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How many possibilities are there for the Last Layer?

qqwref

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How many possibilities are there for the Last Layer, and would you please explain how you found this figure?

Depends, are you counting things that are an AUF away as the same thing? Are you counting mirrors as the same thing? Are you counting positions which are a premove AUF away the same thing (i.e. T-perm versus U T-perm U')? Are you counting inverses as the same thing?
 

Stefan

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I don't understand why you're laughing about that. I suppose for the same reason noncubers think it's funny when they tell us that they used to peel the stickers off?
 

panyan

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I suppose for the same reason noncubers think it's funny when they tell us that they used to peel the stickers off?

i used to laugh before i started speedcubing, now there is just an awkwad silence when they say that. The best way of cheat solving the cube is to disassemble and reassemble in a solved state becuase then you dont damage the stickers!
 

KubeKid73

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I don't understand why their first instinct is to take the stickers off. Why not just take it apart?
 

Weiseguy

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I don't understand why their first instinct is to take the stickers off. Why not just take it apart?

If you think about it, its the same reason why only a small % of people can obtain one face the same color, let alone in the right place.

I know of only 3 or 4 people (outside of the cubers at my school) that can solve the white face, and maybe 1 of those 4 can put the pieces in the right spot.


EDIT: To elaborate on my point, people have no common sense anymore :(
 
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haha, I cannot explain the 43 quintillion, Just another number I memorized!
If you are serious about this one, it's not hard to calculate the the 43*10^18.

First, let's assume a standard orientation such as white on top and green in front. The edge pieces and corner pieces are placed with respect to these.

Edges. The first edge can go into any of 12 locations. Once that is placed, the next edge can go into any of the remaining 11 locations. Once those are placed, the 3rd edge can go into one of the remaining 10 locations. Follow this logic for the remaining edge pieces. This means the edges can be arranged in 12*11*10*9*8*7*6*5*4*3*2*1 = 12! possible ways. Assuming each edge can be flipped or not flipped we have 2^12 ways edges can be flipped. But, we know the number of edges flipped cannot be odd. It must be even, so there are actually only 2^11 ways of flipping the edges.

Corners. By similar logic there are 8! ways to position corners with respect to the centers. A corner can be oriented normally or CW or CCW, and the twists must all cancel to normal. (One CW and CCW cancel and 3 CW cancel and 3CCW cancel to normal.) So there are 3^7 ways the corners can be twisted.

PLL parity. There can never be only 2 edges needing to be swapped without requiring 2 corners to be swapped too. So, we should divide our results by 2.

Results. 12! * 2^11 * 8! * 3^7 / 2 = 43,252,003,274,498,856,000.
 
Last edited:

Weiseguy

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haha, I cannot explain the 43 quintillion, Just another number I memorized!
If you are serious about this one, it's not hard to calculate the the 43*10^18.

First, let's assume a standard orientation such as white on top and green in front. The edge pieces and corner pieces are placed with respect to these.

Edges. The first edge can go into any of 12 locations. Once that is placed, the next edge can go into any of the remaining 11 locations. Once those are placed, the 3rd edge can go into one of the remaining 10 locations. Follow this logic for the remaining edge pieces. This means the edges can be arranged in 12*11*10*9*8*7*6*5*4*3*2*1 = 12! possible ways. Assuming each edge can be flipped or not flipped we have 2^12 ways edges can be flipped. But, we know the number of edges flipped cannot be odd. It must be even, so there are actually only 2^11 ways of flipping the edges.

Corners. By similar logic there are 8! ways to position corners with respect to the centers. A corner can be oriented normally or CW or CCW, and the twists must all cancel to normal. (One CW and CCW cancel and 3 CW cancel and 3CCW cancel to normal.) So there are 3^7 ways the corners can be twisted.

PLL parity. There can never be only 2 edges needing to be swapped without requiring 2 corners to be swapped too. So, we should divide our results by 2.

Results. 12! * 2^11 * 8! * 3^7 / 2 = 43,252,003,274,498,856,000.

I love mathematics at its finest :)

I'm seriously taking this into my AP Calc BC teacher tomorrow and shoving it in his face for saying that 90% I got on my last test was because I cube instead of study. Guy wants me to get 98%'s or better because I did all year in his AP Calc AB class :(
 
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