# Mackey's Blindfold Question



## prafulla.rawal (May 17, 2009)

Hello Cubers,

I have been practicing Mackey's Blindfold method (http://www.cubefreak.net/BLD/M2_guide.html) for like a month now and I am pretty much there but there is one question.

I just cant understand logic for orienting the last piece. This is what the website says:

_Case b: One sticker left
If there were an even number of inactive edges during memorization, orient them just as in the EO step of 3OP edges. If there were an odd number, there are two cases:
(1) If the last sticker is on U, D, or on F/B in the middle layer (recall the UDRLF2B2 EO definition), then orient in addition the edge containing the last sticker.
(2) Otherwise, orient in addition the DF edge._

*Can someone please elaborate this in a simple way? Are there any tips and tricks to quickly figure out, which edge needs orientation DF or the last sticker?
*

Thanks in advance,


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## byu (May 17, 2009)

Simply put:

Even number of inactive edges: Orient like 3OP
Odd number of inactve edges: DF is now inactive, so remember to flip that too.


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## prafulla.rawal (May 17, 2009)

Hi Byu..

Thanks for the reply.. For even number of inactive edges..I agree that we need to orient them like 3OP.. but for Odd Number.. Sometime we need to flip the DF (as you said) OR sometime.. instead of the DF we need to flip the Last Edge of the cycle. I am not too sure.. how to choose which one to flip.

As I quoted above..the website explains this as:

_If there were an odd number, there are two cases:
(1) *If the last sticker is on U, D, or on F/B in the middle layer (recall the UDRLF2B2 EO definition), then orient in addition the edge containing the last sticker.*
(2) Otherwise, orient in addition the DF edge._

but the description in bold doesn't always work for me.. or may be I don't understand it completely..


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## byu (May 17, 2009)

I didn't learn M2 from macky, I learned from Stefan, but Ill check Macky's version now.


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## prafulla.rawal (May 17, 2009)

sure. Let me know if you can figure that out.


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## chimpzi (May 18, 2009)

i think there are algs [commutators i think] that can be used so that you dont have to orient at the end. except if there is one piece in the correct position but is incorrectly oriented.
...if this is what you meant.

those algs are for UF, UB, DB pieces so that it would be placed directly with correct orientation. let me check if i still have them.
i'll post them later.


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## Ellis (May 18, 2009)

chimpzi said:


> i think there are algs [commutators i think] that can be used so that you dont have to orient at the end. except if there is one piece in the correct position but is incorrectly oriented.
> ...if this is what you meant.
> 
> those algs are for UF, UB, DB pieces so that it would be placed directly with correct orientation. let me check if i still have them.
> i'll post them later.



He's not asking about M slice edges, he's asking about misoriented edges that were in the correct location after scramble. 

It's pretty simple, although Macky seems to be making it more confusing than it really is. If you understand an even number orientation with 3OP, there's no reason you shouldn't be able to understand an odd number. An odd number is almost always just one edge, because getting 3 edges misoriented in place is pretty rare. 

So if you have one misoriented edge, then your buffer will end up misoriented also. What I do is set up that edge to the UF position then do an x rotation, then do a 3OP alg which orients UB and UF, then undo everything. So it would be (setup x orient x' setup'). There is really no restrictions on setup moves, and I think it's always a maximum of 2 setup moves. 

Some examples using macky's alg:

If the FL edge was misoriented during memo, and it was the only edge that was misoriented in place, solve edges and then do: L' U' x (M'UM'UM'U2MUMUMU2) x' U L

If the DR edge was misoriented, solve edges and then do: R2 U x (M'UM'UM'U2MUMUMU2) x' U' R2

I think most cases with one misoriented edge are easier than two, and if you understand the two edge cases then just imagine one edge cases being the same thing but just paired with DF as the second edge.

Edit- Although I've never done this with M2, you could just use M2 to solve. Shoot to both positions of the misoriented edge in any order, it's actually probably less moves that way and maybe even easier to understand.


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## prafulla.rawal (May 18, 2009)

Hi Ellis - Thanks for the reply.. I understand everything you said.. except for the fact that..


> So if you have one misoriented edge, then your buffer will end up misoriented also.



Based on what Mackey explained.. and what I have been seeing.. I dont think it is true.. When you have parity..and you are memorizing edge cycles.. you are left with one edge (and of course one buffer).. In this particular case.. Its not always the buffer which is misoriented.. sometime its the last edge..

For example.. lets say in a scramble.. UL was inactive (i.e. Misoriented in correct position). At the end you realise that there is a parity.. and you are left with two edges.. your buffer DF and another edge FR... Now to orient the UL correctly.. (from what Mackey explained).. you will either need to flip FR or DF.. based on the following rule from http://www.cubefreak.net/BLD/M2_guide.html
_(1) If the last sticker is on U, D, or on F/B in the middle layer (recall the UDRLF2B2 EO definition), then orient in addition the edge containing the last sticker.
(2) Otherwise, orient in addition the DF edge_

This is the point I am not sure of.. If someone can make it simple.. and come up with a thumb rule..about.. How to choose DF or the Last edge for orientation.. I would be thankful.

I would be happy to provide a scramble..and solving steps.. if this is not clear enough.. let me know.


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## Ellis (May 18, 2009)

In that example, why not just solve FR then orient UF with the buffer? The way I explained is the way I do it, and I've never run into any problems. I don't think I fully understand macky's way of solving this. If you gave a scramble I could tell you how I would do it. I'm slightly confused and I don't see any big advantage in doing it that way.


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## prafulla.rawal (May 18, 2009)

> In that example, why not just solve FR then orient UF with the buffer?



If there are only two edge left.. i.e. Buffer (DF) and The last edge (FR).. then we cant just solve them.. atleast not using Mackey's method... solving would be actually fixing parity involving corners.. which wont fix the orientation.


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## Ellis (May 18, 2009)

prafulla.rawal said:


> solving would be actually fixing parity involving corners



I don't see what's bad about that. I think standard M2 parity is much easier than 3OP parity if you have DF edge + another random edge.

Edit- I went through Macky's example solve, and I think I may be able to word it a little differently. If the last sticker you need to solve belongs anywhere on the L or R faces, then orient the piece that's currently in the buffer position with the inactive edge, if it doesn't belong to L/R, then orient that position with the inactive edge. So if you take the example that you gave before, with UF misoriented and the FR position unsolved, if the DF position belongs to RF, then orient the buffer position (the actual FR edge) with UF, if that sticker belongs to FR, then orient the buffer piece (currently in FR) with UF. I'm pretty sure that's how it would work. It would be a little different if the unsolved piece were in the M slice. In that case, I think the rule would be that if the sticker at DF belongs on U or D, then orient the unsolved position with the inactive edge, and if DF belongs on F or B, then orient the buffer position with the inactive edge. 

Sorry if that sounded confusing, I'm pretty sure that's what macky's trying to do there. I still would argue that M2 parity is a little easier to deal with. May be a few more moves, but definitely less thinking involved, and the 3OP parity with DF edge is harder to setup anyway.


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## prafulla.rawal (May 18, 2009)

I think I didn't phrase it correctly.. Here is another attempt:

Lets say UL is disoriented in correct position. Corners UFR and UBR are swapped. While solving edges.. Only two edges left in the end are DF (buffer) and FR (last sticker).

Approach: 
1. Setup - Flip UL with either DF or FR.. - undo setup.
2. Solve parity i.e. DF, FR, UFR, UBR 

But the question is.. in step 1.. which edge will be flipped with UL.. is it DF or is it FR.. just to mention again.. in Mackey's method.. its not always the DF that should be flipped. Hence, I need to understand a simple set of rules which allow me to choose between DF / FR (or any other last edge).

I will update this post with a scramble and sample solve.. and then it will be clear I guess..

Thanks for continuing the conversation


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## chimpzi (May 18, 2009)

@Ellis
sorry, missed his point.

@topic
i think it will also depend on how you do corners. not sure. it depends on how you set up the edges and corners in fixing parity.

on your example, if you fix the parity on F face[meaning you set up the corners to F face then x cube rotation then parity alg, undo set-up], then FR piece will be the one that is misoriented. since FR will be UR and DF will be FU because of the x set-up. 

but if you set-up it to U face, example: M2 to setup the buffer to UB then R to set-up F to UR, then parity alg, undo set-up. then that would result to a misoriented buffer.

so it depends on how you set-up cubies for fixing parity. just track where the stickers are going so you would know.


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## Ellis (May 18, 2009)

prafulla.rawal said:


> I think I didn't phrase it correctly.. Here is another attempt:
> 
> Lets say UL is disoriented in correct position. Corners UFR and UBR are swapped. While solving edges.. Only two edges left in the end are DF (buffer) and FR (last sticker).
> 
> ...



Did you catch my edit? It depends on if DF belongs to RF or FR. If DF goes to RF, then flip buffer position with UL. If DF goes to FR, flip UL with FR. Pretty sure that's how it works.

Btw, when I said here to look at how macky solves it, I was only referring to M slice pairs, not the whole thing 
oh well though, If you wanna do it that way, go for it. I still think it's a little harder than it needs to be.


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## prafulla.rawal (May 18, 2009)

Thanks Ellis and Chimpzi..

@Ellis.. your 'Edit' is kinda explanation I was looking for.. would give it a go.. when I have a cube in my hand. About the approach.. I am now long way into the complete Mackey's method.. so no turning back at least in near future ;-)

Thanks again!


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## AvGalen (May 18, 2009)

Mackey has an M2 guide?

I must be getting old because I was ready to comment "Macky does orientation seperate from permutation....blablabla"


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## byu (May 18, 2009)

Macky's site has both a 3OP tutorial and an M2 tutorial.


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## AvGalen (May 18, 2009)

byu said:


> Macky's site has both a 3OP tutorial and an M2 tutorial.


It does now, but Macky's blindfolded guide used to be a synonyme for 3OP. 
To me, this looks like a Petrus tutorial on Jessica Friddrich's site
Or a bigcubes pairing tutorial by AVG 
(Or even FMC by Dan Brown )

(and yes, I copied the use of Mackey's from the topicstarter)


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## deadalnix (May 18, 2009)

yes but Mackey's tutorial is like an 2 piece at time M2 variation.


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## cmhardw (May 18, 2009)

I'm not sure I understand correctly, but I orient the inactive pieces with the buffer piece before I start doing any cycling. I find all inactive pieces, and from here two things can happen. If there are an even number of inactive pieces I orient them like 3OP. If there are an odd number of inactive pieces then I find the buffer *piece*, not the buffer location, and orient the inactive pieces as well as the buffer piece like 3OP. I do the same for corners if there are permuted but disoriented corners. After I have oriented, then I start cycling.

I don't do 3OP, I solve permutation and orientation at the same time just like M2, but this is how I handle permuted but disoriented pieces after the scramble.

Chris


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## Ellis (May 18, 2009)

AvGalen said:


> but Macky's blindfolded guide used to be a synonyme for 3OP.



Yea, well it still is. This thread is kind of misleading if you don't follow the link. "Macky's blindfold guide" will always mean 3OP. But he does have a really nice M2 variation.


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## AvGalen (May 18, 2009)

Ellis said:


> AvGalen said:
> 
> 
> > but Macky's blindfolded guide used to be a synonyme for 3OP.
> ...


That might be how cubing dinosaurs like you and me think about it, but the topicstarter opened with this


> I have been practicing Mackey's Blindfold method


 and means the M2-variation


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## Mike Hughey (May 18, 2009)

cmhardw said:


> I'm not sure I understand correctly, but I orient the inactive pieces with the buffer piece before I start doing any cycling. I find all inactive pieces, and from here two things can happen. If there are an even number of inactive pieces I orient them like 3OP. If there are an odd number of inactive pieces then I find the buffer *piece*, not the buffer location, and orient the inactive pieces as well as the buffer piece like 3OP. I do the same for corners if there are permuted but disoriented corners. After I have oriented, then I start cycling.



That's funny. When you first posted about this a week or two ago when I asked you about it, I misunderstood you and thought you said you oriented with the buffer location. So I tried to figure out how to do it and decided the only way to do that would be to memorize inactive pieces that need orienting first, and since I wanted to memorize them last, I rejected it as a bad idea for me. So then I eventually figured out I could do it by orienting with the buffer piece (not location), and I was all proud of myself for figuring that out. Now it turns out I'm just doing what you do after all. 

This process still gives me headaches sometimes. When I get cases with two inactive pieces twisted clockwise and one twisted counterclockwise, I know it's obvious what to do but it seems like I often wind up getting a DNF, because I get confused about which way one of the pieces needs to go. I suspect it's just all about practice.


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