# Odd parity Algorithms (specifically, single edge "flip")

##### Member
(Rr)' U R U
[(Rr)' U2] * 3
(Rr)2 U R' U' (Rr)2
U' R' U (Rr)'
This alg is awesome. Finally I have an excuse to use Petrus for speedsolving the 4x4x4.
Is it a good way to scramble the cube, or something I don't get ?

EDIT: OK, say no more, I get it. But how the hell did you find this alg ?

Last edited:

#### Christopher Mowla

(Adding R2 U2 R' U2 could qualify, though.)
Instead of that, it might be better to add F' L' U' L F U R2 U R' :

(Rr)' U R U
[(Rr)' U2] * 3
(Rr)2 U R' U' (Rr)2
U' R' U (Rr)'
F' L' U' L F U R2 U R'

from there, all you need to do is perform one corner cycle (with the single incorrect corner in the bottom layer and the two incorrect corners in the top layer) and orient the left and back composite edges.

In summary,
(Rr)' U R U
[(Rr)' U2] * 3
(Rr)2 U R' U' (Rr)2
U' R' U (Rr)'
F' L' U' L F U R2 U R'
[1 3 corner cycle]
[orient left and back composite edges]

This procedure will swap Front composite edge <-->right composite edge and swap the winged edges of the original right composite edge (a double parity fix between the front and right edges, where the original right edge appears to be flipped).

Last edited:

#### Ron

##### Member
(Rr)' U R U
[(Rr)' U2] * 3
(Rr)2 U R' U' (Rr)2
U' R' U (Rr)'

R2 U2 R U R2 U2 R2 U R2

#### Christopher Mowla

(Rr)' U R U
[(Rr)' U2] * 3
(Rr)2 U R' U' (Rr)2
U' R' U (Rr)'

R2 U2 R U R2 U2 R2 U R2
That's even better. How did you find that?

#### Christopher Mowla

Dan Cohen said he used ACube to search for all algs in <r,l,U2,B2,D2,F2> and that there were none in fewer than 25 quarter turns. Theoretically there might be a parity alg shorter than that, but I don't think we'll find one with the conventional approach.
You are right about the theory, qqwref.

For those interested in God's algorithm for big cubes, I have just lowered the move lower bound for the "one edge flip" case (again). At the beginning of the thread, qqwref mentioned the theory that there may exist "one edge flip algorithms" shorter than 25q but not using the conventional approach. The three algorithms below are (as far as I know) the shortest found (and publically known) using that approach.

r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2 (25 q, 15 h)
r' U2 l F2 l' F2 r2 U2 r U2 r' U2 F2 r2 F2 (25 q, 15 h)
r2 B2 r' U2 r U2 F2 r F2 l' U2 l U2 B2 r2 (25 q, 15 h)

I now have just found a 23q pure edge flip algorithm (by hand of course). I previously found a 24q algorithm on August 17, 2009. Since then I have found several more 24qs. I didn't think that there could be a shorter algorithm (in quarter turn moves) than 24, but I proved myself wrong just a few days ago (December 13, 2009).

But as for half turn moves, I still haven't found anything below 15h: the lowest I have come has been 18h. The computers could be right on that one (using the conventional approach).

I am aiming for 21q because I can see based from how I found my 23q that the constraints are strict, but there still is a (great) possibility for less than 23q. I guess only the future will tell (whether I find it or someone else does).

Again, I am mentioning this comment for those interested in the theory--a 23q has been found. For speedcubing purposes, you probably would prefer to execute already known algorithms (25-27q), not the ones I have found.

Last edited:

#### Kenneth

##### Not Alot
If you have an alg that is 25 q to solve a case, then it is impossible to have an alg of 24 q to solve the same case...

Parity comes from a odd number of q-turns, if it is even it is not a parity. So, 22 is not possible either.

I have no idea how your 24 q can solve this case, did you ignore a ending U turn?

#### Christopher Mowla

I have no idea how your 24 q can solve this case, did you ignore a ending U turn?

No, I did not. The 24q and moreover, the 23q, are complete algorithms. They correct the "one edge flip" to a completely solved cube state.

Edit:
It is a privilege to hear from you first. I have read your occupation in your public profile. You and I appear to have the same passion (finding algorithms).

Last edited:

#### Kenneth

##### Not Alot
I have no idea how your 24 q can solve this case, did you ignore a ending U turn?

No, I did not. The 24q and moreover, the 23q, are complete algorithms. They correct the "one edge flip" to a completely solved cube state.

Edit:
It is a privilege to hear from you first. I have read your occupation in your public profile. You and I appear to have the same passion (finding algorithms).
Hehe, well, that was like 2 years ago, then I found you can setup the same cases but for corners on a 3x3 and use a solver to do the work for me, then I lost intrest...

If you claim you got an alg of 24 q then I'm sure you do not know what you are talking about and/or don't know how to count your turns.

All algs that solves this case MUST have a uneven number of quarterturns, there are no hidden secrets we do not know about yet. That's why i'm sure you are wrong somehow...

#### Stefan

##### Member
Kenneth: I'd count R r Rw' as three quarter turns.

#### Christopher Mowla

All algs that solves this case MUST have a uneven number of quarterturns, there are no hidden secrets we do not know about yet. That's why i'm sure you are wrong somehow...
No. I am not miscounting my moves. In order to correct/induce an inner-layer odd permutation (of pure form), an algorithm must have the sum of the inner-layer turns for a pair of complimentary layers to be odd and the sum of the outer-layer moves to be even.

The odd number of inner layer turns induces/fixes the inner layer odd permutation.
The even number of outer layer turns preserves the permutation of the outer layer.

My algorithms satisfy this requirement.

Edit:
Let me ask you a question. Maybe I am wrong on counting my moves.
If you don't count the following as 10q, then, I am counting wrong: (Rr) B r U L D S E F2.

Otherwise, if I am not mistaken, every possible individual move on a 4X4X4 (or NxNxN) is considered a quarter turn and, multiple layer turns which are parallel, touching, and being turned in the same direction one time are considered one quarter turn move.

Edit:
And of course, this can always be proven with move equivalences. For example, F=([all back layers]' z).
Edit:
If what I just mentioned about multiple layer turns is not true, then a 25q algorithm for the “one edge flip” on a 5X5X5 cube would not be 25q anymore on a 7X7X7 cube (if 2 pairs of winged edges needed to be permuted): there would be a whole lot more quarter turn moves. And, using the 7X7X7 as an example, as I just mentioned, if all the winged edges on a single composite edge needed to be fixed on a 7X7X7 and you chose to perform the usual algorithm:
r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2
then, the r and l turns would be touching each other and thus, the procedure would still be considered 25q.

What about if you are solving a 20X20X20 and you have 3 non-touching sets of complimentary winged edges which needed to be swapped. According to my move interpretation, then the move count would be more than 25q because they are not touching. If it is still considered 25q just because the layers are parallel and being turned in the same direction, then my 23 alg is even less moves--but this is not how I count quarter turn moves: I believe they must be touching (and this way of counting can be proven using move equivalences yet again).

In summary, here is how I count quarter turn moves. If I am wrong here, then I am wrong about all of this. If this is correct, then my claim is legitimate (I assure you).

On a 4X4X4,
(Rr): 1q
Fb': 2q
(Ffb'): 1q
FS: 1q
FS':2q
L R': 1q

On a 5X5X5 (when I say M' , I just mean the central slice)
L' M' R: 2qßbecause I can do l r' x instead.
l'M' 1q
etc…

Last edited:

#### Kenneth

##### Not Alot
"If you don't count the following as 10q, then, I am counting wrong: (Rr) B r U L D S E F2."

Really depends, if you count slices as quarter turns, then it is ok but then specify it is in slice metric because I would count that as 13 q-turns (face turns and wide face turns).

#### Christopher Mowla

So in other words, in quarter turn moves,

On 4X4X4,
F B' =2q
S=2q
(Rr)=2q
M=2q
(Ffb')=1q

and on 5X5X5,
l'M'=2q
(l'M'r)=2q

But in slice metric, they are:
On 4X4X4,
F B' =1q
S=1q
(Rr)=1q
M=1q
(Ffb')=1q

and on 5X5X5,
l'M'=1q
(l'M'r)=1q

Am I understanding you correctly?

Edit:
For cubes greater than the 3X3X3, why would a quarter turn move refer to the same thing as on a 3X3X3? At least M, E, and S should be one quarter turn move just like on a 3X3X3, right? Why are they considered multiple quarter turns? This just doesn't make sense to me. I know that many people use a 3X3X3 solver to find the shortest algorithms for big cubes, but why does that make the moves for big cubes considered to be as 3X3X3 moves?
Anyway, according to what you regard as moves, my algorithms are still 24q and 23q, but with slice metric.

Even Stefan counts M, E, and S as quarter turn moves (see his algorithms which he gave earlier in this thread).

Last edited:

#### Kenneth

##### Not Alot
There are for mayor metrics

QTM, single face quarter turns
HTM, single face half turns
SQTM, slice quarter turns
STM, slice turn metric

If you do not specify I (and most people) assumes you are talking about face turns. This because the "normal" metric is HTM.

If you use a M, S or a E in FMC it is always counted as two turns because there turns are counted in the normal metric...

#### Christopher Mowla

There are for mayor metrics
If you use a M, S or a E in FMC it is always counted as two turns because there turns are counted in the normal metric...
Well, for my algorithms, I don't have to use M, S, or E, but how about
(Uu)--on the 4X4X4 or 5X5X5; or (UuE')--on the 5X5X5, where E is just the central row....you can you just send me a link with the specifications if you want.

Last edited:

#### qqwref

##### Member
There are for mayor metrics

QTM, single face quarter turns
HTM, single face half turns
SQTM, slice quarter turns
STM, slice turn metric

If you do not specify I (and most people) assumes you are talking about face turns. This because the "normal" metric is HTM.

If you use a M, S or a E in FMC it is always counted as two turns because there turns are counted in the normal metric...
For cubes greater than 3x3, this post is wrong and completely nonsensical.

Apart from the distinction between half and quarter turn metrics (since every metric can be considered with either quarter or half turns), there are three useful metrics on larger cubes:
- block turn metric, where any group of contiguous slices moving the same way is counted as one move
- layer turn metric, where the only allowed moves are moves of a single slice layer
- multislice turn metric, where the only allowed moves are moves that move slices 1 through n on a given face in the same way
I believe what you are using, cmowla, is the block quarter turn metric. I prefer the block (half) turn metric for pattern-making because it is the least restrictive without adding moves that don't feel like moves. It reduces to STM/SQTM on 3x3, as does the layer turn metric; the multislice metric reduces to HTM/QTM on 3x3.

#### Christopher Mowla

I believe what you are using, cmowla, is the block quarter turn metric.
Your explanation makes a lot more sense, qqwref.

According to how you described the block quarter turn metric, my alg is 23q. According to the block quarter turn metric, I believe the usuall algs that I have listed in this thread (which use the conventinal method) are still 25q.

In cubetwister, my alg is 28qtm, while the usuall algs that I have listed already in this thread (which use the conventional method) are 34qtm.

I can understand why you like block half turn metric, qqwref, but I personally (probably most people don't) feel that a half turn is 2 moves (I guess I like leaning on the math more). Thus, I lean toward block quarter turn metric.

Based on my 24q (in block turn metric) algorithm, I don't believe one can find an algorithm that is less than 15h, as the algorithms found by the computer have. And, I also must say that I am 100% positive that 23q is the optimal algorithm for block quarter turn moves (and my 23q alg happens to be the least in qtm on cube twister--one of my 24s is as well). However, 15h is definitely optimal for block half turn moves.

In short, for the "pure edge flip algorithm", the optimal algorithms, according to my findings, are:

For block half turn moves, 15 (computer-generated algorithms with conventinal approach).
For block quarter turn moves, 23 (my alg).
For qtm, according to cube twister, 28 (my alg).

Edit:
I just found another 23 block quarter turn alg, and it is 26qtm, according to cube twister.

Last edited: