• Welcome to the Speedsolving.com, home of the web's largest puzzle community!
    You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.

    Registration is fast, simple and absolutely free so please, join our community of 40,000+ people from around the world today!

    If you are already a member, simply login to hide this message and begin participating in the community!

Calculating permutations for a 5x5 (almost)supercube

sgtjosh

Member
Joined
Jul 20, 2013
Messages
22
Location
Greenfield, MA, USA
WCA
2012CARN01
YouTube
Visit Channel
Hey guys,

I have an interesting problem I was hoping the greater minds here could help me with.

I'm wondering how I'd go about calculating the number of possible permutations for a 5x5 picture cube... Except this 'super cube' only has pictures on five of its six sides (The sixth side is just a solid color).

So the center pieces of five sides have to go into the same place every time, but the center pieces of the sixth side do not.

Any help would be greatly appreciated... And the more detail, the better!

Thanks a lot,
-Josh
 

Jakube

Member
Joined
Feb 3, 2011
Messages
790
Location
Austria
WCA
2011KOGL01
YouTube
Visit Channel
We get:

8! - possible corner permutation
3^7 - orientation of corners (notice 7 instead of 8, because the last corner orientation is forced by the others)
12! / 2 - midge-permutations, (divide by 2, because the last two midges are forced. It isn't possible too switch two midges alone)
2^11 - midge-orientations (last orientation is forced by the other 11)

=> 8! * 3^7 * 12! * 2^11 / 2

24! - 24 wings, every permutation is possible (it is possible to switch 2 wings without any other pieces, thanks too the face with only 1 color)

We have 2 groups of center pieces:
Group 1: x-centers
Every permutation can be described with only the picture pices, the other 4 will be forced.
=> 24! / 4!
Same with the t-centers: 24! / 4!

=> There are 8! * 3^7 * 12! * 2^11 / 2 * 24! * 24! / 4! * 24! / 4! = 17934974117270551244186247903990858235542284305932202306560963920617537536000000000000000 = 1.79 × 10^88
 
Last edited:

cmhardw

Premium Member
Joined
Apr 5, 2006
Messages
4,115
Location
Orlando, Florida
WCA
2003HARD01
YouTube
Visit Channel
We get:

8! - possible corner permutation
3^7 - orientation of corners (notice 7 instead of 8, because the last corner orientation is forced by the others)
12! / 2 - midge-permutations, (divide by 2, because the last two midges are forced. It isn't possible too switch two midges alone)
2^11 - midge-orientations (last orientation is forced by the other 11)

=> 8! * 3^7 * 12! * 2^11 / 2

24! - 24 wings, every permutation is possible (it is possible to switch 2 wings without any other pieces, thanks too the face with only 1 color)

We have 2 groups of center pieces:
Group 1: x-centers
Every permutation can be described with only the picture pices, the other 4 will be forced.
=> 24! / 4!
Same with the t-centers: 24! / 4!

=> There are 8! * 3^7 * 12! * 2^11 / 2 * 24! * 24! / 4! * 24! / 4! = 17934974117270551244186247903990858235542284305932202306560963920617537536000000000000000 = 1.79 × 10^88

The centralmost centers must be rotated too. I don't see this in your calculation.

Hey there Jakob -

Just out of curiosity, how many permutations would be possible on a 7x7 cube with the same type of sticker arrangement?

You have the explanation from the 5x5x5 cube of the same type.

For the 7x7x7 supercube with no pictures (and no logos) on the 6th side:

Corners and centralmost edges:
\( 8!*3^7*\frac{12!}{2}*2^{11} \)

Wings:
\( (24!)^2 \)
This calculation for wings uses the same reasoning as for the 5x5x5, except with 2 wing orbits now.

Centers:
\( \left(\frac{24!}{4!}\right)^6*4^5 \)

This centers calculation uses the same method as Jakube used only now there are 6 center piece orbits. The 4^5 accounts for the centralmost center pieces since they can noticeably twist in place on 5 sides of the 6.

Putting it all together:
\( 8!*3^7*\frac{12!}{2}*2^{11}*(24!)^2*\left(\frac{24!}{4!}\right)^6*4^5 \)

Using wolframalpha this is approximately:

5.0896051928992777130079600798397771352708855555 × 10^204
 

cmhardw

Premium Member
Joined
Apr 5, 2006
Messages
4,115
Location
Orlando, Florida
WCA
2003HARD01
YouTube
Visit Channel
Could that be fixed by just multiplying by 4 for every centerpiece whose orientation matters in the difference between a solved and unsolved state? (i.e. - 4^5 ?)

I really appreciate your help, guys!

Yes, multiplying Jakube's result by 4^5 will give the number of combinations to the 5x5x5 almost supercube you described.
 
Top