# 43 quintillion possibilities wrong? Am I right?



## Homsar105 (Aug 15, 2009)

There is an equation that shows that there are 43,252,003,274,489,856,000 combinations to the cube. I think I've found out there are really only 1,802,166,803,103,744,000 combinations. Heres my reason...

Its kinda hard to explain but I'll try me best. That equation showing that there are 43 quint. possible combinations to the cube, I believe is stating that each way the cube is facing, is concidered a combination. For example, scramble a cube, now face the white side up, and the green side facing you. Now turn it so the orange side faces you. Those I belive are concidered diferent combinations of the cube in that equation.

Thereare 24 different ways a cube can face (ie: one side faces up, and with that side facing up, there are 4 different possible ways that cube can face in front. Apply that to all 6 sides. 4x6=24, therefore the cube can face 24 different ways)

As stated before, I think the equation that gave 43 qint combinations as an answer is including each way a cube can face in its equation (not exatly sure). So take those 24 different ways the cube can face and divide it by that 43,252,003,274,489,856,000. You get 1,802,166,803,103,744,000.

So there are 43,252,003,274,489,856,000 possible POSITIONS to the cube, but there are only 1,802,166,803,103,744,000 COMBINATIONS.

Am I right? Please correct me if I'm wrong.


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## qqwref (Aug 15, 2009)

lol

The 43 quintillion figure takes into account that the centers are fixed (the same centers are always on up and front). Hence the 'position' a cube is in does not change if you change the orientation. The number of positions on all puzzles is counted in this way.


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## LNZ (Aug 15, 2009)

There are about 4.32E19 possible states of an 3x3 cube. And that is true. And I think combinations and possible states are the same thing.

Here's the formula for the number of states in logarithm (base 10) form for an NxNXN cube:

LOG10(F(N))=C1*(N MOD 2) + C2 + C3*INT((N*N-2*N)/4) + C4*INT(((N-2)^2)/4)

C1=13.07084816 (or 16.38217811 for a NxNXN super cube)
C2=6.565158065
C3=23.79270567
C4=-8.28126745

The INT function is the largest integer less than the argument. Foe example INT(5.7)=5 and INT(-1.5)=-1

I chose to present the LOG10 formula as your average calculator can only handle numbers up to 10^100. A 6x6 cube has F(6) of 2*10^116 possible states and a 5x5 cube has about 2.8*10^74 possible states.


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## qqwref (Aug 15, 2009)

...right. Thanks for the formula with absolutely no explanation or guarantee of accuracy. Very convincing.

Here's an explanation:


Jaap Scherphuis said:


> There are 8 corner pieces with 3 orientations each, 12 edge pieces with 2 orientations each, giving a maximum of 8!·12!·3^8·2^12 positions. This limit is not reached because:
> 
> * The total twist of the cubes is fixed (3)
> * The total number of edge flips is even (2)
> ...


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## nitrocan (Aug 15, 2009)

Where did you get the idea that people who calculated that didn't think that?


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## cuBerBruce (Aug 16, 2009)

The number of symmetrically distinct cube positions, not counting mirror symmetry, is approximately 43,252,003,274,489,856,000 divided by 24, but not exactly that number. The reason is that the there are symmetrical positions that don't have 24 separate equivalent positions. For example the solved position and the superflip positions are not equivalent to any other positions.

In 1994, Dan Hoey calculated the number of symmetrically distinct cube positions to be 901,083,404,981,813,616. Note that it is slightly more than 43,252,003,274,489,856,000/48. See here. I don't know if anyone has done the similar calculation where mirror symmetry is not included.


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## blindalchemist (Aug 22, 2009)

http://kociemba.org/cube.htm


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## Thomas09 (Aug 23, 2009)

What about impossible combinations. Such as when you get to the last layer, ther are 3 correctly flipped edges. Are these added in the equation?


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## endless_akatsuki (Aug 23, 2009)

Thomas09 said:


> What about impossible combinations. Such as when you get to the last layer, ther are 3 correctly flipped edges. Are these added in the equation?



nah, read qqwref's post


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