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Closed loop on a side length 2 cube Question

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Place 4 tiles on each face of a cube so that the arcs on all 24 tiles form a single closed loop.

If you want to try a simpler version first, there is Closed loop on a side length 1 cube.

You may rotate each tile by any multiple of 90 degrees.

The tiles to be used are equal numbers of 2 kinds of tile:

  • 12 single arc tiles: tile with an arc joining 2 adjacent edges
  • 12 double arc tiles: tile with arcs joining 2 pairs of adjacent edges

Here are 12 of each that can be printed and cut out:

12 tiles with 2 adjacent edges joined by an arc, and 12 tiles with 2 pairs of adjacent edges joined by arcs

Here is a net of a cube to place the tiles on:

24 squares that can be folded up to give a cube

Each of the images can be opened in a separate tab (which will display the tiles and the squares on the cube faces at exactly the same size) if you would like to print them for experimenting or to photograph your solution to post in an answer.

Hint

This is not intended as a brute force puzzle where you try all the combinations until you find one that works. There are some insights that will make this puzzle practical to solve even for larger versions.

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1 answer

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Solution

Solution

Process I start with a fairly symmetric placement which doesn't meet the constraints and make simple modifications until it does.

We have two "single" and two "double" tiles per side. We can place them to leave four vertices of the cube clear:

Pick one vertex and then on each face which it touches pick the diagonally opposite vertex

Now place the double tiles in the remaining spaces:

All in the same orientation

The only problem is that we have more than one loop. For each double tile in some arbitrary order, if rotating it 90 degrees will join two loops then do so.

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