A friend was playing with that cute little particle-physics simulator idea I mentioned last week. And encountered a problem. With a little bit of thought, I was able to not solve the problem. But I was able to explain why it was a subtler and more difficult problem than they had realized. These are the moments that make me feel justified calling myself a mathematician.

The proposed simulation was simple enough: imagine a bunch of particles that interact by rules that aren’t necessarily symmetric. Like, the attraction particle A exerts on particle B isn’t the same as what B exerts on A. Or there are multiple species of particles. So (say) red particles are attracted to blue but repelled by green. But green is attracted to red and repelled by blue twice as strongly as red is attracted to blue. Your choice.

Give a mathematician a perfectly good model of something. She’ll have the impulse to try tinkering with it. One reliable way to tinker with it is to change the domain on which it works. If your simulation supposes you have particles moving on the plane, then, what if they were in space instead? Or on the surface of a sphere? Or what if something was strange about the plane? My friend had this idea: what if the particles were moving on the surface of a cube?

And the problem was how to find the shortest distance between two particles on the surface of a cube. The distance matters since most any attraction rule depends on the distance. This may be as simple as “particles more than this distance apart don’t interact in any way”. The obvious approach, or if you prefer the naive approach, is to pretend the cube is a sphere and find distances that way. This doesn’t get it right, not if the two points are on different faces of the cube. If they’re on adjacent faces, ones which share an edge — think the floor and the wall of a room — it seems straightforward enough. My friend got into trouble with points on opposite faces. Think the floor and the ceiling.

This problem was posed (to the public) in January 1905 by Henry Ernest Dudeney. Dudeney was a newspaper columnist with an exhaustive list of mathematical puzzles. A couple of the books collecting them are on Project Gutenberg. The puzzles show their age in spots. Some in language; some in problems that ask to calculate money in pounds-shillings-and-pence. Many of them are chess problems. But many are also still obviously interesting, and worth thinking about. This one, I was able to find, was a variation of The Spider and the Fly, problem 75 in **The Canterbury Puzzles**:

Inside a rectangular room, measuring 30 feet in length and 12 feet in width and height, a spider is at a point on the middle of one of the end walls, 1 foot from the ceiling, as at A; and a fly is on the opposite wall, 1 foot from the floor in the centre, as shown at B. What is the shortest distance that the spider must crawl in order to reach the fly, which remains stationary? Of course the spider never drops or uses its web, but crawls fairly.

(Also I admire Dudeney’s efficient closing off of the snarky, problem-breaking answer someone was sure to give. It suggests experienced thought about how to pose problems.)

What makes this a puzzle, even a paradox, is that the obvious answer is wrong. At least, what seems like the obvious answer is to start at point A, move to one of the surfaces connecting the spider’s and the fly’s starting points, and from that move to the fly’s surface. But, no: you get a shorter answer by using more surfaces. Going on a path that *seems* like it wanders more gets you a shorter distance. The solution’s presented here, along with some follow-up problems. In this case, the spider’s shortest path uses five of the six surfaces of the room.

The approach to finding this is an ingenious one. Imagine the room as a box, and unfold it into something flat. Then find the shortest distance on *that* flat surface. Then fold the box back up. It’s a good trick. It turns out to be useful in many problems. Mathematical physicists often have reason to ponder paths of things on flattenable surfaces like this. Sometimes they’re boxes. Sometimes they’re toruses, the shape of a doughnut. This kind of unfolding often makes questions like “what’s the shortest distance between points” easier to solve.

There are wrinkles to the unfolding. Of course there are. How interesting would it be if there weren’t? The wrinkles amount to this. Imagine you start at the corner of the room, and walk up a wall at a 45 degree angle to the horizon. You’ll get to the far corner eventually, if the room has proportions that allow it. All right. But suppose you walked up at an angle of 30 degrees to the horizon? At an angle of 75 degrees? You’ll wind your way around the walls (and maybe floor and ceiling) some number of times, each path you start with. Probably different numbers of times. Some path will be shortest, and that’s fine. But … like, think about the path that goes along the walls and ceiling and floor three times over. The room, unfolded into a flat panel, has only one floor and one ceiling and each wall once. The straight line you might be walking goes right off the page.

And this is the wrinkle. You might need to tile the room. In a column of blocks (like in Dudeney’s solution) every fourth block might be the floor, with, between any two of them, a ceiling. This is fine, and what’s needed. It can be a bit dizzying to imagine such a state of affairs. But if you’ve ever zoomed a map of the globe out far enough that you see Australia six times over then you’ve understood how this works.

I cannot attest that this has helped my friend in the slightest. I am glad that my friend wanted to think about the surface of the cube. The surface of a dodecahedron would be far, far past my ability to help with.