#### WoowyBaby

##### Member
It is by no means short, but I did find this one a little while back.
[2R2 B2 D2 U2 F2 2R2 D2 F2 2L U2 2R2 F2 2L' U2 2R U2 2R', U2 B2 2R' B2 U2 2R U2 2R' B2 2R2 B2 U2 2R2] 2R

I was just searching using Cube Explorer (manually). I know the question was just to solve the first face, not necessarily the first layer. But even with solving the entire face, I couldn't find a position (with about 50 tries) which required more than 11. You mentioned 13 moves. Do you know of a position which requires 12 or 13?
No I don’t, I just made a wild guess. Good work going through fifty tries. I place my next guess on gods number being either 11 or 12. We’ll see though, I don’t have any proof.

#### xyzzy

##### Member
what is the god's number for one side on a 3x3 cube? i mean side, not layer
(@WoowyBaby)

Distribution for solving a specific face:
Code:
0	1
1	12
2	150
3	1886
4	21916
5	242166
6	2292695
7	14228012
8	25293406
9	2825994
10	162

Restricted to the domino-solvable states (but allowing non-domino moves):
Code:
0	1
1	4
2	22
3	82
4	300
5	1278
6	2305
7	892
8	16

Restricted to the domino-solvable states and domino moves only:
Code:
0	1
1	4
2	22
3	82
4	292
5	986
6	2001
7	1312
8	200

Christopher's 3-flip example can actually be solved in 8 moves: F D B' D2 R' F' U R2 (note that this does rearrange some of the edge pieces)

PS: Hey, @Christopher Mowla, you don't have to treat everything I say as some kind of a personal attack on you. I was just trying to provide some additional information you or another reader might find interesting. What the hell, man.

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#### Filipe Teixeira

##### Member
what is the OO CF method?

#### GodCubing

##### Member
How is the 2x2 solved by a computer and what is God's Number on 2x2

#### MJbaka

##### Member
How many possible permutations are there of the U Layer Corners and the DFR and DBR corners all together, if all of them are all oriented?

#### MJbaka

##### Member
Nevermind I realized it is 47

#### CubeRed

##### Member
I could have done some research but what are the chances of getting pregnant on 4x4 and can you predict it?

EDIT : OMG I MEANT PARITY NOT PREGNANT I HATE AUTOCORRECT

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#### Thom S.

##### Member
I could have done some research but what are the chances of getting pregnant on 4x4 and can you predict it?
I'm gonna assume you mean parity because last time I've gotten pregnant on 4x4 is something I'd rather not talk about.
OLL Parity has a 1/2 chance. There are methods to recognise parity, they are hard. If you are good at ZZ you can of course before doing 3x3 stage count the bad edges and act accordingly but I think noone does this.
Lucas Parity is quite nice I think.

#### Thom S.

##### Member
What is Lucas parity?
The (former) most commonly used algorithm for OLL Parity. You know it, the Rw U2 x Rw U2 one. The Wiki should even have it listed as Lucas Parity somewhere

#### xyzzy

##### Member
The (former) most commonly used algorithm for OLL Parity.
Did it get dethroned by some new zoomer alg?

#### Thom S.

##### Member
Did it get dethroned by some new zoomer alg?
I am pretty sure but I just can't remember which one that would be.

#### xyzzy

##### Member
I am pretty sure but I just can't remember which one that would be.
A quick search on YT (filtered to the year) didn't find anything about a new parity alg (it's just the same few ones; Lucas parity still the most popular), but I also found this funny video, so at least it wasn't a waste of time.

#### Labano

##### Member
With CFOP how many moves does it take to solve on QTM and HTM?

#### Christopher Mowla

With CFOP how many moves does it take to solve on QTM and HTM?
The table on page 41 of this PDF (you can read a little of the background of the author from this post) is showing the mean move counts which were the results of 10,000 solves using different variants of Fridrich (CFOP).

The first sentence of the bottom paragraph on page 8 reads:
Joseph Miller said:
Throughout our treatment, a move is by definition a 90- or 180-degree face-turn of the puzzle, i.e. we use the half-turn metric (HTM) to define the length of a sequence of moves.

So it is about HTM only.

You probably don't know what all the acronyms stand for in the table on page 41 (unless you skim through the text), but I will help out there.

You can see in Figure 1.1 on page 11 that F1-F7 refer to the 7 steps of Fridrich (CFOP):
F1=cross
F2-F5 are each one F3L slot
F6 = OLL
F7 = PLL

Also, here are the meaning to some of the other acronyms:
BPC (best-possible cancellation)
CC (commutative cancellation)
NC (no cancellation)
SC (Simple Cancellation)

So from the results shown in the table, cancellations didn't come up often enough in arbitrary solves to make that much of a difference between solves in which there were no cancellations.

But to answer the question (regarding HTM), it seems that the min from the table is a mean of 42.7896 HTM and a max of 53.6325 HTM. (It all depends on the variants used, etc.)

I think it would mean the most to anyone asking this question what the mean would be for their specific variant as well as the algorithms that they prefer to use for that variant. (That's the only way you would know.) Maybe someone can write software for this?

#### Labano

##### Member
The table on page 41 of this PDF (you can read a little of the background of the author from this post) is showing the mean move counts which were the results of 10,000 solves using different variants of Fridrich (CFOP).

The first sentence of the bottom paragraph on page 8 reads:

So it is about HTM only.

You probably don't know what all the acronyms stand for in the table on page 41 (unless you skim through the text), but I will help out there.

You can see in Figure 1.1 on page 11 that F1-F7 refer to the 7 steps of Fridrich (CFOP):
F1=cross
F2-F5 are each one F3L slot
F6 = OLL
F7 = PLL

Also, here are the meaning to some of the other acronyms:
BPC (best-possible cancellation)
CC (commutative cancellation)
NC (no cancellation)
SC (Simple Cancellation)

So from the results shown in the table, cancellations didn't come up often enough in arbitrary solves to make that much of a difference between solves in which there were no cancellations.

But to answer the question (regarding HTM), it seems that the min from the table is a mean of 42.7896 HTM and a max of 53.6325 HTM. (It all depends on the variants used, etc.)

I think it would mean the most to anyone asking this question what the mean would be for their specific variant as well as the algorithms that they prefer to use for that variant. (That's the only way you would know.) Maybe someone can write software for this?
I actually didn't understand can you simplify it.