# Thread: algs that goes back to solved position after X times

1. Originally Posted by theZcuber
Explain why R U has a factor of 105 then
You don't understand what he's trying to say.

2. Originally Posted by theZcuber
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)

Originally Posted by bwronski
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)

3. 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

4. Originally Posted by nitay6669
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.)

5. 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

6. 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?)

7. Originally Posted by 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?)
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.

8. Originally Posted by 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?)
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.

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

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