Are there positions of the rubik cube which cannot be reached by applying the standard moves starting from the solved cube? If so, how many such positions are there?

  • $\begingroup$ Lots has been written about this question, try here for a start. $\endgroup$ – David Feb 13 '14 at 1:09
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    $\begingroup$ For one, notice the middle blocks are fixed relative to one another. permute them and youll get unreachable positions $\endgroup$ – David P Feb 13 '14 at 1:09
  • $\begingroup$ Rotate a corner clockwise (or anticlockwise) and you won't be able to reach that position. $\endgroup$ – Beni Bogosel Feb 13 '14 at 1:22
  • $\begingroup$ The same for flipping an edge piece. $\endgroup$ – user1729 Feb 13 '14 at 15:43

11 out of 12 random positions are unreachable.

This first part just looks at the positions of cublets, without regard to their orientation.

It is possible to place the corners in all $8!$ position permutations. Separately, it is possible to place all edge pieces in all $12!$ position permutations. But it is not possible to apply a permutation that swaps only two corners or two edges.

This is easily seen. All possible permutations arise from the application of 6 basic permutations corresponding to quarter-turns on each face. Each of these quarter-turns cycles 4 corners (an odd permutation, equivalent to 3 swaps, an odd number) and 4 edges (also an odd permutation). The combination of these two cycles is an even permutation (3+3=6 swaps, an even number), so any series of basic moves also results in an even permutation. This sets an upper bound of $\dfrac{8!12!}{2}$ position permutations.

It is possible to construct permutations that solely cycle the positions of 3 corners or 3 edges, so using these 3-cycles you can independendly reach all $\dfrac{8!}2$ even corner permutations and $\dfrac{12!}2$ even edge permutations, resulting in $\dfrac{8!12!}{4}$ total even position permutations. Turning one face throws both the corners and edges into odd permutations, and from here using the 3-cycle moves you can reach all $\dfrac{8!12!}{4}$ odd position permutations. Add these together and you get a lower bound of $\dfrac{8!12!}{2}$ reachable position permutations, which meets the upper bound.

In the second part here I'm talking about changing the orientations of the cubes, without changing their positions.

Corner rotations:

Moves exist that jointly counter-rotate any two corner pieces without affecting any other changes. This sets a lower bound of $3^7$ possible rotations of the corner pieces.

Consider a rotation labeling system for the corner cubes. Mark one facelet of each corner as special. Each corner cubelet position has a face whose normal points upward or downward, and call this position 0, the one clockwise from it 1, and the next clockwise 2. You will find that each basic move preserves the sum of these orientation numbers modulo 3, which sets an upper bound of $3^7$ possible corner orientations.

Edge rotations:

Move sequences exist that jointly flip any two edge pieces without affecting any other changes. This sets a lower bound of $2^{11}$ possible rotations of the edge pieces.

If you put together an arbitrary orientation numbering system like the corner orientation numbering system above, you'll find that all basic moves preserve the sum of orientations modulo 2. This sets an upper bound of $2^{11}$ possible rotations of the edge pieces.

So the cube in total has $\dfrac{8!12!2^{11}3^{7}}{2}$ permutations reachable through basic moves.

If you disassemble a cube you can place the corners in any of $8!$ position permutations and $3^8$ orientations, and the edges in any of $12!$ position permutations and $2^{12}$ orientations. This totals to $8!12!2^{12}3^{8}$ possible configurations, which means 11 out of 12 potential configurations are not reachable through basic moves.


There are obviously some positions that are unreachable, and one way to get there is to twist a corner once. To elaborate on @NovaDenizen's answer, I tried many ways to make a cube unsolvable. First, there are two ways to get an unreachable position just by twisting a corner, and that is to twist it once clockwise or counterclockwise. This accounts for two 'sets' of unreachable positions. Each set is made up of all of the positions reachable with normal moves from a particular starting position. Another way to make a cube unsolvable, is to flip one edge. This accounts for another set of unreachable positions. Combining a twisted corner and a flipped edge yields two more sets of unreachable positions. The final way is to switch two corners, but this is identical to switching two edges. Doing an algorithm, such as the V-perm, can make two switched corners change to two switched edges. Combine this with the other five unreachable sets, and you will have a total of 11 unreachable sets. Given that the number of possible positions of a 3x3x3 Rubik's Cube is 43,252,003,274,489,856,000, and that there are 11 other sets of unreachable positions, you will have a total of 519,024,039,293,878,272,000 positions. This means that if you take apart the whole cube(except for the centers), you have a 1 in 12 chance that you will be able to solve it using normal turns, and not by taking apart the cube. Hope this helped.

  • $\begingroup$ Funny names: a twisted corner is a quark / antiquark, two twisted corners with opposed orientation are a meson, three twisted corners with the same orientation are a baryon / antibaryon. More in math.niu.edu/~rusin/known-math/95/rubik.physics. $\endgroup$ – Martín-Blas Pérez Pinilla May 7 '14 at 6:24
  • $\begingroup$ Just like the particles in atomic science. $\endgroup$ – Jason Chen May 7 '14 at 23:11

Yes, there are. These arise from to quick observations: pieces move as part of a $3\times 3$ cube, not as individual faces, and center cubes must stay in the center, which is also true with corner pieces and side faces. Without loss of generality you can fix the center pieces, and then there are $8$ corner pieces $12$ side pieces. This gives us a maximum of $$8!12!=19313344512000$$ However I'm sure there are many fewer reachable positions than this, as pointed out in the comments. But this number already shows how many fewer positions this is the the overall number of permutations of 26 cubes: $$\frac{8!12!}{26!}=\frac{1}{20881492632000}$$ which is obviously a pretty small number. Obviously this isn't nearly exhaustive, but I hope this helps (also, I completely ignored changing orientation, which can definitely happen).

  • $\begingroup$ The first maximum should be multiplied by $3^82^{12}$ for orientations of the cubies. Wikipedia has a good discussion. It is in fact a factor $12$ down from this $\endgroup$ – Ross Millikan Feb 15 '14 at 23:39

I think a better way to ask your question would be to ask if there are configurations of the colored stickers, if you peel them off and move them, that don't correspond to cube positions reachable from the solved position by standard moves (or conversely are not solvable starting positions), and the answer is an emphatic "yes" -- for example, if you take the two stickers on one of the edge pieces and reverse them, the resulting cube is unsolvable. I don't have any numbers, but I suspect that the number of unsolvable configurations of the stickers is several orders of magnitude higher than the number of solvable configurations.

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    $\begingroup$ The OP explicitly stated "by applying standard moves", so moving stickers doesn't count. This answer does not address the intent of the OP. $\endgroup$ – Tpofofn Feb 13 '14 at 1:49
  • $\begingroup$ That depends on how you interpret "positions" -- positions of the stickers (which is the interesting question, the one I answered, and how I interpreted the OP) or positions of the cube when scrambled using standard moves (in which case the answer is tautologically and vacuously "no," you can just undo the scrambling moves) -- I apologize for assuming that the OP wanted to ask something interesting instead of something silly. Oh, and both NovaDenizen and dirty derwin took interpretations equivalent to moving the stickers. I can't be wrong if they are right. $\endgroup$ – user128390 Feb 13 '14 at 3:29
  • $\begingroup$ @user128390 I took dirty derwin's answer to be weaker than moving stickers - he seems to be saying that you could make an edge piece a corner piece, and vice-versa. Which is silly! $\endgroup$ – user1729 Feb 13 '14 at 15:45
  • $\begingroup$ @user1729 -- to be fair, there is a difference between moving the stickers and disassembling/reassembling the cube (by moving stickers, for example, you could wind up with a yellow-yellow edge piece, which can't happen if you disassemble without peeling), and NovaDenizen's answer at least could be interpreted as disassembling the cube instead of moving stickers (dirty derwin specifically talks about "moving individual faces," i.e. peeling and moving stickers), but neither of these fit Tpofofn's overly restrictive requirement that the only thing allowed is standard moves. $\endgroup$ – user128390 Feb 13 '14 at 17:12
  • $\begingroup$ @user128390 Actually, I think I mis-interpreted the other answer. But I am not sure any more... $\endgroup$ – user1729 Feb 14 '14 at 9:53

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