CFOP Superman’s clairvoyance (Colors behind the corners guessing)

[Tito_H]

This is the situation.

Please refer to the following cube.

(Source: https://ruwix.com/online-puzzle-simulators/)

Cube position is Yellow Top, Blue Front. Cross and the first F2L (DBL & BL) are finished.

Now I want to fill the Red-Green slot (DBR & BR) because BR edge is on the U layer, and it is normally a good idea to fill the back slots before front ones.

Oh, oh, there are DFL and DBR which have a Red face. But which is the Red-Green-White corner piece I am looking for?

Superman can guess it instantly without peeking to the hidden faces, but regrettably I am not him.

Is there any easy method to know which Red-Green-White corner piece is?

In a real solving, peeking the left side of the DFL to check its color and proceed with F2L may be the choice.

However, if we can spot it instantly like Superman, we may be able to avoid peering to the left.

 

[THE_V01DL0RD]

This is the situation.

Please refer to the following cube.

View attachment 43889
(Source: https://ruwix.com/online-puzzle-simulators/)

Cube position is Yellow Top, Blue Front. Cross and the first F2L (DBL & BL) are finished.

Now I want to fill the Red-Green slot (DBR & BR) because BR edge is on the U layer, and it is normally a good idea to fill the back slots before front ones.

Oh, oh, there are DFL and DBR which have a Red face. But which is the Red-Green-White corner piece I am looking for?

Superman can guess it instantly without peeking to the hidden faces, but regrettably I am not him.

Is there any easy method to know which Red-Green-White corner piece is?

In a real solving, peeking the left side of the DFL to check its color and proceed with F2L may be the choice.

However, if we can spot it instantly like Superman, we may be able to avoid peering to the left.

I'm pretty sure superman can just solve it instantly considering he can move faster than the speed of light(?)

 

[bcube]

But which is the Red-Green-White corner piece I am looking for?

No one knows for sure because UB and RB edges, as well as DLF and DRB corners are ambiguous on your picture.

Here is a simulator with all known stickers, you can solve it (try it) only ambiguously.

 

[Tito_H]

Thank you for your explanation.

To prove the logics you taught me, I tried to solve it in the following 2 ways.

Case 1)
DFL is Blue-Red-White
DBR is Red-Green-White

Case 2)
DFL is Red-Green-White
DBR is Blue-Red-White

And, Yes, I could solve the both cases.

So, it is true that it is impossible to define which is the Red-Green-White corner piece under this situation.
It was a Mission Impossible!

But, then, another question has come arise to me.

What if UB and RB edges are NOT ambiguous?

Let’s suppose that I could track them in the solving, and knew exactly this is the following case 1, not 2, or vice versa.

Case 1)
UB is Red-Yellow
BR is Blue-Red

Case 2)
UB is Blue-Red
BR is Red-Yellow

Is this Mission still Impossible?

 

[Jorian Meeuse]

Thank you for your explanation.

To prove the logics you taught me, I tried to solve it in the following 2 ways.

Case 1)
DFL is Blue-Red-White
DBR is Red-Green-White

Case 2)
DFL is Red-Green-White
DBR is Blue-Red-White

And, Yes, I could solve the both cases.

So, it is true that it is impossible to define which is the Red-Green-White corner piece under this situation.
It was a Mission Impossible!

But, then, another question has come arise to me.

What if UB and RB edges are NOT ambiguous?

Let’s suppose that I could track them in the solving, and knew exactly this is the following case 1, not 2, or vice versa.

Case 1)
UB is Red-Yellow
BR is Blue-Red

Case 2)
UB is Blue-Red
BR is Red-Yellow

Is this Mission still Impossible?

No, in this scenario it is not impossible, because assuming either case leaves only the corners in question unidentified. Since a swap of only two corners is impossible, we can determine the corner case based on the edge case. Notice that the edges in case 1 have odd parity, and in case 2 they don't. The corners need to reflect that. Taking DFL to be red-white-blue puts the corners in a no-parity state, so this must belong to case 2. Therefore, in case 1, DFL is the red-white-green. Not worth thinking about during a speed solve.

Notice that this all hinges on the fact that we can identify the UBL corner as orange-white-blue: all yellow corners are visible, and the only other remaining white corner that also contains orange is already solved. If, for example, you take your scramble (either case 1 or 2) and apply a U move, there are now 3 corners unidentified (UBL, DFL, DBR), and all we know is that they all have red. Since a 3-cycle is always possible without changing the edge state, we cannot determine which corner is which, even if we know all edges.

Ultimately, the trick is just to keep track of all (or most) of your F2L pieces when solving, and the best way to get used to this is to do many solves.

PS just solve the orange blue pair here, it's free! If you are in case 1 you even get red-green as a free pair afterwards.

 

[Caleb Valenzuela]

This is the situation.

Please refer to the following cube.

View attachment 43889
(Source: https://ruwix.com/online-puzzle-simulators/)

Cube position is Yellow Top, Blue Front. Cross and the first F2L (DBL & BL) are finished.

Now I want to fill the Red-Green slot (DBR & BR) because BR edge is on the U layer, and it is normally a good idea to fill the back slots before front ones.

Oh, oh, there are DFL and DBR which have a Red face. But which is the Red-Green-White corner piece I am looking for?

Superman can guess it instantly without peeking to the hidden faces, but regrettably I am not him.

Is there any easy method to know which Red-Green-White corner piece is?

In a real solving, peeking the left side of the DFL to check its color and proceed with F2L may be the choice.

However, if we can spot it instantly like Superman, we may be able to avoid peering to the left.

I had a feeling it was the DBR corner and I was right!

 

[bcube]

What if UB and RB edges are NOT ambiguous?

Just for the sake of completeness, here is another way how you can look at it.



Notice that if magenta stickers are either red or white, you can not determine the blank stickers. If any magenta sticker is blue or green, you can determine all blank stickers.

As said by Jorian, if you know any blank sticker of any of those two edge pieces, you can determine all blank stickers.

 

[Tito_H]

Thank you very much, Jorian and bcube for your explanation. I could learn the theory (although I have to understand what is exactly “odd parity” more thoroughly).

 

[Tito_H]

PS just solve the orange blue pair here, it's free! If you are in case 1 you even get red-green as a free pair afterwards.

Eureka, You really has a Superman’s eyes!

 

[Tito_H]

I had a feeling it was the DBR corner and I was right!

Thanks for your try.

 

[Jorian Meeuse]

although I have to understand what is exactly “odd parity” more thoroughly).

Call a swap of two edges a transposition. We can now try to 'solve' the edges using only transpositions. Of course, this cannot be done on a cube using conventional turns, since a 2-swap of edges only is impossible, but that's not the point. Given an arbitrary edge state, there exist many different sequences of transpositions that result in solved edges. However, it can be proven that the number of transpositions in the sequence will either always be even, or always be odd, depending on the original state. This is what we call the parity of the edge state. That is to say, an edge state with odd parity is one that you cannot solve with an even number of transpositions.

The same applies to the corners, and for a rubik's cube there's the additional constraint that the parity of the corner state and that of the edge state match. This is because a quarter turn is an odd permutation of both edges and corners. You can therefore also calculate the parity of a state by counting the number of quarter turns in the scramble sequence. Notice that on a megaminx, a 'fifth' turn is an even permutation of both corners and edges, and therefore you can never reach an odd-parity state of either corners or edges (even + even = even always).

If this interests you, I recommend studying some group theory, particularly permutation groups. You'll see these same concepts appear there.

 

[bcube]

the parity of the corner state and that of the edge state match. This is because a quarter turn is an odd permutation of both edges and corners.

Not necessarily (I know what you are trying to say, however, a reader doesn´t necessarily have to be operating on a fixed centers model by default).

 

[Tito_H]

Thanks for your explanation, Jorian.

bcube had explained to me the theory “the parity of the corner state and that of the edge state match” by my another article.

But I have just learned by you that a quarter turn is an odd permutation of both edges and corners (because they need 3 swaps each to be returned, and 3 is an odd number).

Thanks again.

 

[Tito_H]

Not necessarily (I know what you are trying to say, however, a reader doesn´t necessarily have to be operating on a fixed centers model by default).

Thanks.
I have just read your linked article.
It was intereting.