On this page I will try to investigate the different ways in which polyhexes can be tiled with the heptiamond. This problem was inspired to me by the remarkable property that have heptiamonds to be bi-tilable with tetrahexes. Since finding all heptiamond shapes that can be tiled with tetrahexes seems out of reach for our computers of today, my idea was to construct at least subsets of such shapes.
Since proving that a particular shape cannot be tiled with heptiamonds is often time consuming, my idea was to avoid this obstacle. There are 7 tetrahexes with a total area of 28 hexes. Because one hex is equivalent to six triangles, but a heptiamond has seven of them, the only polyhexes that could theorically be tiled with the heptiamonds are 7-hexes, 14-hexes and 21-hexes, excluding the obvious 28-hexes. If there is a way to simultaneously tile various polyhexes with the set of heptiamonds, it would be enough to combine these polyhexes in various ways and thus to obtained 28-hexes that are tilable with heptiamonds. At this stage it would be enough to check whether these 28-hexes are tilable with the tetrahexes, which is considerably easier.
There are four theoretical cases: (a) four 7-hexes; (b) two 14-hexes; (c) one 7-hex and one 21-hex; (d) two 7-hexes and one 14-hex; . The most interesting configurations are (a) and (d) because more combinations could be obtained with them. Combining four heptahexes would produce millions of shapes; combining two 14-hexes gives us just a few hundred and things are even worse for one 7-hex and one 21-hex.
Below I provide my results for each of these four cases:
My first hope was to find sets of four heptahexes that can be simultaneously tiled with the heptiamonds. This would have been great, because the total numbers would have been small and working with them would have been possible. I was hoping there would be just a handful of such quadruples. It turned out there is none. Working my way to this disappointing result I found all sets of heptahexes that can be tiled with the heptiamonds (some heptiamond pieces being left out) and here are they in tabular form:
| Number of heptahexes to tile | Possible sets | Visual examples | Notes |
|---|---|---|---|
| One heptahex | 92 | Out of 333 heptahexes. | |
| Two heptahex | 826 |
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| Three heptahex | 44 |
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Only 23 distinct heptahexes are to be found in these triplets and just one can appear twice in the same triplet. The cumulative perimeter of these triplets ranges from 62 to 68. Since the lowest perimeter for a heptahex is 18, this means that if a quadruple was possible, its perimeter would have been at least 80. This is still well below the theorical maximum of 95 but it never happens. |
Thus the most appealing configuration is impossible.
I cannot do an extensive search of 14-hexes tilable with heptiamonds. The total number of 14-hexes is 15 796 897. Upon request, Joseph Myers kindly sent them to me and I did some calculations which showed that:
Thus I'll limit myself to a demonstration of the way how just one pair of heptiamond tilable 14-hexes gives us a hundred bi-tilable shapes. I manually discovered this tiling:
Both shapes are identical which will bring down the number of their combinations. It turned out there is a total of 185 ways to glue the 14-hexes together, 34 symmetric and 151 assymetric.
The next step was to pass them through my tetrahex solver, merely checking whether a solution exists for each shape, not caring about the total number of solutions. It turned out 21 symmetric and 119 assymetric shapes are tilable, which is just a bit more than 75% tilablity rate. In less than 10 minutes I had obtained 140 bi-tilable shapes! Here I publish only the symmetric ones but I would happily send the full package to any one willing to have a look at it. Be reminded that all the heptiamond tilings are just mechanical combinations of the shapes shown above.
| Tetrahex tiling | Heptiamond tiling |
|---|
By clicking around and guessing fancy 14-hexes, I came across a few more cases. Here I provide only the heptiamond tilings and the numbers of bi-tilable shapes.
| Heptiamond tiling | Symmetric bi-tilables | Assymmetric bi-tilables |
|---|---|---|
| 0 | 18 | |
| 0 | 33 | |
| 3 | 11 |
This case is possible and I already provided an example of it which allows to quickly find all enlarged bi-tilable tetrahexes (4 out of a theoretical set of 7). However, conducting an extensive search definitely seems out of reach. In order to narrow things down we could look for all ternary 21-hexes and try to simultaneously tile them with the 92 tilable heptahexes.
Even one set of two 7-hexes and one 14-hex would be of great value, since it would allow us to produce a very big number of bi-tilable shapes. Obviously, since a triplet would produce many combinations of two 14-hexes or one 7-hex and one 21-hex.
I compiled a list of all possible ways to split the heptiamonds into two groups of 12, one of the groups tiling two heptahexes. My idea was to look whether the left over 12 pieces can tile a tetrahex. However, this was working too slowly and I didn't reach a desired triplet that way.I was getting seriously convinced that no such triplet even exists, like there is no quadruple of 7-hexes. But then I calculated the cumulative perimeter of shapes in the triples of heptahexes. The numbers varied around 64, each heptahex contributing 20 or 20 something edges. The biggest perimeter of a single heptahex was 28. Then I calculated perimeters for some 14-hexes and found that it can be 26, which makes for an even lesser value as for a triplet of 7-hexes. I picked one such 14-hex and two heptahexes that occure very often in triplets of heptahexes, each having a perimeter of 20. To my amazement, it worked and I had a 7-7-14-hex triplet tilable with heptiamonds:
This merely means that there are many 7-7-14-hex configurations but the cumulative perimeter must be kept low.
I proceeded to calculating all possible combinations of the three shapes. There are 17,098 of them, 6,353 of which are tilable with tetrahexes and therefore bi-tilable. This is a bit more than one third. None of the shapes is symmetric because of the building blocks used. They all have weird outlines and show the monstruosity of the tetrahexes -- we are used to seeing only the handful of symmetric shapes, that appear on the Internet, but the variety of the sets is very big and very different. I think this situation ressembles the one with butterflies: we imagine them as beautiful and colour because of the day species, whilest these represent only 10% of all butterflies, most of them being little, grey and visually unappealing animals. Here are a few specimens of the bi-tilable shapes (only one tetrahex solution is shown; heptiamond solutions are obtained by joining together the shapes above):
As I was hoping to find some symmetric bi-tilables, I shifted just one hex from the previous pattern and was lucky enough to find a new 7-7-14-hex configuration (notice that the two 7-hexes are tiled exactly in the same way as in the previous case):
Having an assymmetric 14-hex led to two very important consequences: (a) the total number of combinations was much greater; (b) there were some symmetric amongst them.
To be more precise, there were 72,762 combinations of the three polyhexes, 25,229 of which tilable with tetrahexes and therefore bi-tilable (as in the previous case, this is roughly one third). There were just 3 symmetric combinations but they were all tilable. Here I present only them, leaving aside the assymmetric ones:
Later on I found even more examples, but I didn't check their combinations for tetrahex tilability:
I have more results about tiling two heptahexes and one tetrakaidecahex.