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Smiles

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So I was thinking for a CFOP who likes m slices. I think that you could do F2B, COLL/CMLL, fix cross, ELL. It is not a new method and I'm sure A LOT of you have found this out or experimented to find this out but I think it was cool for me to find it out on my own. I believe I've gotten sub 25 without practice on the first try.
isn't that just roux but having LSE broken up into D-layer and U-layer?
im pretty sure there's no way it can compare to roux.
 

elrog

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Yes this has been thought of before and it is a varient of Roux.

im pretty sure there's no way it can compare to roux.
How so? The move count is no greater and neither is the recognition or turn speed? The only thing that is greater is the alg number because L5E is done intuitively except for some certain cases.
 

Smiles

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i was thinking that cause LSE in roux solves 2 edges, then solves the last 4 edges.
this LSE variant solves 2 edges, then takes at least one of them out during ELL.
it would also be harder to maintain a constant look ahead like Roux's LSE, since ELL requires its own recognition.

and im not here to argue about movecount and potential in this variant based on learning different things and doing it differently, im just pointing out that Roux's LSE is better.
 

alevine

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FTP method (or PTF)

The FTP method is good for cubers transitioning from Fridrich to Petrus, as it incorporates many elements from Fridrich, but the F2L is solved in a slightly different way. FTP stands for Fridrich to Petrus, BTW. Basically, you do two adjacent cross colors, and then insert the F2L pair between them, making a 2x2 block as in Petrus. Then you solve another adjacent cross color, and solve the F2L pair in between that. Then solve the last color, and solve the last parts of the F2L Fridrich style. Then it's just OLL/PLL as in Fridrich. Also there's the PTF method (Petrus to Fridrich), which solves Cross/F2L as in Fridrich, but a Petrus-style LL (don't know this, I know the basic steps of Petrus, but not all the algs.) I don't even use Petrus myself.
 

cuBerBruce

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Also there's the PTF method (Petrus to Fridrich), which solves Cross/F2L as in Fridrich, but a Petrus-style LL (don't know this, I know the basic steps of Petrus, but not all the algs.) I don't even use Petrus myself.
Petrus "last layer" assumes edges have already been oriented. Thus, you would seem to need to insert an orient last layer edges step in order to finish the LL Petrus-style. One could use techniques for influencing the LL during F2L, but since this method is apparently promoted as being for someone just learning CFOP F2L, inserting the EOLL step seems to be the way the person would get to the Petrus-style finish.
 

5BLD

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i was thinking that cause LSE in roux solves 2 edges, then solves the last 4 edges.
this LSE variant solves 2 edges, then takes at least one of them out during ELL.
it would also be harder to maintain a constant look ahead like Roux's LSE, since ELL requires its own recognition.

and im not here to argue about movecount and potential in this variant based on learning different things and doing it differently, im just pointing out that Roux's LSE is better.
No thats pretty bad. Anyway 2.5 look lse is great if you use it loosely.
 

ben1996123

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1.Solve the corners like a 2x2.
2. bring 2 edges with an alg to make half a face (i calculated it to be 20 algs, although[/I]may be wrong.(The edges don't have to permuted correctly.)
3. Repeat step 2. to make a face.
4. Bring 1 f2l edge into its correct position while preserving the corners and permuting the first layer edges(? algs.)
5. ring two f2l edges into their correct position simultaneously with one alg(? algs.)
6. ring the last f2l edge into its correct position while influencing ell(if that's possible.)
7. ell (with less algs, better recognition, etc)


no its not
 

Benyó

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please before post your ideas, just practice it a bit and if it works or at least not totally stupid, THEN post it.
noone wants to hear as *****ic methods as 'build 2x2 blocks in a 5x5' or 'i have made an already slow method much more complicated in order to create a method named after me'
 

elrog

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I have had much time for cubing lately, but when I do, I've been trying to solve the 3x3x3 cube using different move sets. I can already do M U, R U, 180 degree turns, slice moves, always having to turn opposite sides in oppossite directions at the same time, and some others. I also do not break the move set at any time hile solving it such as doing U2 M U2 M' while doing 180 degree turns.

I tried doing only Rw and U turns and it turned out more challanging than I thought it would. Doing this moveset is essentially not being able to turn one side and having a 2x2x3 block on the opposite side that you can't break at any time. While doing this moveset, corner permutation is solved after you finish the F2L. I have come up with a solution to this substep, but I want to see what everyone else comes up with as my solution is very lengthy.

After solving the F2L, I solve edge permutation, edge orientation, then do a form of a H-Perm if corner permutation isn't algined with edge permutation, and finally finish with corner orientation. It took me about and hour to come up with a solution that worked consistantly every time. If anyone is interested, I could post my solution here.
 
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elrog

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Alright, but the solutions can sometimes be very, very lengthy. I have found some ways to shorten this, but it would take more algorithms and you can't always do them. There are 3 algorithms (all other algorithms are made up of these 3) you will need to know to solve this moveset and you repeat than many times. I found these by just breaking my F2L and fixing it in a different way and seeing how it affected the pieces. Normally I'd use commutators, but this moveset didn't allow for many commutators (that are easily seen at least).

Here are the 3 algorithms you will use (when I type R I mean Rw):

Alg 1 v1: R U' R' U' R U' R' U R U R'

Alg 1 v2: R' U R U R' U R U' R' U' R

Alg 2 v1: (R U R' U2 R U' R' U R U R') U' (R U' R' U' R U R' U2 R U' R')

Alg 2 v2: (R U R' U2 R U' R' U R U R') U2 (R U' R' U' R U R' U2 R U' R')

Alg 3 v1: R' U R2 U' R2 U' R2 U R'

Alg 3 v2: R U' R2 U R2 U R2 U' R

If you are experienced enough at stardard intuitive F2L, you should be able to get the F2L and use some short cuts for this step. If not, you should start with the BR F2L pair (1x1x2 block composed of the BRD corner and the BR edge). This is made by getting the BRD corner in the top layer unoriented (without the bottom color on top) and moving it into the left side by using U moves. You can then match the edge up with it using R turns and possibly a U move or two. If you can't figure this out, you should probably get more experience before trying this.

You may need to get the pair out of the left side so you can flip it over with an R move. At this point, you may notice that if you put the pair in its correct place in relation to the 2x2x3 block on the left, you will have trouble getting the M layer centers matched up. You will need to get the pair in the U and L layers so you can move the M centers with an R move to the correct position so that the top color center will be on bottom when you place the pair. I know your probably thinking, why the top center? You will realize soon enough.

Your next step is to place the RD edge in the RU position. It must be oriented (has the bottom color on the top).This should also be pretty simple and easy to figure out. You then need to do the moves R' U2 R2 U2 R. To place the DB edge. To do this, you will need to place the edge in the UL position unoriented (having the bottom color not on top) and apply Alg 1 v2. You can flip the edge in the UL position by doing U' R U' R'. If the BD edge is in its correct place and is misoriented, you will need to do Alg 1 v2 to get the edge in the top layer and then solve it like normal.

Next, you will need to solve the FR pair. This can be done similarly to the first pair. If you get the case with the FRD corner in the UFL position with the bottom color on the left side and the FR edge in the UB position with the right color on top, you can use these moves to solve that case: R' U R2 U2 R2 U' R. It isn’t hard to see how this "algorithm" works.

Now you should have the F2L complete except for the FD edge. Get the edge in the UL position misoriented and apply Alg 1 v1 to solve it. If you’re having trouble flipping the edge while it is in the top layer, put it in the UR or UB position and perform Alg 2 v 1. If the edge is in the correct place and is misoriented, you will need to apply Alg 1 v 1 to get it into the top layer.

The first step to solving the top layer is to permute the top 4 edges. This is done using edge 3-cycles. Alg 3 v1 cycles UB --> UL --> UF and alg 3 v 2 cycles UF --> UL --> UB. You should never need to do more than 2 three cycles to solve edge permutation.

After permuting the top layer edges, you orient them using Alg 2. Alg 2 v 1 flips the UB and UR edges while alg 2 v 2 flips the UL and UR edges. If all edges are misoriented, you will need to do 2 algorithms to flip all 4 edges.

At this point, AUF until all edges are solves. Look at your corners to see if they are in their correct positions or if a X-perm (H-perm) is necessary to permute them. If you need to do and X-perm, do [ (Alg 3 v1) U ]2.

You should now only have to orient the last layer corners. All of the cases can be solved using the Pi case and the H case. (* means it is a solution for that case that I found later)

Pi case - AUF so that the headlights are facing the front. Do (Alg 2 v2) U2 (Alg 2 v2)

H case – AUF so that the headlights are on the left and right sides. Do (Alg 1 v1) U (Alg 1 v1)

Headlights case – AUF so that the headlights are in front. Do the H case to convert to the Pi case
* AUF so that the Headlights are in back. Do R U' R' (2nd Bowtie case) U' R U R'

T case – AUF so that the misoriented corners are in front. Do the Pi case to convert to the H case
* AUF so that the misoriented corners are in front. Do R' U R U' (2nd Bowtie case) R' U' R

Bowtie case – AUF so that there is no top color on the front side. Do the Pi case to convert to the Pi case
* AUF so that the misoriented corners are in the UFL and UBR positions and there is no top color on front. Do (X-perm) U' (X-perm)

Sune case – AUF so that you have the correctly oriented corner in the F layer and there is another corner with the top color on the front. Do the Pi case to convert to the Headlights case

I know it is EXTREMELY long, but atleast its a solution.
 
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TheNextFeliks

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4x4 idea.
1. Two centers
2. Edges yau style
a. Four cross edges and place
b. 3-2-3 edge pairing with slightly less restriction
3. 3x3 solve. Before f2l, adjust cross so most corners seem solved.
4. z' plus centers. I used niklas (r U' l' U r' U l U' i think) and variants.
Not bad. Kinda yau-cage blend.
 

Username

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4x4 idea.
1. Two centers
2. Edges yau style
a. Four cross edges and place
b. 3-2-3 edge pairing with slightly less restriction
3. 3x3 solve. Before f2l, adjust cross so most corners seem solved.
4. z' plus centers. I used niklas (r U' l' U r' U l U' i think) and variants.
Not bad. Kinda yau-cage blend.
The centers will take forever
 

TDM

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Centres last

I've been doing some CFOP solves, except solving the white cross not on the white side. It could help when the cross is difficult if you aren't colour neutral; it can be an alternative to colour neutrality to make crosses easier. At the end of the solve, you finish the centres using (M' y)*4, (M' y')*4 or M2 E M2 E'. Here are some example solves because I'm bad at describing things:
x' z
L B F R' F' // 'Cross' (5/5)
U R' U R L' U L // F2L-1 (7/12)
U' R' U R // F2L-2 (4/16)
U R U' R' U' R U R' // F2L-3 (8/24)
y' L' U2 L d' L U L' // F2L-4 (7/31)
U F R U R' U' S R U R' U' f' // OLL (12/43)
R2' u' R U' R U R' u R2 y R U' R' U' // PLL (13/56)
M' E M E' // Centres (4/60)
z' y'
R' L' D L D' // 'Cross' (5/5)
d' R U R' // F2L-1 (4/9)
L' U' L U' L' U L // F2L-2 (7/16)
d' R U' R' y R U R' // F2L-3 (7/23)
U2 y R U R' U2 R U' R' // F2L-4 (8/31)
U R2 D R' U2 R D' R' U2 R' // COLL (10/41)
R2 U R U R' U' R' U' R' U R' U // EPLL (12/53)
M' y' M' y' M' y' M' // Centres (4/57)

I also could've ended with L R' E M E' to avoid rotations.
There are some certain rules when making the cross: the pieces must be correct in relation to each other; if made on the opposite side to normal (Y if you start on W), two opposite pieces must be on their correct face (e.g. RW and OW edges on R and O sides); if made on a side adjacent to W (e.g. G) then the piece that would normally go between that side and W has to be a quarter turn from its correct place (e.g. WG edge goes in GO or GR position).

Opinions? I think it looks like it could be useful with a bit of practise - just like CN.
 
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PoHos1

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NEW metod idea :)

HI just have new idea of metod its bit advenced :)
but I mean that can be fast

1 STEP

Just easy solve 2x2x3 block like whit petrus :)

2 STEP

solve the corner and opposite edge like with keyhole trick :) you will see in example solve :) next

3 STEP A

Next you have to solve last edge to cross and this is advenced you have to twist edges to have orient 3 edges on top

3 STEP B
now you have to like with keyhole trick solve last pair to f2l but you have to look to edges on top and dont destroy it to have cross on top after it

4 STEP

now you have to do all coll alg. to solve last layer and that just do U perms or Z or H perm :) its fast







Example solve :F2 R2 D' U' B2 U' B2 L2 D B2 R2 F' R' D B F2 U R' D F L2 U2


orient: x2 y
2x2x3 block: R Ui Ri yi U2 r Ui ri y Ri Fi R
Corner+Edge: y R2 U Ri U R Ui Ri
Edges orient : y U Mi U M
Keyhole: Di Li U2 L U2 Li U L D
Last layer R U Ri U R U2 Ri and then Z perm

its not best but you can try it I mean it that sometime its good solve like I have on it time like 14 sec
 
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