The Unapologetic Mathematician

Rubik’s Group notation

Today I want to introduce some notation for discussing the cube.

Take a cube — either a real one, the Java version, or just one in your mind’s eye — and hold it with the center cubies fixed in place pointing up, down, left, right, front, and back. Twists of the faces will generate Rubik’s goup $G$. We pick the six generators as follows

• $U$ is a twist of the upper face by a quarter turn clockwise, looking down at the top of the cube.
• $D$ is a twist of the lower face by a quarter turn clockwise, looking up at the bottom of the cube.
• $R$ is a twist of the right face by a quarter turn clockwise, looking left at the right side of the cube.
• $L$ is a twist of the left face by a quarter turn clockwise, looking right at the left of the cube.
• $F$ is a twist of the front face by a quarter turn clockwise, looking at the front of the cube.
• $B$ is a twist of the back face by a quarter turn clockwise, looking at the back of the cube.

For instance, executing $B$ involves turning the whole cube around to look at the back, twisting that face clockwise by 90°, and turning the cube back to the original orientation.

For each twist $T$, the 180° twist of the corresponding face is $T^2$, and the anticlockwist twist is $T^{-1}$. Four quarter-twists is the same as doing nothing, so $T^4$ is the identity.

It’s also useful to label the cubicles. I’ll do this by listing the faces it touches. The face cubicle on the left side of the cube is $l$, the lower edge on the back ls $db$ or $bd$. The upper-right corner on the front of the cube is $urf$, $ufr$, $ruf$, $rfu$, $fur$, or $fru$. The order of the faces doesn’t matter so much for a single cubicle, but becomes important when we start thinking about how maneuvers affect the state of the cube.

For instance, the effect of the maneuver $R^2U^{-1}DB^2UD^{-1}$ has an effect on the cube I’ll write as $(fr\,bl\,br)$. This takes the cubie in the front-right cubicle and puts it in the back-left cubicle, with the facelet that was in the front now in the back and the facelet from the right now on the left. It takes the cubie from the back-left cubicle and puts it in the back-right cubicle with the facelet from the back still on the back and the facelet from the left now on the right. Finally it takes the cubie from the back-right and moves it to the front-right with the right facelet still on the right and the back facelet now on the front.

As another example, $(R^{-1}U^2RB^{-1}U^2B)^2$ takes the cubie from the upper-right-front cubicle and leaves it where it is, but twists it to the right, and similarly twists the cubie in the lower-left-back cubicle. We write this transformation as $(urf\,rfu\,fur)(dlb\,lbd,bdl)$. For shorthand we’ll modify this permutation notation slightly. When a cycle brings a cubie back to its starting cubicle but rotated, we add a sign. In this example we’ll write $(+urf)(-dlb)$, since if we look directly at the upper-right-front cubie it’s been rotated clockwise by 1/3 of a turn and looking directly at the lower-left-back it’s been rotated anticlockwise by 1/3 of a turn.

This notation does make sense. Let’s say that we have a maneuver whose effect contains some cycle on the corners of the cube of length $k$. If we apply that maneuver $k$ times each cubie in the cycle comes back where it started, but the cubies may have been twisted in the process. Each one will be twisted by the same amount: either 1/3 to the right, 1/3 to the left, or untwisted. The sign in the notation tells us what that twist is.

The same notation goes for edges. The maneuver $RUD^{-1}FUD^{-1}LUD^{-1}BUD^{-1}$ has effect $(+ur)(+fr)(+fl)(+bl)$, flipping those four edges. We could write this out as $(ur\,ru)(fr\,rf)(fl\,lf)(bl\,lb)$, but the “twisted permutation” notation is more compact.

From this notation it’s easy to compute the order of a maneuver — what power of the maneuver returns to the identity transformation. A corner cycle of length $k$ has order $k$ if there’s no twist, and order $3k$ if there is a twist. Similarly, an edge $k$-cycle has order $k$ if there’s no flip and order $2k$ if there is a flip. So if we write the effect of a maneuver as a twisted permutation we can find the order of each twisted cycle. The order the the whole maneuver is the least common multiple of those orders.

As an example, consider the maneuver $RU$. This has effect $(+urf)(-ufl\,ulb\,ubr\,bdr\,dfr)(fr\,uf\,ul\,ub\,ur\,br\,dr)$. The three twisted cycles have orders 3, 15, and 7, so the total order is 105. If you actually sit and perform $(RU)^{105}$ you’ll get back exactly where you started.