How do you calculate all the possible combinations on a Rubik's cube?

  • Thread starter Thread starter Nerdydude101
  • Start date Start date
  • Tags Tags
    Combinations Cube
AI Thread Summary
Calculating the total combinations of a Rubik's Cube involves understanding the mechanical constraints of the puzzle rather than simply multiplying the number of faces by the cubes on each face. The permutations include 8! for corner cube locations, 3^8 for their orientations, and additional factors for edge cubes, resulting in an astronomical total of 43 quintillion to 519 quintillion possible configurations. Most of these configurations do not lead to a solved state, as many are not achievable through legal moves. The complexity of the problem suggests that programming may not effectively count permutations due to the sheer size of the number. Understanding these constraints is crucial for accurately calculating the combinations of a Rubik's Cube.
Nerdydude101
Messages
54
Reaction score
0
I thought it would just be the number of faces multiplied by the nine cubes on each face? What am i doing wrong?
 
Mathematics news on Phys.org
Nerdydude101 said:
I thought it would just be the number of faces multiplied by the nine cubes on each face? What am i doing wrong?

Not all combinations are possible mechanically. I would probably try to solve this with a program. Are you comfortable writing a C program (or using some other programming language) to solve this?
 
I know very little programming, a tiny but if Python but that's about it
 
You're not going to be able to count the permutations on a computer. The number is too big.

If you consider the problem of the number of permutations that can be made by pulling a Rubik's cube apart piece by piece and then reassembling it, this is a huge number. There are eight corner cubes which can be placed. That means 8! permutations just based on corner cube location. Each corner cube can be placed in one of three orientations. That's a factor of 38 permutations on top of the 8! location permutations. The twelve corner cubes lead to two more factors, 12! and 212. Altogether, there are ##8! \, 3^8 \, 12! \, 2^{12}## permutations of the ripped apart and resembled cube. That is a *big* number.

Most of these permutations do not lead to the nice all colors on one face arrangement. There are constraints, but the final number is still huge.
 
See here
http://en.wikipedia.org/wiki/Rubik's_Cube
There are
$${8! \times 3^7 \times (12!/2) \times 2^{11}} = 43,252,003,274,489,856,000 \\

{8! \times 3^8 \times 12! \times 2^{12}} = 519,024,039,293,878,272,000. $$
combinations
the larger number is 12 times the smaller as there are 12 orbits
that is any position can reach 1/12 positions though legal moves separating possible moves into 12 orbits
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
2
Views
2K
Replies
23
Views
3K
Replies
7
Views
5K
Replies
33
Views
3K
Replies
3
Views
4K
Replies
1
Views
2K
Back
Top