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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 to the length of its convex hull. 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).

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).

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.

A 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 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 $r$.

Is there a rubber band monotile?

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

Is the equilateral (or any) triangle a 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.

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.)

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 $r$.

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.

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 to the length of its convex hull. 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.

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.

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 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 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 a ball of radius $r$?

Observe that any rubber band monotile admits a partial tiling that fills the entire plane. In 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 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.

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.

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 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 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 rubber band monotile?

Observe that any rubber band monotile admits a partial tiling that fills the entire plane. In 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 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.