algs that goes back to solved position after X times

[riffz]

Explain why R U has a factor of 105 then

You don't understand what he's trying to say.

 

[Cielo]

Explain why R U has a factor of 105 then

105=3x5x7, each is ≤ 12.

As for 105, copy from CubeTwister:
Twists: 3l 6f 12q
Order: 105v 420r

Permutation:
(-ufl,ulb,ubr,bdr,dfr) (+urf)
(fu,lu,bu,ru,rb,rd,rf)
(+r) (+u)

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

Twists: 3l 6f 12q
Order: 315v 1260r

Permutation:
(+ufl) (+dlf) (+urf) (+dfr) (+ubr) (+drb,dbl,bul)
(fu,lf,fd,rf,ru) (dr,db,dl,bl,ul,ub,br)
(+r) (+d) (+l) (+u)

 

[nitay6669]

WOW i didn't knew there was actually a term for this things...
this is neat!
can someone explain again why 13 is not possible?
i didn't get that

 

[cuBerBruce]

can someone explain again why 13 is not possible?
i didn't get that

In short, since 13 is a prime number, you would need to have 13 pieces that can form a 13-cycle. Since the most number of pieces of one type is 12 (and pieces only can be permuted among other pieces of the same type), there are not sufficient pieces to form a 13-cycle. Hence, you can't make a 13-cycle, so no order-13 maneuvers. However, you can get an order-35 position, for instance, since it is possible to make separate cycles of lengths 7 and 5. (LCM (7,5) = 35. LCM means least common multiple.)

 

[Ranzha]

U L D R takes 315 repetitions.
R y takes 1260, and ULDR is a rotation of (R y)4. Therefore, 1260/4 = 315.

I just remembered Joël has an order calculator. Have fun, everybody.
http://solvethecube.110mb.com/tools.html#order

 

[qqwref]

Hey, I have one too http://mzrg.com/rubik/ordercalc.shtml

 

[Cool Frog]

does anyone have a odd number of QTM moves that repeaced an odd number of times returns to a solved state?

(or is this not possible?)

 

[JonWhite]

does anyone have a odd number of QTM moves that repeaced an odd number of times returns to a solved state?

(or is this not possible?)

we could simplify this problem by asking, "is there any algorithm with an odd number of QTM moves that returns the cube to its starting position?" Let's assume there is, call such an algorithm X. Then our solution would be (R U X) *105. (if you don't count X repeated once as a solution)

If there exists no X, then I can confidently state that every legal scramble + solve with no pops on a 3x3x3 cube has an even number of QTM moves.

 

[qqwref]

does anyone have a odd number of QTM moves that repeaced an odd number of times returns to a solved state?

(or is this not possible?)

It isn't possible. Each turn in QTM changes the permutation parity of the corners (whether the permutation can be solved with 3-cycles or not). So after an odd number of moves in QTM, the permutation parity of the corners is changed. If your question was possible, then there would be some sequence with an odd number of moves QTM which goes from solved to solved, and since that sequence must change the corner permutation parity, this can't be done.

 

[JonWhite]

so therefore all legal pop-less scramble+solves ever done on a 3x3x3 were even QTM. interesting.

 

[cuBerBruce]

I have calculated optimal face turn maneuvers for all possible orders within the <U,D,L,R,F,B> group, except for order 11 might not be optimal. I omit the trivial ones with less than two moves.

Spoiler

2 moves:
6: U2 R2
30: U R2
63: U R'
105: U R

3 moves:
8: R2 U D
9: U2 R F
12: U2 F2 R
24: U2 R' L
36: U2 F R
60: U F' R
80: R U F
84: U R F
90: R U D
168: R U D2
180: R U D'
360: R U' D2

4 moves:
3: R2 U2 R2 U2
5: R U R' U
7: R U' F U
10: U2 R F R'
15: U2 R F2 L'
20: R' U2 R D
21: R2 U R2 D
28: U L U' R
33: F R' U' D
35: U2 R U2 L'
40: U2 L U' R
42: R2 U2 R U'
44: F R' U D'
45: R U R D'
48: F U2 R U
56: F R' U2 D'
72: R B U' D'
77: F R U' L
99: R2 U L2 F
120: F2 R2 U' D2
126: R U F D'
132: F2 U' R L2
140: R U' F U'
144: F R U' D2
198: F2 R U2 D'
210: B' L F' R
231: F R' U D
240: R U2 F D'
252: F2 L U' R
315: L F R B
420: L F R' B
504: R' U2 F' D
840: F R U2 D'

5 moves:
16: R F' R' U D'
18: U2 L' F R L
22: U2 D F R' L
66: R2 F' R U' D
70: F2 U F2 D R
165: F2 U' R U' L
280: R U L F' U
330: U2 L D F R'
336: U2 F D' R L'
462: L2 F R U D'
495: L2 B' D L U'
630: U R' D R2 L'
720: F U R U D2
1260: F U' R U' D2

6 moves:
14: U R L' F' R L'
55: U R' L2 D F2 B
112: U' F L' U L' R
154: U F L D R' L
990: D' F L U L R'

7 moves:
110: U L2 F U L D' R

> 7 moves:
11: U B U D' L' F' R' U2 D2 R (may not be optimal)

Of course, if you want to allow inner layer turns, double-layer turns, and x/y/z cube rotations, then shorter algs may exist and other possible orders exist.

I just remembered Joël has an order calculator. Have fun, everybody.
http://solvethecube.110mb.com/tools.html#order

Hey, I have one too http://mzrg.com/rubik/ordercalc.shtml

Personally, I like Joël's better. Joël's gives the correct order for x, which is 4.

 

[JonWhite]

Personally, I like Joël's better. Joël's gives the correct order for x, which is 4.

solved cube --> perform x --> solved cube --> order = 1

 

[cuBerBruce]

solved cube --> perform x --> solved cube --> order = 1

The actual definition of "order" has to do with repeating a move until the identity element reached. x is not an identity, as U x U x is not the same as U U = U2. So x can not possibly have order 1. There is only one identity element in a group, and it is the only element of a group with order 1.

 

[qqwref]

To be fair, I'm more interested in how many times an alg takes to return to solved, not the mathematical order. So that's why my version works that way.

 

[Erdos]

solved cube --> perform x --> solved cube --> order = 1

If you perform x, the cube becomes entirely unsolved.

 

[Lucas Garron]

What's this sudden fascination with repeating algorithms? Everyone keeps making puzzle theory threads about it.

 

[nickcolley]

What's this sudden fascination with repeating algorithms? Everyone keeps making puzzle theory threads about it.

Shexy mouve

 

[cuBerBruce]

If you perform x, the cube becomes entirely unsolved.

The usual definition of solved is that each face of the puzzle is a solid color. x is a whole cube rotation. Applying x does not change whether or not the cube is solved by that definition, even though x is not an identity.

 

[Erdos]

^ That may be true according to our definition of "solved state" in cubing, but that leads to contradictions in mathematics. That is, your definition is not well-defined. You can't claim that x is not the identity if by your definition, it is exactly what the identity is (by definition). After all, any element with order 1 must be equivalent to the identity. Therefore, all rotations are equivalent to the identity element (since the identity element cannot be unique in a group), and therefore all elements in the cube group must commute with the rotations. This is clearly not true.

 

[cuBerBruce]

^ That may be true according to our definition of "solved state" in cubing, but that leads to contradictions in mathematics. That is, your definition is not well-defined. You can't claim that x is not the identity if by your definition, it is exactly what the identity is (by definition). After all, any element with order 1 must be equivalent to the identity. Therefore, all rotations are equivalent to the identity element (since the identity element cannot be unique in a group), and therefore all elements in the cube group must commute with the rotations. This is clearly not true.

You are confusing "the" cube group typically used by mathematicians and the group that is applicable to speedcubing algs. In executing speedcubing algs, the orientation of the cube as a whole does matter. A speedcuber views the move U as moving the layer that is (more or less) facing "up." Sometimes it moves the layer with the white center, sometimes it moves the layer with the green center, sometimes some other color center. It depends on how the cube is currently oriented. As I mentioned before, U x U x is not the same as U U, so clearly the "move" x does have some effect on the cube as far as executing speedcubing algs is concerned. Since x does have an effect that matters, it is not an identity. By definition, an identity has no effect (that matters), regardless of the initial state the cube is in before applying the identity. The definition of identity has nothing really to do with having a solved state, or how many states might be considered solved.

The group used for speedcubing algs has 24*43252003274489856000 = 1038048078587756544000 elements. A simple way to (mathematically) generate the group is <U, x, y>. (y is another whole cube rotation that rotates the cube on a different axis than x does.)