#### AvGalen

##### Premium Member

I did my first ever fewest moves challenge at the US Open 2007 and got a DNF because I didn't write down my solution within the hour (I didn't know xyz-notation). It was a 48 or 49 moves solution and basically just a regular cross, f2l, last layer solve. Somehow I liked fewest moves and when it became a part of the weekly competition I always did it (with mixed results. 31/51 is a wide range). Somehow I always see good beginnings and manage to do a reasonable final part. My results lately have become pretty good with a best official result of 37 and 33 at the Polish Open 2007. Not good enough to win the event (the current WR-holder did 33 and 31), but good enough to make me the number 6 of the world. Not to bad considering I only know about 30 algorithms in total (keyhole and 4 look last layer) and that those were my 11th and 12th attempt ever (including official/unofficial competitions and practise). I don't consider myself an expert, but I think I am good enough to share some tips and techniques with all of you. If you know any other, feel free to share them and don't hesitate to ask me if you don't understand, have extra questions, or think I made a mistake.

OK, here comes the good stuff:

Basically this means that you don't start with a cross but with a 2x2x3 block instead. Sometimes I do this by building a 2x2x2 block and then extend it to a 2x2x3 block. Other times I build a 1x2x3 block and attach it to another 1x2x3 block (just 3 centers with 2 edges). The advantage of block building is that you have a lot of freedom and even after you have build a 2x2x2 block you can extend it in 3 different directions giving you a lot of choice.

Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2

I noticed the two pairs (yellow-red-blue, green-orange-yellow) and after some trying I found that the second pair could be made into a 2x2x2 block (U2 R2 L F2) and then into a 2x2x3 block (R B' R2 B' R) with B2 U2 that would have made Cross + 2 pairs in 11 moves. But I was not satisfied because that wouldn't give me a good continuation. After some more analyzing I found that inserting a B move in the beginning (U2 R2 becomes U2 R B R) would form another pair that could be used to form a 1x2x3 so I made the other 1x2x3 block and connected them (D2 L F2). This gave me a 2x2x3 block in 7 moves (or cross + 2 pairs in 8 moves with U2

Sometimes it is usefull to use keyhole. That means you leave out a corner in the first layer and use it to insert edges in the middle layer with fewer moves. This is especially usefull if you can use it to insert a corner-edgepair.

Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2, perform U2 R B R D2 L F2 U2

As you can see, everything in the F2L is done except for 2 pairs. If you would do them 1 pair at a time it would take a lot of moves. with keyhole you only need 5: F R B2 R' F'. That means full F2L was done in 13 moves!

This is obvious, the more algorithms you know for the last layer the better. It is also important that you know OPTIMAL algs, not fingertrickfriendly2gen agls. I only know 10 OLL's and 6 PLL's (0 COLL) which is a big problem because I have to setup the last layer into

Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2, perform U2 R B R D2 L F2 U2 F R B2 R' F'.

OLL in 10 (N7): x' y R'L2DFD'F'L'FL'R

PLL in 11 (P8): y2 B2 R' U' R B2 L' D L' D' L2 U

This would have made a total of 34 moves, better than the 37 I got with a more complicated ending.

If you have a corner-edgepair you can insert it in 3 ways most of the time. Experiment with the results of each way of inserting it. The results on the last layer (especially the orientation of edges) can be very important.

Scramble with R' F R F'. You could insert the pair with

1) F R' F' R to leave 4 edges oriented

2) U' F' U F to leave 2 opposite edges oriented

3) U2 F' U2 F to leave 2 adjacent edges oriented

Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2, perform U2 R B R D2 L F2 U2 F

Inserting the "pair" with R B2 R' F' is good, but leaves only 2 oriented edges. You could manipulate the last layer and orient more edges before inserting the "pair" by doing B' R B R' (manipulation) B2 R' U R U' F' (edge control). This cost 5 moves more, but it left me with a short OLL and PLL that i knew. (OLL (8) U' L' D' L U L' D L, PLL (10) R2 B' U D' R2 D U' B' R2 B')

Sometimes you do algorithm one, rotate the cube and then do algorithm two. The end of algorithm one might be a turn (or several) of the same face that starts algorithm two. If you write down your solution without cube rotations this is much easier to notice and correct.

Scramble with R2 B2 R F R' B2 R F' R y F U R U' R' F'.

Don't write down your solution like F R U R' U' F' y' R' F R' B2 R F' R' B2 R2 (15 moves)

Write it like this instead F R U R' U' F' F' L F' R2 F L' F' R2 F2 (15 moves) and correct it to F R U R' U' F2 L F' R2 F L' F' R2 F2 (14 moves)

This is very related to the previous point. Basically it means that you try to save moves by blending several steps together. This can be done in very many ways. The following is just one of them. Sometimes you can perform algorithms in different ways U2 R2 U2 R2 U2 R2 is the same as R2 U2 R2 U2 R2 U2. If the previous step ended with a U move then starting with U2 is better. If the next step starts with a U move then starting with R2 is better.

Let's create the H-Perm U2 R2 U2 R2 U2 R2 U U2 R2 U2 R2 U2 R2 U' (14) can be shortened to U2 R2 U2 R2 U2 R2 U' R2 U2 R2 U2 R2 U' (13), but if you create it like R2 U2 R2 U2 R2 U2 U R2 U2 R2 U2 R2 U2 U' (14) it can be shortened to R2 U2 R2 U2 R2 U' R2 U2 R2 U2 R2 U (12). It could even be shortened to R2 U2 R U2 R2 U2 R2 U2 R U2 R2 (11).

If you don't see a good start for the scramble you could scramble it in reverse. That means this scramble U2 F D becomes D' F' U2. Now find a good solution and write it down inverted. It will look very strange if you solve it like this, but it gives you two chances of finding a good start.

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

Perform scramble as: L' D2 U2 L' U' B2 U B' L2 D B' R U' B' F' L2 B L

Find a solution like this

2x2x2 (7): L F2 L' F U R' U2

2x2x3 (7): L2 F2 D L' D2 F D

Cross + 3rd Pair (4): L F L F

4th Pair (5): L2 D F' D' F

OLL (6): B L D L' D' B'

PLL (10): U' F U' B2 U F' U' B2 U2 L2

Write down your solution as L2 U2 B2 U F U' B2 U F' U, B D L D' L' B', F' D F D' L2, F' L' F' L', D' F' D2 L D' F2 L2, U2 R U' F' L F2 L'

Sometimes it is better to do a 3-cycle of edges or corners in the middle of the solve than at the end. A 3-cycle of edges takes 9 moves when they are in the same layer (PLL) and even more when they are not. But if they are on the same slice it can often be done in 6 moves

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

Cross (9): R2 U' R' B' L U L' D' U2

First 2 Pairs (3): B2 L B2

3rd Pair (4): D F D F'

Last 5 corners (7): D2 L' D L D2 L' D'

As you can see, all corners are done and there are 5 edges that need to be cycled. Put numbered stickers on the first color of the following pieces

1 = yellow-red

2 = yellow-blue

3 = yellow-green

4 = yellow-orange

5 = green-orange

This list is the order of the cycle. Piece 1 goes to the location of piece 2, 2->3, 3-> 4, 4-> 5 and 5-> 1.

After that, do the scramble on a solved cube that has the stickers on it. Perform your solve and look for a point where 3 continuously numbered stickers are on the same slice or on the same face (permutation) and all stickers are on the same or opposite faces (orientation). If you find such a moment you can perform an algorithm that cycles these pieces (hopefully you even get cancellations at the beginning or end).

If you look carefully you can see that after the first six moves (R2 U' R' B' L U) pieces 1, 2 and 3 are in the S-slice with the correct orientation. They can be cycled in 6 moves (D2 F B' L2 F' B) so now the solution becomes

Cross Part 1 (6): R2 U' R' B' L U

Edge-Cycle insertion 1 (6): D2 F B' L2 F' B

Cross Part 2 (3): L' D' U2

First 2 Pairs (3): B2 L B2

3rd Pair (4): D F D F'

Last 5 corners (7): D2 L' D L D2 L' D'

Now only pieces 3, 4 and 5 are wrong.

A good moment to insert a cycle that solves pieces 3, 4 and 5 is just before the "Last 5 corners" because at that moment pieces 3, 4 and 5 are all on D-face with the correct orientation. They can be cycled in 9 moves (B2 D' L R' B2 R L' D' B2) so the final solution becomes

Cross Part 1 (6): R2 U' R' B' L U

Edge-Cycle insertion 1 (6): D2 F B' L2 F' B

Cross Part 2 (3): L' D' U2

First 2 Pairs (3): B2 L B2

3rd Pair (4): D F D F'

Edge-Cycle inserstion 2 (9): B2 D' L R' B2 R L' D' B2

Last 5 corners (7): D2 L' D L D2 L' D'

P.S. don't think that these are lucky scrambles or specially prepared examples that never happen in real life.

Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2 was the first scramble at the Polish Open 2007

Scramble2: L' B' L2 F B U R' B D' L2 B U' B2 U L U2 D2 L was my solution to weekly competition 2007-32

OK, here comes the good stuff:

**Block building**.Basically this means that you don't start with a cross but with a 2x2x3 block instead. Sometimes I do this by building a 2x2x2 block and then extend it to a 2x2x3 block. Other times I build a 1x2x3 block and attach it to another 1x2x3 block (just 3 centers with 2 edges). The advantage of block building is that you have a lot of freedom and even after you have build a 2x2x2 block you can extend it in 3 different directions giving you a lot of choice.

**Block building examples**: (Scramble with white on top, green on front)Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2

I noticed the two pairs (yellow-red-blue, green-orange-yellow) and after some trying I found that the second pair could be made into a 2x2x2 block (U2 R2 L F2) and then into a 2x2x3 block (R B' R2 B' R) with B2 U2 that would have made Cross + 2 pairs in 11 moves. But I was not satisfied because that wouldn't give me a good continuation. After some more analyzing I found that inserting a B move in the beginning (U2 R2 becomes U2 R B R) would form another pair that could be used to form a 1x2x3 so I made the other 1x2x3 block and connected them (D2 L F2). This gave me a 2x2x3 block in 7 moves (or cross + 2 pairs in 8 moves with U2

**Keyhole**:Sometimes it is usefull to use keyhole. That means you leave out a corner in the first layer and use it to insert edges in the middle layer with fewer moves. This is especially usefull if you can use it to insert a corner-edgepair.

**Keyhole example:**Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2, perform U2 R B R D2 L F2 U2

As you can see, everything in the F2L is done except for 2 pairs. If you would do them 1 pair at a time it would take a lot of moves. with keyhole you only need 5: F R B2 R' F'. That means full F2L was done in 13 moves!

**Know full OLL, PLL and COLL:**This is obvious, the more algorithms you know for the last layer the better. It is also important that you know OPTIMAL algs, not fingertrickfriendly2gen agls. I only know 10 OLL's and 6 PLL's (0 COLL) which is a big problem because I have to setup the last layer into

**With OLL/PLL knowledge example**:Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2, perform U2 R B R D2 L F2 U2 F R B2 R' F'.

OLL in 10 (N7): x' y R'L2DFD'F'L'FL'R

PLL in 11 (P8): y2 B2 R' U' R B2 L' D L' D' L2 U

This would have made a total of 34 moves, better than the 37 I got with a more complicated ending.

**Edge control / Last Layer manipulation:**If you have a corner-edgepair you can insert it in 3 ways most of the time. Experiment with the results of each way of inserting it. The results on the last layer (especially the orientation of edges) can be very important.

**Edge control example (easy):**Scramble with R' F R F'. You could insert the pair with

1) F R' F' R to leave 4 edges oriented

2) U' F' U F to leave 2 opposite edges oriented

3) U2 F' U2 F to leave 2 adjacent edges oriented

**Last layer manipulation example (harder):**Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2, perform U2 R B R D2 L F2 U2 F

Inserting the "pair" with R B2 R' F' is good, but leaves only 2 oriented edges. You could manipulate the last layer and orient more edges before inserting the "pair" by doing B' R B R' (manipulation) B2 R' U R U' F' (edge control). This cost 5 moves more, but it left me with a short OLL and PLL that i knew. (OLL (8) U' L' D' L U L' D L, PLL (10) R2 B' U D' R2 D U' B' R2 B')

**Write down your solution without cube rotations!**:Sometimes you do algorithm one, rotate the cube and then do algorithm two. The end of algorithm one might be a turn (or several) of the same face that starts algorithm two. If you write down your solution without cube rotations this is much easier to notice and correct.

**No rotation example:**Scramble with R2 B2 R F R' B2 R F' R y F U R U' R' F'.

Don't write down your solution like F R U R' U' F' y' R' F R' B2 R F' R' B2 R2 (15 moves)

Write it like this instead F R U R' U' F' F' L F' R2 F L' F' R2 F2 (15 moves) and correct it to F R U R' U' F2 L F' R2 F L' F' R2 F2 (14 moves)

**Cancellations:**This is very related to the previous point. Basically it means that you try to save moves by blending several steps together. This can be done in very many ways. The following is just one of them. Sometimes you can perform algorithms in different ways U2 R2 U2 R2 U2 R2 is the same as R2 U2 R2 U2 R2 U2. If the previous step ended with a U move then starting with U2 is better. If the next step starts with a U move then starting with R2 is better.

**Cancellations example:**Let's create the H-Perm U2 R2 U2 R2 U2 R2 U U2 R2 U2 R2 U2 R2 U' (14) can be shortened to U2 R2 U2 R2 U2 R2 U' R2 U2 R2 U2 R2 U' (13), but if you create it like R2 U2 R2 U2 R2 U2 U R2 U2 R2 U2 R2 U2 U' (14) it can be shortened to R2 U2 R2 U2 R2 U' R2 U2 R2 U2 R2 U (12). It could even be shortened to R2 U2 R U2 R2 U2 R2 U2 R U2 R2 (11).

**Inverse scramble:**If you don't see a good start for the scramble you could scramble it in reverse. That means this scramble U2 F D becomes D' F' U2. Now find a good solution and write it down inverted. It will look very strange if you solve it like this, but it gives you two chances of finding a good start.

**Inverse scramble example:**Scramble: L' B' L2 F B U R' B D' L2 B U' B2 U L U2 D2 L

Perform scramble as: L' D2 U2 L' U' B2 U B' L2 D B' R U' B' F' L2 B L

Find a solution like this

2x2x2 (7): L F2 L' F U R' U2

2x2x3 (7): L2 F2 D L' D2 F D

Cross + 3rd Pair (4): L F L F

4th Pair (5): L2 D F' D' F

OLL (6): B L D L' D' B'

PLL (10): U' F U' B2 U F' U' B2 U2 L2

Write down your solution as L2 U2 B2 U F U' B2 U F' U, B D L D' L' B', F' D F D' L2, F' L' F' L', D' F' D2 L D' F2 L2, U2 R U' F' L F2 L'

**Insertions:**(This might be the most advanced technique. It helps if you know something about solving a cube blindfolded.)Sometimes it is better to do a 3-cycle of edges or corners in the middle of the solve than at the end. A 3-cycle of edges takes 9 moves when they are in the same layer (PLL) and even more when they are not. But if they are on the same slice it can often be done in 6 moves

**Insertions example:**Scramble: L' B' L2 F B U R' B D' L2 B U' B2 U L U2 D2 L

Cross (9): R2 U' R' B' L U L' D' U2

First 2 Pairs (3): B2 L B2

3rd Pair (4): D F D F'

Last 5 corners (7): D2 L' D L D2 L' D'

As you can see, all corners are done and there are 5 edges that need to be cycled. Put numbered stickers on the first color of the following pieces

1 = yellow-red

2 = yellow-blue

3 = yellow-green

4 = yellow-orange

5 = green-orange

This list is the order of the cycle. Piece 1 goes to the location of piece 2, 2->3, 3-> 4, 4-> 5 and 5-> 1.

After that, do the scramble on a solved cube that has the stickers on it. Perform your solve and look for a point where 3 continuously numbered stickers are on the same slice or on the same face (permutation) and all stickers are on the same or opposite faces (orientation). If you find such a moment you can perform an algorithm that cycles these pieces (hopefully you even get cancellations at the beginning or end).

If you look carefully you can see that after the first six moves (R2 U' R' B' L U) pieces 1, 2 and 3 are in the S-slice with the correct orientation. They can be cycled in 6 moves (D2 F B' L2 F' B) so now the solution becomes

Cross Part 1 (6): R2 U' R' B' L U

Edge-Cycle insertion 1 (6): D2 F B' L2 F' B

Cross Part 2 (3): L' D' U2

First 2 Pairs (3): B2 L B2

3rd Pair (4): D F D F'

Last 5 corners (7): D2 L' D L D2 L' D'

Now only pieces 3, 4 and 5 are wrong.

A good moment to insert a cycle that solves pieces 3, 4 and 5 is just before the "Last 5 corners" because at that moment pieces 3, 4 and 5 are all on D-face with the correct orientation. They can be cycled in 9 moves (B2 D' L R' B2 R L' D' B2) so the final solution becomes

Cross Part 1 (6): R2 U' R' B' L U

Edge-Cycle insertion 1 (6): D2 F B' L2 F' B

Cross Part 2 (3): L' D' U2

First 2 Pairs (3): B2 L B2

3rd Pair (4): D F D F'

Edge-Cycle inserstion 2 (9): B2 D' L R' B2 R L' D' B2

Last 5 corners (7): D2 L' D L D2 L' D'

P.S. don't think that these are lucky scrambles or specially prepared examples that never happen in real life.

Scramble1: D2 U' R D' U2 L' D' R2 B F2 R2 B' F L' D' B2 F2 D' U B2 F' D2 B2 D2 L2 was the first scramble at the Polish Open 2007

Scramble2: L' B' L2 F B U R' B D' L2 B U' B2 U L U2 D2 L was my solution to weekly competition 2007-32

Last edited: