Here are two thick forward and backward slashes, each made of two isosceles right triangles glued along the smaller sides.
For what values of n can you wrap n forward slashes and 6-n backward slashes around a cube with no gaps and no overlaps?
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2 Answers
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3
The n's that work are
2, 3 and 4.
Preliminary remarks.
The shapes are made of 2 half-squares. There are 6 of them, just like the number of faces of the cube. This means that every shape saddles 2 adjacent faces covering half of these faces.
From there you can see that each half-square of a shape is completed by the half-square of another shape.
Suppose you have a solution that wraps the cube. If you start with a shape anywhere and chose one half, there is an unique shape that completes that half. That second shape has another half, which is then completed uniquely by a third shape. Continuing like this you form a chain that eventually loops back to the original shape.
You cannot have multiple disconnected chains, as the smallest one would be of 2 or 3 shapes. That would mean the small chain covers 2 or 3 faces, which is easy to see to be impossible. You must have a single chain of size 6.
From there we can identify which n works.
If an n works, the 6-n also works by just mirroring the solution.
If all shapes are of the same type, they simply run around the cube in 4 steps, missing 2 squares.
This means n=0 or 6 is not possible.
If exactly one shape is different from the other, you will necessarily have a sequence of 5 of the same type in the chain. The 5 identical shapes would have to go around 4 faces and 5th would overlap the 1st.
So n=1 or 5 is impossible.
For n=2, it is possible by making the following loop: \ \ / \ \ /.
It wraps the cube with the following path (using Rubik's cube notation): LFRUBDL. The last L connecting to the first.
So, n=2 is possible and also n=4, by mirroring the solution.
And n=3 is easiest, actually. The sequence of shapes is \ / \ / \ /.
The path to wrap the cube is: LFURBDL. The mirror solution si the same.
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$\begingroup$ Nice! It would be great if you could add some simple figures for those that work. I understood reveal spoilerthe two paths that work, but I don't understand the notations reveal spoiler\/\/ and ///. $\endgroup$Pranay– Pranay2026-04-30 16:41:58 +00:00Commented 17 hours ago
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$\begingroup$ Some \'s got eaten as escape sequences. I will make a picture. $\endgroup$Florian F– Florian F2026-04-30 16:51:25 +00:00Commented 16 hours ago
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$\begingroup$ Ah got it. Makes sense now. $\endgroup$Pranay– Pranay2026-04-30 16:57:19 +00:00Commented 16 hours ago
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@Florian F has given a nice answer. I wanted to show a nice way to think about the solutions found there.
First, take the cube and cut it into two halves along the plane through the red lines (parallel opposite face-diagonals). Each half can be covered with three slashes (two of one type and one of the other) in an essentially unique way (up to reflections). And then the two halves can be glued back together in exactly two ways: directly, or after flipping (i.e., reflecting) one of them. These correspond to the two solutions found by Florian F.