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Call a set of polyhedra free if it is possible to rigidly move the polyhedra, without any polyhedron intersecting any other, so that their pairwise distances are arbitrary large, and locked otherwise. So two linked tori are locked, as is a ship in a bottle.

Can a finite set of convex polyhedra in $\mathbb{R}^3$ ever be locked?

note: We can move these polyhedra simultaneously.

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  • $\begingroup$ This is on M.SE as well. There are useful answers over on that page too. $\endgroup$
    – Milo Brandt
    Commented Jan 10, 2016 at 2:57
  • $\begingroup$ Yes, I asked it on M.SE too. $\endgroup$
    – Minghui Ouyang
    Commented Jan 10, 2016 at 3:53

5 Answers 5

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No locking, even if you restrict to translations. Scale the whole arrangement up by a factor of $c$, then scale each polyhedron down by $c$ around its center of mass. Neither step introduces collisions. Now, that's not quite rigid motion like you asked for, so instead do this recipe $N$ times with the factor $\sqrt[N]{c}$, and let $N\to \infty$.

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    $\begingroup$ I do think that one can make a non-moveable arrangement if one restricts the moves to only one polytope at a time. $\endgroup$
    – Lev Borisov
    Commented Jan 9, 2016 at 14:53
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    $\begingroup$ @LevBorisov, Do you have a proof? (By some reason I feel there are such examples.) $\endgroup$
    – Anton Petrunin
    Commented Jan 10, 2016 at 18:50
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    $\begingroup$ @AntonPetrunin I am not sure what are precise statements, but there is something on the topic here and in references turgor.ru/lktg/2002/problem4.ru $\endgroup$
    – Fedor Petrov
    Commented Jan 10, 2016 at 22:26
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I have already up-voted Allen's answer (see above). His solution is discrete though. I thought I would write a comment but it's simpler to provide an extension of his method right here. I will even generalize his approach slightly. You may up-vote Allen for all this, instead of my answer.

Let's fix a selection function--given an arbitrary convex body $B\subseteq\mathbb R^n$, let $s(B)\in B$.

Let $F$ be a family of pairwise disjoint convex bodies in $\mathbb R^n\,$ (bodies $\,$means$\,$ compact and n-dimensional). $\,$Define

$$\forall_{t\ge 1}\ T_t(B)\ := \{x\in\mathbb R^n\,:\, x-(t+1)\cdot s(B)\in B\}$$

Then $T_t\ $ (for $t\ge 1)\ $ move convex bodies $B\in F$ rigidly, and keep them separated all the time, where the separation goes to $\infty$ when $t\rightarrow \infty$.

In particular, one can select $s(B)$ to be the center of gravity of $B$.
REMARK $\,$ We may decide on $s(B)$ being the center of gravity of $B$, as Allen has done. Then $\ \forall_{t\ u\ge 1}\ T_{t\cdot u} = T_u\circ T_t$.

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I also asked this question in M.SE and at there Igor Pak gives a quite useful reference book which solve this problem completely.

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To supplement your references, this class of problems has been studied since the 1980's:

Toussaint, Godfried. "Movable separability of sets." Computational Geometry. 1985.

         
Google Scholar finds this paper has by now been cited 141 times.

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you can use tetrahedrons to cover a sphere in such a way that the two high points of each tetrahedron cover low points of other tetrahedrons, as if they were in a square lattice. not sure what to do where the lattice breaks though. this kind of stracture should stop them from moving one at a time (allen already solved the case otherwise).

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