The math behind the cube

[goidlon]

Hey guys this is a project for school about the math behind the cube and I have been asked to do some research about it. I thought it would be a good idea to get some help from you guys! If you have any ideas or websites I should check out let me know.

 

[ProStar]

Talk about god's number

 

[BenChristman1]

Talk about god's number

(20 moves) and the number of possible positions a cube can be in (over 43 quintillion aka 43,000,000,000,000,000,000).

 

[brododragon]

Talk about why the equation for the amount of possibilities works. There's a great Popular Science article on it called 'the math behind the Rubik's cube' or something like that.

 

[goidlon]

Talk about god's number

I will I already have a video but I will go more in depth

 

[ProStar]

I will I already have a video but I will go more in depth

God's Number is 20

 

[RiceMan_]

TPS and Movecount

 

[BenChristman1]

Talk about why the equation for the amount of possibilities works. There's a great Popular Science article on it called 'the math behind the Rubik's cube' or something like that.

(8! x 12! x2^12 x 3^8)/12 = 43,252,003,274,489,856,000

 

[ProStar]

Talk about why the equation for the amount of possibilities works.

This is an explanation I wrote of how the equation for calculating the possible solvable cases works.


8! * 12! * 12^2 * 8^3 / 12 is the equation used to calculate the number of permutations on a 3x3. Here's why it works:

8! - there are 8 corners on the 3x3, each of which can be arranged uniquely. To find the number one set of items can be arranged(i.e, how many ways can 8 blocks be arranged), we take the number of items(8 corners) and multiply it with every previous number down to 1. This is called a factorial, and is represented with an !, so 8! is 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. The reason we need to do this is because the first corner can be put in 8 different positions, the second corner can now go in 7 different positions, the third can go in 6, etc.

* 12! - this does the same thing as the previous step, except for the edges

* 2^12 - last step, we calculated how many permutations that can be created with edges, but we didn't account for orientation. Each edge can have 2 orientations, so we multiply the number of orientations by itself for every unique edge. This is represented by an exponent, which can be written like this: x^y. So if we have 2^10, it would be 2 multiplied by itself 10 times, or 2*2*2*2*2*2*2*2*2*2.

* 3^8 - this does the same thing as the previous step, accounting for 8 corners, each with 3 unique orientations

/ 12 - this is to account for impossible cases, sometimes referred to as "parity cases". If you've ever twisted a corner, then you'll know that some cases are impossible to solve. I personally am not sure why exactly the number needed is 12, but I'm sure another member could explain. If we were to remove this from the equation, we would get all the possible combinations, including unsolvable ones.

So this is how we get the large number of 43,252,003,274,489,856,000.

 

[xyzzy]

The reason for that is that this equation counts a solved cube with white top green front and a solved cube as white top red front as two different positions.

Actually, no, the "white top red front" scenario isn't counted at all in the 43 quintillion figure. That figure is working under the assumption that the centres are fixed (i.e. the opposite of what you're trying to say).

I think you're trying to talk about symmetry reduction? (E.g. the sequences "U" and "R" lead to cube states that are essentially identical, up to a rotation.) Symmetry reduction is considerably more complicated because different cube states can have different numbers of equivalent states. For example, the solved state, the checkerboard state (M2 E2 S2) and the superflip state all have maximum symmetry, and hence have only one equivalent state each (which is to say, themselves and only themselves). In contrast, most cube states have only the trivial symmetry, and hence have 24 equivalent states (if you only allow proper rotations) or 48 equivalent states (if you allow reflections as well), or even 96 equivalent states (if you also allow inverting the cube state).

As different states can have different equivalence class sizes, you can't just divide the total number of states by 24 (or 48 or 96) to get the number of equivalence classes (although this will usually be a very good approximation). You need to go through this thing and painstakingly count the number of states with each symmetry subgroup.

 

[ProStar]

Actually, no, the "white top red front" scenario isn't counted at all in the 43 quintillion figure. That figure is working under the assumption that the centres are fixed (i.e. the opposite of what you're trying to say).

I think you're trying to talk about symmetry reduction? (E.g. the sequences "U" and "R" lead to cube states that are essentially identical, up to a rotation.) Symmetry reduction is considerably more complicated because different cube states can have different numbers of equivalent states. For example, the solved state, the checkerboard state (M2 E2 S2) and the superflip state all have maximum symmetry, and hence have only one equivalent state each (which is to say, themselves and only themselves). In contrast, most cube states have only the trivial symmetry, and hence have 24 equivalent states (if you only allow proper rotations) or 48 equivalent states (if you allow reflections as well), or even 96 equivalent states (if you also allow inverting the cube state).

As different states can have different equivalence class sizes, you can't just divide the total number of states by 24 (or 48 or 96) to get the number of equivalence classes (although this will usually be a very good approximation). You need to go through this thing and painstakingly count the number of states with each symmetry subgroup.

Ok, I removed that part from my post. I thought I read about that somewhere, I must have forgotten something. Thanks!

 

[Cuberstache]

/ 12 - this is to account for impossible cases, sometimes referred to as "parity cases". If you've ever twisted a corner, then you'll know that some cases are impossible to solve. I personally am not sure why exactly the number needed is 12, but I'm sure another member could explain. If we were to remove this from the equation, we would get all the possible combinations, including unsolvable ones.

If you randomly assemble the cube, as described in the previous steps:
50% chance of edge flip
67% (2/3) chance of corner twist
50% chance of parity (one edge-edge or corner-corner swap)

Therefore, the solvable permutations are the possible ones divided by 2*3*2, or 12.

 

[brododragon]

If you randomly assemble the cube, as described in the previous steps:
50% chance of edge flip
67% (2/3) chance of corner twist
50% chance of parity (one edge-edge or corner-corner swap)

Therefore, the solvable permutations are the possible ones divided by 2*3*2, or 12.

What are the individual chances of a corner-corner swap or edge-edge swap?

 

[ProStar]

What are the individual chances of a corner-corner swap or edge-edge swap?

He already said it, 50%

 

[brododragon]

He already said it, 50%

I mean individually

 

[ProStar]

I mean individually

Oh then I'm not sure

 

[Cuberstache]

What are the individual chances of a corner-corner swap or edge-edge swap?

They're the same thing. If you have a corner-corner swap, you can do a T-perm (for example) to make it an edge-edge swap.

 

[brododragon]

They're the same thing. If you have a corner-corner swap, you can do a T-perm (for example) to make it an edge-edge swap.

Oh.

 

[abunickabhi]

Interesting treatment to the cube group perceiving it as linear algebra with interesting rules yo