algs that goes back to solved position after X times

[nitay6669]

a challenge was set on the israeli forums, to find algororithms that goes back to solved position after X times of doing them.

X is going from 1 to whatever...
so far we have come out with 1-10 and 12 without 11.

i thought about trying to make an 11 cycle?
and generally its a fun challenge.

here is the list so far:

Spoiler

1 :U4
2 : U2
3 : R U R' U' R' F R F'
4 : R U' R'
5 : R U R' U
6 : R U R' U'
7 : U R U' F
8 : U D R2
9 : U R F2
10 : U R U' F2
11 :
12 : R U2 R' U'

 

[y235]

I have a one that it's need to be repeated 11 times:
B' R' L M' U M U' R L' B z' M' U M U' y x y' (U PERM) z2

 

[nitay6669]

did you check if it works?
and why didnt you wrote it in the israeli forums?

 

[y235]

did you check if it works?
and why didnt you wrote it in the israeli forums?

it works, this is an 11 cycle of edges made of 2 5 cycles and one 3 cycle

 

[James Ludlow]

It seems to work.

How the hell did you find this?

EDIT - this alg takes forever to watch btw

 

[y235]

It seems to work.

How the hell did you find this?

EDIT - this alg takes forever to watch btw

2 5 cycle connected by a 3 -cycle.
B' R' L M' U M U' R L' B - 5 cycle of M and one edge in R
z' M' U M U' - 5 cycle of the R edges and one in L
y x y' (U PERM) z2 - connect them with 3 -cycle and restore to original orientation


I have another one with only UPERMS

 

[Jakube]

B' R' L M' U M U' R L' B z' M' U M U' y x y' (U PERM) z2

Here´s the optimal one (generated with Cube Explorer): F2 L' D' L R2 B' R F D B D2 U' R D' U (15f*)

 

[Kirjava]

Optimal? I seriously doubt that that's the shortest alg that returns to solved after 11 moves.

 

[riffz]

Optimal? I seriously doubt that that's the shortest alg that returns to solved after 11 moves.

I think he means the optimal alg for y235's cycle.

 

[Kirjava]

Of course. I'm just establishing that that doesn't mean it's the shortest for that criteria.

 

[nitay6669]

actually some in the forum made a program that finds algs like that but for 1 the alg was too long for it to work ><
next goal

13
thats gonna be interesting, cos you cant make a 13 cycle can you?
it must involve both corners and edges.

 

[Stefan]

http://mzrg.com/rubik/orders.shtml

 

[cuBerBruce]

13
thats gonna be interesting, cos you cant make a 13 cycle can you?
it must involve both corners and edges.

No, there are no order-13 maneuvers. Using face turns only, only 73 different orders are possible, as listed here. If inner layer turns, double-layer turns, and cube rotations are allowed, a few more orders are possible (but not 13). Stefan gave a link to qqwref's page which gives maneuvers not restricted to face turns for many of the possible orders (as well as for 2x2x2 and 4x4x4).

Doing a breadth-first search to a depth of 7 face turns, I find all 73 of these orders achievable via face turns can be generated in 7 face turns or less, except for an order of 11 requires more. qqwref gives a maneuver corresponding to 10 face turns. An order of 110 requires 7 face turns, and the remaining 71 require no more than 6 face turns.

 

[RNewms27]

Well you can MU to solved in 8 repititions as a shorter alg.

 

[Cielo]

Actually, there are 8 corners, 12 edges and 6 centers.
One can flip edges, twist corners or rotate centers.
So the order of an alg can only have factors which are less than 13.

 

[conor]

F R U R' ( 10 times )

 

[cubernya]

Actually, there are 8 corners, 12 edges and 6 centers.
One can flip edges, twist corners or rotate centers.
So the order of an alg can only have factors which are less than 13.

You're only talking orientation, not permutation

 

[Cielo]

You're only talking orientation, not permutation

Permutation of edge: factor of order <= 12.
Of corner: <=8.
Of center: <=6.

13 is a prime larger than 12, so...

 

[cubernya]

Explain why R U has a factor of 105 then

 

[bwronski]

how long does it take for this one to repeat?
U L D R