Smallest non-space-filling polycube?

Question

The title nearly says it all: what is the fewest number of cubes that can be fused face-to-face into a polyhedron that does not fill space? The smallest that seemed like a sure non-tiler to me was 9: seven in a ring missing one "corner", plus one above one of the "middle" cubes of this ring, and a final one fused to that one, positioned above the hole in the ring. Then there seemed to be no way fill the hole in the ring with another copy of this polycube -- but EDIT: as Georgios points out in a comment to his answer, you can actually interlock a pair of these to get a simply-connected compound. So the smallest truly obvious non-filler is a decacube, consisting of an 8-ring with two attached to a middle cube to "cap" the hole.

But I wouldn't be surprised if there were an octacube or heptacube non-space-filler, and if so, I suspect it's well-known to some folks. However, I can't seem to find a reference, even though the answer to the corresponding question for polyominoes in 2D is easy to find, e.g. MathWorld notes that all but four heptominoes tile the plane, as all smaller polyominoes do.

Answer 1

I recently wrote some code based on Matt Parker's Hypercube Folding Video, and run it with all heptacubes, octacubes and nonocubes. It found tilings for all heptacubes, all but one octacube (the donut shaped one), and all but six five nonocubes.

My code and all tilings found so far are in my whuts-solver github repository.

Please note, my code does not prove that a solution does NOT exist for some polycube, it only verifies ones that exist. Moritz Firsching has done some similar work and to my understanding he has proven that the donut-octacube does indeed not tile space, but I will let him comment on this part.

These are the nonocubes I could not find a tiling for yet:

All other solutions are in the github repository in the directories <n>cubes_solutions, these are as json, with "base_blocks" a list of coordinates of where to position some of the polycubes, and "offsets" showing three directions (and distances), to repeat the pattern to get a full tiling.

I will update later with some visualisations for the tiling produced for the three heptacubes that were mentioned as non-tiling in another answer.

Edit: For reference, the three heptacube claimed to not tile space in a separate answer are numbered 144, 556 and 272 in my enumeration, the tilings for these can be found in the files heptacube_solutions/solution_0NNN.json in the repository.

Edit: In the meantime, my solver (after running for 4 days) found a solution for one more of the nonocubes (number 1345 in my numbering in github), the primitive block is composed of 12 nonocubes! Since this is an exhaustive search, this is the simplest possible solution for this nonocube (assuming a periodically repeating tiling of course). Also, here is an image showing the tiling for the three heptacubes mentioned in another answer:

Edit: Here is an image for the two nonocubes (4921 and 4931) and how they can generate an interlocking pattern each:

Answer 2

Edit: As mentioned in the comments, this answer was incorrect. The linked source's claims are wrong. I've included some updated analysis above the break.

I've verified with my own tiling solver that every heptacube tiles space - at most four copies are needed for any heptacube to form a translationally-tiling patch. Interestingly, all four heptacubes that require more than two copies to form a translationally-tiling patch are planar (the first three in the image below, and a tiling heptomino shown in the first image at this page).

The unique octacube which does not tile space is the 3D version of the "donut" octomino:

* * *
*   *
* * *

It is able to cover a $6\times6\times7$ box, but not a $6\times7\times7$ box.

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Wrong answer below

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From http://www.recmath.com/PolyPages/PolyPages/index.htm?Tiling.htm:

All polycubes up to and including the hexacubes will tile space.

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However, of the four heptominoes that do not tile space (shown below), the latter three still don't tile when you "elevate" them to three dimensions.

The first heptomino pictured above does, however - in fact, it tiles a $2\times 5\times 7$ box:

So the answer is $7$ cubes. (I don't recall if there are nonplanar heptacubes that don't tile space - I would guess so, but don't know of examples offhand.)

As a remark, "elevating" a lower-dimensional failure will always produce tilers of a higher-dimensional space eventually - see Tiling with Arbitrary Tiles for a proof that every polyomino tiles $\mathbb{Z}^d$ for some $d$. (In fact, the result holds even when an arrangement of cells is disconnected!)