cuBerBruce
Member
For a Hamiltonian circuit, the "(Devil's Alg)" would have to be a Hamiltonian circuit leaving out the final quarter-turn. Also (for the above to become a Hamiltonian circuit), you would also require a simultaneous clockwise corner twist and edge swap at the end to get back to the "legal" coset (and the solved state, if that's your starting position). I realize elrog may not have been clear if he meant a Hamiltonian circuit or just a (not necessarily minimal) sequence that visits all illegal cube group states.
My solution:
Let's say the illegal moves are to be limited to:
P = Flip edge at UF
T = Twist corner at UFR clockwise
C = Swap the corners at positions ULF, UFR
Further, let's let H = standard Hamiltonian circuit of quarter-turn moves, leaving out the final move, assuming that move would have been D.
Then, a Hamiltonian circuit for the illegal cube group could be made as:
HTHTHPHCHTHCHTHCHPHTHTHC
