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The surface of a dodecahedron is tiled with 6 of the shown tiles, each tile covering two faces of the dodecahedron. In how many essentially different ways this can be done?

Two tiled dodecahedrons are considered to bear the same tiling if some rotation of the first matches the second.

How many essentially different tilings exist if we allow rotations and reflections for the matching procedure?

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2 Answers 2

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I think I got it without graph theory.

Please forgive the lack of illustations. This would take too much time. You should use a solid model of a dodecahedron to follow the reasoning.

Also, I am not hiding the reasoning, It is a bit tedious to follow, so you shouldn't "accidentally" stumble over a trick that would spoil the solution. Only the final count is spoiler-protected.

First, some definitions.

A tile has 2 neighbours that touch both faces. Let's call close neighbour these 2 tiles.

Close neighbours can touch with 2 or 3 edges. Let's say tiles hug when they touch at 3 edges, and tiles are orthogonal when they touch only at 2 edges. Hugging tiles form a lozenge, orthgonal tiles form kind of a T shape.

Tiles are friends if they hug each other or they are friends transitively via a chain of hugs. Friendship relations partition the tiles into groups of friends.

OK. Now let's enumerate the positions.

  1. If a tile has no friend then both close neighbours are orthogonal. When you place these, the remaining faces form a region that allows only one way to place the remaining tiles. That position has the property that all tiles are orthogonal. This yields one position.

In the remaining positions each tile has at least one friend.

  1. If there is a group of 2 friends having no other friend, then the 2 tiles hug each other and have 2 orthogonal neighbours on the other sides. Placing these on a dodecahedron, the remaining faces form exactly 2 pairs, allowing for a single position. This position is such that it forms 3 groups of 2 friends.

The remaining positions must have groups of at least 3 friends, which boils down to 2 groups of 3 or one group of 6.

  1. If we have a group of only 3 friends, we have one tile in the middle hugging the 2 others. There are essentially 2 ways to place them together. But if you add the 2 orthogonal non-friends to the friends placed on the side, in one case these orthogonal tiles overlap, in the other case they leave 2 disconnected faces for the final tile. So this case yields no solution.

  2. The next case is a connected group of 6 friends. These must form a closed chain of 6 tiles hugging each other. This is easy to explore. Start from a tile (placed vertically). The next tile on the right is either higher or lower. Continue with the next tile until you hopefully complete the chain. The only patterns that work are HLHLHL... and HLLHLL. HHLHHL... also works but it results in the reflexion of the second position. So this case yields 2 solutions.

  3. Update. Who says the 6 friends must form a closed chain. In fact they don't have to. It can be a string of 6 that end on both sides with a friend that happens to be in orthogonal position.
    Starting from one end of the chain, following the notation of point 4:
    HHH... doesn't work.
    HHL... forces HHLHH, which closes a loop already covered.
    HLHHL is the closed loop HHLHHL... again.
    HLHLH is the closed loop HLHLHL...
    HLLH... forces HLLHL which is the closed loop HLLHLL...
    HLLL... forces HLLLH which is a genuine new solution!

Conclusion. The number of essentially different placements of the tiles is ... tadaaa ...

not four but five!

One with all tiles orthogonal, one with 3 groups of friends, two with a closed chain of 6 hugging tiles and one with a non-closed string of 6 friends.

Update 2: the original question asked for the count without reflexions.

Well, we can start from the 5 tilings we have.
Point 1. yields a completely symetric tiling, even by reflection.
Point 2. yields a single chiral tiling. All 3 groups of friends form a lozenge in the same (chiral) way.
Point 4 yields 2 tilings. One chiral one not. HLHLHL is LHLHLH but HHLHHL is different from LLHLLH.
Point 5 yields also a chiral tiling (HLLLH is different from LHHHL).

All chiral solutions must be doubled if reflexions are not considered. There are 3 chiral solutions, that brings a total of 8 ways to tile modulo rotations only.

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  • $\begingroup$ Your number refers to rotational plus reflectional symmetry I suppose. I have a different result for this case but it will take some time to track down where the discrepancy arises. $\endgroup$ Commented Aug 13, 2023 at 13:58
  • $\begingroup$ Yes, mirror images are counted as the same. $\endgroup$
    – Florian F
    Commented Aug 13, 2023 at 15:05
  • $\begingroup$ In theory, can you tile a dodecahedron perfectly symmetrically so that when you flip it, it remains the same? Because that would put a thorn in the side of my 'divide by 120' plan. $\endgroup$ Commented Aug 13, 2023 at 15:08
  • $\begingroup$ Yes, it's not that easy. You have to take account of the symmetries. Use the Polya Burnside Lemma if you want to do this in a correct way. $\endgroup$ Commented Aug 13, 2023 at 15:11
  • $\begingroup$ @FlorianF: I checked your arguments and I am impressed by the clearness and simplicity of the arguments. My reasoning is more complicated. Nevertheless it was easier than I thought to track down the discrepancy. I found a little hole in your argumentation which leads to a wrong result. I will not go into the details until hopefully some other interesting solution attempts are presented by the community. $\endgroup$ Commented Aug 13, 2023 at 15:34
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First of all I made a 2D representation of the dodecahedron which allowed me to place coins etc. to represent the tiles.

enter image description here

I placed a tile on faces 11/12 and then looked for valid configurations if the second tile was placed on faces 1/2. This can be done manually by an obvious backtracking algorithm.There are 5 valid configurations. Placing the second tile on 1/3 also gives 5 cases, on 1/4 again. I see no obvious reason why it is 5 in all cases. Due to axial symmetry we know that the second tile on 1/5 or 1/6 also gives 5 cases. So in total there are 25 tilings with the 11/12 tile fixed.

We can assume that the first tile always covers face 11. But the other part must not necessarily cover face 12. There are 4 other possibilities (faces 10, 5, 4 and 7). These have the same number of valid configurations. So there are 125 possible tilings if we do not allow symmetry reductions.

The dodecahedron has 60 rotational symmetries: 10 axes through opposite corners give 20 three-fold symmetries (20 because we can rotate by 120° or 240°) , 15 axes through the centers of opposite edges give 15 two-fold symmetries and 6 axes through the centers of opposite faces give 6*4=24 five-fold symmetries. 20+15+24+1= 60. 1 is the contributions of the do nothing symmetry. Wenn we allow reflections we can combine these symmetries with a point reflection at the dodecahedron center to get another 60 symmetries.

I manually worked out the number of symmetric positions for each of the symmetry types with the help of my 2D representation. These are the results:

A two-fold axis rotation: 17 symmetric positions, 17$*$15= 255
A three-fold axis rotation: 5 symmetric positions, 5$*$20=100
A five-fold axis rotation: 0$*$24 symmetric positions.
Do nothing symmetry: All 125 cases are invariant under this symmetry.

With the Polya Burnside Lemma we then get for the number of essential tilings using rotational symmetries:

(255+100+0+125)/60 = 8

When we include reflections, for the additional 60 symmetries we get the following numbers:

two-fold rotation $*$ point reflection, 3$*$15 = 45 symmetric positions
three-fold rotation $*$ point reflection, 3$*$20 = 60 symmetric positions
five-fold rotation $*$ point reflection, 0$*$24 = 0 symmetric positions
do nothing $*$ point reflection: 15$*$1 = 15 symmetric positions

With the Polya Burnside Lemma we then get for the number of essential tilings using rotational and reflectional symmetries:

(255+100+0+125 + 45+60+0+15)/120 = 5

I found a nice way to create all possible essential different tilings.

Take one half of the dodecahedron and tile it. There is essentially only one way to do so.

enter image description here

Then complete the dodecahedron with a copy of this half on the other side. There is one way such that the resulting tiling has axial symmetry - the red tiles show in the same direction. Let's call this tiling T1.
But you can add the second part also in 4 different twisted ways:
+72° gives T2a
-72° gives T2b
+144° gives T3a
-144° gives T3b
In total we get 5/8 or 3/5 of the essentially different tilings. In the view of the solution of Florian F these are the six friends solutions where two groups of three friends in all cases form a six friend solution together.

For the remaining cases we tile a differently shaped half of a dodecahedron in such a way that we have three-fold symmetry:

enter image description here

Also create a mirrored tiling, the swirl then goes in the opposite direction:

enter image description here

Then again glue two halfs together. Due to the tree-fold symmetry there is only one way to do so. Then you get tilings

T4 if you glue together two halfs with opposite swirls.
T5a if you glue together two halfs with right swirls.
T5b if you glue together two halfs with left swirls.

In the view of Florin F we have two groups of no friends which are forced into three groups of two friends in the chiral cases.

And that's it!

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