My (non-mathematician) friend asked me a physics/tilings question that maybe someone here is interested in dissecting, or can point to the literature if this problem has been studied.

Does there exist a tile such that when you put a bunch of copies of it on a table and push from all sides, they always form a tiling?

My friend illustrated with physical (uniform density) lozenge tiles that they do not have this property, by throwing some on the table, and pushing them together. More specifically this suggests the stronger property that a typical initial configuration will get stuck. The tiles this was demonstrated on had positive friction.

The informal question as stated is a bit ambiguous. I am not going to try to formalize the physics of the problem, but I'll at least try to specify how the force is applied in a hopefully unambiguous (but somewhat arbitrary) way. You can suggest a better variant in the same spirit if e.g. it's easier to solve or mine misses the point for a "stupid" reason. (edit: I have added a "physics-free" formalization below.)

Let's say a tile is a nice enough subset $P \subset \mathbb{R}^2$, you can pick what that means. E.g. if going for a negative answer, you can choose something like "simply connected convex polygon". If going for a positive answer, I could imagine something like piecewise smooth being helpful. (For physics considerations it's a zero friction rigid body, and let's say of uniform density.)

Let $G = \mathbb{R}^2 \rtimes S^1$ be the rototranslation group (so no flips), which acts on $\mathbb{R}^2$ from the left. A partial tiling is a subset of $T \subset G$ such that the interiors of $t \cdot P$ for distinct $t \in T$ are disjoint. We say a partial tiling $T$ fills $C \subset \mathbb{R}^2$ if $T \cdot P \supset C$.

A physical jam is a finite partial tiling $T \subset G$ such that, assuming the tiles have zero friction and behave according to physics, if you stretch a rubber band around the convex hull of $T \cdot P = \bigcup_{t \in T} \{t \cdot P\}$, the tiles will not budge. Intuitively, jams always exist aplenty, just put some tiles on the table, stretch the band around them and let go (if there's a third dimension available there's a problem with that strategy, but you see what I mean).

Definition. A tile $P$ is a physical rubber band monotile if all $r > 0$, there exists $R > 0$ such that every jam whose convex hull contains the ball of radius $R$ fills the ball of radius $r$.

In terms of this, the question is:

Is there a physical rubber band monotile?

Observe that any physical rubber band monotile admits a partial tiling that fills the entire plane. In the usual terminology, $P$ tiles the plane under rototranslations, and such $P$ is sometimes called a monotile.

In case this question is non-trivial, here's some starters:

Is the equilateral (or any) triangle a physical rubber band monotile? Is the square (or any other rhombus, e.g. the lozenge)? Is the hexagon? Any of the pentagon monotiles?

I'm also interested in higher dimensions of course (my friend may or may not be). In one dimension I was able to solve the problem myself.

Physics-free formulation

Pick a (compatible) metric for $G$ and topologize the set of closed sets of $G$ with the Hausdorff metric, and the set $\mathcal{T}$ of all finite partial tilings with the induced metric. Let $c : \mathcal{T} \to \mathbb{R}_+$ be the (continuous) map that takes a partial tiling $T$ to the length of the boundary curve of the convex hull of $T \cdot P$. Paths in $\mathcal{T}$ starting from a finite partial tiling amount to moving the tiles in a continuous way (adding or removing a tile would necessarily be a jump because interiors must stay disjoint).

A weak jam is a finite tiling $T \in \mathcal{T}$ such that there does not exist a path $p : [0,1] \to \mathcal{T}$ with $p(0) = T$ and $x \mapsto c(p(x))$ strictly decreasing. A strong jam is a finite tiling $T \in \mathcal{T}$ such that there does not exist a path $p : [0,1] \to \mathcal{T}$ with $p(0) = T$, and $x \mapsto c(p(x))$ nonincreasing and $c(p(1)) < c(p(T))$. Every strong jam is a weak jam, obviously. The difference is whether we allow moving tiles so that the rubber band length stays constant.

Definition. A tile $P$ is a strong (resp. weak) rubber band monotile if all $r > 0$, there exists $R > 0$ such that every weak (resp. strong) jam whose convex hull contains the ball of radius $R$ fills the ball of radius $r$.

Every strong rubber band monotile is a weak rubber band monotile, obviously. In terms of these, the question is:

Do strong/weak rubber band monotiles exist?

It is easy to prove that no rectangle is a strong rubber band monotile, by arranging the rectangles into a bigger rectangle and removing all but the boundary tiles. I'd say that's definitely also a physical jam. Perhaps @GerhardPaseman's answer shows that the square is not even a weak rubber band monotile.

  • $\begingroup$ It seems that the definition of "rubber band monotile" merged with the subsequent question... $\endgroup$ Jun 13, 2020 at 23:18
  • $\begingroup$ @PietroMajer: thanks! $\endgroup$
    – Ville Salo
    Jun 13, 2020 at 23:58
  • $\begingroup$ I changed an "a" to a "the" in the definition of rubber band monotile, which may drastically change the problem, but which seems more natural after seeing @GerhardPaseman's suggestion. $\endgroup$
    – Ville Salo
    Jun 14, 2020 at 0:02
  • $\begingroup$ The existence of "maximally random jammed (MRJ) configurations" seems relevant. For example:Atkinson, Steven, Frank H. Stillinger, and Salvatore Torquato. "Existence of isostatic, maximally random jammed monodisperse hard-disk packings." Proceedings of the National Academy of Sciences 111, no. 52 (2014): 18436-18441. PNAS link. $\endgroup$ Jun 14, 2020 at 0:27
  • $\begingroup$ One more small detail: $c(T)$ is the length of the convex hull of $T\cdot P$, isn't it? $\endgroup$ Jun 14, 2020 at 5:19

1 Answer 1


I suspect not, at least for regular polygons. (For nonregular polygons, there are orientation issues which I believe won't be solved by rubber bands.) Let me illustrate with squares.

Consider 8 squares in a three by three arrangement with a central hole. This configuration by itself does not tile when you apply a rubber band, but if you have a larger configuration with more tiles, some of these eight tiles might get pushed in.

Now push four edge squares out, and place a diamond tilted square in the middle, then push the squares back in. Now you have a nontiling arrangement. I suspect you can expand this to get larger nontiling arrangements which resist rubber bands.

Gerhard "Hope This Doesn't Snap Back" Paseman, 2020.06.13.

  • $\begingroup$ This seems to work, and give arbitrarily large configurations with minor modification. However, it's not clear to me that it continues working if force is applied unevenly to the different sides (it looks like it might snap into a 3x3 square of squares if the symmetry of the configuration is broken?) Maybe the question should be extended to "typical" configurations in some sense, or to configurations in which some "external shell" is randomly perturbed. $\endgroup$ Jun 14, 2020 at 0:21
  • $\begingroup$ My guess is also that it's impossible, and your suggestion indeed should work for the square with the uniform rubber band. The comment of @GevaYashfe is also relevant for figuring out whether this also proves the square is not a weak rubber band monotile, i.e. whether this jam is strong. $\endgroup$
    – Ville Salo
    Jun 14, 2020 at 2:15
  • $\begingroup$ In the 9-square configuration, let's rotate the central diamond, and let the N,S,W,E squares move respectively S,N E,W, always touching the corresponding vertices of the diamond, and let the other square follow staying in touch. Then the length of the boundary of the convex hull (the rubber) decreases, because it is made by 4 unit segments plus 4 hypotenuses of 4 right-angled triangles with a unit cathetus, and a cathetus whose length decreases. $\endgroup$ Jun 14, 2020 at 8:20
  • $\begingroup$ A small check also show that one can do the whole 90 degrees rotation of the diamond, till the tiling reaches the 3x3 minimal configuration (that is, in the process the 4 corner squares never overlap the central square) $\endgroup$ Jun 14, 2020 at 8:20
  • $\begingroup$ So if I understand @PietroMajer correctly, this is not a strong jam. I give a weak jam in the question, and this is presumably a physical jam, so so far I suppose we believe the square is neither a physical nor a weak rubber band monotile, but do not yet know whether it is a weak rubber band monotile, i.e. it might not have a strong jam. To me strong jams seem like the most interesting of these, so would be nice to figure out whether they exist. (Sorry about the choice of terminology, it's maybe not the most convenient.) $\endgroup$
    – Ville Salo
    Jun 14, 2020 at 14:21

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