I'm not asking for minimum moves. I know it's 20. I also know that it can be solved using only 1 sufficiently long sequence, but that would take ages to perform and it'd be nearly impossible to remember. I also know that if you know hundreds of sequences you can solve it in 30 seconds or less, but you can achieve the same results as many of those hundreds of sequences but using more moves.

What I'd like to know is, efficiency aside, the least amount of sequences needed to be learned to solve it.

As this sequences have to be memorized it could also be put this way. If I sum up the amount of moves of each sequence, what's the minimum amount of moves I have to learn?.

The sequences of moves should be the same to solve any starting point, of course.

  • $\begingroup$ If you really want the restriction that the sequence should be the same for any starting point, then this is a duplicate of Is it possible to use one sequence of moves to solve the Rubik's cube from any position?. $\endgroup$ May 17, 2016 at 17:57
  • $\begingroup$ I learned to solve a Rubik's cube with something like 7 sequences. Can be done easily in under 5 minutes; without even racing. $\endgroup$ May 17, 2016 at 18:01
  • $\begingroup$ Ian MacDonald - I was thinking something along the lines of that. Let's hear what the mathematical experts here have to say. By the way, how many moves do they sum up?. $\endgroup$
    – Yuta73
    May 17, 2016 at 18:09
  • $\begingroup$ 2012rcampion - I specifically addressed that I was no refering that. That one needs a lot of moves. $\endgroup$
    – Yuta73
    May 17, 2016 at 18:10
  • $\begingroup$ Then I suggest that you remove the restriction that the sequence has to be the same for every starting point. With all reasonable strategies, you have to look at the cube and choose a different order/number of times to apply each sequence. $\endgroup$ May 17, 2016 at 18:16

5 Answers 5


While @2012rcampion is technically correct I will add what I think is a more naively implementable answer that uses both more face turns and more sequences (although still not all that easy to perform):

$2$ sequences totalling $14$ face turns

Firstly one $8$ move sequence to permute three corners for which we ignore the effect on edges, such as:
U' R U' F' R2 F U2 F'
Secondly one $6$ move sequence to permute three edges which does not affect the corners whatsoever, such as:
R2 D' U F2 U' D

Now we can:
Apply the first move in different cube orientations until all the corners are placed correctly (Note: most could probably place the first four corners without learning any algorithms)
Apply the second move sequence in different orientations, possibly prefixing and postfixing with any necessary turns to align the three you wish to permute and undo that change (that is conjugation), until the cube is solved.

Assuming we cannot perform any conjugations:

3 sequences totalling $23$ face turns

Do the same as before for the corners
For the edges also learn a $9$ move sequence to permute three edges on a side rather than in a slice, such as:
F2 D L' R F2 L R' D F2
(Note this is really the result of a conjugation: we move the three into a slice, permute with a reorientation of the $6$ move sequence from before, and then move them back out of the slice, it is also set up in such a way as to cancel a face turn).
If one needs to permute, for example, three edges neither in a single face nor in a single slice one can permute either one unsolved and two solved or two unsolved and one solved to get to a set of states one can directly can solve).

Update: ais523 asked "How does this handle orienting the edges and corners once they're placed?"

This is, indeed, not immediately obvious from the above - the answer is also not all that enlightening on its own - "We can do two things: 1. turn the cube; 2. move those already placed edges".

As such I will run through an example of a typical position one would normally think of as "there are two edges that need orienting" on the last layer.

You can see it in this jsfiddle (Using the work of Lars Petrus's Roofpig. Each time the cube is reoriented it is being set up for one or two sequential edge-permutations using the conjugation-included form, F2 D L' R F2 L R' D F2. I have left in two D F2 F2 D sequences that would, of course, just become D2 for clarity.

Here is a rundown of that:

First one may setup the "scramble" like this: B' L' B2 U2 F2 R' F2 U2 B' (or if you prefer less faces L' R F R' L D2 L' R F R'). This should leave you with U and F fully solved, all the corners in the correct locations, while the remaining three edges on D are incorrect, where one would identify the ones adjoining B and L as "flipped".

Now the (uber-convoluted!) solve using only (1) cube-orientations (x, y, z are turns of the whole cube clockwise if one were to look at R, U, or F respectively) and (2) the edge-permutation above. Steps 1-6 permute edges in such a way that we are left with two faces each requiring only edge permutations, steps 7-10 then solves those two faces:

  1. orient the cube such that R->U & D->F: y x

  2. apply face 3-edge: F2 D L' R F2 L R' D F2

  3. orient the cube such that R->F & D->U: x2 y

  4. apply face 3-edge twice (or reflect the sequence & perform it once): F2 D L' R F2 L R' (D F2 F2 D) L' R F2 L R' D F2

  5. orient the cube such that R->F & D->R: y z'

  6. apply face 3-edge: F2 D L' R F2 L R' D F2

  7. orient the cube such that R->U & D->R: z'

  8. apply face 3-edge twice (or reflect the sequence & perform it once): F2 D L' R F2 L R' (D F2 F2 D) L' R F2 L R' D F2

  9. orient the cube such that R->F & D->U: x2 y

  10. apply face 3-edge: F2 D L' R F2 L R' D F2

  • $\begingroup$ You probably can't just use these sequences alone - you also need setup moves, i.e. you need various conjugates of these sequences. The edge sequence for example keeps the edge pieces in the same slice, so any cube state where an edge is not already in the correct slice is impossible to solve. $\endgroup$ May 19, 2016 at 1:48
  • $\begingroup$ @JaapScherphuis added to the wording to specify this - I suppose one needs to learn the idea of conjugation too :) $\endgroup$ May 19, 2016 at 2:07
  • $\begingroup$ Interesting, 8+6=14. This is a pretty low amount of turns to learn. Nevertheless, I'd consider conjugation kind of cheating. Many, many times I got lost solving a cube for not being able to "come back" to original. I'd like a method where conjugation is not needed. There's no problem in freely rotate the cube before starting the algoritms. $\endgroup$
    – Yuta73
    May 27, 2016 at 21:43
  • $\begingroup$ Added one sequence to allow a non-conjugation-aware solver. $\endgroup$ May 28, 2016 at 1:32
  • $\begingroup$ Your response was the best and better documented. Thank you Jonathan for your time!. $\endgroup$
    – Yuta73
    Jun 9, 2016 at 20:35

1 sequence, 1 move total

The only move you need to learn is:

Rotate face 90 degrees left

Note that a right turn can be achieved by repeating this sequence three times.

Of course, this is assuming that you can rotate the cube between moves, otherwise you will have to learn 6 moves:

Rotate top face 90 degrees left, rotate left face 90 degrees left, rotate right face 90 degrees left, rotate bottom face 90 degrees left, rotate back face 90 degrees left, and rotate front face 90 degrees left.

  • $\begingroup$ Instead of your 6 answer, wouldn't it only be 4? Since the centers don't move, moving 4 faces will allow you to manipulate every necessary piece in the cube. You could get away without rotating the top or bottom faces, although it will take considerable skill to solve it this way. $\endgroup$ May 17, 2016 at 23:39
  • 1
    $\begingroup$ @ChronoD: You need to be able to move 5 faces. If you leave out both top and bottom face turns, then it is impossible to flip any edge pieces, so not all mixed cubes can be solved with 4 faces. You can get away with using only 5 faces because (using standard cube move notation) the move sequence F2R2D2F2U2R2F2 U F2R2U2F2D2R2F2 has the same effect as D. $\endgroup$ May 19, 2016 at 1:43
  • $\begingroup$ @JaapScherphuis I see. I didn't know enough to think about flipping corners. Thanks for the correction. $\endgroup$ May 19, 2016 at 1:44
  • 2
    $\begingroup$ Sorry, but this is not a correct interpretation of the intended objective. There's no problem in rotating the cube. I understand of course that moving one face at a time you can move any piece, but the idea here is to be able to solve the cube following a step by step procedure and knowing no more than a minimum amount of sequences. For example, if piece X is HERE and piece Y is HERE, follow this sequence to achieve THIS desired result (or repeat it the amount of times needed to do so). $\endgroup$
    – Yuta73
    May 27, 2016 at 21:33
  • 3
    $\begingroup$ An issue with the framing of a question should be taken up with a close vote/comment; this answer is just a plain unhelpful response to the question, and as a result, I've downvoted it. $\endgroup$
    – user20
    May 28, 2016 at 2:15

You can do it with just four sequences (and maybe fewer). These are

  1. swap two corners
  2. swap two pairs of edges
  3. rotate two corners, one counterclockwise, one clockwise
  4. flip two edges

Each of these sequences should leave the rest of the cube in the same state as it was. If your two corners swap acts on the top right and top left corners, you can use it to swap any pair. To do this you move the corners you want to swap to the top left and top right, not worrying about how it scrambles the rest but remembering the (usually rather short) sequence of moves you used. Apply your swap sequence, then undo the moves that brought the corners into position. The rest of the cube will be back where it was and the corners of interest will be swapped. I found it easier to do the top layer by inspection, then only use these sequences for the bottom two layers.

  • 1
    $\begingroup$ This is approximately the technique I 'discovered' about thirty-many years ago. Could solve it in about 3-4 minutes after which I got bored and never tried it again. $\endgroup$
    – Penguino
    May 18, 2016 at 23:34
  • $\begingroup$ I found the sequences by experimenting with different simple moves and their inverses. For example A followed by InvA gets you back to the starting point, but if you have two simple sequences A and B, then teh sequence (A B InvA InvB) quite often gives you a corner swap or an edge rotate or similar. $\endgroup$
    – Penguino
    May 18, 2016 at 23:40
  • $\begingroup$ I got the cube soon after it came out, played with it for a while, got frustrated and put it down. There was a Martin Gardner article in Scientific American that talked of the power of commutators, I think early 1981. One of my friends at work and I solved it the same Sunday morning after seeing that. $\endgroup$ May 18, 2016 at 23:43
  • $\begingroup$ OK. How may moves are needed for each sequence?. $\endgroup$
    – Yuta73
    May 27, 2016 at 21:34
  • 1
    $\begingroup$ You can't just swap two corners -- every legal move permutes the 28 visible cubies in an even permutation, since one quarter-turn decomposes as two 4-cycles. So your (1) needs to be "swap two pairs of corners", and then you need something to use if you've gotten into a situation where the corners and edges are both an odd permutation from being solved -- but a single quarter-turn will fix that. $\endgroup$ Jul 2, 2016 at 20:55

Basing on the beginners method I use:

Completing the bottom layer.

I'll assume this to be trivial.

Putting in the edge pieces on the second layer

There's an 8-move sequence (1) that allows you to put any edge from the unfinished third layer into a slot of your choice on the second layer you're trying to complete.

Turning all colors up on the last layer

There's a 6-step routine (2) that'll lead up to creating a 'fish' shape in the top layer, and a 7-step routine (3) to go from 'fish' to 'full same color' top view in that last layer.

Putting the corners in the right places.

This takes the longest routine to memorise (4), involving 9 steps.

Putting the edges from the top in their correct position.

There's a routine that switches around 3 of the edges (and you'll always need either 3 or 4 swapped, or none at all). This routine however is applying the 7 step routine (3) twice.

All in all, as far as learning algorithms goes, I have used a total of


algorithms, assuming completing the first layer is deemed trivial. Not how I'm not spelling them out - basically because I never learnt them in code form.

Also trying to calculate an upper boundary of moves for this method (so absolute bad luck):
Edges on bottom layer: 4 x 2 moves = 8 moves
Corners on bottom layer: 3 x 8 moves + 1 x 5 moves = 29 moves
Edges on middle layer: 4 x (8+8) moves = 64 moves
All colors on top surface: 3 x 6 moves + 2 x 7 moves = 32 moves
Aligning all corners in top layer to correct position: 2 x 9 moves = 18 moves
Alining all edges in top layer to correct position: 2 x (2x7) moves = 28 moves

Grand total: 179 moves worst case.

  • $\begingroup$ Well, the amount of moves to learn really are 8+6+7+9=30 moves. I think this very same algorithms can be used also to complete the bottom layer. You are using more moves than Allan, but apparently there's no need to conjugate in your case. It's a pity you didn't learn the notation so I can probe your claims. $\endgroup$
    – Yuta73
    May 27, 2016 at 21:55
  • $\begingroup$ Well, I think this is what I used as reference: ryanheise.com/cube/beginner.html (well, from the middle layer onwards anyway) $\endgroup$ May 28, 2016 at 18:30

There is a few really difficult but extremely efficient methods that does not require you to learn any algorithms. One example which I've mastered is known as Heise. Heise method is a really efficient method that requires the knowledge of conjugates and commutators. The steps for Heise method:

  1. Build 4 1x2x2 squares with at least 1 color in common and 1 color that isn't in any of them such that you could make it into a 2x3x3 block minus a pair in at most 3 moves.
  2. Match the squares and orient the edges, should be pretty easy if you learn methods like Petrus
  3. Solve all edges and 2 corners. This step requires lots of practice and patience to do efficiently, however, with practice, should be able to do all cases in 10 moves or less
  4. Solve the whole cube with a commutator or a conjugate.

As you can see, this method requires no algorithm whatsoever. It can also easily be transposed to other blockbuilding methods like Petrus or Tripod if you did not master it before a competition. However, this might not be feasible in half an hour if you are clueless about how to perform the steps. Step 3 and 4 could be broken into sub-steps to make it easier, though it will still be really difficult for a beginner. This method, when mastered, will take at most 40 moves!

Much more beginner-friendly method: This method would only require 1 algorithm at least. It is the Ja/b perm(10 moves). The cube could be solved 1 piece at a time by letting one piece be a 'buffer' and solving all the edges/corners through it. This is also a commonly thought beginners bilndfold method, though I highly discourage it. To understand how it works, let's assume we want to sort a series of numbers into ascending order.


Now we could let 4 be a 'buffer'. 4 has to go to 3's position. So we swop them.


Now 3 is swopped with 2


Now 2 and 5


And finally 5 and 1


This is how the method that only requires the Ja/b perm works, except it is for the whole cube. And it occasionally requires you to replace the buffer with a random unsolved piece if the buffer is solved.

If you are planning to just be fast and actually compete with other people, I do not recommend learning the above methods, except for the first if you are competing fewest move, or modify it for one handed and/or feet solving.

A good beginners method will be like:

  1. Solve cross of a face(assume white here, but I recommend to practice on all 6 colors)
  2. Solve the white corners(1 alg, R U R' U', 4 moves)
  3. Solve E slice edges. There are a few algorithms here, and also it can be done intuitively(0-2 algs, optimal is 8 moves)
  4. Orient the yellow edges(1 alg, 5 moves)
  5. Orient the yellow corners(0 algs, done with R U R' U', 4 moves)
  6. Permute the yellow edges and corners(2 algs, 9 each)

It should take about 100 moves, but I didn't really counted it.

If you want examples of the method shown above, you could ask in the comments, and maybe provide a scramble, and tell me where you get stuck at.


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