Inspired by A closed path on the Rubik's cube.

Let $N$ be a positive integer. Consider an $N\times N\times N$ cube, with each face tiled by $N^2$ squares measuring $1\times 1$. A closed path is drawn on the surface of the cube, never entering the same square twice, never passing through the corner of a square.

Each time the path enters a square, necessarily through an edge, it will exit through a different edge. Let $R$ be the number of squares with exit edge immediately to the right of the entry edge, and let $L$ be the number of squares with exit edge immediately to the left of the entry edge. Ignore squares not meeting the path, and squares for which the exit edge is across from the entry edge.

What are the possible values for $R-L$?

  • $\begingroup$ Are you accepting the answer or looking for a better one? $\endgroup$ Jan 18, 2016 at 15:01
  • $\begingroup$ I have a solution which is a pain to write up, showing that any path can be be deformed into a small loop around a single vertex, where R–L only changes when passing a vertex of the cube. Is there a better proof? Something with to do with integration, maybe? $\endgroup$ Dec 16, 2017 at 19:09

1 Answer 1


I'll start with ...

- any integer between $-4$ and $+4$ (for $N \ge 2$)
- any integer between $-3$ and $+3$ (for $N = 1$)

$-4$, $0$ and $+4$ are obvious (left image). We also see that any positive number can be made negative by mirroring or walking in other direction. The right image shows how to construct paths with values $1$, $2$ and $3$.

Changing the paths while staying on a single plane obviously doesn't increase the number because for every added left turn we get another right turn. The same applies if we move to another plane with both ends of our path:

Using other ways to leave the plane only seems to decrease the number of turns like in the right picture at the top.

  • 2
    $\begingroup$ The path divides the cube's surface into two regions. Let's say that the "inside" of the path is the region that the the path travels clockwise around. Then it appears that $R-L$ will always equal $4$ minus the number of the cube's vertices which are inside the path. $\endgroup$ Dec 5, 2015 at 3:57

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