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Number of "Solved" Positions for a 7x7!?

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As we all know, the centers of the 7x7 have one sticker. So I think that it's logical to conclude some may be swapped and the cube still remain solved. Is this correct? And if so, how many total "solved" positions are there?

thnx
 
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What do you mean? You mean I worded it badly? I mean that, say for instance you took two of the corner centers on the blue side of a 7x7 and swapped them. The cube is still solved, but the pieces have been swapped, so this is two "solved" positions. I way "solved" because the cube is solved, but the two pieces are in different spots. So the question is, how many of these "solved" positions are there in total?
 
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#13
Hmm. Interesting question! I have no idea how I might even go about calculating this... Congrats, you stumped me! I need to start learning about bigcube theory.
 

cmhardw

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#14
As we all know, the centers of the 7x7 have one sticker. So I think that it's logical to conclude some may be swapped and the cube still remain solved. Is this correct? And if so, how many total "solved" positions are there?

thnx
I did a calculation of this for the 20x20x20 cube here.

For the 7x7x7 there would be:
(4!)^36 / 2^6 = (4!)^30 * 2^12 * 3^6
That's approximately 7 * 10^47 "solved" states where you could move around centers of the same color and still appear solved.

And for the n x n x n cube it would be:
(4!)^[6*floor((n-2)^2 / 4)] / 2^floor[(n-2)^2 / 4]

or

2^[17*floor((n-2)^2 / 4)] * 3^[6*floor((n-2)^2 / 4)]

"solved" states.

Here floor(x) means the greatest integer less than or equal to x.

Chris
 
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Personally, I got unreasonably huge numbers. Here's my logic, explained with a 5x5 because its easier to explain:

Consider 1 side

On this side, there are 4 outer corner centers, which can be placed 4! ways, 4 inner corner centers, which can also be placed 4! ways, as well as inner and outer center "edge" pieces, if you know what I mean. They both can, too, be calculeted with 4!.

So 1 side has 4! x 4! x 4! x 4! combinations. Mulsiply that by 6 sides and you got it!

PS- I don't think edges can be interchanged. If they were, the stickers would be swapped. Unless you have an even layered cube... hmm...
 
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#17
Personally, I got unreasonably huge numbers. Here's my logic, explained with a 5x5 because its easier to explain:

Consider 1 side

On this side, there are 4 outer corner centers, which can be placed 4! ways, 4 inner corner centers, which can also be placed 4! ways, as well as inner and outer center "edge" pieces, if you know what I mean. They both can, too, be calculeted with 4!.

So 1 side has 4! x 4! x 4! x 4! combinations. Mulsiply that by 6 sides and you got it!

PS- I don't think edges can be interchanged. If they were, the stickers would be swapped. Unless you have an even layered cube... hmm...
S and T aren't even close on the keyboard!
 
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Thread starter #19
As we all know, the centers of the 7x7 have one sticker. So I think that it's logical to conclude some may be swapped and the cube still remain solved. Is this correct? And if so, how many total "solved" positions are there?

thnx
I did a calculation of this for the 20x20x20 cube here.

For the 7x7x7 there would be:
(4!)^36 / 2^6 = (4!)^30 * 2^12 * 3^6
That's approximately 7 * 10^47 "solved" states where you could move around centers of the same color and still appear solved.

And for the n x n x n cube it would be:
(4!)^[6*floor((n-2)^2 / 4)] / 2^floor[(n-2)^2 / 4]

or

2^[17*floor((n-2)^2 / 4)] * 3^[5*floor((n-2)^2 / 4)]

"solved" states.

Here floor(x) means the greatest integer less than or equal to x.

Chris
Hmm. Interesting. I got 80263249920. So I guess that wasn't insanely unreasonable...
 
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