# Thread: Fewest Moves: Tips and Techniques

1. Welcome (back) Per, we seem to meet a lot online lately.

Have you had time to read the whole first post of this thread? I hope you will share some of your ideas with us.

I just noted that I forgot to mention this one (very obvious, but still)

Don't restore the cross:
After/between corner-edge-pair-insertions there is no reason to keep restoring the cross. Most of the time the "solved pieces" are just fine where they are and you will probably save 2 or more moves (1 for restoring the cross, another 1 for repositioning it for the next pair)

Don't restore the cross 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
2 pairs (4 pieces) in 3 moves is much better then 7 (R D2 R') (L B' L' B)

P.S. I would like a brownie, it is lunch time

2. Originally Posted by joey
From speaking to Per Fredlund, one of the masters of FM, there are a few extra things. Per is amazing, and once solved the cube in 19 moves, yes, that probably was the optimal solution!

Another approach is to solve edges first, while trying to gather corners. This is basically solving all edges in the shortest scramble, while trying to solve a few corners aswell. Then using insertions to correct the rest of the corners.

Also, orienting first can help. So orient all the edges first, while try to solve blocks/cross can help, as it will lead to shorter F2L pairs and a most likely easier F2L.

The last thing I can think of right now, is solving like a Domino. For those who don't have a Domino (2x3x3), It is the same as using only <UDR2F2L2B2> on a 3x3. To solve like a domino, you would first solve the E layer while orienting edges. You would then complete as much of the first layer and second layer. Then go back and finish it off with insertions.
Hi

It is a common misconception that a "domino solve" should reduce to <UDR2F2L2B2>. It will also be equivalent to reduce to <U2D2RF2LB2> or <U2D2R2FL2B>. They're rotation invariants ...

My best advice for those just starting to embark on fewest moves solving would be:

- forget about using "standard" speed solving methods with a few optimisations and lucky cases

- think of it more like so: solve it in 2 stages where stage 1 solves as many pieces as possible as efficiently as possible, ie make a good skeleton. The situation now would typically not be good for normal continued solving. The second step is fixing the unsolved pieces, typically by inserting algorithms into the skeleton.

This approach requires some knowledge about what are good skeletons?? I would say a good skeleton should leave no more than 2 easy cycles for insertion, unless the skeleton is VERY short. Leaving 5 unsolved corners (as a pure 5-cycle) is good because it breaks down to 2 3-cycles, with lots of options what cycle to do first. 5 unsolved corners where 1 or more is in correct position but wrongly oriented is typically bad. This would require 3 3-cycles to complete. There is similar considerations for cases where the skeleton solves all corners, not all edges. A skeleton leaving 3 unsolved edges and 3 unsolved corners is also good, though peronally i prefer working with corner cycle insertions. There is too many possibilities for making edge 3-cycle to be sure that what you are doing is anywhere near optimal insertion procedure ;-)

Making corner 3-cycles is INTUITIVE and requires almost no skill whatsoever!! Just apply stickers (numbered) to the corners to know the cycle order and orientation. We are not going to waste moves on doing a 3-cycle in 2 steps: permute then orient:-P This will however require an understanding of commutators. I leave that for a later post

PS! The only bad case with 4 unsolved corners is the case where they all just need to be twisted..

-Per

3. This will however require an understanding of commutators. I leave that for a later post
This is what they call a cliffhanger!

I don't have any problems with edge-cycles, insertions and commutators. I have no idea how to do corner-cycles. The alg I currently use (A-Perm) is pretty short and often has cancellations. However it requires orienting corners first. I only remember two old corner-cycle-commutators: (RUR'U')L' (URU'R') L and (URU'R') L' (RUR'U') L.

About the skeleton: What do you consider the worst acceptable skeleton? 5 corners unsolved, 5 edges unsolved or 5 corners and 5 edges unsolved? (and yes, I understand that that depends on the amount of moves it took to build the skeleton)
Please don't keep me waiting for this to long. I would like to be able to use it at Worlds.

4. Hi :-)

RUR'U' L' URU'R' L is already wasting moves ;-)

Breaking down that alg we have

RUR'U' L' URU'R' L = P Q P' Q' where
P=RUR'U', Q=L'

However, a shorter P would simply be B' R' B, yielding

B'R'B L' B' R B L

Alternatively one could move L-layer first, then it becomes:

L F R' F' L' F R F'

This should give some idea about how commutators work

============

The worst acceptable skeleton would probably be one i didn't mention in my post: edge 4-cycle pluss corner 4-cycle.

>>Arrange all those 8 cubies on same layer so they're solved by a single turn, then undo the setup.

Alternatively, since the setup is bound to be a bit long:

>>Arrange 7 of the cubies on same layer so that a single turn pluss undoing setup either leaves a corner 3-cycle or an edge 3-cycle. Then try to insert that somewhere else in the new skeleton.

============

Note that the order of insertions has an effect. After some insertion we are working with a new skeleton ;-) A pre-emptive search takes time. I normally go for "best insertion first" :-)

2-flip on edges or 2/3 twist on corners can be solved by 2 inserted 3-cycles...

-Per

(edit)
The case with BOTH 5 unsolved edges and 5 corners is of course the worst. Inserting all that is pointless. I would solve the cube onto a better skeleton, by extension or restart from scratch ;-)

5. Originally Posted by AvGalen
I don't have any problems with edge-cycles, insertions and commutators. I have no idea how to do corner-cycles.
Now that's weird. Just use commutators. Ryan has good tutorials at http://www.ryanheise.com/cube/commutators.html.

6. I will look into this on tuesday after I have moved and before Worlds. It will be a hectic week.

7. Talking about fewest moves tricks and tips, I'd also mention pseudo-blocks and premoves.
Knowing whole OLL, PLL, COLL is actually not that important. The shortest solutions are the intuitive ones with some inserted cycles. An algorithm cannot be optimized(aside from cancelling 1-2 moves at the beginning or the ending).

8. Marcell, the corner commutators can be solved in a 9 move average. There aren't really that many corner cycles. It's just seeing everything from the proper angle.

Since you're not blindfolded, Per is right, finding the best insertion and which one to do first can affect the other insertions.

You could have a 5-cycle of corners.
Lets say you took the 9 mover ferris-wheel, for both insertions. 18 moves to finish the corners. Yet, if you cycled differently, rather than cycling from the URB, say you cycled from the FDR, then you could wind up using only 17 moves. A pure 8 mover, and then a ferriswheel.

If you were to conjugate the 2 commutators into an optimized 5 cycle. It could actually work out that the cycle could become 15 moves, because of a cancelation, because both cycles have the same interchangeable slice per se.

I don't have actual examples at the moment.

From a blindfolded perspective, bound by a buffer, you'll get the 9 move average for each cycle. Yet you could pick how to cycle better sighted with an hour of alotted time. You could find the two optimal 8 move commutators.

ABCDE cycling from A for both cycles may take 18 moves (two ferris wheels)
BCDEA could take 17 (a ferris wheel and a pure commutator)
EABCD would be the optimal 16 move solution.

PS.
A Ferris wheel is the nickname of a special conjugation of a commuator, where there is a cancelation with the A or B and the setup (conjugate).

9. When I made this tutorial I didn't know pseudo-blocks and premoves (they are tightly related). I don't have time to make a tutorial right now, so I am going to copy/paste some things from week 42:

Scramble:
1. B F U' B L D' F R2 U2 L F L U2 L2 B' F U (17 moves optimal solution, easier to apply multiple times)

Solution: D' R2 F2 B' L F' D2 L B2 R L' D' R' D' B' D B2 L' D L B R' B' R D U2 B
Explanation: Do Premoves U2 B so the scramble becomes U2 B2 F U' B L D' F R2 U2 L F L U2 L2 B' F U
1x2x3 block (1): D'
2x2x3 block (6): R2 F2 B' L F'
(the rest is not important, but might be interesting anyway so I included it anyway. The premove/pseudo-block is basically finished here, except for the undoing of them at the end)
Create remaining 2 pairs (7): D2
Cross + Last Layer Manipulation (14): R . D R' D' B' D B2
Insert 3rd pair (17): L' D L
Insert 4th pair (21): B R' B' R
Fix last layer leaving a 3 cycle (22): D
Undo premoves (24): U2 B
The 3 cycle can be inserted at the dot between move 8 and 9 as R' L B2 R L' D2. The first move completely cancels with move 8 and the last move changes move 9 from a D to a D'

It is pretty hard to explain this, but I will try:
1. Perform the scramble with white on top and green on front.
2. D' makes a perfect column of orange-blue on the left. It also makes a line of green with red-orange on both sides. Red and orange are opposite colors, so this green line is also correct. Together this makes a pseudo 1x2x3 block.
3. Realize that to turn this pseuse 1x2x3 block into a pseudo 2x2x3 block all that is necessary are two more edges in the S-slice.
4. Realize that the best two positions for those two edges are UL and UR because the pseudo 1x2x3 block can be positioned at Up-Front with the green line correctly placed.
5. R2 puts an edge into the UR position with the correct orientation.
6. F2 moves the pseudo-1x2x3 block out of the way for the next edge.
7. B' L puts an edge into the UL position with the correct orientation and relative position compared to the UR edge.
8. F' finishes the pseudo-2x2x3 block.
9. Now to find out how to do the premoves you need to see that the green line is alread positioned correctly. So all you need to do is find the shortest move sequence that positions the white line and the blue-orange column
10. U2 positions the white line.
11. B positions the blue-orange column
12. Now that we know the premoves, undo all the moves (B' U2 F L' B F2 R2 D and undo the scramble.
13. Perform the premoves U2 B and do step 2 to 8 again. You can see that all pseudo-blocks are now real blocks.

And I want to warn everyone for larger corner-cycles. They can be very usefull, but because they will require quite a lot of moves (15, even with some cancellations) they are probably only usefull if you have a very short beginning that fixes all the edges. I think that is what Per calles a skeleton.

10. Originally Posted by AvGalen
When I made this tutorial I didn't know pseudo-blocks and premoves (they are tightly related). I don't have time to make a tutorial right now, so I am going to copy/paste some things from week 42:

Scramble:
1. B F U' B L D' F R2 U2 L F L U2 L2 B' F U (17 moves optimal solution, easier to apply multiple times)

Solution: D' R2 F2 B' L F' D2 L B2 R L' D' R' D' B' D B2 L' D L B R' B' R D U2 B
Explanation: Do Premoves U2 B so the scramble becomes U2 B2 F U' B L D' F R2 U2 L F L U2 L2 B' F U
1x2x3 block (1): D'
2x2x3 block (6): R2 F2 B' L F'
(the rest is not important, but might be interesting anyway so I included it anyway. The premove/pseudo-block is basically finished here, except for the undoing of them at the end)
Create remaining 2 pairs (7): D2
Cross + Last Layer Manipulation (14): R . D R' D' B' D B2
Insert 3rd pair (17): L' D L
Insert 4th pair (21): B R' B' R
Fix last layer leaving a 3 cycle (22): D
Undo premoves (24): U2 B
The 3 cycle can be inserted at the dot between move 8 and 9 as R' L B2 R L' D2. The first move completely cancels with move 8 and the last move changes move 9 from a D to a D'

It is pretty hard to explain this, but I will try:
1. Perform the scramble with white on top and green on front.
2. D' makes a perfect column of orange-blue on the left. It also makes a line of green with red-orange on both sides. Red and orange are opposite colors, so this green line is also correct. Together this makes a pseudo 1x2x3 block.
3. Realize that to turn this pseuse 1x2x3 block into a pseudo 2x2x3 block all that is necessary are two more edges in the S-slice.
4. Realize that the best two positions for those two edges are UL and UR because the pseudo 1x2x3 block can be positioned at Up-Front with the green line correctly placed.
5. R2 puts an edge into the UR position with the correct orientation.
6. F2 moves the pseudo-1x2x3 block out of the way for the next edge.
7. B' L puts an edge into the UL position with the correct orientation and relative position compared to the UR edge.
8. F' finishes the pseudo-2x2x3 block.
9. Now to find out how to do the premoves you need to see that the green line is alread positioned correctly. So all you need to do is find the shortest move sequence that positions the white line and the blue-orange column
10. U2 positions the white line.
11. B positions the blue-orange column
12. Now that we know the premoves, undo all the moves (B' U2 F L' B F2 R2 D and undo the scramble.
13. Perform the premoves U2 B and do step 2 to 8 again. You can see that all pseudo-blocks are now real blocks.

And I want to warn everyone for larger corner-cycles. They can be very usefull, but because they will require quite a lot of moves (15, even with some cancellations) they are probably only usefull if you have a very short beginning that fixes all the edges. I think that is what Per calles a skeleton.
This is an example of a skeleton yes. Don't underestimate inserting corner 5-cycles. I once did it in 7 turns extra (!!!) after cancellations. First cancellation cancelled 7 turns, the second one cancelled 2 more turns. This is exceptionla though, and probably was due to overlooking something in the skeleton anyway

A skeleton is any start that leaves cubies to be inserted. Proper insertions or just adding at the end ...

-Per

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