Tetracubes, revisited

A while ago I wrote a post exploring polycubes up to and including the tetracubes. And I didn't really go too far in depth with it. I found a few things with the tetracubes then just kind of stuffed them in a cupboard and forgot about them. But recently I dug them out and all it took to rekindle my interest with them was a few minutes scribbling on them in felt tip pen so that the individual pieces could be told apart in a construction. Behold:

It's messy and it looks like a six year old did it, but it's a slight improvement on how they looked before. From a solving point of view anyway.

After doing this I decided I needed to solve a few things to test out how they looked. And this quickly showed me that there was a lot more overlooked potential in this set of shapes than I ever realised.

Scaled-up tetracubes

The total volume of the tetracubes is 32 units, which is just enough to construct a big tetracube scaled up by a factor of two. For four of the five planar tetracubes and two of the non-planar ones this is trivial once we've got the two 2x2x4 blocks above solved, just put them together in whichever configuration. For the two remaining tetracubes - the T-tetromino and the cyan one bottom left in this image - it's a little more tricky.

For the diagram on the right, a dot indicates that that piece extends into the layer below there, and a square indicates that it extends into the layer above.

The solution for the T-tetromino is shown above, but the final tetracube is a challenge for the reader. It's possible, but I can't be bothered to draw out another diagram for it so you'll have to find it yourself.

Almost-cuboids

There are also the two cuboids-with-holes-in-them that can be done. There's the 2x3x6 with a 1x1x4 hole, and the 3x3x4 with a 1x2x2 hole. I imagine there are several ways to solve each, but I've included once possibility for each, in that not immediately readable notation everyone uses for polycube constructions.

Octomino towers

Imagine an octomino. Any octomino you like. Then imagine it made of cubes as opposed to squares, the planar octacube equivalent of the octomino. Now imagine stacking four of these perfectly on top of each other, creating a prism with volume 32 units squared and the octomino as its cross section. There is a chance that this resulting shape can be filled with the tetracubes. I mean, sometimes there's not, the I-octomino in this instance corresponds to a 1x4x8 rectangle which clearly can't fit the non-planar tetracubes. But The tetracubes show a surprising versatility when it comes to most of the other octominoes.

Some highlights shown in the diagrams below.

The stairstep octomino came as a surprise to me, I really didn't think it would be possible.

A useful solution is the one above to the doughnut-shaped one. This can be broken into three small 4-unit high towers as shown below, and these can be pushed together in various ways to make a whole host of octomino stacks.

Which begs the question: which of the octominoes can we solve this way, and which ones can't be? The above 'kit' of three pieces probably covers the majority, but it'd be interesting to see the set that just can't be done, either through the kind of impossibility the I-tetromino demonstrates immediately, or just cases where the pieces won't go despite there not being a clear reason for it.

This felt like the kind of problem where if I posted it to the Puzzle Fun Facebook group someone would get back with the results of an exhaustive search by the end of the day. So I tried that.

Pretty quickly Edo Timmermans had found all of the octominoes that could be created by combining the monomino, triomino and L-tetromino that make up the solution to the ring octomino way up there in the previous bit. These are marked in green in the above image. He also showed that any 'L' shaped octomino with a single bend in it was impossible, as the three nonplanar pieces would all have to occupy that bend.

I then found that my solution for the zigzag octomino could be similarly partitioned into a monomino, triomino and Z-tetromino which allowed solutions for a further five octominoes, those in dark blue.

George Sicherman then found the set of the 48 octominoes for which the prism has no solution, these were marked in red in the diagram. Which left nine octominoes, all of which had solutions, but which hadn't yet been found. I managed to pick off some of these by hand, as shown in the diagrams below, and the remaining five I verified in Aad van de Wetering's 'Poly3D' software. But I'll not put the solutions here, just in case anyone reading has a set of tetracubes themselves and fancies a nice challenge.

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I've been updating this blog and the website sort of in tandem; I prefer the site for its more flexible formatting and the fact I can interlink and break things into pages a bit better, but this blog probably gets more traffic (it's not much, but it's traffic nonetheless) so I'll keep adding stuff here too. I'll figure something out.

One-Sided Hexominoes

It's been a long time since I last posted anything on here. Too long. But sadly there just hasn't been that much in the way of polyomino-related goodness to post. Partly because I've just been too busy doing other things, and partly because there's only so much you can do with polyominoes - the easy stuff isn't interesting enough to write blog posts about, the hard stuff is too hard to do without the assistance of a beefy computer, and the stuff in the sweet spot is hard to come by.

I've never written up anything about the one-sided hexominoes before on here. Basically, earlier this year I got a second set of heptominoes laser cut, which opened me up to the possibility of solving shapes using the set of one-sided heptominoes by combining the sets and then trying not to flip the pieces over. Then I realised the same thing could be done using my two sets of hexominoes too. And the set of one sided hexominoes is a much more versatile set than the plain ol' regular hexominoes.

There are 60 one-sided hexominoes in comparison to the standard set's 35. But this new set doesn't have those pesky parity constraints which means that a lot more shapes are possible to tile - rectangles without unsightly internal holes, for example.

The total area is 60x6 = 360, which means that rectangles of size 4x90, 5x72, 6x60, 8x45, 9x40, 10x36, 12x30, 15x24 and 18x20 should all be possible. Sadly, due to the length of the perimeter compared to the amount of perimeter squares the pieces can provide, 4x90 isn't possible.

Solving manually is a little trickier than the normal hexominoes - each piece having only one accepted 'right side up' means that any given piece is slightly less practical than its two-sided equivalent, and you do get those cases where you're down to one piece left and the hole is the mirror-image of the piece you're holding.

Here's two 9x20's which can be combined to make either a 9x40 or an 18x20, in a two-rectangles-for-one type deal. As 360 divides up really nicely, this gives a lot of possibilities for tiling groups of congruent shapes, but that'll be a blog post for the future. Others have already solved congruent sets of ten or twelve shapes, so go look at those. Scroll about three-quarters of the way down the page for them. In fact just read the entire page, it's all good.

Something else nice you can do with the one-sided hexominoes is square rings, i.e. squares with a centred square hole. Here are three possibilities (these might be the only three actually), with the rings getting progressively thinner, and as a result a little harder to solve manually:

And here's one more shape, a diamond with a central hole and those tricky diagonal edges.

And I'm purposely leaving this post a bit less thorough than usual so I've got an excuse to post 'One sided Hexominoes - Part Deux' in a few weeks. Or a few months, if my recent posting schedule is anything to go by.