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Imagine that we have something like the rubik's cube on $U=I^{3}=[0,1]^{3}$, and that we say that all the points on the xy, yz, and zx faces are black, and that all the points on the other three faces are white.

Define a legal move to be a rotation by an integral multiple of $\frac{\pi}{4}$ around one of the lines going through the center of each pair of opposite faces, for any square subset of $U$ orthogonal to the respective line. Note that there are an uncountable number of them on each axes. Points retain their color after legal moves.

Is there a sequence of legal moves that will produce a circle -- some points one color and some points another color? -- on one of the faces of $U'$? (where $U'$ is meant to denote $U$ after said sequence of moves has been performed?)

addition: I strongly suspect that no single countably infinite set of legal moves works, just on the basis of cardinality arguments -- a circle would have a continuum of points, and any countably infinite sequence of legal moves would only be able to move a countably infinite set of points.

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Take $T$ a black face; two neighbors, $F$ and $S$, are white. Rotate almost every square parallel to $S$ by a quarter turn, such that points of $F$ become points of $T'$; the exception is the plane through the center of the cube. Now $T'$ has white points everywhere except for a segment of black points parallel to $S$. $S$ still has only white points, so rotate almost every square parallel to $F'$ such that the points of $S'$ become points of $T''$; again, the exception is the plane through the center. Now $T''$ is white everywhere except its center: a circle of radius 0! :) (Cheating?) – Blue Nov 7 '10 at 7:15
You might want to say something about the color of points on the edges of the cube. I suggest making them transparent. :) – Blue Nov 7 '10 at 7:20

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What a great question!

Here is my solution. I am aiming for a black circle inscribed on a white face (where the boundary edge points are invisible, as mentioned in the comments).

Pick one of the white faces to be the desired final face F. Now, consider any given horizontal row on that face, which is all white. We want to add two black dots to such a row (except for top/bottom/center, which are handled already by the edge point color condition). These rows come in upper/lower symmetric dual pairs. Call such a pair the current working pair of rows. We make a quarter turn to each of these working rows to the side. Now, operating on that side face S, we make two moves parallel to F, which will not upset F, in order to make the working rows on S have exactly two black dots each in the correct positions. This is possible since the squares containing the working rows have not been moved previously, and the corresponding columns containing the desired black dots have also not been moved previously, since those dots will only be moved for this pair of working rows. That is, it is precisely the black dot opposite the dual white dot that we desire, and by working in dual pairs, we put all four black dots into position with two moves. (Is it clear?) Thus, on the side face, we have the two working rows looking exactly like we want. So we turn them each a quarter turn back to the main face F, and proceed with another pair of working rows. The center row and top/bottom rows do not need adjusting, because of the invisible edge points. So this seems to do it!

That said, there are problematic issues in setting up your problem. This kind of problem, where one wants to describe a task involving infinitely many steps, is known as a supertask, and is open to a number of interesting paradoxes and problematic issue, some of which I describe in this MathOverflow answer.

In this problem, we can imagine carrying out my solution in a transfinite sequence of moves. The fact that each row on the desired face is handled with finitely many moves makes my particular solution less problematic than other situations one can imagine. In general, for example, if a square has been moved infinitely many times, what position is it in at the limit? You haven't said, and it isn't clear what it should be. But solutions in which every given square is moved only finitely often seem to avoid this problem, particularly when the order that the finite sequences of moves are made in doesn't matter.

Edit. I notice now that you didn't say that the diameter of the circle should be the same as the diameter of the square. You can easily modify my solution for any size circle on any face. In this case, you do have to handle the top/bottom/middle rows of the circle, but this is no problem and is handled the same as the other rows. The top/bottom rows of the circle form a dual pair of working rows just like the others, and the center row has no dual, and is handled by itself.

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