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There are many methods that have a 1x2x3 as it's first step. This thread will bring many examples to one place. Also some methods have different orientations of the 1x2x3, mainly Mehta. You may do an example of a Mehta 1x2x3, I won't encourage it, just say that you are doing that at the top of your example. I also ask that you explain your thinking process, explaining what you are doing as you go along.

Scramble: R2 U' B2 D2 R2 D' L2 U L2 F2 L2 B U' R' D' L B L' D F2

FB: I saw the DL edge and the DFL corner can be paired while after that, the FL edge can be moved in the FB area to solved with the DL edge and DFL corner.
R B' L2 D'
That solves a square. While I was planning the square, I saw that the D' from earlier would pair up the back pair. I took use of this and finished the block with
R2 B2

Final Solution: R B' L2 D' R2 B2

Next: B' U B2 U2 F2 U R2 D L2 R2 U' F2 B' L F R2 F D L U B

I tried to follow your thinking and trace your steps, but am confused with the DL and DFL, etc. You still holding the cube with White on top and Green in front?

I tried to follow your thinking and trace your steps, but am confused with the DL and DFL, etc. You still holding the cube with White on top and Green in front?

Yes. This scramble happened to not need a rotation. DFL is the corner that goes in the bottom left in the front. DL edge is the edge that is in the bottom left of the FB.

There are many methods that have a 1x2x3 as it's first step. This thread will bring many examples to one place. Also some methods have different orientations of the 1x2x3, mainly Mehta. You may do an example of a Mehta 1x2x3, just say that you are doing that at the top of your example. I also ask that you explain your thinking process, explaining what you are doing as you go along.

Scramble: R2 U' B2 D2 R2 D' L2 U L2 F2 L2 B U' R' D' L B L' D F2

FB: I saw the DL edge and the DFL corner can be paired while after that, the FL edge can be moved in the FB area to solved with the DL edge and DFL corner.
R B' L2 D'
That solves a square. While I was planning the square, I saw that the D' from earlier would pair up the back pair. I took use of this and finished the block with
R2 B2

Final Solution: R B' L2 D' R2 B2

Next: B' U B2 U2 F2 U R2 D L2 R2 U' F2 B' L F R2 F D L U B

After the initial rotations I first saw that you could do F2 D' F D' to make a square. But that leaves the other pair or pairs in a bad position. So I saw that if R2 U F' D' is done instead, the final edge can be positioned better after the R2 to make a complete block. Leaving R2 B' U F' D'.

Spoiler: Full Nautilus L5E solve with non-matching blocks

FB: z2 x' R2 B' U F' D'
NSB: R' r' U R2
Final pair: U2 R U R' U' R
NCLL: L U' R' U L' U' R
U2 M' U2 M U M' U2 R r2'

Next: F D2 L' B2 R2 D2 B F U2 L2 U2 F D2 B' L' D' F L2 R' U L

Yes. This scramble happened to not need a rotation. DFL is the corner that goes in the bottom left in the front. DL edge is the edge that is in the bottom left of the FB.

Amazing how you did the first square. It's not intuitive to me at all. Nice.

Thanks. I think I understand the notations now.
I was so confused previously because I thought you started by working on the pieces that were in the DL and DFL positions and they looked totally unmatchable. So, DF and DFL are the destination positions, for the pieces that actually came from the back.

After the initial rotations I first saw that you could do F2 D' F D' to make a square. But that leaves the other pair or pairs in a bad position. So I saw that if R2 U F' D' is done instead, the final edge can be positioned better after the R2 to make a complete block. Leaving R2 B' U F' D'.

Spoiler: Full Nautilus L5E solve with non-matching blocks

FB: z2 x' R2 B' U F' D'
NSB: R' r' U R2
Final pair: U2 R U R' U' R
NCLL: L U' R' U L' U' R
U2 M' U2 M U M' U2 R r2'

Next: F D2 L' B2 R2 D2 B F U2 L2 U2 F D2 B' L' D' F L2 R' U L

Mehta FB: 9 STM z2 E r U r' U2 R' U R' D' //Mehta FB(+1)
I saw that the white-red edge can be solved by doing an E and then inserting using r U r'. Then, the green-red edge and the green-red-white corner can be solved by just doing U2 R2 R U R' which is just U2 R' U R'. Then I aligned it using a D'

Roux FB: 9 STM z2 y' L' D' R' U2 R U' R M2 B' //Roux/Nautilus FB
L' D' solves a nice square, R' U2 R sets up the other two pieces to be inserted by U2 B' U' B, U' R keeps the corner in DBR, M2 B' solves the pair

Mehta FB: 9 STM z2 E r U r' U2 R' U R' D' //Mehta FB(+1)
I saw that the white-red edge can be solved by doing an E and then inserting using r U r'. Then, the green-red edge and the green-red-white corner can be solved by just doing U2 R2 R U R' which is just U2 R' U R'. Then I aligned it using a D'

Roux FB: 9 STM z2 y' L' D' R' U2 R U' R M2 B' //Roux/Nautilus FB
L' D' solves a nice square, R' U2 R sets up the other two pieces to be inserted by U2 B' U' B, U' R keeps the corner in DBR, M2 B' solves the pair

U F' solves E-line and sets up back corner with DL edge, then doing r2 D' solves everything but the front corner (now in suboptimal position).
Thus, start with D2 so the front corner is aligned with the DL edge and back corner after the U F', then solve the D-line.
Final solution: x D2 U F' M' r D'
Next: B' L2 D2 U2 F L2 D2 R2 F2 R2 U2 L U' F R' U R2 B2 F D' F

z M2 U2 R B' U' F'
Pretty obvious DL+pair+pair, M2 U2 bring the BL edge to DB, the DFL to RUB and the FL edge to UL. After solving F2 with R B' you solve the other pair and insert that with U' F'. I understand that not everyone could use that block so I made another solution with normal w/y bottom which is:
x D2 M2 U' R B' r F'
D2 solve DL and M2 put both of FL and BL in a nice position for insert, so just solve the back one first with R B' and then followed by r F' to finish

Next: D2 L' B2 D2 L U2 R B2 U2 L2 D2 R2 D R' D' F' D2 R' U' L'

z M2 U2 R B' U' F'
Pretty obvious DL+pair+pair, M2 U2 bring the BL edge to DB, the DFL to RUB and the FL edge to UL. After solving F2 with R B' you solve the other pair and insert that with U' F'. I understand that not everyone could use that block so I made another solution with normal w/y bottom which is:
x D2 M2 U' R B' r F'
D2 solve DL and M2 put both of FL and BL in a nice position for insert, so just solve the back one first with R B' and then followed by r F' to finish

Next: D2 L' B2 D2 L U2 R B2 U2 L2 D2 R2 D R' D' F' D2 R' U' L'

I saw the free pair but the edge that goes with it needs to be moved a little. So I thought about making another pair first to go with it. After the y2, L2 solves a useable pair. After that, if you do R2 it sets up that first pre-built pair and the edge that goes with it into better positions. L2 R2 is pretty much M2. So the first move is M2. Then do D' to place the edge with the center and the remaining moves attach the two pairs.

Next: D2 R D2 U2 L R2 B2 L F2 R' D2 F L D' B F L' D' B' U'

x // This positions a corner/edge pair in dbL (DBL and DLS). green on L, white on D.
R' U F2 // this completes the DL triplet. at this point, i will usually extend to either Roux FB or Mehta FB.

// option 1a: Mehta FB
// Mehta FB looks easier with 1 edge in DBM, the D center in back, and the other edge in BRE.
R' U' // DB/DF is connected but backward. if i just did M' now, DB and DF would be solved in each other's spots.
r' U2 r2 // wide R moves are slightly easier than M moves.
// total moves: 8 moves for MFB.

// option 1b: Mehta FB
// Mehta FB looks easier with 1 edge in DBM, the D center in back, and the other edge in BRE.
r' U2 // this aligns DBM correctly. DFM still misoriented.
r' // begin to insert DBM
U' // insert misoriented DFM along the way.
r' // done
// total moves: 8 moves for MFB. a little more ergonomic than 1a though.

// option 2: Roux FB
// Roux FB: one edge is in URS. one edge is flipped in UBM. the L center is on U.
r' U // reorients the one edge and moves it out of the way.
r2 // when moving an edge line (edge-center-edge) from the U layer to the E slice, i remember it like this: you push the edge towards its matching color in the triplet. so red is on back of the triplet; therefore, i push green/red edge to the back. (this prevents me moving the edge line down backward and having to flip it around.)
u2 R' // connect the other edge
u // align the edge line to form a Roux FB
// total moves: 9 moves for RFB.

NEXT: D' F2 L' D' F' B2 D2 R' U2 D2 F D2 F2 B' L2 D2 B U2 L

x // This positions a corner/edge pair in dbL (DBL and DLS). green on L, white on D.
R' U F2 // this completes the DL triplet. at this point, i will usually extend to either Roux FB or Mehta FB.

// option 1a: Mehta FB
// Mehta FB looks easier with 1 edge in DBM, the D center in back, and the other edge in BRE.
R' U' // DB/DF is connected but backward. if i just did M' now, DB and DF would be solved in each other's spots.
r' U2 r2 // wide R moves are slightly easier than M moves.
// total moves: 8 moves for MFB.

// option 1b: Mehta FB
// Mehta FB looks easier with 1 edge in DBM, the D center in back, and the other edge in BRE.
r' U2 // this aligns DBM correctly. DFM still misoriented.
r' // begin to insert DBM
U' // insert misoriented DFM along the way.
r' // done
// total moves: 8 moves for MFB. a little more ergonomic than 1a though.

// option 2: Roux FB
// Roux FB: one edge is in URS. one edge is flipped in UBM. the L center is on U.
r' U // reorients the one edge and moves it out of the way.
r2 // when moving an edge line (edge-center-edge) from the U layer to the E slice, i remember it like this: you push the edge towards its matching color in the triplet. so red is on back of the triplet; therefore, i push green/red edge to the back. (this prevents me moving the edge line down backward and having to flip it around.)
u2 R' // connect the other edge
u // align the edge line to form a Roux FB
// total moves: 9 moves for RFB.

NEXT: D' F2 L' D' F' B2 D2 R' U2 D2 F D2 F2 B' L2 D2 B U2 L