Help A High Schooler Understand Please!

So when calculating the 43 quintillion possible permutations for a rubiks cube, I understand the part of the equation where its: 8!*3^8*12!*2^12 but I don't understand the part where you divide by 2*3*2. Can someone explain this to me? And if I wanted to learn how to calculate permutations for a 5x5x5 for example, I could calculate the edges and corners, but how would you calculate the centers? I googled it and was unable to find an understandable answer. Please help, im very eager to learn!

You do know if you you have a twisted corner and/or flipped edge it impossible to solve solve along with two flipped edges or corners so there a lot you can't solve. Centres don't have rules like this. They can be in any configuration

okso like, 4x4 centres:

there are 24 centres so there are 24! ways of arranging them, but there are 6 sets of 4 that can be in any order which should only be counted once, so divide by 4!^6 so its 24!/4!^6=3246670537110000. divide by 24 if you arent calculating whole cube permutations (orientation of the cube doesnt matter)

5x5 + centres are the same, but for all 5x5 centres multiply by 24 core orientations if you want to: (24!/4!^6)^2 or (24!/4!^6)^2*24

I think thats correct anywæ

also you divide by 2*3*2 to get 3x3 permutations because groups and stuff

The 2*3*2 you're dividing by are the 12 orbits of the cube. 11 of them are improperly assembled cubes and don't count as valid scrambles. Specifically cubes where a corner piece is twisted, cubes where an edge piece is twisted, cubes were two edge pieces are swapped, cubes were two corner pieces are swapped - and some but not all combinations of these conditions. They are the 12 ways to put the cube together, 11 of them where it can't be solved by just rotating the faces.

One equation for a 3x3 is; possible scrambles = how many ways could you randomly assemble the non-center pieces divided by the orbits.

see http://www.youtube.com/watch?v=QV9k6dRQQe4

The crazy thing is the inverse, that impossible scrambles outnumber possible scrambles 11 to 1.

perhaps someone else will chime in on the specifics of calculating bigger cubes. I don't think we quite fleshed that out yet.

I'm just gonna keep editing this till it is clear & correct.

It's 3^7 and 2^11 because the orientation of the last corner or edge is decided by the orientation of the other 7 or 11
And then dividing by PLL parity is divide by 2

8!*3^7*12!*2^10 divided by 2 I just entered this into the calculator and got 21626001637244930000 so there is something wrong with that, help?

Ok.. I'll try to explain the best I can. And I'll start from the beginning for those who don't make it as far as you did.

There are 8 corners. The first can be placed in 8 places while the second cannot be placed in the same place as the first so it can go in 7 places. This continues all the way down to ! for the l place for the last corner to be in. You would write this 8! or 8x7x6x5x4x3x2x1

The same applies for edges of which there are 12. So you get 12!.

Each corner can be oriented three ways, so you would multiply by 3 for each corner. This is written as 3^8.

The same applies for edges, but they only have 2 orientations each. This gives you 2^12.

When calculating orientation, the last corners orientation is decided by the other 7 because you cannot flip just one corner. Thus you would divide by 3 to ignore the last corner. Please note that instead of dividing by this, you could just change the 3^8 into 3^7.

You do the same with the last edge and divide by 2 or change 2^12 to 2^11.

Finally, you divide by 2 again to eliminate cases where edges and corners are not in an "even" permutation. This is where it gets confusing. If you try to solve just the corners using only 3 cycles (because swapping only 2 pieces is impossible), you will find that it cannot always be done. If you think about it, doing a single turn does a 4 cycle of corners which cannot be solved by 3-cycles. So doing a single turn changes the corners from an "odd" permutation to an "even" one and vice-versa. The same thing applies to edges. So doing a single turn (In QTM!!!) will change both corners and edges permutation to an even/odd one. In the solved state, both corners and edges are in an even permutation btw. Another cool thing about this is that if you mix a cube up with an odd number of moves, you must solve it with an odd number of moves (same thing goes for evens).

So in the end you get one of these two equations:

8! x 12! x 3^8 x 2^12 / 3 x 2 x 2

8! x 12! x 3^7 x 2^11 / 2

You can also use that Even/Odd permutation thing to explain void cube parity. If you do not solve in relation to centers, doing a single M move is a 4-cycle of edges which changes the permutation to even/odd, but doesn't affect corners. Thus, you can have corners and edges permutation acting independently of each other letting you swap only 2 pieces.

In the post right above mine, you wrote 2^10 instead of 2^11. Could this be your source of error?

I am a High Schooler btw.

Pablo17 said:

8!*3^7*12!*2^10 divided by 2 I just entered this into the calculator and got 21626001637244930000 so there is something wrong with that, help?

Click to expand...

Since you already reduced 2^11 to 2^11, you don't have to divide by 2 (you already did)

Thank you everyone for the help! Particularly elrog because its clear he put a decent amount of time and effort into this! I understand it well now, and im still working on understanding how to do it for 5x5x5+

I could make a post for the 4x4 and 5x5 and super cubes if you want me to.

If you could that'd be GREAT!

Pablo17 said:

Thank you everyone for the help! Particularly elrog because its clear he put a decent amount of time and effort into this! I understand it well now, and im still working on understanding how to do it for 5x5x5+

Click to expand...

If you really understand it well it shouldn't take you 4 days to do it for 5x5x5 or any cube of any size really so I would recommend for you to figure this out for yourself instead of Elrog "doing your homework". You learn more by going through the trouble yourself

AvGalen said:

If you really understand it well it shouldn't take you 4 days to do it for 5x5x5 or any cube of any size really so I would recommend for you to figure this out for yourself instead of Elrog "doing your homework". You learn more by going through the trouble yourself

Click to expand...

Well, the centers are something new that you can't apply any concept from the 3x3 to. So its understandable to get stuck on them. You don't necessarily learn more by figuring things out yourself, you just understand it better by the time you finally figure it out. But after being taught you'd start to understand it better and be at the same level of understanding you would be by the time you finally would have figured it out.


I'll start on the 4x4, then 5x5, then the 4x4 supercube, and finally the 5x5 supercube, but first I'm going to add a new way of thinking to that Even/Odd permutation thing. Rather than thinking about corners and edges together, think of them separately. The corners can have 8! permutations. If the corners were solved, the edges would have 12!/2 permutations because the last 2 edges permutations are decided by the rest of the pieces as it is impossible to switch just 2 pieces. You can do this same explanation with edges first.

4x4

The corners on all NxNxN cubes act in the same way. You already know that they are 8! x 3^7.

There are 24 edges with any edge being able to get into any other edges spot. This gives you 24! permutations.

The edges also have no orientation. After studying the edges, you find that there are edge cycles. If you do only U turns, you'll notice that some of the edges are never placed in other edges places. If you force one to be in the other edge cycle by doing a slice move or moving it across all the axis, it will be flipped when it gets there. So the orientation is decided by the permutation.

Finally you have the 24 centers which have 24! permutations and no orientation.

The tricky thing with centers is there are 4 of each color. 4 centers of the same color can be arranged in 4! ways. Since there are 6 colors, you divide by 4!^6

When you try to reduce the cases by keeping everything in an even/odd permutation, you may realize that corners even/odd permutation is completely independent of the edges. The edges do not switch with them because you actually do 2 4-cycles of edges making them not change.

The corners do in fact stay in even/odd permutation with centers because doing a single U turn does a 4-cycle of centers and corners, but centers can be solved in either a even or odd permutation because of multiples being the same color.

If you do a single slice move, you change the edges permutation to even/odd because you do a single 4-cycle of edges. But this doesn't affect the corners. It also doesn't affect the centers because you do 2 4-cycles of the centers with a single slice move.

Because permutations are independent, you can swap only 2 corners (or edges in 3x3 state, but this is really 2 2-swaps) and swap 2 edge pieces. You can also swap just 2 edge pieces. Swapping 2 edge pieces right beside each other makes them both flip and makes it appear as if you flipped one edge in a 3x3 state.

The very hardest part for me to understand was how corners/centers being in an even/odd permutation affects possible permutations of the centers. But because centers can appear solved in an even or odd permutation the corners even/odd permutation seems not to affect the possible number of ways the centers are arranged.

One last thing you need to do is eliminate duplicates of the same case due to cube rotations because there are no fixed centers. There are 6 sides and you can look at each side from 4 angles (Z rotations) giving you 24 ways to view the cube. This applies to all even layered cubes.

As the 4x4s permutation don't simplify, you get: 8! x 3^7 x 24! x 24! / 4!^6 x 24

I'll do the 5x5 in a little bit.

Hey Pablo17,

I apologize for re-posting basically the same thing, but I made some changes to the PowerPoint presentation (mainly for the supercube section, but also I added some more visuals), and I wanted you to notice I made changes before you get too deep into reading the old version (just in case you wouldn't notice my edit).

I have also numbered the slides now, so if you have a specific question about a certain slide, ask away, and either I or someone can answer that for you.

cmowlaMath3900

I hope that my new explanation for supercubes is understandable now (I believe I improved it a lot).

I would have added more by now, but with the expert here (cmowla), I don't see the point.

elrog said:

Finally, you divide by 2 again to eliminate cases where edges and corners are not in an "even" permutation. This is where it gets confusing. If you try to solve just the corners using only 3 cycles (because swapping only 2 pieces is impossible), you will find that it cannot always be done. If you think about it, doing a single turn does a 4 cycle of corners which cannot be solved by 3-cycles. So doing a single turn changes the corners from an "odd" permutation to an "even" one and vice-versa. The same thing applies to edges. So doing a single turn (In QTM!!!) will change both corners and edges permutation to an even/odd one. In the solved state, both corners and edges are in an even permutation btw. Another cool thing about this is that if you mix a cube up with an odd number of moves, you must solve it with an odd number of moves (same thing goes for evens).

Click to expand...

Hey, i have a question regarding this. I need help finding the number of permutations of a cube. From the paragraph i quoted, do we only divide by 2 for odd degree cubes (3x3...5x5...7x7..etc) or mus we also divide by 2 for even degree cubes?

Here's how to find the number of possible permutations of a 3x3.

There are 8 corners, so any one given corner can be in any of 8 places. Any given second corner can go anywhere besides the place where the first one is, so it can go in 7 places. This goes on and on until you end up with 8! (8 factorial) or 8x7x6x5x4x3x2x1.

There are 12 edges, with the same principles applying. This gives you 12! or 12x11x10x9x8x7x6x5x4x3x2x1.

You would now divide by 2 to eliminate cases that have the edges and corners permutations not in synch as described previously.

So this gives you 8!x12!/2 or 9,656,672,256,000 possible permutations.

You would not divide by 2 on an even cube to keep edge and corners in correlation because there is no center edge. Take the 4x4 for example. There are 8 edge on a single side and they are all inter-changeable. Doing a single U turn makes the corners permutation swap to even/odd, but the edges are actually doing 2 4-cycles making them stay in even/odd permutation. You can change the permutation of the edges by doing a single layer slice move which does a single 4-cycle of edges and 2 4-cycles of centers.

Doing a single U turn does do a 4-cycle of the centers, which does change their permutation to even/odd, but it is solvable either way because there are multiple centers with the same color. You would divide by 2 for each separate center cycle on a supercube though, as well as dividing by 2 if it is an odd cube.

All right thank you

elrog said:

The edges also have no orientation. After studying the edges, you find that there are edge cycles. If you do only U turns, you'll notice that some of the edges are never placed in other edges places. If you force one to be in the other edge cycle by doing a slice move or moving it across all the axis, it will be flipped when it gets there. So the orientation is decided by the permutation.

Click to expand...

And is it possible for u to explain this part a little more clearly and in depth? thks

Fieryflamer said:

elrog said:

The edges also have no orientation. After studying the edges, you find that there are edge cycles. If you do only U turns, you'll notice that some of the edges are never placed in other edges places. If you force one to be in the other edge cycle by doing a slice move or moving it across all the axis, it will be flipped when it gets there. So the orientation is decided by the permutation.

Click to expand...

And is it possible for u to explain this part a little more clearly and in depth? thks

Click to expand...

Hi Fieryflamer,

I apologize that the link to my document in post #14 was broken, but now it's fixed. However, a working link has been in my signature for several months:
# of Permutations on the nxnxn

To save elrog some time,
In my document, I answer this GOOD question simply with an image, which can be found on pg 36 of the PDF (slide 71).

My complete explanation of orientations spans only a few pages if you would like to read it in its entirety: from slide 66 on page 33 to slide 71 on page 36.