## Restrictions of Rubik’s Group

Today I’m going to prove the other two constraints I mentioned in the first post about Rubik’s Group: the total corner twist and the total edge flip must both be zero.

First I’ll tackle the corner flipping. Imagine a cube painted with just black and white. All the facelets are black, except for the four corners of the top face and the four corners of the bottom face. If you’ve got a real cube in front of you, tape a little bit of paper onto each of those eight facelets. We’re going to look at how a maneuver twists the corners by looking at how it moves those marked facelets.

Now every maneuver is a composition of the six basic moves , , , , , and . If we can show that these all have a net twist of zero then any composition of them must also have net twist zero. The moves and are easy: they don’t change the marked facelets at all.

Now let’s consider the move . After twisting the right face of the cube, the four marked facelets on the left are left alone. The upper-front corner on the right was marked on the top, but now is marked on the front. That’s an anticlockwise twist of 1/3 if we look directly at that corner. The upper-rear corner is now marked on the back, which is a clockwise twist of 1/3. The lower-front is twisted clockwise by 1/3, and the lower-right is twisted anticlockwise by 1/3. Adding all of these up, we get a total twist of zero.

The moves , , and are similar: each changes which facelet of four corners is marked. Each twists two markings clockwise and two markings anticlockwise, for a total twist of zero. If we do any maneuver and look at which facelet of each corner is marked, the total twist from the starting position will always be zero.

The restriction on edge flips is proved similarly. This time mark the top facelets of the four upper edges, the bottom facelets of the four lower edges, the front facelets of the two front middle edges, and the back facelets of the two rear middle edges. Now the four moves , , , and send marked facelets to marked facelets. The moves and flip the markings on the four edges that they move, and four flips is the same as zero flips, since flipping an edge twice returns it to its original state. If we do any maneuver, the total number of markings that have been flipped from the starting position will always be even, for a net flip of zero.

We can use this to analyze the cycle structure of a maneuver. Let’s say that the cycle notation of a maneuver contains a positively twisted cycle of length on the corners. If we do the maneuver times, it returns those cubies to their original cubicles, each twisted once clockwise. That is, has a total twist of on these cubies. Since each copy of does the same thing, that’s one twist each time we perform . If we look at *all* the corner cycles in , some will have a positive twist, some a negative twist, and some no twist. The number of positively twisted cycles minus the number of negatively twisted cycles must be a multiple of three, since three twists counts as zero total twist.

The same goes for the edges. Each flipped edge cycle in contributes a single net flip, and the total number of flips has to be even. You can check yourself that all the cycle notations I wrote down last time satisfy both of these conditions. You can also see that the parity of the edge permutation and the parity of the corner permutation are equal in each example.

So I’ve established that the total edge flip is zero, the total corner twist is zero, and the parities of the edge and corner permutations are equal. When I come back to the cube I’ll show that we can realize *any* maneuver whose cycle notation satisfies these three conditions can be realized as a composition of basic twists. That will lead us to the structure of Rubik’s Group, and to a solution of the Cube.

## Unapologetic Spring Tour 2007

This Friday after I’m done teaching I’m going to hit the road and spend the next two weeks in Maryland, with a side trip to Ohio.

While in Maryland I’ll be at the College Perk coffeehouse/bar most evenings. It’s a great little place a bit north of the University of Maryland, College Park campus. If you’re in the Maryland area, drop by and look for the underemployed math guy. I know a few regular readers who already hang out there all the time, so it should be a party.

From Thursday the 15th to Sunday the 18th I’ll be in Oxford, Ohio at the Sectional AMS meeting, held at Miami University. I’m closing the Saturday morning session on quantum topology with my talk on Kauffman Bracket extensions, right after Louis Kauffman himself and his (former) student Heather Dye, one of whose papers I reference. If you’re in that area (or will be there for the conference), drop a line and meet the UM.