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#1

By which we mean the cube group can be generated by 2 elements:

G = <g1, g2> = <U B L U L' U' B', R2 F L D' R'>

So, any position of the cube can be written as something like \( g1 g2^2 g1^{-1} g2^3 \), except longer.

Apparently this was known in 1980. Joyner cites this set by "F. Barnes" in a book by Bandelow (which I have yet to see).

I believe this, but I'd like to see a proof by generating each of {B, F, U, D, L, R} as an "alg" in <g1, g2>. (Or at least 5 of them.)

ksolve can't quite handle this. Since 4.3*10^19 is about 65 binary digits, we'd expect the length of an alg for a position (like the regular generators) to have a length of about 65, ~32 if we allow inverses.

Does anyone know a nice (algorithmic?) way to generate {B, F, U, D, L, R} using <g1, g2>, or come up with some other nice set of two generators for which we can do this?

G = <g1, g2> = <U B L U L' U' B', R2 F L D' R'>

So, any position of the cube can be written as something like \( g1 g2^2 g1^{-1} g2^3 \), except longer.

Apparently this was known in 1980. Joyner cites this set by "F. Barnes" in a book by Bandelow (which I have yet to see).

I believe this, but I'd like to see a proof by generating each of {B, F, U, D, L, R} as an "alg" in <g1, g2>. (Or at least 5 of them.)

ksolve can't quite handle this. Since 4.3*10^19 is about 65 binary digits, we'd expect the length of an alg for a position (like the regular generators) to have a length of about 65, ~32 if we allow inverses.

Does anyone know a nice (algorithmic?) way to generate {B, F, U, D, L, R} using <g1, g2>, or come up with some other nice set of two generators for which we can do this?