# Rubik's cube two-person game

I will define the game as follows:

There are two players, the "Solver" and the "Unsolver". The Solver's goal is to return the cube to it's standard solved state. Likewise, the Unsolver's goal is to prevent the Solver from solving the cube.

Start the game by scrambling a standard 3x3 Rubik's cube.

Starting with the Solver, each player is allowed one 90 degree face twist. However, the current player's move may not be the simple inverse of the previous move (e.g. you aren't allowed to simply "undo" the other's).

My question is: are there any legal initial Rubik's cube states such that the Solver can win, assuming the Unsolver plays perfectly? I will exclude "scrambles" that involve only a single twist, since the solver would win in one move.

Bonus question: What is the behavior of this game given symmetry-breaking rule changes? The Solver might be allowed 180 degree twists, for example, or two moves before the Unsolver gets one.

• It would be more interesting according to me if we allow the unsolver to make 1 90/180 degree move and the solver could get 2 90 degree moves. So the additional advantage here is that the solver can make like one clockwise front move and one right, but all the unsolver can do is rotate a face either 90 or 180 degrees in one turn. – Rohinb97 Apr 29 '17 at 22:45

No.

Reasoning:

From every unsolved position, the Unsolver has many moves toward positions that aren't one move away from the solved state. As long as they make such a move, choosing one that isn't undoing the previous move, the Solver can never win.
This doesn't change if the Solver is further allowed 180 degree twists.

Two answers so far, but no proper justification.

If both Solver and Unsolver make one move per turn, Unsolver wins (if Solver didn't on first move).

It suffices to prove that from any unsolved state $S$ the Unsolver can make two possible moves that cause it to go to a state at least $2$ moves away from the solved state.

If $S$ can be solved by one face turn, Unsolver can turn the opposite face in two distinct directions.

If $S$ can be solved by distinct face turns $A,B$ in that order, Unsolver can turn the same face as $A$ but not the same way as $A$, in two distinct ways. (Even if Unsolver is only allowed quarter turns, then the only case where this fails is when $A$ is a quarter turn and Solver did $A$ in the previous move. In that case let $x,y$ be the corner positions affected by $A$ but not $B$, such that $A$ moves the piece from $x$ to its correct position $y$. Then Unsolver can turn the opposite face as $B$ to move the piece from $x$ to $y$ but with the wrong orientation, which obviously takes at least $2$ moves to solve.)

If $S$ cannot be solved by $2$ or less distinct face turns, Unsolver can make any move, which can only bring $S$ closer to solved by $1$ face turn.

If Solver can make more moves than Unsolver in his/her previous turn, Solver wins.

Trivially, Solver undoes whatever Unsolver did and then makes one move to progress towards the solved state.

NO

The cube could be two steps away from the solution. No matter what the Solver does, the Scrambler will endeavour to hamper his progress.

• More precisely, any position one move away from the solved state has 11 moves that bring it 2 moves away. If the Solver brings the cube to a state one move away, the Unsolver is only forbidden one of the 11, he still has 10 possible moves to scramble it. – Florian F Apr 29 '17 at 12:08