Corner Swappings on Rubiks Cube

Question

My brother likes to solve Rubik's cubes. Occasionally I flip a corner or two as an added challenge, and he eventually is able to see where the illegal corner is. The other day, I turned all eight corners and the cube was solvable without any corner turns.

I know that the Rubik's cube is a group and that a corner turn takes the cube out of the group, meaning that we cannot transform the cube back into its fully solved position.

My question is: how many corners need to be turned exactly once for the cube to be a part of the original group that consists of all legal positions? Can I turn just two, or maybe four corners and create a legal position?

Additionally, any good papers or authors to read to learn about the mathematics of the Rubik's Cube?

Answer 1

As you mentioned, turning just one corner (say, clockwise) results in a position that's not in the Rubik's Cube Group, but turning one clockwise and one counterclockwise is!

If we count a clockwise turn as +1 and a counterclockwise turn as -1, then it turns out that as long as our turns add up to 0 mod 3, then that's a legal permutation. (So turning all eight clockwise is not legal, perhaps you turned one counterclockwise?)

My go-to reference for the Rubik's Cube Group is the book Adventures in Group Theory by David Joyner.

Answer 2

It depends on how you turn them. A 1/3 turn clockwise (when looking into the corner from the outside) is called a "quark" and similarly a counter-clockwise rotation is called an "antiquark". In analogy to "real" quarks/antiquarks, you can combine a quark and an antiquark, or you can combine three quarks (or three antiquarks) and still leave the cube in a solvable state. That's IIRC.

Hofstadter's article in Scientific American (ca. 1980) is one reference and David Singmaster's booklet on the cube is another.

Answer 3

If you turn all the corners one step the same way (that is, for example, all clockwise), the resulting configuration is solvable if and only if the number of corners you turned is a multiple of three.

So turning all eight corners one step clockwise each will not produce a solvable configuration.

More generally, the cube will remain solvable iff the difference between how many corners you turned clockwise and how many corners you turned counterclockwise is a multiple of three.

For example, turning equally many corners clockwise as counterclockwise will produce a solvable configuration, because $0$ is a multiple of $3$ (namely, it is $0\times 3$).