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If you subdivide a 2x2 tile into 4 unit squares and then color each unit square either red or green, then there are $2^4=16$ ways you can do this as shown below:

2x2 tiles colored red or green in 16 ways

Can you use all 16 tiles (rotations allowed) to tile an 8x8 square so that whenever a unit square from a tile shares a common edge with a unit square from an adjacent tile, then the two unit squares will have the same color? Flipping the tiles over is not allowed.

BONUS: Can you find a symmetrical solution?

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This puzzle is a slight variant of David Butler’s puzzle: Panda squares.

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    $\begingroup$ Flipping might as well be allowed. All flips can be achieved with rotation. $\endgroup$
    – Daniel Mathias
    Nov 17 at 23:20
  • $\begingroup$ Then you can try it without rotation. $\endgroup$
    – Florian F
    Nov 18 at 0:04
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    $\begingroup$ @FlorianF I have :o) and not only is it possible without rotation, it is toroidally possible without rotation. $\endgroup$
    – Daniel Mathias
    Nov 18 at 1:01
  • $\begingroup$ I too tiled the doughnut... $\endgroup$
    – Florian F
    Nov 18 at 8:58
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    $\begingroup$ The doughnut solution can obviously be repeated to tile the plane, resulting in a really nice tiling pattern. When I was a kid, the seats of my Dad's Citroën CX car had that pattern woven into it. I would get carsick from looking at it, but I still like the pattern. $\endgroup$
    – Jaap Scherphuis
    Nov 18 at 10:05

1 Answer 1

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Time's up. Here is a pattern.

enter image description here

It is just one solution. There are others given the freedom of rotations.
This solution has the extra constraint that no rotation is allowed and it works on a torus. In other words it still works if we repeat the pattern horizontally and vertically.

With these additional constraints this solution is unique except for translations (mod 4) and flipping.

And giving in to the social pressure asking to do more and provide a symmetrical solution, here it is:

Here are reflective-symmetrical solutions.

and



These are the only I found.
You can go quite far actually by deduction alone. For instance the unicolor squares must be in the diagonal and cannot touch. It is quite fun.

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    $\begingroup$ The extra constraints make this a 2x2 De Bruijn torus $\endgroup$
    – 2012rcampion
    Nov 20 at 2:12
  • $\begingroup$ Is there a solution that has reflectional or rotational symmetry? $\endgroup$
    – Will Octagon Gibson
    Nov 20 at 3:42
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    $\begingroup$ Since there is only one red square, rotational symmetry is impossible. Maybe a reflectional symetry along the diagonal is possible. $\endgroup$
    – Florian F
    Nov 20 at 10:25
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    $\begingroup$ Reminds me the pattern used to define tilemaps in some game engines. $\endgroup$
    – 0x5453
    Nov 20 at 14:06
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    $\begingroup$ I think those two symmetric solutions are the only ones, apart from rotations and colour inversions. $\endgroup$
    – Jaap Scherphuis
    Nov 21 at 22:58

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