You are standing in a large, oddly-shaped room, all of whose walls are lined with mirrors. Everything is dark except for a single light bulb somewhere else in the room. The light from this bulb spreads out in all directions and bounces off the mirrors, possibly many times over. Is it possible that, despite all the mirrors, you could still be standing in the dark, unable to see light from this bulb from any direction?
Said differently, consider the room as a large hole of mathematical mini-golf, where the locations of the ball (just one point) and cup (also just a point) are predetermined. Your goal is to pick a direction to hit the ball, hoping that it will reflect off the walls and eventually find its way to the cup. Is a hole-in-one always possible?
We assume that when a ray of light (i.e., the golf ball) hits a mirror, it reflects perfectly, as illustrated above. Let’s also assume that it never loses energy, so it will continue as far as necessary to reach its destination. There’s one corner case to consider, namely, what happens if the ray hits a corner of the room? Well, it’s (usually) ambiguous as to where it should reflect next, so let’s say that the ray just dies if it hits a corner. In the mini-golf formulation, our hole-in-one must avoid the hole’s corners.
With these rules in place, it turns out that it is possible to be left in the dark, or to make a hole-in-one impossible! One of the first examples, provided by Tokarski in 1995, is illustrated below; let’s call this room R.
Let’s see why this room R works, i.e., why point Q is not illuminated by a light source at P.
Notice that R is built from identical square cells. As our ray of light continues through the room, we can keep track of where it is in its current cell—but not which cell it is in—by imagining that the ray is just bouncing around in a single square, as illustrated below. Call this single square S. We are essentially folding up the light ray’s trajectory into S.
Color the grid points with light yellow and dark purple as illustrated above; in particular, P and Q are both yellow. Tokarski’s room has the interesting property that every dark purple point is in fact a corner of the room, so if the ray hits there, it dies. This means that the corresponding ray in square S would die when hitting one of the purple corners. So, in order for the light to reach Q in Tokarski’s room, the corresponding path in S must start at the yellow corner, end back at this same corner, and completely avoid the purple corners of the square. I claim this is impossible. Why?
Just as we folded room R into a single square, let’s consider unfolding square S to cover the whole plane. Because we’ve assumed perfect reflections, our trajectory in cell S unfolds to a straight line!
This means that we would need to find a straight trajectory from the initial yellow point to some other yellow point that does not hit any purple points. But this is impossible: all the yellows are blocked by purples! So the light can’t get from P to Q in Tokarski’s room, as claimed.
This room R is not the only example of an unilluminable room; with similar methods, many such rooms may be constructed. For example, in an effort to find an unilluminable room with the smallest number of edges, Tokarski provided the 26-sided polygon below on the left, which was then improved by Castro to a 24-sided polygon on the right. Both are built from 45-45-90 triangles instead of square cells.
Tokarski also provides the gem below (with no right angles!), and his paper gives recipes for a wide variety of other shapes.
Here’s a challenge: so far, we’ve seen that it is possible to design a mathematical mini-golf hole that requires at least 2 shots. Can we make an even harder hole? Can we design one that needs at least, say, 10 shots? We’ll discuss this next time. See you then!
- George W. Tokarsky. Polygonal Rooms Not Illuminable from Every Point. American Mathathematical Monthly, 102:867–879, 1995. [↩]
- Here’s a proof: given any yellow-yellow segment, its midpoint lies on the grid. If this midpoint is purple, then we’re done. Otherwise, take half of the original segment and repeat this argument. [↩]
- D. Castro. Corrections. Quantum 7:42, 1997. [↩]
- It is not known whether 24 is minimal. [↩]