# A simple blindfold problem



## sgsawant (Sep 4, 2010)

Here's a simple problem for which I did not have a short answer:

The whole cube is solved except EP(1 7)(2 6). The orientation of the edges is alright - by the definition of the 3-cycle method (that's what I use). Now this should have been a simple double transposition problem. But think it over and you will see that setting it up is an involved process.

Please let me know if you think this is problem has a simple answer and if I am overlooking any obvious solution. I don't need the optimal solution (I can look it up myself with a simple A-star program). All I need is a solution which is short and can be implemented by the generic algorithms used for blindfold cubing (the 3-cycle method).

Regards,

-sgsawant


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## Stefan (Sep 4, 2010)

More people would know what you're talking about if you wrote something like UF rather than a number.


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## Mike Hughey (Sep 4, 2010)

Do (1 7 2). That will solve 7, 2 will have 1, and 1 will have 6.

Then, you just do (1 6 2), and it's solved.

Just 2 easy 3-cycles.

Not optimal, but I think it does what you want, doesn't it?



StefanPochmann said:


> More people would know what you're talking about if you wrote something like UF rather than a number.


If I understood what was being asked, this was a generic question where it didn't matter what actual pieces were being discussed. But maybe I just understood the question.


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## Stefan (Sep 4, 2010)

Mike Hughey said:


> Just 2 easy 3-cycles.


Well...


http://cubefreak.net/bld/3op_guide.html#EP2 said:


> 2-cycles of edges can only be solved in pairs (double transpositions)


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## blah (Sep 4, 2010)

D' x H perm U2 x' D


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## Stefan (Sep 4, 2010)

For edges?


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## blah (Sep 4, 2010)

oops


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## riffz (Sep 4, 2010)

I agree with Stefan. I had no clue what those numbers meant.


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## Stefan (Sep 4, 2010)

Well, *assuming* he's using Macky's guide so he's talking about F2B2 orientation and (UF BR)(UL BL), I'd say R'U'LU(Zperm)U'L'UR is obvious and U2lB(Hperm)B'l'U2 is shortest.


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## Mike Hughey (Sep 4, 2010)

StefanPochmann said:


> Mike Hughey said:
> 
> 
> > Just 2 easy 3-cycles.
> ...



Sorry. I guess this was a side effect of me doing BH for so long. When I saw


> All I need is a solution which is short and can be implemented by the generic algorithms used for blindfold cubing (the 3-cycle method).


I was thinking "a solution which can be implemented with just 3-cycles". But it seems much more likely that you're reading it the way he was.

This is a potentially significant drawback of BH. If you're doing pure BH, you always solve pairs of 2-cycles as 2 3-cycles, which is generally significantly more moves than a setup into an H perm or a Z perm or something similar. In fact, I generally don't even notice that I have a case of pairs of 2-cycles, so I'd never even think to optimize it.


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## Micael (Sep 4, 2010)

Still, 2 3-cycles is good when using memory system.


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## mr6768 (Sep 4, 2010)

I would do R' B2 U R' U' Z perm and undo the setup !


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## mr6768 (Sep 4, 2010)

StefanPochmann said:


> Well, *assuming* he's using Macky's guide so he's talking about F2B2 orientation and (UF BR)(UL BL), I'd say R'U'LU(Zperm)U'L'UR is obvious and U2lB(Hperm)B'l'U2 is shortest.



I think (1 6)(2 7) means this : (UF BL)(UL BR)


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## Stefan (Sep 4, 2010)

Mike Hughey said:


> This is a potentially significant drawback of BH. If you're doing pure BH, you always solve pairs of 2-cycles as 2 3-cycles, which is generally significantly more moves than a setup into an H perm or a Z perm or something similar.



Well, at least Z perm isn't so short. How would you solve this particular case with BH?



mr6768 said:


> I would do R' B2 U R' U' Z perm and undo the setup !



Is that supposed to be better than mine in any way?



mr6768 said:


> I think (1 6)(2 7) means this : (UF BL)(UL BR)



Um... he asked for (1 7)(2 6)?


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## mr6768 (Sep 4, 2010)

> Um... he asked for (1 7)(2 6)?


Oh sorry . My bad ! yes he asked for that


> Is that supposed to be better than mine in any way?


I didn't say it's better ! I just said the first idea came into my head !
By the way I don't use Cycle method for edges , I prefer your M2 method Mr.Pochmann !


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## Sakarie (Sep 4, 2010)

With BH using UF as buffer, it could be 
[U': F E' F'] [U': F' E F] or
[R' U' R: E'] [L': B M2 B'] as easiest. If I got the commutator notation right.


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## cmhardw (Sep 4, 2010)

I really don't think I would have seen this during a solve, but if you do two commutator three cycles just right, you get 1 move cancellation. You have to do an 8 mover Drop and Catch followed by a A9, so it's still 16 moves, but the two commutator types use the same set of turns. Do (UF UL BL) (UF BR UL)
x'
[L', U' M' U]
U2 L' U' M' U L U' M U'
x

Which is:
x' (L' U' M' U L U' M *U*) (*U2* L' U' M' U L U' M U') x

And this can be executed as:
r' U' M' U L U' M U' L' U' M' U L U' M U' x = 16 turns STM

---------------------------
I wouldn't be using UF as my buffer, but UB. If I had the FB reflected cycle of (UB FR)(UL FL) on a real solve then I probably would have done:
[L' E2 L, U'] y' [U, L E' L'] y

Which means that for the original cycle of (UF BR)(UL BL) I would most likely have done:
[L E2 L', U] y' [U', R' E R] y

--edit--
Seeing it now, this would be a fast way to do the cycle possibly:
[L E2 L', U] x' U' M' U L' U' M U r
--edit--

Chris


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## Sakarie (Sep 5, 2010)

But it wouldn't be faster than 
[U': F E' F'] [U': F' E F]
with cuberotation, would it?
zx R' U M' U' R U M U' R' U' M U R U' M' U x'z'

It's not faster than Pochmann's, but I guess it's good threecycle way.
Another possibility is:
x' M U M U2 M' U l' U' M' U L U' M U x 14 STM


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## XXGeneration (Sep 5, 2010)

I'm pretty sure that you can solve a 2cycle in 2-3cycles.

Copy pasted from Macky's site: 

Say we need to solve the cycles (AB)(CDEF). Doing (ABC) breaks into a second cycle and leaves us with (ADEFC). Note that C, which we used to break into the second cycle, must be attached to the end of the cycle. In 3-cycle methods, this allows us to avoid leaving 2-cycles unsolved in even cycles. In effect, we replace the double transposition (XY)(ZW) with two 3-cycles, (XYZ)(XWZ). This is especially useful for 4x4, where double transpositions are hard to set up.


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## cmhardw (Sep 5, 2010)

XXGeneration said:


> I'm pretty sure that you can solve a 2cycle in 2-3cycles.
> 
> Copy pasted from Macky's site:
> 
> ...



Keep in mind that sometimes (XZW)(XZY) is also useful.

Chris


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## XXGeneration (Sep 5, 2010)

cmhardw said:


> XXGeneration said:
> 
> 
> > I'm pretty sure that you can solve a 2cycle in 2-3cycles.
> ...



Ah, that would be very helpful; Personally, I haven't tried using two 3-cycles in place of a double transposition, but remembering all the tough ones I've had, I'll try to use them. They seem much easier than awkward double transpositions.


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## reThinking the Cube (Sep 6, 2010)

sgsawant said:


> The whole cube is solved except EP(1 7)(2 6). *The orientation of the edges is alright - by the definition of the 3-cycle method (that's what I use)*. Now this should have been a simple double transposition problem. But think it over and you will see that setting it up is an involved process.



The relative orientation for edges (in this case UF<>BR) can be made different from the F2B2 orientation.



StefanPochmann said:


> Well, *assuming* he's using Macky's guide so he's talking about F2B2 orientation and (UF BR)(UL BL), I'd say R'U'LU(Zperm)U'L'UR is obvious and U2lB(Hperm)B'l'U2 is shortest.



For this problem, I would do the 10-move DTcommutator [zF;M',U':M'] written out is, zF(M'(U'M'U)M(U'MU))F'z'

z F M' U' M' U M U' M U F' z'

Note that the setup (zF) places the pieces that need to be exchanged in the same spots as the common edge orientation alg (M'U')4, but I prefer by force of habit (M'U)4, and usually locate this and all double transpostions at (UL<>UB,DF<>DB), rather than (UR<>UB,DF<>DB). Therefore the alg that I use for simul swapping+ orientation on all 4 edge pieces is [S;M',U':M'] which written out is S(M'(U'M'U)M(U'MU))S'. Leaving out the *S* setup will prevent the top two edges (UR<>UB) from flipping, and is equivalent (less setup zF) to the original DTcommutator given earlier [zF;M',U':M'].


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## XXGeneration (Sep 7, 2010)

I would do:
R' U' L (H-perm) L' U R.
It's not too hard, and pretty quick lol.


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## qqwref (Sep 7, 2010)

Personally I don't like to set up to Z perms. H perms can be setup to on any face (like N perms) and are faster as well as harder to mess up.

Assuming the edges are (UF BR) (UL BL):
- If you are allowed to do R/L quarter turns, I'd do something like U R' U L setup to H-perm on U. Stefan's 3-mover is fewer moves but more confusing.
- If you are allowed to do F/B quarter turns, I'd do B U' B2 setup to H-perm on U.


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## XXGeneration (Sep 7, 2010)

qqwref said:


> Personally I don't like to set up to Z perms. H perms can be setup to on any face (like N perms) and are faster as well as harder to mess up.
> 
> Assuming the edges are (UF BR) (UL BL):
> - If you are allowed to do R/L quarter turns, I'd do something like U R' U L setup to H-perm on U. Stefan's 3-mover is fewer moves but more confusing.
> - If you are allowed to do F/B quarter turns, I'd do B U' B2 setup to H-perm on U.



F/B turns are not allowed for 3OP; they'll mess up the orientation.


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## Stefan (Sep 7, 2010)

reThinking the Cube said:


> z F M' U' M' U M U' M U F' z'



That is quite nice, didn't know that yet.



XXGeneration said:


> I would do:
> R' U' L (H-perm) L' U R.



And you'd DNF.



XXGeneration said:


> F/B turns are not allowed for 3OP; they'll mess up the orientation.



Wrong.


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## Sakarie (Sep 7, 2010)

I looked it up with ACube, and there are a lot of 10 STM algorithms:



Spoiler



U L E2 L' U' B2 L E2 L' B2 . (18q, 12f, 10s)
U R S2 R' U' B2 R S2 R' B2 . (18q, 12f, 10s)
F L' E' L F' E' L' E L E . (14q, 14f, 10s)
F L' E' L E' L' E L E F' . (14q, 14f, 10s)
F E F' E' L' F E' F' L E . (14q, 14f, 10s)
F E' L' E' L E L' E L F' . (14q, 14f, 10s)
*B M B' R' B U M' U' B' R . (12q, 12f, 10s)*
B2 L E2 L' B2 U L E2 L' U' . (18q, 12f, 10s)
B2 R S2 R' B2 U R S2 R' U' . (18q, 12f, 10s)
L2 U' L' E L2 S' L' S D F2 . (16q, 13f, 10s)
L2 U' R E R2 S' R S D F2 . (16q, 13f, 10s)
L2 D' M' B M B2 E' B U B2 . (16q, 13f, 10s)
L2 D' M' F' M F2 E' F' U B2 . (16q, 13f, 10s)
R' B U M U' B' R B M' B' . (12q, 12f, 10s)
E M S D R E2 R' U' S' M . (18q, 16f, 10s)
E' L' F E F' L E F E' F' . (14q, 14f, 10s)
E' L' E' L E F L' E L F' . (14q, 14f, 10s)
S M U F E2 F' D' M' S E' . (18q, 16f, 10s)
S M U F E2 F' U' M S' M2 . (20q, 16f, 10s)
S M U B M2 B' U' S2 M' S . (20q, 16f, 10s)
S2 M S' U F E2 F' U' S' M' . (20q, 16f, 10s)
S' M S2 U B M2 B' U' M' S' . (20q, 16f, 10s)
M S U F E2 F' U' S M' S2 . (20q, 16f, 10s)
M2 S M' U F E2 F' U' M' S' . (20q, 16f, 10s)



The one that seems the best is the one I've bolded, with a cuberotation.
x' U M U' R' U F M' F' U' R x


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## qqwref (Sep 7, 2010)

XXGeneration said:


> qqwref said:
> 
> 
> > Personally I don't like to set up to Z perms. H perms can be setup to on any face (like N perms) and are faster as well as harder to mess up.
> ...



I gave two options for a reason. Some people allow R/L but not F/B; some people allow F/B but not R/L. You choose one and neither is really better. I personally use 3OP and allow F/B.


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## XXGeneration (Sep 8, 2010)

qqwref said:


> XXGeneration said:
> 
> 
> > qqwref said:
> ...



Ah, now that I think about it, that makes sense. Using F/B would be like using R/L except with a y/y' turn.


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