Return to Question

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.

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.

A configuration $f$ of $\mathcal Q$ is an assignment to every cell one of six 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 $\{x=+\infty\} \mapsto r$, $\{x=-\infty\} \mapsto o$, $\{y=+\infty\} \mapsto w$, $\{y=+\infty\} \mapsto y$, $\{z=+\infty\} \mapsto g$, $\{z=-\infty\} \mapsto b$.

Our main interest is merely $\omega$-sequences of twists, for which the above can be stated as follows. After $\omega$ many twists, if any tile had changed color infinitely often, its value is $\perp$, otherwise, it retains its limiting value.

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 $\{x=+\infty\} \mapsto r$, $\{x=-\infty\} \mapsto o$, $\{y=+\infty\} \mapsto w$, $\{y=+\infty\} \mapsto y$, $\{z=+\infty\} \mapsto g$, $\{z=-\infty\} \mapsto b$.

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.

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.

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.

A configuration $f$ of $\mathcal Q$ is an assignment to every cell one of six 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 $\\{x=+\infty\\} \mapsto r$, $\\{x=-\infty\\} \mapsto o$, $\\{y=+\infty\\} \mapsto w$, $\\{y=+\infty\\} \mapsto y$, $\\{z=+\infty\\} \mapsto g$, $\\{z=-\infty\\} \mapsto b$.

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.

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.

A configuration $f$ of $\mathcal Q$ is an assignment to every cell one of six 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 $\{x=+\infty\} \mapsto r$, $\{x=-\infty\} \mapsto o$, $\{y=+\infty\} \mapsto w$, $\{y=+\infty\} \mapsto y$, $\{z=+\infty\} \mapsto g$, $\{z=-\infty\} \mapsto b$.