My 10-year old son asked me this simple question, and I've been unable to answer it.

Suppose we start with two (unlinked) circular wire loops (maybe different sizes), and allow both to be continuously deformed in three-dimensional space, the only constraint being that different parts of the loop mustn't touch. The two loops can be deformed in different ways.

Is it possible to do so in such a way that neither of the loops (now considered rigid) can be passed through the other by rigid motion? Consider the wire to be one-dimensional; the definition of 'passing through' permits the two loops to touch (as is obviously necessary in the case of two identical circles), but not to cross each other.

For a formal definition of 'passing through', suppose the two loops are A (fixed) and B (moving). If there is a point on B that traces a path C through space as we move B, such that C is a closed loop that is linked with A, then we say B has passed through A.

Edit Joseph O'Rourke has given an interesting example where the wires are 'interlocked'. So I'd now like to add a condition to exclude this situation, and say that, prior to the 'passing through', the wires must be separated (occupying different sides of some plane).

Edit Experimenting with garden wire, inspired by some of the comments below, I came up with a configuration that may work: two loops But even with this I am not certain. Ideally I'd like a simple provable example. In particular, could there be an example where both loops are planar?

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    $\begingroup$ @Jan-Christoph Schlage-Puchta If I correctly understand the question, the order of quantifiers is: For all continuous deformations, there exists a choice of one of the two (deformed) loops such that, without any further deformation, the chosen loop can pass through the other one. And the initial deformations would even be allowed to pass one loop through the other, so that we needn't worry about linking. $\endgroup$ Apr 19, 2019 at 14:08
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    $\begingroup$ Also, I interpret "pass through" as allowing rigid motions during the attempt to pass one loop through the other. (So there's a similarity with known questions about moving a couch through a hallway that has corners.) I don't immediately see an obvious solution, so if the people voting to close think the problem is trivial then I'd appreciate a hint. $\endgroup$ Apr 19, 2019 at 14:22
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    $\begingroup$ This is a really interesting question. It seems like if you take a circle, twist it into a tight double helix and then repeatedly twist that double helix like a twisty-tie, it should be possible to make configurations that can't be passed through each other. However, I have no idea how to go about proving that. $\endgroup$
    – Gabe K
    Apr 19, 2019 at 15:07
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    $\begingroup$ "Passing through" may need a definition, though. If you take a loop, make a half twist, and fold (so you have basically two circles on top of each other), is it "passing through" if you send something through the hole in the middle? Because technically you can argue that doing so is passing through the hole twice (in, and then out again) and therefore not really "passing through". $\endgroup$ Apr 19, 2019 at 19:28
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    $\begingroup$ "Passing through" has a natural definition in terms of the intersection number of a point on the first loop with the spanning disc for the 2nd loop. The spanning disc is homotopically-unique, so this definition makes sense for 1-parameter families of 2-component unlinks. $\endgroup$ Apr 20, 2019 at 2:13

3 Answers 3


I think there are quite a few examples. Here is one that is the tip of a family of ideas.

Let's call the two wire loops $L_1$ and $L_2$.

I'll start by describing a slightly more pliable definition of what it means for one loop to pass through another. This can be done without too much technical detail, as we are moving our loops with rigid motions.

Fix an embedded spanning disc $D_1$ for $L_1$ and $D_2$ for $L_2$. As you move $L_1$ and $L_2$ by rigid motions, imagine doing the same rigid motion to $D_1$ and $D_2$.

The loop $L_1$ passes through $L_2$ (once) if, in the family of loops, any point on the spanning disc $D_1$ intersects the spanning disc $D_2$ transversely exactly once (through the motion). To make this definition fully general you'd need to bring in some transversality theory, but let's ignore that detail for now.


$L_1$ is a circle of radius 1, and $L_2$ is a "double wound" helix about another circle of radius 1, as in the picture. Basically this is a gimmick to make the loop $L_2$ behave as if it were a "thick" circle.

enter image description here

If you think about the rigid motions that put the loop $L_1$ so that its centre is on the spanning disc $D_2$, they either force $L_1$ and $L_2$ to intersect, or they link.

I hope that makes some sense.

This example makes me think you should also be able to come up with examples of loops $L_1$ and $L_2$ where $L_1$ can't pass through $L_2$, but $L_2$ can pass through $L_1$ $n$ times for some $n>1$, but not with $n=1$. i.e. to have one loop pass through another, it might have to do it multiple times, but it won't be able to do it just a single time.

  • $\begingroup$ Thanks, this is pretty convincing. $\endgroup$ Apr 22, 2019 at 6:13

It is relatively easy for even flexible chains (rigid links joined by universal joints) to be interlocked—cannot be separated to $\infty$—(although not easy to prove they are interlocked).

Demaine, Erik D., Stefan Langerman, Joseph O'Rourke, and Jack Snoeyink. "Interlocked open linkages with few joints." In Proceedings of the 18th Symposium on Computational geometry, pp. 189-198. ACM, 2002

          Figure 6a.

So it is even easier to arrange for two rigid chains to interlock.

One could reform the "wire loops" into thin ellipses, and treat them as 1D wires bent into the forms of the interlocked linkages displayed. Finally, expand one of these "wires" to a loop enclosing the other nearby wire. Then we have a way-overboard proof that two rigid loops can interlock, and so neither can pass through the other.

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    $\begingroup$ Thanks very much for this! I think it does answer the question as asked. But what if we add the condition that the two pieces of wire are initially separated (in the sense that, say, they occupy distinct half-spaces)? $\endgroup$ Apr 20, 2019 at 5:53
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    $\begingroup$ @AlecEdgington: Good point. Then this technique is not applicable. It seems Jan-Christoph Schlage-Puchta's idea is best: Form one wire into a ball, the other into a twisted, thin needle. $\endgroup$ Apr 20, 2019 at 11:10
  • $\begingroup$ I think you should give donuts a chance. Surely there are two dense periodic trajectories on a torus that will not allow pass through in the sense of making a linked path around one of the tori. One can then approximate these sufficiently with wire loops. Gerhard "Two Donuts Better Than One" Paseman, 2019.04.20. $\endgroup$ Apr 20, 2019 at 15:50

I think the job can be accomplished with two non looped wires. Wrap one in a coiled loop with (say) five turns or more, so it forms a small near circle. Then use the other wire and wrap it around like you are gift wrapping a donut, except you only have ribbon and do not want the color of the donut to show. Arrange this second wire so that both ends are on the outside of the torus.

Wait ! you say. If we are dealing with wires, can't one take the end of the first wire, poke it through the wrap, and with five or more rotations free the inner wire from the enclosing torus? I admit that maybe you can. So to defeat that, use Gabe K's twist idea to form the inner coil. Now any rigid motion to pass the inner wire through has to rotate the circle outside the enclosing torus.


I've decided to weaken the problem to allow a proof to become more clear, and to make more rigorous the notion of "pass through". Let a deformed loop be given a soap bubble, a surface much like a minimal surface with the loop as boundary, but we use it to help define the notion pass through. If an object such as a probe comes from elsewhere and deforms the bubble by epsilon, the bubble breaks, and the probe is said to have passed through the loop. I assert without proof that whatever reasonable notion is present in the original problem, it implies this notion of pass through.

Now for step one. Take a wire loop and bend it around a pipe, and then give it a bubble which is shaped like the pipe. (I know this is not quite minimal as a surface in differential geometry, but it should still help.) The part of the bubble that touches the pipe is the inside, and the other side is the outside. If I have two such bubbles, I can't break one with the other by placing the insides together: I have to rotate one to pierce the other. Now the idea is to build a shape where this does not happen.

As before, bend the loop into a thin ellipse, then wrap it several times around a one inch pipe, and leave the ends tucked on the outside away from the pipe. Take the other and make a torus wrap around it with inner diameter 3/4 inch and outer diameter 1 3/4 inch. Make sure the ends of the torus loop stick out away from the circle.

Now the only parts of the loops that can get close to one another are like two insides. We can change the inner and outer diameters so that no penetration of epsilon/2 is possible without deforming either loop further.

It would be interesting to see if the (differential geometry version of the) minimal surface version also holds.

End Edit.

Gerhard "Next: Shipping A Klein Bottle" Paseman, 2019.04.19.

  • $\begingroup$ The question is about looped wires -- but perhaps you are describing a construction of two looped wires from non-looped wires with their ends glued? In that case, doesn't your torus wrapping link the two loops? $\endgroup$ Apr 19, 2019 at 19:32
  • $\begingroup$ Topologically no, because they are not loops. You can also reduce the previous problem to this by taking each wire loop and squishing it into a "doubled wire". Gerhard "Has Handy Home Repair Tips" Paseman, 2019.04.19. $\endgroup$ Apr 19, 2019 at 20:03
  • $\begingroup$ It just occurred to me that you might mean a soap bubble version of this: allow each loop to have a thin film, which if deformed by epsilon by an outside probe will break. Is there a rigid motion that will break either bubble? I believe a modification of the design gives one where again there is no such rigid motion, primarily because points of curvature (of the appropriate sign) can be constrained to not approach a film of the other loop. Gerhard "Has Handy Cleaning Tips Too" Paseman, 2019.04.19. $\endgroup$ Apr 19, 2019 at 20:18

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