Chance of dissassembling a cube to solve it

[Feryll]

As per the thread title, I saw the thread here and wondered "What about the chance of taking pieces out and randomly placing them in spots? I am in no way sure of if my work is correct, but whatever.

Spoiler

This is first assuming that you can take the middle pieces out. There are 6! different combinations of how to put on the centers from one angle. But many are the same but just from another angle. This reduces it by a factor of 2(6), so the number drops to 60.
There are 8! different places to put the corners on, and that is multiplied by the flip coefficient (or whatever) of 3 different flips and 8 places the flips occur, so 3^8. That gives us 8! x 3^8. But as before, there are 24 clones per the actual number, so it becomes (8! x 3^8)/24, or 7! x 3^7.
There are 12! different places to place the edges, with 2 flips, so it becomes 12! x 2^12. But once again, there are clones in that calculation. There are 24different places to use as your point of view when looking at a cube (Don't say it's only six, because you can rotate any face 4 times), so it's reduced by a factor of 24. So the number becomes (12! x 2 ^ 12)/24, or 11! x 2^11.
Finally, it's time to multiply 60 (centers) by 11,022,480 (corners) by 81,749,606,400 (edges). If you divide it by the 4.3 quintillion possibilities, this yields a "1 in 9.6 chance" or 10.4%.

But the thing is, I thought it was a 1 in twelve chance. If anybody could find an error in my proof (using my method, not the orientated and permeated simple one), I'd be grateful And you get the opportunity to facepalm me!

 

[joey]

Well, it's cos you also counted centres, which "don't come off".

 

[Cyrus C.]

I got:

Spoiler

1/2 chance of EO
1/3 Chance of CO
1/2 Chance of EP
1/1 Chance sf CP
1/12 Chance of solvable cube.

I don't think there's any error in your proof, You just counted centers, & usually people don't take then off when dissambling the cube.

 

[Micael]

I got:

Spoiler

1/2 chance of EO
1/3 Chance of CO
1/2 Chance of EP
1/1 Chance sf CP
1/12 Chance of solvable cube.

I don't think there's any error in your proof, You just counted centers, & usually people don't take then off when dissambling the cube.

I think this is correct. This is 8.3% (1/12). So I do not believe 10.4% is correct because it should reduce the chance if you also remove center caps.

 

[Sa967St]

Spoiler

An interesting fact about Rubik's Cube is that if you take it apart and put it together randomly, there will be only a 1 in 12 chance of it actually being solvable by legal moves (that is, without taking the cube apart again).
...
Combining these laws, only 1/12 (1/2 * 1/2 * 1/3) of the conceivable cube states are reachable by legal moves.

^that

 

[Cyrus C.]

Let me try with centers:

Spoiler

1/12 EO CO EP
1/64 CP
1/768 Total

I feel like that number is way too low, did I do something wrong?

 

[Thomas09]

The answer is 42.

 

[Feryll]

Let me try with centers:

Spoiler

1/12 EO CO EP
1/64 CP
1/768 Total

I feel like that number is way too low, did I do something wrong?

Actually, its 1/60 for centers. The answer probably is 1/12 x 1/60, or 1/720. I just wonder where I went wrong...
Actually, I counted centers because in a truly mathematical problem, the model would probably have removable centers. And if you're 4x4x4 fell apart completely (like my storeRIPOFFbought), it would be useful info. I don't know anything about 4x4 BLD, so I don't know how to simply go about the 4x4's chance of dissassembling.

 

[Logan]

The answer is 42.

yep.

 

[joey]

Chances of a solvable 4x4 = chances that CO is correct.

 

[Stefan]

Actually, its 1/60 for centers.

Try again.

 

[cmhardw]

Spoiler

If the centers fell off and were put back on the cube randomly, you could place them onto the cube in 6! different ways. There are 6 choices for the U center, 5 for the F center, etc.. Now, there is only one coloring scheme that will be solvable with the corners and edges, and it has 24 different rotations considering cube rotations. So that's a:
24/6! = 1/(5*6) = 1/30 probability of putting the centers on the cube in an orientation that will solve with the corners and edges.

We must now place corners on the cube, considering the fact that the centers are in a randomized orientation (but solvable if we place corners and edges in an also solvable positions). This effectively changes the cube to allow 2 different kinds of manipulations:

1) a quarter turn of an outer layer
2) a quarter turn cube rotation

All other manipulations to the cube can be realized as some combination of these two generators. We must know how these manipulations effect the parity of the center, corner, and edge orbits.

A quarter turn of an outer layer toggles the parity of the corners and edges, but not centers. A quarter turn cube rotation toggles the parity of centers and edges, but not corners.

This allows for 4 different parity states by manipulating the cube starting from the solved state:
1) Centers, corners, edges all 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) Corners, centers have odd parity, edges have even parity.

Now back to permuting corners onto our cube with a randomized, but solvable with the corners and edges, center coloring scheme. At this stage the centers either have even or odd parity. Notice that when centers are even corners can be even or odd. Also when centers are odd, corners can be either even or odd. So we can permute the corners into all 8! possible arrangements at this stage without breaking any laws of a solvable cube. The corners now have either even or odd parity as well. This leaves 4 possibilities for center and corner parities considered together. However, based on the parity schemes above, once the parity of the centers and corners are known the edge parity is fixed.

So the edges now have only 12!/2 possible arrangements after permuting centers and corners.

This means that the edges have a 1/2 chance of being permuted onto the cube "correctly" to match the parity they should have considering the parity of corners and centers.

With the corners you can rotate the first 7 independently, and the 8th has a 1/3 chance of having a solvable orientation with regards to the other 7. For the edges the first 11 can be twisted independently and the 12th has a 1/2 chance of having a solvable twist with regards to the other 11.

This means the cube is solvable with a:
1/30 * 1/2 * 1/3 * 1/2 = 1/360 probability assuming it was assembled completely randomly from a naked core.

On a side note:

... so I don't know how to simply go about the 4x4's chance of dissassembling.

I'll give you a hint, it's WAY more likely to assemble a solvable 4x4x4 than a solvable 3x3x3. Try to use the same strategy I used for the 3x3x3 in my spoiler to figure it out.

Chris

 

[TMOY]

And you can try the Square-1 after that

Spoiler

For Square-1, the chance is 1/1. You can disassemble and reassemble it however you want, you will always get a solvable state.

 

[Feryll]

... so I don't know how to simply go about the 4x4's chance of dissassembling.

I'll give you a hint, it's WAY more likely to assemble a solvable 4x4x4 than a solvable 3x3x3. Try to use the same strategy I used for the 3x3x3 in my spoiler to figure it out.

Chris

Thanks, Chris. Maybe now I can understand these 3 dimensional combinations better. I'll probably work on it during school tomarrow.

Actually, its 1/60 for centers.

Try again.

:fp Wow, did I really miss that? Something stupid went wrong in my 2D model I used. Well, I tried to figure out this one first, and so I made mistakes.

And you can try the Square-1 after that

Spoiler

For Square-1, the chance is 1/1. You can disassemble and reassemble it however you want, you will always get a solvable state.

Interesting. That will be useful when mine falls apart

 

[nigtv]

And you can try the Square-1 after that

Spoiler

For Square-1, the chance is 1/1. You can disassemble and reassemble it however you want, you will always get a solvable state.

Not doubting you, but proof?

 

[Lucas Garron]

And you can try the Square-1 after that

Spoiler

For Square-1, the chance is 1/1. You can disassemble and reassemble it however you want, you will always get a solvable state.

Not doubting you, but proof?

If you enumerate all possible ways of arranging the pieces on a Square-1, they shapes turn out to form a connected graph, so you can always get to cube shape no matter how you assemble it. Once you have it in cube shape, there are well-known methods to solve any permutation, including parity.

 

[nigtv]

you can always get to cube shape no matter how you assemble it.

Hmm I probably should have thought of that.