Devil's Number for 2x2x2 Rubik's Cube - Page 2

bcp19 · 2012-03-03, 20:03

I did some looking, while there are a lot of programs you can use to solve a scrambled cube, I did not find any that you could use to repeat the same moves over and over easily. While performing repetitive moves on my 3x3 cube, as expected you can solve a lot faster than a 2x2, but it was kind of amazing how long it would take for 2 or 4 repeating turns. On a 2x2 if you hold the cube steady in space and rotate the front CW and then the right side clockwise, you have to repeat this pattern 15 times (30 twists) on a 2x2 to return back to solved, while you have to perform it 105 times on a 3x3. If you expand this out to 4, going front, right, back, left, it still takes 15 repetitions (60 twists) to return to solved, and I am guessing it would take 105 on the 3x3 except I messed up in the process (was checking the 8 corners every 60) and it became unsolvable. It would be interesting to have a program you could use to see what would give you more repetitions til solved for 3/4/5/etc length patterns.

cuBerBruce · 2012-08-14, 19:09

First of all, I announced a Hamiltonian cycle for the 2x2x2 cube back in December. (I note that it does make use of making the same quarter-turn two or three times in a row, but nothing on the web site referenced in the initial post in this thread prohibits that). A specification of such a cycle can be found at http://www.speedsolving.com/forum/showthread.php?34318.

I note that in that Hamiltonian cycle, the first 787,320 quarter-turn moves are repeated for the next 787,320 quarter-turn moves. Therefore, the last 3,674,160 - 787,320 = 2,886,840 quarter-turn moves satisfies what that author of that web site calls a Devil's Algorithm. Thus, 2,886,840 is an upper bound on the Devil's Number.

I also point out what I consider a flaw in the lower bound proof. The web site claims that the highest order for any sequence of moves is 36. However, there are maneuvers of order 45 if you are allowed to turn all six faces. This web site, http://mzrg.com/rubik/orders.shtml, lists R L2 U' as an example of an order-45 maneuver. Assuming all six faces can be turned, an similar argument proves a lower bound of 81,648. If moves are restricted to three mutually adjacent faces (so that one cubie remains in a fixed position and orientation), then that web site's lower bound of 102,060 would be valid.

I also note that I've found a Hamiltonian cycle for the Rubik's cube. See http://bruce.cubing.net/ham333/rubik...planation.html. Within this Hamiltonian cycle I can find a sequence of 319,987,003,392,004 moves repeated twice in a row. This means that 43,251,683,287,486,463,996 is an upper bound for the Devil's Number for the Rubik's cube.

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2012-03-03, 20:03	#12
bcp19 Oct 2011 7·97 Posts	I did some looking, while there are a lot of programs you can use to solve a scrambled cube, I did not find any that you could use to repeat the same moves over and over easily. While performing repetitive moves on my 3x3 cube, as expected you can solve a lot faster than a 2x2, but it was kind of amazing how long it would take for 2 or 4 repeating turns. On a 2x2 if you hold the cube steady in space and rotate the front CW and then the right side clockwise, you have to repeat this pattern 15 times (30 twists) on a 2x2 to return back to solved, while you have to perform it 105 times on a 3x3. If you expand this out to 4, going front, right, back, left, it still takes 15 repetitions (60 twists) to return to solved, and I am guessing it would take 105 on the 3x3 except I messed up in the process (was checking the 8 corners every 60) and it became unsolvable. It would be interesting to have a program you could use to see what would give you more repetitions til solved for 3/4/5/etc length patterns.

2012-08-14, 19:09	#13
cuBerBruce Aug 2012 Mass., USA 2·3·53 Posts	First of all, I announced a Hamiltonian cycle for the 2x2x2 cube back in December. (I note that it does make use of making the same quarter-turn two or three times in a row, but nothing on the web site referenced in the initial post in this thread prohibits that). A specification of such a cycle can be found at http://www.speedsolving.com/forum/showthread.php?34318. I note that in that Hamiltonian cycle, the first 787,320 quarter-turn moves are repeated for the next 787,320 quarter-turn moves. Therefore, the last 3,674,160 - 787,320 = 2,886,840 quarter-turn moves satisfies what that author of that web site calls a Devil's Algorithm. Thus, 2,886,840 is an upper bound on the Devil's Number. I also point out what I consider a flaw in the lower bound proof. The web site claims that the highest order for any sequence of moves is 36. However, there are maneuvers of order 45 if you are allowed to turn all six faces. This web site, http://mzrg.com/rubik/orders.shtml, lists R L2 U' as an example of an order-45 maneuver. Assuming all six faces can be turned, an similar argument proves a lower bound of 81,648. If moves are restricted to three mutually adjacent faces (so that one cubie remains in a fixed position and orientation), then that web site's lower bound of 102,060 would be valid. I also note that I've found a Hamiltonian cycle for the Rubik's cube. See http://bruce.cubing.net/ham333/rubik...planation.html. Within this Hamiltonian cycle I can find a sequence of 319,987,003,392,004 moves repeated twice in a row. This means that 43,251,683,287,486,463,996 is an upper bound for the Devil's Number for the Rubik's cube.