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

CuBeOrDiE

Member
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

ZamHalen

Member
Because I don't think he asked clearly enough for some of the more brutal posters.(I don't understand the question.)
(And for some reason people don't treat him kindly on this forum.Even though he gave me some pretty good advice while trying to progress last summer.)

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4Chan

Because I don't think he asked clearly enough for some of the more brutal posters.(I don't understand the question.)
>I don't understand the question.

Logan

Alot of Azure
Because I don't think he asked clearly enough for some of the more brutal posters.(I don't understand the question.)
I understand it perfectly. He wants to know the "number of solved positions for a 7x7".

CuBeOrDiE

Member
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?

miniGOINGS

Member
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

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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CuBeOrDiE

Member
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...

MichaelP.

Member
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!

CuBeOrDiE

Member
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...