• Welcome to the Speedsolving.com, home of the web's largest puzzle community!
    You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.

    Registration is fast, simple and absolutely free so please, join our community of 35,000+ people from around the world today!

    If you are already a member, simply login to hide this message and begin participating in the community!

A Hamiltonian circuit for Rubik's Cube!

cuBerBruce

Member
Joined
Oct 8, 2006
Messages
914
Location
Malden, MA, USA
WCA
2006NORS01
YouTube
cuBerBruce
(Devil's Alg) [twist corner clockwise] (Devil's Alg) [twist corner clockwise]
(Devil's Alg) [flip edge]
(Devil's Alg) [twist corner clockwise] (Devil's Alg) [twist corner clockwise]
(Devil's Alg) [swap two edges]
(Devil's Alg) [twist corner clockwise] (Devil's Alg) [twist corner clockwise]
(Devil's Alg) [flip edge]
(Devil's Alg) [twist corner clockwise] (Devil's Alg) [twist corner clockwise]
(Devil's Alg)
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
 
Thread starter Similar threads Forum Replies Date
O Puzzle Theory 17
C Puzzle Theory 30
Top