# algs that goes back to solved position after X times



## nitay6669 (Aug 23, 2011)

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'


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## y235 (Aug 23, 2011)

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


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## nitay6669 (Aug 23, 2011)

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


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## y235 (Aug 23, 2011)

nitay6669 said:


> 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


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## James Ludlow (Aug 23, 2011)

It seems to work.

How the hell did you find this?

EDIT - this alg takes forever to watch btw


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## y235 (Aug 23, 2011)

James Ludlow said:


> 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


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## Jakube (Aug 23, 2011)

y235 said:


> 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*)


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## Kirjava (Aug 23, 2011)

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


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## riffz (Aug 23, 2011)

Kirjava said:


> 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.


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## Kirjava (Aug 23, 2011)

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


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## nitay6669 (Aug 23, 2011)

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.


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## Stefan (Aug 23, 2011)

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


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## cuBerBruce (Aug 23, 2011)

nitay6669 said:


> 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.


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## RNewms27 (Aug 23, 2011)

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


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## Cielo (Aug 23, 2011)

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.


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## conor (Aug 23, 2011)

F R U R' ( 10 times )


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## cubernya (Aug 23, 2011)

Cielo said:


> 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


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## Cielo (Aug 23, 2011)

theZcuber said:


> 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...


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## cubernya (Aug 23, 2011)

Explain why R U has a factor of 105 then


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## bwronski (Aug 23, 2011)

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


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## riffz (Aug 23, 2011)

theZcuber said:


> Explain why R U has a factor of 105 then


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


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## Cielo (Aug 23, 2011)

theZcuber said:


> 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)



bwronski said:


> 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)


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## nitay6669 (Aug 23, 2011)

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


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## cuBerBruce (Aug 23, 2011)

nitay6669 said:


> 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.)


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## Ranzha (Aug 24, 2011)

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


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## qqwref (Aug 24, 2011)

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


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## Cool Frog (Aug 24, 2011)

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?)


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## JonWhite (Aug 24, 2011)

Cool Frog said:


> 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.


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## qqwref (Aug 24, 2011)

Cool Frog said:


> 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.


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## JonWhite (Aug 24, 2011)

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


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## cuBerBruce (Aug 24, 2011)

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.



Ranzha V. Emodrach said:


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


 


qqwref said:


> 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.


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## JonWhite (Aug 24, 2011)

cuBerBruce said:


> 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


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## cuBerBruce (Aug 24, 2011)

JonWhite said:


> 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.


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## qqwref (Aug 24, 2011)

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.


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## Erdos (Aug 24, 2011)

JonWhite said:


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


 
If you perform x, the cube becomes entirely unsolved.


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## Lucas Garron (Aug 24, 2011)

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


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## nickcolley (Aug 24, 2011)

Lucas Garron said:


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


 Shexy mouve


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## cuBerBruce (Aug 24, 2011)

Erdos said:


> 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.


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## Erdos (Aug 24, 2011)

^ 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.


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## cuBerBruce (Aug 24, 2011)

Erdos said:


> ^ 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.)


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## Micael (Aug 24, 2011)

Lol! I did not knew any sequence eventually return to its state. I thought I was going to explose qqwref's calculator by inputting a scramble. After many try, I was very disappointed. But I used qqtimer for scrambles, there should be a ploy...


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## Erdos (Aug 25, 2011)

cuBerBruce said:


> ...


 I don't see what you're getting at. I was also showing that x is not the identity through a proof by contradiction. And since x is not the identity element, we can use the contraposition of the identity's definition: applying x once to a solved state cannot lead to a solved state. So to use the speedcubing definition in terms of order would not make sense because it is not well-defined.


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## cuBerBruce (Aug 25, 2011)

Erdos said:


> ... applying x once to a solved state cannot lead to a solved state.



No, it seems to me you are confusing the concept of a solved state with the concept of a "move" that is an identity. It is perfectly possible for a non-identity "move" of a puzzle to change the "state" of the puzzle from one of its solved states to another one of its solved states, assuming the puzzle has multiple states that are each considered solved. I believe you are equating the term "identity" with the term "solved state," which is the flaw in your logic, it seems to me. Think of group elements as representing "moves" of a puzzle (actions on the puzzle), not static states of the puzzle.


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## qqwref (Aug 25, 2011)

Erdos said:


> ^ 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 [...]


Why not simply decide that "solved states" need not be identities? That is, there are 24 states of the full rotation-included cube group that count as solved.



Micael said:


> I thought I was going to explose qqwref's calculator by inputting a scramble. After many try, I was very disappointed. But I used qqtimer for scrambles, there should be a ploy...


Haha!


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## Erdos (Aug 25, 2011)

cuBerBruce said:


> ...


 
I'm not necessarily "confusing" the concept. I'm defining it in that sense in order to avoid having more than one unique solved state. For example, by your definition, there are 24 solved states (I think).


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## cuBerBruce (Aug 25, 2011)

Erdos said:


> I'm not necessarily "confusing" the concept. I'm defining it in that sense in order to avoid having more than one unique solved state. For example, by your definition, there are 24 solved states (I think).


 
Yes, for the 3x3x3 anyway, I would consider there to be 24 solved states when the orientation of the puzzle is considered to matter (and we ignore the 2048 configurations for the orientation of the center pieces as part of the state).

One of the beautiful things about treating the group elements as actions rather than as states (which are generally recognized by looking at the stickers) is that I don't have pick any one of these 24 states as being the identity. The identity is simply a "null action" rather than any particular state of the stickers. Note that how an action rearranges the pieces of the cube is totally independent of how those pieces are stickered (at least when our grip on the puzzle serves to define our reference frame for the puzzle)!

So I'm not clear what exactly you're trying to claim is not well-defined. Definitely, the <U, x, y> group is a well-defined group, and the order of each of its 1038048078587756544000 elements is well-defined. qqwref's definition of order will not always coincide with the actual group theory definition for these elements, though.


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## mrCage (Aug 27, 2011)

Somewhat related. 

What is the shortest sequence effectively doing nothing?? No internal repetitions allowed, like U2 D2 U2 D2 for instance.
I discovered R' D' R F D2 L D L' F' D2 almost 30 yrs ago. Any better ones??

Per


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## Kirjava (Aug 27, 2011)

10 is the shortest. We had a conversation about this back on twistypuzzles about 6 years ago.


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## mrCage (Aug 27, 2011)

Hmm. Before i joined. I see. Did anyone come up with any kind of proof or other interesting examples??

Per


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## Stefan (Aug 27, 2011)

What about this?



clement said:


> well, U2 D2 R2 L2 U2 D2 R2 L2 is shorter...


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## mrCage (Aug 28, 2011)

Ok, let us switch to quarter turn metric then 

[R',F] [U',R] [F',U] - another 12 qtm candidate ... Any proof 12 qtm is optimal when no repetitive sequences are allowed??

Per


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## Herbert Kociemba (Aug 29, 2011)

mrCage said:


> Ok, let us switch to quarter turn metric then
> 
> [R',F] [U',R] [F',U] - another 12 qtm candidate ... Any proof 12 qtm is optimal when no repetitive sequences are allowed??
> 
> Per


 
Yes, 12q is the shortest even if repetitive sequences which cannot be simplified in a trivial way (by relations like U D = D U or U U' = Id etc.) are allowed. 

U2 R U R' F' U2 L' U' L F (12q*)

was the shortest, my optimal qtm solver gave out for the identity position. Though some consider it as a bug to give out a 12q* maneuver for the identity position....


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## reThinking the Cube (Aug 30, 2011)

M' E2 M2 E2 M'


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## uberCuber (Aug 30, 2011)

reThinking the Cube said:


> M' E2 M2 E2 M'


 
That's 16qtm and 10htm. I don't see how this is at all special?


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