The Mathematics of Doodling

Wondering about a childhood doodle compels us to ask a series of questions, which will lead us on a tour of increasingly sophisticated ideas from many parts of mathematics. Let me tell you about a doodle I did when I was very small, and the mathematics that flows inevitably from it. It looks like play, but in my mind this is what mathematics is really about: finding patterns in nature, explaining them, and extending them. Mathematics is about asking the right questions, and that’s what we will do here, finding interesting questions, and then finding questions behind the questions. These ideas will lead to rather deep mathematics, in lots of fields. I am not an expert in these fields, and you needn’t be an expert in a part of mathematics to let your curiosity pull you in. On a related note: in this article, I will ask a lot of questions, and give relatively few answers. And there are many questions that you will think should be asked, that I don’t—I encourage you to follow up on them, because they may lead you somewhere interesting and unexpected. This article is intended for readers with widely different mathematical backgrounds, so if you come across a notion you have seen, or one you find trivial, please keep reading. This article is based on the first of the Hedrick lectures at the 2009 Mathfest, in Portland, Oregon. THE DOODLE. The doodle involves finding some shape on your piece of paper, and then drawing a curve tightly around it, as close as you can. After you’ve completed the loop, do it again. And again. And again. (See Figures 1 and 2.) Figure 1. The doodle. THE QUESTION. I noticed that if I kept on doing the doodle over and over again, no matter what shape I started from, the doodle got “more and more circular.” So my first question is: is this true? Original Question. If we repeat the doodle a lot, does the resulting shape get very circular? doi:10.4169/amer.math.monthly.118.02.116 116 c © THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 118 Figure 2. The doodle, once. That’s an imprecise question, but that’s how science is done: we observe something qualitative, and then we try to make sense of it mathematically. So we have a prior question: what precisely (mathematically) are we doing? Here is a reasonable definition of what we are doing. Definition. Given a plane set X , and a constant r ≥ 0, define the “radius r neighborhood of X” Nr (X) to be those points within a distance r of some point in X . Formally, define: Nr (X) = {y : |y − x | ≤ r for some x ∈ X}. We can now reword our question: Reworded Question. In some sense does Nr (Nr (· · · (Nr (Nr (X))) · · · )) become more and more circular, as we iterate Nr more and more often? Of course, we still have to make sense of the phrase “more and more circular.” But as always in science, we will let the answer tell us what the question should have been. There is also the matter of the number r . How relevant is it? Prior Question. How does our choice of r affect the problem? For example, which is bigger, N1(N1(X)) or N2(X)? In fact, the two sets are exactly the same! And as you might expect, this isn’t a special fact about 1 + 1 = 2, so we will want the most general statement that has the same proof. Theorem 1. For any X, and any a, b > 0, Na(Nb(X)) = Na+b(X). What I love about this fact is that it comes from a natural question that a five-yearold can (and did) appreciate, and yet the answer is in a precise sense equivalent to a fundamental mathematical fact: the triangle inequality. We will show two inclusions: Na+b(X) ⊂ Na(Nb(X)) and Na(Nb(X)) ⊂ Na+b(X). The first doesn’t need the triangle inequality: suppose we can get to a point p from a point in X by taking a step of size at most a + b. Then by dividing this step into two in a ratio of b to a we see that we can also get there from a point in X by taking a step of size at most b, followed by a step at most a. Conversely suppose there is a point p we can get to from a point in X by a step of size at most b followed by a step of size at most a. Then by the triangle inequality, the February 2011] THE MATHEMATICS OF DOODLING 117 total distance we’ve traveled is at most a + b, so we can get there in a single step at most that large, concluding the proof. You can see that Theorem 1 is equivalent to the triangle inequality. Experts may want to consider different spaces than the plane, where the triangle inequality may or may not hold. This leads to a number of very interesting questions. But let’s return to our original question. If there were n parentheses in the reworded question, then we can reword it yet again as: Reworded question′. In some sense does Nnr (X) become more and more circular as n gets large? We know we’ve made progress, because our question has gotten shorter! And clearly the “right” question to ask is: Reworded question′′. In some sense does NR(X) become more and more circular as R gets large? We are now ready to see that the answer is yes! (And after we have the answer, we will have the proper question.) The key idea is the following. (We will use this simple principle a lot.) Lemma 2. If A is contained in B, then NR(A) is contained in NR(B). A little thought will convince you why it is true: if every point in A is also a point in B, then everything you can get to by a step (of size at most R) from a point in A you can also get to by a step (of size at most R) from a point in B—indeed the same point! Equipped with this insight, let’s answer the question. Fix a point p of X . Let Dt denote the disk of radius t around p. Pick an r such that X ⊂ Dr . Then {p} ⊂ X ⊂ Dr . Applying the lemma, we find that DR = NR({p}) ⊂ NR(X) ⊂ NR(Dr ) = DR+r . So the boundary of the Rth doodle NR(X) is stuck between two circles around p, one of radius R and one of radius R + r . As R →∞, the ratio of these two radii goes