A Hamiltonian Circuit for Rubik's Cube
bruce.cubing.net
bruce.cubing.net
You also need several billion years to do this, so it's not recommended for beginner solvers.
I wonder if there is a single not-too-long rotation sequence that generates the whole cube group. That is, a sequence XYZ that you can perform repeatedly and have that bring you through every cube state. If not, maybe there is some other very simple algorithm that traverses all the states, instead of a zip file that uncompresses to 200MB.
The Hamiltonian cycle sequence from the original post is not a generator, but it visits every state. The question is: Is there a significantly shorter sequence that (when repeated) does the same?
123, 213, 231, 321, 312, 132, 123
I’m sure they could make it go faster but I suspect the reason the robot works at this speed is the friction in the cube only raises the temperature a few degrees before the cube settles and can cool down. And pausing for 60 milliseconds so the human eye has time to register the cube positions probably doesn’t suffice.
You could probably get it down to a billion years without reducing it to just a blur. A couple million if you point a bunch of high speed cameras at it and show the last fifty moves on screens next to the machine. Or 30 machines all showing 1/30th of the cycle.
wouldn't that be the very paper that's shared as article here?
The linked circuit does have a repeating pattern like this at its core, and visits entire cosets of a subgroup at a time (presumably the same way each time) so it seems like there's already a bit more structure than a random 200MB file.
https://en.wikipedia.org/wiki/Rubik%27s_Cube_group#Group_str...
They made a 15 puzzle where instead of numbers the tiles had letters on them, and the goal was to make a particular sentence. That sentence had 13 letters that appeared once and one letter that appeared twice.
That letter that appeared twice was the first letter of the sentence, and its other appearance was at or near the end. Let's call the two tiles with the same letter T1 and T2, where T1 is the tile that is the tile that is first.
They would show someone the solved puzzle, then scramble it, and challenge the person to solve it. When they were scrambling it they made sure to move T2 to the first position and T1 to somewhere near the end.
Nearly everyone who tried to solve it would see the right letter at the first position and never move it, because most people try to solve left to right and/or top to bottom.
On the 15 puzzle exactly half of the possible permutations of the 15 tiles are reachable from a given starting position. Exchanging two tiles in a reachable permutation produces an unreachable permutation.
That means that the right sentence but with T1 and T2 swapped is not reachable and so people who left T2 in front were doomed to failure.
I know it happens on the 3x3x4 cuboid - each of the 3x4 faces has two centers that appear identical, but if you have exactly two such unsolved pieces, it's unsolvable because it's impossible to exchange exactly two of them - what you need to do is cycle 3 of them where 2 appear identical. I remember this happening on some other shapes and sizes as well.
(Pic for reference: https://www.grubiks.com/images/puzzles/30/small.png . This puzzle exists as do many other NxMxH cuboids. Rectangular but non-square faces can only make 180º turns.)
Did you mean flip an edge, or 120°?
If you got to the point where you can routinely solve cubes, i.e. you know your algorithms or you developed a good intuition, then unsolvable cubes should be pretty obvious. Some algorithms are like: if you are in this position, do this, if you are in that position, do that, other positions are illegal. If you are in an illegal position, then you know there is a problem with your cube.
The Rubik's cube is a really hard puzzle unless you know some theory, and the existence of unsolvable cubes is a rather well known fact, meaning that there shouldn't be that many people who are able to solve the cube and don't know about unsolvable cubes.
Or maybe that's just me. I learned on a terrible cube that had a tendency to dismantle itself, so knowing that it had to be put back properly was a requirement for a successful solve.
When I first read the article, the sequence of subgroups that were described evoked that image of a combination lock to me:
< UR >
< U, R >
< U, R, D >
< U, R, D, L >
< U, R, D, L, F >
The behavior of the basic operations on the cube reminds me of the product of quaternion base vectors (i,j,k). For instance, the product of i and j would yield either k or -k depending on the order of i and j. I think the point I wanted to make is that on a combination lock, each operation on a wheel only affect that wheel, not the others, so one cannot produce another operation by combining several of them, like what we see with quaternions. However, on the cube, it is often possible to go from one combination to another by different sequences of different operations.
But that may not matter much, if all we care about is going through every possible combination exactly once, just like what one does when using gray code on binary numbers (which is why I alluded to that in my other post), and that for that purpose we can find a set of sequences of operations - let's call them large operations - that are orthogonal (and thus emulating the rotating wheel aspect of the combination lock). I suppose that these subgroups represent the large operations. The problem you bring up now is that these large operations are not commutative, and so finding a correct way to apply them to build the circuit is more involved than simply spinning the wheels on a lock.
Is that correct?
Edit: I just had a first look at cayley graphs on wikipedia, and they use quaternion rotations as an example!