# Riddle



## gogozerg (Nov 5, 2010)

Hello!

I already asked this question years ago, I think never got the answer. So let's post it again.

When you scramble the cube in a special way, the configurations seem to have a special property, a kind of unusual subgroup. You can be sure that if put 4 corners of a side correctly permuted in a layer using Y-commutators only (RU'R'U like), then the 4 other corners are automatically correctly permuted.

The special scrambling is hard to explain. The best way to understand it is to imagine you are working on ULF-UFR-URB-DRF corners with the six Y-commutators around UFR, and you can move these corners to different places (think pseudo-blocks).

===========================

Example:

--- Scrambling ---

White top / Green front

(RU'R'U)^2
RU'x' -- (Notice that ULF-UFR-URB-DRF corners are still in the same scope)
(RU'R'U)
y'x'
(R'FRF')
RU'y
lF'lF'lF'
Rd'
(FR'F'R)^3
Rd'
lF'lF'lF'
x'y' -- (back to White top / Green front)

-> Cube looks fully scrambled.

--- Positioning white corners using Y-commutators ---

FL'F'L -- (three white corners in a layer)
FD'F'D
F'RFR' -- (fourth white corner added)

-> Yellow corners correctly permuted!

===========================

Question:
- Why?
( - And how to recognize such a cube configuration?)


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## mr. giggums (Nov 5, 2010)

I tried the scramble twice but I ended with green front, yellow top.


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## Lucas Garron (Nov 5, 2010)

gogozerg said:


> The best way to understand it is to imagine you are working on ULF-UFR-URB-DRF corners with the six Y-commutators around UFR, and you can move these corners to different places (think pseudo-blocks).


 
The Y-commutator is irrelevant, the part that matters are the two halves: any symmetry of RU' that keeps those four corners together (after a cube rotation to put them in the same place) effectively does the same to the other four corners.
When you get three corners of a layer, it means one of those sets has to be solved. Once you get the other one into place, the whole cube must be solved.

This reduces the problem to comparison with a tripod finish: A cube is in this state if there exists such a set of four corners such that they and their complement are both tripod-solvable.
(I'm not sure how centers figure into it, but they shouldn't make it too much more complicated.)


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## gogozerg (Nov 6, 2010)

Lucas Garron said:


> symmetry of RU' that keeps those four corners together (after a cube rotation to put them in the same place)


You got the idea, and gave a good description.

But I'm sorry, I still don't understand.

After such a scrambling, you can apply any Y-commutator, anywhere, as much as you want (random cube orientation+DR'D'R FR'F'R x'y L'BLB' ...), and the cube keeps this special property. That's what I find interesting about it.


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## Lucas Garron (Nov 6, 2010)

gogozerg said:


> After such a scrambling, you can apply any Y-commutator, anywhere, as much as you want



Actually, it gets better: You can apply any two smooth moves anywhere, (I use "smooth" for moves that follow each other like gears - such as RU') and it still holds. A Y-commutator is two sets of two smooth moves, of course.
I'm not so sure why, but there's probably a good reason related to the fact that two smooth moves create two 3-cycles.

(The important thing here is to think even more general than Y-commutators.)

(By the way, does this feel like a generalization of the 2-gen corner permutation group, anyone?)


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## Lucas Garron (Nov 6, 2010)

So, 1/30 of states are solvable this way, if my calculations are correct.

I'm trying to find a proof related to http://www.jaapsch.net/puzzles/pgl25.htm that's not heavily computational.

EDIT: Trying to understand the cosets to discern some structure. 1/5 scrambles have the two sets of corners in the proper shapes, but after parity that still leaves 1/3 to be properly accounted for.

EDIT 2: Oh, I'm silly. Problem solved. See next post.


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## Lucas Garron (Nov 6, 2010)

Call the set of four corners adjacent to a given corner (including itself) its tripod.
Call a set of two adjacent corners and their diagonal opposites a rectangle.

Conditions for solvability using two smooth moves at a time:

The pieces of the original UFR-tripod (from the solved position) must be in either a tripod or a rectangle shape.
The same must be true of the original UFL-tripod.
Corner parity must be even.

(I omit the boring details about invariants and completeness, as well as the "intuitive" fact that EO, CO, and EP can be resolved easily.)


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## gogozerg (Nov 6, 2010)

You're right... Smooth Moves© !

I thought such things could be very interesting. A wide family of not-too-constrained sequences (and you can of course choose to move one or two layers anywhere), that you can apply freely whatever the orientation, while preserving a property detected early (or maybe generated by applying a single move after inspection...).


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## Lucas Garron (Nov 6, 2010)

gogozerg said:


> (or maybe generated by applying a single move after inspection...).


A single move is obviously not enough, but I don't have my code to set up to see how many are needed, but I think I have a lower bound of 5 quarter turns.
(In practice, one or two moves and something like a T-perm seems to work for me.)

By the way, the corresponding criterion for what I call "rough" moves (the opposite of smooth, like RU) is simply even corner parity.


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## Ravi (Dec 24, 2010)

Lucas Garron said:


> (I use "smooth" for moves that follow each other like gears - such as RU')


 
So a sequence XY is smooth if and only if X and Y are quarter turns of adjacent sides in opposite (cw vs. ccw) directions, and rough if X and Y are quarter turns of adjacent sides in the same direction? I like this idea--it's a distinction I (and presumably many other speed cubers) have thought of before, but I've never seen a name attached to it. I think this is interesting enough to stop and list some more properties of smooth and rough moves:

There are 48 (= 6*2*4) possible smooth moves and 48 possible rough moves on a 3x3.

When a smooth move (eg. RU') is executed on an idealized cube, the center of the unique edge that moves twice (in this case, FR -> UR -> UB) has a differentiable path from start to finish. In this case, the edge in question has a velocity vector pointing directly towards B both at the end of the R and at the beginning of the U'. By contrast, during a rough move, the velocity vector changes direction by 180 degrees. Consequently, the momentum of the cubies favors smooth moves and opposes rough moves.

If one tries to execute the second half of a smooth move a little early (that is, while finishing the first half), it requires the standard type of corner cutting. Executing the second half of a rough move early, however, requires reverse corner cutting. This, together with the above (but, I'd argue, mostly the above) makes smooth moves much easier to execute quickly than rough moves.

The inverse of a smooth move is a smooth move, and the inverse of a rough move is a rough move.

Smooth moves have order 63. Rough moves have order 105.

Lucas gave some necessary conditions for an arbitrary Rubik's Cube group element to belong to the "smooth move subgroup" S (Lucas: can you prove that those are sufficient? Also, how did you get the number 30?) Rough moves generate all cube group elements with even corner permutation (and therefore even edge permutation). Proof: any sequence of two quarter turns can be effectively done with rough moves using the algorithms U2 = (U F)(F' R')(R U) and U D' = (U R)(R' D'), so any sequence of an even number of quarter turns can be expanded into a sequence of rough moves.


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## Lucas Garron (Dec 25, 2010)

Ravi said:


> (Lucas: can you prove that those are sufficient? Also, how did you get the number 30?)


Yes, but my proof is rather informal right now. Some Mathematica code to verify the corner situation and exhibited cycles for edges.


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## Athefre (Dec 25, 2010)

gogozerg said:


> You're right... Smooth Moves© !


 
Sorry, but that may be trademarked already.

http://www.nintendo.com/games/detail/7vgUzwrkjswZ6wiUXTtZQB8ji6_uPB3v


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