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A question about the math...

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So there are 43 Quintilian possible configuration of a 3x3 cube.

Does this account for that fact that the center cubes never move and that the corner cubes will never take up edge positions? Is there a guide to the math anywhere?
 
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#2
Yes.

The problem is a reasonably basic combinatorics one (if you want to look up what that is).

The Wikipedia page does a fairly good explanation of it https://en.wikipedia.org/wiki/Rubik's_Cube#Mathematics

Essentially you look at the number of positions for edges (as they are in one orbit) and the number of configurations for corners (as they are in a separate orbit) then multiply (then account for parity as well)

Incidentally I think you mean quintillion.
 
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I found a good pdf that does the job and your wiki entry is good too. None of them address the fact that there are 6 stationary locations on the cube....

Ah crap, nevermind, thinking about it with my fingers I figured it out. If they don't move, they don't get factored into the possible combinations. They are static. So each side only has 8 variable objects, not 9. Duh.

Thanks!
 
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#4
I found a good pdf that does the job and your wiki entry is good too. None of them address the fact that there are 6 stationary locations on the cube....

Ah crap, nevermind, thinking about it with my fingers I figured it out. If they don't move, they don't get factored into the possible combinations. They are static. So each side only has 8 variable objects, not 9. Duh.

Thanks!
No problem :)
 
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#5
So let's say there is a certain scramble with green center on the F face. One in 43 odd quintillion.

Now, if you ONLY do an X' (and it's white that is now facing you), so same scramble just different cube orientation, is that a whole different possibility? Or does it still count as the same one in 43 quintillion as before? Always wondered that.
 
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#6
So let's say there is a certain scramble with green center on the F face. One in 43 odd quintillion.

Now, if you ONLY do an X' (and it's white that is now facing you), so same scramble just different cube orientation, is that a whole different possibility? Or does it still count as the same one in 43 quintillion as before? Always wondered that.
It's regarded as the same. If it wasn't, there would be 24 times as many combinations.
 
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It's regarded as the same. If it wasn't, there would be 24 times as many combinations.
It isn't counted as the same. Although you are right that permutation appears in the 43 quintillion figure, the count does not take the rotations of the cube into account.

Basically if you do x alg x', the permutation would appear in the 43 quintillion but so would alg, y alg y' etc.
 
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