#### Christopher Mowla

##### Premium Member

The title is a little general, but it should be, as I have a lot to share about forming several types of 2-cycle inner-layer odd permutation algorithms.

To begin, why not start with the "Pure Edge Flip" (my favorite, obviously).

I would first like to show you how to derive common OLL parity algorithms by hand. To do this, I have taken the time to make tutorial videos for three popular algorithms. Each of the three algorithm's derivations are two videos long.

r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2

Derivation

Part I

Part II

r U2 r U2 r' U2 r U2 l' U2 l x U2 r' U2 x' r' U2 r'

Derivation

Part I

Part II

r U2 r U2 r' U2 r U2 l' U2 r U2 r' U2 l r2 U2 r'

Derivation

Part I

Part II

Please watch the videos before reading this spoiler:
Spoiler

To begin, why not start with the "Pure Edge Flip" (my favorite, obviously).

I would first like to show you how to derive common OLL parity algorithms by hand. To do this, I have taken the time to make tutorial videos for three popular algorithms. Each of the three algorithm's derivations are two videos long.

*Note: I recommend you to watch the derivation on the Standard Algorithm first before you watch the other videos.***The Standard Algorithm**r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2

Derivation

Part I

Part II

**Lucasparity**r U2 r U2 r' U2 r U2 l' U2 l x U2 r' U2 x' r' U2 r'

Derivation

Part I

Part II

**Another algorithm which consists of all U2 face turns**r U2 r U2 r' U2 r U2 l' U2 r U2 r' U2 l r2 U2 r'

Derivation

Part I

Part II

Please watch the videos before reading this spoiler:

Notice that all three algorithms are derived from the same commutator. Each algorithm is very related in structure. Only a few minor internal adjustments differentiates Lucasparity and the other algorithm from the standard algorithm.

In addition, I have chosen to release...

**God's Algorithm?**

For my main method for pure edge flip algorithms, I have made a video on the best/briefest algorithm I have ever found in block quarter turns (BQTM) that works for all cube sizes (Very few low move count algorithms work for all cube sizes).

The Holy Grail

*From now on, I will abbreviate BQTM with just q, and BHTM with h.*

(Also note that BHTM is commonly called btm. So h = BHTM = btm).

On the 4X4X4: 19q/18h

z d' m D l' u' 2R' u l u' l'2 b' 2R' b r' R' 2U y' m' u x2 z'

On the 5X5X5: 19q/18h

z d' 3m D 3l' u' 2R' u 3l u' 3l'2 b' 2R' b r' R' 2-3u y' 3m' u x2 z'

On the inner-orbit of the 6X6X6: 20q/19h

z 3d' 4m D 3l' 3u' 3R' 3u 3l 3u' 3l'2 2R' 3b' 3R' 3b 3r' R' 2-3u y' 4m' 3u x2 z'

On the outer-orbit of the 6X6X6: 19q/19h

z 3d' 4m D 3l' 3u' 2R' 3u 3l 3u' 3l'2 3R' 3b' 2R' 3b 3r' R' 2-3u y' 4m' 3u x2 z'

=

z 3d' 4m D 3l' 3u' 2R' 3u 3l 3u' 4l 3l 3b' 2R' 3b 3r' R' 2-3u y' 4m' 3u x2 z'

On the inner-orbit of the 7X7X7: 20q/19h

3d' 5m D 4l' 3u' 3R' 3u 4l 3u' 4l'2 2R' 3b' 3R' 3b 3r' R' 2-4u y' 5m' 3u x2 z'

On the outer-orbit of the 7X7X7: 19q/19h

z 3d' 5m D 4l' 3u' 2R' 3u 4l 3u' 4l'2 3R' 3b' 2R' 3b 3r' R' 2-4u y' 5m' 3u x2 z'

=

z 3d' 5m D 4l' 3u' 2R' 3u 4l 3u' 5l 4l 3b' 2R' 3b 3r' R' 2-4u y' 5m' 3u x2 z'

I point out in the latter portion in the video that its average between quarter and half turns is also less than all other algorithms which currently exist.

Here is a link to the following formula in Wolfram|Alpha (the simplified formula shown near the end of the video)

\( \frac{\left\lfloor \frac{n}{2} \right\rfloor }{2-2^{\left\lfloor \frac{n}{2} \right\rfloor }}+19.5 \)

Just substitute an integer greater than or equal to 4 for n to obtain the average for a cube of size

*n.*

**Formula Derivation**

In order to explain it, I will explain the original form of the formula:

\( 19.0\left( \left\lfloor \frac{n-2}{2} \right\rfloor -1 \right) \)

This represents the number of cases where there are consecutive orbits which have odd parity in the same composite edge, starting from the corner and working its way inward, but excluding when all orbits are involved (previous case).

These cases are all 19q/19h thus the average is 19.0.

This represents all other cases:

, etc.

These cases are all (proportionally) 20q/19h thus the average is 19.5.

Now, I am going to explain the series.
Spoiler

Spoiler
So our formula simplifies to:

[FONT="]
[/FONT]

[FONT="]

\( 18.5\left( 1 \right) \)

**The First Portion**[/FONT]\( 18.5\left( 1 \right) \)

This represents the case when any big cube size has a 2-wing edge swap between every orbit in the same composite edge. For example, any even cube size reduced to a 4x4x4 form, and any odd cube size reduced to a 5x5x5 form.

This case is 19q/18h thus the average is 18.5.

This case is 19q/18h thus the average is 18.5.

**The Second Portion**

\( 19.0\left( \left\lfloor \frac{n-2}{2} \right\rfloor -1 \right) \)

This represents the number of cases where there are consecutive orbits which have odd parity in the same composite edge, starting from the corner and working its way inward, but excluding when all orbits are involved (previous case).

These cases are all 19q/19h thus the average is 19.0.

**The Third Portion**

This represents all other cases:

, etc.

These cases are all (proportionally) 20q/19h thus the average is 19.5.

It is the total number of cases of odd parity in the orbits of a big cube, ,minus the occurrences of the first two cases.

Next, we divide everything by the total number of possible cases for odd parity in all inner-layer orbits (that same series) to get the overall average.

Now, I am going to explain the series.

Let the total number of inner-layer orbits of wing edges in a big cube \( n\ge 4 \) be \( N=\left\lfloor \frac{n-2}{2} \right\rfloor \)

Then, the total number of cases of odd parity between the orbits is the sum of the combinations of the number of orbits involved:

\( \left( \begin{matrix}

N \\

N \\

\end{matrix} \right)+\left( \begin{matrix}

N \\

N-1 \\

\end{matrix} \right)+\left( \begin{matrix}

N \\

N-2 \\

\end{matrix} \right)+\cdot \cdot \cdot +\left( \begin{matrix}

N \\

N-\left( N-1 \right) \\

\end{matrix} \right) \)

\( =\left( \begin{matrix}

N \\

N \\

\end{matrix} \right)+\left( \begin{matrix}

N \\

N-1 \\

\end{matrix} \right)+\left( \begin{matrix}

N \\

N-2 \\

\end{matrix} \right)+\cdot \cdot \cdot +\left( \begin{matrix}

N \\

1 \\

\end{matrix} \right) \)

\( =\sum\limits_{k=0}^{N-1}{\left( \begin{matrix}

N \\

N-k \\

\end{matrix} \right)} \) \( =\sum\limits_{k=0}^{\left\lfloor \frac{n-2}{2} \right\rfloor -1}{\left( \begin{matrix}

\left\lfloor \frac{n-2}{2} \right\rfloor \\

\left\lfloor \frac{n-2}{2} \right\rfloor -k \\

\end{matrix} \right)} \)

For example, the 10X10X10 has 4 inner-layer orbits.

\( \left( \begin{matrix}

N \\

N \\

\end{matrix} \right) \) or \( \left( \begin{matrix}

4\\

4 \\

\end{matrix} \right) \) represents the case

because we have 4 orbits taking all 4 at once.

\( \left( \begin{matrix}

N \\

N-1 \\

\end{matrix} \right) \) or \( \left( \begin{matrix}

4\\

3\\

\end{matrix} \right) \) represents the cases:

, etc.

because we have 4 orbits taking 3 at once.

\( \left( \begin{matrix}

N \\

N-2 \\

\end{matrix} \right) \) or \( \left( \begin{matrix}

4 \\

2 \\

\end{matrix} \right) \) represents the cases:

,etc.

because we have 4 orbits taking 2 at once.

and finally,

\( \left( \begin{matrix}

N \\

N-3 \\

\end{matrix} \right)=\left( \begin{matrix}

N \\

N-\left( N-1 \right) \\

\end{matrix} \right)=\left( \begin{matrix}

4 \\

4-\left( 4-1 \right) \\

\end{matrix} \right)=\left( \begin{matrix}

4 \\

1 \\

\end{matrix} \right) \) represents the cases

,etc.

because we have 4 orbits, taking only one at a time.

Then, the total number of cases of odd parity between the orbits is the sum of the combinations of the number of orbits involved:

\( \left( \begin{matrix}

N \\

N \\

\end{matrix} \right)+\left( \begin{matrix}

N \\

N-1 \\

\end{matrix} \right)+\left( \begin{matrix}

N \\

N-2 \\

\end{matrix} \right)+\cdot \cdot \cdot +\left( \begin{matrix}

N \\

N-\left( N-1 \right) \\

\end{matrix} \right) \)

\( =\left( \begin{matrix}

N \\

N \\

\end{matrix} \right)+\left( \begin{matrix}

N \\

N-1 \\

\end{matrix} \right)+\left( \begin{matrix}

N \\

N-2 \\

\end{matrix} \right)+\cdot \cdot \cdot +\left( \begin{matrix}

N \\

1 \\

\end{matrix} \right) \)

\( =\sum\limits_{k=0}^{N-1}{\left( \begin{matrix}

N \\

N-k \\

\end{matrix} \right)} \) \( =\sum\limits_{k=0}^{\left\lfloor \frac{n-2}{2} \right\rfloor -1}{\left( \begin{matrix}

\left\lfloor \frac{n-2}{2} \right\rfloor \\

\left\lfloor \frac{n-2}{2} \right\rfloor -k \\

\end{matrix} \right)} \)

For example, the 10X10X10 has 4 inner-layer orbits.

\( \left( \begin{matrix}

N \\

N \\

\end{matrix} \right) \) or \( \left( \begin{matrix}

4\\

4 \\

\end{matrix} \right) \) represents the case

because we have 4 orbits taking all 4 at once.

\( \left( \begin{matrix}

N \\

N-1 \\

\end{matrix} \right) \) or \( \left( \begin{matrix}

4\\

3\\

\end{matrix} \right) \) represents the cases:

, etc.

because we have 4 orbits taking 3 at once.

\( \left( \begin{matrix}

N \\

N-2 \\

\end{matrix} \right) \) or \( \left( \begin{matrix}

4 \\

2 \\

\end{matrix} \right) \) represents the cases:

,etc.

because we have 4 orbits taking 2 at once.

and finally,

\( \left( \begin{matrix}

N \\

N-3 \\

\end{matrix} \right)=\left( \begin{matrix}

N \\

N-\left( N-1 \right) \\

\end{matrix} \right)=\left( \begin{matrix}

4 \\

4-\left( 4-1 \right) \\

\end{matrix} \right)=\left( \begin{matrix}

4 \\

1 \\

\end{matrix} \right) \) represents the cases

,etc.

because we have 4 orbits, taking only one at a time.

**PROOF**

As you can see, there are many more cases for an average of 19.5 than for 18.5 and 19.0.

But, the overall average still is less than 19.5.

Based on the fact that even and odd cubes have the same number of orbits (e.g. both the 6X6X6 and 7X7X7 have two inner-layer orbits), the floor function can be omitted and we can take the limit as the cube size gets large.

\( \underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{n}{2}}{2-2^{\frac{n}{2}}}+19.5=19.5 \)

This means, no matter how large

*n*gets, the average will come arbitrarily close to, but it never reaches 19.50. (This is apparent already without calculus, but I thought a "second opinion" would further verify this statement.)

My 23q/16h "cmowlaparity" obviously has an average of 19.5, but the Holy Grail beats it slightly.

x' r2 U2 l' U2 r U2 l x U2 x' U r U' F2 U r' U r2 x

Last edited: Jun 9, 2011