# Calculating the Number of K4 "OLLs" on an nxnxn Cube



## Christopher Mowla (Jun 18, 2011)

*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.


Spoiler













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)



(b)



(c)



(d)



(e)





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


Spoiler



*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


Spoiler



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



Spoiler



*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)_.


In general though for odd cubes, we take into consideration the number of orbits of wings, so we have:
\( \left( 4\left( 1 \right)+2\left( 1 \right)+1 \right)\text{X}-2\left( 1 \right)\left( 1+6^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+3 \)

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?


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## MaeLSTRoM (Jun 18, 2011)

If this is for K4, wouldn't you ignore corner orienations since you do them in a seperate step.
Also could you post the numbers without corners anyway, just for interest.


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## uberCuber (Jun 18, 2011)

MaeLSTRoM said:


> Also could you post the numbers without corners anyway, just for interest.


 
Isn't this just the 22 listed in the first spoiler?


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## MaeLSTRoM (Jun 18, 2011)

uberCuber said:


> Isn't this just the 22 listed in the first spoiler?


 
Yes but I meant could he provide the formula for other size cubes ignoring corners


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## Kirjava (Jun 18, 2011)

This is cool. I lol'd at the idea of someone considering OLL on 11x11x11.

OLL isn't a system that I think is very good for bigcubes. I don't like OLL skips.


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## Christopher Mowla (Jun 19, 2011)

Kirjava said:


> This is cool. I lol'd at the idea of someone considering OLL on 11x11x11.


Yeah, 27 million+!



MaeLSTRoM said:


> Also could you post the numbers without corners anyway, just for interest.


If I did my calculations correctly in my first post, then we have:

\( \left[ 1-6\left\lfloor \left\lfloor \frac{n}{2} \right\rfloor -\frac{n}{2} \right\rfloor \right]\left( \frac{1}{19}19^{\left\lfloor \frac{n}{2} \right\rfloor } \right)+\left( 2^{\left\lfloor \frac{n}{2} \right\rfloor }-3 \right)\left\lfloor \left\lfloor \frac{n}{2} \right\rfloor -\frac{n}{2} \right\rfloor \)​

Here's a link to it.

4x4x4: 19
5x5x5: 132
6x6x6: 361
7x7x7: 2,522
8x8x8: 6,859
9x9x9: 48,000
10x10x10: 130,321
11x11x11: 912,218
100x100x100: \( 4.55959686821366\times 10^{62} \) (Here)
 


Spoiler



*For Even Cubes,*
*X *= \( 19^{\left\lfloor \frac{n-2}{2} \right\rfloor }=19^{\left\lfloor \frac{n}{2}-1 \right\rfloor }=19^{\left\lfloor \frac{n}{2} \right\rfloor -1}=\frac{1}{19}19^{\left\lfloor \frac{n}{2} \right\rfloor } \)

*For Odd Cubes*
\( \left( 4\left( 1 \right)+2\left( 1 \right)+1 \right)\text{X}-2\left( 1 \right)\left( 2^{\left\lfloor \frac{n-2}{2} \right\rfloor } \right)+3 \)
, where \( 2^{\left\lfloor \frac{n-2}{2} \right\rfloor } \) represents the symmetrical wing edge "OLLs" (6) and (15) (by themselves only). When compared to my equation for the 3x3x3 OLLs, "OLLs" (6) and (15) are analogous to case (e), the corner OLL. The table we form has the wing edge "OLLs" as the column headers, and the central edge pieces are the row headers. Since we have the central edges as the row headers, we still have the (4(1)+2(1)+1), where:
4(1) is case (a),
2(1) is case (b), and
1 is case (c).
_Case (d) is already taken care of because we assumed all central edges to be oriented by the value *X*._

-2(1)
We still subtract half (hence the 2) of the edge case (a) (hence the (1)).

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


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## AvGalen (Jun 19, 2011)

So the conclusion is "too many"?


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## uberCuber (Jun 19, 2011)

AvGalen said:


> So the conclusion is "too many"?


 
I read this, lol'd, and then was starting to think to myself, "but not nearly too many for the edge-only OLLs [on a 4x4 ofc]". And then the related question popped into my head, How many last layer edge permutation cases are there for 4x4 once the edges are "oriented" in this way?

EDIT: Don't take this as me actually thinking about trying to solve this way in timed solves


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## Tim Major (Jun 19, 2011)

How many 'PLLs'? Doesn't matter about edge orientation or corners.


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## vcuber13 (Jun 19, 2011)

do you mean 4x4 ells? wouldnt it be 8!*4=161,280


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## Christopher Mowla (Jun 19, 2011)

Tim Major said:


> How many 'PLLs'? Doesn't matter about edge orientation or corners.


I can't say I can answer that question right now, as it seems to be much more difficult to do by hand (assuming that I did the "OLL" calculations correctly, even at that).

However, I did some more useful calculations lately regarding "PLLs." I have made a document for the number of "PLLs" for one orbit of wings in a big cube. The calculations assume that the corners are solved (or any equivalent of this) and that central edge pieces on odd cubes are not involved. In addition, these PLLs are just for all wing edges "oriented." My first attempt at this was very close to being correct, but cuBerBruce corrected me on some of the numbers for different compositions of disjoint cycles. As a result, I am confident this document is correct.

On the first page of the document, it says that for one orbit of wings, there are a total of 167 cases, but it only takes at most 66 different algorithms to invert and/or take the mirror of to achieve all 167 cases.

And, if there are only 3 composite edges involved, there is a mere 32 cases, needing only 25 algorithms to invert and/or take the mirror of to achieve all 32 cases. In fact, you need even less than 25 very different algorithms to handle all 32 cases. For example, with some algorithms, you can invert a U turn and get another case.

*"OLLs"*
I have the "OLLs" on the last page of the document. 

In 4 composite edges, obviously from reading this thread, there are _19_ "OLLs," needing a maximum of 17 different algorithms.

In 3 or less composite edges, there are 13, needing a maximum of 11 different algorithms.

Of course, this number of "needed" OLLs can be less, as inverting an "OLL" algorithm produces another "OLL" case. It just depends on the algorithm you use to "orient" the wings.

I have been trying to make algorithms for these, but I have a long way to go (assuming that I don't want to resort to a solver for all of them; however, CubeExplorer did produce nice results for some cases, and I have included them in my collection).

EDIT:
I have updated the document (fixing the last page) according to cuBerBruce's post below.


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## cuBerBruce (Jun 19, 2011)

cmowla said:


> 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.
> 
> ...



I would have to say that it appears to me cmowla's numbers are generally too high. 
First of all, I count only 20 cases for one orbital of wings, including the solved case, or 19 if you don't count the solved case. cmowla's cases 14 & 15 are the same, as well as 17 & 18, and finally 20 & 21. That accounts for the difference in his value of 22 and my value of 19 (when excluding the solved case).

There are actually 70 "raw" cases if you don't reduce the cases by considering ones that are equivalent by viewing from a different angle. The eight wings can be divided into two sets of four. Given a position, if a particular wing is "oriented," then every wing in that set will be "oriented" in that position. A wing not "oriented" in that position would have to be in the other set of four cubies.

The total number of permutations for the 8 wings is 8! or 40320. You can permute the four cubies of one set any way you want (24 possible arrangements). Likewise the other set of 4 cubies can be rearranged 24 ways. This gives 24^2 or 576 ways to permute the cubies without changing the "orientation" case. 40320/576 = 70 gives the number of raw "orientation" cases.

The actual number of cases is the number of rotational symmetry equivalence classes for all of the raw cases. Since the last layer is square, the number of elements in each rotational symmetry equivalence class is usually 4. Therefore dividing the number of raw cases by 4 will generally be close to the correct number. Because there are some symmetrical configurations for which the equivalence class only contains 1 or 2 elements instead of 4, simply dividing the raw cases by 4 will slightly undercount the actual value.

For multiple wing orbitals, (70^n)/4 should be close to correct (with n being the number of orbitals). The larger n is, the closer (percentagewise) the result should be since it only takes one unsymmetrical orbital to force an equivalence class to have 4 elements. Corners have 27 raw orientation cases, and central edge orbitals have 8 raw orientation cases.


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## Christopher Mowla (Jun 19, 2011)

cuBerBruce said:


> I would have to say that it appears to me cmowla's numbers are generally too high.
> First of all, I count only 20 cases for one orbital of wings, including the solved case, or 19 if you don't count the solved case. cmowla's cases 14 & 15 are the same, as well as 17 & 18, and finally 20 & 21. That accounts for the difference in his value of 22 and my value of 19 (when excluding the solved case).


Thanks for letting me see that! I can't believe I overlook these equivalences all of the time! I have edited my first two posts with this adjustment. Are the numbers correct now?


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## cuBerBruce (Jun 20, 2011)

Okay, I think I have a way of calculating the correct number for arbitrary size cubes. Instead of a strict formula, I use a recurrence relation. It may be possible to derive a strict formula from the recurrence relation.

Anyway, the values for the number of OLL cases I get for various size cubes are:

4x4x4: 477
5x5x5: 3799
6x6x6: 33103
7x7x7: 264711
8x8x8: 2315415
9x9x9: 18522655
10x10x10: 162068479
11x11x11: 1296543903
100x100x100: 17342341482671474270981274712635904439208472500000000101035159683095507580392731532043747327
or approximtely 1.734*10^91

Note that the solved case is not counted. My derivation is given in the spoiler below.



Spoiler



(Note, I will use curly braces to represent vectors with specific numeric contents, and also for indicating subscripting on vectors, as square brackets may have special interpretation in this forum.)

Let C be a vector giving the number of raw possible corner OLL cases for symmetry equivalence classes of size 4, 2, and 1, respectively. We include the solved case. We have C = {24,2,1};

Let W be a similar vector giving the number of raw possible wing edge OLL cases for a single orbital. We have W = {64,4,2};

Let M be a similar vector giving the number of raw possible middle edge OLL cases. We have M = {4,2,2};

Let's assume we have an OLL case for one orbital with n-fold symmetry, and another OLL case for a different orbital with m-fold symmetry. (n and m have possible values of 1, 2, or 4, since only values that are factors of 4 are possible.) If we combined these two, the overall symmetry (I believe) will be the smaller of m and n. (This rule does not hold for more general symmetry considerations, but as we are dealing with just the rotational symmetries of the square, I think this works.) Note, if a case has n-fold symmetry, there will be 4/n elements in its symmetry equivalence class.

Let's assume v is a vector for some nxnxn cube, defined in a similar way as other vectors above. If we add a new orbital of pieces to this nxnxn cube, and this new orbital is represented by a similar vector x, then the number of cases for the new cube will be represented by a vector y, where the components of y are calculated using the following formulas.

y{1} = v{1}*x{1} + v{1}*x{2} + v{1}*x{3} + v{2}*x{1} + v{3}*x{1}

y{2} = v{2}*x{2} + v{2}*x{3} + v{3}*x{2}

y{3} = v{3}*x{3}

Let's call this operation y = Combine(v,x). This formula reflects how various symmetry equivalence class sizes are affected when combining orbitals, as mentioned above. It also assumes no parity restrictions are involved.

Now the 2x2x2 cube is just corners, so it is represented by the vector C.
If we add a wing edge orbital to make a 4x4x4, we get Combine(C,W) = Combine({24,2,1},{64,4,2}) = {1872,16,2}.

To get the number of reduced cases for the 4x4x4 we calculate 1872/4 + 16/2 + 2/1 = 478. If we want to exclude the solved case, we simply need to subtract one to get 477.

To get the number for the 6x6x6, we simply combine the 4x4x4 vector with the vector W. This gives Combine({1872,16,2},{64,4,2}) = {132192,104,4}. The number of reduced cases excluding solved case is 132192/4 + 104/2 + 4/1 - 1 = 33103.

Likewise, for 8x8x8, we get Combine({132192,104,4},{64,4,2}) = {9260352,640,8}. Thus, the value is 9260352/4 + 640/2 + 8/1 - 1 = 2315415.

For odd size cubes, we simply perform the combine operation with the next smaller even cube and M, the vector for the orbital of middle edges.

For 3x3x3, we get Combine({24,2,1},{4,2,2}) = {204,10,2}. 204/4 + 10/2 + 2/1 - 1 = 57.

For 5x5x5, we get Combine({1872,16,2},{4,2,2}) = {15048,68,4}. 15048/4 + 68/2 + 4/1 - 1 = 3799.

For 7x7x7, we get Combine ({132192,104,4},{4,2,2}) = {1057968,424,8}. 1057968/4 + 424/2 + 8/1 - 1 = 264711.


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## vcuber13 (Jun 20, 2011)

vcuber13 said:


> do you mean 4x4 ells? wouldnt it be 8!*4=161,280


 
This makes no sense


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## Christopher Mowla (Jun 24, 2011)

*The Formula for the Number of OLLs for All Cube Sizes*

Thanks Bruce. I have looked at your explanation, and I agree that you are indeed correct.




cuBerBruce said:


> It may be possible to derive a strict formula from the recurrence relation.


It turns out that such a formula exists. Using your y vector function, I made it. It works for all cube sizes (2x2x2 to nxnxn).




​ 
Here's a link to it.

2x2x2 , 3x3x3 ,4x4x4 ,5x5x5, 6x6x6, 7x7x7, 8x8x8 ,9x9x9 ,10x10x10 ,11x11x11 ,100x100x100 ,1000x1000x1000



Spoiler



The formula piece for even cubes is:


\( \frac{2^{\frac{n}{2}}\left( 27\times 35^{\frac{n}{2}}+35\left( 3^{\frac{n}{2}}+2 \right) \right)}{280} \)​ 
, and the formula for odd cubes is

\( \frac{2^{\left\lfloor \frac{n}{2} \right\rfloor }\left( 54\times 35^{\left\lfloor \frac{n}{2} \right\rfloor }+35\left( 3^{\left\lfloor \frac{n}{2} \right\rfloor }+1 \right) \right)}{70} \)​ [FONT=&quot]
(We subtract 1 from each to get the number of unsolved cases).

[/FONT]It's very interesting that the odd and even cube formulas turned out to be nearly identical! In fact, since that's true, the two can be merged as one. This gives the formula above. 




In addition, I have made a document showing all of the work for Bruce's calculations from his previous post. It also illustrates the raw cases for corners, middle edges, and wings. This should hopefully help everyone understand what Bruce did.


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## Christopher Mowla (Jun 24, 2011)

I have made another document which requires either having understood cuBerBruce's previous post and/or the calculations portion of my "Calculating_Big_Cube_Wing_Edge_OLLs.pdf" document. It is the full derivation of the formula for the number of OLLs for all cube sizes.

View attachment The_Formula_for_the_Number_of_OLLs_for_All_Cube_Sizes.pdf​


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