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**Calculating the Number of K4 "OLLs" on a nxnxn Cube**

Hi everyone,

First of all, I know pretty well that wing edges do not have an orientation on a big cube. Nonetheless, I wanted to find out how many "OLLs" there are for the nxnxn cube.

Second, I am not sure if all of my calculations in this post are correct, but I will explain every number I come up with. (Therefore, if I am wrong, it will be easy to see where I went wrong, what I left out, etc.)

Anyway, I was observing my 3x3x3 OLL "Times Table," which can be found here. (I posted it in this thread).

Obviously there are 7 for the 2x2x2, and thus I want to focus on the 4x4x4 and larger cubes. The other day, I determined the number of wing edge OLL cases for each orbit of a cube. If I did my calculations correct, there are 19. I'll number them too for future reference.

At the same time, I want to show the edge case generators for the 3x3x3 OLLs too, as well as one corner case. I am going to just number the following:

(a)

With that out of the way, I wanted to find out a way to calculate the number of 3x3x3 OLLs to hopefully find a pattern to find a formula for the nxnxn in terms of the wing edges and corners. And, if it's an odd cube, including the central edge pieces as well.

For the 3x3x3, I came up with:

\( \left( 4\left( 1 \right)+2\left( 1 \right)+1+1 \right)7-2\left( 1 \right)+3=57 \), where

**Part I**

4(1)

Represents having 4 rows of (a) as the row header (see the pdf if confused). (a) is not the same when crossed by

*most*of the corner OLLs as it's rotated 90 degrees 4 times. Since (a) is the only non-symmetrical case in terms of rotations, we multiply the 4 by 1.

2(1)

Represents having 2 rows of (b), because (b) is symmetrical when rotated 180 degrees. It's the only one of its category in terms of rotations, hence we multiple the 2 by 1.

1

Represents the unique case of (c) and the other

1

Represents the unique case of (d)

All of these terms are added together and multiplied by 7, the total number of corner OLL cases.

**Part II**

From here, we subtract 2(1), because, if you see the PDF, two of the cells in the table are blank in the second to right most column. The 2 means that we cancel half (since we originally included 4 rows of cells with (a) as the row header) of the cases involving (a) with the symmetrical corner case (e). We multiply by 1 because (a) was the only edge case of this category.

**Part III**

Finally, we add the total number of edge cases (with corners oriented) by adding 3.

If I follow the same example for one orbit of a big cube, let's just say the last layer 4x4x4 wing edges with the corners, I get:

\( \left( 4\left( 16 \right)+2\left( 2 \right)+1+1 \right)7-2\left( 9 \right)+19=491 \)

, where

**Part I**

4(16)

Represents 4(16) = 64 rows consisting of cases crossed with all of the wing edge "OLL" cases, except for the symmetrical ones about the x and y axis: (6), (15) and (19). Note that cases (6) and (15) are analogous to edge case (b), and (19) is analogous to case (c) in the 3x3x3 equation. Hence, the 16 is from 19-3 = 16.

2(2)

Represents having two rows of each of the cases (6) and (15) which are identical if rotated 180 degrees.

And the 1+1 are the unique cases of (19) and (d).

Like the 3x3x3, we multiply all of these terms by 7, the number of corner OLL cases.

**Part II**

-2(9)

Means we cancel half of the cases involving crossing cases (1), (2), (3), (4), (5), (7), (8), (13), and (17) with the symmetrical corner case (e), because they are symmetrical about one axis. When they are crossed with the symmetrical corner case (e), there will be some cancellations. Another way of identifying these symmetrical cases is that we can flip any one of them (horizontally or vertically) and rotate to achieve the original case. Going back to the 3x3x3 example, edge case (a) can be looked at this way as well.

The reason that cases (b), (6), and (15) are not treated this way is basically because, by the way I made the table for the 3x3x3 OLLs, I left out obvious cancellations.

**Part III**

Like the 3x3x3, we lastly add the number of edge OLL cases, or in this case, wing edge "OLL" cases, which is 19.

**Extrapolating to Higher Order Even Cubes**

Since \( \left\lfloor \frac{n-2}{2} \right\rfloor \) is the number of orbits wings in a big cube, adjusting the formula (the brackets mean the floor/greatest integer function),

\( \left( 4\left( 16^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+2\left( 2^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+1+1 \right)7-2\left( 9^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+19^{\left\lfloor \frac{n-2}{2} \right\rfloor } \)

So for the 6x6x6, there's 7,437 "OLL" cases.

**For the 5x5x5 and Other Odd Cubes**

If we want to include the central edge pieces on odd cubes too, we can just imagine setting up a table like the one I did for the 3x3x3 OLL cases. However, this time the number achieved from the formula above are the number of column headers of the table (where the corner OLL cases originally were). The rest of the table is the same as the 3x3x3 one, having the 3 edge cases as row headers except that we do not have case (d) to be the row header of its own row because we have already taken care of it.

*By ignoring the middle edges before, we have already taken care of the case when all middle edges are oriented.*

*That is, the formula above IS that result.*

Hence, let X be the number formed from the above formula for even cubes. Then for the 5x5x5, we have (X = 491)

\( \left( 4\left( 1 \right)+2\left( 1 \right)+1 \right)491-2\left( 1 \right)\left( 3+2\left( 2 \right) \right)+3=3,426 \)

**Part I**

4(1) represents having 4 rows with the row header being rotations of case (a) (just like the 3x3x3 table),

2(1) represents having 2 rows of with the row header being rotations of case (b) (also, just like the 3x3x3 table),

and 1 represents having 1 row with the row header case (c) (just like the 3x3x3 table).

We multiply all of this by X instead of 7 (the number of corner OLLs) because X includes the corner OLLs by themselves, corner OLLs with wing edge "OLLs," and wing edge "OLLs" by themselves.

**Part II**

-2(1)(3+2(2))

Means we cancel 2(1)(3+2(2)) symmetrical OLL cases.

The (1) means that we have one central edge case (a) involved as a row header, and, again, we multiply it by 2 to designate canceling half of the case images with it combined with the symmetrical column headers.

For the (3+2(2)),

3

Represents the symmetrical corner OLL case (e) by itself, and the symmetrical wing edge OLLs (6) and (15) by themselves. (Simply put, cases (e), (6) and (15).)

The 2(2) represents the cases which will be just as symmetrical as (e), (6), and (15):

*Hence, in the 3x3x3 formula, there was an imaginary (1) in the term -2(1), i.e. -2(1)(1), so that the first (1) represented the case (a) and the second (1) represented the corner OLL case (e)*.

\( \left( 4\left( 1 \right)+2\left( 1 \right)+1 \right)\text{X}-2\left( 1 \right)\left( 1+6^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+3 \)

The \( 1+6^{\left\lfloor \frac{n-2}{2} \right\rfloor } \) means that we have only 1 corner OLL case, no matter the cube size (obviously), and 6 is the remaining amount of the original 3+2(2). This number will be dependent on the cube size because each orbit of wings can have each of these.

**A Formula for All Big Cube Sizes (4x4x4 and Greater)**

If we simplify the formula for X and the formula adjustment for odd cubes, that is,

**X**= \( \left[ 4\left( 16^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+2\left( 2^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+1+1 \right]7-2\left( 9^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+19^{\left\lfloor \frac{n-2}{2} \right\rfloor } \)

= \( \left[ 4\left( 16^{\left\lfloor \frac{n}{2}-1 \right\rfloor } \right)+2\left( 2^{\left\lfloor \frac{n}{2}-1 \right\rfloor } \right)+1+1 \right]7-2\left( 9^{\left\lfloor \frac{n}{2}-1 \right\rfloor } \right)+19^{\left\lfloor \frac{n}{2}-1 \right\rfloor } \)

= \( \left[ 4\left( 16^{\left\lfloor \frac{n}{2} \right\rfloor -1} \right)+2\left( 2^{\left\lfloor \frac{n}{2} \right\rfloor -1} \right)+2 \right]7-2\left( 9^{\left\lfloor \frac{n}{2} \right\rfloor -1} \right)+19^{\left\lfloor \frac{n}{2} \right\rfloor -1} \)

= \( 28\left( \frac{1}{16} \right)\left( 16^{\left\lfloor \frac{n}{2} \right\rfloor } \right)+14\left( \frac{1}{2} \right)\left( 2^{\left\lfloor \frac{n}{2} \right\rfloor } \right)+14-2\left( \frac{1}{9} \right)\left( 9^{\left\lfloor \frac{n}{2} \right\rfloor } \right)+\left( \frac{1}{19} \right)19^{\left\lfloor \frac{n}{2} \right\rfloor } \)

= \( \left( \frac{7}{4} \right)16^{\left\lfloor \frac{n}{2} \right\rfloor }+\left( 7 \right)2^{\left\lfloor \frac{n}{2} \right\rfloor }-\left( \frac{2}{9} \right)9^{\left\lfloor \frac{n}{2} \right\rfloor }+\left( \frac{1}{19} \right)19^{\left\lfloor \frac{n}{2} \right\rfloor }+14 \)

And the formula adjustment for odd cubes,

\( \left( 4\left( 1 \right)+2\left( 1 \right)+1 \right)\text{X}-2\left( 1 \right)\left( 1+6^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+3 \)

= \( 7\text{X}-2-2\left( 6^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+3 \)

= \( 7\text{X}-2\left( 6^{\left\lfloor \frac{n}{2} \right\rfloor -1} \right)+1 \)

= \( 7\text{X}-2\left( \frac{1}{6}6^{\left\lfloor \frac{n}{2} \right\rfloor } \right)+1 \)

= \( 7\text{X}-\frac{1}{3}6^{\left\lfloor \frac{n}{2} \right\rfloor }+1 \)

= \( 7\text{X}-\left( \frac{1}{3}6^{\left\lfloor \frac{n}{2} \right\rfloor }-1 \right) \)

we can write a general formula to be:

\( \left[ 1-6\left\lfloor \left\lfloor \frac{n}{2} \right\rfloor -\frac{n}{2} \right\rfloor \right]\left( \frac{7}{4}\left( 16^{\left\lfloor \frac{n}{2} \right\rfloor } \right)+7\left( 2^{\left\lfloor \frac{n}{2} \right\rfloor } \right)-\frac{2}{9}\left( 9^{\left\lfloor \frac{n}{2} \right\rfloor } \right)+\frac{1}{19}\left( 19^{\left\lfloor \frac{n}{2} \right\rfloor } \right)+14 \right) \)\( +\left( \frac{1}{3}\left( 6^{\left\lfloor \frac{n}{2} \right\rfloor }-1 \right) \right)\left\lfloor \left\lfloor \frac{n}{2} \right\rfloor -\frac{n}{2} \right\rfloor \)

Here's a link to it.

So, listing values from the formula,

4x4x4: 491

5x5x5: 3,426

6x6x6: 7,437

7x7x7: 51,988

8x8x8: 120,215

9x9x9: 841,074

10x10x10 1,952,445

11x11x11 13,664,524

100x100x100 \( 4.587718283988192\times 10^{62} \) (Here)

, etc.

Are these results the correct numbers?

Last edited: Jun 19, 2011