# Is it possible to solve the cube with one piece moving at all time?



## Cyrus C. (Aug 19, 2010)

Is it possible to solve the cube with one piece being moved by every turn? If you don't get what I'm saying, green front, white top, R U R2 D L U2 R' F L'. The white, green, red corner is always moving.


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## Kirjava (Aug 19, 2010)

No, a parity exists. 

Positions scrambled with an even number of quarter turns can be solved this way - otherwise an extra quarter turn without the moving piece is required. 

(I'm assuming this is for corners btw - it makes the puzzle <r,u,f>)


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## Rinfiyks (Aug 19, 2010)

Kirjava said:


> No, a parity exists.
> 
> Positions scrambled with an even number of quarter turns can be solved this way - otherwise an extra quarter turn without the moving piece is required.
> 
> (I'm assuming this is for corners btw - it makes the puzzle <r,u,f>)



Huh? The scramble *U* is an odd number of quarter turns. Take any piece on top layer and U'. I don't think I understand.

Can anyone think of a way to solve U D'?


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## ErikJ (Aug 19, 2010)

what if you solve with only deep layer turns. if you restrict yourself to r, u, and b moves then the corner at DFL is always moved relatively to the rest of the cube since a u move is the same as a D move. I think stefan has already done a solve like this a long time ago.


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## Kirjava (Aug 19, 2010)

Rinfiyks said:


> Huh? The scramble *U* is an odd number of quarter turns. Take any piece on top layer and U'. I don't think I understand




For all corners/any given corner. Solving the scramble U while always moving the corner at DBL is not possible.


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## jaap (Aug 19, 2010)

The corner locations can be put into two 'tetrads', {UFR, UBL, DFL, DBR} and {UFL, UBR, DFR, DBL}. Every quarter turn involving a particular corner piece will move that piece from one tetrad to the other.
Therefore any even permutation (which must be an even number of quarter turns) will leave that corner somewhere in its original tetrad, and any odd permutation leaves it in the other tetrad.
In other words, at most half the positions of the normal cube are reachable.


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## TMOY (Aug 19, 2010)

More than half if you get to choose the moving corner instead of picking a random one. But there are still some completely unreachable positions (like the N-perm for example: every corner is in its original tetrad but the permutation is odd).


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## guzman (Aug 19, 2010)

Cyrus C. said:


> Is it possible to solve the cube with one piece being moved by every turn?



Hi Everyone,
I'm new here.

I'll try to give my answer ... (sorry for any english mistakes or not clear sentences).

First of all, this is how I understand your problem:
You give me a scrambled cube and you also tell which piece I allways have to move during my solve (I have to move it at every quarter turn I make).

I think I can answer the question affirmatively: the cube can be solved in that way (allways moving the piece at every quarter turn).

Say you tell me to allways move the yellow-blue-orange corner which happens to be at the beginning in the UFR position,
then 
I first solve the cube by only using the moves 
*l, b, d* (lower case letters) (which I think can be done, correct me if I'm wrong),
this way the piece at UFR does actually never move 
(more or less as in a siamese cube),
then I transcribe the alg I found in al alg where the yellow-blue-orange corner allways moves ... Example:

if I found the solution l, b, d, b'
then I transcribe it as R U R B' and the corner allways moves ...

Hope it makes sense ...
(mathematical idea: you solve the cube using coordinates relative to the choosen corner and the you translate them to absolute coordinates ...)

guzman.


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## uberCuber (Aug 19, 2010)

guzman said:


> Cyrus C. said:
> 
> 
> > Is it possible to solve the cube with one piece being moved by every turn?
> ...



first of all, welcome to the forum

second, you cannot solve the cube using only l, b, and d moves unless the cube was scrambled using only l, b, and d moves. I think.


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## guzman (Aug 19, 2010)

uberCuber said:


> first of all, welcome to the forum
> 
> second, you cannot solve the cube using only l, b, and d moves unless the cube was scrambled using only l, b, and d moves. I think.



Hi, thank you.

So, let's use what I said differently:

the cube can be solved always moving the piece initially in UFR position
if, and only if, it can be solved by only using only l, b, d moves (and actually never moving the piece in UFR position).

In fact if I can solve the cube always moving the piece initially in UFR
I can keep that cubie blocked so that all moves will look like l, b or d's.
On the contrary if I can solve the cube using only l, b, d then I can transcribe the moves in order to always move the piece initially in UFR.

Does it make sense ?

I also think that in 2x2 cube this can actually be done for any scramble.

guzman.

guzman.


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## maggot (Aug 19, 2010)

i think this is an issue of relation... as people were saying using wide layer turns, well, if the piece is not going to move with the next turn, you can x, y, z, x', y', z' and then do a wide layer turn in relation to the turn where the selected piece is always moving. like i guess for example, if your UFL is your selected piece and you have to do an R, well, you can do x' l x, the piece is moving in relation to your new face. to solve the cube with that piece moving without this loophole, jaap post explain this..


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## jaap (Aug 20, 2010)

uberCuber said:


> you cannot solve the cube using only l, b, and d moves unless the cube was scrambled using only l, b, and d moves. I think.



Correct. If, as guzman suggests, you think of the problem not as always moving a layer containing a particular corner, but of always moving the other part of the cube, i.e. the two layers not including the specified corner, then it is indeed just solving the cube using three wide moves, such as l, b, d.

Looking at it in this reference frame you can again deduce that not all scrambled positions are possible.
Every wide quarter turn is an odd permutation of corners.
Every wide quarter turn is also an odd permutation of the centres.
Therefore the parity of the corners always remains the same as the parity of the centres. 
There are (full-cube) scrambles for which this is not the case, for example just do a D move. This therefore is not solvable using only l, b, d.

(Note that any whole cube quarter turn, for example x, breaks the parity equality we have. Imposing a fixed corner prevents us from doing so. This is why allowing all 6 wide moves will make any scramble solvable despite the above parity constraint - you can simply solve it in a different orientation that does have the right parity.)


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## macky (Aug 20, 2010)

This problem has previously been proposed under the name "keychain problem." The idea is that you have to always move the corner with the chain on a keychain cube, as if doing all turns by pulling on the chain. Linking appropriate threads from the Yahoo! forum:
http://games.groups.yahoo.com/group/speedsolvingrubikscube/message/15568
http://games.groups.yahoo.com/group/speedsolvingrubikscube/message/15614

And I remember it came up again a few months later, but I can't the thread. The point there was that solving in <u, f, r> is an equivalent problem.


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## Stefan (Aug 20, 2010)

Here are "recent" threads about <r,u,f> solving:
http://www.speedsolving.com/forum/showthread.php?t=10267
http://www.speedsolving.com/forum/showthread.php?t=10777

Did we ever discuss the version with a keychain *edge*?


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## guzman (Aug 21, 2010)

Two questions.

First. The original question was:


Cyrus C. said:


> Is it possible to solve the cube with one piece being moved by every turn?


Shouldn't we allow wide moves that move the chosen piece ?
In this case the problem would be equivalent to <r, u, f, R, U, F> (without cube rotations), right ? and it should be fairly easy.

Second. If I understand, the problem of the keychain edge is 
<r, f, U, D> (without cube rotations), right ?


guzman.


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## Stefan (Aug 21, 2010)

guzman said:


> Shouldn't we allow wide moves that move the chosen piece ?


No. That would make it trivial and boring.



guzman said:


> If I understand, the problem of the keychain edge is
> <r, f, U, D> (without cube rotations), right ?


In this case I admit it's not as trivial, so yeah, both <r, f, U, D> and <r,f> are interesting (though I prefer the latter (or rather <r,u>) for consistency with the corner version).


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## Swordsman Kirby (Aug 22, 2010)

StefanPochmann said:


> guzman said:
> 
> 
> > If I understand, the problem of the keychain edge is
> ...



Isn't it <r, f, UD'>?


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## Rinfiyks (Aug 22, 2010)

The corner version is <f, u, r> so the edge version should be <f, u> (also, the centre version would be <f>, but that's stupid, it just shows how the consistent pattern goes.)

After having a go with it on a 2 hour car journey, the edge version is sooo much harder. All I could do was get a 2x2x2 block.


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## Stefan (Aug 27, 2010)

Swordsman Kirby said:


> StefanPochmann said:
> 
> 
> > guzman said:
> ...



Right. And now I'm not sure which is more consistent/natural, that or <r,f>.


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