Formula for Calculating Positions of a NxNxN Cube?

[cmhardw]

Well, at least he didn't explain it, so I'll have to try to understand 6!/(2*3)
If I don't get it, could someone explain it?

Hint: scramble in [U,R] then correctly permute the DFR and DBR corners. Look at the permutation of all the U layer corners. Do it again, scramble in [U, R] then permute DFR and DBR then look at the permutation of the U layer corners. Do it again. Notice a pattern? If not, do it again - and again. Notice a pattern?

 

[Stefan]

I like Chris's hint
(And I wasn't all that serious about Oscar, although I do like people trying stuff on their own and try to support that.)

The /2 is because permutation parity, right?

So you covered permutation parity twice? Or why did you /2 for edge permutation as well?

 

[Carrot]

I like Chris's hint
(And I wasn't all that serious about Oscar, although I do like people trying stuff on their own and try to support that.)

(Hence why I didn't explain why it should be that He still needs some reasoning of his own to see why it holds)

 

[G2013]

Hint: scramble in [U,R] then correctly permute the DFR and DBR corners. Look at the permutation of all the U layer corners. Do it again, scramble in [U, R] then permute DFR and DBR then look at the permutation of the U layer corners. Do it again. Notice a pattern? If not, do it again - and again. Notice a pattern?

I scramble with [U, R] then solve the corners you indicated. All the top corners are permuted correctly.
I do it again, now every corner is it its place but with a U2
Again, them are solved with a U' turn applied.
I continue with this (the only thing I see) and they go U, U2, U2, U2, U, U2... I stopped here

I don't see any other "pattern", although the one I saw is not a pattern :S
I see that all corners solve after solving 2 other corners.
Nothing more ...

So you covered permutation parity twice? Or why did you /2 for edge permutation as well?

I started with this thing of combinations just yesterday.
My logic was: A pair of edges can't be switched unless 2 corners are.
With R and U, you can never switch 2 corners.
So, as far as I know, when that happens you have to divide by 2, but I don't know the reason. I divided both corners and edges.
It must be wrong, because if it wasn't you wont have posted that.

But I observed that if 2 edges are switched, the corners are with a U or U' move applied... I didn't see more than that

 

[cmhardw]

I scramble with [U, R] then solve the corners you indicated. All the top corners are permuted correctly.
I do it again, now every corner is it its place but with a U2
Again, them are solved with a U' turn applied.
I continue with this (the only thing I see) and they go U, U2, U2, U2, U, U2... I stopped here

I don't see any other "pattern", although the one I saw is not a pattern :S
I see that all corners solve after solving 2 other corners.
Nothing more ...

When you are permuting the corners, you have 6 possible places for where to place the DFR corner. After that you have 5 possible places for where to place the DBR corner. It's certainly possible that while doing that you end up correctly permuting both corners. When both DFR and DBR end up correctly permuted, how many different possible ways are there to permute the remaining U layer corners, based on your experiment?

I started with this thing of combinations just yesterday.
My logic was: A pair of edges can't be switched unless 2 corners are.
With R and U, you can never switch 2 corners.
So, as far as I know, when that happens you have to divide by 2, but I don't know the reason. I divided both corners and edges.
It must be wrong, because if it wasn't you wont have posted that.

But I observed that if 2 edges are switched, the corners are with a U or U' move applied... I didn't see more than that

That's a great place to start

What is the minimal effect on corners on regular 3x3x3 when you swap just two edges? How is this different from [R,U] only?

Start by answering the question:
Ignoring all information about corners, is it possible to permute all 7 edges on the [R,U] bandaged 3x3x3 to any of the, theoretical, 7! permutations?
-If yes, then what happens when trying to permute corners if the edges have even permutation parity? What happens when trying to permute corners if edges have odd permutation parity?
-If no, then must the edges always have either even or odd permutation parity? Whichever parity the edge permutation has, how does this affect corners when you try to permute them?

 

[G2013]

That's a great place to start

What is the minimal effect on corners on regular 3x3x3 when you swap just two edges? How is this different from [R,U] only?

Start by answering the question:
Ignoring all information about corners, is it possible to permute all 7 edges on the [R,U] bandaged 3x3x3 to any of the, theoretical, 7! permutations? Yes, it is possible
-If yes, then what happens when trying to permute corners if the edges have even permutation parity? 4 of them end in a 4-cycle (1 goes to 2, 2 to 3, 3 to 4 and 4 to 1)
What happens when trying to permute corners if edges have odd permutation parity? They can be sovled
But the concept of "odd permutation parity" and "even permutation parity" is not clear to me

..

 

[cmhardw]

G2013,

Definitely read Ryan's Cube Laws page as a start.

The basic idea is that the permutation parity of the corners must match the permutation parity of the edges. Put more simply: if you swap two corners on a solved cube, then you must also swap two edges is the idea. The idea extends beyond just "swap two corners" and "swap two edges".

A 4-cycle of corners is an odd permutation. So for example, you said in [U,R] that if you swap two edges you must 4-cycle the corners as if you had done U or U'. This means that the permutation parity of the corners is odd, and also the permutation parity of the edges is odd.

Think of permutation parity as the number of two swaps necessary to solve the cube, where swapping the locations of two pieces is a two-swap. The number of two-swaps necessary to solve your cube will either be an even or an odd number. If it's even, you have "even permutation parity" and if it's odd you have "odd permutation parity".

Combine this info with the previous posts and tell us where you're at. We are trying to help you by the way, but realize that if you figure out a concept on your own with only some help, it will stick forever. If you are told a concept, you may forget it in a week.

 

[G2013]

Well...

2 edges are switched = 1 switch
4 corners are wrong permutated, in a 4-cycle = 3 switches

3*1=3
Maybe that is the /3?

Or it is 3+1=4, even parity

?

I'm too dumb for this things

 

[cmhardw]

Here are two methods:

Method 1:

Spoiler

When permuting the corners, they can have 6*5*4 possible permutations. The first corner can be placed into one of six possible locations, the next corner in 5 locations. All 4 of the remaining corners now only have 4 possible permutations. Think of your experiment with solving DFR and DBR. The U layer only ever had 4 possible permutations of the corners.

Now that corners are permuted they can be flipped in 3^5 ways.

Lastly we permute the edges. They have 7!/2 permutations. The reason it's not 7! is because of the Cube Law that the overall corner permutation parity must match the overall edge permutation parity. This is a law for the 3x3x3, so any subset of the 3x3x3 that does not include cube rotations (x, y, z) turns inherits this law as well.

Since the corners have a permutation parity (either even or odd) then half of the 6! edge permutations are impossible, because they have the opposite permutation parity.

There is no concept of edge flip, so we're done with edges.

The overall count is (6*5*4)*3^5*7!/2

Method 2:

Spoiler

Let's permute the edges into their 7! possible permutations. There is no concept of edge flip, so now we're done with edges.

When permuting the corners there are 6*5*4 ways to place the corners. This is because after permuting 2 corners, the last 4 corners can only have 4 possible permutations. Remember your experiment with solving DFR and DBR. However, since the edges are already permuted, and now have a permutation parity (either even or odd) then half the corner permutations are not possible because they have the opposite permutation parity of the edges. Think cube laws again (see my Method 1 spoiler for more explanation of this).

So the corners only have (6*5*4)/2 or 6*5*2 permutations. The corners have 3^5 orientations possible.

Putting it all together we get:
7!*(6*5*4)/2*3^5

 

[G2013]

Ahhh ok, I see.
But what I don't understand is (in the second method): why do you say "Let's permute the edges into their 6! possible permutations.", if the first one can go into 7, the second into 6, etc. resulting 7! ?

I understood the /3 as: Instead of 6*5*4*3*2*1 for corners you use 6*5*4, that is 6!/2*3. So the 3 is actually there to match the number.

----------------------------------------------------------------------

And if I wanted [M,U], I can think of:

Edges: 6!
Corners: 4/2, actually 2, and *3^5
Centers: Have 4 positions but 2 are not possible in some cases, so 2.
Resulting: 6!*2*3^5*2=699840

Am I right?

 

[cmhardw]

Ahhh ok, I see.
But what I don't understand is (in the second method): why do you say "Let's permute the edges into their 6! possible permutations.", if the first one can go into 7, the second into 6, etc. resulting 7! ?

I understood the /3 as: Instead of 6*5*4*3*2*1 for corners you use 6*5*4, that is 6!/2*3. So the 3 is actually there to match the number.

----------------------------------------------------------------------

And if I wanted [M,U], I can think of:

Edges: 6!
Corners: 4/2, actually 2, and *3^5
Centers: Have 4 positions but 2 are not possible in some cases, so 2.
Resulting: 6!*2*3^5*2=699840

Am I right?

You're right, I often make mistakes when posting.

There are 7 edges so the calculations should be:

(6*5*4)*7!/2*3^5

or

7!*(6*5*4)/2*3^5

 

[cmhardw]

And if I wanted [M,U], I can think of:

Edges: 6!
Corners: 4/2, actually 2, and *3^5
Centers: Have 4 positions but 2 are not possible in some cases, so 2.
Resulting: 6!*2*3^5*2=699840

Am I right?

Corners cannot be mis-oriented in [M,U]. Otherwise it looks good.

 

[Jakube]

And if I wanted [M,U], I can think of:

Edges: 6!
Corners: 4/2, actually 2, and *3^5
Centers: Have 4 positions but 2 are not possible in some cases, so 2. Once you fixed the edges and the corners, there are only 2 center cases.

The correct solution should be: 6!*2^5*4*2 = 184320
Resulting: 6!*2*3^5*2=699840

Am I right?

First of the corner orientation is always solved. The edge orientation isn't. So it should be 2^5 instead of 3^5.
And you canceled a 2 too often.

If you have any of the 6! edge permutations, each corner permutation (out of the 4) is possible. Once you fixed the edges and corners, there are only 2 center cases.

Therefore 6!*2^5*4*2 = 184320

 

[tseitsei]

And if I wanted [M,U], I can think of:

Edges: 6!
Corners: 4/2, actually 2, and *3^5
Centers: Have 4 positions but 2 are not possible in some cases, so 2.
Resulting: 6!*2*3^5*2=699840

Am I right?

I think you forgot edge orientation... I'm not an expert tough...


so I would say you still have to multiply that by 2^5

Also there is no corner orientation here so delete that.

Also I think you divided by 2 too many times. One should be enough (again not sure tough). I think you must divide only EITHER for corners OR centers... Because doing M changes parity (4cycle of centers and 4 cycle of centers) and also doing U changes parity (4 cycle of corners and 4 cycle of edges). So doing both M and U "changes" parity 2 times so it doesn't really change parity...

EDIT: Ninja'd. But looks like I was correct

 

[Randomno]

What about for cuboids other than 1x1x2/1x1x3/1x1x4/etc.?

 

[G2013]

First of the corner orientation is always solved. The edge orientation isn't. So it should be 2^5 instead of 3^5.
And you canceled a 2 too often.

If you have any of the 6! edge permutations, each corner permutation (out of the 4) is possible. Once you fixed the edges and corners, there are only 2 center cases.

Therefore 6!*2^5*4*2 = 184320

You are telling me that if I have any of the 6! permutations, each corner permutation is possible. Alright, then:
If I have 2 edges switched, then all corners can be solved.
But that is not real
So there must be a mistake in what you had said (written)...

 

[cmhardw]

You are telling me that if I have any of the 6! permutations, each corner permutation is possible. Alright, then:
If I have 2 edges switched, then all corners can be solved.
But that is not real
So there must be a mistake in what you had said (written)...

M' U2 M U2 M'

This leaves 2 edges swapped and all 4 corners solved.

 

[G2013]

But centers in a third position, if we count the 2 "possible" ones

 

[Jakube]

You can see it this way: there are 4 different cases to consider:

1) No parity at all:

• Edge Permutation: 6! / 2
• Corner Permutation: 2 (solved and U2)
• Center Permutation: 2 (solved and M2)

2) Edge-Corner parity:

• Edge Permutation: 6! / 2
• Corner Permutation: 2 (U and U')
• Center Permutation: 2 (solved and M2)

3) Edge-Center parity:

• Edge Permutation: 6! / 2
• Corner Permutation: 2 (solved and U2)
• Center Permutation: 2 (M and M')

4) Corner-Center parity:

• Edge Permutation: 6! / 2
• Corner Permutation: 2 (U and U')
• Center Permutation: 2 (M and M')

I ignored edge orientation in all cases, it's always 2^5.

Each of these 4 cases have the same amount of combinations, therefore there are 2^5*4*(6!/2*2*2) combinations


edit:

Or you can just say: there are 4*4*6!*2^6 possibilities, divide /2 because of edge orientation and /2 because of of parity.

 

[cmhardw]

There are 4 possible states considering the permutation parity of center, corners, and edges.

1) All orbits have even parity
2) Centers have even parity, corners and edges have odd parity
3) Corners have even parity, centers and edges have odd parity
4) Edges have even parity, corners and centers have odd parity

If 2 edges are swapped and the corners are solved then edges have odd parity while corners have even parity. This means that edges and centers have odd parity and is case number 3.

I would count all possible [M, U] cases as:

(6!*2^5)*4*2

(6!*2^5) is me permuting edges first, then flipping them.

4 is permuting the corners into any of their 4 states.

2 is because I have only 2 choices of center state. Knowing the parity of edges and corners will uniquely determine the parity of the centers. No matter what, there are 2 center states that have the correct parity.

I may have missed something, but I'm pretty sure this is sound.

--edit--

Ninja'd by Jakube