Group structure of the Rubik's Revenge?

[mande]

Hi,
I read up on Wikipedia the article describing the group structure of the Rubik's cube, and I understood it. (I hope). What I am confused about is what the group structure of the 4x4 might be. Because I thought from the solved state, the centers being interchangeable keeping the cube solved might cause complications. Can anyone help me out?
Thanks.

 

[jcuber]

Yes, there are things called parity (can happen on any even-layered cube) which cause either one edge to flip (orientation parity) or two edges to swap places (permutation parity). Niether of these can happen on an odd-layered cube, and there are algorithms to correct these.

 

[TheBB]

Hi,
I read up on Wikipedia the article describing the group structure of the Rubik's cube, and I understood it. (I hope). What I am confused about is what the group structure of the 4x4 might be. Because I thought from the solved state, the centers being interchangeable keeping the cube solved might cause complications. Can anyone help me out?
Thanks.

Group theory isn't my strong point really, but can't you just quotient out the different center piece permutations?

 

[mande]

Group theory isn't my strong point really, but can't you just quotient out the different center piece permutations?

Umm...I'm sure we could, but could you tell me the group of the different center piece permutations?
Will it be (A_4)^6?

 

[TheBB]

I guess so. If that's the one with only even permutations. Can't remember.

 

[TomZ]

The 24 4x4 centers can be arranged in 24! different ways. However, as there are six pairs of four indistinguishable centers, the total figure is only 24!/(4^6).

 

[mande]

I got that the group for the edges is A_24, for centers (A_24)/((A_4)^6), for corners A_8 X (Z_3)^7
(these accounting only for even permutations). Would that be correct?
My main question was how to combine these 3 subgroups using some type of product of groups to get a group structure on the 4x4.

 

[EmersonHerrmann]

Yes, there are things called parity (can happen on any even-layered cube)

What about 2x2?

 

[TomZ]

Yes, there are things called parity (can happen on any even-layered cube)

What about 2x2?

The 2x2 has parity. And so does the 3x3 for that matter. In fact, any cube can have parity.

 

[cmhardw]

Yes, there are things called parity (can happen on any even-layered cube) which cause either one edge to flip (orientation parity) or two edges to swap places (permutation parity). Niether of these can happen on an odd-layered cube, and there are algorithms to correct these.

At the risk of being the nit-picky math guy, this statement is wrong on many levels.

Yes, there are things called parity (can happen on any even-layered cube) which cause either one edge to flip (orientation parity)...

It is not possible to "flip" any edge on an even cube. If you replace that with edge group or even better swap two edges in the same orbit then I might agree with you.

... or two edges to swap places (permutation parity).

Swapping the place of two edges in the same orbit is actually "orientation parity". I think you're thinking about edge groups, or if you want to generalize this two pairs of edges within the same orbit.

Niether of these can happen on an odd-layered cube, and there are algorithms to correct these.

First off you have to use more general definitions for the n x n x n case, and second of all using the more general definitions orientation parity in terms of speed solving would typically be defined as "interchanging two edge pieces in the same orbit" and permutation parity would be defined as "interchanging two pairs of edge pieces in the same orbit, and this may span multiple if not all edge orbits". Using these more general definitions both of these cases are absolutely possible on an odd layered n x n x n cube provided n > 3

Chris

P.S. I'm not trying to be mean or elitist, but your post is spreading untruths, and I felt the need to correct it.

 

[Lord Voldemort]

I think Parity (on a cube) is basically anything scarmbled with an even number of turns and solved with an odd number or vice versa. That's why they call it parity on BLD after you solve edges.

 

[trying-to-speedcube...]

Yes, there are things called parity (can happen on any even-layered cube) which cause either one edge to flip (orientation parity) or two edges to swap places (permutation parity). Niether of these can happen on an odd-layered cube, and there are algorithms to correct these.

Try popping an edge out of your 4x4 and put in back flipped...

 

[qqwref]

The positions in a 4x4 actually do not form a group!

Why not? The basic reason is because of the interchangeable centers. This actually leads to us not being able to decide what the product of any two positions is - the operation itself is ambiguous. In the 3x3 we can just apply the first position and then apply the second to that, but on the 4x4 there are multiple positions that look the same so we can get many different results from this seemingly simple calculation.

Let me explain with an example. Here are two 4x4 patterns that look the same:
r U' l' U r' U' l U
F U' l' U r U' l U r' F'
Call this position "A", and call the position achieved from a simple d move position "B". Now let's *try* to calculate what the product BA is. The surprising result is that, depending on which of the first two algorithms we use, we get different patterns! They both give the same position A, but when we try to tack on a d move at the beginning we actually get different positions. So we actually cannot multiply B and A!

Remember that on 3x3 we always have a unique way to combine two permutations, since every piece is distinct. No matter what sequence of moves we use to set up A and B, we will always get the same result for AB or BA, so we have a valid group structure. But the 4x4 does not give us a valid single-valued way to multiply two positions, and so we can't make a group out of it that would correspond to the way an actual 4x4 works.

 

[JBCM627]

The positions in a 4x4 actually do not form a group!

Why not? The basic reason is because of the interchangeable centers. This actually leads to us not being able to decide what the product of any two positions is - the operation itself is ambiguous. In the 3x3 we can just apply the first position and then apply the second to that, but on the 4x4 there are multiple positions that look the same so we can get many different results from this seemingly simple calculation.

Let me explain with an example. Here are two 4x4 patterns that look the same:
r U' l' U r' U' l U
F U' l' U r U' l U r' F'
Call this position "A", and call the position achieved from a simple d move position "B". Now let's *try* to calculate what the product BA is. The surprising result is that, depending on which of the first two algorithms we use, we get different patterns! They both give the same position A, but when we try to tack on a d move at the beginning we actually get different positions. So we actually cannot multiply B and A!

I don't follow your logic here. This example only illustrates to me that A and B do not commute.

As best I can tell, the 4x4 satisfies all the requirements listed here.

 

[Ton]

The positions in a 4x4 actually do not form a group!

Why not? The basic reason is because of the interchangeable centers. This actually leads to us not being able to decide what the product of any two positions is - the operation itself is ambiguous. In the 3x3 we can just apply the first position and then apply the second to that, but on the 4x4 there are multiple positions that look the same so we can get many different results from this seemingly simple calculation.

Let me explain with an example. Here are two 4x4 patterns that look the same:
r U' l' U r' U' l U
F U' l' U r U' l U r' F'
Call this position "A", and call the position achieved from a simple d move position "B". Now let's *try* to calculate what the product BA is. The surprising result is that, depending on which of the first two algorithms we use, we get different patterns! They both give the same position A, but when we try to tack on a d move at the beginning we actually get different positions. So we actually cannot multiply B and A!

Remember that on 3x3 we always have a unique way to combine two permutations, since every piece is distinct. No matter what sequence of moves we use to set up A and B, we will always get the same result for AB or BA, so we have a valid group structure. But the 4x4 does not give us a valid single-valued way to multiply two positions, and so we can't make a group out of it that would correspond to the way an actual 4x4 works.

In the 3x3 group description we discard the center, -there are more solved states if you would mark the center- For the 4x4 we could mark the centers as Ra Rb Rc Rd etc

 

[Lucas Garron]

The positions in a 4x4 actually do not form a group!

Why not? The basic reason is because of the interchangeable centers. This actually leads to us not being able to decide what the product of any two positions is - the operation itself is ambiguous. In the 3x3 we can just apply the first position and then apply the second to that, but on the 4x4 there are multiple positions that look the same so we can get many different results from this seemingly simple calculation.

Let me explain with an example. Here are two 4x4 patterns that look the same:
r U' l' U r' U' l U
F U' l' U r U' l U r' F'
Call this position "A", and call the position achieved from a simple d move position "B". Now let's *try* to calculate what the product BA is. The surprising result is that, depending on which of the first two algorithms we use, we get different patterns! They both give the same position A, but when we try to tack on a d move at the beginning we actually get different positions. So we actually cannot multiply B and A!

Remember that on 3x3 we always have a unique way to combine two permutations, since every piece is distinct. No matter what sequence of moves we use to set up A and B, we will always get the same result for AB or BA, so we have a valid group structure. But the 4x4 does not give us a valid single-valued way to multiply two positions, and so we can't make a group out of it that would correspond to the way an actual 4x4 works.

In the 3x3 group description we discard the center, -there are more solved states if you would mark the center- For the 4x4 we could mark the centers as Ra Rb Rc Rd etc

Well, yes, the 4x4x4 supercube has a group. But that's a different puzzle!

JCBM: Associativity fails, because, as qq showed, composition is not consistently definable.
Even better, consider
A = [U r U', l']
B = [l, F r' F']

A B = e, but B A ≠e
Breaks inverses.

 

[mande]

The positions in a 4x4 actually do not form a group!

Why not? The basic reason is because of the interchangeable centers. This actually leads to us not being able to decide what the product of any two positions is - the operation itself is ambiguous.

Doesn't quotienting the whole "group" by the group of permutations of the center pieces solve that problem?

 

[Lucas Garron]

The positions in a 4x4 actually do not form a group!

Why not? The basic reason is because of the interchangeable centers. This actually leads to us not being able to decide what the product of any two positions is - the operation itself is ambiguous.

Doesn't quotienting the whole "group" by the group of permutations of the center pieces solve that problem?

If by "quotienting" you mean "ignoring," yes. But that's not the same puzzle...

 

[qqwref]

As I showed, there is no way to construct a group that will really represent the 4x4. Any construction that seems to - like yours - will construct a group that probably won't represent anything like the 4x4 if you look more closely.

 

[TheBB]

I guess you mean to say (A_4)^6 isn't normal in the 4x4 supergroup.