Monday, April 14, 2008

Double-pinned painting

Let's say you have a painting. As it happens, the painting is Magritte's Golconde. You apparently really like Magritte.

For mysterious reasons (hey, this is what you want, not me), you want to hang it up in a special way. Some string is attached to the painting, and you want to hang it on two pins that are in the wall. You want it such that removing either of the two pins will result in the string unwinding and the painting falling down. How can you do it?

After you've figured that out, there is a vastly more difficult problem to solve. See, you have Escher's Knots, and you want to hang it on the wall too...
Can you hang this on 3 pins such that removing any one will result in its fall? In a brief moment of insanity, you wonder, can you find a way to hang it on N pins such that the removal of any M pins will result in its fall?

There will, of course, be some difficulty in conveying a solution to me without pictures, but you're obviously very creative, so you can figure it out. I might add that the solution to any case with more than two pins is too hard to draw anyway.

intrinsicallyknotted said...

Let me make sure I've got the problem straight: I interpret this to mean that both ends of the string are attached to the painting (where on the painting probably doesn't matter) and the string is looped around the pins in some (arbitrarily convoluted) way. When you remove each pin, the string has to be absolutely free of all other pins--it can't just slip down a little bit, it actually has to become unattached. Am I interpreting this correctly?

So basically we're talking about Brunnian links here. I'll have to think about what the solution looks like (I know there is a solution even to the general version of the problem; I think I saw a paper on that recently).

miller said...

Yep, you've got it right!

Ah, so they're called Brunnian links. Thanks for the tip! However, I don't think the general case is a Brunnian link, because in general, dropping the painting requires the removal of M pins, not just 1.

miller said...

Oh, and I do happen to have a solution to the general case. I guess I'm not the first?

intrinsicallyknotted said...

Yep, there is a general version of a Brunnian link--I think the notation is an (n, m)-Brunnian link--where there are n components, and removing fewer than m of them leaves the rest still linked, but removing m or more leaves the rest unlinked. I've seen at least one paper that mentioned them, but I can't remember where now. I'll have to ask my advisor…

Just to be clear, in a Brunnian link (or any link, for that matter) all the components are closed loops, topological circles, unlike the pins described in the problem. But all we have to do is assume the pins stick out a long way (away from the neighborhood of the string) and then loop around so their ends are joined, and then we can talk about the pins as components in a link.

So basically the answer is yes, it's possible. I have no intention of trying to visualize the (n, m) case, however, as that's too much work for me (especially when I have a topology presentation to write!) But the (n, 1) case is not too difficult to describe:

If there are n pins, then the n pins as well as the loop formed by the string and painting form an (n+1)-component link. The goal, then, is to find an embedding of the (n+1)th component such that the entire thing is a Brunnian link, a link which is nontrivial, but which becomes trivial if we remove any one component. Since Brunnian links of arbitrary size (greater than 3, naturally) exist, the problem can be solved.

The solution to the 2-pin problem would be a version of the Borromean rings, although they (and any size Brunnian link) are usually drawn with the components overlapping each other although they are pairwise unlinked. But any Brunnian link can be embedded in such a way that all but one of the components are circles lying in parallel planes, with the last component getting more convoluted as the number of components grows. I have no intention of trying to draw anything but the 2-pin case, however!

intrinsicallyknotted said...

Yikes, that was long-winded. Sorry, I tend to get carried away when I have the opportunity to talk knot theory. You might have noticed it's a favorite subject of mine…

Aha, found the article. It's in the American Mathematical Monthly, February 2008: Brunnian Spheres, by Hugh Nelson Howards. Yay, higher-dimensional linking!

intrinsicallyknotted said...

Okay, have a picture! The top left is the Borromean rings, inspiration for my answer. The bottom left is the Borromean rings with the red and green components separated. The right side is my answer, which you get from the previous picture by rotating it out of the plane and stretching those red and green circles into straight pins on which to hang the string.

miller said...

Hey, cool solution! Personally, I could never solve this one without major hints on where to start.

I love knot theory too, though for me it's more a source of mystery than anything else. I can't say I can imagine higher-dimensional knots.

The general case isn't quite a (n,m)-Brunnian link, because breaking the string-link will always result in the rest unlinked.

intrinsicallyknotted said...

Ooh, you're right. Darn, I'll have to think more carefully about this then. Is there a reasonable way to describe the solution, even if it's not easy to draw?

miller said...

I invented my own naming system to describe exactly the path taken by the string, but you can do whatever works for you.