Self Solving Algs

[philkt731]

This does not include undoing like R R'.

The fastest I know (by HTM) is:
r' U2 R U2 M' U2 R' U2 R (10)

Some others I know are 12 by HTM:
M2 U2 M2 U2 M2 U2 M2 U2
R2 U2 R2 U2 R2 U2 R2 U2 R2 U2 R2 U2
R' F R F' U' R U R' F' U F U'
R' D R' D' R2 F' U' R U R2 F R

obviously all inversion/refletios/ roations are also valid?

Are ther any others that are 12 or less that you can think of?

 

[Lucas Garron]

"Self Solving Algs"
You mean identities? Lemme go find the twistypuzzles.com thread...

 

[povatix]

Isn't it true that if you repeat any algorithm ( R U etc.), the cube will solve itself?

 

[joey]

Isn't it true that if you repeat any algorithm ( R U etc.), the cube will solve itself?

Yes, but thats boring!

 

[Johannes91]

Are ther any others that are 12 or less that you can think of?

Last summer I wrote a program for finding these. The shortest is (M2 S2)2, 8 moves FTM. There are many different 10-movers, like F R' F' R2 U' B' R B R2 U and R U R' F' U2 L' U' L F U2.

 

[Kenneth]

All 2-cycles self solves.

 

[Jason]

you can dream up any alg, if you repeat it enough times, you'll return to the solved state. The amount of times you have to repeat the alg before returning to the solved state is called "the order" of the algorithm. Gernerally, all the factors of the order have interesting properties. Say the order of an alg is 63, if you do the alg 21 times, you'll notice that maybe only 4 or 5 cubies have moved aroud and that the rest of the cube is solved

 

[magicsquares]

for R U I think you have to do it around 63 or 64 times...

 

[Johannes91]

for R U I think you have to do it around 63 or 64 times...

It's 105, but R U' is 63.

 

[Lucas Garron]

Say the order of an alg is 63, if you do the alg 21 times, you'll notice that maybe only 4 or 5 cubies have moved aroud and that the rest of the cube is solved

Could we please get the math right? All you could have is independent 3-cycles and misoriented corners. Actually, I'm not sure the former is even possible.

 

[Jason]

So, you're saying that factors of any alg of order 63 can only cycle 3 corners and twist 2? That's a bit of a brave statement! What about edge flips/ permutations? Show us the maths then!!!!

 

[Lucas Garron]

So, you're saying that factors of any alg of order 63 can only cycle 3 corners and twist 2? That's a bit of a brave statement! What about edge flips/ permutations? Show us the maths then!!!!

I didn't write that, and I'm not saying (or claiming) it either. I'm not limiting the number of pieces involved, either. (RU')x21 rotates 6 corners...

if you do the alg 21 times

If its order is 63, then doing it 21 times has order 3. This cannot flip an edge, since doing it 63 times would leave the edge flipped. Same thing for 2-cycles.

The only divisor of 63 that does not divide 21 is 9. So, a nine-cycle can give three 3-cycles. This is only possible with edges, requiring 7 corners for the 7. So, either [three 3-edge-cycles], or [misoriented corners] after 21 repetitions (for an alg of order 63).
RU' represents the latter, B2 U' L2 D' R2 B2 U2 L D L2 B2 R' F2 D2 U' R U2 F the former

 

[philkt731]

Ah! on my R' D R' D' R2 F' U' R U R2 F R, you can just get rid of the R' at the begininng and the R at the end and it becomes 10 and the same as Johannes's just rotated.

I got this from the two common V perms:
R' U R' U' y R' F' R2 U' R' U R' F R F
R' U R' U' y R' D R' D' R2 F' R' F R F

Here's a random 16 one that I found
F2 U B2 U' F2 U B2 U' F2 R' B' R F2 R' B R

 

[Derrick Eide17]

wow all these are so weird lol. some i already knew cause they are pretty much like 2 cycles and when done twice it solves it again. Today i found a FRIGGING LOOOOOOOOOOOOOOOOOOONG one today that takes FOREVER to FINALLY restore the cube back. if anyone wants to find out how many times it takes to solve the cube again BE MY GUEST but it took me like 25 STRAIGHT minutes of doing this to finally resolve. The alg is RU'B. SOOO DAMN LONG...

 

[philkt731]

I got that to be 360. Any larger ones?

 

[Lucas Garron]

I got that to be 360. Any larger ones?

http://mzrg.com/rubik/orders.shtml
He still hasn't added the multiples of 11 I sent him...

 

[Derrick Eide17]

all i have to say is WOW......

 

[fw]

I got that to be 360. Any larger ones?

There is a permutation in the cube group which has order 1260 and it is proven that there is no cyclic subgroup of the cube group which has a higher order

see "Adventures in Group theory" if you are interested in the details. There is some interesting stuff about the orders which actually can occur, for example, for every prime factor of |G| there is a permutation which has exactly this order.. if you take any non-prime number which does NOT divide the size of the cube group, than there is no element of this order, for example, there is no algorithm which has to be executed 123 times to "solve itself", because 123 is not prime and does not divide |G| (see "Lagrange's Theorem")

 

[Sam Kennedy]

the shortest one I can think of is r L'
does that count?

 

[Johannes91]

the shortest one I can think of is r L'
does that count?

This is shorter: