Is there a state in Rubik's cube which can be considered to have the highest degree of randomness (maximum entropy?) asssuming that the solved Rubik's cube has the lowest?

4How do you define "most random"? Perhaps by the number of moves it takes to solve the cube? – Alex Becker May 28 '12 at 20:13

I don't know exactly what you mean by *random*, but you could consider a state that requires the most amount of moves to solve. – mrf May 28 '12 at 20:14

Even most amount of moves might be hard. You would need to know a 'most efficient' algorithm. – john w. May 28 '12 at 20:19

9It is known that [the maximum required number of moves is 20](http://www.cube20.org/), but the state that requires 20 moves is not very "random" in the intuitive sense of random. – May 28 '12 at 20:30

2I would suggest the interpretation of most randoms should be most probable number of moves to solve. The site Rahul Narain linked to says this is 18 moves from start. But no single position can be considered random. – Ross Millikan May 28 '12 at 21:51

@johnw. There are optimal algorithms which are computationally useful, and have been since at least 2006. The one I use on my 10yearold laptop rarely takes more than an hour on random positions, but it seems no one can prove that the algorithms are much better than bruteforce searching in the worstcase. – Logan M May 29 '12 at 20:43

6You offer a bounty on this question, but you have yet to define what you mean by "most random". How can we possibly address your question? – Austin Mohr May 31 '12 at 01:46

The idea of maximal entropy is valid. It was first proposed (as far as I know), by Douglas Hofstadter in GEB. I've never tackled the problem completely, but I even imagined to find an algorithm diminishing the entropy until solved. I realized that the cube can be in local minima that are not solved, i.e., whatever side you turn, the entropy will increase.  A similar question was asked here (forums.xkcd.com/viewtopic.php?t=86624), but no answer was forthcoming.  Here you can find a genetic algorithm approach with a download. francocube.com/cyril/genetic_alg. – Cuc Jul 08 '17 at 01:05
5 Answers
Assuming 'most random' means 'takes the most number of moves to solve the cube', then the answer is 20. This site also has an example of a state that requires at least 20 moves to solve. This result is from 2010, and is a computerassisted proof.
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12A state requiring $20$ can be described easily: the corner cubes are all correct, and the $12$ edge cubes are all on their correct edges but are all flipped. – Brian M. Scott May 28 '12 at 20:34

9"a computerassisted proof" is a bit of an understatement. IIRC, the computer doing the assistance was a $>1$ petaFLOP Google supercomputer, and even it took a while. – Alex Becker May 28 '12 at 21:23

4It should be stated that this is with the halfturn metric (i.e. both quarter and halfturns are allowed in the generating set and count for 1 turn). The quarterturn metric is still unsolved and thought to be very hard compared to the halfturn metric AFAIK, and harder positions to solve than the superflip (which Brian described) are known. – Logan M May 29 '12 at 20:36

2@Brian, if that state has such a nice description, it cannot be called *random*, can it? :) – lhf May 29 '12 at 22:35

2nice point, another one definition of randomness for this problem  the longest sentence to describe the state, kind of Kolmogorov complexity. – Yrogirg May 31 '12 at 06:28


1The quarterturn metric has since been solved; God's number is 26 qt. More interestingly, unlike the HTM case, there is exactly one state that is 26 qt away from solved: Superflip composed with 4spot. – Mark Reed Mar 28 '15 at 05:41
Just as shuffling a deck of cards several times (say 7 or more) gives a very good approximation of a randomly chosen ordering of cards, mixing up a cube with 50 or 60 turns will give a very good approximation of a randomly chosen cube state.
A randomly chosen cube state might be what you want there is no single "most random" state. If there were, then in some sense, it wouldn't be very random; it would be very special! However, if you take a randomly chosen cube state, there's over a 99.75% chance it is at least 16 moves away from being solved; so we expect a randomly chosen cube to be very far away from a solved one.
Yes, it's possible to randomly mix up a cube and end up with something only a few moves from a solved cube, or even to end up with a perfectly solved cube. But the odds against it are astronomical. You could shuffle cards until the sun burns out (~5 billion years), and it's unlikely you would have put them back in order even once. Likewise, you could jumble the cube until the sun burns out, it's unlikely you'll just happen to solve the cube in all that time.
The nice thing about a randomly chosen state is that it's easy to approximate just randomly make turns for a while. I don't know how many turns you should do, I would guess 60 turns is plenty. I've seen people spend a long time, doing hundreds and hundreds of turns, trying to make the cube really hard to solve but all those extra turns don't accomplish much, it's still just another randomly chosen cube state, and we expect it to be just as hard to solve.
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3"I've seen people spend a long time, doing hundreds and hundreds of turns, trying to make the cube _really_ hard to solve"  Or perhaps they are just aware that humans are fairly lousy generators of randomness? When I try to just scramble a cube aimlessly, I tend to settle into repeated cycles of moves, and each move in such a cycle contributes much less randomness that an actual (equidistributed) randomly chosen move would have. – hmakholm left over Monica May 31 '12 at 19:53

2@HenningMakholm: Yes, that's possible. But I certainly don't know how to exploit such failures of randomness to solve the cube more quickly. In fact, even if you just use 15 or even 12 turns, I would start my cubesolving algorithm from scratch, and won't solve the cube any faster than usual, even though it's "closer" to a solved state than 99.75% of all cube states. – Jonas Kibelbek Jun 01 '12 at 00:51
If you're considering maximal entropy as I think your question alludes to, you need to maximize the degeneracy of the states since $S \equiv k \ln \Omega$, where $\Omega$ is the number microstates.
It doesn't make sense to consider a single state of the cube to be most random...we could just as well consider our chosen state to be "solved" and that would make it the least "random" in that sense. It makes more sense to consider an ensemble of cubes to be the most random state. In this case, the ensemble of cubes with the highest degeneracy is 18 moves away from being solved, with degeneracy of roughly 29 quintillion. [1]
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There are many many positions that require 20 moves to solve, so I'd say the position that qualifies as "most scrambled" among them is the one that has fewest alternative paths to the sorted state. I have downloaded Herbert Kociemba's Cube Explorer (halfturn version) from http://kociemba.org/cube.htm and have been using it to track some 20move positions. The position illustrated here was put to me by a Stephen Baxter (not me but an interesting coincidence of names) as requiring 25 moves. It doesn't, of course; it only needs 20. But Cube Explorer takes nearly two hours running on my quadcore 3.8 GHz AMD Windows 7 desktop to reduce its initial 21move solution to one with 20 moves. The position with all 12 edges flipped ("superflip") solves to 20 much more quickly (26 seconds). I assume this is because the illustrated configuration has very few paths to a solution. So I think it has some claim to be at least one of the most scrambled positions.
I think there will always be more than one path to a solution, because (a) trivial differences such as the order of successive noninterfering moves such as FB or U2D', which could be executed as BF or D'U2; and because of the sheer number of possible 20move sequences (18^20 = more than 12 septillion), which exceeds the number of configurations by a factor of some 294,000.
Baxter's "hard to solve" configuration doesn't look at all "random" as most people would intuitively think of the term; it has four regular pairs of top & bottom colours on the side faces, and on top and bottom faces the sidefacecolour pairs are all a chessknight's move apart.
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Wouldn't the cube with the fewest minimumlengtb paths to the solved cube be the already solved cube? – Deusovi May 25 '16 at 04:25

Yes, but I'm dealing only with configurations requiring the maximum number of moves (20) to solve, and asking for which of these configurations is there only one (or possibly only a few) distinct solutions. – Steve B May 27 '16 at 16:36
I feel like a "most randomized cube" state that this question is asking if one where the least number of individual blocks one each side are In the correct position. Not necessarily one that takes a lot of moves to solve, but where the most amount of blocks need to be moved for it to be solved. I think that's a better, and more productive for the purpose of conversation, definition then the amount of turns because that has sort of hit a conversation dead end.
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Or you could even think of it as one with the least amount of symmetry could be a good definition as well – anthony crone Oct 23 '16 at 03:25

It is easy to create a notatallrandom state where _every_ cubie is in a wrong position  e.g. $U^2D^2R^2L^2$. – hmakholm left over Monica Oct 23 '16 at 03:36