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Given a polyomino $P$ with $n$ cells, we can ask about its maximal packing density $\delta_P$ in the plane (perhaps the limsup if we are concerned about convergence issues, though I don't think this actually comes up).

If $\delta_P=1$, then $P$ tiles the plane: having arbitrarily good packings implies that $P$ can cover arbitrarily large $N\times N$ squares, and from there the result follows via a compactness argument.

We can then ask: among polyominoes on $n\ge 7$ cells that do not tile the plane, which one achieves the highest density? Call this maximal value $\Delta_n$. For each $n$, $\Delta_n<1$, though of course in the limit for large $n$ it approaches $1$.

As a simple lower bound, we always have $\Delta_n\ge n/(n+1)$, as exhibited by the polyomino given by taking all but the first square in a spiral of length $n+1$ around the origin; once we plug the hole, it tiles the plane without gaps, but the hole cannot be filled. For instance, with $n=7$:

enter image description here

In contrast, putting remotely nontrivial upper bounds on $\Delta_n$ in general should be exceedingly difficult or impossible, since any computable upper bound less than 1 would yield an algorithm to decide the tiling problem for polyominoes, which is suspected not to exist. (Non-computable upper bounds on the order of $1-1/\text{BB}(k\cdot n)$ can be done, of course, but these are rather silly.)

However, improved lower bounds and exact values for small $n$ seem pretty tractable, and I'm curious what is known in this direction. Some questions:

  • Is the sequence $\Delta_7,\Delta_8,\ldots$ monotonically increasing? I suspect not.

  • When does $\Delta_n$ first exceed $n/(n+1)$? I don't actually know that it does, but I strongly suspect so for reasons described above. When $n=7$ I have not found any packings of density greater than $7/8$, although only for one of the four non-tiling heptominoes (the one pictured above) have I proven this is optimal. (All four obtain $7/8$ by adding one square to yield a tiling octomino.)

  • Can $\Delta_7$ be proven to equal $7/8$, if indeed it does so? Exhausting all tilings of an $N\times N$ square by each of the four non-tiling polyominoes and counting the max density therein would at least yield an upper bound; if $\Delta_7$ exceeds $7/8$, I would guess it does so via finding a tiling $15$-omino which is the union of two copies of a non-tiling heptomino and an additional cell.

  • [Subjective] What are some examples of interesting "near-miss" tilings?

RavenclawPrefect
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    Can you elaborate a bit on the compactness argument that shows that $\Delta_n$ can't approach $1$? – Carl Schildkraut Dec 24 '20 at 01:14
  • Yes! Suppose, for a fixed $n$, there is a polyomino $P$ that achieves arbitrarily good packings. Then for every $N$, that polyomino can cover an $N\times N$ square (if not, its density would be at most $1-1/N^2$). We can then construct a tiling of the plane by this polyomino as follows. Cover a $k\times k$ square by copies of $P$ in a way that can be extended arbitrarily far around the square (such a covering must exist, since there are arbitrarily large squares containing a $k\times k$ square in the center and only finitely many coverings to choose from - they can't all be bounded). (1/2) – RavenclawPrefect Dec 24 '20 at 01:54
  • Then, among the fintely many ways to extend this to a covering of a $(k+2)\times (k+2)$ square centered at the first square, we know at least one must extend arbitrarily far in all directions; take that extension. Repeat, at each stage fixing the previous while allowing for arbitrary further extension, and we get a tiling of the plane. (Importantly, we're requiring that the extensions be able to fully surround the current tiling in all directions at once for as far as we like; if we can extend in one direction, but are stuck in another, that's not permitted.) (2/2) – RavenclawPrefect Dec 24 '20 at 01:55
  • What does "in a way that can be extended arbitrarily far around the square" mean? Does it mean that, for all $n$, there is a way to extend it to a covering of a $(k+2n)\times (k+2n)$ square centered at that $k\times k$ square? – Carl Schildkraut Dec 24 '20 at 02:02
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    Yes, exactly. Sorry if that was ambiguous! – RavenclawPrefect Dec 24 '20 at 02:02
  • Ah -- having thought about this, it now makes perfect sense. No worries! – Carl Schildkraut Dec 24 '20 at 02:03

3 Answers3

7

Here's a tiling of the plane using a $29$-omino that can be partitioned into $4$ heptominoes plus a single cell, giving that $\Delta_7\geq 28/29$.

tiling

Carl Schildkraut
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    Fantastic! I used some tiling software to confirm (I think) that this particular hexomino cannot cover the $37$-celled region given by a $7\times 7$ chessboard with all corners and cells edge-adjacent to a corner removed. So that bounds the density of this polyomino above by $36/37$, which annoyingly doesn't rule out a $36$-omino construction using five copies and a single hole. (I didn't try to optimize this very hard, though; it's possible much better could be done.) – RavenclawPrefect Dec 24 '20 at 03:09
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    @RavenclawPrefect Nice! It's also theoretically possible, say, for nine copies to tile a $65$-celled region with two holes to give something between $28/29$ and $35/36$, right? – Carl Schildkraut Dec 24 '20 at 05:15
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    Yeah, I think any bound of this form short of exactly $28/29$ (which might well be possible?) will leave open the possibility that any intermediate density could be achieved. – RavenclawPrefect Dec 24 '20 at 06:00
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    @RavenclawPrefect Could you attach/give a link to your code? It might be useful to have others tinker with it. – Carl Schildkraut Dec 24 '20 at 08:34
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    The code is not mine; I used the open-source software BurrTools, which allows one to build and solve tiling puzzles in 2D or 3D. I constructed a puzzle with "variable" voxels in a thick border around the shape, which can be left unfilled in a solution. The built-in algorithm was very slow without any guaranteed empty cells to go off of, however, so I manually deleted cells in the target for each of the possible positions of a piece touching the center, and checked that each of these restricted puzzles had no solutions. – RavenclawPrefect Dec 24 '20 at 08:53
4

Here's a very simple heptomino tiling that is indeed "a tiling 15-omino which is the union of two copies of a non-tiling heptomino and an additional cell."

enter image description here

nickgard
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4

After getting frustrated with existing tiling software, I coded up some simple algorithms myself, from which I can establish an upper bound of $64/65$ on $\Delta_7$.

Note that if copies of a polyomino $P$ cannot cover a region $R$ of size $k$ with no overlaps, then any packing of $P$ has density at most $1-1/k$.

Label the four non-tiling heptominoes $H_1,H_2,H_3,H_4$ in the order below:

enter image description here

Then we have (as shown via exhaustive computer search):

  • $H_1$ does not cover a $7\times 11$ rectangle with L-trominoes removed at the corners (image), and so $\delta(H_1)\le64/65$. Fascinatingly, $H_1$ does cover a $10\times 10$ square, as shown in the following rotationally symmetric configuration:

                                               enter image description here

  • $\delta(H_2)=7/8$, as discussed in the OP.

  • $H_3$ does not cover a size-$33$ region given by deleting $2\times 2$ squares from the corners of a $7\times 7$ grid, so $\delta(H_3)\le 32/33$.

  • $H_4$ does not cover a $6\times 7$ rectangle with the corners removed, so $\delta(H_4)\le 37/38$.

For all of these regions, I tried a few ways to shrink them and found tilings for every smaller region investigated. While the bounds on the other three polyominoes are reasonably close to the $28/29$ obtained by Carl Schildkraut, $H_1$ seems to present a serious obstacle. Perhaps this is because it really can tile the plane in a very dense manner not yet discovered? Either way, it seems like further investigation of $H_1$ tilings is the route to good bounds on $\Delta_7$.

RavenclawPrefect
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