Shortest or interesting identical moves?

I was playing around with a 2x2, and discovered that these two sequences are identical:

R U' R' U2 R' U R2
and
R2 U R' U2 R' U' R

 Twizzle link 

 Setup 

R U' R' U2 R' U R2

 Moves 

 Twizzle link 

 Setup 

R2 U R' U2 R' U' R

 Moves 

It's even funnier when you notice that the order of the moves are reversed, but not the moves themselves.

I also know from before that on a 3x3 the following two sequences are equal:

R U R' F' U2
and
U2 F' L' U L

 Twizzle link 

 Setup 

R U R' F' U2

 Moves 

 Twizzle link 

 Setup 

U2 F' L' U L

 Moves 

I suspect this is one of the shortest such pair. Obviously it works on 2x2 too, but is there any shorter for 2x2? What other interesting, non-trivially equal sequences out there for 2x2 and 3x3?

R2 U2 R2 = U2 R2 U2 is the shortest on 2x2
R U R U R = U R U R U is the shortest on 2x2 in qtm
U2 D2 L2 R2 = L2 R2 U2 D2 is the shortest on 3x3
R U R' F' U2 = U2 F' L' U L is the shortest on 3x3 in qtm

You can find the length of the shortest such alg by looking at the start of the God's algorithm distribution and finding the smallest n such that the number of positions at distance n is less than the number of canonical sequences of length n. When this happens, (at least) two of the canonical sequences map to the same permutation.

Ben Whitmore said:

R2 U2 R2 = U2 R2 U2 is the shortest on 2x2
R U R U R = U R U R U is the shortest on 2x2 in qtm
U2 D2 L2 R2 = L2 R2 U2 D2 is the shortest on 3x3
R U R' F' U2 = U2 F' L' U L is the shortest on 3x3 in qtm

You can find the length of the shortest such alg by looking at the start of the God's algorithm distribution and finding the smallest n such that the number of positions at distance n is less than the number of canonical sequences of length n. When this happens, (at least) two of the canonical sequences map to the same permutation.

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Hmmm. Although I appreciate it, I don't find some of these identities satisfying.
For example, U2 D2 L2 R2 = L2 R2 U2 D2 can be considered a trivial solution, since it's basically saying E2 M2 = M2 E2, and double slice moves are commutative. It's almost the same as saying R L = L R (not exactly, since the latter are actually identical if we consider only canonical moves*).

I'm struggling to articulate why I don't like the R2 U2 R2 = U2 R2 U2 identity. It's just looks way too symmetric; so naturally a mirrored version of one algorithm would result an identical state, again, trivializing the problem.

On the other hand, I think R U R U R = U R U R U is way more surprising, and R U R' F' U2 = U2 F' L' U L is absolutely fascinating.

* = canonical moves where there are no consecutive moves of the same side, like U2 U', and if opposite sides turns are next to each other, they have a strict order. Like U D2 is preferred instead of D2 U.

Renslay said:

 Twizzle link 

 Setup 

R U' R' U2 R' U R2

 Moves 

 Twizzle link 

 Setup 

R2 U R' U2 R' U' R

 Moves 

It's even funnier when you notice that the order of the moves are reversed, but not the moves themselves.

Click to expand...

Thread hijack!

Off-topic, but I recall that Kåre Krig's 4x4x4 solver found OLL parity algorithms of that structure which gave me the same feeling:

Christopher Mowla said:

Speaking of which, I found at least one algorithm that is equal to its reverse, but it is not equal to its inverse!
Rw U' Rw U2 R' U2 Rw' U' Rw U2 Rw R U2 Rw U' Rw' U2 R' U2 Rw U' Rw (28,22)

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But since length is a constraint, here's another one of his (a "Petrus Parity" alg): Rw U' R2 U2 R2 U r U2 r U2 r U R2 U2 R2 U' Rw (17). And of course, r U2 r U2 r U2 r U2 r (9) and ultimately f2 r E2 r E2 r f2 (7) and Rw2 F2 U2 r2 U2 F2 Rw2 (7).

But actual useful 4x4x4 equivalent algorithms of the same length (to me) are algs like:
F' r' F U2 F' r F U2 = [F' r' F, U2] (8)
F r' F' U2 F r F' U2 = [F r' F', U2] (8)
U2 l U2 r' U2 r U2 l' = [U2, l U2 r'] (8)
U2 l U2 l' B2 r B2 r' (8)
U2 r B2 r' B2 l U2 l' (8)
(because if you substitute any of them into where A is in (r2 B2) A (l F2 r F2 l')(B2 r2), you get different single dedge flip algs).

Algs like r' U2 F' U2 F U2 r' U2 r' U2 r' F' U2 F r' vs l U2 F' U2 F U2 l U2 l U2 l F' U2 F l were always interesting to me as well (identical, even on the supercube), etc.

Renslay said:

Hmmm. Although I appreciate it, I don't find some of these identities satisfying.

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Although I agree, that's typically the case for the shortest of algorithms. (Nothin' 2 em.)

Like Rw2 U2 r E2 r' E2 r U2 Rw2 vs Lw2 U2 l E2 l' E2 l U2 Lw2.

Christopher Mowla said:

Although I agree, that's typically the case for the shortest of algorithms. (Nothin' 2 em.)

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I assume requiring them to be short enforces them to by symmetric in some way, which naturally spawns more than one solution to them, yet somehow they all belong to the same "family".

Maybe I should drop the requirement of being the shortest, and just ask for "interesting pairs"? (How do you define "interesting pairs"? I guess when you cannot point out a trivial relationship between the two?)

But of course it's easy to find such pairs if we allow lengthy movements: Do some random scrambling movements and solve the cube. Together they form a series of movements that basically "do nothing" (you start from a solved cube, end at a solved cube). Cut them into two at any point, let's say A B, and invert one of them. Voila, A = B'.

I still find that R U R' F' U2 = U2 F' L' U L is somehow beautiful, mesmerizing, and elegant in this regard.

Unsurprisingly, this is an interesting cube state with its 4416 optimal 20-move canonical sequences:

acrystal said:

Spoiler: The Classic 12

(R2 U2)3 = (U2 R2)3
(B2 U2)3 = (U2 B2)3
(L2 U2)3 = (U2 L2)3
(F2 U2)3 = (U2 F2)3
(F2 R2)3 = (R2 F2)3
(R2 B2)3 = (B2 R2)3
(B2 L2)3 = (L2 B2)3
(L2 F2)3 = (F2 L2)3
(R2 D2)3 = (D2 R2)3
(B2 D2)3 = (D2 B2)3
(L2 D2)3 = (D2 L2)3
(F2 D2)3 = (D2 F2)3

Click to expand...

Likewise, since (R' F R U')6 = id, we can say (R' F R U')3 = (U R' F' R)3.
Or because (R U R' U')6 = id, it is true that (R U R' U')3 = (U R U' R')3.
There are many cycles with short repetition number, and you can break them at any point into two (see my previous post), not particularly interesting.

Also, did you know that (R U R2)31 = (R2 U' R')32 ? Can you figure out why?

Also-also, (R F U L B D)30 = (D' B' L' U' F' R')30 = (E' L')4 M' (M' U')4 M. But these are quite forced.

Yeah I know, but it still works as what you are complying.