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Fewest Moves: Tips and Techniques

Discussion in 'How-to's, Guides, etc.' started by AvGalen, Sep 23, 2007.

  1. 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:

    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

    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)

    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: Sep 23, 2007
    DaveGraff likes this.
  2. joey

    joey Member

    Apr 8, 2007
    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.
  3. ExoCorsair

    ExoCorsair Member

    Dec 18, 2006
    Very helpful post! *would +rep if the option were available here*
  4. joey, a few examples would be very helpful. Could you provide some?

    Exo, what is +rep?
  5. joey

    joey Member

    Apr 8, 2007
    Well, this is his 19 move solve:
    Scramble: D U' R' F B2 (R B2 R' U2)*4 B2 F2 U (D F D' B2)*4 F' D B' F R'
    (Don't ask me why the scramble is like that, its how it was given)

    His solution: R' B U' D F L' F2 L D2 L2 F2 D F2 D L2 U' R2 D2 R2

    The thing is, it is hard to look back at a solution, because of cancelations / insertions.
    I believe the first 8 moves orient edges and corners, which can then be solved like a Domino.(Althought its is a slightly non-pure domino finish apparently)

    If you want to see more of his solutions, head on over to http://www.cubestation.co.uk/cs2/index.php?page=fmc/fmcnewsystem where you can look back at the archives to see his old solutions. He averages around 24moves most of the time. And he uses all the tricks arnaud says, and is a master at commutators!

    I will see if he will join this forum!
  6. Do you have a crush on Per? Or is he your father? And are you sure that he averages 24 moves when there is a 1 hour time limit? And 19 moves was really bad because it could have been 17 according to Cube Explorer :)

    Seriously, people like Per, ZZ, Guus and probably Ryan Heise are so much better than I am that I wouldn't dream of calling myself an expert. I do seem to be really good at finding efficient beginnings. It would be great to have Per on this forum.

    Oh, and a scramble like D U' R' F B2 (R B2 R' U2)*4 B2 F2 U (D F D' B2)*4 F' D B' F R' is given in some competitions to provide a long scramble with less room for mistakes. A long scramble supposedly gives more random scrambles and it is easier to see if competitors obey this rule: "E2e) The solution of the competitor must not be in any way related to the scrambling algorithm". If you have a 45 move scramble and a 37 move solution it is pretty easy to convince a judge that you did obey rule E2e.
    Last edited: Sep 24, 2007
  7. ExoCorsair

    ExoCorsair Member

    Dec 18, 2006
    +rep is just a representation of how many good/useful posts a user has made, determined by other users, as seen on some other vBulletin forums.

    I'm pretty sure he's already registered here.
  8. joey

    joey Member

    Apr 8, 2007
    I didn't want people to know XD

    If so, he hasn't said anything yet :p

    I think he does try stay within the 1 hour, but will somestimes get obsessed and do it for many more hours. I think he could sub-30 in an hour, not entirely sure.

    The next time he is on MSN, I'll speak to him.
  9. pjk

    pjk Administrator Staff Member

    Mar 13, 2006
    Very nice Arnaud, thanks! You have definitely improved very quickly with this stuff.

    Per's ID on here I believe is "mrcage".

    "+rep" is basically reputation. Users can give other users reputation points for basically anything they want, such as a helpful post. However, I didn't enable that on here since I didn't think it was necessary.
  10. mrCage

    mrCage Member

    Jun 17, 2006
    Brownie points and sledge hammers don't belong in a friendly nice forum like this :)

  11. 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 :)
  12. mrCage

    mrCage Member

    Jun 17, 2006
    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 :D

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

  13. 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.
    Last edited: Sep 27, 2007
  14. mrCage

    mrCage Member

    Jun 17, 2006
    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 :D


    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...


    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 ;-)
    Last edited: Sep 27, 2007
  15. Johannes91

    Johannes91 Member

    Mar 28, 2006
    Now that's weird. Just use commutators. Ryan has good tutorials at http://www.ryanheise.com/cube/commutators.html.
    Robert Marik likes this.
  16. I will look into this on tuesday after I have moved and before Worlds. It will be a hectic week.
  17. Marcell

    Marcell Member

    Sep 1, 2007
    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).
  18. dbeyer

    dbeyer Member

    Sep 12, 2006
    Harrington, Delaware
    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.

    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).
  19. 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:

    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.
    Last edited: Dec 12, 2007
  20. mrCage

    mrCage Member

    Jun 17, 2006
    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 :D

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


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