Schmidt
Veteran
If you do F R U R' U' F' on a solved 2x2x2 cube it gets to a position where you need to do y2 and the same alg. again.
If you do f r u r' u' f' on a 4x4x4 cube it is the same.
So why can't I solve the 4x4x4 centers on a scrambled cube like I would solve a scrambled 2x2x2 ( using the algs. used in the Ortega method) ?
If you do f r u r' u' f' on a 4x4x4 cube it is the same.
So why can't I solve the 4x4x4 centers on a scrambled cube like I would solve a scrambled 2x2x2 ( using the algs. used in the Ortega method) ?