# FB 1x2x3 Example Solve Thread



## voidrx (Aug 21, 2021)

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


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## Dan the Beginner (Aug 22, 2021)

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?


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## voidrx (Aug 22, 2021)

Dan the Beginner said:


> 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.


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## Athefre (Aug 22, 2021)

Rouxvolutionist said:


> 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
> 
> ...



Full solution: z2 x' R2 B' U F' D'

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


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## Dan the Beginner (Aug 22, 2021)

Rouxvolutionist said:


> 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.


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## Dan the Beginner (Aug 22, 2021)

Athefre said:


> Full solution: z2 x' R2 B' U F' D'
> 
> 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'.
> 
> ...


That was brilliant. So much for me to digest and, if it's possible, to learn here. I think CN is necessary for that.


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## qwr (Aug 22, 2021)

I legit thought you were talking about example solves for this and wondered why anyone would bother with example solves


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## BenChristman1 (Aug 22, 2021)

qwr said:


> I legit thought you were talking about example solves for this and wondered why anyone would bother with example solves
> 
> View attachment 16728


So did I!


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## Cubing Forever (Aug 22, 2021)

Athefre said:


> 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



Spoiler: Full Mehta solve(53 STM)



z2
E r U r' U2 R' U R' D' //Mehta FB(+1)
U u R2 U R u2 //3QB
F' U2 F U2 S' U S //EOLE
R' U R' U R U2 R //6CO
U2 R2 U R2 U2 R2 U R2 //APDR
D2 U R U R' F' R U R' U' R' F R2 U' R' U //PLL





Spoiler: Full Roux solve(37 STM)



z2 y'
L' D' R' U2 R U' R M2 B' //Roux/Nautilus FB
R' U R' U' R' U' M' U r' //SB
S' L' U' L U L F' L' f //CMLL
M2 U' M U2 M' U' M U2 M' U' //LSE



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


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## stanleydesu (Aug 22, 2021)

Cubing Forever said:


> 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'
> ...


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


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## abunickabhi (Aug 22, 2021)

I only know 1x2x3 as the Roux first block and second block dimension.


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## UNO_FASY (Aug 22, 2021)

stanleydesu said:


> 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'


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## voidrx (Aug 22, 2021)

UNO_FASY said:


> 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
> ...



Inspection: x
Full Solution: L U2 L D R' D

First I see that I can make a line with L U2 L, and that puts the DBL corner in position to be solved with the rest of the bottom line with D R' D.

Next: F' U' D F' U' F B L D B R2 L2 D2 R2 B R2 B' D2 B


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## BenChristman1 (Aug 23, 2021)

Rouxvolutionist said:


> Next: F' U' D F' U' F B L D B R2 L2 D2 R2 B R2 B' D2 B


x2 y D L' U2 L2 D R2 D' F2

Next: D2 R' D2 L' B2 F2 L2 U2 R U2 R' F2 U' F R' B' R' F U L U'


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## Cubing Forever (Aug 26, 2021)

BenChristman1 said:


> Next: D2 R' D2 L' B2 F2 L2 U2 R U2 R' F2 U' F R' B' R' F U L U'


8 STM:
z2 y2 // orientation 
D B' L2 R U2 D R D //Roux block
//so, D B' L2 solves the 2 edges, I put the pair in the D Layer using R, U2 D R D solves the rest of the block
//very inefficient 

Next: L D U2 L2 D F2 D L2 F2 U2 F2 R B' D2 U' F R' U2 L F


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## Athefre (Sep 1, 2021)

Cubing Forever said:


> 8 STM:
> z2 y2 // orientation
> D B' L2 R U2 D R D //Roux block
> //so, D B' L2 solves the 2 edges, I put the pair in the D Layer using R, U2 D R D solves the rest of the block
> ...


y2 M2 D' F2 L F

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'


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## Cubing Forever (Sep 1, 2021)

Athefre said:


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


Roux NMC FB
M U R' F M B2 R' U2 B //Roux block with non-matching centers
//This one is actually a really fun and fancy block. I noticed that the green-yellow edge is "solved". M U R' pairs up the green-red-yellow pair, F solves it M B2 R' pairs up the green-orange yellow pair and U2 B solves it.

Mehta FB:
F' L' U2 L u2 U R2 F R F' D' //Mehta FB
//F' L' U2 L forms a square, u2 U R' solves the blue-yellow edge and I inserted the blue-red-yellow corner with a sledge but you can cancel the last move of u2 U R' and sledge which makes it u2 U R2 F R F'. 11 moves, not so efficient.

Next: B D' B R' F U R F2 L' F2 B2 U L2 D2 R2 U2 F2 U F2 D


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## Cubing Forever (Sep 19, 2021)

Cubing Forever said:


> Next: B D' B R' F U R F2 L' F2 B2 U L2 D2 R2 U2 F2 U F2 D


Roux:
B L' R U F' //square 
R U' R' M' B' //pair
(this also gives a free SB pair LOL)

Mehta:
R B L' U F' //Square
R B2 R U r U2 r' //FB+1
I actually influenced 3QB here. Hence the high movecount.

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


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## tsmosher (Nov 25, 2021)

Cubing Forever said:


> Roux:
> B L' R U F' //square
> R U' R' M' B' //pair
> (this also gives a free SB pair LOL)
> ...



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


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## voidrx (Nov 26, 2021)

tsmosher said:


> 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
> ...



z2 y' // this places the DLS edge in position and gives an easy corner edge pair.

M2 F' // this pairs the corner edge pair and solves it to make a 1x2x2 square in dfL.

R // this places the DBL corner in position to be solved in the back with the BLE edge.

U' M2 B' // this moves the DLE edge to be paired with the DBL corner and then that corner edge pair is inserted with a B'. 

6 moves.

Next: D F' L2 F' R2 D' F2 U R F' L2 U2 F' R2 B R2 D2 R2 F' U2


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## tsmosher (Nov 26, 2021)

voidrx said:


> z2 y' // this places the DLS edge in position and gives an easy corner edge pair.
> 
> M2 F' // this pairs the corner edge pair and solves it to make a 1x2x2 square in dfL.
> 
> ...



// This is a tough scramble for me, and I will be using a color-neutral FB as a result.

y2 z // In this orientation, I see that F' will join together a corner/edge pair in DfL.
S // But I can do better! Instead of bringing down the corner, I can bring up the edge in DLS!
r' // this completes a cute little square, which we can then put back on our L layer with...
f' // dfR complete.

// we just need the back pair. I notice that my D center and B center are also aligned at this point. (dumb luck)
// The other corner is misoriented in UBL. It's gonna take me ~3 moves to insert it into the triplet.
// Since it will take a lot of moves, I'd like to join the corner with its matching edge while I am preparing it for insertion.

// The matching edge is in DFM, so pairing is pretty easy with:
U M2 // pair formed in UBr. But it's not oriented correctly, so I will have to mess up some centers to get it in quickly.
r B' // inserted into FB. FB is complete!

// without verbose comments:
y2 z S r' f' U M2 r B' // Total moves: 7 (Roux FB)

NEXT:
D2 R' F2 D2 L F2 D2 R D2 R B2 D U' R' D2 L' U2 F U2 B2 R


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## voidrx (Nov 26, 2021)

tsmosher said:


> // This is a tough scramble for me, and I will be using a color-neutral FB as a result.
> 
> y2 z // In this orientation, I see that F' will join together a corner/edge pair in DfL.
> S // But I can do better! Instead of bringing down the corner, I can bring up the edge in DLS!
> ...



You missed a very good FB in that scramble.

z y M' (B' F2) R F
Technically four moves. 
Now to the current scramble.

y2 // I see that the FLE edge is paired with its center so I decide to go with that.

(U' D') // these two moves are done at the same time, so it is one move. This pairs the dBL pair and gets the DfL pair in position to be solved.

M U2 R2' // this pairs the DfL pair and moves the dBL pair to uFR.

D' // this solves the dfL square.

R U' B // this solves the remaining pair and finishs the FB.

8 moves.

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


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## voidrx (Nov 27, 2021)

voidrx said:


> You missed a very good FB in that scramble.
> 
> z y M' (B' F2) R F
> Technically four moves.
> ...



I like doing these and I don't like to wait for the next person, so imma do this one too.

y' x' // there is a non-matching centers square so I position it in dbL.

R' U' // this pairs the other pair.

R' U F' // this solves the non matching center FB.

5 STM

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


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## tsmosher (Nov 27, 2021)

voidrx said:


> I like doing these and I don't like to wait for the next person, so imma do this one too.
> 
> y' x' // there is a non-matching centers square so I position it in dbL.
> 
> ...


Back with another terrible FB 

x' // At this angle, I see that B will form a corner/edge pair in DbL.
// Let's pair up the corner with its matching edge before inserting it.
R E // This pairs them up. This also dictates that I will be forming Mehta FB (Dl).
r // Influencing the D center just makes things easier for us later. This move puts the D center on the D layer.
B // Dbl square has been formed. Time for FBLS (DFl).
R2 // R prepares the corner for insertion into the Mehta FB. But the matching edge is in UBM. So better to go one quarter turn further and pair up corner and edge.
U R' F // inserts the misoriented pair into the front of the Mehta FB.

// total moves: 8

NEXT:
D' U2 R' U2 L' D2 F2 D2 B2 R2 U2 R D2 U R' B2 U L2 F' D' B'


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## Cubing Forever (Nov 27, 2021)

tsmosher said:


> NEXT:
> D' U2 R' U2 L' D2 F2 D2 B2 R2 U2 R D2 U R' B2 U L2 F' D' B'


Mehta FB: 6 STM
x2 z //inspection: I see a pair and 2 edges solved in relation to each other and the other corner is in DBR
S' //solve white center
L' U //move the pair
R' //pair up the other corner with the pair with the pair to make a line
U R2 //solve the line
//final solution: S' L' U R' U R2

wow this one's ingenious

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


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## ray5 (Dec 23, 2021)

y' x'

R' D'
L' U'
R U R'
L2


(8h)

Next: U' R2 F2 D F2 U B2 U' B2 U2 L2 B2 R' B F D B2 R' D2 F L'


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## tsmosher (Dec 23, 2021)

ray5 said:


> y' x'
> 
> R' D'
> L' U'
> ...



x // seeking to take advantage of the pre-formed corner-edge pair in DbL
M // this aligns the D center and solves DBM (thus solving a square in Dbl)
F // move the orange yellow edge up to be joined with DFL
R' // joins DFL with orange yellow edge (i.e., DFM)
F2 // completes Mehta FB (Dl)

// 4 STM, x2/y neutral Mehta FB

NEXT:
L2 U L' U2 F' D R U' F' L2 D2 F2 R2 U2 F U2 F L2 B D2


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## ray5 (Dec 24, 2021)

y' // orange
B2 // pair up
D F' // align on bottom
U D M // connect

(7h)

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


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## V Achyuthan (Dec 24, 2021)

ray5 said:


> U2 B2 U' F2 L F' U D B' R2 D F2 D' L2 D' B2 L2 U2 F2 L2


y x2 D2 F' L2 F' // Square
L F' // Pair
7 STM

Next : 
L B2 L2 F2 D' R2 U2 F2 D2 R2 U' B D2 U F D R' F L F'


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## ray5 (Dec 25, 2021)

[R':U'] L
B2
U L2
L' F2 L F'

(11h)

I really struggled with this one. I could not find a good solution.

Next: L2 R2 B' L2 D2 R2 U2 B2 F2 R2 U2 F D' F2 R F D' F' U B R'


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## voidrx (Dec 26, 2021)

ray5 said:


> [R':U'] L
> B2
> U L2
> L' F2 L F'
> ...




So I see the blue red edge is connected to the blue center and I can pair the DLS edge with the DFL corner by doing z y' D'. Then I do an R' to bring the DfL pair to the D layer.
Now, I'd I just did a D2 to solve the square, it would leave the last pair in a difficult spot, so instead, I will do a D R2 to move the BLE edge to a better spot. Then I do another D to solve the square and it happens to pair the last pair together. So then I finish with an R2 B2.

Full Solution:
z y' 
D' R' D R2 D R2 B2

7 STM

Next: L2 F' U L' U2 L2 D R U D2 F R2 F2 U2 D2 L2 D2 L2 B L2 F


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## tsmosher (Dec 26, 2021)

voidrx said:


> So I see the blue red edge is connected to the blue center and I can pair the DLS edge with the DFL corner by doing z y' D'. Then I do an R' to bring the DfL pair to the D layer.
> Now, I'd I just did a D2 to solve the square, it would leave the last pair in a difficult spot, so instead, I will do a D R2 to move the BLE edge to a better spot. Then I do another D to solve the square and it happens to pair the last pair together. So then I finish with an R2 B2.
> 
> Full Solution:
> ...



z2 // taking advantage of corner/edge pair in DfL
S' // displace orange center (and temporarily DLS)
E S // move flipped FLE edge over and attach to orange center (orienting it)
R2 // correctly permute FLE
u2 // E2 completes the square in dfL -- doing U2 at the same time sets up DBR to be joined with BLE after M'
M' // joins DBR and BLE (inadvertently solves DFM as well)
U2 B // inserts back pair, completing FB

z2 S' E S R2 u2 M' U2 B // Roux FB, 8 STM, also solves DFM (inadvertent)

NEXT: 
U2 B2 D R2 U2 R2 B2 D R2 D' L2 B2 R B U' F2 U' L B' L D2


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## V Achyuthan (Dec 26, 2021)

tsmosher said:


> U2 B2 D R2 U2 R2 B2 D R2 D' L2 B2 R B U' F2 U' L B' L D2


L D' B2 // setting up LF and LB edges
R' L2 // Build a line 
D' // Finish FB

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


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## voidrx (Dec 26, 2021)

V Achyuthan said:


> L D' B2 // setting up LF and LB edges
> R' L2 // Build a line
> D' // Finish FB
> 
> ...


So first I see the orange green yellow pair and the orange green edge can be solved into a square by doing U' r' u. But i want to put the last pair in a better position, so I do an R' first. so it would be R' U' r'. Now if I did a u, it would move the last pair edge into a worse spot, so I am going to do an E'. then to finish it off, I do R U2 B
Full Solution:
z 
R' U' r' E' R U2 B 

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


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## tsmosher (Dec 27, 2021)

voidrx said:


> So first I see the orange green yellow pair and the orange green edge can be solved into a square by doing U' r' u. But i want to put the last pair in a better position, so I do an R' first. so it would be R' U' r'. Now if I did a u, it would move the last pair edge into a worse spot, so I am going to do an E'. then to finish it off, I do R U2 B
> Full Solution:
> z
> R' U' r' E' R U2 B
> ...



EDIT: Improvement suggested by @voidrx .

z2 y
F' // forming pair in DfR
R F2 // inserting pair and forming square in dfL
R // forming pair in DBr
B' // inserting pair to finish FB

z2 y F' R F2 R B' // Roux FB, 5 STM



Spoiler: old solution



R2 S' b // forms triplet in DL
U // moves the white/green edge out of the way
M' // attach white center to white/blue edge
u2 R // attach white/green edge
u' // align FB

R2 S' b U M' u2 R u' // Roux FB, 8 STM



NEXT:
B R U2 R F' D' R F' L F2 D2 F' D2 L2 D2 F2 D2 F' R2 U2


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## voidrx (Dec 27, 2021)

tsmosher said:


> R2 S' b // forms triplet in DL
> U // moves the white/green edge out of the way
> M' // attach white center to white/blue edge
> u2 R // attach white/green edge
> ...



So I see the red blue yellow pair can be paired with the red blue edge by doing y2 x E2 R E2. However. I want to influence the last pair which is already made by doing a u2 in place of the last E2. So it would be y2 x E2 R u2. Then I finish the FB with r B'.

Full Solution
y2 x 
E2 R u2 r B' 

5 STM

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


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## ray5 (Dec 28, 2021)

z' y
D [R:U']
L' U B'

(7h, 7q, 7s, 9e)

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


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## voidrx (Dec 29, 2021)

ray5 said:


> z' y
> D [R:U']
> L' U B'
> 
> ...



I see that I can make a square with the premade pair by doing y2 D' r B'.
Then I can make the other pair by doing U' R' and solving the pair with U2 F'

Full Solution:
y2 
D' r B' U' R U2 F' 

7 STM

Next: R2 D' R2 U L2 B2 R2 F2 D' R2 D2 U2 B U2 R2 F' D B2 L U R2 D2


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## V Achyuthan (Dec 30, 2021)

voidrx said:


> Next: R2 D' R2 U L2 B2 R2 F2 D' R2 D2 U2 B U2 R2 F' D B2 L U R2 D2


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


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