Fixing a Pop

JTW2007

BattsMan
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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?
 
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.
 
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.
 
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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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.
 
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.
 
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 :D 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.
 
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.
 
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.
 
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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