# Fixing a Pop



## JTW2007 (Mar 12, 2009)

Is there some bordering-on-magical group theory trick that would make me able to tell how to insert popped pieces correctly? I would think it could be done mathematically... Any ideas?


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## *LukeMayn* (Mar 12, 2009)

it can easily be done, it's just that recognizing it will take some time :/
for edges, just count how many are mis oriented and insert the poped edge to make the total orientation an even number.
A corners uses the same kind of concept but it's a bit more complex.


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## IamWEB (Mar 12, 2009)

This wouldn't be efficient for competitions unless a single edge just kind falls out a little, otherwise just put it back it in and change if necessary.

In practice, yes this would be nice to know.


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## Asheboy (Mar 12, 2009)

For Edges:

Surely you could look at the edges and see how many are mis-orientated and if it's an odd number you need to put it in mis-orientated and if it's even you need to put it in orientated.

For corners it's going to be something about the number of orientated, clockwise and counter-clockwise corners.


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## AvGalen (Mar 12, 2009)

Most blindfolded solvers can do this relatively fast, so if you want to know how to do that, just learn a blindfolded method.

If you don't want to learn blindfolded, http://www.ryanheise.com/cube/cube_laws.html might help

But in competition, just push the pieces back, hope for the best (1/12 chance) or fix the problems when you can clearly see them at the end of your solve. Pay attention to these rules though: http://www.worldcubeassociation.org/regulations/#puzzledefects


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## JLarsen (Mar 12, 2009)

If I'm anywhere around bad edges or in step 4 or later I can put in the edges correctly just because it's what I do.


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## Tyson (Apr 8, 2009)

Generally in a competition, it would cost you more time to figure out which way to put it back in then it would for you to randomly put it back in and continue solving. Also, in many pop situations, it's pretty obvious which way the piece needs to be put back in. This is more likely if you have significant parts of the cube solved.


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## Ryanrex116 (Apr 26, 2009)

My very unstable pyraminx popped twice in one solve. That is because one of the tips/corners is too loose. The three edges under it were always about to pop. On one of my pops, the edge I couldn't see popped out. I put it in wrong, and it was very hard to pop and put it back in. If I don't get a better one next time, I will count the edges.


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## dougbenham (Apr 27, 2009)

For corners you could do something like this:

Keep a value in your mind. Start with 0. Check a corner (eg. UFR):
-If it is correctly oriented (U or D color), then don't change your number.
-If it is clockwise oriented, then add 1 to your number.
-If it is counter-clockwise oriented, then subtract 1 from your number.

Do this for all the corners and if your "internal" number isn't 0 in the end, you have an incorrect corner  and if your internal number is 1, you know that you need to rotate any corner counter-clockwise to fix it. If you internal number is -1, you know that you need to rotate any corner clockwise to fix it.


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## byu (Apr 27, 2009)

Wrong. It can be any of the following numbers:

-24, -21, -18, -15, -12, -9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21, 24

Not just zero. For example, do a Sune and look at it and you'll see what I mean.


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## dougbenham (Apr 27, 2009)

byu said:


> Wrong. It can be any of the following numbers:
> 
> -24, -21, -18, -15, -12, -9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21, 24
> 
> Not just zero. For example, do a Sune and look at it and you'll see what I mean.



Haha my mind thought about that but I was too lazy to actually check in-depth.


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## byu (Apr 27, 2009)

More clarification on corner orientation.
Using dougbenham's numbering system above (+1 for clockwise, -1 for counter clockwise):

Let y equal the resulting number.

x = y/3

If x is an integer (no decimal) then you have a correct orientation. If x leaves a decimal of .333..., then you need a counter clockwise corner. If x leaves a decimal of .666... then you need a clockwise corner.

EDIT: As mentioned by jason9000, it's reversed if y is negative.


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## jason9000 (Apr 27, 2009)

byu's scheme is reversed, of course, if y is negative.


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## byu (Apr 27, 2009)

Yeah, I forgot about that.


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## rahulkadukar (May 25, 2009)

Learn BLD and you will automatically be able to do it. Though it will take some time


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## JTW2007 (Jun 4, 2009)

rahulkadukar said:


> Learn BLD and you will automatically be able to do it. Though it will take some time



I use Classic Pochmann, so I approach orientation differently in a blindsolve than I do in a normal solve.


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