Given a suitable infinitary analogue $\mathcal Q$ of the Rubik's cube (as developed below), consider the two player game played between the Scrambler and the Solver wherein the Scrambler scrambles the cube by a (possibly infinite) sequence of twists (without the Solver watching), presents the scrambled cube to the Solver, and the Solver attempts to unscramble it by a sequence of twists. The Solver wins after $\omega$ many twists if $\mathcal Q$ is returned to its original "solved" configuration, otherwise the Scrambler wins. Can the Solver win in general?
The cube
Let $L$ be a linear order with no greatest element, write its reverse as $-L$. Consider the order $L^\dagger = -L + \hat1 + L$ where $\hat 1$ is the singleton order $1$ but its element, say $\hat 0 \in L^\dagger$, should be distinguished among the elements of $L^\dagger$. Having a distinguished center element in the order is analogous to ordinary Rubik's cubes of odd side length. The foremost examples I have in mind are $L = \omega$, $L = \mathbb Q^+$, and $L = \mathbb R^+$. In these cases, $L^\dagger$ as an order is just $\mathbb Z$, $\mathbb Q$, or $\mathbb R$, respectively, but again, there is a distinguished element (in particular, there are no automorphic translations of the structure $\langle L^\dagger, \hat 0\rangle$).
Now, we imagine the $L^\dagger \times L^\dagger \times L^\dagger$ cube $\mathcal Q$ as six $L^\dagger \times L^\dagger$ "grids" arranged like the faces of a cube centered a the origin in $x,y,z$ coordinates with $(\hat 0,\hat 0)$ on the axes. The point is that we have corresponding positions in opposite faces and we can refer to particular "planes" within $\mathcal Q$—such as the $x = \alpha$ plane for $\alpha \in L^\dagger$—in which the twists will take place. To have some notation, let's introduce two new elements $\pm\infty$, thought of as bookending $L^\dagger$. We can think of $\mathcal Q$ as embedded in $[-\infty,+\infty]^3 = (\{-\infty\} + L^\dagger + \{+\infty\})^3$ with the faces in the $\pm\infty$ planes. A cell is a point in one of the faces.
Twists
Define a positive orientation in each axis plane, call it clockwise. A twist $T_{i,\alpha}$ where $i \in \{x,y,z\}$ is a permutation of the cells in the $i=\alpha$ plane given by a clockwise quarter turn. For example, the twist $T_{x,\hat 0}$ takes $(\hat 0,\alpha,+\infty)$ to $(\hat 0,-\infty,\alpha)$ and similarly for the other three edges. Clockwise quarter turns suffice to do any manipulation. Given a twist $T$, we'll write $T^{-1} = T^3$ for the inverse (counterclockwise) twist, whenever useful.
In a standard Rubik's cube the individual pieces that make up the puzzle are called cubies. We will refer to the (exposed) faces of a cubies as tiles. Nota bene, cells and tiles are not to be confused. Cells are particular locations on $\mathcal Q$, as seen from $[-\infty,+\infty]^3$, they do not move with twists, whereas tiles move between cells.
A key property of a standard Rubik's cube is that a twist in any of the outermost planes permutes not only the tiles in the four rows at that coordinate, but also all the tiles in the adjacent face. We definitely would like to be able to make such a face twist, and so we also define $T_{i,\pm\infty}$ to do just this. For instance $T_{x,+\infty}$ permutes the tiles in the $x = +\infty$ face by a clockwise quarter turn about $(\hat0,\hat0)$. There are no corner or edge cubies, so this twist acts non-trivially on only these cells, quite different from an ordinary Rubik's cube.
Configurations and legal twist sequences
A configuration $f$ of $\mathcal Q$ is an assignment to every cell one of seven colors among $\Gamma = \{r,g,b,o,w,y,\perp\}$ (red, green, blue, orange, white, yellow, not a color). The solved configuration $f_{\text{solved}}$ is the particular constant color assignment $\{x=+\infty\} \mapsto r$, $\{x=-\infty\} \mapsto o$, $\{y=+\infty\} \mapsto w$, $\{y=+\infty\} \mapsto y$, $\{z=+\infty\} \mapsto g$, $\{z=-\infty\} \mapsto b$.
Twists act on the space of configurations in the obvious way. Namely, $f' = T\cdot f$ is the configuration $f'(c) = f(T^{-1}c)$ for every cell $c$. We will be pushed to consider infinite sequences of twists, and this is where the $\perp$ color comes into play. Namely, we'd like to have a well defined configuration after executing infinitely many twists. Given a sequence of twists $\langle T_\eta : \eta < \theta\rangle$ for some ordinal $\theta$ and and initial configuration $f_0$, we can define subsequent configurations $f_{\eta+1} = T_\eta\cdot f_\eta$ and at limit stages $\lambda$, set $f_\lambda(c) = \gamma$ if the value of $\langle f_\eta(c) : \eta < \lambda\rangle$ (i.e., the history of the colors of cell $c$ up to stage $\lambda$) had eventually stabilized on $\gamma$, otherwise, we set $f_\lambda(c) = {\perp}$.
Our main interest is merely $\omega$-sequences of twists, for which the above can be stated as follows. After $\omega$ many twists, if any cell had changed color infinitely often, its value is $\perp$, otherwise, it retains its limiting value.
We'll say a configuration is legal if and only if no cell has color $\perp$. Given a legal initial configuration and a sequence of twists (in particular, an $\omega$-sequence), we'll say the sequence of twists is legal if every obtained configuration is legal including the final limiting one if there is one.
The Scrambler-Solver game
The Scrambler scrambles the cube by performing a legal $\le\omega$ sequence of twists (hidden from the solver) starting from the solved configuration, resulting in some legal "scrambled" configuration $g_0$. The Solver sees the configuration $g_0$ and may execute any $\le \omega$ sequence of twists. The Solver wins if the solved configuration is attained at or before stage $\omega$, otherwise the Scrambler wins.
Does the Solver have a winning strategy? When can they win before stage $\omega$? Certain instances are clear. For example, if we only allow finite scrambles and $L$ is countable, then there are only countably many possible twists, and so the Solver can systematically execute and undo every finite sequence of twists, eventually exactly undoing the scramble, and this will happen at a finite stage. This argument completely fails if $L$ is uncountable or if the scramble is infinite. Indeed, even if the Solver is allowed to observe the infinite scramble, she cannot simply "undo" it, since the reverse sequence is not well-ordered. This invites the question, are some (legal) scrambles unsolvable even in principal? If so, does it help to allow the Solver $> \omega$ many moves?
If the scramble is computable, can the Solver win? Can she win computably?
