Can a (finite) collection of disjoint circle arcs in $\mathbb{R}^3$ be interlocked in the sense in that they cannot be separated, i.e. each moved arbitrarily far from one another while remaining disjoint (or at least never crossing) throughout? (Imagine the arcs are made of rigid steel; but infinitely thin.) The arcs may have different radii; each spans strictly less than $2 \pi$ in angle, so each has a positive "gap" through which arcs may pass:
Of course, if one could prove that in any such collection, one arc can be removed to infinity, the result would follow by induction. But an impediment to that approach is that sometimes there is no arc than can be removed while all the others remain fixed.

Another approach would be to reduce the piercing number of the configuration: the number of intersections of an arc with the disks on whose boundary the arcs lie. If the piercing number could always be reduced in any configuration, then it would "only" remain to prove that if there are no disk-arc piercings at all, the configuration can be separated.

Intuitively it seems that no such collection can interlock, but I am not seeing a proof. I'd appreciate any proof ideas—or interlocked configurations!


I believe there is no such locked configuration. The proof is by induction, as you suggest.

Pick any arc and imagine moving it to infinity. Of course, to do this, it will have to pass through some other arcs, and thus this is not a valid motion. We can, however, by picking our motion "generically", ensure that there are just finitely many times when our arc passes through another arc, and that at each of these times, it passes through exactly one other arc at exactly one point. But now if we rotate (in the plane of the circle) the arc during the motion, we can ensure that it's "gap" is moved to each of the points where it used to pass through another arc. Thus we have turned our invalid motion into a valid one.

  • $\begingroup$ Of course, I might be missing something since you claim such an inductive strategy will not work. $\endgroup$ – John Pardon Aug 10 '13 at 22:21
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    $\begingroup$ I think it is an excellent and very clean solution. $\endgroup$ – Misha Aug 10 '13 at 23:25
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    $\begingroup$ Beautiful proof, immediately convincing! In fact one arc can be removed while all the others remain rigid, as you detail. Great! $\endgroup$ – Joseph O'Rourke Aug 11 '13 at 0:53
  • $\begingroup$ This raises a natural question: The same, but with arcs of ellipses. Then your clever proof fails, as the rotation of an arc within its plane could bang into other arcs. $\endgroup$ – Joseph O'Rourke Aug 11 '13 at 1:03
  • $\begingroup$ Now posted in a separate question. $\endgroup$ – Joseph O'Rourke Aug 11 '13 at 1:11

There is already a beautiful correct answer.

Mine is just an illustrated comment, for the case $\mathbb R^2$, although the question was about $\mathbb R^3$.

Here is how three arcs can be locked, in $\mathbb R^2$:

enter image description here

Here is why two can't be locked, in $\mathbb R^2$:

enter image description here


Ooops! Actually, both configurations with 3 circles can be unlocked.

enter image description here

So the problem remains open for the $\mathbb R^2$ case too. I made it into a question.

The question got answered.

  • $\begingroup$ Very clever variation! $\endgroup$ – Joseph O'Rourke Aug 27 '13 at 13:26

@Cristi: Perhaps your first example needs an argument to exclude the following motion, which might only be possible with my generous arc-gap?

  • $\begingroup$ In fact, in the meantime I realized they can be unlocked. And it works no matter how small is the gap. $\endgroup$ – Cristi Stoica Aug 27 '13 at 14:40
  • $\begingroup$ Tricky! ${}{}{}$ $\endgroup$ – Joseph O'Rourke Aug 27 '13 at 15:20

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