's Cube be defeated?

Q thinks for a bit and says, possibly not. But he doesn't know. It's up to us to prove him right or wrong.

But Q does asks the following:

Generalizing this, if you can make $n<20$ moves, and the cube can make $m\le n$ moves, then for what $n$ and $m$ is the cube defeatable, if at all?

A related problem can be found here

  • $\begingroup$ I think you should specify which definition of move we are using. Since the question seems to be related to God's number, I will asume we are considering the half turn metric. (eg. 180° counts as one turn, moving middle layer counts as two) $\endgroup$
    – BoltKey
    Commented May 5, 2020 at 22:02
  • $\begingroup$ This seems very difficult to solve - do you have any reason to believe this is feasibly attackable without heavy computer search? $\endgroup$
    – Deusovi
    Commented May 5, 2020 at 22:04
  • $\begingroup$ @BoltKey Yes, we are using the HTM. $\endgroup$ Commented May 5, 2020 at 22:09
  • $\begingroup$ Also, it is very unclear what is meant by "cannot reverse or make any of the moves". Does it mean that when the cube makes a F R, I cannot make a F, F', R nor R' as any of the 18 turns? That seems to be implied. But then, since I have 18 turns available, I can make all possible moves there are, so the cube cannot make any moves at all. $\endgroup$
    – BoltKey
    Commented May 5, 2020 at 22:11
  • $\begingroup$ @Deusovi I am basing this on an intuition. Optimally solving the cube till the 18th move obviously lands us in a non-random stage ( at least unshuffled, that is, there is some pattern ). Given 2 moves to destroy the pattern still leaves "some" pattern, and thus maybe, just maybe the cube would be solvable in under 18 moves. Maybe it won't. $\endgroup$ Commented May 5, 2020 at 22:12

1 Answer 1


The cube is definitely defeatable if

$n > m+2$

The way to beat the cube is to

undo the cube's moves, but to stick in two moves that essentially do nothing just so that you are not directly reversing the cube's moves as that is against the rules. If for example the cube's last move is R, so you would like to do R' to undo it, do L R' L' instead. For any other first move a similar thing can be done using the move's opposite face.
So you can undo the cube's $m$ moves using $m+2$ moves, and then still have at least one move left available to you to get the cube closer to being solved.

If this method is considered too trivial and against the spirit of the rules, you can still do it for

$n > m+7 $

by making use the non-trivial identity

F2 B2 R2 L2 F2 B2 R2 L2
If you want to do R' as your first move, do F2 B2 R2 L2 F2 B2 L2 R instead. This uses 7 extra moves to get around the restrictions. Because of the symmetry of the cube a similar thing can be done for any first move.


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