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Given a polyomino $P$, we can ask if it is possible for disjoint copies of $P$ to surround a single cell in the square grid - i.e., for the complement of their union to have a connected component of size $1$.

We can further refine this into polyominoes that weakly surround a cell (just covering the four edge-adjacent cells) and those that strongly surround a cell (covering all $8$ squares which share a vertex with the hole). Thanks to Julian Rosen for clarifying this distinction.

My original intuition was that this is always possible, but I was having trouble proving it; after enough struggling to show it was true, I started searching for counterexamples. Here is a polyomino with $48$ cells which does not even weakly surround a one-celled region, which is not too difficult to verify by hand:

                                                    enter image description here

After some modifications, I've reduced this down to a simply-connected solution with $26$ cells:

                                                    enter image description here

I have a size-$23$ example of a polyomino which does not strongly surround a hole:

                                                    enter image description here

What is the smallest polyomino that cannot surround a single-celled hole? I am interested in this question for both the weak and strong cases.

I've written some code to explore this, and have confirmed that all of the $1,227,708$ free polyominoes on at most $14$ cells can strongly surround a hole. How much can we tighten these bounds?

RavenclawPrefect
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  • For that second one, you're definitely missing more than a single hole. There's no way that polyomino tiles the plane. – Rivers McForge Jan 15 '21 at 16:44
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    I am not asking whether polyominoes tile the plane; I'm asking whether the complement of disjoint copies of them can have connected components of size $1$. Here is an image of a polyomino surrounding a $1\times 1$ hole, for reference. – RavenclawPrefect Jan 15 '21 at 16:46
  • @RiversMcForge: I think you may be interpreting the question as asking about polyominoes which tile the whole plane except for one square? I'm asking about a much weaker condition, as evidenced by the computational data mentioned in the post; the former property implies tiling the whole plane, as sketched out in a comment here. Let me know if there's a way I can make the wording of this post clearer! – RavenclawPrefect Jan 15 '21 at 17:40
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    Is it correct that to surround a square, we should cover the 8 adjacent squares (including the diagonally adjacent ones)? – Julian Rosen Jan 15 '21 at 22:24
  • @JulianRosen: I was thinking just about the $4$ edge-adjacent squares, but that's also an interesting question! For all $8$ adjacent, I’ve checked that every polyomio on up to $13$ cells works; of course the $26$-cell counterexample given in the OP is an upper bound for this question, too. I’ll edit in this variant to the OP. – RavenclawPrefect Jan 16 '21 at 00:36
  • Am I missing something, or is this quite susceptible to computer analysis? It seems like, for a fixed $N$-polyomino, you can determine whether it can weakly surround a single-celled hole for certain in $O(N^5)$, and in practice very much faster. When you checked up to $13$ cells, presumably you used a computer program to do it? Was it not fast enough to extend the check further, or did you just not have the full list of free polyominoes for larger $N$? – mjqxxxx Feb 01 '21 at 06:27
  • @mjqxxxx: Yes, my program's bottleneck was generating all $14$-ominoes more than having to check them. If you know of nicely computer-parseable lists of polyominoes, that would be great! In the meantime, I think giving it a longer runtime might let me extend the $13$-cell results a bit further (perhaps even finding a counterexample!) before it becomes intractable. – RavenclawPrefect Feb 01 '21 at 14:03
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    This is just a idle thought, but what about looking at the problem in reverse; i.e., given a unit square hole (either weakly or strongly surrounded), establish some kind of constraint on how its neighbors may be connected, which in turn limits the features of a polyomino that cannot leave such a hole. Perhaps this is too nebulous or difficult but it was just something that seemed to offer a different perspective than generating polyominoes and then seeing if they can be arranged to leave such a hole. – heropup Feb 01 '21 at 20:54
  • Do rotated copies of $P$ count as copies? – Good Boy Feb 02 '21 at 19:19
  • @GoodBoy: Yes, any rotation or flip of $P$ counts. If only translation were allowed, the $V$-pentomino would not strongly surround a hole, and neither of these two octominoes would weakly surround a hole. – RavenclawPrefect Feb 02 '21 at 19:36
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    Your size-$23$ upper bound is nicely symmetric. It would be easier to verify (if true) that there are no smaller examples with that symmetry. – mjqxxxx Feb 02 '21 at 22:07
  • In your $23$-cell example, can't you hollow it out to reduce cell count? – Fimpellizzeri Feb 04 '21 at 16:12
  • @Fimpellizieri: Yes, but then it would not be a polyomino, because it would not be edge-connected. (One can get down to $12$ cells by relaxing connectivity - take the borders of $5\times 5$ square except for the four corners.) – RavenclawPrefect Feb 04 '21 at 16:17
  • But can't you remove the 7 interior cells and just add 3 at the corners of the resulting 5x3 empty rectangle, enough to connect the outer edges to each other? That gets you down to 19, right? – mjqxxxx Feb 04 '21 at 16:48
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    @mjqxxxx: That can strongly surround a hole as shown in this image. – RavenclawPrefect Feb 04 '21 at 17:06
  • Do you have a lower bound? For example, have you brute-forced up to $n = 10$ or anything? – Peter Kagey Feb 05 '21 at 01:55
  • @PeterKagey: Yes - see the last paragraph of the original post. I've checked all polyominoes up to $14$ cells; $15$ might be in reach, though it would take a while to run. – RavenclawPrefect Feb 05 '21 at 02:49
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    I just wanted to comment on what a nicely formulated problem this is (can't +1 more than once!) It's simply stated but the answer is non-trivial. Also, in a sense this is the hardest problem of this type. A natural generalization is to look for minimal polyominoes that cannot (strongly) surround a hole of shape $P$, for some polyomino $P$. For all $P$ with size > $1$ that I've checked, the minimal polyominoes have size $\le 11$, making them easy enough to find through exhaustive search. – mjqxxxx Feb 05 '21 at 18:54
  • @mjqxxxx: I think I can actually prove that every polyomino of size at least $2$ fails to be strongly surrounded by at least one of these three polyominoes - there is a finite list of configurations of the form "this cell must be filled, this one must be unfilled, etc", at least one of which must appear in every polyomino's strongly-surrounding configuration, and none of which are satisfied by all three of these polyominoes. – RavenclawPrefect Feb 06 '21 at 19:04

1 Answers1

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This is by no means a complete answer, but as I mentioned in my last comment, we can make some headway by restricting a computer search to particular symmetry classes. I've been able to check all the free polyominoes of size $N \le 23$ that have a horizontal or vertical symmetry axis (a class that contains OP's size-$23$ non-strongly-surrounding example). Within this symmetry class, there are no non-weakly-surrounding examples of size $N \le 23$, and the smallest non-strongly-surrounding examples have size $N=23$; so OP's example is minimal. Of the $464188$ size-$23$ free polyominoes with a horizontal or vertical axis of symmetry, there are just two that cannot strongly surround a single cell: the example given in the question, and the polyomino pictured below.

                                                    non-strongly-surrounding symmetric polyomino

Heuristically, one might expect minimal examples to have some symmetry, because symmetric polyominoes have the fewest "different-looking local regions", and hence the fewest different possibilities for packing local regions around a hole. Next steps, then, might be to check the other two large symmetry classes for free polyominoes: $180^\circ$ rotational symmetry around a point, and reflection symmetry across a diagonal.

Update: Checking the polyominoes with rotational symmetry around a point turned up the following minimal (within that class) non-strongly-surrounding example, with size $20$:

                                                    non-strongly-surrounding rotationally symmetric polyomino

Second update: Checking the polyominoes with a diagonal reflection symmetry, there are exactly two minimal (within that class) non-strongly-surrounding examples with size $19$. One is formed by removing an interior cell from the previous example (as already pointed out in a comment), and the other is new:

                                 enter image description here enter image description here

Third update: In order to make progress on the non-weakly-surrounding case (which is a more stringent criterion, so the minimal examples are at least as large), I looked at polyominoes with both vertical and horizontal reflection symmetries. The size-$25$ example below provides an improved upper bound for this case.

                                                    enter image description here

mjqxxxx
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  • Thanks for this investigation! I notice that both your example and the one in the OP have the line of reflection passing through a cell; did you also check cases where the line coincides with edges between cells? (Obviously this doesn't have $N=23$ examples, since the number of cells must be even, but perhaps it does for lower even $N$.) – RavenclawPrefect Feb 04 '21 at 15:56
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    @RavenclawPrefect: Yes, the reflection-symmetric cases do include both types of line; I checked polyominoes of both even and odd width. – mjqxxxx Feb 04 '21 at 16:29
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    You can remove one of the central squares in your last picture to get a size-19 example. It loses its rotational symmetry, of course, but it's still symmetric about one of its diagonals. – TonyK Feb 04 '21 at 16:38
  • Would this size-16 version not work, then? It also has 180 degree rotational symmetry, but I'm struggling to see how it fails. – Misha Lavrov Feb 04 '21 at 16:51
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    @MishaLavrov: It can strongly surround a hole as shown here. – RavenclawPrefect Feb 04 '21 at 17:09
  • I went through polyominoes that are centrally symmetric about a central cell, rather than a vertex; the three minimal examples are all of size 23, pictured here. – RavenclawPrefect Feb 05 '21 at 00:16